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Complete Solutions Manual to accompany Chemical Principles Sixth Edition Steven S. Zumdahl Thomas J. Hummel Steven S. Zumdahl University of Illinois at Urbana-Champaign HOUGHTON MIFFLIN COMPANY Boston New York Vice President and Publisher: Charles Hartford Senior Marketing Manager: Nicole Moore Senior Development Editor: Rebecca Berardy Schwartz Supplements Editor: Kathryn White Editorial Associate: Henry Cheek Ancillary Coordinator: Sean McGann Marketing Associate: Kris Bishop Copyright © 2009 by Houghton Mifflin Company. All rights reserved. No part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying and recording, or by any information storage or retrieval system without the prior written permission of Houghton Mifflin Company unless such copying is expressly permitted by federal copyright law. Address inquiries to College Permissions, Houghton Mifflin Company, 222 Berkeley Street, Boston, MA 02116. Printed in the U.S.A. ISBN-13: 978-0-618-94822-2 ISBN-10: 0-618-94822-8 TABLE OF CONTENTS Page How to Use This Guide....................................................................................................................v Chapter 2 Atoms, Molecules, and Ions.....................................................................................1 Chapter 3 Stoichiometry.........................................................................................................16 Chapter 4 Types of Chemical Reactions and Solution Stoichiometry ...................................51 Chapter 5 Gases ......................................................................................................................96 Chapter 6 Chemical Equilibrium..........................................................................................149 Chapter 7 Acids and Bases ...................................................................................................192 Chapter 8 Applications of Aqueous Equilibria.....................................................................254 Chapter 9 Energy, Enthalpy, and Thermochemisty..............................................................341 Chapter 10 Spontaneity, Entropy, and Free Energy ...............................................................369 Chapter 11 Electrochemistry ..................................................................................................413 Chapter 12 Quantum Mechanics and Atomic Theory ............................................................455 Chapter 13 Bonding: General Concepts .................................................................................490 Chapter 14 Covalent Bonding: Orbitals ................................................................................540 Chapter 15 Chemical Kinetics................................................................................................576 Chapter 16 Liquids and Solids................................................................................................624 Chapter 17 Properties of Solutions.........................................................................................668 Chapter 18 The Representative Elements...............................................................................707 Chapter 19 Transition Metals and Coordination Chemistry...................................................730 Chapter 20 The Nucleus: A Chemist’s View .........................................................................756 Chapter 21 Organic Chemistry ...............................................................................................773 iii HOW TO USE THIS GUIDE Solutions to all of the end of chapter exercises are in this manual. This “Solutions Guide” can be very valuable if you use it properly. The way NOT to use it is to look at an exercise in the book and then immediately check the solution, often saying to yourself, “That’s easy, I can do it.” Developing problem solving skills takes practice. Don’t look up a solution to a problem until you have tried to work it on your own. If you are completely stuck, see if you can find a similar problem in the Sample Exercises in the chapter. Only look up the solution as a last resort. If you do this for a problem, look for a similar problem in the end of chapter exercises and try working it. The more problems you do, the easier chemistry becomes. It is also in your self interest to try to work as many problems as possible. Most exams that you will take in chemistry will involve a lot of problem solving. If you have worked several problems similar to the ones on an exam, you will do much better than if the exam is the first time you try to solve a particular type of problem. No matter how much you read and study the text, or how well you think you understand the material, you don’t really understand it until you have taken the information in the text and applied the principles to problem solving. You will make mistakes, but the good students learn from their mistakes. In this manual we have worked problems as in the textbook. We have shown intermediate answers to the correct number of significant figures and used the rounded answer in later calculations. Thus, some of your answers may differ slightly from ours. When we have not followed this convention, we have usually noted this in the solution. The most common exception is when working with the natural logarithm (ln) function, where we usually carried extra significant figures in order to reduce round-off error. In addition, we tried to use constants and conversion factors reported to at least one more significant figure as compared to numbers given in the problem. For some problems, this required the use of more precise atomic masses for H, C, N, and O as given in Chapter 3. This practice of carrying one extra significant figure in constants helps minimize round-off error. We are grateful to Claire Szoke for her outstanding effort in preparing the manuscript of this manual. We also thank Estelle Lebeau for her careful accuracy review and Jim Madru for his thorough copyediting of the Solutions Manual. We also are grateful to Don DeCoste for his assistance in creating solutions to some of the problems in this Solutions Guide. TJH SSZ v CHAPTER 2 ATOMS, MOLECULES, AND IONS Development of the Atomic Theory 18. Law of conservation of mass: mass is neither created nor destroyed. The total mass before a chemical reaction always equals the total mass after a chemical reaction. Law of definite proportion: a given compound always contains exactly the same proportion of elements by mass. For example, water is always 1 g hydrogen for every 8 g oxygen. Law of multiple proportions: When two elements form a series of compounds, the ratios of the mass of the second element that combine with 1 g of the first element always can be reduced to small whole numbers. For CO2 and CO discussed in section 2.2, the mass ratios of oxygen that react with 1 g carbon in each compound are in a 2 : 1 ratio. 19. From Avogadro’s hypothesis (law), volume ratios are equal to molecule ratios at constant temperature and pressure. Therefore, we can write a balanced equation using the volume data, Cl2 + 3 F2 → 2 X. Two molecules of X contain 6 atoms of F and two atoms of Cl. The formula of X is ClF3 for a balanced reaction. 20. a. The composition of a substance depends on the numbers of atoms of each element making up the compound (depends on the formula of the compound) and not on the composition of the mixture from which it was formed. b. Avogadro’s hypothesis (law) implies that volume ratios are equal to molecule ratios at constant temperature and pressure. H2 + Cl2 → 2 HCl. From the balanced equation, the volume of HCl produced will be twice the volume of H2 (or Cl2) reacted. 21. Hydrazine: 1.44 × 10 −1 g H/g N; Ammonia: 2.16 × 10 −1 g H/g N; Hydrogen azide: 2.40 × 10 −2 g H/g N; Let's try all of the ratios: 0.144 0.216 0.216 3 = 6.00; = 9.00; = 1.50 = 0.0240 0.0240 0.144 2 All the masses of hydrogen in these three compounds can be expressed as simple wholenumber ratios. The g H/g N in hydrazine, ammonia, and hydrogen azide are in the ratios 6 : 9 : 1. 1 2 22. CHAPTER 2 ATOMS, MOLECULES, AND IONS The law of multiple proportions does not involve looking at the ratio of the mass of one element with the total mass of the compounds. To illustrate the law of multiple proportions, we compare the mass of carbon that combines with 1.0 g of oxygen in each compound: Compound 1: 27.2 g C and 72.8 g O (100.0 - 27.2 = mass O) Compound 2: 42.9 g C and 57.1 g O (100.0 - 42.9 = mass O) The mass of carbon that combines with 1.0 g of oxygen is: Compound 1: 27.2 g C = 0.374 g C/g O 72.8 g O Compound 2: 42.9 g C = 0.751 g C/g O 57.1 g O 0.751 2 = ; this supports the law of multiple proportions as this carbon ratio is a whole 0.374 1 number. 23. To get the atomic mass of H to be 1.00, we divide the mass that reacts with 1.00 g of oxygen by 0.126, that is, 0.126/0.126 = 1.00. To get Na, Mg, and O on the same scale, we do the same division. Na: 2.875 1.500 1.00 = 22.8; Mg: = 11.9; O: = 7.94 0.126 0.126 0.126 H O Na Mg Relative value 1.00 7.94 22.8 11.9 Accepted value 1.0079 15.999 22.99 24.31 The atomic masses of O and Mg are incorrect. The atomic masses of H and Na are close. Something must be wrong about the assumed formulas of the compounds. It turns out that the correct formulas are H2O, Na2O, and MgO. The smaller discrepancies result from the error in the assumed atomic mass of H. The Nature of the Atom 24. Deflection of cathode rays by magnetic and electrical fields led to the conclusion that they were negatively charged. The cathode ray was produced at the negative electrode and repelled by the negative pole of the applied electrical field. β particles are electrons. A cathode ray is a stream of electrons (β particles). 25. Density of hydrogen nucleus (contains one proton only): CHAPTER 2 ATOMS, MOLECULES, AND IONS Vnucleus = 3 4 3 4 π r = (3.14) (5 × 10 −14 cm) 3 = 5 × 10 − 40 cm 3 3 3 d = density = 1.67 × 10 −24 g = 3 × 1015 g/cm 3 − 40 3 5 × 10 cm Density of H atom (contains one proton and one electron): Vatom = d= 26. 4 (3.14) (1 × 10 −8 cm) 3 = 4 × 10 − 24 cm 3 3 1.67 × 10 −24 g + 9 × 10 −28 g = 0.4 g/cm 3 4 × 10 − 24 cm 3 From Fig. 2.13 of the text, the average diameter of the nucleus is ≈ 10 −13 cm, and the average diameter of the volume where the electrons roam about is ≈ 10 −8 cm. 10 −8 cm = 105; −13 10 cm 1 mile 5280 ft 63,360 in = = 1 grape 1 grape 1 grape Because the grape needs to be 105 times smaller than a mile, the diameter of the grape would need to be 63,360/(1 × 105) ≈ 0.6 in. This is a reasonable size for a grape. 27. First, divide all charges by the smallest quantity, 6.40 × 10−13. 2.56 × 10 −12 = 4.00; 6.40 × 10 −13 7.68 = 12.00; 0.640 3.84 = 6.00 0.640 Because all charges are whole-number multiples of 6.40 × 10−13 zirkombs, the charge on one electron could be 6.40 × 10−13 zirkombs. However, 6.40 × 10−13 zirkombs could be the charge of two electrons (or three electrons, etc.). All one can conclude is that the charge of an electron is 6.40 × 10−13 zirkombs or an integer fraction of 6.40 × 10−13. 28. The proton and neutron have similar mass, with the mass of the neutron slightly larger than that of the proton. Each of these particles has a mass approximately 1800 times greater than that of an electron. The combination of the protons and the neutrons in the nucleus makes up the bulk of the mass of an atom, but the electrons make the greatest contribution to the chemical properties of the atom. 29. If the plum pudding model were correct (a diffuse positive charge with electrons scattered throughout), then α particles should have traveled through the thin foil with very minor deflections in their path. This was not the case because a few of the α particles were deflected at very large angles. Rutherford reasoned that the large deflections of these α particles could be caused only by a center of concentrated positive charge that contains most of the atom’s mass (the nuclear model of the atom). 4 CHAPTER 2 ATOMS, MOLECULES, AND IONS Elements and the Periodic Table 30. a. A molecule has no overall charge (an equal number of electrons and protons are present). Ions, on the other hand, have electrons added to form anions (negatively charged ions) or electrons removed to form cations (positively charged ions). b. The sharing of electrons between atoms is a covalent bond. An ionic bond is the force of attraction between two oppositely charged ions. c. A molecule is a collection of atoms held together by covalent bonds. A compound is composed of two or more different elements having constant composition. Covalent and/or ionic bonds can hold the atoms together in a compound. Another difference is that molecules do not necessarily have to be compounds. H2 is two hydrogen atoms held together by a covalent bond. H2 is a molecule, but it is not a compound; H2 is a diatomic element. d. An anion is a negatively charged ion, for example, Cl−, O2−, and SO42− are all anions. A cation is a positively charged ion, for example, Na+, Fe3+, and NH4+ are all cations. 31. The atomic number of an element is equal to the number of protons in the nucleus of an atom of that element. The mass number is the sum of the number of protons plus neutrons in the nucleus. The atomic mass is the actual mass of a particular isotope (including electrons). As we will see in Chapter 3, the average mass of an atom is taken from a measurement made on a large number of atoms. The average atomic mass value is listed in the periodic table. 32. a. Metals: Mg, Ti, Au, Bi, Ge, Eu, and Am. Nonmetals: Si, B, At, Rn, and Br. b. Si, Ge, B, and At. The elements at the boundary between the metals and the nonmetals are B, Si, Ge, As, Sb, Te, Po, and At. Aluminum has mostly properties of metals, so it is generally not classified as a metalloid. 33. a. The noble gases are He, Ne, Ar, Kr, Xe, and Rn (helium, neon, argon, krypton, xenon, and radon). Radon has only radioactive isotopes. In the periodic table, the whole number enclosed in parentheses is the mass number of the longest-lived isotope of the element. b. promethium (Pm) and technetium (Tc) 34. Carbon is a nonmetal. Silicon and germanium are called metalloids as they exhibit both metallic and nonmetallic properties. Tin and lead are metals. Thus metallic character increases as one goes down a family in the periodic table. The metallic character decreases from left to right across the periodic table. 35. a. Five: F, Cl, Br, I, and At b. Six: Li, Na, K, Rb, Cs, and Fr (H is not considered an alkali metal.) c. 14: Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, and Lu d. 40: All elements in the block defined by Sc, Zn, Uub, and Ac are transition metals. CHAPTER 2 36. 5 a. Cobalt is element 27. A = mass number = 27 + 31 = 58; 58 Co 27 b. 37. ATOMS, MOLECULES, AND IONS 10 B 5 c. 23 Mg 12 d. 132 I 53 19 F 9 e. 65 Cu 29 f. a. 24 Mg: 12 b. 24 Mg2+: 12 12 p, 12 n, 10 e c. 59 Co2+: 27 d. 59 Co3+: 27 27 p, 32 n, 24 e e. 59 Co: 27 f. 79 Se: 34 34 p, 45 n, 34 e g. 79 2− Se : 34 34 p, 45 n, 36 e h. 63 Ni: 28 28 p, 35 n, 28 e i. 59 2+ Ni : 28 28 p, 31 n, 26 e 12 protons, 12 neutrons, 12 electrons 27 p, 32 n, 25 e 27 p, 32 n, 27 e 38. Number of Protons in Nucleus Number of Neutrons in Nucleus Number of Electrons Net Charge 238 92 U 92 146 92 0 40 2+ Ca 20 20 20 18 2+ 51 3+ V 23 23 28 20 3+ 89 Y 39 39 50 39 0 79 − Br 35 35 44 36 1− 31 3− P 15 15 16 18 3− Symbol 39. Atomic number = 63 (Eu); net charge = +63 − 60 = 3+; mass number = 63 + 88 = 151; 3+ symbol: 151 63 Eu Atomic number = 50 (Sn); mass number = 50 + 68 = 118; net charge = +50 − 48 = 2+; 2+ symbol: 118 50 Sn . 6 40. CHAPTER 2 ATOMS, MOLECULES, AND IONS Atomic number = 16 (S); net charge = +16 − 18 = 2−; mass number = 16 + 18 = 34; symbol: 34 S2− 16 Atomic number = 16 (S); net charge = +16 −18 = 2−; Mass number = 16 + 16 = 32; symbol: 32 S2− 16 41. 42. In ionic compounds, metals lose electrons to form cations, and nonmetals gain electrons to form anions. Group 1A, 2A and 3A metals form stable 1+, 2+ and 3+ charged cations, respectively. Group 5A, 6A, and 7A nonmetals form 3!, 2! and 1! charged anions, respectively. a. Lose 2 e − to form Ra2+. b. Lose 3 e − to form In3+. c. Gain 3 e − to form P 3− . d. Gain 2 e − to form Te 2− . e. Gain 1 e − to form Br!. f. Lose 1 e − to form Rb+. See Exercise 41 for a discussion of charges various elements form when in ionic compounds. a. Element 13 is Al. Al forms 3+ charged ions in ionic compounds. Al3+ b. Se2− c. Ba2+ d. N3− e. Fr+ f. Br− Nomenclature 43. AlCl3, aluminum chloride; CrCl3, chromium(III) chloride; ICl3, iodine trichloride; AlCl3 and CrCl3 are ionic compounds following the rules for naming ionic compounds. The major difference is that CrCl3 contains a transition metal (Cr) that generally exhibits two or more stable charges when in ionic compounds. We need to indicate which charged ion we have in the compound. This is generally true whenever the metal in the ionic compound is a transition metal. ICl3 is made from only nonmetals and is a covalent compound. Predicting formulas for covalent compounds is extremely difficult. Because of this, we need to indicate the number of each nonmetal in the binary covalent compound. The exception is when there is only one of the first species present in the formula; when this is the case, mono- is not used (it is assumed). 44. a. Dinitrogen monoxide is correct. N and O are both nonmetals resulting in a covalent compound. We need to use the covalent rules of nomenclature. The other two names are for ionic compounds. b. Copper(I) oxide is correct. With a metal in a compound, we have an ionic compound. Because copper, like most transition metals, forms at least a couple of different stable charged ions, we must indicate the charge on copper in the name. Copper oxide could be CuO or Cu2O, hence why we must give the charge of most transition metal compounds. Dicopper monoxide is the name if this were a covalent compound, which it is not. c. Lithium oxide is correct. Lithium forms 1+ charged ions in stable ionic compounds. Because lithium is assumed to form 1+ ions in compounds, we do not need to indicate the charge of the metal ion in the compound. Dilithium monoxide would be the name if Li2O were a covalent compound (a compound composed of only nonmetals). CHAPTER 2 45. 46. 47. 48. 49. 50. ATOMS, MOLECULES, AND IONS a. sulfur difluoride b. dinitrogen tetroxide c. iodine trichloride d. tetraphosphorus hexoxide a. sodium perchlorate b. magnesium phosphate c. aluminum sulfate d. sulfur difluoride e. sulfur hexafluoride f. g. sodium dihydrogen phosphate h. lithium nitride i. sodium hydroxide j. magnesium hydroxide k. aluminum hydroxide l. silver chromate sodium hydrogen phosphate a. copper(I) iodide b. copper(II) iodide d. sodium carbonate e. sodium hydrogen carbonate or sodium bicarbonate f. tetrasulfur tetranitride g. sulfur hexafluoride i. barium chromate j. c. cobalt(II) iodide h. sodium hypochlorite ammonium nitrate a. acetic acid b. ammonium nitrite c. colbalt(III) sulfide d. iodine monochloride e. lead(II) phosphate f. potassium iodate g. sulfuric acid h. strontium nitride i. aluminum sulfite j. k. sodium chromate l. hypochlorous acid tin(IV) oxide a. SO2 b. SO3 c. Na2SO3 d. KHSO3 e. Li3N f. g. Cr(C2H3O2)2 h. SnF4 Cr2(CO3)3 i. NH4HSO4: composed of NH4+ and HSO4− ions j. (NH4)2HPO4 k. KClO4 l. NaH m. HBrO n. HBr a. Na2O b. Na2O2 c. KCN e. SiCl4 f. PbO g. PbO2 d. Cu(NO3)2 h. CuCl i. GaAs: We would predict the stable ions to be Ga and As3−. j. CdSe 3+ m. HNO2 51. 7 k. ZnS l. Hg2Cl2: Mercury(I) exists as Hg22+. n. P2O5 a. Pb(C2H3O2)2; lead(II) acetate b. CuSO4; copper(II) sulfate c. CaO; calcium oxide d. MgSO4; magnesium sulfate e. Mg(OH)2; magnesium hydroxide f. CaSO4; calcium sulfate g. N2O; dinitrogen monoxide or nitrous oxide (common name) 8 52. CHAPTER 2 ATOMS, MOLECULES, AND IONS a. Iron forms 2+ and 3+ charged ions; we need to include a Roman numeral for iron. Iron(III) chloride is correct. b. This is a covalent compound so use the covalent rules. Nitrogen dioxide is correct. c. This is an ionic compound, so use the ionic rules. Calcium oxide is correct. Calcium only forms stable 2+ ions when in ionic compounds, so no Roman numeral is needed. d. This is an ionic compound, so use the ionic rules. Aluminum sulfide is correct. e. This is an ionic compound, so use the ionic rules. Mg is magnesium. Magnesium acetate is correct. f. Because phosphate has a 3− charge, the charge on iron is 3+. Iron(III) phosphate is correct. g. This is a covalent compound, so use the covalent rules. Diphosphorus pentasulfide is correct. h. Because each sodium is 1+ charged, we have the O22− (peroxide) ion present. Sodium peroxide is correct. Note that sodium oxide would be Na2O. 53. i. HNO3 is nitric acid, not nitrate acid. Nitrate acid does not exist. j. H2S is hydrosulfuric acid or dihydrogen sulfide or just hydrogen sulfide (common name). H2SO4 is sulfuric acid. a. nitric acid, HNO3 b. perchloric acid, HClO4 d. sulfuric acid, H2SO4 e. phosphoric acid, H3PO4 c. acetic acid, HC2H3O2 Additional Exercises 54. a. The smaller parts are electrons and the nucleus. The nucleus is broken down into protons and neutrons, which can be broken down into quarks. For our purpose, electrons, neutrons, and protons are the key smaller parts of an atom. b. All atoms of hydrogen have 1 proton in the nucleus. Different isotopes of hydrogen have 0, 1, or 2 neutrons in the nucleus. Because we are talking about atoms, this implies a neutral charge, which dictates 1 electron present for all hydrogen atoms. If charged ions were included, then different ions/atoms of H could have different numbers of electrons. c. Hydrogen atoms always have 1 proton in the nucleus, and helium atoms always have 2 protons in the nucleus. The number of neutrons can be the same for a hydrogen atom and a helium atom. Tritium (3H) and 4He both have 2 neutrons. Assuming neutral atoms, then the number of electrons will be 1 for hydrogen and 2 for helium. CHAPTER 2 ATOMS, MOLECULES, AND IONS 9 d. Water (H2O) is always 1 g hydrogen for every 8 g of O present, whereas H2O2 is always 1 g hydrogen for every 16 g of O present. These are distinctly different compounds, each with its own unique relative number and types of atoms present. e. A chemical equation involves a reorganization of the atoms. Bonds are broken between atoms in the reactants, and new bonds are formed in the products. The number and types of atoms between reactants and products do not change. Because atoms are conserved in a chemical reaction, mass is also conserved. 55. A compound will always have a constant composition by mass. From the initial data given, the mass ratio of H : S : O in sulfuric acid (H2SO4) is: 2.02 32.07 64.00 : : = 1 : 15.9 : 31.7 2.02 2.02 2.02 If we have 7.27 g H, then we will have 7.27 × 15.9 = 116 g S and 7.27 × 31.7 = 230. g O in the second sample of H2SO4. 56. Mass is conserved in a chemical reaction. Mass: Chromium(III) oxide + aluminum → chromium + aluminum oxide 34.0 g 12.1 g 23.3 ? Mass aluminum oxide produced = (34.0 + 12.1) ! 23.3 = 22.8 g 57. From the Na2X formula, X has a 2! charge. Because 36 electrons are present, X has 34 protons, 79 ! 34 = 45 neutrons, and is selenium. a. True. Nonmetals bond together using covalent bonds and are called covalent compounds. b. False. The isotope has 34 protons. c. False. The isotope has 45 neutrons. d. False. The identity is selenium, Se. 58. a. Fe2+: 26 protons (Fe is element 26.); protons − electrons = charge, 26 − 2 = 24 electrons; FeO is the formula because the oxide ion has a 2− charge. b. Fe3+: 26 protons; 23 electrons; Fe2O3 c. Ba2+: 56 protons; 54 electrons; BaO d. Cs+: 55 protons; 54 electrons; Cs2O e. S2−: 16 protons; 18 electrons; Al2S3 f. P3−: 15 protons; 18 electrons; AlP g. Br−: 35 protons; 36 electrons; AlBr3 h. N3−: 7 protons; 10 electrons; AlN 59. From the XBr2 formula, the charge on element X is 2+. Therefore, the element has 88 protons, which identifies it as radium, Ra. 230 − 88 = 142 neutrons. 10 CHAPTER 2 ATOMS, MOLECULES, AND IONS 60. The solid residue must have come from the flask. 61. In the case of sulfur, SO42− is sulfate, and SO32− is sulfite. By analogy: SeO42−: selenate; SeO32−: selenite; TeO42−: tellurate; TeO32−: tellurite 62. The polyatomic ions and acids in this problem are not named in the text. However, they are all related to other ions and acids named in the text that contain a same group element. Because HClO4 is perchloric acid, HBrO4 would be perbromic acid. Because ClO3− is the chlorate ion, KIO3 would be potassium iodate. Since ClO2− is the chlorite ion, NaBrO2 would be sodium bromite. And finally, because HClO is hypochlorous acid, HIO would be hypoiodous acid. 63. If the formula is InO, then one atomic mass of In would combine with one atomic mass of O, or: 4.784 g In A = , A = atomic mass of In = 76.54 16.00 1.000 g O If the formula is In2O3, then two times the atomic mass of In will combine with three times the atomic mass of O, or: 2A 4.784 g In = , A = atomic mass of In = 114.8 (3)16.00 1.000 g O The latter number is the atomic mass of In used in the modern periodic table. 64. a. Ca2+ and N3−: Ca3N2, calcium nitride b. K+ and O2−: K2O, potassium oxide c. Rb+ and F−: RbF, rubidium fluoride d. Mg2+ and S2−: MgS, magnesium sulfide e. Ba2+ and I−: BaI2, barium iodide f. Al3+ and Se2−: Al2Se3, aluminum selenide g. Cs+ and P3−: Cs3P, cesium phosphide h. In3+ and Br−: InBr3, indium(III) bromide; In also forms In+ ions, but you would predict In3+ ions from its position in the periodic table. 65. The cation has 51 protons and 48 electrons. The number of protons corresponds to the atomic number. Thus this is element 51, antimony. There are 3 fewer electrons than protons. Therefore, the charge on the cation is 3+. The anion has one-third the number of protons of the cation which corresponds to 17 protons; this is element 17, chlorine. The number of electrons in this anion of chlorine is 17 + 1 = 18 electrons. The anion must have a charge of 1−. The formula of the compound formed between Sb3+ and Cl– is SbCl3. The name of the compound is antimony(III) chloride. The Roman numeral is used to indicate the charge of Sb because the predicted charge is not obvious from the periodic table. 66. a. This is element 52, tellurium. Te forms stable 2− charged ions in ionic compounds (like other oxygen family members). CHAPTER 2 67. ATOMS, MOLECULES, AND IONS b. Rubidium. Rb, element 37, forms stable 1+ charged ions. c. Argon. Ar is element 18. d. Astatine. At is element 85. 11 Because this is a relatively small number of neutrons, the number of protons will be very close to the number of neutrons present. The heavier elements have significantly more neutrons than protons in their nuclei. Because this element forms anions, it is a nonmetal and will be a halogen because halogens form stable 1! charged ions in ionic compounds. From the halogens listed, chlorine, with an average atomic mass of 35.45, fits the data. The two isotopes are 35Cl and 37Cl, and the number of electrons in the 1! ion is 18. Note that because the atomic mass of chlorine listed in the periodic table is closer to 35 than 37, we can assume that 35Cl is the more abundant isotope. This is discussed in Chapter 3. Challenge Problems 68. Because the gases are at the same temperature and pressure, the volumes are directly proportional to the number of molecules present. Let’s consider hydrogen and oxygen to be monatomic gases and that water has the simplest possible formula (HO). We have the equation: H + O → HO But the volume ratios are also equal to the molecule ratios, which correspond to the coefficients in the equation: 2 H + O → 2 HO Because atoms cannot be created nor destroyed in a chemical reaction, this is not possible. To correct this, we can make oxygen a diatomic molecule: 2 H + O2 → 2 HO This does not require hydrogen to be diatomic. Of course, if we know water has the formula H2O, we get: 2 H + O2 → 2 H2O The only way to balance this is to make hydrogen diatomic: 2 H2 + O2 → 2 H2O 69. a. Both compounds have C2H6O as the formula. Because they have the same formula, their mass percent composition will be identical. However, these are different compounds with different properties because the atoms are bonded together differently. These compounds are called isomers of each other. b. When wood burns, most of the solid material in wood is converted to gases, which escape. The gases produced are most likely CO2 and H2O. c. The atom is not an indivisible particle but is instead composed of other smaller particles, for example, electrons, neutrons, and protons. 12 CHAPTER 2 ATOMS, MOLECULES, AND IONS d. The two hydride samples contain different isotopes of either hydrogen and/or lithium. Although the compounds are composed of different isotopes, their properties are similar because different isotopes of the same element have similar properties (except, of course, their mass). 70. For each experiment, divide the larger number by the smaller. In doing so, we get: experiment 1 experiment 2 experiment 3 X = 1.0 Y = 1.4 X = 1.0 Y = 10.5 Z = 1.0 Y = 3.5 Our assumption about formulas dictates the rest of the solution. For example, if we assume that the formula of the compound in experiment 1 is XY and that of experiment 2 is YZ, we get relative masses of: X = 2.0; Y = 21; Z = 15 (= 21/1.4) and a formula of X3Y for experiment 3 [three times as much X must be present in experiment 3 as compared to experiment 1 (10.5/3.5 = 3)]. However, if we assume the formula for experiment 2 is YZ and that of experiment 3 is XZ, then we get: X = 2.0; Y = 7.0; Z = 5.0 (= 7.0/1.4) and a formula of XY3 for experiment 1. Any answer that is consistent with your initial assumptions is correct. The answer to part d depends on which (if any) of experiments 1 and 3 have a formula of XY in the compound. If the compound in expt. 1 has formula XY, then: 21 g XY × 4.2 g Y = 19.2 g Y (and 1.8 g X) (4.2 + 0.4) g XY If the compound in experiment 3 has the XY formula, then: 21 g XY H 7 .0 g Y = 16.3 g Y (and 4.7 g X) (7.0 + 2.0) g XY Note that it could be that neither experiment 1 nor experiment 3 has XY as the formula. Therefore, there is no way of knowing an absolute answer here. 71. Compound I: 14.0 g R 4.67 g R 7.00 g R 1.56 g R = ; Compound II: = 3.00 g Q 1.00 g Q 4.50 g Q 1.00 g Q The ratio of the masses of R that combines with 1.00 g Q is 4.67 = 2.99 ≈ 3. 1.56 CHAPTER 2 ATOMS, MOLECULES, AND IONS 13 As expected from the law of multiple proportions, this ratio is a small whole number. Because compound I contains three times the mass of R per gram of Q as compared with compound II (RQ), the formula of compound I should be R3Q. 72. Most of the mass of the atom is due to the protons and the neutrons in the nucleus, and protons and neutrons have about the same mass (1.67 × 10−24 g). The ratio of the mass of the molecule to the mass of a nuclear particle will give a good approximation of the number of nuclear particles (protons and neutrons) present. 7.31 × 10 −23 g = 43.8 ≈ 44 nuclear particles 1.67 × 10 −24 g Thus there are 44 protons and neutrons present. If the number of protons equals the number of neutrons, we have 22 protons in the molecule. One possibility would be the molecule CO2 [6 + 2(8) = 22 protons]. 73. Avogadro proposed that equal volumes of gases (at constant temperature and pressure) contain equal numbers of molecules. In terms of balanced equations, Avogadro’s hypothesis (law) implies that volume ratios will be identical to molecule ratios. Assuming one molecule of octane reacting, then 1 molecule of CxHy produces 8 molecules of CO2 and 9 molecules of H2O. CxHy + n O2 → 8 CO2 + 9 H2O. Because all the carbon in octane ends up as carbon in CO2, octane must contain 8 atoms of C. Similarly, all hydrogen in octane ends up as hydrogen in H2O, so one molecule of octane must contain 9 × 2 = 18 atoms of H. Octane formula = C8H18 and the ratio of C:H = 8:18 or 4:9. 74. Let Xa be the formula for the atom/molecule X, Yb be the formula for the atom/molecule Y, XcYd be the formula of compound I between X and Y, and XeYf be the formula of compound II between X and Y. Using the volume data, the following would be the balanced equations for the production of the two compounds. Xa + 2 Yb → 2 XcYd; 2 Xa + Yb → 2 XeYf From the balanced equations, a = 2c = e and b = d = 2f. Substituting into the balanced equations: X2c + 2 Y2f → 2 XcY2f 2 X2c + Y2f → 2 X2cYf For simplest formulas, assume that c = f = 1. Thus: X2 + 2 Y2 → 2 XY2 and 2 X2 + Y2 → 2 X2Y Compound I = XY2: If X has relative mass of 1.00, 1.00 = 0.3043, y = 1.14. 1.00 + 2 y 14 CHAPTER 2 ATOMS, MOLECULES, AND IONS Compound II = X2Y: If X has relative mass of 1.00, 2.00 = 0.6364, y = 1.14. 2.00 + y The relative mass of Y is 1.14 times that of X. Thus if X has an atomic mass of 100, then Y will have an atomic mass of 114. Marathon Problem 75. a. For each set of data, divide the larger number by the smaller number to determine relative masses. 0.602 = 2.04; A = 2.04 when B = 1.00 0.295 0.401 = 2.33; C = 2.33 when B = 1.00 0.172 0.374 = 1.17; C = 1.17 when A = 1.00 0.320 To have whole numbers, multiply the results by 3. Data set 1: A = 6.1 and B = 3.0 Data set 2: C = 7.0 and B = 3.0 Data set 3: C = 3.5 and A = 3.0 or C = 7.0 and A = 6.0 Assuming 6.0 for the relative mass of A, the relative masses would be A = 6.0, B = 3.0, and C = 7.0 (if simplest formulas are assumed). b. Gas volumes are proportional to the number of molecules present. There are many possible correct answers for the balanced equations. One such solution that fits the gas volume data is: 6 A2 + B4 → 4 A3B B4 + 4 C3 → 4 BC3 3 A2 + 2 C3 → 6 AC In any correct set of reactions, the calculated mass data must match the mass data given initially in the problem. Here, the new table of relative masses would be: 6 ( mass A 2 ) 0.602 = ; mass A2 = 0.340(mass B4) mass B 4 0.295 4 ( mass C 3 ) 0.401 = ; mass C3 = 0.583(mass B4) mass B 4 0.172 2 (mass C 3 ) 0.374 = ; mass A2 = 0.570(mass C3) 3 (mass A 2 ) 0.320 CHAPTER 2 ATOMS, MOLECULES, AND IONS 15 Assume some relative mass number for any of the masses. We will assume that mass B = 3.0, so mass B4 = 4(3.0) = 12. Mass C3 = 0.583(12) = 7.0, mass C = 7.0/3 Mass A2 = 0.570(7.0) = 4.0, mass A = 4.0/2 = 2.0 When we assume a relative mass for B = 3.0, then A = 2.0 and C = 7.0/3. The relative masses having all whole numbers would be A = 6.0, B = 9.0, and C = 7.0. Note that any set of balanced reactions that confirms the initial mass data is correct. This is just one possibility. CHAPTER 3 STOICHIOMETRY Atomic Masses and the Mass Spectrometer 23. A = 0.0800(45.95269) + 0.0730(46.951764) + 0.7380(47.947947) + 0.0550(48.947841) + 0.0540(49.944792) = 47.88 amu This is element Ti (titanium). 24. Let x = % of 151Eu and y = % of 153Eu, then x + y = 100 and y = 100 − x. 151.96 = x(150.9196) + (100 − x)(152.9209) 100 15196 = (150.9196)x + 15292.09 − (152.9209)x, −96 = −(2.0013)x x = 48%; 48% 151Eu and 100 − 48 = 52% 153Eu 25. 186.207 = 0.6260(186.956) + 0.3740(A), 186.207 - 117.0 = 0.3740(A) A= 26. 69.2 = 185 amu (A = 184.95 amu without rounding to proper significant figures) 0.3740 A = 0.0140(203.973) + 0.2410(205.9745) + 0.2210(206.9759) + 0.5240(207.9766) A = 2.86 + 49.64 + 45.74 + 109.0 = 207.2 amu; from the periodic table, the element is Pb. 27. There are three peaks in the mass spectrum, each 2 mass units apart. This is consistent with two isotopes, differing in mass by two mass units. The peak at 157.84 corresponds to a Br2 molecule composed of two atoms of the lighter isotope. This isotope has mass equal to 157.84/2, or 78.92. This corresponds to 79Br. The second isotope is 81Br with mass equal to 161.84/2 = 80.92. The peaks in the mass spectrum correspond to 79Br2, 79Br81Br, and 81Br2 in order of increasing mass. The intensities of the highest and lowest masses tell us the two isotopes are present at about equal abundance. The actual abundance is 50.68% 79Br and 49.32% 81Br. 16 CHAPTER 3 28. STOICHIOMETRY 17 Scaled Intensity Compound Mass Intensity Largest Peak = 100 __________________________________________________________________ H2120Te 121.92 0.09 0.3 H2122Te 123.92 2.46 7.1 H2123Te 124.92 0.87 2.5 H2124Te 125.92 4.61 13.4 H2125Te 126.92 6.99 20.3 H2126Te 127.92 18.71 54.3 H2128Te 129.92 31.79 92.2 H2130Te 131.93 34.48 100.0 100 50 0 122 29. 124 126 128 130 132 134 GaAs can be either 69GaAs or 71GaAs. The mass spectrum for GaAs will have two peaks at 144 (= 69 + 75) and 146 (= 71 + 75) with intensities in the ratio of 60 : 40 or 3 : 2. 144 146 Ga2As2 can be 69Ga2As2, 69Ga71GaAs2, or 71Ga2As2. The mass spectrum will have three peaks at 288, 290, and 292 with intensities in the ratio of 36 : 48 : 16 or 9 : 12 : 4. We get this ratio from the following probability table: 18 CHAPTER 3 69 Ga (0.60) 71 Ga (0.40) 69 Ga (0.60) 0.36 0.24 71 Ga (0.40) 0.24 0.16 288 290 STOICHIOMETRY 292 Moles and Molar Masses 30. 4.24 g C6H6 × 1 mol = 5.43 × 10 −2 mol C6H6 78.11 g 6.022 × 10 23 molecules = 3.27 × 1022 molecules C6H6 mol 5.43 × 10 −2 mol C6H6 × Each molecule of C6H6 contains 6 atoms C + 6 atoms H = 12 total atoms. 3.27 × 1022 molecules C6H6 × 12 atoms total = 3.92 × 1023 atoms total molecule 0.224 mol H2O × 18.02 g = 4.04 g H2O mol 0.224 mol H2O × 6.022 × 10 23 molecules = 1.35 × 1023 molecules H2O mol 1.35 × 1023 molecules H2O × 3 atoms total = 4.05 × 1023 atoms total molecule 2.71 × 1022 molecules CO2 × 1 mol = 4.50 × 10 −2 mol CO2 6.022 × 10 23 molecules 4.50 × 10 −2 mol CO2 × 44.01 g = 1.98 g CO2 mol 2.71 × 1022 molecules CO2 × 3.35 × 1022 atoms total × 3 atoms total = 8.13 × 1022 atoms total molecule CO 2 1 molecule = 5.58 × 1021 molecules CH3OH 6 atoms total CHAPTER 3 STOICHIOMETRY 5.58 × 1021 molecules CH3OH × 9.27 × 10 −3 mol CH3OH × 31. a. 20.0 mg C8H10N4O2 × 19 1 mol = 9.27 × 10 −3 mol CH3OH 6.022 × 10 23 molecules 32.04 g = 0.297 g CH3OH mol 1g 1 mol × = 1.03 × 10 −4 mol C8H10N4O2 1000 mg 194.20 g b. 2.72 × 1021 molecules C2H5OH × c. 1.50 g CO2 × 32. 1 mol = 4.52 × 10 −3 mol C2H5OH 23 6.022 × 10 molecules 1 mol = 3.41 × 10 −2 mol CO2 44.01 g a. A chemical formula gives atom ratios as well as mole ratios. We will use both ideas to show how these conversion factors can be used. Molar mass of C2H5O2N = 2(12.01) + 5(1.008) + 2(16.00) + 14.0l = 75.07 g/mol 5.00 g C2H5O2N × 1 mol C 2 H 5 O 2 N 6.022 × 10 23 molecules C 2 H 5 O 2 N × 75.07 g C 2 H 5 O 2 N mol C 2 H 5 O 2 N 1 atom N = 4.01 × 1022 atoms N molecule C 2 H 5 O 2 N H b. Molar mass of Mg3N2 = 3(24.31) + 2(14.01) = 100.95 g/mol 5.00 g Mg3N2 × 1 mol Mg 3 N 2 6.022 × 10 23 formula units Mg 3 N 2 × 100.95 g Mg 3 N 2 mol Mg 3 N 2 × 2 atoms = 5.97 × 1022 atoms N mol Mg 3 N 2 c. Molar mass of Ca(NO3)2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol 5.00 g Ca(NO3)2 × 1 mol Ca ( NO 3 ) 2 164.10 g Ca ( NO 3 ) 2 × 2 mol N mol Ca ( NO 3 ) 2 × 6.022 × 10 23 atoms N = 3.67 × 1022 atoms N mol N 20 CHAPTER 3 STOICHIOMETRY d. Molar mass of N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol 5.00 g N2O4 × 6.022 × 10 23 atoms N 1 mol N 2 O 4 2 mol N × × mol N 92.02 g N 2 O 4 mol N 2 O 4 = 6.54 × 1022 atoms N 33. 4.0 g H2 × 1 mol H 2 2 mol H 6.022 × 10 23 atoms H × × = 2.4 × 1024 atoms 2.016 g H 2 1 mol H 2 1 mol H 1 mol He 6.022 × 10 23 atoms He × = 6.0 × 1023 atoms 4.003 g He 1 mol He 4.0 g He × 1.0 mol F2 H 44.0 g CO2 × 146 g SF6 × 2 mol F 6.022 × 10 23 atoms F × = 1.2 × 1024 atoms 1 mol F2 1 mol F 1 mol CO 2 3 mol atoms (1 C + 2 O) 6.022 × 10 23 atoms × × 44.01 g CO 2 1 mol CO 2 1 mol atoms = 1.81 × 1024 atoms 1 mol SF6 7 mol atoms (1 S + 6 F) 6.022 ×10 23 atoms × × = 4.21 × 1024 atoms 146.07 g SF6 1 mol SF6 1 mol atoms The order is: 4.0 g He < 1.0 mol F2 < 44.0 g CO2 < 4.0 g H2 < 146 g SF6 34. a. 14 mol C H 12.011 g 1.0079 g 14.007 g + 18 mol H × + 2 mol N × mol C mol H mol N + 5 mol O × b. 10.0 g C14H18N2O5 × c. 1.56 mol × d. 5.0 mg × e. 15.999 g = 294.305 g/mol mol O 1 mol C14 H18 N 2 O 5 = 3.40 × 10−2 mol C14H18N2O5 294.3 g C14 H18 N 2 O 5 294.3 g = 459 g C14H18N2O5 mol 1g 1 mol 6.02 × 10 23 molecles × × = 1.0 × 1019 molecules C14H18N2O5 1000 mg 294.3 g mol 1.2 g C14H18N2O5 × 1 mol C14 H18 N 2 O 5 2 mol N 6.02 × 10 23 atoms N × × 294.3 g C14 H18 N 2 O 5 mol C14 H18 N 2 O 5 mol N = 4.9 × 1021 atoms N CHAPTER 3 f. STOICHIOMETRY 1.0 × 109 molecules × g. 1 molecule × 35. a. 1 mol 294.3 g × = 4.9 × 10−13 g 23 mol 6.02 × 10 atoms 1 mol 294.305 g × = 4.887 × 10−22 g C14H18N2O5 23 mol 6.022 × 10 atoms 2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol 1 mol = 3.023 mol C2H3Cl3O2 165.39 g b. 500.0 g × c. 21 2.0 × 10-2 mol × 165.39 g = 3.3 g C2H3Cl3O2 mol d. 5.0 g C2H3Cl3O2 × 1 mol 6.02 × 10 23 molecules 3 atoms Cl × × 165.39 g mol molecule = 5.5 × 1022 atoms of chlorine 36. 1 mol Cl 1 mol C 2 H 3Cl 3O 2 165.39 g C 2 H 3Cl 3O 2 = 1.6 g chloral hydrate × × 35.45 g 3 mol Cl mol C 2 H 3Cl 3O 2 e. 1.0 g Cl × f. 500 molecules × 1.0 lb flour × 1 mol 165.39 g × = 1.373 × 10−19 g 23 mol 6.022 × 10 molecules 454 g flour 30.0 × 10 −9 g C 2 H 4 Br2 1 mol C 2 H 4 Br2 × × lb flour g flour 187.9 g C 2 H 4 Br2 × 6.02 × 10 23 molecules = 4.4 × 1016 molecules C2H4Br2 mol C 2 H 4 Br2 Percent Composition 37. Molar mass = 20(12.01) + 29(1.008) + 19.00 + 3(16.00) = 336.43 g/mol Mass % C = 20(12.01) g C × 100 = 71.40% C 336.43 g compound Mass % H = 29(1.008) g H × 100 = 8.689% H 336.43 g compound 22 CHAPTER 3 Mass % F = STOICHIOMETRY 19.00 g F × 100 = 5.648% F 336.43 g compound Mass % O = 100.00 ! (71.40 + 8.689 + 5.648) = 14.26% O or: Mass % O = 38. a. 3(16.00) g O × 100 = 14.27% O 336.43 g compound C8H10N4O2: Molar mass = 8(12.01) + 10(1.008) + 4(14.0l) + 2(16.00) = 194.20 g/mol 8(12.01) g C 96.08 × 100 = × 100 = 49.47% C 194.20 g C8 H10 N 4 O 2 194.20 Mass % C = b. C12 H22O11: Molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol Mass % C = c. 12(12.01) g C × 100 = 42.10% C 342.30 g C12 H 22 O11 C2H5OH: Molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol Mass % C = 2(12.01) g C × 100 = 52.14% C 46.07 g C 2 H 5 OH The order from lowest to highest mass percentage of carbon is: sucrose (C12H22O11) < caffeine (C8H10N4O2) < ethanol (C2H5OH) 39. For each compound, determine the mass of 1 mole of compound, and then calculate the mass percentage of N in 1 mole of that compound. a. NO: % N = b. NO2: % N = 14.007 g N × 100 = 30.447% N 46.005 g NO 2 c. 28.014 g N × 100 = 30.447% N 92.010 g N 2 O 4 N2O4: % N = d. N2O: % N = 40. 14.007 g N × 100 = 46.681% N 30.006 g NO 28.014 g N × 100 = 63.649% N 44.013 g N 2 O Assuming 100.00 g cyanocobalamin: mol cyanocobalamin = 4.34 g Co × 1 mol Co 1 mol cyanocobalamin × 58.93 g Co mol Co = 7.36 × 10−2 mol cyanocobalamin CHAPTER 3 STOICHIOMETRY 23 x g cyanocobalamin 100.00 g , x = molar mass = 1360 g/mol = 1 mol cyanocobalamin 7.36 × 10 − 2 mol 41. There are 0.390 g Cu for every 100.000 g of fungal laccase. Let’s assume 100.000 g fungal laccase. Mol fungal laccase = 0.390 g Cu × 1 mol Cu 1 mol fungal laccase × = 1.53 × 10−3 mol 63.55 g Cu 4 mol Cu x g fungal laccase 100.000 g = , x = molar mass = 6.54 × 104 g/mol 1 mol fungal laccase 1.53 × 10 −3 mol 42. If we have 100.0 g of Portland cement, we have 50. g Ca3SiO5, 25 g Ca2SiO4, 12 g Ca3Al2O6, 8.0 g Ca2AlFeO5, and 3.5 g CaSO4•2H2O. Mass percent Ca: 1 mol Ca 3SiO 5 3 mol Ca 40.08 g Ca × × = 26 g Ca 228.33 g Ca 3SiO 5 1 mol Ca 3SiO 5 1 mol Ca 50. g Ca3SiO5 × 25 g Ca2SiO4 × 80.16 g Ca = 12 g Ca 172.25 g Ca 2SiO 4 12 g Ca3Al2O6 × 120.24 g Ca = 5.3 g Ca 270.20 g Ca 3 Al2 O 6 8.0 g Ca2AlFeO5 × 80.16 g Ca = 2.6 g Ca 242.99 g Ca 2 AlFeO5 3.5 g CaSO4•2H2O × 40.08 g Ca = 0.81 g Ca 172.18 g CaSO 4 • 2 H 2 O Mass of Ca = 26 + 12 + 5.3 + 2.6 + 0.81 = 47 g Ca % Ca = 47 g Ca × 100 = 47% Ca 100.0 g cement Mass percent Al: 12 g Ca3 Al2O6 × 53.96 g Al = 2.4 g Al 270.20 g Ca 3 Al 2 O 6 8.0 g Ca2AlFeO5 × 26.98 g Al = 0.89 g Al 242.99 g Ca 2 AlFeO5 24 CHAPTER 3 % Al = STOICHIOMETRY 2.4 g + 0.89 g × 100 = 3.3% Al 100.0 g Mass percent Fe: 8.0 g Ca2AlFeO5 × 1.8 g 55.85 g Fe = 1.8 g Fe; % Fe = × 100 = 1.8% Fe 242.99 g Ca 2 AlFeO5 100.0 g Empirical and Molecular Formulas 43. ⎛ 12.011 g ⎞ ⎛ 1.0079 g ⎞ ⎟⎟ + 2 mol H ⎜⎜ ⎟⎟ a. Molar mass of CH2O = 1 mol C ⎜⎜ ⎝ mol C ⎠ ⎝ mol H ⎠ ⎛ 15.999 g ⎞ ⎟⎟ = 30.026 g/mol + 1 mol O ⎜⎜ ⎝ mol O ⎠ %C= %O= 12.011 g C 2.0158 g H × 100 = 40.002% C; % H = × 100 = 6.7135% H 30.026 g CH 2 O 30.026 g CH 2 O 15.999 g O × 100 = 53.284% O or % O = 100.000 − (40.002 + 6.7135) 30.026 g CH 2 O = 53.285% b. Molar mass of C6H12O6 = 6(12.011) + 12(1.0079) + 6(15.999) = 180.155 g/mol %C= 72.066 g C × 100 = 40.002%; 180.155 g C 6 H12 O 6 %H= 12(1.0079) g × 100 = 6.7136% 180.155 g % O = 100.00 − (40.002 + 6.7136) = 53.284% c. Molar Mass of HC2H3O2 = 2(12.011) + 4(1.0079) + 2(15.999) = 60.052 g/mol %C= 24.022 g × 100 = 40.002%; 60.052 g %H= 4.0316 g × 100 = 6.7135% 60.052 g % O = 100.000 − (40.002 + 6.7135) = 53.285% All three compounds have the same empirical formula, CH2O, and different molecular formulas. The composition of all three in mass percent is also the same (within rounding differences). Therefore, elemental analysis will give us only the empirical formula. 44. a. The molecular formula is N2O4. The smallest whole number ratio of the atoms (the empirical formula) is NO2. CHAPTER 3 STOICHIOMETRY 25 b. Molecular formula: C3H6; empirical formula: CH2 c. Molecular formula: P4O10; empirical formula: P2O5 d. Molecular formula: C6H12O6; empirical formula: CH2O 45. Compound I: mass O = 0.6498 g HgxOy − 0.6018 g Hg = 0.0480 g O 0.6018 g Hg × 0.0480 g O × 1 mol Hg = 3.000 × 10−3 mol Hg 200.6 g Hg 1 mol O = 3.00 × 10−3 mol O 16.00 g O The mole ratio between Hg and O is 1 : 1, so the empirical formula of compound I is HgO. Compound II: mass Hg = 0.4172 g HgxOy − 0.016 g O = 0.401 g Hg 0.401 g Hg × 1 mol Hg 1 mol O = 2.00 × 10−3 mol Hg; 0.016 g O × = 1.0 × 10−3 mol O 200.6 g Hg 16.00 g O The mole ratio between Hg and O is 2 : 1, so the empirical formula is Hg2O. 46. Out of 100.00 g of adrenaline, there are: 56.79 g C × 1 mol C 1 mol H = 4.728 mol C; 6.56 g H × = 6.51 mol H 12.011 g C 1.008 g H 28.37 g O H 1 mol O 1 mol N = 1.773 mol O; 8.28 g N H = 0.591 mol N 15.999 g O 14.01 g N Dividing each mole value by the smallest number: 4.728 6.51 1.773 0.591 = 8.00; = 11.0; = 3.00; = 1.00 0.591 0.591 0.591 0.591 This gives adrenaline an empirical formula of C8H11O3N. 47. First, we will determine composition in mass percent. We assume that all the carbon in the 0.213 g CO2 came from the 0.157 g of the compound and that all the hydrogen in the 0.0310 g H2O came from the 0.157 g of the compound. 0.213 g CO2 × 12.01 g C 0.0581 g C = 0.0581 g C; % C = × 100 = 37.0% C 44.01 g CO 2 0.157 g compound 26 CHAPTER 3 0.0310 g H2O × STOICHIOMETRY 2.016 g H 3.47 × 10 −3 g = 3.47 × 10−3 g H; % H = × 100 = 2.21% H 18.02 g H 2 O 0.157 g We get the mass percent of N from the second experiment: 0.0230 g NH3 H %N= 14.01 g N = 1.89 × 10−2 g N 17.03 g NH 3 1.89 × 10 −2 g × 100 = 18.3% N 0.103 g The mass percent of oxygen is obtained by difference: % O = 100.00 − (37.0 + 2.21 + 18.3) = 42.5% O So, out of 100.00 g of compound, there are: 37.0 g C × 1 mol C 1 mol H = 3.08 mol C; 2.21 g H × = 2.19 mol H 12.01 g C 1.008 g H 18.3 g N × 1 mol N 1 mol O = 1.31 mol N; 42.5 g O × = 2.66 mol O 14.01 g N 16.00 g O Lastly, and often the hardest part, we need to find simple whole number ratios. Divide all mole values by the smallest number: 3.08 2.19 1.31 2.66 = 2.35; = 1.67; = 1.00; = 2.03 1.31 1.31 1.31 1.31 Multiplying all these ratios by 3 gives an empirical formula of C7H5N3O6. 48. Assuming 100.00 g of compound (mass oxygen = 100.00 g − 41.39 g C − 3.47 g H = 55.14 g O): 41.39 g C × 1 mol C 1 mol H = 3.446 mol C; 3.47 g H × = 3.44 mol H 12.011 g C 1.008 g H 55.14 g O × 1 mol O = 3.446 mol O 15.999 g O All are the same mole values, so the empirical formula is CHO. The empirical formula mass is 12.01 + 1.008 + 16.00 = 29.02 g/mol. Molar mass = 15.0 g = 116 g/mol 0.129 mol CHAPTER 3 STOICHIOMETRY 27 Molar mass 116 = = 4.00; molecular formula = (CHO)4 = C4H4O4 29.02 Empirical mass 49. Assuming 100.00 g of compound (mass hydrogen = 100.00 g − 49.31 g C − 43.79 g O = 6.90 g H): 49.31 g C × 1 mol C 1 mol H = 4.105 mol C; 6.90 g H × = 6.85 mol H 12.011 g C 1.008 g H 43.79 g O × 1 mol O = 2.737 mol O 15.999 g O Dividing all mole values by 2.737 gives: 4.105 6.85 2.737 = 1.500; = 2.50; = 1.000 2.737 2.737 2.737 Because a whole number ratio is required, the empirical formula is C3H5O2. Empirical formula mass ≈ 3(12.0) + 5(1.0) +2(16.0) = 73.0 g/mol Molar mass 146.1 = = 2.00; molecular formula = (C3H5O2)2 = C6H10O4 Empirical formula mass 73.0 50. 41.98 mg CO2 × 6.45 mg H2O × 12.011 mg C 11.46 mg = 11.46 mg C; % C = × 100 = 57.85% C 44.009 mg CO 2 19.81 mg 2.016 mg H 0.722 mg = 0.722 mg H; % H = × 100 = 3.64% H 18.02 mg H 2 O 19.81 mg % O = 100.00 − (57.85 + 3.64) = 38.51% O Out of 100.00 g terephthalic acid, there are: 57.85 g C × 1 mol C 1 mol H = 4.816 mol C; 3.64 g H × = 3.61 mol H 12.011 g C 1.008 g H 38.51 g O × 1 mol O = 2.407 mol O 15.999 g O 4.816 = 2.001; 2.407 3.61 2.407 = 1.50; = 1.000 2.407 2.407 The C : H : O mole ratio is 2 : 1.5 : 1 or 4 : 3 : 2. The empirical formula is C4H3O2. 28 CHAPTER 3 STOICHIOMETRY Mass of C4H3O2 ≈ 4(12) + 3(1) + 2(16) = 83 Molar mass = 51. 41.5 g 166 = 166 g/mol; = 2.0; the molecular formula is C8H6O4. 0.250 mol 83 First, we will determine composition by mass percent: 16.01 mg CO2 × %C= × 12.011 g C 1000 mg × = 4.369 mg C 44.009 g CO 2 g 4.369 mg C × 100 = 40.91% C 10.68 mg compound 4.37 mg H2O × %H= 1g 1000 mg 1g 1000 mg × 2.016 g H 1000 mg × = 0.489 mg H 18.02 g H 2 O g 0.489 mg × 100 = 4.58% H; % O = 100.00 - (40.91 + 4.58) = 54.51% O 10.68 mg So, in 100.00 g of the compound, we have: 40.91 g C × 1 mol H 1 mol C = 3.406 mol C; 4.58 g H × = 4.54 mol H 12.011 g C 1.008 g H 1 mol O = 3.407 mol O 15.999 g O 4.54 4 Dividing by the smallest number: = 1.33 . ; the empirical formula is C3H4O3. 3.406 3 54.51 g O × The empirical formula mass of C3H4O3 is ≈ 3(12) + 4(1) + 3(16) = 88 g. Because 52. 176.1 = 2.0, the molecular formula is C6H8O6. 88 a. Only acrylonitrile contains nitrogen. If we have 100.00 g of polymer: 8.80 g N × % C3H3N = 1 mol C 3 H 3 N 53.06 g C 3 H 3 N × = 33.3 g C3H3N 14.01 g N 1 mol C 3 H 3 N 33.3 g C 3 H 3 N = 33.3% C3H3N 100.00 g polymer Only butadiene in the polymer reacts with Br2: 0.605 g Br2 × 1 mol C 4 H 6 54.09 g C 4 H 6 1 mol Br2 × × = 0.205 g C4H6 159.8 g Br2 mol Br2 mol C 4 H 6 CHAPTER 3 STOICHIOMETRY % C4H6 = 29 0.205 g × 100 = 17.1% C4H6 1.20 g b. If we have 100.0 g of polymer: 33.3 g C3H3N × 1 mol C 3 H 3 N = 0.628 mol C3H3N 53.06 g 17.1 g C4H6 × 1 mol C 4 H 6 = 0.316 mol C4H6 54.09 g C 4 H 6 49.6 g C8H8 × 1 mol C8 H 8 = 0.476 mol C8H8 104.14 g C8 H 8 Dividing by 0.316: 0.628 0.316 0.476 = 1.99; = 1.00; = 1.51 0.316 0.316 0.316 This is close to a mole ratio of 4 : 2 : 3. Thus there are 4 acrylonitrile to 2 butadiene to 3 styrene molecules in the polymer, or (A4B2S3)n. Balancing Chemical Equations 53. When balancing reactions, start with elements that appear in only one of the reactants and one of the products, and then go on to balance the remaining elements. a. C6H12O6(s) + O2(g) → CO2(g) + H2O(g) Balance C atoms: C6H12O6 + O2 → 6 CO2 + H2O Balance H atoms: C6H12O6 + O2 → 6 CO2 + 6 H2O Lastly, balance O atoms: C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g) b. Fe2S3(s) + HCl(g) → FeCl3(s) + H2S(g) Balance Fe atoms: Fe2S3 + HCl → 2 FeCl3 + H2S Balance S atoms: Fe2S3 + HCl → 2 FeCl3 + 3 H2S There are 6 H and 6 Cl on right, so balance with 6 HCl on left: Fe2S3(s) + 6 HCl(g) → 2 FeCl3(s) + 3 H2S(g) c. CS2(l) + NH3(g) → H2S(g) + NH4SCN(s) C and S are balanced; balance N: 30 CHAPTER 3 STOICHIOMETRY CS2 + 2 NH3 → H2S + NH4SCN H is also balanced. CS2(l) + 2 NH3(g) → H2S(g) + NH4SCN(s). 54. An important part to this problem is writing out correct formulas. If the formulas are incorrect, then the balanced reaction is incorrect. a. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) b. 3 Pb(NO3)2(aq) + 2 Na3PO4(aq) → Pb3(PO4)2(s) + 6 NaNO3(aq) 55. a. 16 Cr(s) + 3 S8(s) → 8 Cr2S3(s) b. 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) c. 2 KClO3(s) → 2 KCl(s) + 3 O2(g) d. 2 Eu(s) + 6 HF(g) → 2 EuF3(s) + 3 H2(g) e. 2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) 56. a. 2 KO2(s) + 2 H2O(l) → 2 KOH(aq) + O2(g) + H2O2(aq) or 4 KO2(s) + 6 H2O(l) → 4 KOH(aq) + O2(g) + 4 H2O2(aq) b. Fe2O3(s) + 6 HNO3(aq) → 2 Fe(NO3)3(aq) + 3 H2O(l) c. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) d. PCl5(l) + 4 H2O(l) → H3PO4(aq) + 5 HCl(g) e. 2 CaO(s) + 5 C(s) → 2 CaC2(s) + CO2(g) f. 2 MoS2(s) + 7 O2(g) → 2 MoO3(s) + 4 SO2(g) g. FeCO3(s) + H2CO3(aq) → Fe(HCO3)2(aq) Reaction Stoichiometry 57. 1.000 kg Al × 3 mol NH 4 ClO 4 117.49 g NH 4 ClO 4 1000 g Al 1 mol Al × × × kg Al 26.98 g Al 3 mol Al mol NH 4 ClO 4 = 4355 g NH4ClO4 58. 1.0 × 106 kg HNO3 × 1000 g HNO 3 1 mol HNO 3 × = 1.6 × 107 mol HNO3 kg HNO 3 63.0 g HNO 3 We need to get the relationship between moles of HNO3 and moles of NH3. We have to use all three equations: CHAPTER 3 STOICHIOMETRY 31 2 mol HNO 3 16 mol HNO 3 2 mol NO 2 4 mol NO × × = 3 mol NO 2 2 mol NO 4 mol NH 3 24 mol NH 3 Thus we can produce 16 mol HNO3 for every 24 mol NH3 that we begin with: 1.6 × 107 mol HNO3 × 24 mol NH 3 17.0 g NH 3 × = 4.1 × 108 g or 4.1 × 105 kg NH3 16 mol HNO 3 mol NH 3 This is an oversimplified answer. In practice, the NO produced in the third step is recycled back continuously into the process in the second step. If this is taken into consideration, then the conversion factor between mol NH3 and mol HNO3 turns out to be 1 : 1; that is, 1 mol of NH3 produces 1 mol of HNO3. Taking into consideration that NO is recycled back gives an answer of 2.7 × 105 kg NH3 reacted. 59. Fe2O3(s) + 2 Al(s) → 2 Fe(l) + Al2O3(s) 15.0 g Fe × 60. 1 mol Fe 2 mol Al 26.98 g Al = 0.269 mol Fe; 0.269 mol Fe × × = 7.26 g Al 55.85 g Fe 2 mol Fe mol Al 0.269 mol Fe × 1 mol Fe 2 O 3 159.70 g Fe 2 O 3 × = 21.5 g Fe2O3 2 mol Fe mol Fe 2 O 3 0.269 mol Fe × 1 mol Al 2 O 3 101.96 g Al2 O 3 × = 13.7 g Al2O3 2 mol Fe mol Al2 O 3 10 KClO3(s) + 3 P4(s) → 3 P4O10(s) + 10 KCl(s) 1 mol KClO3 3 mol P4 O10 283.88 g P4 O10 × × = 36.8 g P4O10 122.55 g KClO3 10 mol KClO3 mol P4 O10 52.9 g KClO3 × 61. 2 LiOH(s) + CO2(g) → Li2CO3(aq) + H2O(l) The total volume of air exhaled each minute for the 7 astronauts is 7 × 20. = 140 L/min. 25,000 g LiOH × 1 mol LiOH 1 mol CO 2 44.01 g CO 2 100 g air × × × × 23.95 g LiOH 2 mol LiOH mol CO 2 4.0 g CO 2 1 mL air 1L 1 min 1h × × × = 68 h = 2.8 days 0.0010 g air 1000 mL 140 L air 60 min 62. 1.0 × 104 kg waste × + + 1 mol C 5 H 7 O 2 N 3.0 kg NH 4 1 mol NH 4 1000 g × × × × + + 100 kg waste kg 18.04 g NH 4 55 mol NH 4 113.1 g C 5 H 7 O 2 N = 3.4 × 104 g tissue if all NH4+ converted mol C 5 H 7 O 2 N 32 CHAPTER 3 STOICHIOMETRY Because only 95% of the NH4+ ions react: mass of tissue = (0.95)(3.4 × 104 g) = 3.2 × 104 g or 32 kg bacterial tissue 63. 1.0 × 103 g phosphorite × 75 g Ca 3 (PO 4 ) 2 1 mol Ca 3 (PO 4 ) 2 × × 100 g phosphorite 310.18 g Ca 3 (PO 4 ) 2 1 mol P4 2 mol Ca 3 (PO 4 ) 2 64. × 123.88 g P4 = 150 g P4 mol P4 Total mass of copper used: 10,000 boards × (8.0 cm × 16.0 cm × 0.060 cm) 8.96 g × = 6.9 × 105 g Cu 3 board cm Amount of Cu to be recovered = 0.80 × (6.9 × 105 g) = 5.5 × 105 g Cu 5.5 × 105 g Cu × 1 mol Cu ( NH 3 ) 4 Cl 2 202.6 g Cu ( NH 3 ) 4 Cl 2 1 mol Cu × × 63.55 g Cu mol Cu mol Cu ( NH 3 ) 4 Cl 2 = 1.8 × 106 g Cu(NH3)4Cl2 5.5 × 105 g Cu × 4 mol NH 3 17.03 g NH 3 1 mol Cu × × = 5.9 × 105 g NH3 63.55 g Cu mol Cu mol NH 3 Limiting Reactants and Percent Yield 65. The product formed in the reaction is NO2; the other species present in the product representtation is excess O2. Therefore, NO is the limiting reactant. In the pictures, 6 NO molecules react with 3 O2 molecules to form 6 NO2 molecules. 6 NO(g) + 3 O2(g) → 6 NO2(g) For smallest whole numbers, the balanced reaction is: 2 NO(g) + O2(g) → 2 NO2(g) 66. In the following table we have listed three rows of information. The “Initial” row is the number of molecules present initially, the “Change” row is the number of molecules that react to reach completion, and the “Final” row is the number of molecules present at completion. To determine the limiting reactant, let’s calculate how much of one reactant is necessary to react with the other. 10 molecules O2 × 4 molecules NH 3 = 8 molecules NH3 to react with all the O2 5 molecules O 2 CHAPTER 3 STOICHIOMETRY 33 Because we have 10 molecules of NH3 and only 8 molecules of NH3 are necessary to react with all the O2, O2 is limiting. Initial Change Final 4 NH3(g) 10 molecules !8 molecules 2 molecules + 5 O2(g) → 4 NO(g) 10 molecules 0 !10 molecules +8 molecules 0 8 molecules + 6 H2O(g) 0 +12 molecules 12 molecules The total number of molecules present after completion = 2 molecules NH3 + 0 molecules O2 + 8 molecules NO + 12 molecules H2O = 22 molecules. 67. 1.50 g BaO2 × 25.0 mL × 1 mol BaO 2 = 8.86 × 10−3 mol BaO2 169.3 g BaO 2 0.0272 g HCl 1 mol HCl × = 1.87 × 10−2 mol HCl mL 36.46 g HCl The required mole ratio from the balanced reaction is 2 mol HCl to 1 mol BaO2. The actual mole ratio is: 1.87 × 10 −2 mol HCl = 2.11 8.86 × 10 −3 mol BaO 2 Because the actual mole ratio is larger than the required mole ratio, the denominator (BaO2) is the limiting reagent. 8.86 × 10−3 mol BaO2 H 1 mol H 2 O 2 34.02 g H 2 O 2 × = 0.301 g H2O2 mol BaO 2 mol H 2 O 2 The amount of HCl reacted is: 8.86 × 10−3 mol BaO2 × 2 mol HCl = 1.77 × 10−2 mol HCl mol BaO 2 Excess mol HCl = 1.87 × 10−2 mol − 1.77 × 10−2 mol = 1.0 × 10−3 mol HCl Mass of excess HCl = 1.0 × 10−3 mol HCl × 68. 25.0 g Ag2O × 36.46 g HCl = 3.6 × 10−2 g HCl mol HCl 1 mol = 0.108 mol Ag2O 231.8 g 50.0 g C10H10N4SO2 × 1 mol = 0.200 mol C10H10N4SO2 250.29 g 34 CHAPTER 3 STOICHIOMETRY Mol C10 H10 N 4SO 2 0.200 = 1.85 (actual) = Mol Ag 2 O 0.108 The actual mole ratio is less than the required mole ratio (2), so C10H10N4SO2 is limiting. 0.200 mol C10H10N4SO2 × 2 mol AgC10 H 9 N 4SO 2 357.18 g × 2 mol C10 H10 N 4SO 2 mol AgC10 H 9 N 4SO 2 = 71.4 g AgC10H9N4SO2 produced 1 mol Cu 3 FeS3 1000 kg 1000 g 3 mol Cu × × × × metric ton kg 342.71 g 1 mol Cu 3 FeS3 63.55 g = 1.39 × 106 g Cu (theoretical) mol Cu 86.3 g Cu (actual) = 1.20 × 106 g Cu = 1.20 × 103 kg Cu 1.39 × 106 g Cu (theoretical) × 100. g Cu ( theoretical) = 1.20 metric tons Cu (actual) 69. 2.50 metric tons Cu3FeS3 × 70. a. 1142 g C6H5Cl × 1 mol C 6 H 5 Cl = 10.15 mol C6H5Cl 112.55 g C 6 H 5 Cl 1 mol C 2 HOCl3 = 3.29 mol C2HOCl3 147.38 g C 2 HOCl3 2 mol C 6 H 5 Cl From the balanced equation, the required mole ratio is = 2. The actual 1 mol C 2 HOCl 3 485 g C2HOCl3 × mole ratio present is 10.15 mol C 6 H 5 Cl = 3.09. The actual mole ratio is greater than 3.29 mol C 2 HOCl3 the required mole ratio, so the denominator of the actual mole ratio (C2HOCl3) is limiting. 3.29 mol C2HOCl3 × 1 mol C14 H 9 Cl 5 354.46 g C14 H 9 Cl 5 × = 1170 g C14H9Cl5 (DDT) mol C 2 HOCl3 mol C14 H 9 Cl 5 b. C2HOCl3 is limiting, and C6H5Cl is in excess. c. 3.29 mol C2HOCl3 × 2 mol C 6 H 5 Cl 112.55 g C 6 H 5 Cl × = 741 g C6H5Cl reacted mol C 2 HOCl3 mol C 6 H 5 Cl 1142 g ! 741 g = 401 g C6H5Cl in excess d. Percent yield = 200.0 g DDT × 100 = 17.1% 1170 g DDT CHAPTER 3 71. STOICHIOMETRY 35 An alternative method to solve limiting-reagent problems is to assume that each reactant is limiting and then calculate how much product could be produced from each reactant. The reactant that produces the smallest amount of product will run out first and is the limiting reagent. 5.00 × 106 g NH3 × 5.00 × 106 g O2 × 1 mol NH 3 2 mol HCN × = 2.94 × 105 mol HCN 17.03 g NH 3 2 mol NH 3 1 mol O 2 2 mol HCN × = 1.04 × 105 mol HCN 32.00 g O 2 3 mol O 2 5.00 × 106 g CH4 × 1 mol CH 4 2 mol HCN × = 3.12 × 105 mol HCN 16.04 g CH 4 2 mol CH 4 O2 is limiting because it produces the smallest amount of HCN. Although more product could be produced from NH3 and CH4, only enough O2 is present to produce 1.04 × 105 mol HCN. The mass of HCN that can be produced is: 1.04 × 105 mol HCN × 5.00 × 106 g O2 × 72. 27.03 g HCN = 2.81 × 106 g HCN mol HCN 1 mol O 2 6 mol H 2 O 18.02 g H 2 O × × = 5.63 × 106 g H2O 32.00 g O 2 3 mol O 2 1 mol H 2 O 2 C3H6(g) + 2 NH3(g) + 3 O2(g) → 2 C3H3N(g) + 6 H2O(g) a. We will solve this limiting reagent problem using the same method as described in Exercise 3.71. 1.00 × 103 g C3H6 × 1.50 × 103 g NH3 × 2.00 × 103 g O2 × 1 mol C 3 H 6 42.08 g C 3 H 6 × 2 mol C 3 H 3 N = 23.8 mol C3H3N 2 mol C 3 H 6 1 mol NH 3 2 mol C 3 H 3 N × = 88.1 mol C3H3N 17.03 g NH 3 2 mol NH 3 2 mol C 3 H 3 N 1 mol O 2 × = 41.7 mol C3H3N 32.00 g O 2 3 mol O 2 C3H6 is limiting because it produces the smallest amount of product, and the mass of acrylonitrile that can be produced is: 23.8 mol × 53.06 g C 3 H 3 N = 1.26 × 103 g C3H3N mol 36 CHAPTER 3 b. 23.8 mol C3H3N × STOICHIOMETRY 6 mol H 2 O 18.02 g H 2 O × = 1.29 × 103 g H2O 2 mol C 3 H 3 N mol H 2 O Amount NH3 needed: 23.8 mol C3H3N × 2 mol NH 3 17.03 g NH 3 × = 405 g NH3 2 mol C 3 H 3 N mol NH 3 Amount NH3 in excess = 1.50 × 103 g − 405 g = 1.10 × 103 g NH3 Amount O2 needed: 23.8 mol C3H3N × 3 mol O 2 32.00 g O 2 × = 1.14 × 103 g O2 2 mol C 3 H 3 N mol O 2 Amount O2 in excess = 2.00 × 103 g − 1.14 × 103 g = 860 g O2 1.10 × 103 g NH3 and 860 g O2 are in excess. 73. P4(s) + 6 F2(g) → 4 PF3(g); the theoretical yield of PF3 is: 120. g PF3 (actual) × 154 g PF3 × 100.0 g PF3 ( theoretical) = 154 g PF3 (theoretical) 78.1 g PF3 (actual) 1 mol PF3 6 mol F2 38.00 g F2 × × = 99.8 g F2 87.97 g PF3 4 mol PF3 mol F2 99.8 g F2 are needed to actually produce 120. g of PF3 if the percent yield is 78.1%. 74. a. From the reaction stoichiometry we would expect to produce 4 mol of acetaminophen for every 4 mol of C6H5O3N reacted. The actual yield is 3 mol of acetaminophen compared with a theoretical yield of 4 mol of acetaminophen. Solving for percent yield by mass (where M = molar mass acetaminophen): percent yield = 3 mol × M × 100 = 75% 4 mol × M b. The product of the percent yields of the individual steps must equal the overall yield, 75%. (0.87)(0.98)(x) = 0.75, x = 0.88; step III has a percent yield of 88%. CHAPTER 3 STOICHIOMETRY 37 Additional Exercises 75. 17.3 g H H 1 mol H 1 mol C = 17.2 mol H; 82.7 g C × = 6.89 mol C 1.008 g H 12.01 g C 17.2 = 2.50; the empirical formula is C2H5. 6.89 The empirical formula mass is ~29 g, so two times the empirical formula would put the compound in the correct range of the molar mass. Molecular formula = (C2H5)2 = C4H10 2.59 × 1023 atoms H × 1 molecule C 4 H10 1 mol C 4 H10 × 10 atoms H 6.022 × 10 23 molecules = 4.30 × 10−2 mol C4H10 58.12 g = 2.50 g C4H10 mol C 4 H10 4.30 × 10−2 mol C4H10 × 76. Assuming 100.00 g E3H8: Mol E = 8.73 g H × 77. 1 mol H 3 mol E × = 3.25 mol E 1.008 g H 8 mol H xgE 91.27 g E = , x = molar mass of E = 28.1 g/mol; atomic mass of E = 28.1 amu 1 mol E 3.25 mol E 0.105 g = 70.9 g/mol Molar mass X2 = 1 mol 20 8.92 × 10 molecules × 6.022 × 10 23 molecules The mass of X = 1/2(70.9 g/mol) = 35.5 g/mol. This is the element chlorine. Assuming 100.00 g of MX3 compound: 54.47 g Cl × 1 mol = 1.537 mol Cl 35.45 g 1.537 mol Cl × Molar mass of M = 1 mol M = 0.5123 mol M 3 mol Cl 45.53 g M = 88.87 g/mol M 0.5123 mol M M is the element yttrium (Y), and the name of YCl3 is yttrium(III) chloride. The balanced equation is 2 Y + 3 Cl2 → 2 YCl3. Assuming Cl2 is limiting: 38 CHAPTER 3 1.00 g Cl2 × STOICHIOMETRY 2 mol YCl3 195.26 g YCl3 1 mol Cl 2 × × = 1.84 g YCl3 70.90 g Cl 2 3 mol Cl 2 1 mol YCl3 Assuming Y is limiting: 1.00 g Y × 2 mol YCl3 195.26 g YCl3 1 mol Y × × = 2.20 g YCl3 88.91 g Y 2 mol Y 1 mol YCl3 Because Cl2, when it all reacts, produces the smaller amount of product, Cl2 is the limiting reagent, and the theoretical yield is 1.84 g YCl3. 78. Empirical formula mass = 12.01 + 1.008 = 13.02 g/mol; because 104.14/13.02 = 7.998 ≈ 8, the molecular formula for styrene is (CH)8 = C8H8. 2.00 g C8H8 × 79. 1 mol C8 H 8 8 mol H 6.022 × 10 23 atoms × × = 9.25 × 1022 atoms H 104.14 g C8 H 8 mol C8 H 8 mol H Mass of H2O = 0.755 g CuSO4•xH2O − 0.483 g CuSO4 = 0.272 g H2O 0.483 g CuSO4 × 0.272 g H2O × 1 mol CuSO 4 = 0.00303 mol CuSO4 159.62 g CuSO 4 1 mol H 2 O = 0.0151 mol H2O 18.02 g H 2 O 0.0151 mol H 2 O 4.98 mol H 2 O = ; compound formula = CuSO4•5H2O, x = 5 0.00303 mol CuSO 4 1 mol CuSO 4 80. In 1 hour, the 1000. kg of wet cereal contains 580 kg H2O and 420 kg of cereal. We want the final product to contain 20.% H2O. Let x = mass of H2O in final product. x = 0.20, x = 84 + (0.20)x, x = 105 ≈ 110 kg H2O 420 + x The amount of water to be removed is 580 − 110 = 470 kg/h. 81. Consider the case of aluminum plus oxygen. Aluminum forms Al3+ ions; oxygen forms O2− anions. The simplest compound of the two elements is Al2O3. Similarly, we would expect the formula of a Group 6A element with Al to be Al2X3. Assuming this, out of 100.00 g of compound, there are 18.56 g Al and 81.44 g of the unknown element, X. Let’s use this information to determine the molar mass of X, which will allow us to identify X from the periodic table. 18.56 g Al × 1 mol Al 3 mol X = 1.032 mol X × 26.98 g Al 2 mol Al 81.44 g of X must contain 1.032 mol of X. CHAPTER 3 STOICHIOMETRY Molar mass of X = 39 81.44 g X = 78.91 g/mol 1.032 mol X From the periodic table, the unknown element is selenium, and the formula is Al2Se3. 82. The reaction is BaX2(aq) + H2SO4(aq) → BaSO4(s) + 2 HX(aq). 0.124 g BaSO4 × 137.3 g Ba 0.0729 g Ba = 0.0729 g Ba; % Ba = × 100 = 46.1% 233.4 g BaSO 4 0.158 g BaX 2 The formula is BaX2 (from positions of the elements in the periodic table), and 100.0 g of compound contains 46.1 g Ba and 53.9 g of the unknown halogen. There must also be: 1 mol Ba 2 mol X = 0.672 mol of the halogen in 53.9 g of halogen × 137.3 g Ba mol Ba 53.9 g Therefore, the molar mass of the halogen is = 80.2 g/mol. 0.672 mol 46.1 g Ba × This molar mass is close to that of bromine. Thus the formula of the compound is BaBr2. 83. 1.20 g CO2 × 1 mol C 24 H 30 N 3O 1 mol CO 2 1 mol C 376.51 g × × × 44.01 g mol CO 2 24 mol C mol C 24 H 30 N 3O = 0.428 g C24H30N3O 0.428 g C 24 H 30 N 3O × 100 = 42.8% C24H30N3O (LSD) 1.00 g sample 84. Ca3(PO4)2(s) + 3 H2SO4(aq) → 3 CaSO4(s) + 2 H3PO4(aq) 1.0 × 103 g Ca3(PO4)2 × 1 mol Ca 3 (PO 4 ) 2 = 3.2 mol Ca3(PO4)2 310.2 g Ca 3 (PO 4 ) 2 1.0 × 103 g conc. H2SO4 × 98 g H 2SO 4 1 mol H 2SO 4 = 10. mol H2SO4 × 100 g conc. H 2SO 4 98.1 g H 2SO 4 The required mole ratio from the balanced equation is 3 mol H2SO4 to 1 mol Ca3(PO4)2. The 10. mol H 2SO 4 actual ratio is = 3.1. 3.2 mol Ca 3 (PO 4 ) 2 This is larger than the required mole ratio, so Ca3(PO4)2 is the limiting reagent. 3.2 mol Ca3(PO4)2 × 3 mol CaSO 4 136.2 g CaSO 4 × = 1300 g CaSO4 produced mol Ca 3 (PO 4 ) 2 mol CaSO 4 40 CHAPTER 3 3.2 mol Ca3(PO4)2 × 85. 453 g Fe × 2 mol H 3 PO 4 98.0 g H 3 PO 4 = 630 g H3PO4 produced × mol Ca 3 (PO 4 ) 2 mol H 3 PO 4 1 mol Fe 2 O 3 159.70 g Fe 2 O 3 1 mol Fe × × = 648 g Fe2O3 55.85 g Fe 2 mol Fe mol Fe 2 O 3 Mass % Fe2O3 = 86. STOICHIOMETRY 648 g Fe 2 O 3 × 100 = 86.2% 752 g ore 12 C21H6: 2(12.000000) + 6(1.007825) = 30.046950 amu 12 C1H216O: 1(12.000000) + 2(1.007825) + 1(15.994915) = 30.010565 amu 14 N16O: 1(14.003074) + 1(15.994915) = 29.997989 amu The peak results from 12C1H216O. 85 87. 87 Rb atoms = 2.591; Rb atoms If we had exactly 100 atoms, x = number of 85Rb atoms and 100 − x = number of 87Rb atoms. x 259.1 = 2.591, x = 259.1 − (2.591)x, x = = 72.15; 72.15% 85Rb 3.591 100 − x 0.7215(84.9117) + 0.2785(A) = 85.4678, A = 88. 85.4678 − 61.26 = 86.92 amu 0.2785 Assuming 100.00 g of tetrodotoxin: 41.38 g C × 5.37 g H × 1 mol C 1 mol N = 3.445 mol C; 13.16 g N × = 0.9395 mol N 12.011 g C 14.007 g N 1 mol H 1 mol O = 5.33 mol H; 40.09 g O × = 2.506 mol O 1.008 g H 15.999 g O Divide by the smallest number: 3.445 5.33 2.506 = 3.667; = 5.67; = 2.667 0.9395 0.9395 0.9395 To get whole numbers for each element, multiply through by 3. Empirical formula: (C3.667H5.67NO2.667)3 = C11H17N3O8; the mass of the empirical formula is 319.3 g/mol. CHAPTER 3 STOICHIOMETRY Molar mass tetrodotoxin = 41 1.59 × 10 −21 g = 319 g/mol 1 mol 3 molecules × 6.022 × 10 23 molecules Because the empirical mass and molar mass are the same, the molecular formula is the same as the empirical formula, C11H17N3O8. 165 lb × 1 kg 10. μg 1 × 10 −6 g 1 mol 6.022 × 10 23 molecules × × × × 2.2046 lb kg μg 319.3 g 1 mol = 1.4 × 1018 molecules tetrodotoxin is the LD50 dosage 89. The volume of a gas is proportional to the number of molecules of gas. Thus the formulas are: I: NH3 II: N2H4 III: HN3 The mass ratios are: I: 4.634 g N gH II: 6.949 g N gH III: 41.7 g N gH If we set the atomic mass of H equal to 1.008, then the atomic mass for nitrogen is: I: 14.01 II: 14.01 For example, for compound I: 90. III. 14.0 A 4.634 = , A = 14.01 3(1.008) 1 PaO2 + O2 → PaxOy (unbalanced) 0.200 g PaO2 × 231 g Pa = 0.1757 g Pa (We will carry an extra significant figure.) 263 g PaO 2 0.2081 g PaxOy − 0.1757 g Pa = 0.0324 g O; 0.0324 g O H 0.1757 g Pa H 1 mol O = 2.025 × 10−3 mol O 16.00 g O 1 mol Pa = 7.61 × 10−4 mol Pa 231 g Pa 2.025 × 10 −3 mol O Mol O 2 8 mol O = 2.66 . 2 = ; empirical formula: Pa3O8 = −4 Mol Pa 3 3 mol Pa 7.61 × 10 mol Pa 91. 1.375 g AgI × 1 mol AgI = 5.856 × 10−3 mol AgI = 5.856 × 10−3 mol I 234.8 g AgI 1.375 g AgI × 126.9 g I = 0.7431 g I; XI2 contains 0.7431 g I and 0.257 g X. 234.8 g AgI 42 CHAPTER 3 5.856 × 10−3 mol I × Molar mass = 92. STOICHIOMETRY 1 mol X = 2.928 × 10−3 mol X 2 mol I 0.257 g X 87.8 g = ; atomic mass = 87.8 amu (X is Sr.) −3 mol 2.928 × 10 mol X 40.0/A x (40.0)A z mol X =2= = or Az = 3Ax X2Z: 40.0% X and 60.0% Z by mass; mol Z 60.0/A z (60.0)A x where A = molar mass For XZ2, molar mass = Ax + 2Az = Ax + 2(3Ax) = 7Ax. Mass % X = 93. Ax × 100 = 14.3% X; % Z = 100.0 − 14.3 = 85.7% Z 7A x Assuming 1 mole of vitamin A (286.4 g vitamin A): mol C = 286.4 g vitamin A × 0.8396 g C 1 mol C × = 20.00 mol C g vitamin A 12.011 g C mol H = 286.4 g vitamin A × 0.1056 g H 1 mol H × = 30.01 mol H g vitamin A 1.0079 g H Because 1 mole of vitamin A contains 20 mol C and 30 mol H, the molecular formula of vitamin A is C20H30E. To determine E, lets calculate the molar mass of E: 286.4 g = 20(12.01) + 30(1.008) + molar mass E, molar mass E = 16.0 g/mol From the periodic table, E = oxygen, and the molecular formula of vitamin A is C20H30O. 94. We would see the peaks corresponding to: 10 B35Cl3 [mass ≈ 10 + 3(35) = 115 amu], 10B35Cl237Cl (117), 10B35Cl37Cl2 (119), 10 B37Cl3 (121), 11B35Cl3 (116), 11B35Cl237Cl (118), 11B35Cl37Cl2 (120), 11B37Cl3 (122) We would see a total of eight peaks at approximate masses of 115, 116, 117, 118, 119, 120, 121, and 122. Challenge Problems 95. When the discharge voltage is low, the ions present are in the form of molecules. When the discharge voltage is increased, the bonds in the molecules are broken, and the ions present are in the form of individual atoms. Therefore, the high discharge data indicate that the ions 16 + 18 + O , O , and 40Ar+ are present. The only combination of these individual ions that can explain the mass data at low discharge is 16O16O+ (mass = 32), 16O18O+ (mass = 34), and 40Ar+ (mass = 40). Therefore, the gas mixture contains 16O16O, 16O18O, and 40Ar. To determine the CHAPTER 3 STOICHIOMETRY 43 percent composition of each isotope, we use the relative intensity data from the high discharge data to determine the percentage that each isotope contributes to the total relative intensity. For 40Ar: 1.0000 1.0000 × 100 = × 100 = 57.094% 40Ar 1.7515 0.7500 + 0.0015 + 1.0000 0.7500 0.0015 × 100 = 42.82% 16O; for 18O: × 100 = 8.6 × 10−2% 18O 1.7515 1.7515 For 16O: Note: 18F instead of 18O could also explain the data. However, OF(g) is not a stable compound. This is why 18O is the best choice because O2(g) does form. 96. Fe(s) + 1 2 O 2 (g) → FeO(s) ; 2 Fe(s) + 20.00 g Fe × 3 2 O 2 (g ) → Fe 2 O 3 (s) 1 mol Fe = 0.3581 mol 55.85 g (11.20 − 3.24) g O 2 × 1 mol O 2 = 0.2488 mol O2 consumed (1 extra sig. fig.) 32.00 g Assuming x mol of FeO is produced from x mol of Fe, so that 0.3581 – x mol of Fe reacts to form Fe2O3: 1 x O 2 → x FeO 2 3 ⎛ 0.3581 − x ⎞ ⎛ 0.3581 − x ⎞ (0.3581 − x) mol Fe + ⎜ ⎟ mol O 2 → ⎜ ⎟ mol Fe 2 O 3 2 ⎝ 2 2 ⎠ ⎝ ⎠ x Fe + Setting up an equation for total moles of O2 consumed: 1 2 x + 3 4 (0.3581 − x) = 0.2488 mol O 2 , 0.079 mol FeO × 71.85 g FeO = 5.7 g FeO produced mol Mol Fe2O3 produced = 0.140 mol Fe2O3 × 97. x = 0.0791 = 0.079 mol FeO 0.3581 − 0.079 = 0.140 mol Fe2O3 2 159.70 g Fe 2 O 3 = 22.4 g Fe2O3 produced mol 10.00 g XCl2 + excess Cl2 → 12.55 g XCl4; 2.55 g Cl reacted with XCl2 to form XCl4. XCl4 contains 2.55 g Cl and 10.00 g XCl2. From mole ratios, 10.00 g XCl2 must also contain 2.55 g Cl; mass X in XCl2 = 10.00 ! 2.55 = 7.45 g X. 44 CHAPTER 3 2.55 g Cl × STOICHIOMETRY 1 mol XCl 2 1 mol Cl 1 mol X × × = 3.60 × 10 −2 mol X 35.45 g Cl 2 mol Cl mol XCl 2 So, 3.60 × 10 −2 mol X has a mass equal to 7.45 g X. The molar mass of X is: 7.45 g X = 207 g/mol X; atomic mass = 207 amu, so X is Pb. 3.60 × 10 − 2 mol X 98. The two relevant equations are: Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) and Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) Let x = mass Mg, so 10.00 ! x = mass Zn. Zn + moles Mg. Mol H2 = 0.5171 g H2 × 0.2565 = From the balanced equations, moles H2 = moles 1 mol H 2 = 0.2565 mol H2 2.0158 g H 2 x 10.00 − x + ; solving, x = 4.008 g Mg. 24.31 65.38 4.008 g × 100 = 40.08% Mg 10.00 g 99. For a gas, density and molar mass are directly proportional to each other. Molar mass XHn = 2.393(32.00) = 2 mol H 76.58 g ; 0.803 g H2O × mol 18.02 g H 2 O = 8.91 × 10−2 mol H 8.91 × 10 −2 mol H 4 mol H = −2 mol XH n 2.23 × 10 mol XH n Molar mass X = 76.58 − 4(1.008 g) = 72.55 g/mol; the element is Ge. 100. The balanced equation is 2 Sc(s) + 2x HCl(aq) → 2 ScClx(aq)+ x H2(g). The mol ratio of Sc to H2 = Mol Sc = 2.25 g Sc × 2 . x 1 mol Sc = 0.0500 mol Sc 44.96 g Sc Mol H2 = 0.1502 g H2 × 1 mol H 2 = 0.07451 mol H2 2.0158 g H 2 2 0.0500 = , x = 3; the formula is ScCl3. x 0.07451 CHAPTER 3 101. STOICHIOMETRY 45 4.000 g M2S3 → 3.723 g MO2 There must be twice as many moles of MO2 as moles of M2S3 in order to balance M in the reaction. Setting up an equation for 2(mol M2S3) = mol MO2 where A = molar mass M: ⎛ ⎞ 4.000 g 3.723 g 8.000 3.723 ⎟⎟ = = 2⎜⎜ , + 2 A 3 ( 32 . 07 ) A + 2 ( 16 . 00 ) 2 A + 96 . 21 A + 32.00 ⎝ ⎠ (8.000)A + 256.0 = (7.446)A + 358.2, (0.554)A = 102.2, A = 184 g/mol; atomic mass = 184 amu 102. We know that water is a product, so one of the elements in the compound is hydrogen. XaHb + O2 → H2O + ? To balance the H atoms, the mole ratio between XaHb and H2O = Mol compound = 2 . b 1.39 g 1.21 g = 0.0224 mol; mol H2O = = 0.0671 mol 62.09 g / mol 18.02 g / mol 2 0.0224 = , b = 6; XaH6 has a molar mass of 62.09 g/mol. b 0.0671 62.09 = a(molar mass of X) + 6(1.008), a(molar mass of X ) = 56.04 Some possible identities for X could be Fe (a = 1), Si (a = 2), N (a = 4), and Li (a = 8). N fits the data best so N4H6 is the most likely formula. 103. The balanced equations are: C(s) + 1/2 O2(g) → CO(g) and C(s) + O2(g) → CO2(g) If we have 100.0 mol of products, then we have 72.0 mol CO2, 16.0 mol CO, and 12.0 mol O2. The initial moles of C equals 72.0 (from CO2) + 16.00 (from CO) = 88.0 mol C and the initial moles of O2 equals 72.0 (from CO2) + 16.0/2 (from CO) + 12.0 (unreacted O2) = 92.0 mol O2. The initial reaction mixture contained: 92.0 mol O 2 = 1.05 mol O2/mol C 88.0 mol C 104. CxHyOz + oxygen → x CO2 + y/2 H2O 2.20 g CO 2 × Mass % C in aspirin = 1 mol C 1 mol CO 2 12.01 g C × × 44.01 g CO 2 mol CO 2 mol C = 60.0% C 1.00 g aspirin 46 CHAPTER 3 0.400 g H 2 O × Mass % H in aspirin = STOICHIOMETRY 1 mol H 2 O 2 mol H 1.008 g H × × 18.02 g H 2 O mol H 2 O mol H = 4.48% H 1.00 g aspirin Mass % O = 100.00 ! (60.0 + 4.48) = 35.5% O Assuming 100.00 g aspirin: 60.0 g C × 1 mol C 1 mol H = 5.00 mol C; 4.48 g H × = 4.44 mol H 12.01 g C 1.008 g H 1 mol O = 2.22 mol O 16.00 g O 5.00 4.44 Dividing by the smallest number: = 2.25; = 2.00 2.22 2.22 35.5 g O × Empirical formula: (C2.25 H2.00O)4 = C9H8O4. Empirical mass ≈ 9(12) + 8(1) + 4(16) = 180 g/mol; this is in the 170–190 g/mol range, so the molecular formula is also C9H8O4. Balance the aspirin synthesis reaction to determine the formula for salicylic acid. CaHbOc + C4H6O3 → C9H8O4 + C2H4O2, CaHbOc = salicylic acid = C7H6O3 105. LaH2.90 is the formula. If only La3+ is present, LaH3 would be the formula. If only La2+ is present, LaH2 would be the formula. Let x = mol La2+ and y = mol La3+: (La2+)x(La3+)yH(2x + 3y) where x + y = 1.00 and 2x + 3y = 2.90 Solving by simultaneous equations: 2x + 3y = 2.90 −2x − 2y = −2.00 y = 0.90 and x = 0.10 LaH2.90 contains 106. 1 9 La2+, or 10.% La2+, and La3+, or 90.% La3+. 10 10 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l); C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) 30.07 g/mol 44.09 g/mol Let x = mass C2H6, so 9.780 ! x = mass C3H8. Use the balanced reaction to set up an equation for the moles of O2 required. x 7 9.780 − x 5 × + × = 1.120 mol O2 30.07 2 44.09 1 CHAPTER 3 STOICHIOMETRY 3.7 g × 100 = 38% C2H6 by mass 9.780 g Solving: x = 3.7 g C2H6; 107 . 47 Let x = mass KCl and y = mass KNO3. Assuming 100.0 g of mixture, x + y = 100.0 g. Molar mass KCl = 74.55 g/mol; molar mass KNO3 = 101.11 g/mol Mol KCl = x y ; mol KNO3 = 74.55 101.11 Knowing that the mixture is 43.2% K, then in the 100.0 g mixture: y ⎞ ⎛ x 39.10 ⎜ + ⎟ = 43.2 ⎝ 74.55 101.11 ⎠ We have two equations and two unknowns: (0.5245)x + (0.3867)y = 43.2 x + y = 100.0 Solving, x = 32.9 g KCl; 108. 32.9 g × 100 = 32.9% KCl 100.0 g Let M = unknown element Mass % M = mass M 2.077 × 100 = × 100 = 56.01% M total mass compound 3.708 100.00 – 56.01 = 43.99% O Assuming 100.00 g compound: 43.99 g O × 1 mol O = 2.750 mol O 15.999 g O 56.01 g M = 20.37 g/mol. 2.750 mol M This is too low for the molar mass. We must have fewer moles of M than moles O present in If MO is the formula of the oxide, then M has a molar mass of the formula. Some possibilities are MO2, M2O3, MO3, etc. It is a guessing game as to which to try. Let’s assume an MO2 formula. Then the molar mass of M is: 56.01 g M = 40.73 g/mol 1 mol M 2.750 mol O × 2 mol O This is close to calcium, but calcium forms an oxide having the CaO formula, not CaO2. 48 CHAPTER 3 STOICHIOMETRY If MO3 is assumed to be the formula, then the molar mass of M calculates to be 61.10 g/mol which is too large. Therefore, the mol O to mol M ratio must be between 2 and 3. Some reasonable possibilities are 2.25, 2.33, 2.5, 2.67, and 2.75 (these are reasonable because they will lead to whole number formulas). Trying a mol O to mol M ratio of 2.5 to 1 gives a molar mass of: 56.01 g M = 50.92 g/mol 1 mol M 2.750 mol O × 2.5 mol O This is the molar mass of vanadium and V2O5 is a reasonable formula for an oxide of vanadium. The other choices for the O : M mole ratios between 2 and 3 do not give as reasonable results. Therefore, M is vanadium, and the formula is V2O5. 109. The balanced equations are: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) and 4 NH3(g) + 7 O2(g) → 4 NO2(g) + 6 H2O(g) Let 4x = number of moles of NO formed, and let 4y = number of moles of NO2 formed. Then: 4x NH3 + 5x O2 → 4x NO + 6x H2O and 4y NH3 + 7y O2 → 4y NO2 + 6y H2O All the NH3 reacted, so 4x + 4y = 2.00. 10.00 − 6.75 = 3.25 mol O2 reacted, so 5x + 7y = 3.25. Solving by the method of simultaneous equations: 20x + 28y = 13.0 −20x − 20y = −10.0 8y = 3.0, y = 0.38; 4x + 4 × 0.38 = 2.00, x = 0.12 Mol NO = 4x = 4 × 0.12 = 0.48 mol NO formed 110. a N2H4 + b NH3 + (10.00 − 4.062) O2 → c NO2 + d H2O Setting up four equations to solve for the four unknowns: 2a + b = c (N mol balance) 2c + d = 2(10.00 − 4.062) (O mol balance) 4 a + 3b = 2d (H mol balance) a(32.05) + b(17.03) = 61.00 (mass balance) Solving the simultaneous equations gives a = 1.12 = 1.1 mol N2H4. 1.1 mol N 2 H 4 × 32.05 g / mol N 2 H 4 × 100 = 58% N2H4 61.00 g CHAPTER 3 STOICHIOMETRY 49 Marathon Problems 111. To solve the limiting-reagent problem, we must determine the formulas of all the compounds so that we can get a balanced reaction. a. 40 million trillion = (40 × 106) × 1012 = 4.000 × 1019 (assuming 4 sig. figs.) 4.000 × 1019 molecules A × Molar mass of A = 1 mol A = 6.642 × 10-5 mol A 23 6.022 × 10 molecules A 4.26 × 10 −3 g A = 64.1 g/mol 6.642 × 10 −5 mol A Mass of carbon in 1 mol of A is: 64.1 g A × 37.5 g C = 24.0 g carbon = 2 mol carbon in substance A 100.0 g A The remainder of the molar mass (64.1 g - 24.0 g = 40.1 g) is due to the alkaline earth metal. From the periodic table, calcium has a molar mass of 40.08 g/mol. The formula of substance A is CaC2. b. 5.36 g H + 42.5 g O = 47.9 g; substance B only contains H and O. Determining the empirical formula of B: 5.36 g H × 1 mol H 5.32 = 5.32 mol H; = 2.00 1.008 g H 2.66 42.5 g O × 1 mol O 2.66 = 2.66 mol O; = 1.00 2.66 16.00 g O Empirical formula: H2O; the molecular formula of substance B could be H2O, H4O2, H6O3, etc. The most reasonable choice is water (H2O) for substance B. c. Substance C + O2 → CO2 + H2O; substance C must contain carbon and hydrogen and may contain oxygen. Determining the mass of carbon and hydrogen in substance C: 33.8 g CO2 × 1 mol CO 2 1 mol C 12.01 g C × × = 9.22 g carbon 44.01 g CO 2 mol CO 2 mol C 6.92 g H2O × 1 mol H 2 O 2 mol H 1.008 g H × × = 0.774 g hydrogen 18.02 g H 2 O mol H 2 O mol H 9.22 g carbon + 0.774 g hydrogen = 9.99 g; because substance C initially weighed 10.0 g, there is no oxygen present in substance C. Determining the empirical formula for substance C: 50 CHAPTER 3 9.22 g × STOICHIOMETRY 1 mol C = 0.768 mol carbon 12.01 g C 0.774 g H × 1 mol H = 0.768 mol hydrogen 1.008 g H Mol C/mol H = 1.00; the empirical formula is CH which has an empirical formula mass ≈ 13. Because the mass spectrum data indicate a molar mass of 26 g/mol, the molecular formula for substance C is C2H2. d. Substance D is Ca(OH)2. Now we can answer the question. The balanced equation is: CaC2(s) + 2 H2O(l) → C2H2(g) + Ca(OH)2(aq) 45.0 g CaC2 × 23.0 g H2O × 1 mol CaC 2 = 0.702 mol CaC2 64.10 g CaC 2 1 mol H 2 O mol H 2 O 1.28 = = 1.28 mol H2O; = 1.82 18.02 g H 2 O mol CaC 2 0.702 Because the actual mole ratio present is smaller than the required 2 : 1 mole ratio from the balanced equation, H2O is limiting. 1.28 mol H2O × 112. a. i. 1 mol C 2 H 2 26.04 g C 2 H 2 × = 16.7 g C2H2 = mass of product C 2 mol H 2 O mol C 2 H 2 If the molar mass of A is greater than the molar mass of B, then we cannot determine the limiting reactant because, while we have a fewer number of moles of A, we also need fewer moles of A (from the balanced reaction). ii. If the molar mass of B is greater than the molar mass of A, then B is the limiting reactant because we have a fewer number of moles of B and we need more B (from the balanced reaction). b. A + 5 B → 3 CO2 + 4 H2O To conserve mass: 44.01 + 5(B) = 3(44.01) + 4(18.02); solving: B = 32.0 g/mol Because B is diatomic, the best choice for B is O2. c. We can solve this without mass percent data simply by balancing the equation: A + 5 O2 → 3 CO2 + 4 H2O A must be C3H8 (which has a similar molar mass to CO2). This is also the empirical formula. Note: 3(12.01) × 100 = 81.71% C. So this checks. 3(12.01) + 8(1.008) CHAPTER 4 TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY Aqueous Solutions: Strong and Weak Electrolytes 10. Only statement b is true. A concentrated solution can also contain a nonelectrolyte dissolved in water, for example, concentrated sugar water. Acids are either strong or weak electrolytes. Some ionic compounds are not soluble in water, so they are not labeled as a specific type of electrolyte. 11. a. Polarity is a term applied to covalent compounds. Polar covalent compounds have an unequal sharing of electrons in bonds that results in unequal charge distribution in the overall molecule. Polar molecules have a partial negative end and a partial positive end. These are not full charges as in ionic compounds but are charges much smaller in magnitude. Water is a polar molecule and dissolves other polar solutes readily. The oxygen end of water (the partial negative end of the polar water molecule) aligns with the partial positive end of the polar solute, whereas the hydrogens of water (the partial positive end of the polar water molecule) align with the partial negative end of the solute. These opposite charge attractions stabilize polar solutes in water. This process is called hydration. Nonpolar solutes do not have permanent partial negative and partial positive ends; nonpolar solutes are not stabilized in water and do not dissolve. b. KF is a soluble ionic compound, so it is a strong electrolyte. KF(aq) actually exists as separate hydrated K+ ions and hydrated F− ions in solution. C6H12O6 is a polar covalent molecule that is a nonelectrolyte. C6H12O6 is hydrated as described in part a. c. RbCl is a soluble ionic compound, so it exists as separate hydrated Rb+ ions and hydrated Cl− ions in solution. AgCl is an insoluble ionic compound so the ions stay together in solution and fall to the bottom of the container as a precipitate. d. HNO3 is a strong acid and exists as separate hydrated H+ ions and hydrated NO3− ions in solution. CO is a polar covalent molecule and is hydrated as explained in part a. 12. MgSO4(s) → Mg2+(aq) + SO42−(aq); NH4NO3(s) → NH4+(aq) + NO3−(aq) 13. a. Ba(NO3)2(aq) → Ba2+(aq) + 2 NO3−(aq); picture iv represents the Ba2+ and NO3− ions present in Ba(NO3)2(aq). b. NaCl(aq) → Na+(aq) + Cl−(aq); picture ii represents NaCl(aq). 51 52 CHAPTER 4 SOLUTION STOICHIOMETRY c. K2CO3(aq) → 2 K+(aq) + CO32−(aq); picture iii represents K2CO3(aq). d. MgSO4(aq) → Mg2+(aq) + SO42−(aq); picture i represents MgSO4(aq). Solution Concentration: Molarity 0.79 g 1 mol 1.3 mol × = 5.2 M C2H5OH = 1.3 mol C2H5OH; molarity = mL 46.1 g 0.250 L 14. 75.0 mL × 15. a. 2.00 L × 0.250 mol NaOH 40.00 g NaOH × = 20.0 g NaOH L mol Place 20.0 g NaOH in a 2-L volumetric flask; add water to dissolve the NaOH, and fill to the mark with water, mixing several times along the way. b. 2.00 L × 0.250 mol NaOH 1 L stock = 0.500 L × L 1.00 mol NaOH Add 500. mL of 1.00 M NaOH stock solution to a 2-L volumetric flask; fill to the mark with water, mixing several times along the way. c. 2.00 L × 0.100 mol K 2 CrO 4 194.20 g K 2 CrO 4 × = 38.8 g K2CrO4 L mol K 2 CrO 4 Similar to the solution made in part a, instead using 38.8 g K2CrO4. d. 2.00 L × 0.100 mol K 2 CrO 4 1 L stock × = 0.114 L L 1.75 mol K 2 CrO 4 Similar to the solution made in part b, instead using 114 mL of the 1.75 M K2CrO4 stock solution. 16. a. M Ca ( NO3 ) 2 = 0.100 mol Ca ( NO3 ) 2 = 1.00 M 0.100 L Ca(NO3)2(s) → Ca2+(aq) + 2 NO3−(aq); M Ca 2 + = 1.00 M; M NO − = 2(1.00) = 2.00 M 3 b. M Na 2SO 4 = 2.5 mol Na 2SO 4 = 2.0 M 1.25 L Na2SO4(s) → 2 Na+(aq) + SO42−(aq); M Na + = 2(2.0) = 4.0 M; M SO c. 5.00 g NH4Cl × 1 mol NH 4 Cl = 0.0935 mol NH4Cl 53.49 g NH 4 Cl 4 2− = 2.0 M CHAPTER 4 SOLUTION STOICHIOMETRY M NH 4Cl = 0.0935 mol NH 4 Cl = 0.187 M 0.5000 L NH4Cl(s) → NH4+(aq) + Cl−(aq); M NH d. 1.00 g K3PO4 × M K 3PO 4 = 53 4 + = M Cl − = 0.187 M 1 mol K 3 PO 4 = 4.71 × 10−3 mol K3PO4 212.27 g 4.71 × 10 −3 mol = 0.0188 M 0.2500 L K3PO4(s) → 3 K+(aq) + PO43−(aq); M K + = 3(0.0188) = 0.0564 M; M PO 17. 4 3− = 0.0188 M 3.0 mol Na 2 CO 3 = 0.21 mol Na2CO3 L Na2CO3(s) → 2 Na+(aq) + CO32−(aq); mol Na+ = 2(0.21) = 0.42 mol Mol Na2CO3 = 0.0700 L × 1.0 mol NaHCO 3 = 0.030 mol NaHCO3 L NaHCO3(s) → Na+(aq) + HCO3−(aq); mol Na+ = 0.030 mol Mol NaHCO3 = 0.0300 L × M Na + = 18. total mol Na + total volume 25.0 g (NH4)2SO4 × Molarity = = 0.42 mol + 0.030 mol 0.45 mol = = 4.5 M Na+ 0.0700 L + 0.0300 L 0.1000 L 1 mol = 1.89 × 10−1 mol (NH4)2SO4 132.15 g 1.89 × 10 −1 mol 1000 mL × = 1.89 M (NH4)2SO4 100.0 mL L Moles of (NH4)2SO4 in final solution = 10.00 × 10−3 L × Molarity of final solution = 1.89 × 10 −2 mol 1000 mL × = 0.315 M (NH4)2SO4 (10.00 + 50.00) mL L (NH4)2SO4(s) → 2 NH4+(aq) + SO42−(aq); M NH 19. Stock solution = 1.89 mol L = 1.89 × 10−2 mol (NH4)2SO4 4 + = 2(0.315) = 0.630 M; M SO 10.0 mg 10.0 × 10 −3 g 2.00 × 10 −5 g steroid = = 500.0 mL 500.0 mL mL 4 2− = 0.315 M 54 CHAPTER 4 100.0 × 10-6 L stock × SOLUTION STOICHIOMETRY 1000 mL 2.00 × 10 −5 g steroid × = 2.00 × 10−6 g steroid L mL This is diluted to a final volume of 100.0 mL. 2.00 × 10 −6 g steroid 1000 mL 1 mol steroid × × = 5.95 × 10−8 M steroid 100.0 mL L 336.4 g steroid 20. Stock solution: 1.584 g Mn2+ × 1 mol Mn 2 + 2.883 × 10 −2 mol −2 2+ = 2.883 × 10 mol Mn ; M = 1.000 L 54.94 g Mn 2 + = 2.883 × 10−2 M Mn2+ Solution A contains: 50.00 mL × Molarity = 1L 2.883 × 10 −2 mol × = 1.442 × 10−3 mol Mn2+ 1000 mL L 1.442 × 10 −3 mol 1000 mL × = 1.442 × 10-3 M Mn2+ 1000.0 mL L Solution B contains: 10.00 mL × Molarity = 1L 1.442 × 10 −3 mol × = 1.442 × 10−5 mol Mn2+ 1000 mL L 1.442 × 10 −5 mol = 5.768 × 10−5 M Mn2+ 0.2500 L Solution C contains: 10.00 × 10−3 L × Molarity = 21. 5.768 × 10 −5 mol = 5.768 × 10−7 mol Mn2+ L 5.768 × 10 −7 mol = 1.154 × 10−6 M Mn2+ 0.5000 L a. 5.0 ppb Hg in water = 5.0 ng Hg 5.0 × 10 −9 g Hg = mL H 2 O mL H 2 O 5.0 × 10 −9 g Hg 1 mol Hg 1000 mL × × = 2.5 × 10−8 M Hg mL 200.6 g Hg L CHAPTER 4 b. SOLUTION STOICHIOMETRY 55 1.0 × 10 −9 g CHCl 3 1 mol CHCl 3 1000 mL × × = 8.4 × 10−9 M CHCl3 mL 119.4 g CHCl 3 L c. 10.0 ppm As = 10.0 μg As 10.0 × 10 −6 g As = mL mL 10.0 × 10 −6 g As 1 mol As 1000 mL × × = 1.33 × 10−4 M As mL 74.92 g As L d. 22. 0.10 × 10 −6 g DDT 1 mol DDT 1000 mL × × = 2.8 × 10−7 M DDT mL 354.5 g DDT L We want 100.0 mL of each standard. To make the 100. ppm standard: 100. μg Cu × 100.0 mL solution = 1.00 × 104 µg Cu needed mL 1.00 × 104 µg Cu × 1 mL stock = 10.0 mL of stock solution 1000.0 μg Cu Therefore, to make 100.0 mL of 100. ppm solution, transfer 10.0 mL of the 1000.0 ppm stock solution to a 100-mL volumetric flask, and dilute to the mark. Similarly: 75.0 ppm standard, dilute 7.50 mL of the 1000.0 ppm stock to 100.0 mL. 50.0 ppm standard, dilute 5.00 mL of the 1000.0 ppm stock to 100.0 mL. 25.0 ppm standard, dilute 2.50 mL of the 1000.0 ppm stock to 100.0 mL. 10.0 ppm standard, dilute 1.00 mL of the 1000.0 ppm stock to 100.0 mL. Precipitation Reactions 23. Use the solubility rules in Table 4.1. Some soluble bromides by Rule 2 would be NaBr, KBr, and NH4Br (there are others). The insoluble bromides by Rule 3 would be AgBr, PbBr2, and Hg2Br2. Similar reasoning is used for the other parts to this problem. Sulfates: Na2SO4, K2SO4, and (NH4)2SO4 (and others) would be soluble, and BaSO4, CaSO4, and PbSO4 (or Hg2SO4) would be insoluble. Hydroxides: NaOH, KOH, Ca(OH)2 (and others) would be soluble, and Al(OH)3, Fe(OH)3, and Cu(OH)2 (and others) would be insoluble. 56 CHAPTER 4 SOLUTION STOICHIOMETRY Phosphates: Na3PO4, K3PO4, (NH4)3PO4 (and others) would be soluble, and Ag3PO4, Ca3(PO4)2, and FePO4 (and others) would be insoluble. Lead: PbCl2, PbBr2, PbI2, Pb(OH)2, PbSO4, and PbS (and others) would be insoluble. Pb(NO3)2 would be a soluble Pb2+ salt. 24. Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) (molecular equation) Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 K+(aq) + 2 NO3−(aq) (complete ionic equation) The 1.0 mol of Pb2+ ions would react with the 2.0 mol of I− ions to form 1.0 mol of the PbI2 precipitate. Even though the Pb2+ and I− ions are removed, the spectator ions K+ and NO3− are still present. The solution above the precipitate will conduct electricity because there are plenty of charge carriers present in solution. 25. For the following answers, the balanced molecular equation is first, followed by the complete ionic equation, and then the net ionic equation. a. (NH4)2SO4(aq) + Ba(NO3)2(aq) → 2 NH4NO3(aq) + BaSO4(s) 2 NH4+(aq) + SO42−(aq) + Ba2+(aq) + 2 NO3−(aq) → 2 NH4+(aq) + 2 NO3−(aq) + BaSO4(s) Ba2+(aq) + SO42−(aq) → BaSO4(s) is the net ionic equation (spectator ions omitted). b. Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq) Pb2+(aq) + 2 NO3−(aq) + 2 Na+(aq) + 2 Cl−(aq) → PbCl2(s) + 2 Na+(aq) + 2 NO3−(aq) Pb2+(aq) + 2 Cl−(aq) → PbCl2(s) c. The possible products, potassium phosphate and sodium nitrate, are both soluble in water. Therefore, no reaction occurs. d. No reaction occurs because all possible products are soluble. e. CuCl2(aq) + 2 NaOH(aq) → Cu(OH)2(s) + 2 NaCl(aq) Cu2+(aq) + 2 Cl−(aq) + 2 Na+(aq) + 2 OH−(aq) → Cu(OH)2(s) + 2 Na+(aq) + 2 Cl−(aq) Cu2+(aq) + 2 OH−(aq) → Cu(OH)2(s) 26. The following schemes show reagents to add in order to precipitate one ion at a time. In each scheme, NaOH can be added to precipitate the last remaining ion. CHAPTER 4 SOLUTION STOICHIOMETRY + + 57 + Ag , Ba2 , Cr3 a. Na2SO4 BaSO4(s) + Ag , Cr3+ NaCl Cr3+ AgCl(s) + 2+ 2+ Ag , Pb , Cu b. NaCl Na2SO4 Ag+, Cu2+ + + Hg22 , Ni2 c. PbSO4(s) + Ni2 Hg2Cl2(s) NaCl + Cu2 27. AgCl(s) a. When CuSO4(aq) is added to Na2S(aq), the precipitate that forms is CuS(s). Therefore, Na+ (the gray spheres) and SO42− (the bluish green spheres) are the spectator ions. CuSO4(aq) + Na2S(aq) → CuS(s) + Na2SO4(aq); Cu2+(aq) + S2−(aq) → CuS(s) b. When CoCl2(aq) is added to NaOH(aq), the precipitate that forms is Co(OH)2(s). Therefore, Na+ (the gray spheres) and Cl- (the green spheres) are the spectator ions. CoCl2(aq) + 2 NaOH(aq) → Co(OH)2(s) + 2 NaCl(aq) Co2+(aq) + 2 OH−(aq) → Co(OH)2(s) c. When AgNO3(aq) is added to KI(aq), the precipitate that forms is AgI(s). Therefore, K+ (the red spheres) and NO3− (the blue spheres) are the spectator ions. AgNO3(aq) + KI(aq) → AgI(s) + KNO3(aq); Ag+(aq) + I−(aq) → AgI(s) 58 CHAPTER 4 SOLUTION STOICHIOMETRY 28. There are many acceptable choices for spectator ions. We will generally choose Na+ and NO3− as the spectator ions because sodium salts and nitrate salts are usually soluble in water. a. Fe(NO3)3(aq) + 3 NaOH(aq) → Fe(OH)3(s) + 3 NaNO3(aq) b. Hg2(NO3)2(aq) + 2 NaCl(aq) → Hg2Cl2(s) + 2 NaNO3(aq) c. Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2 NaNO3(aq) d. BaCl2(aq) + Na2CrO4(aq) → BaCrO4(s) + 2 NaCl(aq) 29. Because a precipitate formed with Na2SO4, the possible cations are Ba2+, Pb2+, Hg22+, and Ca2+ (from the solubility rules). Because no precipitate formed with KCl, Pb2+ and Hg22+ cannot be present. Because both Ba2+ and Ca2+ form soluble chlorides and soluble hydroxides, both these cations could be present. Therefore, the cations could be Ba2+ and Ca2+ (by the solubility rules in Table 4.1). For students who do a more rigorous study of solubility, Sr2+ could also be a possible cation (it forms an insoluble sulfate salt, whereas the chloride and hydroxide salts of strontium are soluble). 30. 2 Na3PO4(aq) + 3 Pb(NO3)2(aq) → Pb3(PO4)2(s) + 6 NaNO3(aq) 0.1500 L × 0.250 mol Pb( NO 3 ) 2 2 mol Na 3 PO 4 1 L Na 3 PO 4 × × = 0.250 L L 3 mol Pb( NO 3 ) 2 0.100 mol Na 3 PO 4 = 250. mL Na3PO4 31. 2 AgNO3(aq) + CaCl2(aq) → 2 AgCl(s) + Ca(NO3)2(aq) Mol AgNO3 = 0.1000 L × Mol CaCl2 = 0.1000 L × 0.20 mol AgNO3 = 0.020 mol AgNO3 L 0.15 mol CaCl 2 = 0.015 mol CaCl2 L The required mol AgNO3 to mol CaCl2 ratio is 2 : 1 (from the balanced equation). The actual mole ratio present is 0.020/0.015 = 1.3 (1.3 : 1). Therefore, AgNO3 is the limiting reagent. Mass AgCl = 0.020 mol AgNO3 × 1 mol AgCl 143.4 g AgCl × = 2.9 g AgCl 1 mol AgNO3 mol AgCl The net ionic equation is Ag+(aq) + Cl−(aq) → AgCl(s). The ions remaining in solution are the unreacted Cl− ions and the spectator ions, NO3− and Ca2+ (all Ag+ is used up in forming AgCl). The moles of each ion present initially (before reaction) can be easily determined from the moles of each reactant. 0.020 mol AgNO3 dissolves to form 0.020 mol Ag+ and 0.020 mol NO3−. 0.015 mol CaCl2 dissolves to form 0.015 mol Ca2+ and 2(0.015) = 0.030 mol Cl−. Mol unreacted Cl− = 0.030 mol Cl− initially − 0.020 mol Cl− reacted CHAPTER 4 SOLUTION STOICHIOMETRY 59 Mol unreacted Cl− = 0.010 mol Cl− unreacted M Cl− = 0.010 mol Cl − 0.010 mol Cl − = = 0.050 M Cl− total volume 0.1000 L + 0.1000 L The molarity of the spectator ions are: M NO − = 3 32. 0.020 mol NO 3 0.2000 L − = 0.10 M NO3−; M Ca + = 2 0.015 mol Ca 2 + = 0.075 M Ca2+ 0.2000 L a. Cu(NO3)2(aq) + 2 KOH(aq) → Cu(OH)2(s) + 2 KNO3(aq) Solution A contains 2.00 L × 2.00 mol/L = 4.00 mol Cu(NO3)2, and solution B contains 2.00 L × 3.00 mol/L = 6.00 mol KOH. Let’s assume in our picture that we have 4 formula units of Cu(NO3)2 (4 Cu2+ ions and 8 NO3− ions) and 6 formula units of KOH (6 K+ ions and 6 OH− ions). With 4 Cu2+ ions and 6 OH− ions present, then OH− is limiting. One Cu2+ ion remains as 3 Cu(OH)2(s) formula units form as precipitate. The following drawing summarizes the ions that remain in solution and the relative amount of precipitate that forms. Note that K+ and NO3− ions are spectator ions. In the drawing, V1 is the volume of solution A or B and V2 is the volume of the combined solutions with V2 = 2V1. The drawing exaggerates the amount of precipitate that would actually form. V2 NO 3 V1 K - + K NO3 + NO 3 + NO 3 - 2+ NO 3 - K - K + - Cu NO3 NO 3 - K - K + + NO 3 - Cu(OH)2 Cu(OH)2 Cu(OH)2 b. The spectator ion concentrations will be one-half the original spectator ion concentrations in the individual beakers because the volume was doubled. Or using moles, M K + = − 8.00 mol NO 3 6.00 mol K + = 1.50 M and M NO − = = 2.00 M. The concentration of 3 4.00 L 4.00 L OH− ions will be zero because OH− is the limiting reagent. From the drawing, the number of Cu2+ ions will decrease by a factor of four as the precipitate forms. Because the volume of solution doubled, the concentration of Cu2+ ions will decrease by a factor of eight after the two beakers are mixed: ⎛1⎞ M Cu + = 2.00 ⎜ ⎟ = 0.250 M ⎝8⎠ 60 CHAPTER 4 SOLUTION STOICHIOMETRY Alternately, one could certainly use moles to solve for M Cu 2+ : Mol Cu2+ reacted = 2.00 L × 3.00 mol OH − 1 mol Cu 2 + × = 3.00 mol Cu2+ reacted − L 2 mol OH Mol Cu2+ present initially = 2.00 L × 2.00 mol Cu 2+ = 4.00 mol Cu2+ present initially L Excess Cu2+ present after reaction = 4.00 mol − 3.00 mol = 1.00 mol Cu2+ excess M Cu 2+ = 1.00 mol Cu 2 + = 0.250 M 2.00 L + 2.00 L Mass of precipitate = 6.00 mol KOH × 1 mol Cu (OH) 2 97.57 g Cu (OH) 2 × 2 mol KOH mol Cu (OH) 2 Mass of precipitate = 293 g Cu(OH)2 33. 2 AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2 NaNO3(aq) 0.0750 L × 34. 0.100 mol AgNO 3 1 mol Na 2 CrO 4 161.98 g Na 2 CrO 4 × × = 0.607 g Na2CrO4 L 2 mol AgNO 3 mol Na 2 CrO 4 XCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + X(NO3)2(aq) 1.38 g AgCl × 1 mol AgCl 1 mol XCl 2 × = 4.81 × 10 −3 mol XCl2 143.4 g 2 mol AgCl 1.00 g XCl 2 = 208 g/mol; 4.91 × 10 −3 mol XCl 2 x + 2(35.45) = 208, x = 137 g/mol The metal X is barium (Ba). 35. Use aluminum in the formulas to convert from mass of Al(OH)3 to mass of Al2(SO4)3 in the mixture. 1 mol Al(OH) 3 1 mol Al 2 (SO 4 ) 3 1 mol Al3+ 0.107 g Al(OH)3 × × × × 78.00 g mol Al(OH) 3 2 mol Al3+ Mass % Al2(SO4)3 = 36. 0.235 g × 100 = 16.2% 1.45 g 342.17 g Al2 (SO 4 ) 3 = 0.235 g Al2(SO4)3 mol Al2 (SO 4 ) 3 All the Tl in TlI came from Tl in Tl2SO4. The conversion from TlI to Tl2SO4 uses the molar masses and formulas of each compound. 0.1824 g TlI × 204.4 g Tl 504.9 g Tl 2SO 4 × = 0.1390 g Tl2SO4 331.3 g TlI 408.8 g Tl CHAPTER 4 SOLUTION STOICHIOMETRY Mass % Tl2SO4 = 37. 61 0.1390 g Tl 2SO 4 × 100 = 1.465% Tl2SO4 9.486 g pesticide All the sulfur in BaSO4 came from the saccharin. The conversion from BaSO4 to saccharin uses the molar masses and formulas of each compound. 0.5032 g BaSO4 × 32.07 g S 183.9 g saccharin × = 0.3949 g saccharin 233.4 g BaSO 4 32.07 g S Average mass 0.3949 g 3.949 × 10 −2 g 39.49 mg = = = Tablet 10 tablets tablet tablet 0.3949 g saccharin × 100 = 67.00% saccharin by mass 0.5894 g Use the silver nitrate data to calculate the mol Cl− present, then use the formula of douglasite to convert from Cl− to douglasite. The net ionic reaction is Ag+ + Cl− → AgCl(s). Average mass % = 38. 0.03720 L × 0.1000 mol Ag + 1 mol Cl − 1 mol douglasite 311.88 g douglasite × × × L mol mol Ag + 4 mol Cl − = 0.2900 g douglasite Mass % douglasite = 39. 0.2900 g × 100 = 63.74% 0.4550 g M2SO4(aq) + CaCl2(aq) → CaSO4(s) + 2 MCl(aq) 1.36 g CaSO4 × 1 mol CaSO 4 1 mol M 2SO 4 = 9.99 × 10−3 mol M2SO4 × 136.15 g CaSO 4 mol CaSO 4 From the problem, 1.42 g M2SO4 was reacted, so: molar mass = 1.42 g M 2SO 4 = 142 g/mol 9.99 × 10 −3 mol M 2SO 4 142 amu = 2(atomic mass M) + 32.07 + 4(16.00), atomic mass M = 23 amu From periodic table, M is Na(sodium). Acid-Base Reactions 40. a. NH3(aq) + HNO3(aq) → NH4NO3(aq) (molecular equation) NH3(aq) + H+(aq) + NO3−(aq) → NH4+(aq) + NO3−(aq) (complete ionic equation) NH3(aq) + H+(aq) → NH4+(aq) b. (net ionic equation) Ba(OH)2(aq) + 2 HCl(aq) → 2 H2O(l) + BaCl2(aq) Ba2+(aq) + 2 OH−(aq) + 2 H+(aq) + 2 Cl−(aq) → Ba2+(aq) + 2 Cl−(aq) + 2 H2O(l) OH−(aq) + H+(aq) → H2O(l) 62 CHAPTER 4 SOLUTION STOICHIOMETRY 3 HClO4(aq) + Fe(OH)3(s) → 3 H2O(l) + Fe(ClO4)3(aq) c. 3 H+(aq) + 3 ClO4-(aq) + Fe(OH)3(s) → 3 H2O(l) + Fe3+(aq) + 3 ClO4−(aq) 3 H+(aq) + Fe(OH)3(s) → 3 H2O(l) + Fe3+(aq) AgOH(s) + HBr(aq) → AgBr(s) + H2O(l) d. AgOH(s) + H+(aq) + Br−(aq) → AgBr(s) + H2O(l) AgOH(s) + H+(aq) + Br−(aq) → AgBr(s) + H2O(l) 41. a. Perchloric acid reacted with potassium hydroxide is a possibility. HClO4(aq) + KOH(aq) → H2O(l) + KClO4(aq) b. Nitric acid reacted with cesium hydroxide is a possibility. HNO3(aq) + CsOH(aq) → H2O(l) + CsNO3(aq) c. Hydroiodic acid reacted with calcium hydroxide is a possibility. 2 HI(aq) + Ca(OH)2(aq) → 2 H2O(l) + CaI2(aq) 42. We get the empirical formula from the elemental analysis. Out of 100.00 g carminic acid there are: 53.66 g C × 1 mol C 1 mol H = 4.468 mol C; 4.09 g H H = 4.06 mol H 12.011 g C 1.008 g H 42.25 g O × 1 mol O = 2.641 mol O 15.999 g O Dividing the moles by the smallest number gives: 4.468 = 1.692; 2.641 4.06 = 1.54 2.641 These numbers don’t give obvious mole ratios. Let’s determine the mol C to mol H ratio: 4.468 11 = 1.10 = 4.06 10 So let's try 4.468 4.06 2.641 4.06 = 0.406 as a common factor: = 11.0; = 10.0; = 6.50 0.406 0.406 0.406 10 Therefore, C22H20O13 is the empirical formula. We can get molar mass from the titration data. The balanced reaction is HA(aq) + OH−(aq) → H2O(l) + A−(aq) where HA is an abbreviation for carminic acid, an acid with one acidic proton (H+). CHAPTER 4 SOLUTION STOICHIOMETRY 18.02 × 10−3 L soln × Molar mass = 63 0.0406 mol NaOH 1 mol carminic acid × L soln mol NaOH = 7.32 × 10−4 mol carminic acid 0.3602 g 492 g = −4 mol 7.32 × 10 mol The empirical formula mass of C22H20O13 ≈ 22(12) + 20(1) + 13(16) = 492 g. Therefore, the molecular formula of carminic acid is also C22H20O13. 43. If we begin with 50.00 mL of 0.100 M NaOH, then: 50.00 × 10−3 L × 0.100 mol = 5.00 × 10−3 mol NaOH to be neutralized. L a. NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) 5.00 × 10−3 mol NaOH × 1 mol HCl 1 L soln × = 5.00 × 10−2 L or 50.0 mL mol NaOH 0.100 mol b. 2 NaOH(aq) + H2SO3(aq) → 2 H2O(l) + Na2SO3(aq) 5.00 × 10−3 mol NaOH × 1 mol H 2SO 3 1 L soln × = 2.50 × 10−2 L or 25.0 mL 2 mol NaOH 0.100 mol H 2SO 3 c. 3 NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3 H2O(l) 5.00 × 10−3 mol NaOH × 1 mol H 3 PO 4 1 L soln × = 8.33 × 10−3 L or 8.33 mL 3 mol NaOH 0.200 mol H 3 PO 4 d. HNO3(aq) + NaOH(aq) → H2O(l) + NaNO3(aq) 5.00 × 10−3 mol NaOH × 1 mol HNO 3 1 L soln × = 3.33 × 10−2 L or 33.3 mL mol NaOH 0.150 mol HNO 3 e. HC2H3O2(aq) + NaOH(aq) → H2O(l) + NaC2H3O2(aq) 5.00 × 10−3 mol NaOH × f. 1 mol HC 2 H 3O 2 1 L soln × = 2.50 × 10−2 L mol NaOH 0.200 mol HC 2 H 3O 2 or 25.0 mL H2SO4(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2SO4(aq) 5.00 × 10−3 mol NaOH × 1 mol H 2SO 4 1 L soln × = 8.33 × 10−3 L or 8.33 mL 2 mol NaOH 0.300 mol H 2SO 4 64 44. CHAPTER 4 SOLUTION STOICHIOMETRY Strong bases contain the hydroxide ion, OH−. The reaction that occurs is H+ + OH− → H2O. 0.0120 L × 0.150 mol H + 1 mol OH − × = 1.80 × 10 −3 mol OH− L mol H + The 30.0 mL of the unknown strong base contains 1.80 × 10 −3 mol OH− . 1.8 × 10 −3 mol OH − = 0.0600 M OH− 0.0300 L The unknown base concentration is one-half the concentration of OH− ions produced from the base, so the base must contain 2 OH− in each formula unit. The three soluble strong bases that have 2 OH− ions in the formula are Ca(OH)2, Sr(OH)2, and Ba(OH)2. These are all possible identities for the strong base. 45. The acid is a diprotic acid (H2A) meaning that it has two H+ ions in the formula to donate to a base. The reaction is H2A(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2A(aq), where A2− is what is left over from the acid formula when the two protons (H+ ions) are reacted. For the HCl reaction, the base has the ability to accept two protons. The most common examples are Ca(OH)2, Sr(OH)2, and Ba(OH)2. A possible reaction would be 2 HCl(aq) + Ca(OH)2(aq) → 2 H2O(l) + CaCl2(aq). 46. Because KHP is a monoprotic acid, the reaction is (KHP is an abbreviation for potassium hydrogen phthalate): NaOH(aq) + KHP(aq) → NaKP(aq) + H2O(l) 0.1082 g KHP × 1 mol KHP 1 mol NaOH × = 5.298 × 10−4 mol NaOH 204.22 g KHP mol KHP There is 5.298 × 10−4 mol of sodium hydroxide in 34.67 mL of solution. Therefore, the concentration of sodium hydroxide is: 5.298 × 10 −4 mol = 1.528 × 10−2 M NaOH −3 34.67 × 10 L 47. The pertinent reactions are: 2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Amount of NaOH added = 0.0500 L × 0.213 mol = 1.07 × 10−2 mol NaOH L Amount of NaOH neutralized by HCl: 0.01321 L HCl H 0.103 mol HCl 1 mol NaOH × = 1.36 × 10−3 mol NaOH L HCl mol HCl CHAPTER 4 SOLUTION STOICHIOMETRY 65 The difference, 9.3 × 10−3 mol, is the amount of NaOH neutralized by the sulfuric acid. 9.3 × 10−3 mol NaOH × 1 mol H 2SO 4 = 4.7 × 10−3 mol H2SO4 2 mol NaOH Concentration of H2SO4 = 48. 4.7 × 10 −3 mol = 4.7 × 10−2 M H2SO4 0.1000 L 2 H3PO4(aq) + 3 Ba(OH)2(aq) → 6 H2O(l) + Ba3(PO4)2(s) 0.01420 L × 0.141 mol H 3 PO 4 3 mol Ba (OH) 2 1 L Ba (OH) 2 × × = 0.0576 L L 2 mol H 3 PO 4 0.0521 mol Ba (OH) 2 = 57.6 mL Ba(OH)2 49. HC2H3O2(aq) + NaOH(aq) → H2O(l) + NaC2H3O2(aq) 0.5062 mol NaOH 1 mol acetic acid × L so ln mol NaOH = 8.393 × 10−3 mol acetic acid 8.393 × 10 −3 mol = 0.8393 M HC2H3O2 Concentration of acetic acid = 0.01000 L 1.006 g b. If we have 1.000 L of solution: total mass = 1000. mL × = 1006 g solution mL 60.052 g = 50.40 g HC2H3O2 Mass of HC2H3O2 = 0.8393 mol × mol a. 16.58 × 10−3 L soln H Mass % acetic acid = 50. 50.40 g × 100 = 5.010% 1006 g Because KHP is a monoprotic acid, the reaction is: NaOH(aq) + KHP(aq) → H2O(l) + NaKP(aq) Mass KHP = 0.02046 L NaOH × 0.1000 mol NaOH 1 mol KHP 204.22 g KHP × × L NaOH mol NaOH mol KHP = 0.4178 g KHP 51. HCl and HNO3 are strong acids; Ca(OH)2 and RbOH are strong bases. The net ionic equation that occurs is H+(aq) + OH−(aq) → H2O(l). Mol H+ = 0.0500 L × 0.1000 L × 0.100 mol HCl 1 mol H + × + L mol HCl 0.200 mol HNO 3 1 mol H + × = 0.00500 + 0.0200 = 0.0250 mol H+ L mol HNO 3 66 CHAPTER 4 Mol OH− = 0.5000 L × 0.2000 L × SOLUTION STOICHIOMETRY 0.0100 mol Ca (OH) 2 2 mol OH − × + L mol Ca (OH) 2 0.100 mol RbOH 1 mol OH − × = 0.0100 + 0.0200 = 0.0300 mol OH− L mol RbOH We have an excess of OH−, so the solution is basic (not neutral). The moles of excess OH− = 0.0300 mol OH− initially − 0.0250 mol OH− reacted (with H+) = 0.0050 mol OH− excess. M OH − = 52. 0.0050 mol OH − 0.0050 mol = = 5.9 × 10 −3 M (0.05000 + 0.1000 + 0.5000 + 0.2000) L 0.8500 L 39.47 × 10−3 L HCl H Molarity of NH3 = 53. 0.0984 mol HCl 1 mol NH 3 × = 3.88 × 10−3 mol NH3 L mol HCl 3.88 × 10 −3 mol = 0.0776 M NH3 50.00 × 10 −3 L Ba(OH)2(aq) + 2 HCl(aq) → BaCl2(aq) + 2 H2O(l); H+(aq) + OH−(aq) → H2O(l) 75.0 × 10−3 L × 0.250 mol HCl = 1.88 × 10−2 mol HCl = 1.88 × 10−2 mol H+ + L 1.88 × 10−2 mol Cl− 225.0 × 10−3 L × 0.0550 mol Ba (OH) 2 = 1.24 × 10−2 mol Ba(OH)2 = 1.24 × 10−2 mol Ba2+ + L 2.48 × 10−2 mol OH− The net ionic equation requires a 1 : 1 mol ratio between OH− and H+. The actual mol OH− to mol H+ ratio is greater than 1 : 1, so OH− is in excess. Because 1.88 × 10−2 mol OH− will be neutralized by the H+, we have (2.48 − 1.88) × 10−2 = 0.60 × 10−2 mol OH− in excess. M OH − = 54. mol OH − excess 6.0 × 10 −3 mol OH − = = 2.0 × 10−2 M OH− total volume 0.0750 L + 0.2250 L Let HA = unknown acid; HA(aq) + NaOH(aq) → NaA(aq) + H2O(l) Mol HA present = 0.0250 L × 0.500 mol NaOH 1 mol HA × = 0.0125 mol HA L 1 mol NaOH 2.20 g HA x g HA , x = molar mass of HA = 176 g/mol = mol HA 0.0125 mol HA Empirical formula mass ≈ 3(12) + 4(1) + 3(16) = 88 g/mol. Because 176/88 = 2.0, the molecular formula is (C3H4O3)2 = C6H8O6. CHAPTER 4 SOLUTION STOICHIOMETRY 67 Oxidation-Reduction Reactions 55. Apply rules in Table 4.3. a. KMnO4 is composed of K+ and MnO4− ions. Assign oxygen an oxidation state value of −2, which gives manganese a +7 oxidation state because the sum of oxidation states for all atoms in MnO4− must equal the 1− charge on MnO4−. K, +1; O, −2; Mn, +7. b. Assign O a −2 oxidation state, which gives nickel a +4 oxidation state. Ni, +4; O, −2. c. K4Fe(CN)6 is composed of K+ cations and Fe(CN)64− anions. Fe(CN)64− is composed of iron and CN− anions. For an overall anion charge of 4−, iron must have a +2 oxidation state. d. (NH4)2HPO4 is made of NH4+ cations and HPO42− anions. Assign +1 as oxidation state of H and −2 as the oxidation state of O. For N in NH4+: x + 4(+1) = +1, x = −3 = oxidation state of N. For P in HPO42−: +1 + y + 4(−2) = -2, y = +5 = oxidation state of P. e. O, −2; P, +3 f. g. O, −2; F, −1; Xe, +6 h. F, −1; S, +4 i. 56. O, −2; C, +2 j. O, −2; Fe, + 8/3 H, +1; O, −2; C, 0 a. UO22+: O, −2; for U: x + 2(−2) = +2, x = +6 b. As2O3: O, −2; for As: 2(x) + 3(−2) = 0, x = +3 c. NaBiO3: Na, +1; O, −2; for Bi: +1 + x + 3(−2) = 0, x = +5 d. As4: As, 0 e. HAsO2: Assign H = +1 and O = -2; for As: +1 + x + 2(−2) = 0, x = +3 f. Mg2P2O7: Composed of Mg2+ ions and P2O74− ions. Mg, +2; O, −2; P, +5 g. Na2S2O3: Composed of Na+ ions and S2O32- ions. Na, +1; O, −2; S, +2 h. Hg2Cl2: Hg, +1; Cl, −1 i. 57. Ca(NO3)2: Composed of Ca2+ ions and NO3− ions. Ca, +2; O, −2; N, +5 a. SrCr2O7: Composed of Sr2+ and Cr2O72− ions. Sr, +2; O, !2; Cr, 2x + 7(!2) = !2, x = +6 b. Cu, +2; Cl, !1 c. O, 0 d. H, +1; O, !1 e. Mg2+ and CO32− ions present. Mg, +2; O, !2; C, +4; f. g. Pb2+ and SO32− ions present. Pb, +2; O, !2; S, +4; h. O, !2; Pb, +4 Ag, 0 i. Na+ and C2O42− ions present. Na, +1; O, -2; C, 2x + 4(!2) = !2, x = +3 j. O, −2; C, +4 68 CHAPTER 4 SOLUTION STOICHIOMETRY k. Ammonium ion has a 1+ charge (NH4+), and sulfate ion has a 2− charge (SO42−). Therefore, the oxidation state of cerium must be +4 (Ce4+). H, +1; N, −3; O, −2; S, +6 l. 58. O, −2; Cr, +3 To determine if the reaction is an oxidation-reduction reaction, assign oxidation states. If the oxidation states change for some elements, then the reaction is a redox reaction. If the oxidation states do not change, then the reaction is not a redox reaction. In redox reactions, the species oxidized (called the reducing agent) shows an increase in the oxidation states, and the species reduced (called the oxidizing agent) shows a decrease in oxidation states. Redox? Oxidizing Agent Reducing Agent Substance Oxidized Substance Reduced __________________________________________________________________________________________ a. b. c. d. e. f. g. h. i. Yes Yes No Yes Yes Yes No No Yes O2 HCl − O3 H2O2 CuCl − − SiCl4 CH4 Zn − NO H2O2 CuCl − − Mg CH4 (C) Zn − NO (N) H2O2 (O) CuCl (Cu) − − Mg O2 (O) HCl (H) − O3 (O) H2O2 (O) CuCl (Cu) − − SiCl4 (Si) In c, g, and h, no oxidation states change from reactants to products. 59. a. Al(s) + 3 HCl(aq) → AlCl3(aq) + 3/2 H2(g) or 2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g) Hydrogen is reduced (goes from the +1 oxidation state to the 0 oxidation state), and aluminum Al is oxidized (0 → +3). b. Balancing S is most complicated because sulfur is in both products. Balance C and H first; then worry about S. CH4(g) + 4 S(s) → CS2(l) + 2 H2S(g) Sulfur is reduced (0 → !2), and carbon is oxidized (!4 → +4). c. Balance C and H first; then balance O. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) Oxygen is reduced (0 → −2), and carbon is oxidized (−8/3 → +4). d. Although this reaction is mass balanced, it is not charge balanced. We need 2 mol of silver on each side to balance the charge. Cu(s) + 2 Ag+(aq) → 2 Ag(s) + Cu2+(aq) Silver is reduced (+1 → 0), and copper is oxidized (0 → +2). CHAPTER 4 60. SOLUTION STOICHIOMETRY 69 a. The first step is to assign oxidation states to all atoms (see numbers above the atoms). −3 +1 0 +4 −2 +1 −2 C2H6 + O2 → CO2 + H2O Each carbon atom changes from −3 to +4, an increase of seven. Each oxygen atom changes from 0 to −2, a decrease of 2. We need 7/2 O atoms for every C atom. C2H6 + 7/2 O2 → CO2 + H2O Balancing the remainder of the equation by inspection: or C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(g) 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) b. The oxidation state of magnesium changes from 0 to +2, an increase of 2. The oxidation state of hydrogen changes from +1 to 0, a decrease of 1. We need 2 H atoms for every Mg atom. The balanced equation is: Mg(s) + 2 HCl(aq) → Mg2+(aq) + 2 Cl−(aq) + H2(g) c. The oxidation state of copper increases by 2 (0 to +2) and the oxidation state of silver decreases by 1 (+1 to 0). We need 2 Ag atoms for every Cu atom. The balanced equation is: Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s) d. The equation is balanced. Each hydrogen atom gains one electron (+1 → 0), and each zinc atom loses two electrons (0 → +2). We need 2 H atoms for every Zn atom. This is the ratio in the given equation: Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) 61. a. Review Section 4.11 of the text for rules on balancing by the half-reaction method. The first step is to separate the reaction into two half-reactions, and then balance each halfreaction separately. NO3− → NO + 2 H2O (3 e + 4 H + NO3− → NO + 2 H2O) × 2 (Cu → Cu2+ + 2 e−) × 3 − + Adding the two balanced half-reactions so electrons cancel: 3 Cu → 3 Cu2+ + 6 e− 6 e + 8 H + 2 NO3− → 2 NO + 4 H2O ___________________________________________________________________ − + 3 Cu(s) + 8 H+(aq) + 2 NO3−(aq) → 3 Cu2+(aq) + 2 NO(g) + 4 H2O(l) b. (2 Cl− → Cl2 + 2 e−) × 3 Cr2O72− → 2 Cr3+ + 7 H2O 6 e + 14 H + Cr2O72− → 2 Cr3+ + 7 H2O − + 70 CHAPTER 4 SOLUTION STOICHIOMETRY Add the two balanced half-reactions with six electrons transferred: 6 Cl− → 3 Cl2 + 6 e− 6 e + 14 H + Cr2O72− → 2 Cr3+ + 7 H2O ___________________________________________________________ 14 H+(aq) + Cr2O72−(aq) + 6 Cl−(aq) → 3 Cl2(g) + 2 Cr3+(aq) + 7 H2O(l) − c. + Pb → PbSO4 Pb + H2SO4 → PbSO4 + 2 H+ Pb + H2SO4 → PbSO4 + 2 H+ + 2 e− PbO2 → PbSO4 PbO2 + H2SO4 → PbSO4 + 2 H2O 2 e− + 2 H+ + PbO2 + H2SO4 → PbSO4 + 2 H2O Add the two half-reactions with two electrons transferred: 2 e− + 2 H+ + PbO2 + H2SO4 → PbSO4 + 2 H2O Pb + H2SO4 → PbSO4 + 2 H+ + 2 e− _____________________________________________ Pb(s) + 2 H2SO4(aq) + PbO2(s) → 2 PbSO4(s) + 2 H2O(l) This is the reaction that occurs in an automobile lead storage battery. d. Mn2+ → MnO4− (4 H2O + Mn2+ → MnO4− + 8 H+ + 5 e−) × 2 NaBiO3 → Bi3+ + Na+ 6 H + NaBiO3 → Bi3+ + Na+ + 3 H2O − (2 e + 6 H+ + NaBiO3 → Bi3+ + Na+ + 3 H2O) × 5 + 8 H2O + 2 Mn2+ → 2 MnO4− + 16 H+ + 10 e− 10 e + 30 H+ + 5 NaBiO3 → 5 Bi3+ + 5 Na+ + 15 H2O ___________________________________________________________________ 8 H2O + 30 H+ + 2 Mn2+ + 5 NaBiO3 → 2 MnO4− + 5 Bi3+ + 5 Na+ + 15 H2O + 16 H+ − Simplifying: 14 H+(aq) + 2 Mn2+(aq) + 5 NaBiO3(s) → 2 MnO4-(aq) + 5 Bi3+(aq) + 5 Na+(aq) + 7 H2O(l) e. H3AsO4 → AsH3 H3AsO4 → AsH3 + 4 H2O 8 e− + 8 H+ + H3AsO4 → AsH3 + 4 H2O (Zn → Zn2+ + 2 e−) × 4 8 e− + 8 H+ + H3AsO4 → AsH3 + 4 H2O 4 Zn → 4 Zn2+ + 8 e− _______________________________________________________ 8 H+(aq) + H3AsO4(aq) + 4 Zn(s) → 4 Zn2+(aq) + AsH3(g) + 4 H2O(l) f. As2O3 → H3AsO4 As2O3 → 2 H3AsO4 (5 H2O + As2O3 → 2 H3AsO4 + 4 H+ + 4 e−) × 3 CHAPTER 4 SOLUTION STOICHIOMETRY 71 NO3− → NO + 2 H2O 4 H + NO3− → NO + 2 H2O (3 e− + 4 H+ + NO3− → NO + 2 H2O) × 4 + 12 e− + 16 H+ + 4 NO3− → 4 NO + 8 H2O 15 H2O + 3 As2O3 → 6 H3AsO4 + 12 H+ + 12 e− ______________________________________________________________ 7 H2O(l) + 4 H+(aq) + 3 As2O3(s) + 4 NO3−(aq) → 4 NO(g) + 6 H3AsO4(aq) g. (2 Br- → Br2 + 2 e−) × 5 MnO4− → Mn2+ + 4 H2O (5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O) × 2 10 Br− → 5 Br2 + 10 e− 10 e− + 16 H+ + 2 MnO4− → 2 Mn2+ + 8 H2O ____________________________________________________________ 16 H+(aq) + 2 MnO4−(aq) + 10 Br−(aq) → 5 Br2(l) + 2 Mn2+(aq) + 8 H2O(l) h. CH3OH → CH2O (CH3OH → CH2O + 2 H+ + 2 e−) × 3 Cr2O72− → Cr3+ 14 H + Cr2O72− → 2 Cr3+ + 7 H2O 6 e− + 14 H+ + Cr2O72− → 2 Cr3+ + 7 H2O + 3 CH3OH → 3 CH2O + 6 H+ + 6 e− 6 e + 14 H + Cr2O72− → 2 Cr3+ + 7 H2O _______________________________________________________________ 8 H+(aq) + 3 CH3OH(aq) + Cr2O72−(aq) → 2 Cr3+(aq) + 3 CH2O(aq) + 7 H2O(l) − 62. + Use the same method as with acidic solutions. After the final balanced equation, convert H+ to OH− as described in section 4.11 of the text. The extra step involves converting H+ into H2O by adding equal moles of OH− to each side of the reaction. This converts the reaction to a basic solution while keeping it balanced. a. Al → Al(OH)4− 4 H2O + Al → Al(OH)4− + 4 H+ 4 H2O + Al → Al(OH)4− + 4 H+ + 3 e- MnO4− → MnO2 3 e + 4 H + MnO4− → MnO2 + 2 H2O − + 4 H2O + Al → Al(OH)4− + 4 H+ + 3 e− 3 e + 4 H+ + MnO4− → MnO2 + 2 H2O ______________________________________________ 2 H2O(l) + Al(s) + MnO4−(aq) → Al(OH)4−(aq) + MnO2(s) − Because H+ doesn’t appear in the final balanced reaction, we are done. b. Cl2 → Cl− 2 e− + Cl2 → 2 Cl− Cl2− → ClO− 2 H2O + Cl2− → 2 ClO− + 4 H+ + 2 e− 2 e− + Cl2 → 2 Cl− 2 H2O + Cl2 → 2 ClO− + 4 H+ + 2 e− ________________________________ 2 H2O + 2 Cl2 → 2 Cl− + 2 ClO− + 4 H+ 72 CHAPTER 4 SOLUTION STOICHIOMETRY Now convert to a basic solution. Add 4 OH− to both sides of the equation. The 4 OH− will react with the 4 H+ on the product side to give 4 H2O. After this step, cancel identical species on both sides (2 H2O). Applying these steps gives 4 OH− + 2 Cl2 → 2 Cl− + 2 ClO− + 2 H2O, which can be further simplified to: 2 OH− (aq) + Cl2(g) → Cl−(aq) + ClO−(aq) + H2O(l) c. NO2− → NH3 6 e + 7 H + NO2− → NH3 + 2 H2O − Al → AlO2− (2 H2O + Al → AlO2− + 4 H+ + 3 e−) × 2 + Common factor is a transfer of 6 e−. 6e− + 7 H+ + NO2− → NH3 + 2 H2O 4 H2O + 2 Al → 2 AlO2− + 8 H+ + 6 e− _______________________________________________ OH− + 2 H2O + NO2− + 2 Al → NH3 + 2 AlO2− + H+ + OH− Reducing gives: OH−(aq) + H2O(l) + NO2−(aq) + 2 Al(s) → NH3(g) + 2 AlO2−(aq) d. MnO4− →MnS MnO4− + S2− → MnS − + ( 5 e + 8 H + MnO4− + S2− → MnS + 4 H2O) × 2 S2− → S (S2− → S + 2 e−) × 5 Common factor is a transfer of 10 e−. 5 S2− → 5 S + 10 e− + 2 S2− → 2 MnS + 8 H2O 10 e + 16 H + 2 ________________________________________________________ 16 OH− + 16 H+ + 7 S2− + 2 MnO4− → 5 S + 2 MnS + 8 H2O + 16 OH− − MnO4− + 16 H2O + 7 S2− + 2 MnO4− → 5 S + 2 MnS + 8 H2O + 16 OH− Reducing gives: 8 H2O(l) + 7 S2−(aq) + 2 MnO4−(aq) → 5 S(s) + 2 MnS(s) + 16 OH−(aq) e. CN− → CNO− (H2O + CN− → CNO− + 2 H+ + 2 e−) × 3 MnO4− → MnO2 (3 e + 4 H + MnO4− → MnO2 + 2 H2O) × 2 Common factor is a transfer of 6 electrons. − + 3 H2O + 3 CN− → 3 CNO- + 6 H+ + 6 e− 6 e + 8 H+ + 2 MnO4− → 2 MnO2 + 4 H2O ___________________________________________________________ 2 OH− + 2 H+ + 3 CN− + 2 MnO4− → 3 CNO− + 2 MnO2 + H2O + 2 OH− − Reducing gives: H2O(l) + 3 CN−(aq) + 2 MnO4−(aq) → 3 CNO−(aq) + 2 MnO2(s) + 2 OH−(aq) CHAPTER 4 63. SOLUTION STOICHIOMETRY 73 a. HCl(aq) dissociates to H+(aq) + Cl−(aq). For simplicity, let's use H+ and Cl− separately. H+ → H2 (2 H+ + 2 e− → H2) × 3 Fe → HFeCl4 (H+ + 4 Cl− + Fe → HFeCl4 + 3 e−) × 2 6 H+ + 6 e− → 3 H2 2 H+ + 8 Cl− + 2 Fe → 2 HFeCl4 + 6 e− _________________________________ 8 H+ + 8 Cl− + 2 Fe → 2 HFeCl4 + 3 H2 or b. 8 HCl(aq) + 2 Fe(s) → 2 HFeCl4(aq) + 3 H2(g) I− → I3− (3 I− → I3− + 2 e−) × 8 IO3− → I3− 3 IO3− → I3− 3 IO3− → I3− + 9 H2O 16 e− + 18 H+ + 3 IO3− → I3− + 9 H2O 16 e− + 18 H+ + 3 IO3− → I3− + 9 H2O 24 I− → 8 I3− + 16 e− _______________________________ 18 H+ + 24 I− + 3 IO3− → 9 I3− + 9 H2O Reducing: 6 H+(aq) + 8 I−(aq) + IO3−(aq) → 3 I3−(aq) + 3 H2O(l) c. (Ce4+ + e− → Ce3+) × 97 Cr(NCS)64− → Cr3+ + NO3− + CO2 + SO42− 54 H2O + Cr(NCS)64− → Cr3+ + 6 NO3− + 6 CO2 + 6 SO42− + 108 H+ Charge on left = −4. Charge on right = +3 + 6(−1) + 6(−2) + 108(+1) = +93. Add 97 e− to the product side, and then add the two balanced half-reactions with a common factor of 97 e− transferred. 54 H2O + Cr(NCS)64− → Cr3+ + 6 NO3− + 6 CO2 + 6 SO42− + 108 H+ + 97 e− 97 e− + 97 Ce4+ → 97 Ce3+ ___________________________________________________________________________ 97 Ce4+(aq) + 54 H2O(l) + Cr(NCS)64−(aq) → 97 Ce3+(aq) + Cr3+(aq) + 6 NO3−(aq) + 6 CO2(g) + 6 SO42−(aq) + 108 H+(aq) This is very complicated. A check of the net charge is a good check to see if the equation is balanced. Left: charge = 97(+4) −4 = +384. Right: charge = 97(+3) + 3 + 6(−1) + 6(−2) + 108(+1) = +384. d. CrI3 → CrO42− + IO4− (16 H2O + CrI3 → CrO42− + 3 IO4− + 32 H+ + 27 e−) × 2 Common factor is a transfer of 54 e−. Cl2 → Cl− (2 e− + Cl2 → 2 Cl−) × 27 74 CHAPTER 4 SOLUTION STOICHIOMETRY 54 e− + 27 Cl2 → 54 Cl− 32 H2O + 2 CrI3 → 2 CrO42− + 6 IO4− + 64 H+ + 54 e− ___________________________________________________ 32 H2O + 2 CrI3 + 27 Cl2 → 54 Cl− + 2 CrO42− + 6 IO4− + 64 H+ Add 64 OH− to both sides and convert 64 H+ into 64 H2O. 64 OH− + 32 H2O + 2 CrI3 + 27 Cl2 → 54 Cl− + 2 CrO42− + 6 IO4− + 64 H2O Reducing gives: 64 OH−(aq) + 2 CrI3(s) + 27 Cl2(g) → 54 Cl−(aq) + 2 CrO42− (aq) + 6 IO4−(aq) + 32 H2O(l) e. Ce4+ → Ce(OH)3 (e + 3 H2O + Ce4+ → Ce(OH)3 + 3 H+) × 61 − Fe(CN)64− → Fe(OH)3 + CO32− + NO3− Fe(CN)64− → Fe(OH)3 + 6 CO32− + 6 NO3− There are 39 extra O atoms on right. Add 39 H2O to left, then add 75 H+ to right to balance H+. 39 H2O + Fe(CN)64− → Fe(OH)3 + 6 CO32− + 6 NO3− + 75 H+ net charge = 4− net charge = 57+ Add 61 e− to the product side, and then add the two balanced half-reactions with a common factor of 61 e− transferred. 39 H2O + Fe(CN)64− → Fe(OH)3 + 6 CO3− + 6 NO3− + 75 H+ + 61 e− 61 e− + 183 H2O + 61 Ce4+ → 61 Ce(OH)3 + 183 H+ _______________________________________________________________________ 222 H2O + Fe(CN)64− + 61 Ce4+ → 61 Ce(OH)3 + Fe(OH)3 + 6 CO32− + 6 NO3− + 258 H+ Adding 258 OH− to each side, and then reducing gives: 258 OH−(aq) + Fe(CN)64−(aq) + 61 Ce4+(aq) → 61 Ce(OH)3(s) + Fe(OH)3(s) + 6 CO32−(aq) + 6 NO3−(aq) + 36 H2O(l) 64. Mn → Mn2+ + 2 e− HNO3 → NO2 HNO3 → NO2 + H2O (e− + H+ + HNO3 → NO2 + H2O) × 2 Mn → Mn2+ + 2 e− 2 e + 2 H + 2 HNO3 → 2 NO2 + 2 H2O ________________________________________________________ 2 H+(aq) + Mn(s) + 2 HNO3(aq) → Mn2+(aq) + 2 NO2(g) + 2 H2O(l) or − + 4 H+(aq) + Mn(s) + 2 NO3−(aq) → Mn2+(aq) + 2 NO2(g) + 2 H2O(l) (HNO3 is a strong acid.) CHAPTER 4 SOLUTION STOICHIOMETRY (4 H2O + Mn2+ → MnO4− + 8 H+ + 5 e−) × 2 75 (2 e− + 2 H+ + IO4− → IO3− + H2O) × 5 8 H2O + 2 Mn2+ → 2 MnO4− + 16 H+ + 10 e− 10 e + 10 H+ + 5 IO4− → 5 IO3− + 5 H2O _____________________________________________________________ 3 H2O(l) + 2 Mn2+(aq) + 5 IO4−(aq) → 2 MnO4−(aq) + 5 IO3−(aq) + 6 H+(aq) − 65. (5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O) × 2 (H2C2O4 → 2 CO2 + 2 H+ + 2 e−) × 5 5 H2C2O4 → 10 CO2 + 10 H+ + 10 e− 10 e + 16 H + 2 MnO4− → 2 Mn2+ + 8 H2O ________________________________________________________________ 6 H+(aq) + 5 H2C2O4(aq) + 2 MnO4−(aq) → 10 CO2(g) + 2 Mn2+(aq) + 8 H2O(l) − + − 0.1058 g H2C2O4 × 1 mol H 2 C 2 O 4 2 mol MnO 4 × = 4.700 × 10−4 mol MnO4− 90.034 g 5 mol H 2 C 2 O 4 − Molarity = 66. 4.700 × 10 −4 mol MnO 4 1000 mL × = 1.622 × 10−2 M MnO4− 28.97 mL L a. (Fe2+ → Fe3+ + e−) × 5 5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O The balanced equation is: 8 H+(aq) + MnO4−(aq) + 5 Fe2+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) 20.62 × 10−3 L soln × Molarity = − 0.0216 mol MnO 4 5 mol Fe 2+ = 2.23 × 10−3 mol Fe2+ × − L so ln mol MnO 4 2.23 × 10 −3 mol Fe 2+ = 4.46 × 10−2 M Fe2+ 50.00 × 10 −3 L b. (Fe2+ → Fe3+ + e−) × 6 6 e− + 14 H+ + Cr2O72− → 2 Cr3+ + 7 H2O The balanced equation is: 14 H+(aq) + Cr2O72−(aq) + 6 Fe2+(aq) → 6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l) 50.00 × 10−3 L × 2− 4.46 × 10 −2 mol Fe 2+ 1 mol Cr2 O 7 × L 6 mol Fe 2+ × 1L 0.0150 mol Cr2 O 7 2− = 2.48 × 10−2 L or 24.8 mL 76 CHAPTER 4 SOLUTION STOICHIOMETRY (Fe2+ → Fe3+ + e−) × 5 5 e + 8 H + MnO4− → Mn2+ + 4 H2O __________________________________________________________ 8 H+(aq) + MnO4−(aq) + 5 Fe2+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) 67. − + From the titration data we can get the number of moles of Fe2+. We then convert this to a mass of iron and calculate the mass percent of iron in the sample. 38.37 × 10−3 L MnO4− × 3.80 × 10−3 mol Fe × Mass % Fe = 68. − 0.0198 mol MnO 4 5 mol Fe 2 + = 3.80 × 10−3 mol Fe2+ × − L mol MnO 4 = 3.80 × 10−3 mol Fe present 55.85 g Fe = 0.212 g Fe mol Fe 0.212 g × 100 = 34.6% Fe 0.6128 g The unbalanced reaction is: VO2+ + MnO4− → V(OH)4+ + Mn2+ This is a redox reaction in acidic solution and must be balanced accordingly. The two halfreactions to balance are: VO2+ → V(OH)4+ and MnO4− → Mn2+ Balancing by the half-reaction method gives: MnO4−(aq) + 5 VO2+(aq) + 11 H2O(l) → 5 V(OH)4+(aq) + Mn2+(aq) + 2 H+(aq) − 0.02645 L × 0.581 = 69. 0.1516 g V , 0.1516/0.581 = 0.261 g ore sample mass of ore sample Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) 3.00 g Mg × 70. 0.02250 mol MnO 4 5 mol VO 2+ 1 mol V 50.94 g V × × × = 0.1516 g V − 2+ L mol V mol VO mol MnO 4 a. 1 mol Mg 2 mol HCl 1 L HCl × × = 0.0494 L = 49.4 mL HCl 24.31 g Mg mol Mg 5.0 mol HCl 16 e− + 18 H+ + 3 IO3− → I3− + 9 H2O (3 I− → I3− + 2 e−) × 8 24 I− → 8 I3− + 16 e− 16 e− + 18 H+ + 3 IO3− → I3− + 9 H2O ______________________________ 18 H+ + 24 I− + 3 IO3− → 9 I3− + 9 H2O Reducing: 6 H+(aq) + 8 I−(aq) + IO3−(aq) → 3 I3−(aq) + 3 H2O(l) CHAPTER 4 SOLUTION STOICHIOMETRY b. 0.6013 g KIO3 H 77 1 mol KIO3 = 2.810 × 10−3 mol KIO3 214.0 g KIO3 2.810 × 10−3 mol KIO3 × 8 mol KI 166.0 g KI × = 3.732 g KI mol KIO3 mol KI 2.810 × 10−3 mol KIO3 × 6 mol HCl 1L × = 5.62 × 10−3 L = 5.62 mL HCl mol KIO3 3.00 mol HCl c. I3− + 2 e− → 3 I− 2 S2O32− → S4O62− + 2 e− Adding the balanced half-reactions gives: 2 S2O32−(aq) + I3−(aq) → 3 I−(aq) + S4O62−(aq) d. 25.00 × 10−3 L KIO3 × − 0.0100 mol KIO3 3 mol I 3 2 mol Na 2S2 O 3 × × = − L mol KIO3 mol I 3 1.50 × 10−3 mol Na2S2O3 M Na 2S2O3 = e. 0.5000 L H 1.50 × 10 −3 mol = 0.0468 M Na2S2O3 32.04 × 10 −3 L 0.0100 mol KIO3 214.0 g KIO3 = 1.07 g KIO3 × L mol KIO3 Place 1.07 g KIO3 in a 500-mL volumetric flask; add water to dissolve the KIO3; continue adding water to the 500.00-mL mark, with mixing along the way. Additional Exercises 71. 0.275 mol CaCl 2 = 6.33 × 10−2 mol CaCl2 L CaCl 2 The volume of CaCl2 solution after evaporation is: Mol CaCl2 present = 0.230 L CaCl2 × 6.33 × 10−2 mol CaCl2 × 1 L CaCl 2 = 5.75 × 10−2 L = 57.5 mL CaCl2 1.10 mol CaCl 2 Volume H2O evaporated = 230. mL − 57.5 mL = 173 mL H2O evaporated 72. There are other possible correct choices for the following answers. We have listed only three possible reactants in each case. a. AgNO3, Pb(NO3)2, and Hg2(NO3)2 would form precipitates with the Cl− ion. Ag+(aq) + Cl−(aq) → AgCl(s); Pb2+(aq) + 2 Cl-(aq) → PbCl2(s) Hg22+(aq) + 2 Cl−(aq) → Hg2Cl2(s) 78 CHAPTER 4 SOLUTION STOICHIOMETRY b. Na2SO4, Na2CO3, and Na3PO4 would form precipitates with the Ca2+ ion. Ca2+(aq) + SO42−(aq) → CaSO4(s); Ca2+(aq) + CO32−(aq) → CaCO3(s) 3 Ca2+(aq) + 2 PO43−(aq) → Ca3(PO4)2(s) c. NaOH, Na2S, and Na2CO3 would form precipitates with the Fe3+ ion. Fe3+(aq) + 3 OH−(aq) → Fe(OH)3(s); 2 Fe3+(aq) + 3 S2−(aq) → Fe2S3(s) 2 Fe3+(aq) + 3 CO32−(aq) → Fe2(CO3)3(s) d. BaCl2, Pb(NO3)2, and Ca(NO3)2 would form precipitates with the SO42− ion. Ba2+(aq) + SO42−(aq) → BaSO4(s); Pb2+(aq) + SO42−(aq) → PbSO4(s) Ca2+(aq) + SO42−(aq) → CaSO4(s) e. Na2SO4, NaCl, and NaI would form precipitates with the Hg22+ ion. Hg22+ (aq) + SO42−(aq) → Hg2SO4(s); Hg22+(aq) + 2 Cl−(aq) → Hg2Cl2(s) Hg22+ (aq) + 2 I−(aq) → Hg2I2(s) f. NaBr, Na2CrO4, and Na3PO4 would form precipitates with the Ag+ ion. Ag+(aq) + Br−(aq) → AgBr(s); 2 Ag+(aq) + CrO42−(aq) → Ag2CrO4(s) 3 Ag+(aq) + PO43−(aq) → Ag3PO4(s) 73. a. MgCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Mg(NO3)2(aq) 0.641 g AgCl × 1 mol MgCl 2 1 mol AgCl 95.21 g = 0.213 g MgCl2 × × 143.4 g AgCl 2 mol AgCl mol MgCl 2 0.213 g MgCl 2 × 100 = 14.2% MgCl2 1.50 g mixture b. 0.213 g MgCl2 × 2 mol AgNO3 1 mol MgCl 2 1L 1000 mL × × × 95.21 g mol MgCl 2 0.500 mol AgNO3 1L = 8.95 mL AgNO3 74. Al(NO3)3(aq) + 3 KOH(aq) → Al(OH)3(s) + 3 KNO3(aq) 0.0500 L × 0.200 mol Al( NO 3 ) 3 = 0.0100 mol Al(NO3)3 L 0.2000 L × 0.100 mol KOH = 0.0200 mol KOH L From the balanced equation, 3 moles of KOH are required to react with 1 mole of Al(NO3)3 (3 : 1 mole ratio). The actual KOH to Al(NO3)3 mole ratio present is 0.0200/0.0100 = 2 (2 : 1). CHAPTER 4 SOLUTION STOICHIOMETRY 79 Because the actual mole ratio present is less than the required mole ratio, KOH is the limiting reagent. 75. 0.0200 mol KOH × 1 mol Al(OH ) 3 78.00 g Al(OH) 3 × = 0.520 g Al(OH)3 3 mol KOH mol Al(OH) 3 a. 0.308 g AgCl × 35.45 g Cl 0.0761 g = 0.0761 g Cl; % Cl = × 100 = 29.7% Cl 143.4 g AgCl 0.256 g Cobalt(III) oxide, Co2O3: 2(58.93) + 3(16.00) = 165.86 g/mol 117.86 g Co 0.103 g = 0.103 g Co; % Co = × 100 = 24.8% Co 165.86 g Co 2 O 3 0.416 g 0.145 g Co2O3 × The remainder, 100.0 - (29.7 + 24.8) = 45.5%, is water. Assuming 100.0 g of compound: 45.5 g H2O × 2.016 g H 5.09 g H = 5.09 g H; % H = × 100 = 5.09% H 18.02 g H 2 O 100.0 g compound 45.5 g H2O × 16.00 g O 40.4 g O = 40.4 g O; % O = × 100 = 40.4% O 18.02 g H 2 O 100.0 g compound The mass percent composition is 24.8% Co, 29.7% Cl, 5.09% H, and 40.4% O. b. Out of 100.0 g of compound, there are: 24.8 g Co H 5.09 g H × 1 mol 1 mol = 0.421 mol Co; 29.7 g Cl × = 0.838 mol Cl 58.93 g Co 35.45 g Cl 1 mol 1 mol = 5.05 mol H; 40.4 g O × = 2.53 mol O 1.008 g H 16.00 g O Dividing all results by 0.421, we get CoCl2•6H2O for the empirical formula, which is also the molecular formula. c. CoCl2•6H2O(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Co(NO3)2(aq) + 6 H2O(l) CoCl2•6H2O(aq) + 2 NaOH(aq) → Co(OH)2(s) + 2 NaCl(aq) + 6 H2O(l) Co(OH)2 → Co2O3 This is an oxidation-reduction reaction. Thus we also need to include an oxidizing agent. The obvious choice is O2. 4 Co(OH)2(s) + O2(g) → 2 Co2O3(s) + 4 H2O(l) 80 76. CHAPTER 4 SOLUTION STOICHIOMETRY a. Fe3+(aq) + 3 OH−(aq) → Fe(OH)3(s) Fe(OH)3: 55.85 + 3(16.00) + 3(1.008) = 106.87 g/mol 0.107 g Fe(OH)3 × 55.85 g Fe = 0.0559 g Fe 106.9 g Fe(OH) 3 b. Fe(NO3)3: 55.85 + 3(14.01) + 9(16.00) = 241.86 g/mol 0.0559 g Fe × 241.9 g Fe( NO 3 ) 3 = 0.242 g Fe(NO3)3 55.85 g Fe c. Mass % Fe(NO3)3 = 77. 0.242 g × 100 = 53.1% 0.456 g Ag+(aq) + Cl−(aq) → AgCl(s); let x = mol NaCl and y = mol KCl. (22.90 × 10−3 L) × 0.1000 mol/L = 2.290 × 10−3 mol Ag+ = 2.290 × 10−3 mol Cl− total x + y = 2.290 × 10−3 mol Cl−, x = 2.290 × 10−3 − y Because the molar mass of NaCl is 58.44 g/mol and the molar mass of KCl is 74.55 g/mol: (58.44)x + (74.55)y = 0.1586 g 58.44(2.290 × 10−3 − y) + (74.55)y = 0.1586, (16.11)y = 0.0248, y = 1.54 × 10−3 mol KCl Mass % KCl = 1.54 × 10 −3 mol × 74.55 g / mol × 100 = 72.4% KCl 0.1586 g % NaCl = 100.0 − 72.4 = 27.6% NaCl 78. a. Assume 100.00 g of material. 42.23 g C × 1 mol C 1 mol F = 3.516 mol C; 55.66 g F × = 2.929 mol F 12.011 g C 19.00 g F 1 mol B = 0.195 mol B 10.81 g B 3.516 2.929 = 18.0; = 15.0 Dividing by the smallest number: 0.195 0.195 2.11 g B × The empirical formula is C18F15B. b. 0.3470 L × 0.01267 mol = 4.396 × 10 −3 mol BARF L CHAPTER 4 SOLUTION STOICHIOMETRY Molar mass of BARF = 81 2.251 g = 512.1 g/mol 4.396 × 10 −3 mol The empirical formula mass of BARF is 511.99 g. Therefore, the molecular formula is the same as the empirical formula, C18F15B. 79. Cr(NO3)3(aq) + 3 NaOH(aq) → Cr(OH)3(s) + 3 NaNO3(aq) Mol NaOH used = 2.06 g Cr(OH)3 × to form precipitate 1 mol Cr (OH) 3 3 mol NaOH = 6.00 × 10−2 mol × 103.02 g mol Cr (OH) 3 NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) Mol NaOH used = 0.1000 L × to react with HCl MNaOH = 80. 0.400 mol HCl 1 mol NaOH = 4.00 × 10−2 mol × L mol HCl 6.00 × 10 −2 mol + 4.00 × 10 −2 mol total mol NaOH = = 2.00 M NaOH volume 0.0500 L 3 (NH4)2CrO4(aq) + 2 Cr(NO2)3(aq) → 6 NH4NO2(aq) + Cr2(CrO4)3(s) 0.203 L × 0.307 mol = 6.23 × 10 −2 mol (NH4)2CrO4 L 0.137 L × 0.269 mol = 3.69 × 10 −2 mol Cr(NO2)3 L 0.0623 mol = 1.69 (actual); the balanced reaction requires a 3/2 = 1.5 to 1 mole ratio 0.0369 mol between (NH4)2CrO4 and Cr(NO2)3. Actual > required, so Cr(NO2)3 (the denominator) is limiting. 3.69 × 10 −2 mol Cr(NO2)3 × 0.880 = 81. 1 mol Cr2 (CrO 4 ) 3 452.00 g Cr2 (CrO 4 ) 3 × = 8.34 g Cr2(CrO4)3 2 mol Cr ( NO 2 ) 3 mol Cr2 (CrO 4 ) 3 actual yield , actual yield = (8.34 g)(0.880) = 7.34 g Cr2(CrO4)3 isolated 8.34 g The amount of KHP used = 0.4016 g × 1 mol = 1.967 × 10−3 mol KHP. 204.22 g Because 1 mole of NaOH reacts completely with 1 mole of KHP, the NaOH solution contains 1.967 × 10−3 mol NaOH. 82 CHAPTER 4 Molarity of NaOH = SOLUTION STOICHIOMETRY 1.967 × 10 −3 mol 7.849 × 10 −2 mol NaOH = L 25.06 × 10 −3 L Maximum molarity = 1.967 × 10 −3 mol 7.865 × 10 −2 mol NaOH = L 25.01 × 10 −3 L Minimum molarity = 1.967 × 10 −3 mol 7.834 × 10 −2 mol NaOH = L 25.11 × 10 −3 L We can express this as 0.07849 ±0.00016 M. An alternate way is to express the molarity as 0.0785 ±0.0002 M. This second way shows the actual number of significant figures in the molarity. The advantage of the first method is that it shows that we made all our individual measurements to four significant figures. 82. Desired uncertainty is 1% of 0.02, or ±0.0002. So we want the solution to be 0.0200 ± 0.0002 M, or the concentration should be between 0.0198 and 0.0202 M. We should use a 1L volumetric flask to make the solution. They are good to ±0.1%. We want to weigh out between 0.0198 mol and 0.0202 mol of KIO3. Molar mass of KIO3 = 39.10 + 126.9 + 3(16.00) = 214.0 g/mol 0.0198 mol × 214.0 g 214.0 g = 4.237 g; 0.0202 mol × = 4.323 g (carrying extra sig. figs.) mol mol We should weigh out between 4.24 and 4.32 g of KIO3. We should weigh it to the nearest milligram or 0.1 mg. Dissolve the KIO3 in water, and dilute to the mark in a 1-liter volumetric flask. This will produce a solution whose concentration is within the limits and is known to at least the fourth decimal place. 83. Mol C6H8O7 = 0.250 g C6H8O7 × 1 mol C 6 H 8 O 7 = 1.30 × 10−3 mol C6H8O7 192.1 g C 6 H 8 O 7 Let HxA represent citric acid, where x is the number of acidic hydrogens. The balanced neutralization reaction is: HxA(aq) + x OH−(aq) → x H2O(l) + Ax−(aq) Mol OH− reacted = 0.0372 L × x= 0.105 mol OH − = 3.91 × 10−3 mol OH− L mol OH − 3.91 × 10 −3 mol = = 3.01 mol citric acid 1.30 × 10 −3 mol Therefore, the general acid formula for citric acid is H3A, meaning that citric acid has three acidic hydrogens per citric acid molecule (citric acid is a triprotic acid). CHAPTER 4 SOLUTION STOICHIOMETRY 83 Challenge Problems 84. 2(6 e− + 14 H+ + Cr2O72− → 2 Cr3+ + 7 H2O) 3 H2O + C2H5OH → 2 CO2 + 12 H+ + 12 e− ___________________________________________________________________________ 16 H+ + 2 Cr2O72− + C2H5OH → 4 Cr3+ + 2 CO2 + 11 H2O ⎛ 0.0600 mol Cr2 O 7 2− 0.03105 L ⎜ ⎜ L ⎝ ⎞ ⎛ 1 mol C 2 H 5 OH ⎞ ⎛ 46.07 g ⎞ ⎟⎜ ⎟⎜ ⎟ = 0.0429 g C2H5OH ⎟ ⎜ 2 mol Cr O 2 − ⎟ ⎜ mol C 2 H 5 OH ⎟ ⎠ 2 7 ⎠⎝ ⎠⎝ 0.0429 g C 2 H 5 OH × 100 = 0.143% C2H5OH 30.0 g blood 85. a. Let x = mass of Mg, so 10.00 − x = mass of Zn. Ag+(aq) + Cl−(aq) → AgCl(s). From the given balanced equations, there is a 2 : 1 mole ratio between mol Mg and mol Cl−. The same is true for Zn. Because mol Ag+ = mol Cl− present, one can setup an equation relating mol Cl− present to mol Ag+ added. x g Mg × 1 mol Mg 2 mol Cl − 1 mol Zn 2 mol Cl − × + (10.00 − x) g Zn × × 24.31 g Mg mol Mg 65.38 g Zn mol Zn = 0.156 L × 3.00 mol Ag + 1 mol Cl − × = 0.468 mol Cl− L mol Ag + 20.00 − 2 x 2x 2(10.00 − x) ⎞ ⎛ 2x + = 0.468, 24.31 × 65.38 ⎜ = 0.468 ⎟ + 24.31 65.38 65.38 ⎠ ⎝ 24.31 (130.8)x + 486.2 ! (48.62)x = 743.8 (carrying 1 extra sig. fig.) (82.2)x = 257.6, x = 3.13 g Mg; b. 0.156 L × MHCl = 86. % Mg = 3.13 g Mg × 100 = 31.3% Mg 10.00 g mixture 3.00 mol Ag + 1 mol Cl − × = 0.468 mol Cl− = 0.468 mol HCl added L mol Ag + 0.468 mol = 6.00 M HCl 0.0780 L Let x = mass of NaCl, and let y = mass K2SO4. So x + y = 10.00. Two reactions occur: Pb2+(aq) + 2 Cl−(aq) → PbCl2(s) and Pb2+(aq) + SO42−(aq) → PbSO4(s) Molar mass of NaCl = 58.44 g/mol; molar mass of K2SO4 = 174.27 g/mol Molar mass of PbCl2 = 278.1 g/mol; molar mass of PbSO4 = 303.3 g/mol 84 CHAPTER 4 SOLUTION STOICHIOMETRY x y = moles NaCl; = moles K2SO4 58.44 174.27 mass of PbCl2 + mass PbSO4 = total mass of solid x y (1/2)(278.1) + (303.3) = 21.75 58.44 174.27 We have two equations: (2.379)x + (1.740)y = 21.75 and x + y = 10.00. Solving: x = 6.81 g NaCl; 87. 6.81 g NaCl × 100 = 68.1% NaCl 10.00 g mixture Zn(s) + 2 AgNO2(aq) → 2 Ag(s) + Zn(NO2)2(aq) Let x = mass of Ag and y = mass of Zn after the reaction has stopped. Then x + y = 29.0 g. Because the moles of Ag produced will equal two times the moles of Zn reacted: (19.0 − y) g Zn × 1 mol Ag 1 mol Zn 2 mol Ag × = x g Ag × 107.9 g Ag 65.38 g Zn 1 mol Zn Simplifying: 3.059 × 10−2(19.0 − y) = (9.268 × 10−3)x Substituting x = 29.0 − y into the equation gives: 3.059 × 10−2(19.0 − y) = 9.268 × 10−3(29.0 − y) Solving: 0.581 − (3.059 × 10−2)y = 0.269 − (9.268 × 10−3)y, (2.132 × 10−2)y = 0.312, y = 14.6 g Zn 14.6 g Zn are present, and 29.0 − 14.6 = 14.4 g Ag are also present after the reaction is stopped. 88. 0.2750 L × 0.300 mol/L = 0.0825 mol H+; let y = volume (L) delivered by Y and z = volume (L) delivered by Z. H+(aq) + OH−(aq) → H2O(l); y(0.150 mol/L) + z(0.250 mol/L) = 0.0825 mol H+ mol OH− 0.2750 L + y + z = 0.655 L, y + z = 0.380, z = 0.380 ! y y(0.150) + (0.380 ! y)(0.250) = 0.0825, solving: y = 0.125 L, z = 0.255 L CHAPTER 4 SOLUTION STOICHIOMETRY Flow rate for Y = 89. 85 125 mL 255 mL = 2.06 mL/min; flow rate for Z = = 4.20 mL/min 60.65 min 60.65 min Molar masses: KCl, 39.10 + 35.45 = 74.55 g/mol; KBr, 39.10 + 79.90 = 119.00 g/mol AgCl, 107.9 + 35.45 = 143.4 g/mol; AgBr, 107.9 + 79.90 = 187.8 g/mol Let x = number of moles of KCl in mixture and y = number of moles of KBr in mixture. Ag+ + Cl− → AgCl and Ag+ + Br− → AgBr; so, x = moles AgCl and y = moles AgBr. Setting up two equations from the given information: 0.1024 g = (74.55)x + (119.0)y and 0.1889 g = (143.4)x + (187.8)y Multiply the first equation by 187.8 , and then subtract from the second. 119.0 0.1889 = (143.4)x + (187.8)y −0.1616 = −(117.7)x − (187.8)y x = 1.06 × 10−3 mol KCl 0.0273 = (25.7)x, 1.06 × 10−3 mol KCl × Mass % KCl = 90. 74.55 KCl = 0.0790 g KCl mol KCl 0.0790 g × 100 = 77.1%, % KBr = 100.0 − 77.1 = 22.9% 0.1024 g Pb2+(aq) + 2 Cl−(aq) → PbCl2(s) 3.407 g 3.407 g PbCl2 × 1 mol PbCl 2 1 mol Pb 2+ × = 0.01225 mol Pb2+ 278.1 g PbCl 2 1 mol PbCl 2 0.01225 mol = 6.13 M Pb2+ (evaporated concentration) 2.00 × 10 −3 L Original concentration = 91. 0.0800 L × 6.13 mol / L = 4.90 M 0.100 L a. C12H10-nCln + n Ag+ → n AgCl; molar mass of AgCl = 143.4 g/mol Molar mass of PCB = 12(12.01) + (10 − n)(1.008) + n(35.45) = 154.20 + (34.44)n Because n mol AgCl are produced for every 1 mol PCB reacted, n(143.4) g of AgCl will be produced for every [154.20 + (34.44)n] g of PCB reacted. 86 CHAPTER 4 SOLUTION STOICHIOMETRY Mass of AgCl (143.4)n = or massAgCl[154.20 + (34.44)n] = massPCB(143.4)n Mass of PCB 154.20 + (34.44)n b. 0.4971[154.20 + (34.44)n] = 0.1947(143.4)n, 76.65 + (17.12)n = (27.92)n 76.65 = (10.80)n, n = 7.097 92. Mol CuSO4 = 87.6 mL × Mol Fe = 2.00 g × 1L 0.500 mol × = 0.0439 mol 1000 mL L 1 mol Fe = 0.0358 mol 55.85 g The two possible reactions are: I. CuSO4(aq) + Fe(s) → Cu(s) + FeSO4(aq) II. 3 CuSO4(aq) + 2 Fe(s) → 3 Cu(s) + Fe2(SO4)3(aq) If reaction I occurs, Fe is limiting, and we can produce: 0.0358 mol Fe × 1 mol Cu 63.55 g Cu × = 2.28 g Cu 1 mol Fe mol Cu If reaction II occurs, CuSO4 is limiting, and we can produce: 0.0439 mol CuSO4 × 3 mol Cu 63.55 g Cu × = 2.79 g Cu 3 mol CuSO 4 mol Cu Assuming 100% yield, reaction I occurs because it fits the data best. 93. a. Flow rate = 5.00 × 104 L/s + 3.50 × 103 L/s = 5.35 × 104 L/s b. CHCl = 3.50 × 103 (65.0) = 4.25 ppm HCl 5.35 × 10 4 c. 1 ppm = 1 mg/kg H2O = 1 mg/L (assuming density = 1.00 g/mL) 8.00 h × 60 min 60 s 1.80 × 10 4 L 4.25 mg HCl 1g × × × × = 2.20 × 106 g HCl h min s L 1000 mg 2.20 × 106 g HCl × 1 mol HCl 1 mol CaO 56.08 g Ca × × = 1.69 × 106 g CaO 36.46 g HCl 2 mol HCl mol CaO d. The concentration of Ca2+ going into the second plant was: CHAPTER 4 SOLUTION STOICHIOMETRY 87 5.00 × 10 4 (10.2) = 9.53 ppm 5.35 × 10 4 The second plant used: 1.80 × 104 L/s × (8.00 × 60 × 60) s = 5.18 × 108 L of water. 1.69 × 106 g CaO × 40.08 g Ca 2 + = 1.21 × 106 g Ca2+ was added to this water. 56.08 g CaO C Ca 2+ (plant water) = 9.53 + 1.21 × 109 mg = 9.53 + 2.34 = 11.87 ppm 5.18 × 108 L Because 90.0% of this water is returned, (1.80 × 104) × 0.900 = 1.62 × 104 L/s of water with 11.87 ppm Ca2+ is mixed with (5.35 - 1.80) × 104 = 3.55 × 104 L/s of water containing 9.53 ppm Ca2+. C Ca 2+ (final) = 94. a. (1.62 × 10 4 L / s)(11.87 ppm) + (3.55 × 10 4 L / s)(9.53 ppm) = 10.3 ppm 1.62 × 10 4 L / s + 3.55 × 10 4 L / s 7 H2O + 2 Cr3+ → Cr2O72− + 14 H+ + 6 e− (2 e- + S2O82- → 2 SO42−) × 3 ________________________________________________________________ 7 H2O(l) + 2 Cr3+(aq) + 3 S2O82−(aq) → Cr2O72−(aq) + 14 H+(aq) + 6 SO42−(aq) (Fe2+ → Fe3+ + e−) × 6 6 e + 14 H + Cr2O72− → 2 Cr3+ + 7 H2O ____________________________________________________________ 14 H+(aq) + 6 Fe2+(aq) + Cr2O72−(aq) → 2 Cr3+(aq) + 6 Fe3+(aq) + 7 H2O(l) − b. 8.58 × 10−3 L × + 0.0520 mol Cr2 O 72 − 6 mol Fe 2+ = 2.68 × 10−3 mol of excess Fe2+ × 2− L mol Cr2 O 7 Fe2+ (total) = 3.000 g Fe(NH4)2(SO4)2•6H2O × 1 mol = 7.650 × 10−3 mol Fe2+ 392.17 g 7.650 × 10−3 − 2.68 × 10−3 = 4.97 × 10−3 mol Fe2+ reacted with Cr2O72− generated from the Cr plating. The Cr plating contained: 4.97 × 10−3 mol Fe2+ × 1.66 × 10−3 mol Cr × 1 mol Cr2 O 72 − 2 mol Cr 3+ = 1.66 × 10−3 mol Cr3+ × 2+ 2− 6 mol Fe mol Cr2 O 7 = 1.66 × 10−3 mol Cr 52.00 g Cr = 8.63 × 10−2 g Cr mol Cr 88 CHAPTER 4 Volume of Cr plating = 8.63 × 10−2 g × Thickness of Cr plating = 95. SOLUTION STOICHIOMETRY 1 cm 3 = 1.20 × 10−2 cm3 = area × thickness 7.19 g 1.20 × 10 −2 cm 3 = 3.00 × 10−4 cm = 300. µm 2 40.0 cm a. YBa2Cu3O6.5: +3 + 2(+2) + 3x + 6.5(−2) = 0 7 + 3x − 13 = 0, 3x = 6, x = +2 Only Cu2+ present. YBa2Cu3O7: +3 + 2(+2) + 3x + 7(−2) = 0, x = +2 1/3 or 2.33 This corresponds to two Cu2+ and one Cu3+ present. YBa2Cu3O8: +3 + 2(+2) + 3x + 8(−2) = 0, x = +3; Only Cu3+ present. b. (e− + Cu2+ + I− → CuI) × 2 3I− → I3− + 2 e− 2 e− + Cu3+ + I− → CuI 3I− → I3− + 2 e− ______________________________ Cu3+(aq) + 4 I−(aq) → CuI(s) + I3−(aq) 2 Cu2+(aq) + 5 I−(aq) → 2 CuI(s) + I3−(aq) 2 S2O32− → S4O62− + 2 e− 2 e− + I3− → 3 I− ____________________________________ 2 S2O32−(aq) + I3−(aq) → 3 I−(aq) + S4O62−(aq) c. Step II data: All Cu is converted to Cu2+. Note: Superconductor abbreviated as "123." 22.57 × 10−3 L × − 0.1000 mol S2 O 32 − 1 mol I 3 2 mol Cu 2 + × × − L 2 mol S2 O 32− mol I 3 = 2.257 × 10−3 mol Cu2+ 1 mol "123" = 7.523 × 10−4 mol "123" 3 mol Cu 0.5402 g Molar mass of YBa2Cu3Ox = = 670.2 g/mol 7.523 × 10 − 4 mol 2.257 × 10−3 mol Cu × 670.2 = 88.91 + 2(137.3) + 3(63.55) + x(16.00), 670.2 = 554.2 + x(16.00) x = 7.250; formula is YBa2Cu3O7.25. Check with Step I data: Both Cu2+ and Cu3+ present. CHAPTER 4 SOLUTION STOICHIOMETRY 37.77 × 10−3 L × 89 − 0.1000 mol S2 O 32 − 1 mol I 3 = 1.889 × 10−3 mol I3− × 2− L 2 mol S2 O 3 We get 1 mol I3− per mol Cu3+ and 1 mol I3− per 2 mol Cu2+. Let nCu 3+ = mol Cu3+ and nCu 2+ = mol Cu2+, then: nCu 3+ + nCu 2+ 2 = 1.889 × 10−3 mol 0.5625 g = 8.393 × 10−4 mol "123" 670.2 g / mol In addition: This amount of "123" contains: 3(8.393 × 10−4) = 2.518 × 10−3 mol Cu total = nCu 3+ + nCu 2+ Solving by simultaneous equations: nCu 3+ + nCu 2+ = 2.518 × 10−3 − nCu 3+ − nCu 2+ 2 = −1.889 × 10−3 ___________________________________________ nCu 2+ 2 = 6.29 × 10−4 nCu 2+ = 1.26 × 10−3 mol Cu2+; nCu 3+ = 2.518 × 10−3 − 1.26 × 10−3 = 1.26 × 10−3 mol Cu3+ This sample of superconductor contains equal moles of Cu2+ and Cu3+. Therefore, 1 mole of YBa2Cu3Ox contains 1.50 mol Cu2+ and 1.50 mol Cu3+. Solving for x using oxidation states: +3 + 2(+2) + 1.50(+2) + 1.50(+3) + x(−2) = 0, 14.50 = 2x, x = 7.25 The two experiments give the same result, x = 7.25 with formula YBa2Cu3O7.25. Average oxidation state of Cu: +3 + 2(+2) + 3(x) + 7.25(-2) = 0, 3x = 7.50, x = +2.50 As determined from Step I data, this superconductor sample contains equal moles of Cu2+ and Cu3+, giving an average oxidation state of +2.50. 2− 96. 0.298 g BaSO4 × 96.07 g SO 4 0.123 g SO 4 = 0.123 g SO42−; % sulfate = 233.4 g BaSO 4 0.205 g Assume we have 100.0 g of the mixture of Na2SO4 and K2SO4. There are: 2− = 60.0% 90 CHAPTER 4 SOLUTION STOICHIOMETRY 1 mol = 0.625 mol SO42− 96.07 g There must be 2 × 0.625 = 1.25 mol of 1+ cations to balance the 2! charge of SO42−. 60.0 g SO42− × Let x = number of moles of K+ and y = number of moles of Na+; then x + y = 1.25. The total mass of Na+ and K+ must be 40.0 g in the assumed 100.0 g of mixture. Setting up an equation: 39.10 g 22.99 g x mol K+ × + y mol Na+ × = 40.0 g mol mol So, we have two equations with two unknowns: x + y = 1.25 and (39.10)x + (22.99)y = 40.0 x = 1.25 ! y, so 39.10(1.25 ! y) + (22.99)y = 40.0 48.9 ! (39.10)y + (22.99)y = 40.0, !(16.11)y = !8.9 y = 0.55 mol Na+ and x = 1.25 ! 0.55 = 0.70 mol K+ Therefore: 0.70 mol K+ × 1 mol K 2 SO 4 2 mol K + = 0.35 mol K2SO4; 0.35 mol K2SO4 × 174.27 g mol = 61 g K2SO4 We assumed 100.0 g; therefore, the mixture is 61% K2SO4 and 39% Na2SO4. 97. There are three unknowns so we need three equations to solve for the unknowns. Let x = mass AgNO3, y = mass CuCl2, and z = mass FeCl3. Then x + y + z = 1.0000 g. The Cl− in CuCl2 and FeCl3 will react with the excess AgNO3 to form the precipitate AgCl(s). Assuming silver has an atomic mass of 107.90: Mass of Cl in mixture = 1.7809 g AgCl × Mass of Cl from CuCl2 = y g CuCl2 × Mass of Cl from FeCl3 = z g FeCl3 × 35.45 g Cl = 0.4404 g Cl 143.35 g AgCl 2(35.45) g Cl = (0.5273)y 134.45 g CuCl 2 3(35.45) g Cl = (0.6557)z 162.20 g FeCl3 The second equation is: 0.4404 g Cl = (0.5273)y + (0.6557)z Similarly, let’s calculate the mass of metals in each salt. Mass of Ag in AgNO3 = x g AgNO3 × 107.9 g Ag = (0.6350)x 169.91 g AgNO3 CHAPTER 4 SOLUTION STOICHIOMETRY 91 For CuCl2 and FeCl3, we already calculated the amount of Cl in each initial amount of salt; the remainder must be the mass of metal in each salt. Mass of Cu in CuCl2 = y − (0.5273)y = (0.4727)y Mass of Fe in FeCl3 = z − (0.6557)z = (0.3443)z The third equation is: 0.4684 g metals = (0.6350)x + (0.4727)y + (0.3443)z We now have three equations with three unknowns. Solving: x + y + z) −0.6350 (1.0000 = 0.4684 = (0.6350)x + (0.4727)y + (0.3443)z _____________________________________________ −0.1666 = −(0.1623)y − (0.2907)z 0.5273 [−0.1666 = −(0.1623)y − (0.2907)z] 0.1623 0.4404 = (0.5273)y + (0.6557)z −0.1009 = −(0.2888)z, ____________________________________________________________ z= 0.1009 = 0.3494 g FeCl3 0.2888 0.4404 = (0.5273)y + 0.6557(0.3494), y = 0.4007 g CuCl2 x = 1.0000 − y − z = 1.0000 − 0.4007 − 0.3494 = 0.2499 g AgNO3 Mass % AgNO3 = Mass % CuCl2 = 98. 0.2499 g × 100 = 24.99% AgNO3 1.0000 g 0.4007 g × 100 = 40.07% CuCl2; mass % FeCl3 = 34.94% 1.0000 g Mol BaSO4 = 0.2327 g × 1 mol = 9.970 × 10 −4 mol BaSO4 233.4 g The moles of the sulfate salt depends on the formula of the salt. The general equation is: Mx(SO4)y(aq) + y Ba2+(aq) → y BaSO4(s) + x Mz+ Depending on the value of y, the mole ratio between the unknown sulfate salt and BaSO4 varies. For example, if Pat thinks the formula is TiSO4, the equation becomes: TiSO4(aq) + Ba2+(aq) → BaSO4(s) + Ti2+(aq) 92 CHAPTER 4 SOLUTION STOICHIOMETRY Because there is a 1 : 1 mole ratio between mol BaSO4 and mol TiSO4, you need 9.970 × 10 −4 mol of TiSO4. Because 0.1472 g of salt was used, the compound would have a molar mass of (assuming the TiSO4 formula): 0.1472 g/9.970 × 10 −4 mol = 147.6 g/mol From atomic masses in the periodic table, the molar mass of TiSO4 is 143.95 g/mol. From just these data, TiSO4 seems reasonable. Chris thinks the salt is sodium sulfate, which would have the formula Na2SO4. The equation is: Na2SO4(aq) + Ba2+(aq) → BaSO4(s) + 2 Na+(aq) As with TiSO4, there is a 1:1 mole ratio between mol BaSO4 and mol Na2SO4. For sodium sulfate to be a reasonable choice, it must have a molar mass of about 147.6 g/mol. Using atomic masses, the molar mass of Na2SO4 is 142.05 g/mol. Thus Na2SO4 is also reasonable. Randy, who chose gallium, deduces that gallium should have a 3+ charge (because it is in column 3A), and the formula of the sulfate would be Ga2(SO4)3. The equation would be: Ga2(SO4)3(aq) + 3 Ba2+(aq) → 3 BaSO4(s) + 2 Ga3+(aq) The calculated molar mass of Ga2(SO4)3 would be: 0.1472 g Ga 2 (SO 4 ) 3 3 mol BaSO 4 = 442.9 g/mol × −4 mol Ga 2 (SO 4 ) 3 9.970 × 10 mol BaSO 4 Using atomic masses, the molar mass of Ga2(SO4)3 is 427.65 g/mol. Thus Ga2(SO4)3 is also reasonable. Looking in references, sodium sulfate (Na2SO4) exists as a white solid with orthorhombic crystals, whereas gallium sulfate Ga2(SO4)3 is a white powder. Titanium sulfate exists as a green powder, but its formula is Ti2(SO4)3. Because this has the same formula as gallium sulfate, the calculated molar mass should be around 443 g/mol. However, the molar mass of Ti2(SO4)3 is 383.97 g/mol. It is unlikely, then, that the salt is titanium sulfate. To distinguish between Na2SO4 and Ga2(SO4)3, one could dissolve the sulfate salt in water and add NaOH. Ga3+ would form a precipitate with the hydroxide, whereas Na2SO4 would not. References confirm that gallium hydroxide is insoluble in water. Marathon Problems 99. M(CHO2)2(aq) + Na2SO4(aq) → MSO4(s) + 2 NaCHO2(aq) From the balanced molecular equation, the moles of M(CHO2)2 present initially must equal the moles of MSO4(s) formed. Because moles = mass/molar mass and letting AM = the atomic mass of M: CHAPTER 4 SOLUTION STOICHIOMETRY mol MSO4 = 93 mass MSO 4 9.9392 g = molar mass of MSO 4 A M + 96.07 mol M(CHO2)2 = mass M (CHO 2 ) 2 9.7416 g = molar mass of M (CHO 2 ) 2 A M + 90.04 Because mol MSO4 = mol M(CHO2)2: 9.9392 9.7416 = , (9.9392)AM + 894.9 = (9.7416)AM + 935.9 A M + 96.07 A M + 90.04 AM = 41.0 = 207; from the periodic table, the unknown element M is Pb. 0.1976 From the information in the second paragraph, we can determine the concentration of the KMnO4 solution. Using the half-reaction method, the balanced reaction between MnO4- and C2O42- is: 5 C2O42−(aq) + 2 MnO4−(aq) + 16 H+(aq) → 10 CO2(g) + 2 Mn2+(aq) + 8 H2O(l) − 1 mol Na 2 C 2 O 4 1 mol C 2 O 24 − 2 mol MnO 4 Mol MnO4 = 0.9234 g Na2C2O4 × × × 134.00 g 1 mol Na 2 C 2 O 4 5 mol C 2 O 24− − = 2.756 × 10−3 mol MnO4− M KMnO 4 = M MnO − = 4 mol MnO 4 volume − = 2.756 × 10 −3 mol = 0.1486 mol/L 0.01855 L From the third paragraph, the standard KMnO4 solution reacts with formate ions from the filtrate. We must determine the moles of CHO2− ions present in order to determine volume of KMnO4 solution. The moles of CHO2− ions present initially are: − mol Pb(CHO 2 ) 2 2 mol CHO 2 9.7416 g Pb(CHO2)2 × × = 6.556 × 10−2 mol CHO2− 297.2 g 1 mol Pb(CHO 2 ) 2 The moles of CHO2− present in 10.00 mL of diluted solution are: − 0.01000 L × 6.556 × 10 −2 mol CHO 2 = 2.622 × 10−3 mol CHO2− 0.2500 L Using the half-reaction method in basic solution, the balanced reaction between CHO2- and MnO4− is: 3 CHO2−(aq) + 2 MnO4−(aq) + OH−(aq) → 3 CO32−(aq) + 2 MnO2(s) + 2 H2O(l) 94 CHAPTER 4 SOLUTION STOICHIOMETRY Determining the volume of MnO4− solution: −3 − 2.622 × 10 mol CHO2 × 2 mol MnO 4 3 mol CHO 2 − − × 1L 0.1486 mol MnO 4 − = 1.176 × 10−2 L = 11.76 mL The titration requires 11.76 mL of the standard KMnO4 solution. 100. a. Compound A = M(NO3)x; in 100.00 g of compd.: 8.246 g N × 48.00 g O = 28.25 g O 14.01 g N Thus the mass of nitrate in the compound = 8.246 + 28.25 g = 36.50 g (if x = 1). If x = 1: mass of M = 100.00 ! 36.50 g = 63.50 g Mol M = mol N = 8.246 g = 0.5886 mol 14.01 g / mol Molar mass of metal M = 63.50 g = 107.9 g/mol (This is silver, Ag.) 0.5886 mol If x = 2: mass of M = 100.00 ! 2(36.50) = 27.00 g 0.5886 mol = 0.2943 mol 2 27.00 g = 91.74 g/mol Molar mass of metal M = 0.2943 mol Mol M = ½ mol N = This is close to Zr, but Zr does not form stable 2+ ions in solution; it forms stable 4+ ions. Because we cannot have x = 3 or more nitrates (three nitrates would have a mass greater than 100.00 g), compound A must be AgNO3. Compound B: K2CrOx is the formula. This salt is composed of K+ and CrOx2− ions. Using oxidation states, 6 + x(!2) = !2, x = 4. Compound B is K2CrO4 (potassium chromate). b. The reaction is: 2 AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2 KNO3(aq) The blood red precipitate is Ag2CrO4(s). c. 331.8 g Ag2CrO4 formed; this is equal to the molar mass of Ag2CrO4, so 1 mole of precipitate formed. From the balanced reaction, we need 2 mol AgNO3 to react with 1 mol K2CrO4 to produce 1 mol (331.8 g) of Ag2CrO4. 2.000 mol AgNO3 × 169.9 g = 339.8 g AgNO3 mol 1.000 mol K2CrO4 × 194.2 g = 194.2 g K2CrO4 mol CHAPTER 4 SOLUTION STOICHIOMETRY 95 The problem says that we have equal masses of reactants. Our two choices are 339.8 g AgNO3 + 339.8 g K2CrO4 or 194.2 g AgNO3 + 194.2 g K2CrO4. If we assume the 194.2 g quantities are correct, then when 194.2 g K2CrO4 (1 mol) reacts, 339.8 g AgNO3 (2.0 mol) must be present to react with all the K2CrO4. We only have 194.2 g AgNO3 present; this cannot be correct. Instead of K2CrO4 limiting, AgNO3 must be limiting, and we have reacted 339.8 g AgNO3 and 339.8 g K2CrO4. Solution A: 2.000 mol NO 3 2.000 mol Ag + = 4.000 M Ag+ ; 0.5000 L 0.5000 L Solution B: 339.8 g K2CrO4 × − = 4.000 M NO3− 1 mol = 1.750 mol K2CrO4 194.2 g 1.750 mol CrO 4 2 × 1.750 mol K + = 7.000 M K+; 0.5000 L 0.5000 L 2− = 3.500 M CrO42− d. After the reaction, moles of K+ and moles of NO3− remain unchanged because they are spectator ions. Because Ag+ is limiting, its concentration will be 0 M after precipitation is complete. 2 Ag+(aq) Initial Change After rxn 2.000 mol !2.000 mol 0 + CrO42−(aq) → Ag2CrO4(s) 1.750 mol !1.000 mol 0.750 mol 0 +1.000 mol 1.000 mol M K+ = 2 × 1.750 mol 2.000 mol = 3.500 M K+; M NO − = = 2.000 M NO3− 3 1.0000 L 1.0000 L M CrO = 4 2− 0.750 mol = 0.750 M CrO42−; M Ag + = 0 M (the limiting reagent) 1.0000 L CHAPTER 5 GASES Pressure 21. 4.75 cm × 10 mm 1 atm = 47.5 mm Hg or 47.5 torr; 47.5 torr × = 6.25 × 10−2 atm cm 760 torr 6.25 × 10−2 atm × 22. 1.013 × 105 Pa = 6.33 × 103 Pa atm If the levels of mercury in each arm of the manometer are equal, then the pressure in the flask is equal to atmospheric pressure. When they are unequal, the difference in height in millimeters will be equal to the difference in pressure in millimeters of mercury between the flask and the atmosphere. Which level is higher will tell us whether the pressure in the flask is less than or greater than atmospheric. a. Pflask < Patm; Pflask = 760. − 118 = 642 torr 642 torr × 1 atm = 0.845 atm 760 torr 0.845 atm × 1.013 × 105 Pa = 8.56 × 104 Pa atm b. Pflask > Patm; Pflask = 760. torr + 215 torr = 975 torr c. 975 torr × 1 atm = 1.28 atm 760 torr 1.28 atm × 1.013 × 105 Pa = 1.30 × 105 Pa atm Pflask = 635 − 118 = 517 torr; Pflask = 635 + 215 = 850. torr 96 CHAPTER 5 23. GASES 97 Suppose we have a column of mercury 1.00 cm × 1.00 cm × 76.0 cm = V = 76.0 cm3: mass = 76.0 cm3 × 13.59 g/cm3 = 1.03 × 103 g × 1 kg = 1.03 kg 1000 g F = mg = 1.03 kg × 9.81 m/s2 = 10.1 kg m/s2 = 10.1 N 2 Force 10.1 N ⎛ 100 cm ⎞ 5 N × ⎜ = or 1.01 × 105 Pa ⎟ = 1.01 × 10 2 2 Area m cm m ⎝ ⎠ (Note: 76.0 cm Hg = 1 atm = 1.01 × 105 Pa.) To exert the same pressure, a column of water will have to contain the same mass as the 76.0cm column of mercury. Thus the column of water will have to be 13.59 times taller or 76.0 cm × 13.59 = 1.03 × 103 cm = 10.3 m. 24. a. The pressure is proportional to the mass of the fluid. The mass is proportional to the volume of the column of fluid (or to the height of the column, assuming the area of the column of fluid is constant). d = density = V= mass ; the volume of silicon oil is the same as the volume of mercury in volume Exercise 22. m Hg m d m m = oil , moil = Hg oil ; VHg = Voil; d d Hg d oil d Hg Because P is proportional to the mass of liquid: ⎛d Poil = PHg ⎜ oil ⎜ d Hg ⎝ ⎞ ⎟ = PHg ⎛⎜ 1.30 ⎞⎟ = (0.0956)PHg ⎟ ⎝ 13.6 ⎠ ⎠ This conversion applies only to the column of silicon oil. a. Pflask = 760. torr − (118 × 0.0956) torr = 760. − 11.3 = 749 torr 749 torr × 1.013 × 105 Pa 1 atm = 0.986 atm; 0.986 atm × = 9.99 × 104 Pa 760 torr atm b. Pflask = 760. torr + (215 × 0.0956) torr = 760. + 20.6 = 781 torr 781 torr × 1.013 × 105 Pa 1 atm = 1.03 atm; 1.03 atm × = 1.04 × 105 Pa atm 760 torr 98 CHAPTER 5 GASES b. If we are measuring the same pressure, the height of the silicon oil column would be 13.6/1.30 = 10.5 times the height of a mercury column. The advantage of using a less dense fluid than mercury is in measuring small pressures. The height difference measured will be larger for the less dense fluid. Thus the measurement will be more precise. 25. a. 4.8 atm × 760 mm Hg 1 torr = 3.6 × 103 mm Hg; b. 3.6 × 103 mm Hg × atm mm Hg = 3.6 × 103 torr c. 4.8 atm × 1.013 × 105 Pa 14.7 psi = 4.9 × 105 Pa; d. 4.8 atm × = 71 psi atm atm Gas Laws 26. a. PV = nRT PV = constant b. PV = nRT c. PV = nRT ⎛ nR ⎞ P= ⎜ ⎟ × T = const × T ⎝ V ⎠ P PV T V d. PV = nRT e. P = nR constant = V V P = constant × P V V T PV = constant P ⎛ P ⎞ T= ⎜ ⎟ × V = const × V ⎝ nR ⎠ f. 1 V PV = nRT PV = nR = constant T PV T 1/V P CHAPTER 5 GASES 99 Note: The equation for a straight line is y = mx + b where y is the y axis and x is the x axis. Any equation that has this form will produce a straight line with slope equal to m and a y intercept equal to b. Plots b, c, and e have this straight-line form. 27. Treat each gas separately, and use the relationship P1V1 = P2V2 (n and T are constant). For H2: P2 = P1V1 2.00 L = 475 torr × = 317 torr V2 3.00 L For N2: P2 = 0.200 atm × 760 torr 1.00 L = 0.0667 atm; 0.0667 atm × = 50.7 torr atm 3.00 L Ptotal = PH 2 + PN 2 = 317 + 50.7 = 368 torr 28. For H2: P2 = P1V1 2.00 L = 360. torr × = 240. torr V2 3.00 L Ptotal = PH 2 + PN 2 , PN 2 = Ptotal − PH 2 = 320. torr − 240. torr = 80. torr For N2: P1 = 29. PV = nRT, P2 V2 3.00 L = 80. torr × = 240 torr V1 1.00 L nT V nT n T = = constant, 1 1 = 2 2 ; moles × molar mass = mass P R P1 P2 n1 (molar mass)T1 n ( molar mass)T2 mass1 × T1 mass 2 × T2 , = 2 = P1 P2 P1 P2 mass2 = 30. mass1 × T1P2 1.00 × 103 g × 291 K × 650. psi = = 309 g T2 P1 299 K × 2050. psi PV = nRT, n is constant. V2 = 1.00 L × 31. P = PCO 2 PV PV PV VPT = nR = constant, 1 1 = 2 2 , V2 = 1 1 2 T T1 T2 P2 T1 760. torr (273 − 31) K = 2.82 L; ΔV = 2.82 − 1.00 = 1.82 L × 220. torr (273 + 23) K ⎛ 1 mol ⎞ 0.08206 L atm ⎜⎜ 22.0 g × ⎟× × 300. K n CO 2 RT 44.01 g ⎟⎠ K mol ⎝ = = = 3.08 atm V 4.00 L 100 CHAPTER 5 GASES With air present, the partial pressure of CO2 will still be 3.08 atm. The total pressure will be the sum of the partial pressures. ⎛ 1 atm ⎞ ⎟ = 3.08 + 0.974 = 4.05 atm Ptotal = PCO 2 + Pair = 3.08 atm + ⎜⎜ 740. torr × 760 torr ⎟⎠ ⎝ 32. PV PV PV = nR = constant, 1 1 = 2 2 T T1 T2 P1V1T2 5.0 × 10 2 mL (273 + 820) K = 710 torr × × = 5.1 × 104 torr V2 T1 25 mL (273 + 30.) K P2 = 33. 34. n= PV 135 atm × 200.0 L = = 1.11 × 103 mol 0.08206 L atm RT × (273 + 24) K K mol For He: 1.11 × 103 mol × 4.003 g He = 4.44 × 103 g He mol For H2: 1.11 × 103 mol × 2.016 g H 2 = 2.24 × 103 g H2 mol Because the container is flexible, P is assumed constant. The moles of gas present are also constant. V P1V1 PV V = 2 2 , 1 = 2 ; Vsphere = 4/3 πr3 T2 n1T1 n 2 T2 T1 35. V2 = V1T2 4/3 π (1.00 cm) 3 × 361 K , 4/3 π (r2 ) 3 = T1 280. K r23 = 361 K = 1.29, r2 = (1.29)1/3 = 1.09 cm = radius of sphere after heating 280. K PV = R; for a gas at two conditions: nT P1V1 PV P P = 2 2 ; because n and V are constant: 1 = 2 n1T1 n 2 T2 T1 T2 T2 = 36. P2 T1 2500 torr × 294.2 K = = 970 K = 7.0 × 102EC P1 758 torr For a gas at two conditions: P1V1 PV = 2 2 n1T1 n 2 T2 CHAPTER 5 GASES Because V is constant: n2 = 101 P1 P nPT = 2 , n2 = 1 2 1 n1T1 n 2 T2 P1T2 1.50 mol × 800. torr × 298 K = 2.77 mol 400. torr × 323 K Moles of gas added = n2 – n1 = 2.77 – 1.50 = 1.27 mol For two-condition problems, units for P and V just need to be the same units for both conditions, not necessarily atm and L. The unit conversions from other P or V units would cancel when applied to both conditions. However, temperature always must be converted to the Kelvin scale. The temperature conversions between other units and Kelvin will not cancel each other. 37. As NO2 is converted completely into N2O4, the moles of gas present will decrease by a factor of one-half (from the 2 : 1 mol ratio in the balanced equation). Using Avogadro’s law: 1 n V1 V = 2 , V2 = V1 × 2 = 25.0 mL × = 12.5 mL 2 n1 n1 n2 N2O4(g) will occupy one-half the original volume of NO2(g). 38. PV = nRT, n is constant. V2 = (1.040)V1, P2 = 39. PV PV PV = nR = constant, 1 1 = 2 2 T T1 T2 V1 1.000 = V2 1.040 P1V1T2 1.000 (273 + 58) K = 75 psi × × = 82 psi V2 T1 1.040 (273 + 19) K PV = nRT, P is constant. nT n T nT P = = constant, 1 1 = 2 2 V R V1 V2 n2 TV 294 K 4.20 × 103 m 3 × = 1 2 = = 0.921 n1 T2 V1 335 K 4.00 × 103 m 3 40. We can use the ideal gas law to calculate the partial pressure of each gas or to calculate the total pressure. There will be less math if we calculate the total pressure from the ideal gas law. n O 2 = 1.80 × 102 mg O2 × 1 mol O 2 1g × = 5.63 × 10−3 mol O2 1000 mg 32.00 g O 2 102 CHAPTER 5 nNO = 1.00 L × GASES 1000 cm 3 9.00 × 1018 molecules NO 1 mol NO × × 3 L cm 6.022 × 10 23 molecules NO = 1.49 × 10−2 mol NO ntotal = n N 2 + n O 2 + n NO = 2.00 × 10−2 + 5.63 × 10−3 + 1.49 × 10−2 = 4.05 × 10−2 mol Ptotal = n total RT = V PN 2 = χ N 2 Ptotal = PO 2 = 0.08206 L atm × 273 K K mol = 0.907 atm 1.00 L 4.05 × 10 − 2 mol × 2.00 × 10 −2 mol N 2 × 0.907 atm = 0.448 atm; χ N 2 = mole fraction of N2 4.05 × 10 − 2 mol total 5.63 × 10 −3 1.49 × 10 −2 × 0.907 atm = 0.126 atm; = × 0.907 atm = 0.334 atm P NO 4.05 × 10 − 2 4.05 × 10 − 2 41. At constant T and P, Avogadro’s law applies; that is, equal volumes contain equal moles of molecules. In terms of balanced equations, we can say that mole ratios and volume ratios between the various reactants and products will be equal to each other. Br2 + 3 F2 → 2 X; two moles of X must contain 2 moles of Br and 6 moles of F; X must have the formula BrF3. 42. Ptotal = 1.00 atm = 760. torr = PN 2 + PH 2O = PN 2 + 17.5 torr, PN 2 = 743 torr n N2 = PN 2 × V RT (743 torr × = 1.02 × 10−2 mol N2 × 43. 28.02 g N 2 = 0.286 g N2 mol N 2 PHe + PH 2O = 1.00 atm = 760. torr = PHe + 23.8 torr, PHe = 736 torr nHe = 0.586 g × V= 44. 1 atm 1L ) × (2.50 × 10 2 mL × ) 760 torr 1000 mL 0.08206 L atm × 293 K K mol = 1.02 × 10−2 mol N2 n He RT = PHe 1 mol = 0.146 mol He 4.003 g 0.08206 L atm × 298 K K mol = 3.69 L 1 atm 736 torr × 760 torr 0.146 mol × For O2, n and T are constant, so P1V1 = P2V2. CHAPTER 5 P1 = GASES 103 P2 V2 1.94 L = 785 torr × = 761 torr = PO 2 V1 2.00 L Ptotal = PO 2 + PH 2 O , PH 2 O = 785 ! 761 = 24 torr 45. a. Mole fraction CH4 = χ CH 4 = PCH 4 Ptotal = 0.175 atm = 0.412 0.175 atm + 0.250 atm χ O 2 = 1.000 – 0.412 = 0.588 b. PV = nRT, ntotal = c. χ CH 4 = n CH 4 n total Ptotal × V 0.425 atm × 10.5 L = = 0.161 mol 0.08206 L atm RT × 338 K K mol , n CH 4 = χ CH 4 × n total = 0.412 × 0.161 mol = 6.63 × 10−2 mol CH4 6.63 × 10−2 mol CH4 × 16.04 g CH 4 = 1.06 g CH4 mol CH 4 n O 2 = 0.588 × 0.161 mol = 9.47 × 10−2 mol O2; 9.47 × 10−2 mol O2 × 32.00 g O 2 mol O 2 = 3.03 g O2 46. If we had 100.0 g of the gas, we would have 50.0 g He and 50.0 g Xe. χ He = n He n He + n Xe 50.0 g 12.5 mol He 4.003 g/mol = = = 0.970 50.0 g 50.0 g 12.5 mol He + 0.381 mol Xe + 4.003 g/mol 131.3 g/mol PHe = χHePtotal = 0.970 × 600. torr = 582 torr; PXe = 600. − 582 = 18 torr 47. The partial pressures can be determined by using the mole fractions. Pmethane = Ptotal × χmethane = 1.44 atm × 0.915 = 1.32 atm; Pethane = 1.44 – 1.32 = 0.12 atm Determining the number of moles of natural gas combusted: nnatural gas = PV 1.44 atm × 15.00 L = = 0.898 mol natural gas 0.08206 L atm RT × 293 K K mol nmethane = nnatural gas × χmethane = 0.898 mol × 0.915 = 0.822 mol methane nethane = 0.898 ! 0.822 = 0.076 mol ethane 104 CHAPTER 5 CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l); 0.822 mol CH4 × 0.076 mol C2H6 × GASES 2 C2H6 + 7 O2(g) → 4 CO2(g) + 6 H2O(l) 2 mol H 2 O 18.02 g H 2 O × = 29.6 g H2O 1 mol CH 4 mol H 2 O 6 mol H 2 O 18.02 g H 2 O × = 4.1 g H2O 2 mol C 2 H 6 mol H 2 O The total mass of H2O produced = 29.6 g + 4.1 g = 33.7 g H2O. Gas Density, Molar Mass, and Reaction Stoichiometry 48. Molar mass = dRT where d = density of gas in units of g/L. P 3.164 g/L × Molar mass = 0.08206 L atm × 273.2 K K mol = 70.98 g/mol 1.000 atm The gas is diatomic, so the atomic mass = 70.93/2 = 35.47. This is chlorine, and the identity of the gas is Cl2. 49. Out of 100.0 g of compound, there are: 87.4 g N × 1 mol N 6.24 = 6.24 mol N; = 1.00 6.24 14.01 g N 12.6 g H × 1 mol H 12.5 = 12.5 mol H; = 2.00 1.008 g H 6.24 Empirical formula is NH2. P × (molar mass) = dRT, where d = density. 0.977 g 0.08206 L atm × × 373 K dRT L K mol Molar mass = = = 32.0 g/mol 1 atm P 710. torr × 760 torr Empirical formula mass of NH2 = 16.0 g. Therefore, the molecular formula is N2H4. 50. P × (molar mass) = dRT, d = mass mass , P × (molar mass) = × RT volume V CHAPTER 5 GASES 105 0.08206 L atm × 373 K mass × RT K mol Molar mass = = = 96.9 g/mol 1 atm PV (750. torr × ) × 0.256 L 760 torr 0.800 g × Mass of CHCl ≈ 12.0 + 1.0 + 35.5 = 48.5; 51. 96.9 = 2.00; molecular formula is C2H2Cl2. 48.5 If Be3+, the formula is Be(C5H7O2)3 and molar mass ≈ 13.5 + 15(12) + 21(1) + 6(16) = 311 g/mol. If Be2+, the formula is Be(C5H7O2)2 and molar mass ≈ 9.0 + 10(12) + 14(1) + 4(16) = 207 g/mol. Data set I (molar mass = dRT/P and d = mass/V): 0.08206 L atm 0.2022 g × × 286 K mass × RT K mol = = 209 g/mol molar mass = 1 atm PV −3 (765.2 torr × ) × (22.6 × 10 L) 760 torr Data set II: 0.08206 L atm × 290. K mass × RT K mol molar mass = = = 202 g/mol 1 atm PV (764.6 torr × ) × (26.0 × 10 −3 L) 760 torr 0.2224 g × These results are close to the expected value of 207 g/mol for Be(C5H7O2)2. Thus we conclude from these data that beryllium is a divalent element with an atomic weight (mass) of 9.0 g/mol. 52. We assume that 28.01 g/mol is the true value for the molar mass of N2. The value of 28.15 g/mol is the average molar mass of the amount of N2 and Ar in air. Assume 100.00 mol of total gas present, and let x = the number of moles that are N2 molecules. Then 100.00 − x = the number of moles that are Ar atoms. Solving: 28.15 = x(28.01) + (100.00 − x)(39.95) 100.00 2815 = (28.01)x + 3995 − (39.95)x, (11.94)x = 1180. x = 98.83% N2; % Ar = 100.00 − x = 1.17% Ar Ratio of moles of Ar to moles of N2 = 1.17 = 1.18 × 10−2. 98.83 106 53. CHAPTER 5 GASES 0.70902 g 0.08206 L atm × × 273.2 K dRT L K mol Molar mass = = = 15.90 g/mol P 1.000 atm 15.90 g/mol is the average molar mass of the mixture of methane and helium. Assume 100.00 mol of total gas present, and let x = mol of CH4 in the 100.00 mol mixture. This value of x is also equal to the volume percentage of CH4 in 100.00 L of mixture because T and P are constant. 15.90 = x(16.04) + (100.00 − x)(4.003) , 1590. = (16.04)x + 400.3 − (4.003)x 100.00 1190. = (12.04)x, x = 98.84% CH4 by volume; % He = 100.00 − x = 1.16% He by volume 54. 1.00 × 103 kg Mo × 1000 g 1 mol Mo × = 1.04 × 104 mol Mo kg 95.94 g Mo 1.04 × 104 mol Mo × VO 2 = n O 2 RT P 1 mol MoO 3 7/2 mol O 2 × = 3.64 × 104 mol O2 mol Mo mol MoO 3 0.08206 L atm × 290. K K mol = 8.66 × 105 L of O2 1.00 atm 3.64 × 10 4 mol × = 8.66 × 105 L O2 × 100 L air = 4.1 × 106 L air 21 L O 2 1.04 × 104 mol Mo × 3 mol H 2 = 3.12 × 104 mol H2 mol Mo 3.12 × 10 4 mol × VH 2 = 55. n H2 0.08206 L atm × 290. K K mol = 7.42 × 105 L of H2 1.00 atm 3 ⎡ 1L ⎤ ⎛ 100 cm ⎞ 1.0 atm × ⎢4800 m 3 × ⎜ ⎟ × 3⎥ ⎝ m ⎠ 1000 cm ⎦⎥ ⎢ PV ⎣ = = 0.08206 L atm RT × 273 K K mol = 2.1 × 105 mol 2.1 × 105 mol H2 are in the balloon. This is 80.% of the total amount of H2 that had to be generated: 0.80(total mol H2) = 2.1 × 105, total mol H2 = 2.6 × 105 mol H2 CHAPTER 5 56. GASES 107 2.6 × 105 mol H2 × 1 mol Fe 55.85 g Fe × = 1.5 × 107 g Fe mol H 2 mol Fe 2.6 × 105 mol H2 × 1 mol H 2SO 4 98.09 g H 2SO 4 100 g reagent × × = 2.6 × 107 g of 98% mol H 2 mol H 2SO 4 98 g H 2SO 4 sulfuric acid For ammonia (in 1 minute): n NH 3 = PNH 3 × VNH 3 RT = 90. atm × 500. L = 1.1 × 103 mol NH3 0.08206 L atm × 496 K K mol NH3 flows into the reactor at a rate of 1.1 × 103 mol/min. For CO2 (in 1 minute): n CO 2 = PCO 2 × VCO 2 RT = 45 atm × 600. L = 6.6 × 102 mol CO2 0.08206 L atm × 496 K K mol CO2 flows into the reactor at 6.6 × 102 mol/min. To react completely with 1.1 × 103 mol NH3/min, we need: 1.1 × 103 mol NH 3 1 mol CO 2 × = 5.5 × 102 mol CO2/min min 2 mol NH 3 Because 660 mol CO2/min are present, ammonia is the limiting reagent. 1.1 × 103 mol NH 3 1 mol urea 60.06 g urea × × = 3.3 × 104 g urea/min min 2 mol NH 3 mol urea 57. Because P and T are constant, V and n are directly proportional. The balanced equation requires 2 L of H2 to react with 1 L of CO (2 : 1 volume ratio due to 2 : 1 mole ratio in the balanced equation). The actual volume ratio present in 1 minute is 16.0 L/25.0 L = 0.640 (0.640 : 1). Because the actual volume ratio present is smaller than the required volume ratio, H2 is the limiting reactant. The volume of CH3OH produced at STP will be one-half the volume of H2 reacted due to the 1 : 2 mole ratio in the balanced equation. In 1 minute, 16.0 L/2 = 8.00 L CH3OH are produced (theoretical yield). n CH 3OH = PV 1.00 atm × 8.00 L = 0.357 mol CH3OH in 1 minute = 0.08206 L atm RT × 273 K K mol 108 CHAPTER 5 0.357 mol CH3OH × Percent yield = 58. GASES 32.04 g CH 3OH = 11.4 g CH3OH (theoretical yield per minute) mol CH 3OH 5.30 g actual yield × 100 = × 100 = 46.5% yield theoretical yield 11.4 g CH3OH + 3/2 O2 → CO2 + 2 H2O or 2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(g) 50.0 mL × n O2 = 0.850 g 1 mol × = 1.33 mol CH3OH(l) available mL 32.04 g PV 2.00 atm × 22.8 L = 1.85 mol O2 available = 0.08206 L atm RT × 300. K K mol 1.33 mol CH3OH × 3 mol O 2 = 2.00 mol O2 2 mol CH 3OH 2.00 mol O2 are required to react completely with all the CH3OH available. We only have 1.85 mol O2, so O2 is limiting. 1.85 mol O2 × 59. 4 mol H 2 O = 2.47 mol H2O 3 mol O 2 150 g (CH3)2N2H2 × PN 2 = nRT = V 1 mol (CH 3 ) 2 N 2 H 2 3 mol N 2 × = 7.5 mol N2 produced 60.10 g mol (CH 3 ) 2 N 2 H 2 7.5 mol × 0.08206 L atm × 300. K K mol = 0.74 atm 250 L We could do a similar calculation for PH 2O and PCO 2 and then calculate Ptotal (= PN 2 + PH 2O + PCO 2 ) . Or we can recognize that 9 total moles of gaseous products form for every mole of (CH3)2N2H2 reacted. This is three times the moles of N2 produced. Therefore, Ptotal will be three times larger than PN 2 . Ptotal = 3 × PN 2 = 3 × 0.74 atm = 2.2 atm. 60. 2 NaN3(s) → 2 Na(s) + 3 N2(g) n N2 = PV 1.00 atm × 70.0 L = = 3.12 mol N2 are needed to fill the air bag. 0 . 08206 L atm RT × 273 K K mol Mass NaN3 reacted = 3.12 mol N2 × 2 mol NaN 3 65.02 g NaN 3 × = 135 g NaN3 3 mol N 2 mol NaN 3 CHAPTER 5 61. GASES 109 2 NaClO3(s) → 2 NaCl(s) + 3 O2(g) Ptotal = PO 2 + PH 2O , PO 2 = Ptotal − PH 2O = 734 torr – 19.8 torr = 714 torr n O2 = PO 2 × V RT ⎛ 1 atm ⎞ ⎟ × 0.0572 L ⎜⎜ 714 torr × 760 torr ⎟⎠ ⎝ = = 2.22 × 10−3 mol O2 0.08206 L atm × (273 + 22) K K mol Mass NaClO3 decomposed = 2.22 × 10−3 mol O2 × 2 mol NaClO3 106.44 g NaClO3 × 3 mol O 2 mol NaClO3 = 0.158 g NaClO3 0.158 g × 100 = 18.0% Mass % NaClO3 = 0.8765 g 62. 10.10 atm − 7.62 atm = 2.48 atm is the pressure of the amount of F2 reacted. PV = nRT, V and T are constant. P = constant, n P1 P P n = 2 or 1 = 1 n1 n2 P2 n2 Mol F2 reacted 2.48 atm = = 2.00; so: Xe + 2 F2 → XeF4 Mol Xe reacted 1.24 atm 63. Ptotal = PN 2 + PH 2O , PN 2 = 726 torr – 23.8 torr = 702 torr H n N2 = PN 2 × V RT = 0.924 atm × 31.8 × 10 −3 L = 1.20 × 10−3 mol N2 0.08206 L atm × 298 K K mol Mass of N in compound = 1.20 × 10−3 mol N2 H Mass % N = 64. 1 atm = 0.924 atm 760 torr 28.02 g N 2 = 3.36 × 10−2 g nitrogen mol 3.36 × 10 −2 g × 100 = 13.3% N 0.253 g 0.2766 g CO2 × 7.549 × 10 −2 g 12.011 g C = 7.549 × 10−2 g C; % C = × 100 = 73.79% C 0.1023 g 44.009 g CO 2 0.0991 g H2O × 1.11 × 10 −2 g 2.016 g H = 1.11 × 10−2 g H; % H = × 100 = 10.9% H 0.1023 g 18.02 g H 2 O 110 CHAPTER 5 PV = nRT, n N 2 = PV 1.00 atm × 27.6 × 10 −3 L = = 1.23 × 10−3 mol N2 0.08206 L atm RT × 273 K K mol 1.23 × 10−3 mol N2 × Mass % N = GASES 28.02 g N 2 = 3.45 × 10−2 g nitrogen mol N 2 3.45 × 10 −2 g × 100 = 7.14% N 0.4831 g Mass % O = 100.00 − (73.79 + 10.9 + 7.14) = 8.2% O Out of 100.00 g of compound, there are: 73.79 g C × 10.9 g H × 1 mol 1 mol = 6.144 mol C; 7.14 g N × = 0.510 mol N 12.011 g 14.01 g 1 mol 1 mol = 10.8 mol H; 8.2 g O H = 0.51 mol O 1.008 g 16.00 g Dividing all values by 0.51 gives an empirical formula of C12H21NO. 4.02 g 0.08206 L atm × × 400. K dRT L K mol Molar mass = = = 392 g/mol 1 atm P 256 torr × 760 torr 392 Empirical formula mass of C12H21NO ≈ 195 g/mol and ≈ 2. 195 Thus the molecular formula is C24H42N2O2. 65. For NH3: P2 = For O2: P2 = 2.00 L P1V1 = 0.500 atm × = 0.333 atm V2 3.00 L P1V1 1.00 L = 1.50 atm × = 0.500 atm V2 3.00 L After the stopcock is opened, V and T will be constant, so P % n. The balanced equation requires: n O2 n NH 3 = PO 2 PNH 3 = 5 = 1.25 4 CHAPTER 5 GASES The actual ratio present is: 111 PO 2 PNH 3 = 0.500 atm = 1.50 0.333 atm The actual ratio is larger than the required ratio, so NH3 in the denominator is limiting. Because equal moles of NO will be produced as NH3 reacted, the partial pressure of NO produced is 0.333 atm (the same as PNH 3 reacted). 66. PV = nRT, V and T are constant. P n P1 P = 2 or 1 = 1 P2 n2 n1 n2 When V and T are constant, then pressure is directly proportional to moles of gas present, and pressure ratios are identical to mole ratios. At 25°C: 2 H2(g) + O2(g) → 2 H2O(l), H2O(l) is produced at 25°C. The balanced equation requires 2 mol H2 for every mol O2 reacted. The same ratio (2 : 1) holds true for pressure units. The actual pressure ratio present is 2 atm H2 to 3 atm O2, well below the required 2 : 1 ratio. Therefore, H2 is the limiting reagent. The only gas present at 25°C after the reaction goes to completion will be the excess O2. PO 2 (reacted) = 2.00 atm H2 × 1 atm O 2 = 1.00 atm O2 2 atm H 2 PO 2 (excess) = PO 2 (initial) − PO 2 (reacted) = 3.00 atm - 1.00 atm = 2.00 atm O2 At 125°C: 2 H2(g) + O2(g) → 2 H2O(g), H2O(g) is produced at 125°C. The major difference in the problem is that gaseous water is now a product, which will increase the total pressure. PH 2O (produced) = 2.00 atm H2 × 2 atm H 2 O = 2.00 atm H2O 2 atm H 2 Ptotal = PO 2 (excess) + PH 2O (produced) = 2.00 atm O2 + 2.00 atm H2O = 4.00 atm 67. Rigid container (constant volume): As reactants are converted to products, the moles of gas particles present decrease by one-half. As n decreases, the pressure will decrease (by onehalf). Density is the mass per unit volume. Mass is conserved in a chemical reaction, so the density of the gas will not change because mass and volume do not change. Flexible container (constant pressure): Pressure is constant because the container changes volume in order to keep a constant pressure. As the moles of gas particles decrease by a factor of 2, the volume of the container will decrease (by one-half). We have the same mass of gas in a smaller volume, so the gas density will increase (is doubled). 112 68. CHAPTER 5 GASES Rigid container: As temperature is increased, the gas molecules move with a faster average velocity. This results in more frequent and more forceful collisions, resulting in an increase in pressure. Density = mass/volume; the moles of gas are constant, and the volume of the container is constant, so density must be temperature-independent (density is constant). Flexible container: The flexible container is a constant-pressure container. Therefore, the internal pressure will be unaffected by an increase in temperature. The density of the gas, however, will be affected because the container volume is affected. As T increases, there is an immediate increase in P inside the container. The container expands its volume to reduce the internal pressure back to the external pressure. We have the same mass of gas in a larger volume. Gas density will decrease in the flexible container as T increases. 69. 2 NH3(g) → N2(g) + 3 H2(g); as reactants are converted into products, we go from 2 moles of gaseous reactants to 4 moles of gaseous products (1 mol N2 + 3 mol H2). Because the moles of gas doubles as reactants are converted into products, the volume of the gases will double (at constant P and T). ⎛ RT ⎞ PV = nRT, P = ⎜ ⎟n = (constant)n; pressure is directly related to n at constant T and V. ⎝ V ⎠ As the reaction occurs, the moles of gas will double, so the pressure will double. Because 1 o mol of N2 is produced for every 2 mol of NH3 reacted, PN 2 = 1/2 PNH . Owing to the 3 to 2 3 o mole ratio in the balanced equation, PH 2 = 3/2 PNH 3 . o o o + 1/2 PNH = 2 PNH . As we said earlier, the total presNote: Ptotal = PH 2 + PN 2 = 3/2 PNH 3 3 3 sure will double from the initial pressure of NH3 as reactants are completely converted into products. Kinetic Molecular Theory and Real Gases 70. Boyle's law: P % 1/V at constant n and T In the kinetic molecular theory (KMT), P is proportional to the collision frequency which is proportional to 1/V. As the volume increases, there will be fewer collisions per unit area with the walls of the container, and pressure will decrease (Boyle's law). Charles's law: V % T at constant n and P When a gas is heated to a higher temperature, the velocities of the gas molecules increase and thus hit the walls of the container more often and with more force. In order to keep the pressure constant, the volume of the container must increase (this increases surface area, which decreases the number of collisions per unit area, which decreases the pressure). Therefore, volume and temperature are directly related at constant n and P (Charles’s law). Avogadro’s law: V % n at constant P and T CHAPTER 5 GASES 113 As gas is added to a container (n increases), there will be an immediate increase in the number of gas particle collisions with the walls of the container. This results in an increase in pressure in the container. However, the container is such that it wants to keep the pressure constant. In order to keep pressure constant, the volume of the container increases in order to reduce the collision frequency, which reduces the pressure. V is directly related to n at constant P and T. Dalton’s law of partial pressure: Ptotal = P1 + P2 + P3 + … The KMT assumes that gas particles are volumeless and that they exert no interparticle forces on each other. Gas molecules all behave the same way. Therefore, a mixture of gases behaves as one big gas sample. You can concentrate on the partial pressures of the individual components of the mixture, or you can collectively group all the gases together to determine the total pressure. One mole of an ideal gas behaves the same whether it is a pure gas or a mixture of gases. P versus n relationship at constant V and T. This is a direct relationship. As gas is added to a container, there will be an increase in the collision frequency, resulting in an increase in pressure. P and n are directly related at constant V and T. P versus T relationship at constant V and n. This is a direct relationship. As the temperature of the gas sample increases, the gas molecules move with a faster average velocity. This increases the gas collision frequency as well as increases the force of each gas particle collision. Both these result in an increase in pressure. Pressure and temperature are directly related at constant V and n. 71. V, T, and P are all constant, so n must be constant. Because we have equal moles of gas in each container, gas B molecules must be heavier than gas A molecules. a. Both gas samples have the same number of molecules present (n is constant). b. Because T is constant, KEave must be the same for both gases (KEave = 3/2 RT). c. The lighter gas A molecules will have the faster average velocity. d. The heavier gas B molecules do collide more forcefully, but gas A molecules, with the faster average velocity, collide more frequently. The end result is that P is constant between the two containers. 72. a. Containers ii, iv, vi, and viii have volumes twice those of containers i, iii, v, and vii. Containers iii, iv, vii, and viii have twice the number of molecules present than containers i, ii, v, and vi. The container with the lowest pressure will be the one which has the fewest moles of gas present in the largest volume (containers ii and vi both have the lowest P). The smallest container with the most moles of gas present will have the highest pressure (containers iii and vii both have the highest P). All the other containers (i, iv, v and viii) will have the same pressure between the two extremes. The order is ii = vi < i = iv = v = viii < iii = vii. b. All have the same average kinetic energy because the temperature is the same in each container. Only temperature determines the average kinetic energy. 114 CHAPTER 5 GASES c. The least dense gas will be in container ii because it has the fewest of the lighter Ne atoms present in the largest volume. Container vii has the most dense gas because the largest number of the heavier Ar atoms are present in the smallest volume. To figure out the ordering for the other containers, we will calculate the relative density of each. In the table below, m1 equals the mass of Ne in container i, V1 equals the volume of container i, and d1 equals the density of the gas in container i. Container mass, volume density ⎛ ⎞ ⎜ ⎟ ⎝ volume ⎠ mass i m1, V1 m1 V1 ii iii m1, 2V1 2m1, V1 1 m1 = d1 2 2 V1 = d1 2 m1 V1 = 2d1 iv 2m1, 2V1 2 m1 2 V1 v vi 2m1, V1 = d1 2 m1 V1 = 2d1 2m1, 2V1 2 m1 2 V1 = d1 vii 4m1, V1 4 m1 V1 = 4d1 viii 4m1, 2V1 4 m1 2 V1 = 2d1 From the table, the order of gas density is ii < i = iv = vi < iii = v = viii < vii. d. µrms = (3RT/M)1/2; the root mean square velocity only depends on the temperature and the molar mass. Because T is constant, the heavier argon molecules will have a slower root mean square velocity than the neon molecules. The order is v = vi = vii = viii < i = ii = iii = iv. 73. (KE)avg = 3/2 RT; KE depends only on temperature. At each temperature CH4 and N2 will have the same average KE. For energy units of joules (J), use R = 8.3145 J K−1 mol−1. To determine average KE per molecule, divide by Avogadro’s number, 6.022 × 1023 molecules/mol. At 273 K: (KE)avg = 3 8.3145 J × × 273 K = 3.40 × 103 J/mol = 5.65 × 10−21 J/molecule 2 K mol At 546 K: (KE)avg = 3 8.3145 J × × 546 K = 6.81 × 103 J/mol = 1.13 × 10−20 J/molecule 2 K mol 1/ 2 74. ⎛ 3RT ⎞ urms = ⎜ ⎟ ⎝ M ⎠ , where R = 8.3145 J and M = molar mass in kg K mol For CH4: M = 1.604 × 10−2 kg and for N2, M = 2.802 × 10−2 kg. 8.3145 J ⎛ × 273 K ⎜3 × K mol ⎜ For CH4 at 273 K: urms = ⎜ 1.604 × 10 − 2 kg/mol ⎜ ⎝ 1/ 2 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = 652 m/s CHAPTER 5 GASES 115 At 546 K: urms for CH4 is 921 m/s. For N2: urms= 493 m/s at 273 K and 697 m/s at 546 K. 75. No; the numbers calculated in Exercise 73 are the average kinetic energies at the various temperatures. At each temperature, there is a distribution of energies. Similarly, the numbers calculated in Exercise 74 are a special kind of average velocity. There is a distribution of velocities as shown in Figs. 5.15 to 5.17 of the text. Note that the major reason there is a distribution of kinetic energies is because there is a distribution of velocities for any gas sample at some temperature. 76. a. All the gases have the same average kinetic energy because they are all at the same temperature [KEave = (3/2)RT]. b. At constant T, the lighter the gas molecule, the faster is the average velocity [uave % (1/M)1/2]. Xe (131.3 g/mol) < Cl2 (70.90 g/mol) < O2 (32.00 g/mol) < H2 (2.016 g/mol) slowest fastest c. At constant T, the lighter H2 molecules have a faster average velocity than the heavier O2 molecules. As temperature increases, the average velocity of the gas molecules increases. Separate samples of H2 and O2 can only have the same average velocities if the temperature of the O2 sample is greater than the temperature of the H2 sample. 77. a. They will all have the same average kinetic energy because they are all at the same temperature. Average kinetic energy depends only on temperature. b. Flask C; at constant T, urms % (1/M)1/2. In general, the lighter the gas molecules, the greater is the root mean square velocity (at constant T). c. Flask A: collision frequency is proportional to average velocity × n/V (as the average velocity doubles, the number of collisions will double, and as the number of molecules in the container doubles, the number of collisions again doubles). At constant T and V, n is proportional to P, and average velocity is proportional to (1/M)1/2. We use these relationships and the data in the exercise to determine the following relative values. A B C n (relative) uavg (relative) 1.0 0.33 0.13 1.0 1.0 3.7 Coll. Freq. (relative) = n × uavg 1.0 0.33 0.48 116 CHAPTER 5 78. a b c d Avg. KE (KEavg % T) inc dec same same u rms (u 2rms ∝ T) inc dec same same Coll. freq. gas inc dec inc inc Coll. freq. wall inc dec inc inc Impact E (impact E % KE % T) inc dec same same GASES Both collision frequencies are proportional to the root mean square velocity (as velocity increases, it takes less time to move to the next collision) and the quantity n/V (as molecules per volume increases, collision frequency increases). 1/ 2 79. Graham’s law of effusion: ⎛M ⎞ Rate1 = ⎜⎜ 2 ⎟⎟ Rate 2 ⎝ M1 ⎠ Let Freon-12 = gas 1 and Freon-11 = gas 2: 1/ 2 1.07 ⎛ 137.4 ⎞ ⎟ =⎜ 1.00 ⎜⎝ M1 ⎟⎠ , 1.14 = 137.4 , M1 = 121 g/mol M1 The molar mass of CF2Cl2 is equal to 121 g/mol, so Freon-12 is CF2Cl2. 1/ 2 80. ⎛M ⎞ Rate1 = ⎜⎜ 2 ⎟⎟ Rate 2 ⎝ M1 ⎠ 1/ 2 12 , C17 O ⎛ 30.0 ⎞ =⎜ ⎟ 12 18 C O ⎝ 29.0 ⎠ 12 = 1.02; 1/ 2 C16 O ⎛ 30.0 ⎞ =⎜ ⎟ 12 18 C O ⎝ 28.0 ⎠ = 1.04 The relative rates of effusion of 12C16O : 12C17O : 12C18O are 1.04 : 1.02 : 1.00. Advantage: CO2 isn't as toxic as CO. Major disadvantages of using CO2 instead of CO: 1. Can get a mixture of oxygen isotopes in CO2. 2. Some species, for example, 12C16O18O and 12C17O2, would effuse (gaseously diffuse) at about the same rate because the masses are about equal. Thus some species cannot be separated from each other. 1/ 2 81. ⎛M ⎞ Rate1 = ⎜⎜ 2 ⎟⎟ Rate 2 ⎝ M1 ⎠ ; rate1 = 16.04 g 24.0 mL 47.8 mL ; M2 = ; rate 2 = ; M1 = ? min min mol 1/ 2 24.0 ⎛ 16.04 ⎞ ⎟ =⎜ 47.8 ⎜⎝ M1 ⎟⎠ = 0.502, 16.04 = (0.502)2 H M1, M1 = 16.04 63.7 g = 0.252 mol CHAPTER 5 GASES 117 1/ 2 82. ⎛M ⎞ Rate1 = ⎜⎜ 2 ⎟⎟ Rate 2 ⎝ M1 ⎠ where M = molar mass; let gas (1) = He, gas (2) = Cl2. 1.0 L 1/ 2 4.5 min ⎛ 70.90 ⎞ =⎜ ⎟ , 1.0 L ⎝ 4.003 ⎠ t 83. t = 4.209, t = 19 min 4.5 min a. PV = nRT P= b. nRT = V 0.5000 mol × 0.08206 L atm × (25.0 + 273.2) K K mol = 12.24 atm 1.0000 L 2 ⎡ ⎛n⎞ ⎤ 2 2 ⎢P + a⎜ ⎟ ⎥ (V − nb) = nRT; for N2: a = 1.39 atm L /mol and b = 0.0391 L/mol V ⎝ ⎠ ⎥⎦ ⎢⎣ 2 ⎡ ⎤ ⎛ 0.5000 ⎞ ⎢P + 1.39⎜ ⎟ atm ⎥ (1.0000 L − 0.5000 × 0.0391 L) = 12.24 L atm ⎝ 1.0000 ⎠ ⎥⎦ ⎣⎢ (P + 0.348 atm)(0.9805 L) = 12.24 L atm P= 12.24 L atm − 0.348 atm = 12.48 − 0.348 = 12.13 atm 0.9805 L c. The ideal gas law is high by 0.11 atm, or 84. 0.11 × 100 = 0.91%. 12.13 a. PV = nRT P= b. nRT = V 0.5000 mol × 0.08206 L atm × 298.2 K K mol = 1.224 atm 10.000 L 2 ⎡ ⎛n⎞ ⎤ 2 2 ⎢P + a⎜ ⎟ ⎥ (V – nb) = nRT; for N2: a = 1.39 atm L /mol and b = 0.0391 L/mol V ⎝ ⎠ ⎣⎢ ⎦⎥ 2 ⎡ ⎤ ⎛ 0.5000 ⎞ ⎢P + 1.39⎜ ⎟ atm ⎥ (10.000 L − 0.5000 × 0.0391 L) = 12.24 L atm ⎝ 10.000 ⎠ ⎥⎦ ⎣⎢ (P + 0.00348 atm)(10.000 L − 0.0196 L) = 12.24 L atm 118 CHAPTER 5 P + 0.00348 atm = GASES 12.24 L atm = 1.226 atm, P = 1.226 − 0.00348 = 1.223 atm 9.980 L c. The results agree to ±0.001 atm (0.08%). d. In Exercise 83 the pressure is relatively high and there is significant disagreement. In Exercise 84 the pressure is around 1 atm and both gas laws show better agreement. The ideal gas law is valid at relatively low pressures. 85. The kinetic molecular theory assumes that gas particles do not exert forces on each other and that gas particles are volumeless. Real gas particles do exert attractive forces for each other, and real gas particles do have volumes. A gas behaves most ideally at low pressures and high temperatures. The effect of attractive forces is minimized at high temperatures because the gas particles are moving very rapidly. At low pressure, the container volume is relatively large (P and V are inversely related), so the volume of the container taken up by the gas particles is negligible. 86. a. At constant temperature, the average kinetic energy of the He gas sample will equal the average kinetic energy of the Cl2 gas sample. In order for the average kinetic energies to be the same, the smaller He atoms must move at a faster average velocity than Cl2 molecules. Therefore, plot A, with the slower average velocity, would be for the Cl2 sample, and plot B would be for the He sample. Note the average velocity in each plot is a little past the top of the peak. b. As temperature increases, the average velocity of a gas will increase. Plot A would be for O2(g) at 273 K, and plot B, with the faster average velocity, would be for O2(g) at 1273 K. Because a gas behaves more ideally at higher temperatures, O2(g) at 1273 K would behave most ideally. 87. The pressure measured for real gases is too low compared to ideal gases. This is due to the attractions gas particles do have for each other; these attractions “hold” them back from hitting the container walls as forcefully. To make up for this slight decrease in pressure for real gases, a factor is added to the measured pressure. The measured volume is too large. A fraction of the space of the container volume is taken up by the volume of the molecules themselves. Therefore, the actual volume available to real gas molecules is slightly less than the container volume. A term is subtracted from the container volume to correct for the volume taken up by real gas molecules. 88. The van der Waals constant b is a measure of the size of the molecule. Thus C3H8 should have the largest value of b because it has the largest molar mass (size). 89. The values of a are: H2, 0.244 atm L2 mol 2 ; CO2, 3.59; N2, 1.39; CH4, 2.25 Because a is a measure of intermolecular attractions, the attractions are greatest for CO2. CHAPTER 5 90. GASES ⎛ m ⎞ ⎟⎟ f(u) = 4π ⎜⎜ ⎝ 2πk BT ⎠ 119 3/ 2 u 2 e ( − mu 2 / 2 k BT ) 2 As u → 0, e ( − mu / 2 k BT ) → e0 = 1; at small values of u, the u2 term causes the function to increase. At large values of u, the exponent term, -mu2/2kBT, is a large negative number, and e raised to a large negative number causes the function to decrease. As u → ∞, e-∞ → 0. 1/ 2 91. ⎛ 3 RT ⎞ urms = ⎜ ⎟ ⎝ M ⎠ ⎛ 2 RT ⎞ ump = ⎜ ⎟ ⎝ M ⎠ ⎡ ⎛ 8.3145 kg m 2 ⎞ ⎤ ⎟ (500. K ) ⎥ ⎢ 2⎜⎜ 2 ⎟ s K mol ⎠ ⎥ = ⎢⎢ ⎝ 28.02 × 10 −3 kg/mol ⎥ ⎢ ⎥ ⎢⎣ ⎥⎦ ⎛ 8 RT ⎞ uavg = ⎜ ⎟ ⎝ πM ⎠ ⎡ ⎛ 8.3145 kg m 2 ⎞ ⎤ ⎟ ( 500 . K ) ⎢ 8⎜⎜ ⎥ s 2 K mol ⎟⎠ ⎝ ⎢ ⎥ =⎢ π (28.02 × 10 −3 kg/mol) ⎥ ⎢ ⎥ ⎣⎢ ⎦⎥ 1/ 2 1/ 2 92. ⎡ ⎛ 8.3145 kg m 2 ⎞ ⎤ ⎟ (227 + 273) K ⎥ ⎢ 3⎜⎜ 2 ⎟ s K mol ⎠ ⎥ = ⎢⎢ ⎝ ⎥ 28.02 × 10 −3 kg/mol ⎢ ⎥ ⎣⎢ ⎦⎥ 1/ 2 = 667 m/s 1/ 2 = 545 m/s 1/ 2 = 615 m/s KEave = 3/2 RT per mol; KEave = 3/2 kBT per molecule KEtotal = 3/2 × (1.3807 × 10−23 J/K) × 300. K × (1.00 × 1020 molecules) = 0.621 J 93. The force per impact is proportional to Δ(mu) = 2mu. Because m % M, the molar mass, and u % (1/M)1/2 at constant T, the force per impact at constant T is proportional to M × (1/M)1/2 = M. Impact force (H 2 ) = Impact force (He) 94. Diffusion rate Diffusion rate 235 235 238 M H2 M He = 2.016 = 0.7097 4.003 UF6 = 1.0043 (See Section 5.7 of the text.) UF6 UF6 1526 × (1.0043)100 = , 238 UF6 1.000 × 105 − 1526 235 238 235 238 UF6 1526 × 1.5358 = 98500 UF6 UF6 = 1.01 × 10−2 = initial 235U to 238U atom ratio UF6 120 95. CHAPTER 5 GASES Δ(mu) = 2mu = change in momentum per impact. Because m is proportional to M, the molar mass, and u is proportional to (T/M)1/2: Δ(mu ) O 2 ⎛ T ∝ 2M O 2 ⎜ ⎜ MO 2 ⎝ 1/ 2 ⎞ ⎟ ⎟ ⎠ 1/ 2 ⎛ T ⎞ ⎟⎟ ∝ 2M He ⎜⎜ ⎝ M He ⎠ and Δ (mu ) He 1/ 2 Δ(mu ) O 2 Δ(mu ) He ⎛ T ⎞ ⎟ 2M O 2 ⎜ ⎜ MO ⎟ M O2 2 ⎠ ⎝ = = 1/ 2 M He ⎛ T ⎞ ⎟⎟ 2M He ⎜⎜ ⎝ M He ⎠ ⎛ M He ⎜ ⎜ MO 2 ⎝ 1/ 2 ⎞ ⎟ ⎟ ⎠ = 1/ 2 31.998 ⎛ 4.003 ⎞ ⎜ ⎟ 4.003 ⎝ 31.998 ⎠ = 2.827 The change in momentum per impact is 2.827 times larger for O2 molecules than for He atoms. 1/ 2 ZA = A ZO2 Z He N ⎛ RT ⎞ ⎜ ⎟ V ⎝ 2πM ⎠ = collision rate ⎛ N ⎞⎛ RT A⎜ ⎟⎜ ⎝ V ⎠⎜⎝ 2 πM O 2 1/ 2 1/ 2 ⎞ ⎛ 1 ⎞ ⎟ ⎜ ⎟ ⎟ ⎜ MO ⎟ Z 2 ⎠ ⎠ = = ⎝ = 0.3537; He = 2.827 1/ 2 1/ 2 ZO2 ⎛ 1 ⎞ ⎛ N ⎞⎛ RT ⎞⎟ ⎜⎜ ⎟⎟ A⎜ ⎟⎜⎜ ⎟ ⎝ V ⎠⎝ 2 πM He ⎠ ⎝ M He ⎠ There are 2.827 times as many impacts per second for He as for O2. 1/ 2 96. Δ(mu) = 2mu and u % (T/M)1/2; Δ (mu ) 77 Δ (mu ) 27 ⎛ 350. K ⎞ 2 m⎜ ⎟ 1/ 2 M ⎠ ⎛ 350. ⎞ ⎝ = =⎜ ⎟ = 1.08 1/ 2 ⎝ 300. ⎠ ⎛ 300. K ⎞ 2 m⎜ ⎟ ⎝ M ⎠ The change in momentum is 1.08 times greater for Ar at 77°C than for Ar at 27°C. 1/ 2 1/ 2 N ⎛ RT ⎞ ZA = A ⎜ ⎟ V ⎝ 2 πM ⎠ ; ⎛T ⎞ Z 77 = ⎜⎜ 77 ⎟⎟ Z 27 ⎝ T27 ⎠ 1/ 2 , Z 77 ⎛ 350. ⎞ =⎜ ⎟ Z 27 ⎝ 300. ⎠ = 1.08 There are 1.08 times as many impacts per second for Ar at 77°C as for Ar at 27°C. 1/ 2 97. Intermolecular collision frequency = Z = 4 N 2 ⎛ πRT ⎞ d ⎜ ⎟ V ⎝ M ⎠ n P 3.0 atm = = = 0.12 mol/L 0.08206 L atm V RT × 300. K K mol , where d = diameter of He atom CHAPTER 5 GASES 121 N 0.12 mol 6.022 × 10 23 molecules 1000 L 7.2 × 10 25 molecules × × = = V L mol m3 m3 1/ 2 ⎛ π(8.3145)(300.) ⎞ 7.2 × 10 25 molecules ⎟ Z= 4× × (50. × 10 −12 m) 2 × ⎜⎜ 3 −3 ⎟ m ⎝ 4.00 × 10 ⎠ = 1.0 H 109 collisions/s u avg Mean free path = λ = Z 1/ 2 ; u avg ⎛ 8 RT ⎞ =⎜ ⎟ ⎝ πM ⎠ = 1260 m/s; λ = 1260 m/s = 1.3 × 10−6 m 9 −1 1.0 × 10 s Atmospheric Chemistry 98. χHe = 5.24 × 10 −6 from Table 5.4. PHe = χHe × Ptotal = 5.24 × 10 −6 × 1.0 atm = 5.2 × 10 −6 atm n P 5.2 × 10 −6 atm = = 0.08206 L atm V RT × 298 K K mol = 2.1 × 10 −7 mol He/L 2.1 × 10 −7 mol 1L 6.022 × 10 23 atoms = 1.3 × 1014 atoms He/cm3 × × L mol 1000 cm 3 99. a. If we have 1.0 × 106 L of air, then there are 3.0 × 102 L of CO. PCO = χCOPtotal; χCO = b. nCO = PCO V ; RT nCO VCO 3.0 × 10 2 because V % n; PCO = × 628 torr = 0.19 torr Vtotal 1.0 × 10 6 Assuming 1.0 m3 air, 1 m3 = 1000 L: 0.19 atm × (1.0 × 103 L) 760 = = 1.1 × 10−2 mol CO 0.08206 L atm × 273 K K mol 1.1 × 10−2 mol × 6.02 × 10 23 molecules = 6.6 × 1021 CO molecules in 1.0 m3 of air mol 3 c. 6.6 × 10 21 molecules ⎛ 1 m ⎞ 6.6 × 1015 molecules CO ⎜ ⎟ = × ⎜ 100 cm ⎟ m3 cm 3 ⎝ ⎠ 122 100. CHAPTER 5 GASES N2(g) + O2(g) → 2 NO(g), automobile combustion or formed by lightning 2 NO(g) + O2(g) → 2 NO2(g), reaction with atmospheric O2 2 NO2(g) + H2O(l) → HNO3(aq) + HNO2(aq), reaction with atmospheric H2O S(s) + O2(g) → SO2(g), combustion of coal 2 SO2(g) + O2(g) → 2 SO3(g), reaction with atmospheric O2 H2O(l) + SO3(g) → H2SO4(aq), reaction with atmospheric H2O 2 HNO3(aq) + CaCO3(s) → Ca(NO3)2(aq) + H2O(l) + CO2(g) H2SO4(aq) + CaCO3(s) → CaSO4(aq) + H2O(l) + CO2(g) 101. For benzene: 89.6 × 10-9 g × Vbenzene = 1 mol = 1.15 × 10−9 mol benzene 78.11 g n benzene RT = P Mixing ratio = or ppbv = 0.08206 L atm × 296 K K mol = 2.84 × 10−8 L 1 atm 748 torr × 760 torr 1.15 × 10 −9 mol × 2.84 × 10 −8 L × 106 = 9.47 × 10−3 ppmv 3.00 L vol. of X × 109 2.84 × 10 −8 L = × 109 = 9.47 ppbv total vol. 3.00 L 1.15 × 10 −9 mol benzene 1L 6.022 × 10 23 molecules × × 3.00 L mol 1000 cm 3 = 2.31 × 1011 molecules benzene/cm3 For toluene: 153 × 10−9 g C7H8 × 1 mol = 1.66 × 10−9 mol toluene 92.13 g CHAPTER 5 GASES 123 n toluene RT = P Vtoluene = Mixing ratio = 0.08206 L atm × 296 K K mol 1 atm 748 torr × 760 torr 1.66 × 10 −9 mol × = 4.10 × 10−8 L 4.10 × 10 −8 L × 106 = 1.37 × 10−2 ppmv (or 13.7 ppbv) 3.00 L 1.66 × 10 −9 mol toluene 1L 6.022 × 10 23 molecules × × 3.00 L mol 1000 cm 3 = 3.33 × 1011 molecules toluene/cm3 Additional Exercises 3 102. 14.1 × 102 in Hg C in3 × 2.54 cm 10 mm 1 atm 1L ⎛ 2.54 cm ⎞ × × × ⎜ ⎟ × in 1 cm 760 mm ⎝ in ⎠ 1000 cm 3 = 0.772 atm L Boyle’s law: PV = k, where k = nRT; from Example 5.1 of the text, the k values are around 22 atm L. Because k = nRT, we can assume that Boyle’s data and the Example 5.1 data were taken at different temperatures and/or had different sample sizes (different moles). 103. 0.050 mL × V= 104. nRT = P 1.149 g 1 mol O 2 × = 1.8 × 10−3 mol O2 mL 32.0 g 1.8 × 10 −3 mol × 750. mL juice × 0.08206 L atm × 310. K K mol = 4.6 × 10−2 L = 46 mL 1.0 atm 12 mL C 2 H 5 OH = 90. mL C2H5OH present 100 mL juice 90. mL C2H5OH × 0.79 g C 2 H 5 OH 1 mol C 2 H 5 OH 2 mol CO 2 × × = 1.5 mol CO2 mL C 2 H 5 OH 46.07 C 2 H 5 OH 2 mol C 2 H 5 OH The CO2 will occupy (825 − 750. =) 75 mL not occupied by the liquid (headspace). PCO 2 = n CO 2 RT V 1.5 mol × = 0.08206 L atm × 298 K K mol = 490 atm 75 × 10 −3 L Actually, enough CO2 will dissolve in the wine to lower the pressure of CO2 to a much more reasonable value. 124 105. CHAPTER 5 Mn(s) + x HCl(g) → MnClx(s) + n H2 = x H 2 (g) 2 PV 0.951 atm × 3.22 L = = 0.100 mol H2 0.08206 L atm RT × 373 K K mol Mol Cl in compound = mol HCl = 0.100 mol H2 × x mol Cl = 0.200 mol Cl x mol H 2 2 0.200 mol Cl 0.200 mol Cl Mol Cl = = = 4.00 1 mol Mn Mol Mn 0 . 05000 mol Mn 2.747 g Mn × 54.94 g Mn The formula of compound is MnCl4. 106. a. Volume of hot air: V = 4 3 4 πr = π( 2.50 m) 3 = 65.4 m3 3 3 (Note: Radius = diameter/2 = 5.00/2 = 2.50 m) 3 1L ⎛ 10 dm ⎞ 65.4 m3 × ⎜ = 6.54 × 104 L ⎟ × 3 m dm ⎝ ⎠ ⎛ 1 atm ⎞ ⎜⎜ 745 torr × ⎟⎟ × 6.54 × 10 4 L 760 torr ⎠ PV = ⎝ = 2.31 × 103 mol air n= 0 . 08206 L atm RT × (273 + 65) K K mol Mass of hot air = 2.31 × 103 mol × 29.0 g = 6.70 × 104 g mol 745 atm × 6.54 × 10 4 L PV 760 = = 2.66 × 103 mol air Air displaced: n = 0.08206 L atm RT × ( 273 + 21) K K mol Mass of air displaced = 2.66 × 103 mol × 29.0 g = 7.71 × 104 g mol Lift = 7.71 × 104 g − 6.70 × 104 g = 1.01 × 104 g GASES CHAPTER 5 GASES 125 b. Mass of air displaced is the same, 7.71 × 104 g. Moles of He in balloon will be the same as moles of air displaced, 2.66 × 103 mol, because P, V, and T are the same. Mass of He = 2.66 × 103 mol × 4.003 g = 1.06 × 104 g mol Lift = 7.71 × 104 g − 1.06 × 104 g = 6.65 × 104 g 630. atm × (6.54 × 10 4 L) PV 760 c. Hot air: n = = = 1.95 × 103 mol air 0 . 08206 L atm RT × 338 K K mol 1.95 × 103 mol × 29.0 g = 5.66 × 104 g of hot air mol 630. atm × (6.54 × 10 4 L) PV 760 = = 2.25 × 103 mol air Air displaced: n = 0 . 08206 L atm RT × 294 K K mol 2.25 × 103 mol × 29.0 g = 6.53 × 104 g of air displaced mol Lift = 6.53 × 104 g − 5.66 × 104 g = 8.7 × 103 g d. Mass of hot air = 6.70 × 104 g (from part a) 745 atm × (6.54 × 10 4 L) PV 760 Air displaced: n = = = 2.95 × 103 mol air 0.08206 L atm RT × 265 K K mol 29.0 g = 8.56 × 104 g of air displaced 2.95 × 103 mol × mol Lift = 8.56 × 104 g - 6.70 × 104 g = 1.86 × 104 g 107. PV = nRT, V and T are constant. P1 P P n = 2, 2 = 2 n1 n 2 P1 n1 We will do this limiting-reagent problem using an alternative method than described in Chapter 3. Let's calculate the partial pressure of C3H3N that can be produced from each of the starting materials assuming each reactant is limiting. The reactant that produces the smallest amount of product will run out first and is the limiting reagent. 126 CHAPTER 5 PC3H 3 N = 0.500 MPa × 2 MPa C 3 H 3 N = 0.500 MPa if C3H6 is limiting 2 MPa C 3 H 6 PC3H 3 N = 0.800 MPa × 2 MPa C 3 H 3 N = 0.800 MPa if NH3 is limiting 2 MPa NH 3 GASES 2 MPa C 3 H 3 N = 1.000 MPa if O2 is limiting 3 MPa O 2 PC3H 3 N = 1.500 MPa × C3H6 is limiting. Although more product could be produced from NH3 and O2, there is only enough C3H6 to produce 0.500 MPa of C3H3N. The partial pressure of C3H3N in atmospheres after the reaction is: 0.500 × 106 Pa × n = PV 4.94 atm × 150. L = = 30.3 mol C3H3N 0 . 08206 L atm RT × 298 K K mol 30.3 mol × 108. 1 atm = 4.94 atm 1.013 × 105 Pa 53.06 g = 1.61 × 103 g C3H3N can be produced. mol The partial pressure of CO2 that reacted is 740. - 390. = 350. torr. Thus the number of moles of CO2 that reacts is given by: 350. atm × 3.00 L PV 760 n= = = 5.75 × 10−2 mol CO2 0.08206 L atm RT × 293 K K mol 5.75 × 10−2 mol CO2 × Mass % MgO = 109. 1 mol MgO 40.31 g MgO × = 2.32 g MgO 1 mol CO 2 mol MgO 2.32 g × 100 = 81.4% MgO 2.85 g Ptotal = PH 2 + PH 2O , 1.032 atm = PH 2 + 32 torr × n H2 = PH 2 V RT = 1 atm , 1.032 − 0.042 = 0.990 atm = PH 2 760 torr 0.990 atm × 0.240 L = 9.56 × 10−3 mol H2 0.08206 L atm × 303 K K mol 9.56 × 10−3 mol H2 × 1 mol Zn 65.38 g Zn × = 0.625 g Zn mol H 2 mol Zn CHAPTER 5 110. GASES 127 a. Initially PN 2 = PH 2 = 1.00 atm and the total pressure is 2.00 atm (Ptotal = PN 2 + PH 2 ). The total pressure after reaction will also be 2.00 atm because we have a constant-pressure container. Because V and T are constant before the reaction takes place, there must be equal moles of N2 and H2 present initially. Let x = mol N2 = mol H2 that are present initially. From the balanced equation, N2(g) + 3 H2(g) → 2 NH3(g), H2 will be limiting because three times as many moles of H2 are required to react as compared to moles of N2. After the reaction occurs, none of the H2 remains (it is the limiting reagent). Mol NH3 produced = x mol H2 × Mol N2 reacted = x mol H2 × 2 mol NH 3 = 2x/3 3 mol H 2 1 mol N 2 = x/3 3 mol H 2 Mol N2 remaining = x mol N2 present initially − x/3 mol N2 reacted = 2x/3 mol N2 After the reaction goes to completion, equal moles of N2(g) and NH3(g) are present (2x/3). Because equal moles are present, the partial pressure of each gas must be equal (PN 2 = PNH 3 ). Ptotal = 2.00 atm = PN 2 + PNH 3 ; solving: PN 2 = 1.00 atm = PNH 3 b. V % n because P and T are constant. The moles of gas present initially are: n N 2 + n H 2 = x + x = 2x mol After reaction, the moles of gas present are: n N 2 + n NH 3 = 2x 2x + = 4x/3 mol 3 3 Vafter n 4 x/ 3 2 = after = = 2x 3 Vinitial n initial The volume of the container will be two-thirds the original volume, so: V = 2/3(15.0 L) = 10.0 L 111. P1V1 = P2V2; the total volume is 1.00 L + 1.00 L + 2.00 L = 4.00 L. For He: P2 = P1 V1 1.00 L = 200. torr H = 50.0 torr He V2 4.00 L 128 CHAPTER 5 GASES 760 torr 1.00 L = 76.0 torr Ne = 0.100 atm; 0.100 atm H atm 4.00 L 2.00 L 1 atm 760 torr For Ar: P2 = 24.0 kPa × = 12.0 kPa; 12.0 kPa × × 4.00 L 101.3 kPa atm = 90.0 torr Ar For Ne: P2 = 0.400 atm H Ptotal = 50.0 + 76.0 + 90.0 = 216.0 torr 112. 2 H2(g) + O2(g) → 2 H2O(g); because P and T are constant, volume ratios will equal mole ratios (Vf /Vi = nf /ni). Let x = mol H2 = mol O2 present initially. H2 will be limiting because a 2 : 1 H2 to O2 mole ratio is required by the balanced equation, but only a 1 : 1 mole ratio is present. Therefore, no H2 will be present after the reaction goes to completion. However, excess O2(g) will be present as well as the H2O(g) produced. Mol O2 reacted = x mol H2 × 1 mol O 2 = x/2 mol O2 2 mol H 2 Mol O2 remaining = x mol O2 initially − x/2 mol O2 reacted = x/2 mol O2 Mol H2O produced = x mol H2 × 2 mol H 2 O = x mol H2O 2 mol H 2 Total moles gas initially = x mol H2 + x mol O2 = 2x Total moles gas after reaction = x/2 mol O2 + x mol H2O = (1.5)x nf V (1.5) x 1.5 = f = = = 0.75; Vf/Vi = 0.75 : l or 3 : 4 ni Vi 2x 2 113. a. 156 mL × nHCl = 1.34 g = 209 g HSiCl3 = actual yield of HSiCl3 mL PV 10.0 atm × 15.0 L = = 5.93 mol HCl 0.08206 L atm RT × 308 K K mol 5.93 mol HCl × 1 mol HSiCl 3 135.45 g HSiCl 3 × = 268 g HSiCl3 3 mol HCl mol HSiCl 3 Percent yield = actual yield 209 g × 100 = × 100 = 78.0% theoretical yield 268 g b. 209 g HiSCl3 × 1 mol HSiCl 3 1 mol SiH 4 × = 0.386 mol SiH4 135.45 g HSiCl 3 4 mol HSiCl 3 This is the theoretical yield. If the percent yield is 93.1%, then the actual yield is: CHAPTER 5 GASES 129 0.386 mol SiH4 × 0.931 = 0.359 mol SiH4 VSiH 4 = 114. nRT = P 0.359 mol × 0.08206 L atm × 308 K K mol 10.0 atm = 0.907 L = 907 mL SiH4 33.5 mg CO2 × 12.01 mg C 9.14 mg = 9.14 mg C; % C = × 100 = 26.1% C 44.01 mg CO 2 35.0 mg 41.1 mg H2O × 2.016 mg H 4.60 mg × 100 = 13.1% H = 4.60 mg H; % H = 18.02 mg H 2 O 35.0 mg n N2 740. atm × 35.6 × 10 −3 L 760 = = = 1.42 × 10−3 mol N2 0 . 08206 L atm RT × 298 K K mol PN 2 V 1.42 × 10-3 mol N2 × Mass % N = 28.02 g N 2 = 3.98 × 10−2 g nitrogen = 39.8 mg nitrogen mol N 2 39.8 mg × 100 = 61.0% N 65.2 mg Or we can get % N by difference: % N = 100.0 - (26.1 + 13.1) = 60.8% Out of 100.0 g: 2.17 = 1.00 2.17 26.1 g C × 1 mol = 2.17 mol C; 12.01 g 13.1 g H × 13.0 1 mol = 5.99 = 13.0 mol H; 2.17 1.008 g 60.8 g N × 1 mol 4.34 = 4.34 mol N; = 2.00 14.01 g 2.17 Empirical formula is CH6N2. 1/ 2 Rate1 ⎛ M ⎞ =⎜ ⎟ Rate 2 ⎝ 39.95 ⎠ = 26.4 = 1.07, M = (1.07)2 × 39.95 = 45.7 g/mol 24.6 Empirical formula mass of CH6N2 ≈ 12 + 6 + 28 = 46. Thus the molecular formula is also CH6N2. 130 CHAPTER 5 115. GASES We will apply Boyle’s law to solve. PV = nRT = constant, P1V1 = P2V2 Let condition (1) correspond to He from the tank that can be used to fill balloons. We must leave 1.0 atm of He in the tank, so P1 = 200. − 1.00 = 199 atm and V1 = 15.0 L. Condition (2) will correspond to the filled balloons with P2 = 1.00 atm and V2 = N(2.00 L), where N is the number of filled balloons, each at a volume of 2.00 L. 199 atm × 15.0 L = 1.00 atm × N(2.00 L), N = 1492.5; we can't fill 0.5 of a balloon, so N = 1492 balloons, or to 3 significant figures, 1490 balloons. 116. Average velocity % (1/M)1/2 at constant T; the pressure in container A will increase initially because the lighter H2 molecules will effuse into container A faster than air will escape container A. However, the pressures will eventually equalize once the gases have had time to mix thoroughly. 117. nAr = n CH 4 n CH 4 228 g = 5.71 mol Ar; χ CH 4 = , 0.650 = 39.95 g/mol n CH 4 + n Ar n CH 4 + 5.71 0.650( n CH 4 + 5.71) = n CH 4 , 3.71 = (0.350)n CH 4 , n CH 4 = 10.6 mol CH4 KEavg = 3 RT for 1 mol 2 Thus KEtotal = (10.6 + 5.71 mol) × 3/2 × 8.3145 J K−1 mol−1 × 298 K = 6.06 × 104 J = 60.6 kJ 1/ 2 PV a. = α + βP n (straight line, y = b + mx) 118. PV n b. slope = m = β α Δ (mu ) ⎛T⎞ = 2mu ∝ M⎜ ⎟ impact ⎝M⎠ = M at constant T Δmu impact y intercept = b = α P c. TK = TEC + 273; P = M nR (T° C + 273) nR = constant (T°C + 273), where constant = . V V This is in the form of the straight line equation, y = mx + b. CHAPTER 5 GASES 131 P slope = m = nR V 273 nR V y intercept = b = 273 nR V T(oC) 119. PV 1.00 atm × (1.75 × 10 −3 L) = = 7.16 × 10−5 mol He 0.08206 L atm RT × 298 K K mol Mol of He removed = In the original flask, 7.16 × 10−5 mol of He exerted a partial pressure of 1.960 − 1.710 = 0.250 atm. V= 120. nRT (7.16 × 10 −5 mol) × 0.08206 L atm K −1 mol −1 × 298 K = = 7.00 × 10−3 L V 0.250 atm = 7.00 mL a. Out of 100.00 g of Z, we have: 34.38 g Ni × 1 mol = 0.5858 mol Ni 58.69 g 28.13 g C × 1 mol 2.342 = 2.342 mol C; = 3.998 0.5858 12.011 g 37.48 g O × 1 mol 2.343 = 2.343 mol O; = 4.000 15.999 g 0.5858 The empirical formula is NiC4O4. 1/ 2 b. 1/ 2 ⎛ 39.95 ⎞ ⎟⎟ = ⎜⎜ ⎝ MZ ⎠ ⎛M ⎞ Rate Z = ⎜⎜ Ar ⎟⎟ Rate Ar ⎝ M Z ⎠ ; because initial mol Ar = mol Z: 1/ 2 ⎛ 39.95 ⎞ ⎟⎟ 0.4837 = ⎜⎜ ⎝ MZ ⎠ , Mz = 170.8 g/mol c. NiC4O4: M = 58.69 + 4(12.01) + 4(16.00) = 170.73 g/mol Molecular formula is also NiC4O4. d. Each effusion step changes the concentration of Z in the gas by a factor of 0.4837. The original concentration of Z molecules to Ar atoms is a 1 : 1 ratio. After 5 stages: nZ/nAr = (0.4837)5 = 2.648 × 10-2 132 121. CHAPTER 5 GASES a. 2 CH4(g) + 2 NH3(g) + 3 O2(g) → 2 HCN(g) + 6 H2O(g) b. Volumes of gases are proportional to moles at constant T and P. Using the balanced equation, methane and ammonia are in stoichiometric amounts and oxygen is in excess. In 1 second: n CH 4 = PV 1.00 atm × 20.0 L = = 0.576 mol CH4 RT 0.08206 L atm K −1 mol −1 × 423 K 0.576 mol CH 4 2 mol HCN 27.03 g HCN × × = 15.6 g HCN/s s 2 mol CH 4 mol HCN Challenge Problems 122. a. We assumed a pressure of 1.0 atm and a temperature of 25°C (298 K). 50. lb × 0.454 kg/lb = 23 kg n= PV 1.0 atm × 10. L = = 0.41 mol gas 0 . 08206 L atm RT × 298 K K mol The lift of one balloon is: 0.41 mol(29 g/mol − 4.003 g/mol) = 10. g. To lift 23 kg = 23,000 g, we need at least 23,000/10 = 2300 balloons. This is a lot of balloons. b. The balloon displaces air as it is filled. The displaced air has mass, as does the helium in the balloon, but the displaced air has more mass than the helium. The difference in this mass is the lift of the balloon. Because volume is constant, the difference in mass is directly related to the difference in density between air and helium. 123. Initially we have 1.00 mol CH4 (16.0 g/mol = molar mass) and 2.00 mol O2 (32.0 g/mol = molar mass). CH4(g) + a O2(g) → b CO(g) + c CO2(g) + d H2O(g) b + c = 1.00 (C balance); 2a = b + 2c + d (O balance) 2d = 4 (H balance), d = 2 = 2.00 mol H2O Vinitial = nRT 3.00 mol × 0.08206 L atm K −1 mol −1 × 425 K = = 104.6 L (1 extra sig .fig.) P 1.00 atm CHAPTER 5 GASES Densityinitial = 133 80.0 g = 0.7648 g/L (1 extra sig. fig.) 104.6 L Because mass is constant: mass = Vinitial × dinitial = Vfinal × dfinal, Vfinal = Vinitial × d initial 0.7648 g/L = 104.6 L × d final 0.7282 g/L Vfinal = 109.9 L (1 extra sig. fig.) nfinal = PV 1.00 atm × 109.9 L = = 3.15 total moles of gas 0.08206 L atm RT × 425 K K mol Assuming an excess of O2 is present after reaction, an expression for the total moles of gas present at completion is: b + c + 2.00 + (2.00 – a) = 3.15; Note: d = 2.00 mol H2O was determined previously. Because b + c = 1.00, solving gives a = 1.85 mol O2 reacted. Indeed, O2 is in excess. From the O balance equation: 2a = 3.70 = b + 2c + 2.00, b + 2c = 1.70 Because b + c = 1.00, solving gives b = 0.30 mol CO and c = 0.70 mol CO2. The fraction of methane that reacts to form CO is 0.30 mol CO/1.00 mol CH4 = 0.30 (or 30.% by moles of the reacted methane forms CO). 124. a. The number of collisions of gas particles with the walls of the container is proportional to: ZA ∝ N V T M where N = number of gas particles, V= volume of container, T = temperature (Kelvin), and M = molar mass of gas particles in kilograms. Because both He samples are in separate containers of the same volume, V and M are constant. Because pressure and volume are constant, P ∝ nT (also, n ∝ N). Thus: ZA ∝ N T N T Z1 = 1 1 = 2, and N1T1 = N2T2 Z2 N 2 T2 134 CHAPTER 5 Thus: GASES 2 T2 N1 T 2T1 T = = 2, = 2 , 2 T1 = T2 N2 T1 T1 T1 T2 Solving: 4T1 = T2, T1 = 1/4 T2; because P ∝ nT, and P is constant, n1 = 4n2. Although the number of collisions in container 1 is twice as high, the temperature is onefourth that of container 2. This is so because there are four times the number of moles of helium gas in container 1. b. There are two times the number of collisions, but because the temperature is lower, the gas particles are hitting with less forceful collisions. Overall, the pressure is the same in each container. 125. Cr(s) + 3 HCl(aq) → CrCl3(aq) + 3/2 H2(g); Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) ⎛ 1 atm ⎞ ⎟ × 0.225 L ⎜⎜ 750. torr × 760 torr ⎟⎠ PV ⎝ Mol H2 produced = n = = = 9.02 × 10−3 mol H2 0.08206 L atm RT × (273 + 27) K K mol 9.02 × 10−3 mol H2 = mol H2 from Cr reaction + mol H2 from Zn reaction From the balanced equation: 9.02 × 10−3 mol H2 = mol Cr × (3/2) + mol Zn × 1 Let x = mass of Cr and y = mass of Zn, then: x + y = 0.362 g and 9.02 × 10−3 = y (1.5) x + 52.00 65.38 We have two equations and two unknowns. Solving by simultaneous equations: 9.02 × 10−3 = (0.02885)x + (0.01530)y −0.01530 × 0.362 = −(0.01530)x − (0.01530)y 3.48 × 10−3 = (0.01355)x, x = mass of Cr = y = mass of Zn = 0.362 g − 0.257 g = 0.105 g Zn; mass % Zn = 126. 3.48 × 10 −3 = 0.257 g 0.01355 0.105 g × 100 0.362 g = 29.0% Zn a. When the balloon is heated, the balloon will expand (P and n remain constant). The mass of the balloon is the same, but the volume increases, so the density of the argon in the balloon decreases. When the density is less than that of air, the balloon will rise. b. Assuming the balloon has no mass, when the density of the argon equals the density of air, the balloon will float in air. Above this temperature, the balloon will rise. CHAPTER 5 dair = GASES 135 P • MM air , where MMair = average molar mass of air RT MMair = 0.790 × 28.02 g/mol + 0.210 × 32.00 g/mol = 28.9 g/mol dair = 1.00 atm × 28.9 g/mol = 1.18 g/L 0.08206 L atm × 298 K K mol dargon = 1.00 atm × 39.95 g/mol = 1.18 g/L, T = 413 K 0.08206 L atm ×T K mol Heat the Ar above 413 K or 140.°C and the balloon will float. 127. Molar mass = dRT P × molar mass , P and molar mass are constant; dT = = constant P R d = constant(1/T) or d1T1 = d2T2, where T is in kelvin (K). T = x + °C; 1.2930(x + 0.0) = 0.9460(x + 100.0) (1.2930)x = (0.9460)x + 94.60, (0.3470)x = 94.60, x = 272.6 From these data, absolute zero would be −272.6°C. The actual value is −273.15°C. 128. BaO(s) + CO2(g) → BaCO3(s); CaO(s) + CO2(g) → CaCO3(s) 750. atm × 1.50 L Pi V 760 = initial moles of CO2 = = 0.0595 mol CO2 ni = 0.08206 L atm RT × 303.2 K K mol 230. atm × 1.50 L Pf V 760 nf = = final moles of CO2 = = 0.0182 mol CO2 0.08206 L atm RT × 303.2 K K mol 0.0595 − 0.0182 = 0.0413 mol CO2 reacted Because each metal reacts 1 : 1 with CO2, the mixture contains 0.0413 mol of BaO and CaO. The molar masses of BaO and CaO are 153.3 and 56.08 g/mol, respectively. Let x = mass of BaO and y = mass of CaO, so: x + y = 5.14 g and x y + = 0.0413 mol or x + (2.734)y = 6.33 153.3 56.08 136 CHAPTER 5 GASES Solving by simultaneous equations: x + (2.734)y = 6.33 −y = −5.14 −x (1.734)y = 1.19, y = 1.19/1.734 = 0.686 y = 0.686 g CaO and 5.14 − y = x = 4.45 g BaO Mass % BaO = 129. 4.45 g BaO × 100 = 86.6% BaO; % CaO = 100.0 − 86.6 = 13.4% CaO 5.14 g PV n = 1 + βP; × molar mass = d nRT V P RT βRTP molar mass P = 1 + βP , = × + d molar mass molar mass RT d This is in the equation for a straight line: y = b + mx. If we plot P/d versus P and extrapolate to P = 0, we get a y intercept = b = 1.398 = RT/molar mass. At 0.00°C, molar mass = 0.08206 × 273.15 = 16.03 g/mol. 1.398 1/ 2 130. ZA 1/ 2 ⎛ N ⎞ ⎛ RT ⎞ = A⎜ ⎟⎜ ⎟ ; ⎝ V ⎠ ⎝ 2πM ⎠ ⎛ T1 ⎞ ⎜⎜ ⎟ M1 ⎟⎠ Z1 ⎝ = 1/ 2 Z2 ⎛ T2 ⎞ ⎜⎜ ⎟⎟ ⎝ M2 ⎠ 1/ 2 ⎛M T ⎞ = ⎜⎜ 2 1 ⎟⎟ ⎝ M1T2 ⎠ = 1.00, M1T2 = M2T1 M UF6 T2 M TUF6 352.0 = 2; = = = 87.93 T1 M1 THe M He 4.003 131. Figure 5.16 shows the effect of temperature on the Maxwell-Boltzmann distribution of velocities of molecules. Note that as temperature increases, the probability that a gas particle has the most probable velocity decreases. Thus, since the probability of the gas particle with the most probable velocity decreased by one-half, then the temperature must be higher than 300. K. The equation that determines the probability that a gas molecule has a certain velocity is: ⎛ m ⎞ ⎟⎟ f(u) = 4π ⎜⎜ ⎝ 2 πk B T ⎠ 3/ 2 u 2 e − mu 2 / 2 k BT Let Tx = the unknown temperature, then: CHAPTER 5 GASES 137 3/ 2 ⎛ m ⎞ − mu 2 / 2k T ⎟⎟ u 2mp, x e mp , x B x 4π⎜⎜ f (u mp, x ) 1 ⎝ 2πk B Tx ⎠ = = 3/ 2 f (u mp, 300 ) 2 ⎛ ⎞ m − mu 2 / 2k T ⎟⎟ u 2mp, 300 e mp , 300 B 300 4π⎜⎜ ⎝ 2πk BT300 ⎠ Because ump = ⎛ 1 ⎜⎜ ⎝ Tx 2k B T , the equation reduces to: m 3/ 2 ⎞ ⎟⎟ (Tx ) 1/ 2 ⎛ T300 ⎞ 1 ⎠ ⎟⎟ = = ⎜⎜ 3/ 2 2 ⎛ 1 ⎞ ⎝ Tx ⎠ ⎜⎜ ⎟⎟ (T300 ) ⎝ T300 ⎠ Note that the overall exponent term cancels from the expression when 2kBT/m is substituted for u 2mp in the exponent term; the temperatures cancel. Solving for Tx: ⎛ 300. K ⎞ 1 ⎟⎟ , Tx = 1.20 × 103 K = ⎜⎜ 2 T x ⎝ ⎠ As expected, Tx is higher than 300. K. 132. Dalton’s law states: Ptotal = P1 + P2 + ... + Pk, for k different types of gas molecules in a mixture. The postulates of the kinetic molecular theory are: 1. the volume of the individual particles can be assumed to be negligible. 2. the collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. 3. the particles assert no forces on each other. 4. the average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas. The derivation is very similar to the ideal gas law derivation covered in Section 5.6 of the text. For a mixture of gases in a cube, there exist k different types of gas molecules. For each type (i) of gas molecule, the force on the cube = Fi = (2mi/L)ui2. Because the gas particles are assumed noninteracting, the total force for all the gas molecules in the mixture is: Ftotal = k 2m i 2 (u i ) i =1 L ∑ 138 CHAPTER 5 GASES Now we want the average force for each type of gas particle, which is: Ftotal = k 2m i 2 (u i ) i =1 L ∑ Pressure due to the average particle in this gas mixture of k types is the average total force divided by the total area. The expression for pressure is: k P= 2m i 2 (u i ) i =1 L ∑ 2 6L k ∑ m i (u i2 ) i =1 = , where V is the volume of the cube 3V Total pressure due to the number of moles of different gases is: k Ptotal = ∑ n i N a m i (u i2 ) i =1 , where NA = Avogadro’s number 3V ⎞ ⎛1 Because molar KEi, avg = N a ⎜ m i u i2 ⎟, the expression for total pressure can be written as: ⎠ ⎝2 Ptotal = k ∑ 2 3 n i N a ( 12 m i u i2 ) V i =1 = k ∑ i =1 2 3 n i KE i , avg V Assuming molar KEi, avg is proportional to T and is equal to Ptotal = k ∑ i =1 k n i RT = V ∑ Pi i =1 because Pi = 3 RT , then: 2 n i RT V This is Dalton’s law of partial pressure. Note that no additional assumptions are necessary other than the postulates of the kinetic molecular theory and the conclusions drawn from the ideal gas law derivation. 133. From the problem, we want ZA/Z = 1.00 × 1018 where ZA is the collision frequency of the gas particles with the walls of the container and Z is the intermolecular collision frequency. From the text: ZA = Z A 4 N RT V 2π M N 2 π RT d V M = 1.00 × 1018, 1.00 × 1018 = A 2 4d π 2 If R = length of the cube edge container, then the area A of one cube face is R2 and the total area in the cube is 6R2 (6 faces/cube). He diameter = d = 2(3.2 × 10−11 m) = 6.4 × 10−11 m. Solving the above expression for A, and then for R gives R = 0.11 m = 1.1 dm. CHAPTER 5 GASES 139 Volume = R3 = (1.1 dm)3 = 1.3 dm3 = 1.3 L 134. Assuming 1.000 L of the hydrocarbon (CxHy), then the volume of products will be 4.000 L and the mass of products (H2O + CO2) will be: 1.391 g/L × 4.000 L = 5.564 g products Mol CxHy = n C x H y = Mol products = np = PV 0.959 atm × 1.000 L = = 0.0392 mol RT 0.08206 L atm × 298 K K mol PV 1.51 atm × 4.000 L = = 0.196 mol 0 . 08206 L atm RT × 375 K K mol CxHy + oxygen → x CO2 + y/2 H2O Setting up two equations: (0.0392)x + 0.0392(y/2) = 0.196 (moles of products) (0.0392)x(44.01 g/mol) + 0.0392(y/2)(18.02 g/mol) = 5.564 g (mass of products) Solving: x = 2 and y = 6, so the formula of the hydrocarbon is C2H6. 135. The reactions are: C(s) + 1/2 O2(g) → CO(g) and C(s) + O2(g) → CO2(g) ⎛ RT ⎞ PV = nRT, P = n ⎜ ⎟ = n(constant) ⎝ V ⎠ Because the pressure has increased by 17.0%, the number of moles of gas has also increased by 17.0%. nfinal = (1.170)ninitial = 1.170(5.00) = 5.85 mol gas = n O 2 + n CO + n CO 2 n CO + n CO 2 = 5.00 (balancing moles of C). Solving by simultaneous equations: n O 2 + n CO + n CO 2 = 5.85 − (n CO + n CO 2 = 5.00) _______________________ n O2 = 0.85 If all C were converted to CO2, no O2 would be left. If all C were converted to CO, we would get 5 mol CO and 2.5 mol excess O2 in the reaction mixture. In the final mixture, moles of CO equals twice the moles of O2 present ( n CO = 2n O 2 ). 140 CHAPTER 5 GASES n CO = 2n O 2 = 1.70 mol CO; 1.70 + n CO 2 = 5.00, n CO 2 = 3.30 mol CO2 χ CO = 136. 1.70 3.30 = 0.291; χ CO 2 = = 0.564; 5.85 5.85 χ O2 = 0.85 = 0.145 ≈ 0.15 5.85 Let x = moles SO2 = moles O2 and z = moles He. a. P • MM where MM = molar mass RT 1.924 g/L = 1.000 atm × MM , MMmixture = 43.13 g/mol 0.08206 L atm × 273.2 K K mol Assuming 1.000 total moles of mixture is present, then: x + x + z = 1.000 and: 64.07 g/mol × x + 32.00 g/mol × x + 4.003 g/mol × z = 43.13 g 2x + z = 1.000 and (96.07)x + (4.003)z = 43.13 Solving: x = 0.4443 mol and z = 0.1114 mol Thus: χHe = 0.1114 mol/1.000 mol = 0.1114 b. 2 SO2(g) + O2(g) → 2 SO3(g) Initially, assume 0.4443 mol SO2, 0.4443 mol O2 and 0.1114 mol He. Because SO2 is limiting, we end up with 0.2222 mol O2, 0.4443 mol SO3, and 0.1114 mol He in the gaseous product mixture. This gives: ninitial = 1.0000 mol and nfinal = 0.7779 mol. In a reaction, mass is constant. d = mass 1 and V % n at constant P and T, so d % . V n n initial d 1.0000 ⎛ 1.0000 ⎞ = = final , d final = ⎜ ⎟ × 1.924 g/L, dfinal = 2.473 g/L n final 0.7779 d initial ⎝ 0.7779 ⎠ 137. a. The reaction is: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) PV = nRT, PCH 4 VCH 4 P V PV = RT = constant, = air air n n CH 4 n air The balanced equation requires 2 mol O2 for every mol of CH4 that reacts. For three times as much oxygen, we would need 6 mol O2 per mol of CH4 reacted (n O 2 = 6n CH 4 ). Air is 21% mole percent O2, so n O 2 = (0.21)nair. Therefore, the moles of air we would need to delivery the excess O2 are: CHAPTER 5 GASES 141 n O 2 = (0.21)nair = 6n CH 4 , nair = 29n CH 4 , n air = 29 n CH 4 In 1 minute: Vair = VCH 4 × PCH 4 1.50 atm n air × = 200. L × 29 × = 8.7 × 103 L air/min 1.00 atm n CH 4 Pair b. If x mol of CH4 were reacted, then 6x mol O2 were added, producing (0.950)x mol CO2 and (0.050)x mol of CO. In addition, 2x mol H2O must be produced to balance the hydrogens. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g); CH4(g) + 3/2 O2(g) → CO(g) + 2 H2O(g) Amount O2 reacted: (0.950)x mol CO2 × 2 mol O 2 = (1.90)x mol O2 mol CO 2 (0.050)x mol CO × 1.5 mol O 2 = (0.075)x mol O2 mol CO Amount of O2 left in reaction mixture = (6.00)x − (1.90)x − (0.075)x = (4.03)x mol O2 Amount of N2 = (6.00)x mol O2 × 79 mol N 2 = (22.6)x ≈ 23x mol N2 21 mol O 2 The reaction mixture contains: (0.950)x mol CO2 + (0.050)x mol CO + (4.03)x mol O2 + (2.00)x mol H2O + 23x mol N2 = (30.)x mol of gas total (0.050) x (0.950) x (4.03) x = 0.0017; χ CO 2 = = 0.032; χ O 2 = = 0.13 (30.) x (30.) x (30.) x χ CO = χ H 2O = (2.00) x = 0.067; (30.) x χ N2 = 23 x = 0.77 (30.) x c. The partial pressures are determined by P = χPtotal. Because Ptotal = 1.00 atm, PCO = 0.0017 atm, PCO 2 = 0.032 atm, PO 2 = 0.13 atm, PH 2O = 0.067 atm, and PN 2 = 0.77 atm. 138. ntotal = total number of moles of gas that have effused into the container: ntotal = PV (1.20 × 10 −6 atm) × 1.00 L = = 4.87 × 10-8 mol 0.08206 L atm RT × 300. K K mol 142 CHAPTER 5 GASES This amount has entered over a time span of 24 hours: 24 h × Thus: 60 min 60 s = 8.64 × 104 s × 1h 1 min 4.87 × 10 −8 mol = 5.64 × 10-13 mol/s have entered the container. 8.64 × 10 4 s 5.64 × 10 −13 mol 6.022 × 10 23 molecules × = 3.40 × 1011 molecules/s s mol The frequency of collisions of the gas with a given area is: ⎛N Z = A ⎜⎜ ⎝V 1/ 2 ⎞ ⎛ RT ⎞ ⎟⎟ ⎜⎜ ⎟⎟ ⎠ ⎝ 2 πM ⎠ ; Z total = 3.40 × 1011 molecules = Z N 2 + ZO2 s n P 1.00 atm = = = 4.06 × 10−2 mol/L 0.08206 L atm V RT × 300. K K mol N 4.06 × 10 −2 mol 6.022 × 10 23 molecules 1000 L = × × = 2.44 × 1025 molecules/m3 3 V L mol m N = (0.78)(2.44 × 1025) = 1.9 × 1025 molecules/m3 V For N2 : For O2: N = (0.22)(2.44 × 1025) = 5.4 × 1024 molecules/m3 V Ztotal = 3.40 × 1011 molecules/s = Z N 2 + Z O 2 3.40 × 10 11 1/ 2 1/ 2 ⎡ ⎤ 25 ⎛ 8.3145 × 300. ⎞ 24 ⎛ 8.3145 × 300. ⎞ ⎥ ⎟ ⎜ ⎟ 5 . 4 10 + × = A ⎢1.9 × 10 ⎜⎜ ⎜ 2 π(32.0 × 10 −3 ) ⎟ ⎥ 2 π(28.0 × 10 −3 ) ⎟⎠ ⎢ ⎝ ⎝ ⎠ ⎣ ⎦ ⎡ 2.3 × 10 27 molecules 6.0 × 10 26 molecules ⎤ 3.40 × 1011 molecules + = A⎢ ⎥ s m2 s m2 s ⎣ ⎦ 3.40 × 1011 2 m = 1.2 × 10−16 m2 = πr2, r = A= 2.9 × 10 27 1/ 2 ⎛ 1.2 × 10 −16 m 2 ⎞ ⎜ ⎟ ⎜ ⎟ π ⎝ ⎠ Diameter of hole = 2r = 2(6.2 × 10−9 m) = 1.2 × 10−8 m = 12 nm = 6.2 × 10−9 m = 6.2 nm CHAPTER 5 139. GASES 143 Each stage will give an enrichment of: Diffusion rate Diffusion rate Because 12 ⎛ M 13 CO CO 2 2 ⎜ = 13 ⎜ M 12 CO 2 CO 2 ⎝ 12 1/ 2 ⎞ ⎟ ⎟ ⎠ ⎛ 45.001 ⎞ ⎟ = 1.0113 ⎝ 43.998 ⎠ =⎜ CO2 moves slightly faster, each successive stage will have less 13CO2. 99.990 12 CO 2 99.90 12 CO 2 N × = 1 . 0113 0.010 13 CO 2 0.10 13 CO 2 1.0113N = 9,999.0 = 10.009 999.00 (carrying extra significant figures) N log(1.0113) = log(10.009), N = 140. 1.000391 = 2.05 × 102 ≈ 2.1 × 102 stages are needed. −3 4.88 × 10 After the hole develops, assume each He that collides with the hole goes into the Rn side and that each Rn that collides with the hole goes into the He side. Assume no molecules return to the side in which they began. Initial moles of each gas: n = PV (2.00 × 10 −6 atm) × 1.00 L = = 8.12 × 10−8 mol 0.08206 L atm RT × 300. K K mol Z He = A × 1/ 2 N ⎛ RT ⎞ × ⎜ ⎟ V ⎝ 2 πM ⎠ ZHe = π(1.00 × 10−6 m)2 × , N P = × NA × 1000 L/m3 and A = πr2 V RT 2.00 × 10 −6 × (6.022 × 1023) × 1000 0.08206 × 300. 1/ 2 ⎛ 8.3145 × 300. ⎞ ⎟ × ⎜⎜ −3 ⎟ ⎝ 2 π (4.003 × 10 ) ⎠ = 4.84 × 1010 collisions/s Therefore, 4.84 × 1010 atoms/s leave the He side. 10.0 h × or: 60 min 60 s 4.84 × 1010 atoms × × = 1.74 × 1015 atoms 1h 1 min s 1.74 × 1015 atoms = 2.89 × 10−9 mol He leave in 10.0 h. 6.022 × 10 23 atoms/mol ZRn = π(1.00 × 10−6 m)2 × 2.00 × 10 −6 × (6.022 × 1023) × 1000 0.08206 × 300. 1/ 2 ⎛ 8.3145 × 300. ⎞ ⎟ × ⎜⎜ −3 ⎟ ⎝ 2 π (222 × 10 ) ⎠ = 6.50 × 109 collisions/s 144 CHAPTER 5 GASES 6.50 × 109 atoms/s leave Rn side. 3.60 × 104 s × 6.50 × 109 atoms 1 mol × = 3.89 × 10−10 mol Rn leave in 23 s 6.022 × 10 atoms 10.0 h Side that began with He now contains: 8.12 × 10−8 − 2.89 × 10−9 = 7.83 × 10−8 mol He + 3.89 × 10−10 mol Rn = 7.87 × 10−8 moles total The pressure in the He side is: P= nRT (7.87 × 10 −8 mol) × 0.08206 L atm K −1 mol −1 × 300. K = = 1.94 × 10−6 atm V 1.00 L We can determine the pressure in the Rn chamber two ways. Because no gas has escaped, and because the initial pressures were equal and the pressure in one of the sides decreased by 0.06 × 10−6 atm, P in the second side must increase by 0.06 × 10−6 atm. Thus the pressure on the side that originally contained Rn is 2.06 × 10−6 atm. Or we can calculate P the same way as with He. The Rn side contains: 8.12 × 10−8 − 3.89 × 10−10 = 8.08 × 10−8 mol Rn + 2.89 × 10−9 mol He = 8.37 × 10−8 mol total P= 141. a. nRT (8.87 × 10 −8 mol) × 0.08206 L atm K −1 mol −1 × 300. K = = 2.06 × 10−6 atm V 1.00 L Average molar mass of air = 0.790 × 28.02 g/mol + 0.210 × 32.00 g/mol = 28.9 g/mol; molar mass of helium = 4.003 g/mol A given volume of air at a given set of conditions has a larger density than helium at those conditions. We need to heat the air to a temperature greater than 25°C in order to lower the air density (by driving air out of the hot air balloon) until the density is the same as that for helium (at 25°C and 1.00 atm). b. To provide the same lift as the helium balloon (assume V = 1.00 L), the mass of air in the hot-air balloon (V = 1.00 L) must be the same as that in the helium balloon. Let MM = molar mass: PCMM = dRT, mass = Mass air = 0.164 g = MM • PV ; solving: mass He = 0.164 g RT 28.9 g/mol × 1.00 atm × 1.00 L , T = 2150 K (a very high 0.08206 L atm ×T temperature) K mol CHAPTER 5 142. GASES 145 a. The formula of the compound AxBy depends on which gas is limiting, A2 or B2. We need to determine both possible products. The procedure we will use is to assume one reactant is limiting, and then determine what happens to the initial total moles of gas as it is converted into the product. Because P and T are constant, volume % n. Because mass is conserved in a chemical reaction, any change in density must be due to a change in volume of the container as the reaction goes to completion. Density = d % d n 1 and V % n, so: after = initial V d initial n after Assume the molecular formula of the product is AxBy where x and y are whole numbers. First, let’s consider when A2 is limiting with x moles each of A2 and B2 in our equimolar mixture. Note that the coefficient in front of AxBy in the equation must be 2 for a balanced reaction. x A2(g) Initial Change Final + y B2(g) x mol −x mol 0 → 2 AxBy(g) x mol −y mol (x − y) mol 0 mol +2 mol 2 mol d after n 2x = 1.50 = initial = d initial n after x− y +2 (1.50)x − (1.50)y + 3.00 = 2x, 3.00 − (1.50)y = (0.50)x Because x and y are whole numbers, y must be 1 because the above equation does not allow y to be 2 or greater. When y = 1, x = 3 giving a formula of A3B if A2 is limiting. Assuming B2 is limiting with y moles in the equimolar mixture: x A2(g) Initial Change After y −x y−x + y B2(g) → y −y 0 2 AxBy(g) 0 +2 2 density after n 2y = 1.50 = initial = density before n after y−x+2 Solving gives x = 1 and y = 3 for a molecular formula of AB3 when B2 is limiting. b. In both possible products, the equations dictated that only one mole of either A or B had to be present in the formula. Any number larger than 1 would not fit the data given in the problem. Thus the two formulas determined are both molecular formulas and not just empirical formulas. 146 143. CHAPTER 5 GASES d = molar mass(P/RT); at constant P and T, the density of gas is directly proportional to the molar mass of the gas. Thus the molar mass of the gas has a value which is 1.38 times that of the molar mass of O2. Molar mass = 1.38(32.00 g/mol) = 44.2 g/mol Because H2O is produced when the unknown binary compound is combusted, the unknown must contain hydrogen. Let AxHy be the formula for unknown compound. Mol AxHy = 10.0 g AxHy × Mol H = 16.3 g H2O × 1 mol A x H y 44.2 g = 0.226 mol AxHy 1 mol H 2 O 2 mol H × = 1.81 mol H 18.02 g mol H 2 O 1.81 mol H = 8 mol H/mol AxHy ; AxHy = AxH8 0.226 mol A x H y The mass of the x moles of A in the AxH8 formula is: 44.2 g − 8(1.008 g) = 36.1 g From the periodic table and by trial and error, some possibilities for AxH8 are ClH8, F2H8, C3H8, and Be4H8. C3H8 and Be4H8 fit the data best and because C3H8 (propane) is a known substance, C3H8 is the best possible identity from the data in this problem. Marathon Problem 144. We must determine the identities of element A and compound B in order to answer the questions. Use the first set of data to determine the identity of element A. Mass N2 = 659.452 g − 658.572 g = 0.880 g N2 0.880 g N2 × nRT V = = P 1 mol N 2 = 0.0314 mol N2 28.02 g N 2 0.08206 L atm × 288 K K mol = 0.714 L 1 atm 790. torr × 760 torr 0.0314 mol × ⎛ 1 atm ⎞ ⎜⎜ 745 torr × ⎟ × 0.714 L 760 torr ⎟⎠ ⎝ Moles of A = n = = 0.0285 mol A 0.08206 L atm K −1 mol −1 × (273 + 26) K CHAPTER 5 GASES 147 Mass of A = 660.59 − 658.572 g = 2.02 g A Molar mass of A = 2.02 g A = 70.9 g/mol 0.0285 mol A The only element that is a gas at 26°C and 745 torr and has a molar mass close to 70.9 g/mol is chlorine = Cl2 = element A. The remainder of the information is used to determine the formula of compound B. Assuming 100.00 g of B: 85.6 g C × 7.13 1 mol C = 1.00 = 7.13 mol C; 7.13 12.01 g C 14.4 g H × 1 mol H 14.13 = 14.3 mol H; = 2.01 1.008 g H 7.13 Empirical formula of B = CH2; molecular formula = CxH2x where x is a whole number. The balanced combustion reaction of CxH2x with O2 is: CxH2x(g) + 3x/2 O2(g) → x CO2(g) + x H2O(l) To determine the formula of CxH2x, we need to determine the actual moles of all species present. Mass of CO2 + H2O produced = 846.7 g − 765.3 g = 81.4 g Because mol CO2 = mol H2O = x (see balanced equation): 81.4 g = x mol CO2 × 44.01 g CO 2 18.02 g H 2 O + x mol H2O × , x = 1.31 mol mol CO 2 mol H 2 O Mol O2 reacted = 1.31 mol CO2 × 1.50 mol O 2 = 1.97 mol O2 mol CO 2 From the data, we can calculate the moles of excess O2 because only O2(g) remains after the combustion reaction has gone to completion. n O2 = PV 6.02 atm × 10.68 L = = 2.66 mol excess O2 RT 0.08206 L atm K −1 mol −1 × (273 + 22) K Mol O2 present initially = 1.97 mol + 2.66 mol = 4.63 mol O2 Total moles gaseous reactants before reaction = PV 11.98 atm × 10.68 L = = 5.29 mol RT 0.08206 × 295 K 148 CHAPTER 5 GASES Mol CxH2x = 5.29 mol total − 4.63 mol O2 = 0.66 mol CxH2x Summarizing: 0.66 mol CxH2x + 1.97 mol O2 → 1.31 mol CO2 + 1.31 mol H2O Dividing all quantities by 0.66 gives: CxH2x + 3 O2 → 2 CO2 + 2 H2O To balance the equation, CxH2x must be C2H4 = compound B. a. Now we can answer the questions. The reaction is: C2H4(g) + Cl2(g) → C2H4Cl2(g) B + A C Mol Cl2 = n = PV 1.00 atm × 10.0 L = = 0.446 mol Cl2 RT 0.08206 L atm K −1 mol −1 × 273 K Mol C2H4 = n = PV 1.00 atm × 8.60 L = = 0.384 mol C2H4 RT 0.08206 L atm K −1 mol −1 × 273 K Because a 1 : 1 mol ratio is required by the balanced reaction, C2H4 is limiting. Mass C2H4Cl2 produced = 0.384 mol C2H4 × 1 mol C 2 H 4 Cl 2 98.95 g × mol C 2 H 4 mol C 2 H 4 Cl 2 = 38.0 g C2H4Cl2 b. Excess mol Cl2 = 0.446 mol Cl2 − 0.384 mol Cl2 reacted = 0.062 mol Cl2 Ptotal = n total RT ; ntotal = 0.384 mol C2H4Cl2 produced + 0.062 mol Cl2 excess V V = 10.0 L + 8.60 L = 18.6 L Ptotal = 0.446 mol × 0.08206 L atm K −1 mol −1 × 273 K = 0.537 atm 18.6 L = 0.446 mol CHAPTER 6 CHEMICAL EQUILIBRIUM Characteristics of Chemical Equilibrium 10. a. The rates of the forward and reverse reactions are equal. b. There is no net change in the composition (as long as temperature is constant). 11. 2 NOCl(g) ⇌ 2 NO(g) + Cl2(g) K = 1.6 × 10−5 mol/L The expression for K is the product concentrations divided by the reactant concentrations. When K has a value much less than one, the product concentrations are relatively small, and the reactant concentrations are relatively large. 2 NO(g) ⇌N2(g) + O2(g) K = 1 × 1031 When K has a value much greater than one, the product concentrations are relatively large, and the reactant concentrations are relatively small. In both cases, however, the rate of the forward reaction equals the rate of the reverse reaction at equilibrium (this is a definition of equilibrium). 12. No, equilibrium is a dynamic process. Both reactions: H2O + CO → H2 + CO2 and H2 + CO2 → H2O + CO are occurring at equal rates. Thus 14C atoms will be distributed between CO and CO2. 13. No, it doesn't matter which direction the equilibrium position is reached. Both experiments will give the same equilibrium position because both experiments started with stoichiometric amounts of reactants or products. 14. H2O(g) + CO(g) ⇌ H2(g) + CO2(g) K= [H 2 ][CO 2 ] = 2.0 [H 2 O][CO] K is a unitless number because there is an equal number of moles of product gases as moles of reactant gases in the balanced equation. Therefore, we can use units of molecules per liter instead of moles per liter to determine K. 149 150 CHAPTER 6 CHEMICAL EQUILIBRIUM We need to start somewhere, so let’s assume 3 molecules of CO react. If 3 molecules of CO react, then 3 molecules of H2O must react, and 3 molecules each of H2 and CO2 are formed. We would have 6 ! 3 = 3 molecules of CO, 8 ! 3 = 5 molecules of H2O, 0 + 3 = 3 molecules of H2, and 0 + 3 = 3 molecules of CO2 present. This will be an equilibrium mixture if K = 2.0: ⎛ 3 molecules H 2 ⎞ ⎛ 3 molecules CO 2 ⎞ ⎜ ⎟⎜ ⎟ L L ⎝ ⎠ ⎝ ⎠ = 3 K= 5 ⎛ 5 molecules H 2 O ⎞ ⎛ 3 molecules CO ⎞ ⎜ ⎟⎜ ⎟ L L ⎝ ⎠⎝ ⎠ Because this mixture does not give a value of K = 2.0, this is not an equilibrium mixture. Let’s try 4 molecules of CO reacting to reach equilibrium. Molecules CO remaining = 6 ! 4 = 2 molecules of CO Molecules H2O remaining = 8 ! 4 = 4 molecules of H2O Molecules H2 present = 0 + 4 = 4 molecules of H2 Molecules CO2 present = 0 + 4 = 4 molecules of CO2 ⎛ 4 molecules H 2 ⎞ ⎛ 4 molecules CO 2 ⎞ ⎜ ⎟⎜ ⎟ L L ⎝ ⎠ ⎝ ⎠ = 2 .0 K= ⎛ 4 molecules H 2 O ⎞ ⎛ 2 molecules CO ⎞ ⎜ ⎟⎜ ⎟ L L ⎝ ⎠⎝ ⎠ Because K = 2.0 for this reaction mixture, we are at equilibrium. 15. When equilibrium is reached, there is no net change in the amount of reactants and products present because the rates of the forward and reverse reactions are equal to each other. The first diagram has 4 A2B molecules, 2 A2 molecules, and 1 B2 molecule present. The second diagram has 2 A2B molecules, 4 A2 molecules, and 2 B2 molecules. Therefore, the first diagram cannot represent equilibrium because there was a net change in reactants and products. Is the second diagram the equilibrium mixture? That depends on whether there is a net change between reactants and products when going from the second diagram to the third diagram. The third diagram contains the same number and type of molecules as the second diagram, so the second diagram is the first illustration that represents equilibrium. The reaction container initially contained only A2B. From the first diagram, 2 A2 molecules and 1 B2 molecule are present (along with 4 A2B molecules). From the balanced reaction, these 2 A2 molecules and 1 B2 molecule were formed when 2 A2B molecules decomposed. Therefore, the initial number of A2B molecules present equals 4 + 2 = 6 molecules A2B. The Equilibrium Constant 16. The equilibrium constant is a number that tells us the relative concentrations (pressures) of reactants and products at equilibrium. An equilibrium position is a set of concentrations that satisfies the equilibrium constant expression. More than one equilibrium position can satisfy the same equilibrium constant expression. CHAPTER 6 CHEMICAL EQUILIBRIUM 151 Table 6.1 of the text illustrates this nicely. Each of the three experiments in Table 6.1 has different equilibrium positions; that is, each experiment has different equilibrium concentrations. However, when these equilibrium concentrations are inserted into the equilibrium constant expression, each experiment gives the same value for K. The equilibrium position depends on the initial concentrations one starts with. Since there are an infinite number of initial conditions, there are an infinite number of equilibrium positions. However, each of these infinite equilibrium positions will always give the same value for the equilibrium constant (assuming temperature is constant). 17. K and Kp are equilibrium constants as determined by the law of mass action. For K, concentration units of mol/L are used, and for Kp, partial pressures in units of atm are used (generally). Q is called the reaction quotient. Q has the exact same form as K or Kp, but instead of equilibrium concentrations, initial concentrations are used to calculate the Q value. We use Q to determine if a reaction is at equilibrium. When Q = K (or when Qp = Kp), the reaction is at equilibrium. When Q ≠ K, the reaction is not at equilibrium, and one can deduce the net change that must occur for the system to get to equilibrium. 18. When reactants and products are all in the same phase, these are homogeneous equilibria. Heterogeneous equilibria involve more than one phase. In general, for a homogeneous gasphase equilibrium, all reactants and products are included in the K expression. In heterogeneous equilibria, equilibrium does not depend on the amounts of pure solids or liquids present. The amounts of solids and liquids present are not included in K expressions; they just have to be present. On the other hand, gases and solutes are always included in K expressions. Solutes have (aq) written after them. 19. Solids and liquids do not appear in equilibrium expressions. Only gases and dissolved solutes appear in equilibrium expressions. a. K = PH 2O [ H 2 O] ; Kp = 2 2 [ NH 3 ] [CO 2 ] PNH 3 × PCO 2 c. K = [O2]3; Kp = PO3 2 3 b. K = [N2][Br2]3; Kp = PN 2 × PBr 2 d. K = PH 2O [ H 2 O] ; Kp = [H 2 ] PH 2 20. Kp = K(RT)Δn, where Δn equals the difference in the sum of the coefficients between gaseous products and gaseous reactants (Δn = mol gaseous products ! mol gaseous reactants). When Δn = 0, then Kp = K. In Exercise 19, only reaction d has Δn = 0, so only reaction d has Kp = K. 21. Kp = K(RT)Δn, where Δn = sum of gaseous product coefficients ! sum of gaseous reactant coefficients. For this reaction, Δn = 3 ! 1 = 2. K= [CO][H 2 ]2 (0.24)(1.1) 2 = = 1.9 mol2/L2 (0.15) [CH 3OH] Kp = K(RT)2 = 1.9(0.08206 L atm K−1 mol−1 × 600. K)2 = 4.6 × 103 atm2 152 22. CHAPTER 6 H2(g) + Br2(g) ⇌ 2 HBr(g) Kp = a. HBr ⇌ 1/2 H2 + 1/2 Br2 K 'p CHEMICAL EQUILIBRIUM 2 PHBr = 3.5 × 104 ( PH 2 ) ( PBr2 ) = (PH 2 )1/ 2 (PBr2 )1/ 2 PHBr 1/ 2 ⎛ 1 ⎞ ⎟ =⎜ ⎜ Kp ⎟ ⎝ ⎠ 1/ 2 ⎛ ⎞ 1 ⎟ = ⎜⎜ 4 ⎟ × 3 . 5 10 ⎝ ⎠ = 5.3 × 10−3 b. 2 HBr ⇌ H2 + Br2 K 'p' = c. 1/2 H2 + 1/2 Br2 ⇌ HBr 23. [N2O] = (PH 2 )(PBr2 ) 2 PHBr K 'p'' = = 1 1 = = 2.9 × 10−5 4 Kp 3.5 × 10 PHBr 1/ 2 (PH 2 ) (PBr2 )1/ 2 = (K p )1/ 2 = 190 2.00 × 10 −2 mol 2.80 × 10 −4 mol 2.50 × 10 −5 mol ; [N2] = ; [O2] = 2.00 L 2.00 L 2.00 L 2 ⎛ 2.00 × 10 − 2 ⎞ ⎜ ⎟ ⎜ ⎟ 2.00 [ N 2 O ]2 (1.00 × 10 − 2 ) 2 ⎝ ⎠ K= = = [ N 2 ]2 [O 2 ] ⎛ 2.80 × 10 − 4 ⎞ 2 ⎛ 2.50 × 10 −5 ⎞ (1.40 × 10 − 4 ) 2 (1.25 × 10 −5 ) ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2.00 2.00 ⎝ ⎠ ⎝ ⎠ = 4.08 × 108 L/mol If the given concentrations represent equilibrium concentrations, then they should give a value of K = 4.08 × 108. (0.200) 2 = 4.08 × 108 (2.00 × 10 − 4 ) 2 (0.00245) Because the given concentrations when plugged into the equilibrium constant expression give a value equal to K (4.08 × 108), this set of concentrations is a system at equilibrium 24. Kp = 2 PNH 3 PN 2 × PH3 2 = (3.1 × 10 −2 ) 2 = 3.8 × 104 atm−2 (0.85)(3.1 × 10 −3 ) 3 (0.167) 2 = 1.21 × 103 3 (0.525) (0.00761) When the given partial pressures in atmospheres are plugged into the Kp expression, the value does not equal the Kp value of 3.8 × 104. Therefore, one can conclude that the given set of partial pressures does not represent a system at equilibrium. CHAPTER 6 25. Kp = CHEMICAL EQUILIBRIUM PH4 2 PH4 2O 153 ; Ptotal = PH 2O + PH 2 , 36.3 torr = 15.0 torr + PH 2 , PH 2 = 21.3 torr 4 ⎛ 1 atm ⎞ ⎜⎜ 21.3 torr × ⎟ 760 torr ⎟⎠ ⎝ Because l atm = 760 torr: Kp = = 4.07 4 ⎛ 1 atm ⎞ ⎜⎜15.0 torr × ⎟ 760 torr ⎟⎠ ⎝ Note: Solids and pure liquids are not included in K expressions. 26. N2(g) + 3 H2(g) ⇌ 2 NH3(g); with only reactants present initially, the net change that must occur to reach equilibrium is a conversion of reactants into products. At constant volume and temperature, n ∝ P. Thus if x atm of N2 reacts to reach equilibrium, then 3x atm of H2 must also react to form 2x atm of NH3 (from the balanced equation). Let’s summarize the problem in a table that lists what is present initially, what change in terms of x that occurs to reach equilibrium, and what is present at equilibrium (initial + change). This table is typically called an ICE table for initial, change, and equilibrium. 2 PNH 3 N2(g) + 3 H2(g) ⇌ 2 NH3(g) Kp = PN 2 × PH3 2 Initial Change Equil. 1.00 atm 2.00 atm 0 x atm of N2 reacts to reach equilibrium !x !3 x → +2x 2.00 ! 3x 2x 1.00 ! x From the setup: Ptotal = 2.00 atm = PN 2 + PH 2 + PNH 3 2.00 atm = (1.00 – x) + (2.00 – 3x) + 2x = 3.00 – 2x, x = 0.500 atm PH 2 = 2.00 − 3x = 2.00 – 3(0.500) = 0.50 atm Kp = 27. (1.00) 2 (2 x) 2 [2(0.500)]2 = 16 atm−2 = = (1.00 − x)(2.00 − 3 x) 3 (1.00 − 0.500)[2.00 − 3(0.500)]3 (0.50)(0.50) 3 PCl5(g) ⇌ PCl3(g) + Cl2(g) Kp = PPCl3 × PCl 2 PPCl5 To determine Kp, we must determine the equilibrium partial pressures of each gas. Initially, PPCl5 = 0.50 atm and PPCl3 = PCl 2 = 0 atm. To reach equilibrium, some of the PCl5 reacts to produce some PCl3 and Cl2, all in a 1 : 1 mole ratio. We must determine the change in partial pressures necessary to reach equilibrium. Because moles % P at constant V and T, if we let x = atm of PCl5 that reacts to reach equilibrium, this will produce x atm of PCl3 and x atm of Cl2 at equilibrium. The equilibrium partial pressures of each gas will be the initial partial pressure of each gas plus the change necessary to reach equilibrium. The equilibrium partial pressures are: 154 CHAPTER 6 CHEMICAL EQUILIBRIUM PPCl5 = 0.50 atm − x, PPCl3 = PCl 2 = x Now we solve for x using the information in the problem: Ptotal = PPCl5 + PPCl3 + PCl 2 , 0.84 atm = 0.50 − x + x + x, 0.84 atm = 0.50 + x, x = 0.34 atm The equilibrium partial pressures are: PPCl 5 = 0.50 − 0.34 = 0.16 atm, PPCl 3 = PCl 2 = 0.34 atm Kp = K= 28. PPCl 3 × PCl 2 PPCl 5 Kp (RT ) Δn = (0.34)(0.34) = 0.72 atm (0.16) , Δn = 2 − 1 = 1; K p = S8(g) ⇌ 4 S2(g) Kp = 0.72 = 0.017 mol/L (0.08206)(523) PS42 PS8 Initially: PS8 = 1.00 atm and PS2 = 0 atm Change: Because 0.25 atm of S8 remain at equilibrium, 1.00 atm − 0.25 atm = 0.75 atm of S8 must have reacted in order to reach equilibrium. Because there is a 4 : 1 mole ratio between S2 and S8 (from the balanced reaction), 4(0.75 atm) = 3.0 atm of S2 must have been produced when the reaction went to equilibrium (moles and pressure are directly related at constant T and V). Equilibrium: PS8 = 0.25 atm, PS2 = 0 + 3.0 atm = 3.0 atm; Solving for Kp: Kp = 29. (3.0) 4 = 3.2 × 102 atm3 0.25 When solving equilibrium problems, a common method to summarize all the information in the problem is to set up a table. We commonly call this table an ICE table because it summarizes initial concentrations, changes that must occur to reach equilibrium, and equilibrium concentrations (the sum of the initial and change columns). For the change column, we will generally use the variable x, which will be defined as the amount of reactant (or product) that must react to reach equilibrium. In this problem, the reaction must shift right to reach equilibrium because there are no products present initially. Therefore, x is defined as the amount of reactant SO3 that reacts to reach equilibrium, and we use the coefficients in the balanced equation to relate the net change in SO3 to the net change in SO2 and O2. The general ICE table for this problem is: CHAPTER 6 CHEMICAL EQUILIBRIUM 2 SO3(g) Initial Change Equil. ⇌ 2 SO2(g) 155 + O2(g) K= [SO 2 ]2 [O 2 ] [SO 3 ]2 12.0 mol/3.0 L 0 0 Let x mol/L of SO3 react to reach equilibrium. −x → +x +x/2 4.0 − x x x/2 From the problem, we are told that the equilibrium SO2 concentration is 3.0 mol/3.0 L = 1.0 M ([SO2]e = 1.0 M). From the ICE table setup, [SO2]e = x, so x = 1.0. Solving for the other equilibrium concentrations: [SO3]e = 4.0 − x = 4.0 − 1.0 = 3.0 M; [O2] = x/2 = 1.0/2 = 0.50 M. K= [SO 2 ]2 [O 2 ] (1.0 M ) 2 (0.50 M ) = 0.056 mol/L = [SO 3 ]2 (3.0 M ) 2 Alternate method: Fractions in the change column can be avoided (if you want) be defining x differently. If we were to let 2x mol/L of SO3 react to reach equilibrium, then the ICE table setup is: [SO 2 ]2 [O 2 ] O2(g) K= 2 SO3(g) ⇌ 2 SO2(g) + [SO 3 ]2 Initial Change Equil. 0 0 4.0 M Let 2x mol/L of SO3 react to reach equilibrium. −2x → +2x +x 4.0 − 2x 2x x Solving: 2x = [SO2]e = 1.0 M, x = 0.50 M; [SO3]e = 4.0 − 2(0.50) = 3.0 M; [O2]e = x = 0.50 M These are exactly the same equilibrium concentrations as solved for previously, thus K will be the same (as it must be). The moral of the story is to define x in a manner that is most comfortable for you. Your final answer is independent of how you define x initially. Equilibrium Calculations 30. 2 NO(g) ⇋ N2(g) + O2(g) K = [ N 2 ][O 2 ] = 2.4 × 103; Because Δn = 0, K = Kp. [ NO]2 Use the reaction quotient Q to determine which way the reaction shifts to reach equilibrium. For the reaction quotient, initial concentrations given in a problem are used to calculate the value for Q. If Q < K, then the reaction shifts right to reach equilibrium. If Q > K, then the reaction shifts left to reach equilibrium. If Q = K, then the reaction does not shift in either direction because the reaction is at equilibrium. 156 CHAPTER 6 a. [N2] = Q= CHEMICAL EQUILIBRIUM 2.0 mol 2.6 mol 0.024 mol = 2.0 M; [O2] = = 2.6 M; [NO] = = 0.024 M 1.0 L 1.0 L 1.0 L [ N 2 ]0 [ O 2 ]0 (2.0 M ) (2.6 M ) = = 9.0 × 103 2 [ NO]0 (0.024 M ) 2 Q > K, so the reaction shifts left to produce more reactants in order to reach equilibrium. b. [N2] = Q= (0.31 M ) (2.0 M ) = 2.4 × 103 = K; at equilibrium 2 (0.016 M ) c. [N2] = Q= d. Q = 0.62 mol 4.0 mol 0.032 mol = 0.31 M; [O2] = = 2.0 M; [NO] = = 0.016 M 2.0 L 2.0 L 2.0 L 2.4 mol 1.7 mol 0.060 mol = 0.80 M; [O2] = = 0.57 M; [NO] = = 0.020 M 3 .0 L 3.0 L 3 .0 L (0.80 M )(0.57 M ) = 1.1 × 103 < K; reaction shifts right to reach equilibrium. (0.020 M ) 2 PN 2 × PO 2 2 PNO = (0.11 atm)(2.0 atm) = 2.2 × 103 (0.010 atm) 2 Q < Kp (2.4 × 103), so the reaction shifts right to reach equilibrium. 31. e. Q = (0.36 atm)(0.67 atm) = 4.0 × 103 > Kp; reaction shifts left to reach equilibrium. (0.0078 atm) 2 f. (0.51 atm)(0.18 atm) = 2.4 × 103 = Kp; at equilbrium (0.0062 atm) 2 Q= CaCO3(s) ⇌ CaO(s) + CO2(g) Kp = PCO 2 = 1.04 atm We only need to calculate the initial partial pressure of CO2 and compare this value to 1.04 atm. At this temperature, all CO2 will be in the gas phase. a. PV = nRT, Q = PCO 2 = n CO 2 RT V 58.4 g CO 2 0.08206 L atm × × 1173 K 44.01 g/mol K mol = = 50.0 L 2.55 atm > Kp Reaction will shift to the left because Q > Kp; the mass of CaO will decrease. CHAPTER 6 CHEMICAL EQUILIBRIUM b. Q = PCO 2 = 157 (23.76) (0.08206)(1173) = 1.04 atm = Kp (44.01)(50.0) At equilibrium because Q = Kp; mass of CaO will not change. c. Mass of CO2 is the same as in part b. P = 1.04 atm = KP. At equilibrium; mass of CaO will not change. d. Q = PCO 2 = (4.82) (0.08206)(1173) = 0.211 atm < Kp (44.01)(50.0) Reaction will shift to the right because Q < Kp; the mass of CaO will increase. 32. Q = 1.00, which is less than K. The reaction shifts to the right to reach equilibrium. Summarizing the equilibrium problem in a table: SO2(g) Initial Change Equil. + NO2(g) ⇌ SO3(g) + NO(g) K = 3.75 0.800 M 0.800 M 0.800 M 0.800 M x mol/L of SO2 reacts to reach equilibrium !x !x → +x +x 0.800 ! x 0.800 + x 0.800 + x 0.800 ! x Plug the equilibrium concentrations into the equilibrium constant expression: K= [SO 3 ][ NO] (0.800 + x) 2 , 3.75 = ; take the square root of both sides and solve [SO 2 ][ NO 2 ] (0.800 − x) 2 for x: 0.800 + x = 1.94, 0.800 + x = 1.55 ! (1.94)x, (2.94)x = 0.75, x = 0.26 M 0.800 − x The equilibrium concentrations are: [SO3] = [NO] = 0.800 + x = 0.800 + 0.26 = 1.06 M; [SO2] = [NO2] = 0.800 ! x = 0.54 M 33. H2O(g) + Cl2O(g) ⇌ 2 HOCl(g) K = 0.090 = [HOCl]2 [H 2 O][Cl 2 O] a. The initial concentrations of H2O and Cl2O are: 1.0 g H 2 O 1 mol × = 5.6 × 10−2 mol/L; 1 .0 L 18.0 g 2.0 g Cl 2 O 1 mol × = 2.3 × 10−2 mol/L 1 .0 L 86.9 g Because only reactants are present initially, the reaction must proceed to the right to reach equilibrium. Summarizing the problem in a table: 158 CHAPTER 6 H2O(g) + ⇌ Cl2O(g) CHEMICAL EQUILIBRIUM 2 HOCl(g) 5.6 × 10−2 M 2.3 × 10−2 M 0 x mol/L of H2O reacts to reach equilibrium −x −x → +2x 2.3 × 10−2 − x 2x 5.6 × 10−2 − x Initial Change Equil. K = 0.090 = (5.6 × 10 − 2 (2 x) 2 − x)(2.3 × 10 − 2 − x) 1.16 × 10−4 − (7.11 × 10−3)x + (0.090)x2 = 4x2 (3.91)x2 + (7.11 × 10−3)x − 1.16 × 10−4 = 0 (We carried extra significant figures.) Solving using the quadratic formula (see Appendix 1 of the text): x= − 7.11 × 10 −3 ± (5.06 × 10 −5 + 1.81 × 10 −3 )1/ 2 = 4.6 × 10−3 M or −6.4 × 10−3 M 7.82 A negative answer makes no physical sense; we can't have less than nothing. Thus x = 4.6 ×10−3 M. [HOCl] = 2x = 9.2 × 10−3 M; [Cl2O] = 2.3 × 10−2 − x = 0.023 − 0.0046 = 1.8 × 10−2 M [H2O] = 5.6 × 10−2 − x = 0.056 − 0.0046 = 5.1 × 10−2 M b. H2O(g) + Cl2O(g) ⇌ 2 HOCl(g) 0 0 1.0 mol/2.0 L = 0.50 M 2x mol/L of HOCl reacts to reach equilibrium +x ← −2x +x x x 0.50 − 2x Initial Change Equil. K = 0.090 = [HOCl]2 (0.50 − 2 x) 2 = [H 2 O][Cl 2 O] x2 The expression is a perfect square, so we can take the square root of each side: 0.30 = 0.50 − 2 x , (0.30)x = 0.50 − 2x, (2.30)x = 0.50 x x = 0.217 M (We carried extra significant figures.) x = [H2O] = [Cl2O] = 0.217 = 0.22 M; [HOCl] = 0.50 − 2x = 0.50 − 0.434 = 0.07 M CHAPTER 6 34. K= CHEMICAL EQUILIBRIUM 159 [ HF]2 (0.400 M ) 2 = = 320.; 0.200 mol F2/5.00 L = 0.0400 M F2 added [H 2 ][F2 ] (0.0500 M )(0.0100 M ) After F2 has been added, the concentrations of species present are [HF] = 0.400 M, [H2] = [F2] = 0.0500 M. Q = (0.400)2/(0.0500)2 = 64.0; because Q < K, the reaction will shift right to reestablish equilibrium. H2(g) Initial Change Equil. K = 320. = 17.9 = + F2(g) ⇋ 2 HF(g) 0.0500 M 0.400 M 0.0500 M x mol/L of F2 reacts to reach equilibrium −x −x → +2x 0.0500 − x 0.0500 − x 0.400 + 2x (0.400 + 2 x) 2 ; taking the square root of each side: (0.0500 − x) 2 0.400 + 2 x , 0.895 − (17.9)x = 0.400 + 2x, (19.9)x = 0.495, x = 0.0249 mol/L 0.0500 − x [HF] = 0.400 + 2(0.0249) = 0.450 M; [H2] = [F2] = 0.0500 − 0.0249 = 0.0251 M 35. 2 SO2(g) Initial Change Equil. Kp = 0.25 = + O2(g) ⇋ 2 SO3(g) Kp = 0.25 0.50 atm 0.50 atm 0 2x atm of SO2 reacts to reach equilibrium −2x −x → +2x 0.50 − x 2x 0.50 − 2x 2 PSO 3 2 PSO × PO 2 2 = (2 x) 2 (0.50 − 2 x) 2 (0.50 − x) This will give a cubic equation. Graphing calculators can be used to solve this expression. If you don’t have a graphing calculator, an alternative method for solving a cubic equation is to use the method of successive approximations (see Appendix 1 of the text). The first step is to guess a value for x. Because the value of K is small (K < 1), not much of the forward reaction will occur to reach equilibrium. This tells us that x is small. Let’s guess that x = 0.050 atm. Now we take this estimated value for x and substitute it into the equation everywhere that x appears except for one. For equilibrium problems, we will substitute the estimated value for x into the denominator, and then solve for the numerator value of x. We continue this process until the estimated value of x and the calculated value of x converge on the same number. This is the same answer we would get if we were to solve the cubic equation exactly. Applying the method of successive approximations and carrying extra significant figures: 4x2 4x2 = = 0.25 , x = 0.067 [0.50 − 2(0.050)]2 − [0.50 − (0.050)] (0.40) 2 (0.45) 160 CHAPTER 6 CHEMICAL EQUILIBRIUM 4x2 4x2 = = 0.25 , x = 0.060 [0.50 − 2(0.067)]2 [0.50 − (0.067)] (0.366) 2 (0.433) 4x2 4x2 = 0.25, x = 0.063; = 0.25, (0.38) 2 (0.44) (0.374) 2 (0.437) x = 0.062 The next trial gives the same value for x = 0.062 atm. We are done except for determining the equilibrium concentrations. They are: PSO 2 = 0.50 − 2x = 0.50 − 2(0.062) = 0.376 = 0.38 atm PO 2 = 0.50 − x = 0.438 = 0.44 atm; PSO 3 = 2x = 0.124 = 0.12 atm 36. Because only reactants are present initially, the reaction must proceed to the right to reach equilibrium. Summarizing the problem in a table: N2(g) Initial Change Equil. Kp = 0.050 = + O2(g) ⇌ 2 NO(g) Kp = 0.050 0.80 atm 0.20 atm 0 x atm of N2 reacts to reach equilibrium −x −x +2x 0.20 − x 2x 0.80 − x 2 PNO (2 x) 2 = , 0.050[0.16 − (1.00)x + x2] = 4x2 (0.80 − x)(0.20 − x) PN 2 × PO 2 4x2 = 8.0 × 10−3 − (0.050)x + (0.050)x2, (3.95)x2 + (0.050)x − 8.0 × 10−3 = 0 Solving using the quadratic formula (see Appendix 1 of the text): x= − b ± (b 2 − 4ac)1/ 2 − 0.050 ± [(0.050)]2 − 4(3.95)(−8.0 × 10 −3 )]1/ 2 = 2a 2(3.95) x = 3.9 × 10−2 atm or x = −5.2 × 10−2 atm; only x = 3.9 × 10−2 atm makes sense (x cannot be negative), so the equilibrium NO concentration is: PNO = 2x = 2(3.9 × 10−2 atm) = 7.8 × 10−2 atm 37. The assumption comes from the value of K being much less than 1. For these reactions, the equilibrium mixture will not have a lot of products present; mostly reactants are present at equilibrium. If we define the change that must occur in terms of x as the amount (molarity or partial pressure) of a reactant that must react to reach equilibrium, then x must be a small number because K is a very small number. We want to know the value of x in order to solve the problem, so we don’t assume x = 0. Instead, we concentrate on the equilibrium row in the ICE table. Those reactants (or products) that have equilibrium concentrations in the form of 0.10 – x or 0.25 + x or 3.5 – 3x, etc., is where an important assumption can be made. The CHAPTER 6 CHEMICAL EQUILIBRIUM 161 assumption is that because K << 1, x will be small (x << 1), and when we add x or subtract x from some initial concentration, it will make little or no difference. That is, we assume that 0.10 – x ≈ 0.10 or 0.25 + x ≈ 0.25 or 3.5 – 3x ≈ 3.5, etc.; we assume that the initial concentration of a substance is equal to the final concentration. This assumption makes the math much easier and usually gives a value of x that is well within 5% of the true value of x (we get about the same answer with a lot less work). We check the assumptions for validity using the 5% rule. From doing a lot of these calculations, it is found that when an assumption such as 0.20 – x ≈ 0.20 is made, if x is less than 5% of the number the assumption was made against, then our final answer is within acceptable error limits of the true value of x (as determined when the equation is solved exactly). For our example above (0.20 – x ≈ 0.20), if (x/0.20) × 100 ≤ 5%, then our assumption is valid by the 5% rule. If the error is greater than 5%, then we must solve the equation exactly or use a math trick called the method of successive approximations. See Appendix 1 for details regarding the method of successive approximations, as well as for a review in solving quadratic equations exactly. 38. a. The reaction must proceed to products to reach equilibrium because only reactants are present initially. Summarizing the problem in a table: 2 NOCl(g) Initial Change Equil. ⇌ 2 NO(g) + Cl2(g) K = 1.6 × 10−5 2.0 mol 0 0 = 1.0 M 2.0 L 2x mol/L of NOCl reacts to reach equilibrium −2x → +2x +x 1.0 − 2x 2x x −5 K = 1.6 × 10 = [ NO]2 [Cl 2 ] [ NOCl]2 = (2 x) 2 ( x) (1.0 − 2 x) 2 If we assume that 1.0 − 2x ≈ 1.0 (from the small size of K, we know that the product concentrations will be small, so x will be small), then: 1.6 × 10−5 = 4x 3 , x = 1.6 × 10−2 M; now we must check the assumption. 1 .0 2 1.0 − 2x = 1.0 − 2(0.016) = 0.97 = 1.0 (to proper significant figures) Our error is about 3%, that is, 2x is 3.2% of 1.0 M. Generally, if the error we introduce by making simplifying assumptions is less than 5%, we go no further, the assumption is said to be valid. We call this the 5% rule. Solving for the equilibrium concentrations: [NO] = 2x = 0.032 M; [Cl2] = x = 0.016 M; [NOCl] = 1.0 − 2x = 0.97 M ≈ 1.0 M Note: If we were to solve this cubic equation exactly (a longer process), we get x = 0.016. This is the exact same answer we determined by making a simplifying assumption. We saved time and energy. Whenever K is a very small value, always make 162 CHAPTER 6 CHEMICAL EQUILIBRIUM the assumption that x is small. If the assumption introduces an error of less than 5%, then the answer you calculated making the assumption will be considered the correct answer. b. There is a little trick we can use to solve this problem in order to avoid solving a cubic equation. Because K for this reaction is very small (K << 1), the reaction will contain mostly reactants at equilibrium (the equilibrium position lies far to the left). We will let the products react to completion by the reverse reaction, and then we will solve the forward equilibrium problem to determine the equilibrium concentrations. Summarizing these steps in a table: 2 NOCl(g) Before Change After Change Equil. ⇌ 2 NO(g) + Cl2(g) K = 1.6 × 10−5 1.0 M 0 2.0 M (K is small, reactants dominate.) Let 1.0 mol/L Cl2 react completely. +2.0 ← −2.0 −1.0 React completely 2.0 0 0 New initial conditions 2x mol/L of NOCl reacts to reach equilibrium −2x → +2x +x 2.0 − 2x 2x x K = 1.6 × 10−5 = (2 x) 2 ( x) 4x3 ≈ ( 2.0 − 2 x) 2 2.0 2 (assuming 2.0 − 2x ≈ 2.0) x3 = 1.6 × 10−5, x = 2.5 × 10−2 M; assumption good by the 5% rule (2x is 2.5% of 2.0). [NOCl] = 2.0 − 0.050 = 1.95 M = 2.0 M; [NO] = 0.050 M; [Cl2] = 0.025 M Note: If we do not break this problem into two parts (a stoichiometric part and an equilibrium part), then we are faced with solving a cubic equation. The setup would be: Initial Change Equil. 1.6 × 10−5 = 2 NOCl ⇌ 2 NO 0 +2y 2y ← 2.0 M −2y 2.0 − 2y + Cl2 1.0 M −y 1.0 − y (2.0 − 2 y ) 2 (1.0 − y ) (2 y ) 2 If we say that y is small to simplify the problem, then: 1.6 × 10−5 = 2 .0 2 ; We get y = 250. This is impossible! 4y2 To solve this equation, we cannot make any simplifying assumptions; we have to find a way to solve a cubic equation. Or we can use some chemical common sense and solve the problem the easier way. CHAPTER 6 CHEMICAL EQUILIBRIUM c. 2 NOCl(g) Initial Change Equil. 1.6 × 10−5 = ⇌ 2 NO(g) 163 + Cl2(g) 1.0 M 0 1.0 M 2x mol/L NOCl reacts to reach equilibrium −2x → +2x +x 1.0 + 2x x 1.0 − 2x (1.0 + 2 x) 2 ( x) (1.0) 2 (x) ≈ (1.0 − 2 x) 2 (2.0) 2 (assuming 2x << 1.0) x = 1.6 × 10−5 M; Assumptions are great (2x is 3.2 × 10−3% of 1.0). [Cl2] = 1.6 × 10−5 M and [NOCl] = [NO] = 1.0 M d. 2 NOCl(g) Before Change After Change Equil. 1.6 × 10−5 = ⇌ 2 NO(g) + Cl2(g) 1.0 M 0 3.0 M Let 1.0 mol/L Cl2 react completely. +2.0 ← −2.0 −1.0 React completely 2.0 1.0 0 New initial 2x mol/L NOCl reacts to reach equilibrium −2x → +2x +x 1.0 + 2x x 2.0 − 2x x (1.0 + 2 x) 2 ( x) ≈ ; solving: x = 6.4 × 10−5 M 2 4.0 (2.0 − 2 x) Assumptions are great (2x is 1.3 × 10−2% of 1.0). [Cl2] = 6.4 × 10−5 M; [NOCl] = 2.0 M; [NO] = 1.0 M e. 2 NOCl(g) Before Change After Change Equil. 1.6 × 10−5 = ⇌ 2 NO(g) + Cl2(g) 2.0 M 1.0 M 2.0 M Let 1.0 mol/L Cl2 react completely. +2.0 ← −2.0 −1.0 React completely 4.0 0 0 New initial 2x mol/L NOCl reacts to reach equilibrium −2x → +2x +x 2x x 4.0 − 2x (2 x) 2 ( x) 4 x3 ≈ , x = 4.0 × 10−2 M; assumption good (2% error). 16 ( 4 .0 − 2 x ) 2 [Cl2] = 0.040 M; [NO] = 0.080 M; [NOCl] = 4.0 − 2(0.040) = 3.92 M ≈ 3.9 M 164 CHAPTER 6 f. 2 NOCl(g) 2 NO(g) + Cl2(g) 1.00 M 1.00 M 1.00 M Let 1.00 mol/L NO react completely (the limiting reagent). +1.00 ← −1.00 −0.500 React completely 2.00 0 0.50 New initial 2x mol/L NOCl reacts to reach equilibrium → +2x +x −2x 2.00 − 2x 2x 0.50 + x Before Change After Change Equil. K= ⇌ CHEMICAL EQUILIBRIUM (2 x) 2 (0.50 + x) 4 x 2 (0.50) ≈ = 1.6 × 10−5, x = 5.7 × 10−3 M 2 2 (2.00 − 2 x) (2.00) Assumptions are good (x is 1.1% of 0.50). [NO] = 2x = 1.1 × 10−2 M; [Cl2] = 0.50 + 0.0057 = 0.51 M [NOCl] = 2.00 − 2(0.0057) = 1.99 M 39. 2 CO2(g) Initial Change Equil. ⇌ 2 CO(g) + O2(g) K= [CO] 2 [O 2 ] [CO 2 ] 2 = 2.0 × 10−6 2.0 mol/5.0 L 0 0 2x mol/L of CO2 reacts to reach equilibrium −2x → +2x +x 2x x 0.40 − 2x K = 2.0 × 10−6 = 2.0 × 10-6 ≈ [CO] 2 [O 2 ] [CO 2 ] 2 = (2 x) 2 ( x) ; assuming 2x << 0.40 (K is small, so x is (0.40 − 2 x) 2 small.) 4x3 4 x3 −6 , 2 .0 × 10 = , x = 4.3 × 10−3 M 2 0 . 16 (0.40) Checking assumption: 2(4.3 × 10 −3 ) × 100 = 2.2%; assumption is valid by the 5% rule. 0.40 [CO2] = 0.40 − 2x = 0.40 − 2(4.3 × 10−3) = 0.39 M [CO] = 2x = 2(4.3 × 10−3) = 8.6 × 10−3 M; [O2] = x = 4.3 × 10−3 M 40. a. The reaction must proceed to products to reach equilibrium because no product is present initially. Summarizing the problem in a table where x atm of N2O4 reacts to reach equilibrium: Initial Change Equil. N2O4(g) ⇌ 4.5 atm −x 4.5 − x → 2 NO2(g) 0 +2x 2x Kp = 0.25 CHAPTER 6 CHEMICAL EQUILIBRIUM Kp = 2 PNO 2 PN 2O 4 165 (2 x) 2 = 0.25, 4x2 = 1.125 − (0.25)x, 4x2 + (0.25)x − 1.125 = 0 4 .5 − x = We carried extra significant figures in this expression (as will be typical when we solve an expression using the quadratic formula). Solving using the quadratic formula (see Appendix 1 of text): x= − 0.25 ± [(0.25) 2 − 4(4)(−1.125)]1/ 2 − 0.25 ± 4.25 = , x = 0.50 2( 4) 8 (Other value is negative.) PNO 2 = 2x = 1.0 atm; PN 2O 4 = 4.5 − x = 4.0 atm b. The reaction must shift to reactants (shift left) to reach equilibrium. N2O4(g) Initial Change Equil. Kp = 0 +x x ⇌ 2 NO2(g) ← 9.0 atm −2x 9.0 − 2x (9.0 − 2 x) 2 = 0.25, 4x2 − (36.25)x + 81 = 0 (carrying extra sig. figs.) x Solving: x = − (−36.25) ± [(−36.25) 2 − 4(4)(81)]1/ 2 , x = 4.0 atm 2( 4) The other value, 5.1, is impossible. PN 2O 4 = x = 4.0 atm; PNO 2 = 9.0 − 2x = 1.0 atm c. No, we get the same equilibrium position starting with either pure N2O4 or pure NO2 in stoichiometric amounts. d. From part a, the equilibrium partial pressures are PNO 2 = 1.0 atm and PN 2O 4 = 4.0 atm. Halving the container volume will increase each of these partial pressures by a factor of 2. Q = (2.0)2/8.0 = 0.50. Because Q > Kp, the reaction will shift left to reestablish equilibrium. N2O4(g) ⇌ 2 NO2(g) Initial New Initial Change Equil. Kp = 4.0 atm 8.0 +x 8.0 + x ← 1.0 atm 2.0 −2x 2.0 − 2x ( 2.0 − 2 x) 2 = 0.25, 4x2 − (8.25)x + 2.0 = 0 (carrying extra sig. figs.) 8 .0 + x 166 CHAPTER 6 CHEMICAL EQUILIBRIUM Solving using the quadratic formula: x = 0.28 atm PN 2O 4 = 8.0 + x = 8.3 atm; PNO 2 = 2.0 − 2x = 1.4 atm 41. CaCO3(s) ⇌ CaO(s) + CO2(g) Kp = 1.16 = PCO 2 Some of the 20.0 g of CaCO3 will react to reach equilibrium. The amount that reacts is the quantity of CaCO3 required to produce a CO2 pressure of 1.16 atm (from the Kp expression). PCO 2 V 1.16 atm × 10.0 L = 0.132 mol CO2 0.08206 L atm RT × 1073 K K mol 1 mol CaCO3 100.09 g Mass CaCO3 reacted = 0.132 mol CO2 × × = 13.2 g CaCO3 mol CO 2 mol CaCO3 13.2 g Mass % CaCO3 reacted = × 100 = 66.0% 20.0 g n CO 2 = 42. = This is a typical equilibrium problem except that the reaction contains a solid. Whenever solids and liquids are present, we basically ignore them in the equilibrium problem. NH4OCONH2(s) ⇌ 2 NH3(g) + CO2(g) Kp = 2.9 × 10−3 − 0 0 Some NH4OCONH2 decomposes to produce 2x atm of NH3 and x atm of CO2. Change − → +2x +x Equil. − 2x x Initial 2 × PCO 2 = (2x)2(x) = 4x3 Kp = 2.9 × 10−3 = PNH 3 ⎛ 2.9 × 10 −3 ⎞ ⎟ x = ⎜⎜ ⎟ 4 ⎝ ⎠ 1/ 3 = 9.0 × 10−2 atm; PNH 3 = 2x = 0.18 atm; PCO 2 = x = 9.0 × 10−2 atm Ptotal = PNH3 + PCO 2 = 0.18 atm + 0.090 atm = 0.27 atm −1 43. ⎞ 4.5 × 109 L ⎛ 0.08206 L atm ⎜⎜ × 373 K ⎟⎟ , where Δn = 1 − 2 = −1 a. Kp = K(RT) = mol K mol ⎝ ⎠ Δn Kp = 1.5 × 108 atm−1 b. Kp is so large that at equilibrium we will have almost all COCl2. Assume Ptotal ≈ PCOCl 2 ≈ 5.0 atm. CHAPTER 6 CHEMICAL EQUILIBRIUM CO(g) + Cl2(g) Initial ⇌ 167 COCl2(g) Kp = 1.5 × 108 0 0 5.0 atm x atm COCl2 reacts to reach equilibrium +x +x ← −x x x 5.0 − x Change Equil. Kp = 1.5 × 108 = 5.0 − x 5.0 ≈ 2 x2 x (Assuming 5.0 − x ≈ 5.0.) Solving: x = 1.8 × 10-4 atm. Check assumptions: 5.0 − x = 5.0 − 1.8 × 10−4 = 5.0 atm. Assumptions are good (well within the 5% rule). PCO = PCl2 = 1.8 × 10−4 atm and PCOCl2 = 5.0 atm Fe3+(aq) 44. Before Change After Change Equil. + SCN−(aq) ⇌ FeSCN2+(aq) K = 1.1 × 103 0.10 M 0 0.020 M Let 0.020 mol/L Fe3+ react completely (K is large; products dominate). −0.020 −0.020 → +0.020 React completely 0 0.08 0.020 New initial x mol/L FeSCN2+ reacts to reach equilibrium +x +x ← −x x 0.08 + x 0.020 − x K = 1.1 × 103 = [FeSCN 2 + ] 0.020 − x 0.020 = ≈ 3+ − ( x)(0.08 + x) (0.08) x [Fe ][SCN ] x = 2 × 10−4 M; x is 1% of 0.020. Assumptions are good by the 5% rule. x = [Fe3+] = 2 × 10−4 M; [SCN-] = 0.08 + 2 × 10−4 = 0.08 M [FeSCN2+] = 0.020 − 2 × 10−4 = 0.020 M Le Chatelier's Principle 45. Only statement d is correct. Addition of a catalyst has no effect on the equilibrium position; the reaction just reaches equilibrium more quickly. Statement a is false for reactants that are either solids or liquids (adding more of these has no effect on the equilibrium). Statement b is false always. If temperature remains constant, then the value of K is constant. Statement c is false for exothermic reactions where an increase in temperature decreases the value of K. 46. A change in volume will change the partial pressure of all reactants and products by the same factor. The shift in equilibrium depends on the number of gaseous particles on each side. An increase in volume will shift the equilibrium to the side with the greater number of particles in the gas phase. A decrease in volume will favor the side with lesser gas-phase particles. If there are the same number of gas-phase particles on each side of the reaction, a change in volume will not shift the equilibrium. 168 CHAPTER 6 CHEMICAL EQUILIBRIUM When we change the pressure by adding an unreactive gas, we do not change the partial pressures (or concentrations) of any of the substances in equilibrium with each other since the volume of the container did not change. If the partial pressures (and concentrations) are unchanged, the reaction is still at equilibrium. 47. a. No effect; adding more of a pure solid or pure liquid has no effect on the equilibrium position. b. Shifts left; HF(g) will be removed by reaction with the glass. As HF(g) is removed, the reaction will shift left to produce more HF(g). c. Shifts right; as H2O(g) is removed, the reaction will shift right to produce more H2O(g). 48. a. Doubling the volume will decrease all concentrations by a factor of one-half. 1 [FeSCN 2 + ] eq 2 = 2 K, Q > K Q= ⎛1 ⎞⎛ 1 ⎞ − 3+ ⎜ [Fe ] eq ⎟⎜ [SCN ] eq ⎟ ⎝2 ⎠⎝ 2 ⎠ The reaction will shift to the left to reestablish equilibrium. b. Adding Ag+ will remove SCN- through the formation of AgSCN(s). The reaction will shift to the left to produce more SCN-. c. Removing Fe3+ as Fe(OH)3(s) will shift the reaction to the left to produce more Fe3+. d. Reaction shifts to the right as Fe3+ is added. 49. H+ + OH− → H2O; sodium hydroxide (NaOH) will react with the H+ on the product side of the reaction. This effectively removes H+ from the equilibrium, which will shift the reaction to the right to produce more H+ and CrO42-. Because more CrO42− is produced, the solution turns yellow. 50. a. The moles of SO3 will increase because the reaction will shift left to use up some of the added O2(g). b. Increase; because there are fewer reactant gas molecules than product gas molecules, the reaction shifts left with a decrease in volume. c. No effect; the partial pressures of sulfur trioxide, sulfur dioxide, and oxygen are unchanged, so the reaction is still at equilibrium. d. Increase; Heat + 2 SO3 ⇌ 2 SO2 + O2; decreasing T will remove heat, shifting this endothermic reaction to the left. e. Decrease CHAPTER 6 51. CHEMICAL EQUILIBRIUM a. Right b. Right 169 c. No effect; He(g) is neither a reactant nor a product. d. Left; because the reaction is exothermic, heat is a product: CO(g) + H2O(g) → H2(g) + CO2(g) + heat Increasing T will add heat. The equilibrium shifts to the left to use up the added heat. e. No effect; because the moles of gaseous reactants equals the moles of gaseous products (2 mol versus 2 mol), a change in volume will have no effect on the equilibrium. 52. a. Shift to left b. Shift to right; because the reaction is endothermic (heat is a reactant), an increase in temperature will shift the equilibrium to the right. c. No effect d. Shift to right e. Shift to right; because there are more gaseous product molecules than gaseous reactant molecules, the equilibrium will shift right with an increase in volume. 53. a. Left b. Right c. Left d. No effect (reactant and product concentrations are unchanged) e. No effect; because there are equal numbers of product and reactant gas molecules, a change in volume has no effect on the equilibrium position. f. 54. Right; a decrease in temperature will shift the equilibrium to the right because heat is a product in this reaction (as is true in all exothermic reactions). When the volume of a reaction container is increased, the reaction itself will want to increase its own volume by shifting to the side of the reaction that contains the most molecules of gas. When the molecules of gas are equal on both sides of the reaction, then the reaction will remain at equilibrium no matter what happens to the volume of the container. a. Reaction shifts left (to reactants) because the reactants contain 4 molecules of gas compared with 2 molecules of gas on the product side. b. Reaction shifts right (to products) because there are more product molecules of gas (2) than reactant molecules (1). c. No change because there are equal reactant and product molecules of gas. d. Reaction shifts right. e. Reaction shifts right to produce more CO2(g). One can ignore the solids and only concentrate on the gases because gases occupy a relatively huge volume compared with solids. We make the same assumption when liquids are present (only worry about the gas molecules). 170 CHAPTER 6 CHEMICAL EQUILIBRIUM 55. An endothermic reaction, where heat is a reactant, will shift right to products with an increase in temperature. The amount of NH3(g) will increase as the reaction shifts right, so the smell of ammonia will increase. 56. As temperature increases, the value of K decreases. This is consistent with an exothermic reaction. In an exothermic reaction, heat is a product, and an increase in temperature shifts the equilibrium to the reactant side (as well as lowering the value of K). Additional Exercises 57. a. N2(g) + O2(g) ⇌ 2 NO(g) Kp = 1 × 10−31 = 2 2 PNO PNO = (0.8)(0.2) PN 2 × PO 2 PNO = 1 × 10−16 atm In 1.0 cm3 of air: nNO = PV (1 × 10 −16 atm)(1.0 × 10 −3 L) = = 4 × 10−21 mol NO RT ⎛ 0.08206 L atm ⎞ ⎟⎟ (298 K ) ⎜⎜ K mol ⎠ ⎝ 4 × 10 −21 mol NO 6.02 × 10 23 molecules 2 × 103 molecules NO × = mol NO cm 3 cm 3 b. There is more NO in the atmosphere than we would expect from the value of K. The answer must lie in the rates of the reaction. At 25°C, the rates of both reactions: N2 + O2 → 2 NO and 2 NO → N2 + O2 are so slow that they are essentially zero. Very strong bonds must be broken; the activation energy is very high. Therefore, the reaction essentially doesn’t occur at low temperatures. Nitric oxide, however, can be produced in high-energy or hightemperature environments because the production of NO is endothermic. In nature, some NO is produced by lightning and the primary manmade source is automobiles. At these high temperatures, K will increase, and the rates of the reaction will also increase, resulting in a higher production of NO. Once the NO gets into a more normal temperature environment, it doesn’t go back to N2 and O2 because of the slow rate. 58. a. Na2O(s) ⇌ 2 Na(l) + 1/2 O2(g) K1 2 Na(l) + O2(g) ⇌ Na2O2(s) 1/K3 ___________________________________________________ Na2O(s) + 1/2 O2(g) ⇌ Na2O2(s) K= 2 × 10 −25 = 4 × 103 L1/2/mol1/2 − 29 5 × 10 K = (K1)(1/K3) CHAPTER 6 CHEMICAL EQUILIBRIUM 171 NaO(g) ⇌ Na(l) + 1/2 O2(g) b. Na2O(s) ⇌ 2 Na(l) + 1/2 O2(g) K2 K1 1/K3 2 Na(l) + O2(g) ⇌ Na2O2(s) ________________________________________________________________ K = K2(K1)(1/K3) = 8 × 10−2 L/mol NaO(g) + Na2O(s) ⇌ Na2O2(s) + Na(l) 2 NaO(g) ⇌ 2 Na(l) + O2(g) c. K 22 1/K3 2 Na(l) + O2(g) ⇌ Na2O2(s) ________________________________________________________________ K = K 22 (1/K 3 ) = 8 × 1018 L2/mol2 2 NaO(g) ⇌ Na2O2(s) 59. 60. O(g) + NO(g) ⇌ NO2(g) K = 1/6.8 × 10−49 = 1.5 × 1048 O2(g) + O(g) ⇌ O3(g) K = (1.5 × 1048)(1.7 × 1033) = 2.6 × 1081 L/mol NO2(g) + O2(g) ⇌ NO(g) + O3(g) K = 1/5.8 × 10−34 = 1.7 × 1033 __________________________________________________________________________ ⇌ 2 A(g) + D(g) ⇌ K1 = (1/3.50)2 = 8.16 × 10 −2 2 A(g) + 2 B(g) 2 C(g) C(g) K2 = 7.10 ____________________________________________________________ C(g) + D(g) ⇌ 2 B(g) K = K1 × K2 = 0.579 Kp = K(RT)Δn, Δn = 2 – (1 + 1) = 0; because Δn = 0, Kp = K = 0.579. C(g) + D(g) Initial 1.50 atm Equil. 1.50 – x 0.579 = K = 1.50 atm 1.50 – x ⇌ 2 B(g) 0 2x (2 x) 2 (2 x) 2 = (1.50 − x)(1.50 − x) (1.50 − x) 2 2x = (0.579)1/2 = 0.761, x = 0.413 atm 1.50 − x PB (at equilibrium) = 2x = 2(0.413) = 0.826 atm Ptotal = PC + PD + PB = 2(1.50 – 0.413) + 0.826 = 3.00 atm PB = χBPtotal, χB = PB 0.826 atm = = 0.275 Ptotal 3.00 atm 172 CHAPTER 6 61. 3 H2(g) Initial + N2(g) ⇌ CHEMICAL EQUILIBRIUM 2 NH3(g) [H2]0 [N2]0 0 x mol/L of N2 reacts to reach equilibrium −3x −x → +2x [H2]0 − 3x [N2]0 −x 2x Change Equil. From the problem: [NH3]e = 4.0 M = 2x, x = 2.0 M; [H2]e = 5.0 M = [H2]0 −3x; [N2]e = 8.0 M = [N2]0 − x 5.0 M = [H2]0 − 3(2.0 M), [H2]0 = 11.0 M; 8.0 M = [N2]0 − 2.0 M, [N2]0 = 10.0 M N2(g) 62. + 3 H2(g) Kp = 5.3 × 105 2 NH3(g) P0 = initial pressure of NH3 0 0 P0 2x atm of NH3 reacts to reach equilibrium +x +3x ← −2x x 3x P0 − 2x Initial Change Equil. From problem, P0 − 2x = Kp = ⇌ P0 , so P0 = (4.00)x 2.00 [(4.00) x − 2 x]2 [(2.00) x]2 (4.00) x 2 4.00 = = = = 5.3 × 105, x = 5.3 × 10−4 atm 3 3 4 ( x)(3x) ( x)(3x) 27 x 27 x 2 P0 = (4.00)x = 4.00(5.3 × 10−4 atm) = 2.1 × 10−3 atm 63. a. Initial Change Equil. PCl5(g) ⇌ PCl3(g) P0 −x P0 − x → 0 +x x + Cl2(g) 0 +x x Kp = (PPCl3 × PCl 2 ) / PPCl5 P0 = initial PCl5 pressure Ptotal = P0 − x + x + x = P0 + x = 358.7 torr P0 = n PCl5 RT V 2.4156 g 0.08206 L atm × × 523.2 K 208.22 g/mol K mol = = 0.2490 atm (or 189.2 torr) 2.000 L x = Ptotal − P0 = 358.7 − 189.2 = 169.5 torr PPCl 3 = PCl 2 = 169.5 torr × 1 atm = 0.2230 atm 760 torr CHAPTER 6 CHEMICAL EQUILIBRIUM PPCl5 = 189.2 − 169.5 = 19.7 torr × Kp = b. 173 1 atm = 0.0259 atm 760 torr (0.2230) 2 = 1.92 atm 0.0259 PCl 2 = n Cl 2 RT V = 0.250 × 0.08206 × 523.2 = 5.37 atm Cl2 added 2.000 PCl5(g) ⇌ PCl3(g) + Cl2(g) Initial 0.0259 atm 0.2230 atm 0.2230 atm Adding 0.250 mol Cl2 increases PCl 2 by 5.37 atm. Initial' Change After Change Equil. 0.0259 +0.2230 ← 0.2489 −x → 0.2489 − x 0.2230 −0.2230 0 +x x 5.59 −0.2230 5.37 +x 5.37 + x (from a) React completely New initial (5.37 + x)( x) = 1.92, x2 + (7.29)x − 0.478 = 0 (0.2489 − x) Solving using the quadratic formula: x = 0.0650 atm PPCl3 = 0.0650 atm; PPCl5 = 0.2489 − 0.0650 = 0.1839 atm; PCl 2 = 5.37 + 0.0650 = 5.44 atm 64. Initial Change Equil. SO2Cl2(g) ⇌ Cl2(g) P0 −x P0 - x → 0 +x x + SO2(g) 0 +x x P0 = initial pressure of SO2Cl2 Ptotal = 0.900 atm = P0 − x + x + x = P0 + x x × 100 = 12.5, P0 = (8.00)x P0 Solving: 0.900 = P0 + x = (9.00)x, x = 0.100 atm x = 0.100 atm = PCl 2 = PSO 2 ; P0 − x = 0.800 − 0.100 = 0.700 atm = PSO 2Cl 2 Kp = PCl 2 × PSO 2 PSO 2Cl 2 = (0.100) 2 = 1.43 × 10−2 atm 0.700 174 65. CHAPTER 6 N2O4(g) ⇌ 2 NO2(g) Kp = 2 PNO 2 PN 2O 4 = CHEMICAL EQUILIBRIUM (1.20) 2 = 4.2 atm 0.34 Doubling the volume decreases each partial pressure by a factor of 2 (P = nRT/V). PNO 2 = 0.600 atm and PN 2O 4 = 0.17 atm are the new partial pressures. Q= (0.600) 2 = 2.1, Q < K; equilibrium will shift to the right. 0.17 N2O4(g) ⇌ Initial 0.17 atm Equil. 0.17 − x Kp = 4.2 = 2 NO2(g) 0.600 atm 0.600 + 2x (0.600 + 2 x) 2 , 4x2 + (6.6)x − 0.354 = 0 (carrying extra sig. figs.) (0.17 − x) Solving using the quadratic formula: x = 0.052 atm PNO 2 = 0.600 + 2(0.052) = 0.704 atm; PN 2O 4 = 0.17 - 0.052 = 0.12 atm 66. a. PPCl5 = b. n PCl5 RT V PCl5(g) 2.450 g PCl 5 0.08206 L atm × × 600. K 208.22 g/mol K mol = = 1.16 atm 0.500 L ⇌ PCl3(g) + Cl2(g) Kp = Initial 1.16 atm 0 0 x atm of PCl5 reacts to reach equilibrium Change −x → +x +x Equil. 1.16 − x x x Kp = x2 = 11.5, x2 + (11.5)x − 13.3 = 0 1.16 − x Using the quadratic formula: x = 1.06 atm PPCl5 = 1.16 − 1.06 = 0.10 atm c. PPCl3 = PCl 2 = 1.06 atm; PPCl5 = 0.10 atm Ptotal = PPCl5 + PPCl3 + PCl 2 = 0.10 + 1.06 + 1.06 = 2.22 atm d. Percent dissociation = x 1.06 × 100 = × 100 = 91.4% 1.16 1.16 PPCl3 × PCl 2 PPCl5 = 11.5 CHAPTER 6 67. CHEMICAL EQUILIBRIUM a. 2 NaHCO3(s) ⇌ Na2CO3(s) 175 + CO2(g) + H2O(g) Kp = 0.25 − − 0 0 NaHCO3(s) decomposes to form x atm each of CO2(g) and H2O(g) at equilibrium. +x Change − → − +x Equil. − − x x Initial Kp = 0.25 = PCO 2 × PH 2O , 0.25 = x2, x = PCO 2 = PH 2O = 0.50 atm b. n CO 2 = PCO 2 V RT = (0.50 atm)(1.00 L) = 1.5 × 10−2 mol CO2 (0.08206 L atm K −1 mol −1 )(398 K ) Mass of Na2CO3 produced: 1.5 × 10−2 mol CO2 × 1 mol Na 2 CO 3 106.0 g Na 2 CO 3 × = 1.6 g Na2CO3 mol CO 2 mol Na 2 CO 3 Mass of NaHCO3 reacted: 1.5 × 10−2 mol CO2 × 2 mol NaHCO3 84.01 g NaHCO3 × = 2.5 g NaHCO3 mol CO 2 mol Mass of NaHCO3 remaining = 10.0 − 2.5 = 7.5 g c. 10.0 g NaHCO3 × 1 mol NaHCO 3 1 mol CO 2 × = 5.95 × 10−2 mol CO2 84.01 g NaHCO 3 2 mol NaHCO 3 When all the NaHCO3 has just been consumed, we will have 5.95 × 10−2 mol CO2 gas at a pressure of 0.50 atm (from a). V = 68. nRT (5.95 × 10 −2 mol)(0.08206 L atm K −1 mol −1 )(398 K ) = = 3.9 L P 0.50 atm a. 2 AsH3(g) Initial Equil. 392.0 torr 392.0 - 2x ⇌ 2 As(s) − − + 3 H2(g) 0 3x Ptotal = 488.0 torr = 392.0 − 2x + 3x, x = 96.0 torr PH 2 = 3x = 3(96.0) = 288 torr; PAsH3 = 392.0 − 2(96.0) = 200.0 torr b. Kp = (PH 2 ) 3 (PAsH3 ) 2 = ( 288) 3 1 atm = 0.786 atm = 597 torr × 2 760 torr (200.0) 176 CHAPTER 6 69. NH3(g) + H2S(g) ⇌ NH4HS(s) CHEMICAL EQUILIBRIUM K = 400. = 1 [ NH 3 ][H 2S] 2.00 mol 2.00 mol − 5.00 L 5.00 L x mol/L of NH3 reacts to reach equilibrium Change !x !x − 0.400 ! x − Equil. 0.400 ! x Initial 1 ⎛ 1 ⎞ K = 400. = , 0.400 ! x = ⎜ ⎟ (0.400 − x)(0.400 − x) ⎝ 400. ⎠ Mol NH4HS(s) produced = 5.00 L × 1/ 2 = 0.0500, x = 0.350 M 0.350 mol NH 3 1 mol NH 4 HS × = 1.75 mol L mol NH 3 Total mol NH4HS(s) = 2.00 mol initially + 1.75 mol produced = 3.75 mol total 3.75 mol NH4HS × 51.12 g NH 4 HS = 192 g NH4HS mol NH 4 HS [H2S]e = 0.400 M – x = 0.400 M – 0.350 M = 0.050 M H2S PH 2S = 70. n H 2S RT V = n H 2S V × RT = 0.050 mol 0.08206 L atm × × 308.2 K = 1.3 atm L K mol Assuming 100.00 g naphthalene: 93.71 C × 1 mol C = 7.802 mol C 12.011 g 6.29 g H × 1 mol H = 6.24 mol H; 1.008 g 7.802 = 1.25 6.24 Empirical formula = (C1.25H) × 4 = C5H4; molar mass = 32.8 g = 128 g/mol 0.256 mol Because the empirical mass (64.08 g/mol) is one-half of 128, the molecular formula is C10H8. C10H8(s) Initial Equil. − − ⇌ C10H8(g) 0 x K = 4.29 × 10 −6 = [C10H8] = x K = 4.29 × 10 −6 = [C10H8] CHAPTER 6 CHEMICAL EQUILIBRIUM 177 Mol C10H8 sublimed = 5.00 L × 4.29 × 10 −6 mol/L = 2.15 × 10 −5 mol C10H8 sublimed Mol C10H8 initially = 3.00 g × 2.15 × 10 −5 mol × 100 = 0.0919% 2.34 × 10 − 2 mol Percent C10H8 sublimed = 71. 5.63 g C5H6O3 × 1 mol C 5 H 6 O 3 = 0.0493 mol C5H6O3 initially 114.10 g Total moles of gas = ntotal = at equilibrium C5H6O3(g) Initial 1 mol C10 H 8 = 2.34 × 10 −2 mol C10H8 initially 128.16 g Ptotal V 1.63 atm × 2.50 L = 0.105 mol = 0.08206 L atm RT × 473 K K mol ⇌ C2H6(g) + 3 CO(g) 0.0493 mol 0 0 Let x mol C5H6O3 react to reach equilibrium. !x → +x +3x x 3x 0.0493 ! x Change Equil. 0.105 mol total = 0.0493 – x + x + 3x = 0.0493 + 3x, x = 0.0186 mol 3 ⎡ 0.0186 mol C 2 H 6 ⎤ ⎡ 3(0.0186) mol CO ⎤ ⎢ ⎥⎢ ⎥ 3 2.50 L 2.50 L [C H ][CO] ⎦⎣ ⎦ = 6.74 × 10 −6 mol3/L3 K= 2 6 = ⎣ [C 5 H 6 O 3 ] ⎡ (0.0493 − 0.0186) mol C 5 H 6 O 3 ⎤ ⎢ ⎥ 2.50 L ⎣ ⎦ Challenge Problems 72. N2O4(g) a. Initial Change Equil. ⇌ 2 NO2(g) 0 +(0.32)x (0.32)x x !(0.16)x (0.84)x (0.84)x + (0.32)x = 1.5 atm, x = 1.3 atm; Kp = N2O4 b. Equil. x ⇌ 2 NO2 ; x + y = 1.0 atm; y (0.42) 2 = 0.16 atm 1.1 y2 = 0.16 x Solving: x = 0.67 atm (= PN 2O 4 ) and y = 0.33 atm (= PNO 2 ) 178 CHAPTER 6 c. N2O4 Initial Change Equil. ⇌ 2 NO2 0 +2x 0.33 atm P0 !x 0.67 atm CHEMICAL EQUILIBRIUM P0 = initial pressure of N2O4 2x = 0.33, x = 0.165 (using extra sig. figs.) P0 ! x = 0.67, P0 = 0.67 + 0.165 = 0.84 atm; 73. a. 2 NO(g) Initial Change Equil. + Br2(g) ⇌ 0.165 × 100 = 20.% dissociated 0.84 2 NOBr(g) 98.4 torr 41.3 torr 0 2x torr of NO reacts to reach equilibrium −2x −x → +2x 98.4 − 2x 41.3 − x 2x Ptotal = PNO + PBr2 + PNOBr = (98.4 - 2x) + (41.3 − x) + 2x = 139.7 − x Ptotal = 110.5 = 139.7 − x, x = 29.2 torr; PNO = 98.4 - 2(29.2) = 40.0 torr = 0.0526 atm PBr2 = 41.3 − 29.2 = 12.1 torr = 0.0159 atm; PNOBr = 2(29.2) = 58.4 torr = 0.0768 atm Kp = 2 PNOBr (0.0768 atm) 2 = = 134 atm−1 2 2 PNO × PBr2 (0.0526 atm) (0.0159 atm) b. 2 NO(g) Initial Change Equil. + Br2(g) ⇌ 2 NOBr(g) 0.30 atm 0.30 atm 0 2x atm of NO reacts to reach equilibrium −2x −x → +2x 0.30 − 2x 0.30 − x 2x This would yield a cubic equation, which can be difficult to solve unless you have a graphing calculator. Because Kp is pretty large, let’s approach equilibrium in two steps: Assume the reaction goes to completion, and then solve the back equilibrium problem. 2 NO Before Change After Change Equil. + Br2 ⇌ 2 NOBr 0.30 atm 0.30 atm 0 Let 0.30 atm NO react completely. −0.30 −0.15 → +0.30 React completely 0 0.15 0.30 New initial 2y atm of NOBr reacts to reach equilibrium +2y +y ← −2y 0.15 + y 0.30 − 2y 2y CHAPTER 6 CHEMICAL EQUILIBRIUM 179 K p = 134 = (0.30 − 2 y ) 2 (0.30 − 2 y ) 2 = 134 × 4y2 = 536y2 , 2 ( 0 . 15 + y ) ( 2 y ) (0.15 + y ) If y << 0.15: (0.30) 2 ≈ 536y2 and y = 0.034; assumptions are poor (y is 23% of 0.15). 0.15 Use 0.034 as an approximation for y, and solve by successive approximations (see Appendix 1): (0.30 − 0.068) 2 = 536y2, y = 0.023; 0.15 + 0.034 (0.30 − 0.046) 2 = 536y2, y = 0.026 0.15 + 0.023 (0.30 − 0.052) 2 = 536y2, y = 0.026 atm (We have converged on the correct 0.15 + 0.026 answer.) So: PNO = 2y = 0.052 atm; PBr2 = 0.15 + y = 0.18 atm; PNOBr = 0.30 − 2y = 0.25 atm 74. a. If the volume is increased, equilibrium will shift to the right, so the mole percent of N2O5 decomposed will be greater than 0.50%. ⇌ 4 NO2 1.000 atm −0.0050 → 0.995 0 +0.010 0.010 2 N2O5 b. Initial Change Equil. Kp = + O2 0 +0.0025 0.0025 (0.010) 4 (0.0025) = 2.5 × 10−11 atm3 2 (0.995) The new volume is 10.0 times the old volume. Therefore, the initial partial pressure of N2O5 will decrease by a factor of 10.0. PN 2O5 = 1.00 atm × 2 N2O5 Initial Change Equil. 1.00 = 0.100 atm 10.0 ⇌ 4 NO2 + 0.100 atm 0 −2x → +4x 4x 0.100 − 2x 2.5 × 10−11 = O2 0 +x x (4 x) 4 ( x) (4 x) 4 ( x) ≈ , 2x = 2.0 × 10−3 atm = PN 2O5 decomposed (0.100 − 2 x) 2 (0.100) 2 2.0 × 10 −3 × 100 = 2.0% N2O5 decomposed (moles and P are directly related) 0.100 180 75. CHAPTER 6 P4(g) ⇌ 2 P2(g) Kp = 0.100 = PP22 PP4 CHEMICAL EQUILIBRIUM ; PP4 + PP2 = Ptotal = 1.00 atm, PP4 = 1.00 − PP2 Let y = PP2 at equilibrium, then Kp = y2 = 0.100 1.00 − y Solving: y = 0.270 atm = PP2 ; PP4 = 1.00 − 0.270 = 0.73 atm To solve for the fraction dissociated, we need the initial pressure of P4. ⇌ P4(g) Initial Change Equil. 2 P2(g) P0 0 P0 = initial pressure of P4 x atm of P4 reacts to reach equilibrium −x → +2x P0 − x 2x Ptotal = P0 − x + 2x = 1.00 atm = P0 + x Solving: 0.270 atm = PP2 = 2x, x = 0.135 atm; P0 = 1.00 − 0.135 = 0.87 atm Fraction dissociated = 76. 0.135 x = = 0.16, or 16% of P4 is dissociated to reach equilibrium. P0 0.87 Equilibrium lies to the right (Kp values are very large). Let each reaction go to completion initially, and then solve the back equilibrium problems. CH4 + 2 O2 Before 1.50 15.00 After 0 12.00 → 2 C2H6 + 7 O2 → Before 2.50 After 0 12.00 3.25 CO2 0 1.50 + 2 H2O 0 3.00 atm 4 CO2 + 6 H2O 1.50 6.50 3.00 10.50 atm Thus after we let the two reactions go to completion, we have 3.25 atm O2, 6.50 atm CO2, and 10.50 atm H2O. Let’s solve the back equilibrium problems to determine the equilibrium concentrations. CH4 Initial Change Equil. 0 +x x + 2 O2 3.25 atm +2x 3.25 + 2x K p = 1.0 × 10 = 4 PCO 2 × PH2 2O PCH 4 × PO2 2 = ⇌ CO2 ← 6.50 atm −x 6.50 – x + 2 H2O 10.50 atm −2x 10.50 – 2x (6.50 − x)(10.50 − 2 x) 2 6.50(10.50 ) 2 ≈ x(3.25 + 2 x) 2 x(3.25) 2 CHAPTER 6 CHEMICAL EQUILIBRIUM 181 x = PCH 4 = 6.8 × 10−3 atm; assumptions are good. 2 C2H6(g) + 7 O2(g) ⇌ 4 CO2(g) + 6 H2O(g) 0 +2x 2x 3.25 atm +7x 3.25 + 7x ← 6.50 atm −4x 6.50 – 4x 10.50 atm −6x 10.50 – 6x Initial Change Equil. Kp = 1.0 × 108 = 4 PCO × PH6 2O 2 PC22 H 6 × PO7 2 = (6.50 − 4 x) 4 (10.50 − 6 x) 6 (6.50) 4 (10.50 ) 6 ≈ ( 2 x) 2 (3.25 + 7 x) 7 4 x 2 (3.25) 7 2x = PC 2 H 6 = 0.079 atm; assumptions are good. 77. PO ( molar mass O 2 ) + PO3 (molar mass O3 ) P × (molar mass) = 2 RT RT d = density = 0.168 g/L = PO 2 (32.00 g / mol) + PO3 (48.00 g / mol) , (32.00)PO 2 + (48.00)PO3 = 6.18 0.08206 L atm × 448 K K mol Ptotal = PO 2 + PO3 = 128 torr × 1 atm = 0.168 atm 760 torr We have two equations in two unknowns. Solving using simultaneous equations: (32.00)PO 2 + (48.00)PO3 = 6.18 − (32.00)PO 2 − (32.00)PO3 = −5.38 ____________________________ (16.00)PO3 = 0.80 PO3 = Kp = 0.80 = 0.050 atm and PO 2 = 0.118 atm 16.00 PO2 3 PO3 2 (0.050) 2 = 1.5 atm−1 (0.118) 3 = O2(g) 78. Initial Change Equil. 100 !83 17 Thus: χO = ⇌ 2 O(g) Assume exactly 100 O2 molecules. 0 +166 166 166 = 0.9071 and χ O 2 = 0.0929 183 182 CHAPTER 6 CHEMICAL EQUILIBRIUM PO 2 = χ O 2 Ptotal ; and PO = χOPtotal Because initially Ptotal = 1.000 atm, PO 2 = 0.0929 atm and PO = 0.9071 atm. Kp = PO2 (0.9071) 2 = 8.86 atm = 0.0929 PO 2 ⇌ O2 Initial Change Equil. 2O 0 +2y 2y x !y → x!y (2 y) 2 y × 100 = 95.0; we have two equations and two unknowns. = 8.86; x− y x Solving: x = 0.123 atm and y = 0.117 atm; Ptotal = (x − y) + 2y = 0.240 atm 79. 4.72 g CH3OH × 1 mol = 0.147 mol CH3OH initially 32.04 g Rate1 = Rate 2 Graham’s law of effusion: Rate H 2 Rate CH 3OH = M CH 3OH M H2 M2 M1 32.04 = 3.987 2.016 = The effused mixture has 33.0 times as much H2 as CH3OH. When the effusion rate ratio is multiplied by the equilibrium mole ratio of H2 to CH3OH, the effused mixture will have 33.0 times as much H2 as CH3OH. Let n H 2 and n CH 3OH equal the equilibrium moles of H2 and CH3OH, respectively. 33.0 = 3.987 × n H2 n CH 3OH CH3OH(g) Initial Change Equil. 0.147 mol −x 0.147 − x n CH 3OH = 8.28 ⇌ CO(g) → 0 +x x From the ICE table, 8.28 = Solving: x = 0.118 mol n H2 , n H2 n CH 3OH + 2 H2(g) 0 +2x 2x = 2x 0.147 − x CHAPTER 6 CHEMICAL EQUILIBRIUM 183 ⎛ 0.118 mol ⎞ ⎛ 2(0.118 mol) ⎞ ⎜⎜ ⎟⎜ ⎟⎟ 2 1.00 L ⎟⎠ ⎜⎝ 1.00 L [CO][H 2 ] ⎝ ⎠ K= = (0.147 − 0.118) mol [CH 3OH] 1.00 L 80. a. 89.7 g SbCl5 × = 0.23 mol2/L2 1 mol = 0.300 mol SbCl5 initially 299.1 g SbCl5(g) Initial Change Equil. 2 ⇌ 0.300 mol −(0.292)(0.300) 0.212 SbCl3(g) + 0 +0.0876 0.0876 Cl2(g) 0 +0.0876 0.0876 mol ⎛ 0.0876 mol ⎞ ⎛ 0.0876 mol ⎞ ⎜⎜ ⎟⎜ ⎟ 15.0 L ⎟⎠ ⎜⎝ 15.0 L ⎟⎠ ⎝ K= = 2.41 × 10−3 mol/L ⎛ 0.212 mol ⎞ ⎜⎜ ⎟⎟ ⎝ 15.0 L ⎠ b. Let x = moles Cl2 added; reaction shifts left after Cl2 is added. Equilibrium moles SbCl3 = 0.0876 = 0.0438 mol 2 Therefore, 0.0438 mol SbCl3 was reacted. Equilibrium moles Cl2 = x + 0.0876 − 0.0438 = x + 0.0438 Equilibrium moles SbCl5 = 0.212 + 0.0438 = 0.256 mol 2.41 × 10−3 81. a. PPCl5 = ⎛ x + 0.0438 ⎞ ⎛ 0.0438 ⎞ ⎜ ⎟⎜ ⎟ 15.0 15.0 ⎠ ⎝ ⎠ ⎝ = ; solving: x = 0.168 mol Cl2 added ⎛ 0.256 ⎞ ⎜ ⎟ ⎝ 15.0 ⎠ n PCl5 RT V = PCl5(g) Initial Change Equil. 0.100 mol × ⇌ 0.328 atm → −x 0.328 − x 0.08206 L atm × 480. K K mol = 0.328 atm 12.0 L PCl3(g) 0 +x x + Cl2(g) 0 +x x Kp = 0.267 atm 184 CHAPTER 6 Kp = CHEMICAL EQUILIBRIUM x2 = 0.267, x2 + (0.267)x − 0.08758 = 0 (carrying extra sig. figs.) 0.328 − x Solving using the quadratic formula: x = 0.191 atm PPCl3 = PCl 2 = 0.191 atm; PPCl5 = 0.328 − 0.191 = 0.137 atm b. Initial Change Equil. PCl5(g) ⇌ P0 −x P0 − x → PCl3(g) + 0 +x x Cl2(g) 0 +x x P0 = initial pressure of PCl5 Ptotal = 2.00 atm = (P0 − x) + x + x = P0 + x, P0 = 2.00 − x Kp = x2 x2 = 0.267; = 0.267, x2 = 0.534 − (0.534)x P0 − x 2.00 − 2 x x2 + (0.534)x − 0.534 = 0; solving using the quadratic formula: − 0.534 ± (0.534) 2 + 4(0.534) = 0.511 atm 2 P0 = 2.00 − x = 2.00 − 0.511 = 1.49 atm; the initial pressure of PCl5 was 1.49 atm. x= n PCl5 = PPCl5 V RT = (1.49 atm)(5.00 L) = 0.189 mol PCl5 (0.08206 L atm K −1 mol −1 )(480. K ) 0.189 mol PCl5 × 208.22 g PCl5/mol = 39.4 g PCl5 was initially introduced. 82. P0 (for O2) = n O 2 RT / V = (6.400 g × 0.08206 × 684 K)/(32.00 g/mol × 2.50 L) = 4.49 atm 2 O2(g) → CO2(g) + 2 H2O(g) −2x → +x +2x Change CH4(g) + −x Change CH4(g) + 3/2 O2(g) → CO(g) + 2 H2O(g) −y −3/2 y → +y +2y Amount of O2 reacted = 4.49 atm − 0.326 atm = 4.16 atm O2 2x + 3/2 y = 4.16 atm O2 and 2x + 2y = 4.45 atm H2O Solving using simultaneous equations: 2x + 2y = 4.45 −2x − (3/2)y = −4.16 (0.50)y = 0.29, y = 0.58 atm = PCO CHAPTER 6 CHEMICAL EQUILIBRIUM 2x + 2(0.58) = 4.45, x = 83. N2(g) Equil. 4.45 − 1.16 = 1.65 atm = PCO 2 2 + O2(g) (3.7)p 185 ⇌ 2 NO(g) p Let: equilibrium PO 2 = p x equilibrium PN 2 = (78/21)PO 2 = (3.7)p equilibrium PNO = x equilibrium PNO 2 = y Kp = 1.5 × 10−4 = 2 PNO PO 2 × PN 2 N2 Equil. + 2 O2 (3.7)p Kp = 1.0 × 10−5 = p ⇌ 2 NO2 y 2 PNO 2 PO2 2 × PN 2 We want PNO 2 = PNO at equilibrium, so x = y. Taking the ratio of the two Kp expressions: 2 PNO PO 2 × PN 2 2 PNO 2 = 1.5 × 10 −4 1.5 × 10 −4 = P : P = ; because P = 15 atm NO NO 2 O2 1.0 × 10 −5 1.0 × 10 −5 PO2 2 × PN 2 Air is 21 mol % O2, so: PO 2 = (0.21)Ptotal, Ptotal = 15 atm = 71 atm 0.21 To solve for the equilibrium concentrations of all gases (not required to answer the question), solve one of the Kp expressions where p = PO 2 = 15 atm. 1.5 × 10−4 = x2 , x = PNO = PNO 2 = 0.35 atm 15[3.7(15)] Equilibrium pressures: PO 2 = 15 atm; PN 2 = 3.7(15) = 55.5 = 56 atm; PNO = PNO 2 = 0.35 atm 186 CHAPTER 6 84. Initial Change Equil. CCl4(g) ⇌ C(s) + 2 Cl2(g) Kp = 0.76 P0 −x P0- x → − − − 0 +2x 2x CHEMICAL EQUILIBRIUM P0 = initial pressure of CCl4 Ptotal = P0 − x + 2x = P0 + x = 1.20 atm Kp = (2 x) 2 4 x 2 + (0.76) x = 0.76, 4x2 = (0.76)P0 − (0.76)x, P0 = ; substituting into P0 − x 0.76 P0 + x = 1.20: 2 4x + x + x = 1.20 atm, (5.3)x2 + 2x − 1.20 = 0; solving using the quadratic formula: 0.76 x= 85. Initial Change Equil. − 2 ± (4 + 25.4)1/ 2 = 0.32 atm; P0 + 0.32 = 1.20, P0 = 0.88 atm 2(5.3) SO3(g) ⇌ P0 −x P0 − x → SO2(g) + 1/2 O2(g) 0 +x x 0 +x/2 x/2 P0 = initial pressure of SO3 Average molar mass of the mixture is: average molar mass = dRT (1.60 g / L) (0.08206 L atm K −1 mol −1 ) (873 K ) = P 1.80 atm = 63.7 g/mol The average molar mass is determined by: average molar mass = n SO3 (80.07 g/mol) + n SO 2 (64.07 g/mol) + n O 2 (32.00 g/mol) n total Because χA = mol fraction of component A = nA/ntotal = PA/Ptotal: 63.7 g/mol = = PSO 3 (80.07) + PSO 2 (64.07) + PO 2 (32.00) Ptotal Ptotal = P0 − x + x + x/2 = P0 + x/2 = 1.80 atm, P0 = 1.80 - x/2 63.7 = (P0 − x) (80.07) + x (64.07) + 1.80 x (32.00) 2 CHAPTER 6 CHEMICAL EQUILIBRIUM 187 (1.80 − 3/2 x) (80.07) + x (64.07) + 63.7 = x (32.00) 2 1.80 115 = 144 − (120.1)x + (64.07)x + (16.00)x, (40.0)x = 29, x = 0.73 atm PSO 3 = P0 − x = 1.80 − (3/2)x = 0.71 atm; PSO 2 = 0.73 atm; PO 2 = x/2 = 0.37 atm Kp = 86. PSO 2 × PO1/22 PSO3 = (0.73) (0.37)1/ 2 = 0.63 atm1/2 (0.71) The first reaction produces equal amounts of SO3 and SO2. Using the second reaction, calculate the SO3, SO2, and O2 partial pressures at equilibrium. SO3(g) Initial P0 Change Equil. −x P0 − x ⇌ SO2(g) + 1/2 O2(g) P0 → 0 +x P0 + x P0 = initial pressure of SO3 and SO2 after first reaction occurs. +x/2 x/2 Ptotal = P0 − x + P0 + x + x/2 = 2P0 + x/2 = 0.836 atm PO 2 = x/2 = 0.0275 atm, x = 0.0550 atm 2P0 + x/2 = 0.836 atm; 2P0 = 0.836 − 0.0275 = 0.809 atm, P0 = 0.405 atm PSO 3 = P0 − x = 0.405 − 0.0550 = 0.350 atm; PSO 2 = P0 + x = 0.405 + 0.0550 = 0.460 atm For 2 FeSO4(s) ⇌ Fe2O3(s) + SO3(g) + SO2(g): Kp = PSO 2 × PSO 3 = (0.460)(0.350) = 0.161 atm2 For SO3(g) ⇌ SO2(g) + 1/2 O2(g): Kp = 87. PSO 2 × PO1/22 PSO3 N2(g) + 3 H2 (g) = (0.460) (0.0275)1/ 2 = 0.218 atm1/2 (0.350) ⇌ 2 NH3(g) 1.0 atm N2(g) Initial Equil. 0.25 atm 0.25 − x + Kp = 3 H2(g) 0.75 atm 0.75 − 3x ⇌ 2 PNH 3 PN 2 × PH3 2 2 NH3(g) 0 2x = 6.5 × 10−3 188 CHAPTER 6 CHEMICAL EQUILIBRIUM (2 x) 2 = 6.5 × 10−3; using successive approximations: 3 (0.75 − 3x) (0.25 − x) x = 1.2 × 10−2 atm; PNH 3 = 2x = 0.024 atm 10 atm N2(g) Initial Equil. 2.5 atm 2.5 − x + 3 H2(g) ⇌ 2 NH3(g) 7.5 atm 7.5 − 3x 0 2x (2 x) 2 = 6.5 × 10−3; using successive approximations: 3 (7.5 − 3 x) (2.5 − x) x = 0.69 atm; PNH 3 = 1.4 atm 100 atm Using the same setup as above: 4x2 = 6.5 × 10−3 3 (75 − 3x) (25 − x) Solving by successive approximations: x = 16 atm; PNH 3 = 32 atm 1000 atm N2(g) Initial New initial Equil. + 3 H2(g) ⇌ 2 NH3(g) 250 atm 750 atm 0 Let 250 atm N2 react completely. 0 0 5.0 × 102 x 3x 5.0 × 102 − 2x (5.0 × 10 2 − 2 x) 2 = 6.5 × 10−3; Using successive approximations: 3 (3x) x x = 32 atm; PNH 3 = 5.0 × 102 − 2x = 440 atm CHAPTER 6 CHEMICAL EQUILIBRIUM 189 The results are plotted as log PNH 3 versus log Ptotal. Notice that as Ptotal increases, a larger fraction of N2 and H2 is converted to NH3, that is, as Ptotal increases (V decreases), the reaction shifts further to the right, as predicted by LeChatelier’s principle. 88. a. Because density (mass/volume) decreases while the mass remains constant (mass is conserved in a chemical reaction), volume must increase. The volume increases because the number of moles of gas increases (V ∝ n at constant T and P). V n Density (initial) 4.495 g/L = = 1.100 = equil. = equil. Density (equil.) 4.086 g/L Vinitial n initial Assuming an initial volume of 1.000 L: 4.495 g NOBr × Initial Change Equil. n equil. n initial = 1 mol NOBr = 0.04090 mol NOBr initially 109.91 g 2 NOBr(g) ⇌ 0.04090 mol −2x 0.04090 − 2x → 2 NO(g) + Br2(g) 0 +2x 2x 0 +x x 0.04090 − 2 x + 2 x + x = 1.100; solving: x = 0.00409 mol 0.04090 If the initial volume is 1.000 L, then the equilibrium volume will be 1.110(1.000 L) = 1.110 L. Solving for the equilibrium concentrations: [NOBr] = 0.03272 mol 0.00818 mol = 0.02975 M ; [NO] = = 0.00744 M 1.100 L 1.100 L 190 CHAPTER 6 [Br2] = K= CHEMICAL EQUILIBRIUM 0.00409 mol = 0.00372 M 1.100 L (0.00744) 2 (0.00372) = 2.33 × 10−4 mol/L 2 (0.02975) b. The argon gas will increase the volume of the container. This is because the container is a constant-pressure system, and if the number of moles increases at constant T and P, the volume must increase. An increase in volume will dilute the concentrations of all gaseous reactants and gaseous products. Because there are more moles of product gases versus reactant gases (3 mol versus 2 mol), the dilution will decrease the numerator of K more than the denominator will decrease. This causes Q < K and the reaction shifts right to get back to equilibrium. Because temperature was unchanged, the value of K will not change. K is a constant as long as temperature is constant. 89. 2.00 g = 0.0121 mol XY (initially) 165 g/mol (0.350)(0.0121 mol) = 4.24 × 10−3 mol XY dissociated → XY Initial Change Equil. 0.0121 mol −0.00424 0.0079 mol X + Y 0 0 +0.00424 +0.00424 0.00424 mol 0.00424 mol → Total moles of gas = 0.0079 + 0.00424 + 0.00424 = 0.0164 mol V ∝ n, so: Vinitial = Vfinal n 0.0164 mol = final = = 1.36 Vinitial n initial 0.0121 mol (0.0121 mol)(0.008206 L atm K −1 mol −1 )(298 K ) nRT = = 0.306 L 0.967 atm P Vfinal = 0.306 L(1.36) = 0.416 L; Because mass is conserved in a chemical reaction: density (final) = mass 2.00 g = = 4.81 g/L volume 0.416 L 2 ⎛ 0.00424 mol ⎞ ⎜ ⎟ [X ][Y] ⎜⎝ 0.416 L ⎟⎠ K= = = 5.5 × 10 −3 mol/L [XY] ⎛ 0.0079 mol ⎞ ⎜⎜ ⎟⎟ ⎝ 0.416 L ⎠ CHAPTER 6 CHEMICAL EQUILIBRIUM 191 Marathon Problem 90. Concentration units involve both moles and volume and since both quantities are changing at the same time, we have a complicated system. Let’s simplify the setup of the problem initially by only worrying about the changes that occur to the moles of each gas. A(g) Initial Change Equil. + B(g) ⇌ C(g) K = 130. 0 0 0.406 mol Let x mol of C(g) react to reach equilibrium +x +x ← −x x x 0.406 − x Let Veq = the equilibrium volume of the container, so: [A]eq = [B]eq = x 0.406 − x ; [C]eq = Veq Veq 0.046 − x Veq (0.406 − x) Veq [ C] K = = = x x [A ][B] x2 × Veq Veq From the ideal gas equation: V = nRT/P. To calculate the equilibrium volume from the ideal gas law, we need the total moles of gas present at equilibrium. At equilibrium: ntotal = mol A(g) + mol B(g) + mol C(g) = x + x + 0.406 - x = 0.406 + x Therefore: Veq = n total RT (0.406 + x)(0.08206 L atm K −1 mol −1 )(300.0 K ) = P 1.00 atm Veq = (0.406 + x)24.6 L/mol Substituting into the equilibrium expression for Veq: K = 130. = (0.406 − x)(0.406 + x)24.6 x2 Solving for x (we will carry one extra significant figure): (130.)x2 = (0.1648 − x2)24.6, (154.6)x2 = 4.054, x = 0.162 mol Solving for the volume of the container at equilibrium: Veq = (0.406 + 0.162 mol)(0.08206)(300.0 K ) = 14.0 L 1.00 atm CHAPTER 7 ACIDS AND BASES Nature of Acids and Bases 16. a. H2O(l) + H2O(l) ⇌ H3O+(aq) + OH−(aq) or H2O(l) ⇌ H+(aq) + OH−(aq) K = Kw = [H+][OH−] b. HF(aq) + H2O(l) ⇌ F−(aq) + H3O+(aq) or [H + ][F − ] HF(aq) ⇌ H+(aq) + F−(aq) K = Ka = [ HF] c. C5H5N(aq) + H2O(l) ⇌ C5H5NH+(aq) + OH−(aq) 17. [C 5 H 5 NH + ][OH − ] [C 5 H 5 N ] An acid is a proton (H+) donor, and a base is a proton acceptor. A conjugate acid-base pair differs by only a proton (H+) in the formulas. a. b. c. 18. K = Kb = Acid Base H2CO3 C5H5NH+ C5H5NH+ H2O H2O HCO3− Conjugate Base of Acid HCO3− C5H5N C5H5N Conjugate Acid of Base H3O+ H3O+ H2CO3 a. HClO4(aq) + H2O(l) → H3O+(aq) + ClO4−(aq). Only the forward reaction is indicated because HClO4 is a strong acid and is basically 100% dissociated in water. For acids, the dissociation reaction is commonly written without water as a reactant. The common abbreviation for this reaction is HClO4(aq) → H+(aq) + ClO4−(aq). This reaction is also called the Ka reaction because the equilibrium constant for this reaction is designated as Ka. b. Propanoic acid is a weak acid, so it is only partially dissociated in water. The dissociation reaction is CH3CH2CO2H(aq) + H2O(l) ⇌ H3O+(aq) + CH3CH2CO2−(aq) or CH3CH2CO2H(aq) ⇌ H+(aq) + CH3CH2CO2−(aq). c. NH4+ is a weak acid. Similar to propanoic acid, the dissociation reaction is: NH4+(aq) + H2O(l) ⇌ H3O+(aq) + NH3(aq) or NH4+(aq) ⇌ H+(aq) + NH3(aq) 192 CHAPTER 7 19. 20. − a. HC2H3O2(aq) ⇌ H+(aq) + C2H3O2−(aq) Ka = [H + ][C 2 H 3O 2 ] [HC 2 H 3O 2 ] b. Co(H2O)63+(aq) ⇌ H+(aq) + Co(H2O)5(OH)2+(aq) Ka = [H + ][Co(H 2 O) 5 (OH) 2+ ] [Co(H 2 O) 36+ ] c. CH3NH3+(aq) ⇌ H+(aq) + CH3NH2(aq) Ka = [H + ][CH 3 NH 2 ] + [CH 3 NH 3 ] Strong acids have a Ka >> 1 and weak acids have Ka < 1. Table 7.2 in the text lists some Ka values for weak acids. Ka values for strong acids are hard to determine so they are not listed in the text. However, there are only a few common strong acids so if you memorize the strong acids, then all other acids will be weak acids. The strong acids to memorize are HCl, HBr, HI, HNO3, HClO4, and H2SO4. HClO4 is a strong acid. HOCl is a weak acid (Ka = 3.5 × 10−8). H2SO4 is a strong acid. H2SO3 is a weak diprotic acid because the Ka1 and Ka2 values are less than 1. The beaker on the left represents a strong acid in solution; the acid HA is 100% dissociated into the H+ and A− ions. The beaker on the right represents a weak acid in solution; only a little bit of the acid HB dissociates into ions, so the acid exists mostly as undissociated HB molecules in water. a. b. c. d. e. 22. 193 The dissociation reaction (the Ka reaction) of an acid in water commonly omits water as a reactant. We will follow this practice. All dissociation reactions produce H+ and the conjugate base of the acid that is dissociated. a. b. c. d. 21. ACIDS AND BASES HNO2: weak acid beaker HNO3: strong acid beaker HCl: strong acid beaker HF: weak acid beaker HC2H3O2: weak acid beaker All Kb reactions refer to the base reacting with water to produce the conjugate acid of the base and OH−. + a. NH3(aq) + H2O(l) ⇌ NH4 (aq) + OH (aq) [OH − ][ NH 4 ] Kb = [ NH 3 ] b. CN−(aq) + H2O(l) ⇌ HCN(aq) + OH−(aq) Kb = [OH − ][HCN] [CN − ] c. C5H5N(aq) + H2O(l) ⇌ C5H5NH+(aq) + OH−(aq) Kb = [OH − ][C 5 H 5 NH + ] [C 5 H 5 N ] + − 194 CHAPTER 7 d. C6H5NH2(aq) + H2O(l) ⇌ C6H5NH3+(aq) + OH−(aq) 23. ACIDS AND BASES + Kb = [OH − ][C 6 H 5 NH 3 ] [C 6 H 5 NH 2 ] The Ka value is directly related to acid strength. As Ka increases, acid strength increases. For water, use Kw when comparing the acid strength of water to other species. The Ka values are: HClO4: strong acid (Ka >> 1); HClO2: Ka = 1.2 × 10 −2 NH4+: Ka = 5.6 × 10 −10 ; H2O: Ka = Kw = 1.0 × 10 −14 From the Ka values, the ordering is: HClO4 > HClO2 > NH4+ > H2O 24. Except for water, these are the conjugate bases of the acids in the preceding exercise. In general, the weaker the acid, the stronger the conjugate base. ClO4− is the conjugate base of a strong acid; it is a terrible base (worse than water). The ordering is NH3 > ClO2− > H2O > ClO4− . 25. a. H2SO4 is a strong acid and water is a very weak acid with Ka = Kw = 1.0 × 10−14. H2SO4 is a much stronger acid than H2O. b. H2O, Ka = Kw = 1.0 × 10−14; HOCl, Ka = 3.5 × 10−8; HOCl is a stronger acid than H2O because Ka for HOCl > Ka for H2O. c. NH4+ , Ka = 5.6 × 10−10; HC2H2ClO2, Ka = 1.35 × 10−3; HC2H2ClO2 is a stronger acid than NH4+ because Ka for HC2H2ClO2 > Ka for NH4+. 26. a. H2O; the conjugate bases of strong acids are terrible bases (Kb < 10−14). b. OCl− ; the conjugate bases of weak acids are weak bases (10−14 < Kb < 1); even though they are designated as weak bases, the conjugate bases of weak acids are all better bases than H2O. c. 27. NH3; for a conjugate acid-base pair, Ka × Kb = Kw. From this relationship, the stronger the acid, the weaker is the conjugate base (Kb decreases as Ka increases). Because HC2H2ClO2 is a stronger acid than NH4+ (Ka for HC2H2ClO2 > Ka for NH4+), NH3 will be a stronger base than C2H2ClO2− . a. H2O and CH3CO2− b. An acid-base reaction can be thought of as a competition between two opposing bases. Because this equilibrium lies far to the left (Ka < 1), CH3CO2− is a stronger base than H2O. c. The acetate ion is a better base than water and produces basic solutions in water. When we put acetate ion into solution as the only major basic species, the reaction is: CH3CO2− + H2O ⇌ CH3CO2H + OH− CHAPTER 7 ACIDS AND BASES 195 Now the competition is between CH3CO2− and OH− for the proton. Hydroxide ion is the strongest base possible in water. The preceding equilibrium lies far to the left resulting in a Kb value of less than 1. Those species we specifically call weak bases (10−14 < Kb < 1) lie between H2O and OH− in base strength. Weak bases are stronger bases than water but are weaker bases than OH−. 28. The NH4+ ion is a weak acid because it lies between H2O and H3O+ (H+) in terms of acid strength. Weak acids are better acids than water, thus their aqueous solutions are acidic. They are weak acids because they are not as strong as H3O+ (H+). Weak acids only partially dissociate in water and have Ka values of between 10−14 and 1. 29. In deciding whether a substance is an acid or a base, strong or weak, you should keep in mind a couple ideas: 1. There are only a few common strong acids and strong bases all of which should be memorized. Common strong acids = HCl, HBr, HI, HNO3, HClO4, and H2SO4. Common strong bases = LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2. 2. All other acids and bases are weak and will have Ka and Kb values of less than 1 but greater than Kw (10−14). Reference Table 7.2 for Ka values for some weak acids and Table 7.3 for Kb values for some weak bases. There are too many weak acids and weak bases to memorize them all. Therefore, use the tables of Ka and Kb values to help you identify weak acids and weak bases. Appendix 5 contains more complete tables of Ka and Kb values. a. c. e. g. i. weak acid (Ka = 4.0 × 10−4) weak base (Kb = 4.38 × 10−4) weak base (Kb = 1.8 × 10−5) weak acid (Ka = 1.8 × 10−4) strong acid b. d. f. h. strong acid strong base weak acid (Ka = 7.2 × 10−4) strong base Autoionization of Water and pH Scale 30. a. Because the value of the equilibrium constant increases as the temperature increases, the reaction is endothermic. In endothermic reactions, heat is a reactant, so an increase in temperature (heat) shifts the reaction to produce more products and increases K in the process. b. H2O(l) ⇌ H+(aq) + OH−(aq) Kw = 5.47 × 10−14 = [H+][OH−] In pure water [H+] = [OH−], so 5.47 × 10−14 = [H+]2, [H+] = 2.34 × 10−7 M pH = −log[H+] = −log(2.34 × 10−7) = 6.631 A neutral solution of water at 50.°C has: [H+] = [OH−]; [H+] = 2.34 × 10−7 M; pH = 6.631 Obviously, the condition that [H+] = [OH−] is the most general definition of a neutral solution. 196 CHAPTER 7 c. Temp (°C) 0 25 35 40. 50. 1/T (K−1) Kw ln Kw 3.66 × 10−3 3.36 × 10−3 3.25 × 10−3 3.19 × 10−3 3.10 × 10−3 1.14 × 10−15 1.00 × 10−14 2.09 × 10−14 2.92 × 10−14 5.47 × 10−14 −34.408 −32.236 −31.499 −31.165 −30.537 Temp (K) 273 298 308 313 323 ACIDS AND BASES From the graph: 37°C = 310. K; 1/T = 3.23 × 10−3 K−1 ln Kw = −31.38, Kw = e−31.38 = 2.35 × 10−14 d. At 37°C, Kw = 2.35 × 10−14 = [H+][OH−] = [H+]2, [H+] = 1.53 × 10−7 M pH = −log[H+] = −log(1.53 × 10−7) = 6.815 31. a. H2O(l) ⇌ H+(aq) + OH−(aq) Kw = 2.92 × 10 −14 = [H+][OH−] In pure water: [H+] = [OH−], 2.92 × 10 −14 = [H+]2, [H+] = 1.71 × 10 −7 M = [OH−] b. pH = !log[H+] = !log(1.71 × 10 −7 ) = 6.767 c. [H+] = Kw/[OH−] = (2.92 × 10 −14 )/0.10 = 2.9 × 10 −13 M; pH = !log(2.9 × 10 −13 ) = 12.54 32. At 25°C, the relationship [H+][OH−] = Kw = 1.0 × 10 −14 always holds for aqueous solutions. When [H+] is greater than 1.0 × 10 −7 M, the solution is acidic; when [H+] is less than 1.0 × 10 −7 M, the solution is basic; when [H+] = 1.0 × 10 −7 M, the solution is neutral. In terms of [OH−], an acidic solution has [OH−] < 1.0 × 10 −7 M, a basic solution has [OH−] > 1.0 × 10 −7 M, and a neutral solution has [OH−] = 1.0 × 10 −7 M. At 25°C, pH + pOH = 14.00. CHAPTER 7 ACIDS AND BASES a. [OH−] = 197 Kw 1.0 × 10 −14 = = 1.0 × 10-7 M; the solution is neutral. + −7 [H ] 1.0 × 10 pH = −log[H+] = −log(1.0 × 10 −7 ) = 7.00; pOH = 14.00 − 7.00 = 7.00 b. [OH−] = 1.0 × 10 −14 = 12 M; the solution is basic. 8.3 × 10 −16 pH = −log(8.3 × 10 −16 ) = 15.08; pOH = 14.00 − 15.08 = −1.08 c. [OH−] = 1.0 × 10 −14 = 8.3 × 10 −16 M; the solution is acidic. 12 pH = −log(12) = −1.08; pOH = 14.00 − (−1.08) = 15.08 d. [OH−] = 1.0 × 10 −14 = 1.9 × 10 −10 M; the solution is acidic. −5 5.4 × 10 pH = −log(5.4 × 10 −5 ) = 4.27; pOH = 14.00 − 4.27 = 9.73 33. a. [H+] = 10−pH, [H+] = 10−7.40 = 4.0 × 10−8 M pOH = 14.00 − pH = 14.00 − 7.40 = 6.60; [OH−] = 10−pOH = 10−6.60 = 2.5 × 10−7 M or [OH−] = Kw 1.0 × 10 −14 = = 2.5 × 10−7 M; this solution is basic since pH > 7.00. + −8 [H ] 4.0 × 10 b. [H+] = 10−15.3 = 5 × 10−16 M; pOH = 14.00 − 15.3 = −1.3; [OH−] = 10− (−1.3) = 20 M; basic c. [H+] = 10− (−1.0) = 10 M; pOH = 14.0 − (−1.0) = 15.0; [OH−] = 10-15.0 = 1 × 10−15 M; acidic d. [H+] = 10−3.20 = 6.3 × 10−4 M; pOH = 14.00 − 3.20 = 10.80; [OH−] = 10−10.80 = 1.6 × 10−11 M; acidic e. [OH−] = 10−5.0 = 1 × 10−5 M; pH = 14.0 − pOH = 14.0 − 5.0 = 9.0; [H+] = 10−9.0 = 1 × 10−9 M; basic f. 34. [OH−] = 10−9.60 = 2.5 × 10−10 M; pH = 14.00 − 9.60 = 4.40; [H+] = 10−4.40 = 4.0 × 10−5 M; acidic a. pOH = 14.00 – 9.63 = 4.37; [H+] = 10 −9.63 = 2.3 × 10 −10 M [OH−] = 10 −4.37 = 4.3 × 10 −5 M; basic b. [H+] = 1.0 × 10 −14 = 2.6 × 10 −9 M; pH = −log(2.6 × 10 −9 ) = 8.59 3.9 × 10 −6 pOH = 14.00 – 8.59 = 5.41; basic 198 CHAPTER 7 ACIDS AND BASES c. pH = –log(0.027) = 1.57; pOH = 14.00 – 1.57 = 12.43 [OH−] = 10 −12.43 = 3.7 × 10 −13 M; acidic d. pH = 14.0 – 12.2 = 1.8; [H+] = 10 −1.8 = 1.6 × 10 −2 M [OH−] = 10 −12.2 = 6 × 10 −13 M; acidic Solutions of Acids 35. Strong acids are assumed to completely dissociate in water, for example, HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq) or HCl(aq) → H+(aq) + Cl−(aq). a. A 0.10 M HCl solution gives 0.10 M H+ and 0.10 M Cl− because HCl completely dissociates. The amount of H+ from H2O will be insignificant. pH = −log[H+] = −log(0.10) = 1.00 b. 5.0 M H+ is produced when 5.0 M HClO4 completely dissociates. The amount of H+ from H2O will be insignificant. pH = −log(5.0) = −0.70 (Negative pH values just indicate very concentrated acid solutions.) c. 1.0 × 10−11 M H+ is produced when 1.0 × 10−11 M HI completely dissociates. If you take the negative log of 1.0 × 10−11, this gives pH = 11.00. This is impossible! We dissolved an acid in water and got a basic pH. What we must consider in this problem is that water by itself donates 1.0 × 10−7 M H+. We can normally ignore the small amount of H+ from H2O except when we have a very dilute solution of an acid (as in the case here). Therefore, the pH is that of neutral water (pH = 7.00) because the amount of HI present is insignificant. 36. 50.0 mL conc. HCl soln × 1.19 g 38 g HCl 1 mol HCl × × = 0.62 mol HCl mL 100 g conc. HCl soln 36.5 g 20.0 mL conc. HNO3 soln × 70. g HNO 3 1 mol HNO 3 1.42 g = 0.32 mol HNO3 × × mL 100 g soln 63.0 g HNO 3 HCl(aq) → H+(aq) + Cl−(aq) and HNO3(aq) → H+(aq) + NO3−(aq) (Both are strong acids.) So we will have 0.62 + 0.32 = 0.94 mol of H+ in the final solution. [H+] = 0.94 mol = 0.94 M; pH = −log[H+] = −log(0.94) = 0.027 = 0.03 1.00 L [OH−] = Kw 1.0 × 10 −14 = = 1.1 × 10−14 M 0.94 [H + ] CHAPTER 7 37. ACIDS AND BASES 199 HCl is a strong acid. [H+] = 10 −1.50 = 3.16 × 10 −2 M (carrying one extra sig. fig.) M 2 V2 3.16 × 10 −2 mol/L × 1.6 L = = 4.2 × 10 −3 L M1 12 mol/L M1V1 = M2V2, V1 = 4.2 mL of 12 M HCl with enough water added to make 1600 mL of solution will result in a solution having [H+] = 3.2 × 10 −2 M and pH = 1.50. 38. [H+] = 10 −5.10 = 7.9 × 10 −6 M; HNO3(aq) → H+(aq) + NO3−(aq) Because HNO3 is a strong acid, we have a 7.9 × 10 −6 M HNO3 solution. 0.2500 L × 39. 7.9 × 10 −6 mol HNO 3 63.02 g HNO 3 × = 1.2 × 10 −4 g HNO3 L mol HNO 3 a. Major species: H+(aq), Br−(aq), and H2O(l). (HBr is a strong acid.) [H+] = 0.250 M pH = −log[H+] = −log(0.250) = 0.602 b. H+(aq), ClO4−(aq), and H2O(l). (HClO4 is a strong acid.) pH = 0.602 c. H+(aq), NO3−(aq), and H2O(l). (HNO3 is a strong acid.) pH = 0.602 d. HNO2 (Ka = 4.0 × 10−4) and H2O (Ka = Kw = 1.0 × 10−14) are the major species. HNO2 is much stronger acid than H2O, so it is the major source of H+. However, HNO2 is a weak acid (Ka < 1), so it only partially dissociates in water. We must solve an equilibrium problem to determine [H+]. In this Solutions Guide, we will summarize the initial, change and equilibrium concentrations into one table called an ICE table. Solving the weak acid problem: HNO2(aq) Initial Change Equil. ⇌ H+(aq) + NO2−(aq) 0.250 M ~0 0 x mol/L HNO2 dissociates to reach equilibrium −x → +x +x 0.250 − x x x − Ka = x2 [H + ][ NO 2 ] , 4.0 × 10-4 = ; if we assume x << 0.250, then: [HNO 2 ] 0.250 − x 4.0 × 10-4 ≈ x2 , 0.250 x = We must check the assumption: 4.0 × 10 − 4 (0.250) = 0.010 M x 0.010 × 100 = × 100 = 4.0% 0.250 0.250 200 CHAPTER 7 ACIDS AND BASES All the assumptions are good. The H+ contribution from water (10−7 M) is negligible and x is small compared to 0.250 (percent error = 4.0%). If the percent error is less than 5% for an assumption, we will consider it a valid assumption (called the 5% rule). Finishing the problem: x = 0.010 M = [H+]; pH = −log[H+] = −log(0.010) = 2.00 e. CH3CO2H (Ka = 1.8 × 10−5) and H2O (Ka = Kw = 1.0 × 10−14) are the major species. CH3CO2H is the major source of H+. Solving the weak acid problem: ⇌ CH3CO2H Initial Change Equil. H+ + CH3CO2− 0.250 M ~0 0 x mol/L CH3CO2H dissociates to reach equilibrium −x → +x +x 0.250 − x x x − Ka = [H + ][CH 3CO 2 ] x2 x2 ≈ (assuming x << 0.250) , 1.8 × 10−5 = [CH 3CO 2 H] 0.250 − x 0.250 x = 2.1 × 10−3 M; checking assumption: 2.1 × 10 −3 × 100 = 0.84%. Assumptions good. 0.250 [H+] = x = 2.1 × 10−3 M; pH = −log(2.1 × 10−3) = 2.68 f. HCN (Ka = 6.2 × 10−10) and H2O are the major species. HCN is the major source of H+. ⇌ HCN Initial Change Equil. H+ + CN− 0.250 M ~0 0 x mol/L HCN dissociates to reach equilibrium −x → +x +x 0.250 − x x x Ka = 6.2 × 10−10 = [H + ][CN − ] x2 x2 = ≈ (assuming x << 0.250) [HCN] 0.250 − x 0.250 x = [H+] = 1.2 × 10−5 M; checking assumption: x is 0.0048% of 0.250. Assumptions good. pH = −log(1.2 × 10−5) = 4.92 40. At pH = 2.000, [H+] = 10−2.000 = 1.00 × 10−2 M; at pH = 4.000, [H+] = 10−4.000 Mol H+ present = 0.0100 L × 0.0100 mol H + = 1.00 × 10−4 mol H+ L Let V = total volume of solution at pH = 4.000: 1.00 ×10−4 mol/L = 0.0100 mol H + , V = 1.00 L V Volume of water added = 1.00 L − 0.0100 L = 0.99 L = 990 mL = 1.00 × 10−4 M CHAPTER 7 41. ACIDS AND BASES 201 a. Major species: HOCl (Ka = 3.5 × 10−8) and water; major source of H+ = HOCl. Because Ka for HOCl is less than 1, HOCl is a weak acid and we must solve an equilibrium problem to determine [H+]. The setup is: HOCl (aq) Initial Change Equil. ⇌ H+(aq) + OCl−(aq) ~0 0 0.20 M x mol/L HOCl dissociates to reach equilibrium −x → +x +x 0.20 − x x x Ka = 3.5 × 10−8 = x2 x2 [H + ][OCl − ] = ≈ [HOCl] 0.20 − x 0.20 (assuming x << 0.20) x = [H+] = 8.4 × 10−5 M We have made two assumptions that we must check. 1. 0.20 − x ≈ 0.20 (x/0.20) × 100 = (8.4 × 10−5/0.20) × 100 = 0.042%. Great assumption. If the percent error in the assumption is less than 5%, then the assumption is valid. 2. HOCl is the major source of H+, that is, we can ignore 1.0 × 10−7 M H+ already present in neutral H2O. [H+] from HOCl = 8.4 × 10−5 >> 1.0 × 10−7; this assumption is valid. In future problems we will always begin the problem solving process by making these assumptions, and we will always check them. However, we may not explicitly state that the assumptions are valid. We will always state when the assumptions are not valid and we have to use other techniques to solve the problem. Remember, anytime we make an assumption, we must check its validity before the solution to the problem is complete. Answering the question: [H+] = [OCl−] = 8.4 × 10−5 M; [OH−] = Kw/[H+] = 1.2 × 10−10 M [HOCl] = 0.20 − x = 0.20 M; pH = −log[H+] = −log(8.4 × 10−5) = 4.08 b. HOC6H5 (Ka = 1.6 × 10−10) is the dominant producer of H+. Solving the weak acid problem: HOC6H5 Initial Change Equil. ⇌ H+ + OC6H5− Ka = 1.6 × 10−10 ~0 0 1.5 M x mol/L HOC6H5 dissociates to reach equilibrium −x → +x +x 1.5 − x x x 202 CHAPTER 7 Ka = 1.6 × 10−10 = ACIDS AND BASES − [H + ][OC 6 H 5 ] x2 x2 = ≈ [HOC6 H 5 ] 1 .5 − x 1 .5 (assuming x << 1.5) x = [H+]= 1.5 × 10−5 M; assumptions good: 1.0 × 10−7 << 1.5 × 10−5 << 1.5 [H+] = [OC6H5− ] = 1.5 × 10−5 M; [OH−] = 6.7 × 10−10 M [HOC6H5] = 1.5 − x = 1.5 M; pH = −log(1.5 × 10−5) = 4.82 c. This is a weak acid in water. Solving the weak acid problem: HF Initial Change Equil. ⇌ H+ + F− Ka = 7.2 × 10−4 ~0 0 0.020 M x mol/L HF dissociates to reach equilibrium −x → +x +x 0.020 − x x x x2 x2 [H + ][F − ] = ≈ [HF] 0.020 − x 0.020 x = [H+] = 3.8 × 10−3 M; Check assumptions: Ka = 7.2 × 10−4 = (assuming x << 0.020) x 3.8 × 10 −3 × 100 = H 100 = 19% 0.020 0.020 The assumption x << 0.020 is not good (x is more than 5% of 0.020). We must solve x2/(0.020 − x) = 7.2 × 10−4 exactly by using either the quadratic formula or the method of successive approximations (see Appendix 1 of the text). Using successive approximations, we let 0.016 M be a new approximation for [HF]. That is, in the denominator try x = 0.0038 (the value of x we calculated making the normal assumption) so that 0.020 − 0.0038 = 0.016; then solve for a new value of x in the numerator. x2 x2 ≈ = 7.2 × 10−4, x = 3.4 × 10−3 0.020 − x 0.016 We use this new value of x to further refine our estimate of [HF], that is, 0.020 − x = 0.020 − 0.0034 = 0.0166 (carrying an extra sig. fig.). x2 x2 ≈ = 7.2 × 10−4, x = 3.5 × 10−3 0.020 − x 0.0166 We repeat, until we get a self-consistent answer. This would be the same answer we would get solving exactly using the quadratic equation. In this case it is, x = 3.5 × 10−3. Thus: [H+] = [F−] = x = 3.5 × 10−3 M; [OH−] = Kw/[H+] = 2.9 × 10−12 M CHAPTER 7 ACIDS AND BASES 203 [HF] = 0.020 − x = 0.020 − 0.0035 = 0.017 M; pH = 2.46 Note: When the 5% assumption fails, use whichever method you are most comfortable with to solve exactly. The method of successive approximations is probably fastest when the percent error is less than ~25% (unless you have a graphing calculator). 42. Major species: HC2H2ClO2 (Ka = 1.35 × 10−3) and H2O; major source of H+: HC2H2ClO2 HC2H2ClO2 Initial Change Equil. ⇌ + C2H2ClO2− H+ ~0 0 0.10 M x mol/L HC2H2ClO2 dissociates to reach equilibrium −x → +x +x 0.10 − x x x Ka = 1.35 × 10−3 = x2 x2 ≈ , x = 1.2 × 10−2 M 0.10 − x 0.10 Checking the assumptions finds that x is 12% of 0.10 which fails the 5% rule. We must solve 1.35 × 10−3 = x2/(0.10 − x) exactly using either the method of successive approximations or the quadratic equation. Using either method gives x = [H+] = 1.1 × 10−2 M. pH = −log[H+] = −log(1.1 × 10−2) = 1.96. 43. Major species: HIO3, H2O; major source of H+: HIO3 (a weak acid, Ka = 0.17) ⇌ HIO3 H+ + IO3− Change ~0 0 0.010 M x mol/L HIO3 dissociates to reach equilibrium −x → +x +x Equil. 0.010 − x Ka = 0.17 = [H + ][IO3 ] x2 x2 = ≈ , x = 0.041; check assumption. [HIO3 ] 0.010 − x 0.010 Initial x x − Assumption is horrible (x is more than 400% of 0.010). When the assumption is this poor, it is generally quickest to solve exactly using the quadratic formula (see Appendix 1 in text). Using the quadratic formula and carrying extra significant figures: 0.17 = x= x2 , x2 = 0.17(0.010 − x), x2 + (0.17)x − 1.7 × 10−3 = 0 0.010 − x − 0.17 ± [(0.17) 2 − 4(1)(−1.7 × 10 −3 )]1/ 2 − 0.17 ± 0.189 = , x = 9.5 × 10−3 M 2(1) 2 (x must be positive) x = 9.5 × 10−3 M = [H+]; pH = −log(9.5 × 10−3) = 2.02 204 44. CHAPTER 7 ACIDS AND BASES HC3H5O2 (Ka = 1.3 × 10−5) and H2O (Ka = Kw = 1.0 × 10−14) are the major species present. HC3H5O2 will be the dominant producer of H+ because HC3H5O2 is a stronger acid than H2O. Solving the weak acid problem: ⇌ HC3H5O2 H+ + C3H5O2− ~0 0 0.100 M x mol/L HC3H5O2 dissociates to reach equilibrium −x → +x +x 0.100 − x x x Initial Change Equil. Ka = 1.3 × 10−5 = − [H + ][C3 H 5 O 2 ] x2 x2 = ≈ [HC3 H 5 O 2 ] 0.100 − x 0.100 x = [H+] = 1.1 × 10−3 M; pH = −log(1.1 × 10−3) = 2.96 Assumption follows the 5% rule (x is 1.1% of 0.100). [H+] = [C3H5O2−] = 1.1 × 10−3 M; [OH−] = Kw/[H+] = 9.1 × 10−12 M [HC3H5O2] = 0.100 − 1.1 × 10−3 = 0.099 M Percent dissociation = 45. [H + ] 1.1 × 10 −3 × 100 = × 100 = 1.1% [HC3 H 5 O 2 ]0 0.100 This is a weak acid in water. We must solve a weak acid problem. Let HBz = C6H5CO2H. 0.56 g HBz × 1 mol HBz = 4.6 × 10−3 mol; [HBz]0 = 4.6 × 10−3 M 122.1 g HBz Initial Change Equil. ⇌ H+ + Bz- 4.6 × 10−3 M ~0 0 x mol/L HBz dissociates to reach equilibrium −x → +x +x 4.6 × 10−3 − x x x Ka = 6.4 × 10−5 = x2 x2 [H + ][Bz − ] ≈ = [HBz] (4.6 × 10 −3 − x) 4.6 × 10 −3 x = [H+] = 5.4 × 10−4; check assumptions: x 5.4 × 10 −4 × 100 = × 100 = 12% 4.6 × 10 −3 4.6 × 10 −3 Assumption is not good (x is 12% of 4.6 × 10−3). When assumption(s) fail, we must solve exactly using the quadratic formula or the method of successive approximations (see Appendix 1 of text). Using successive approximations: CHAPTER 7 ACIDS AND BASES 205 x2 = 6.4 × 10−5, x = 5.1 × 10−4 −3 −4 (4.6 × 10 ) − (5.4 × 10 ) x2 = 6.4 × 10−5, x = 5.1 × 10−4 M (consistent answer) (4.6 × 10 −3 ) − (5.1 × 10 − 4 ) Thus: x = [H+] = [Bz−] = [C6H5CO2−] = 5.1 × 10−4 M [HBz] = [C6H5CO2H] = 4.6 × 10−3 − x = 4.1 × 10−3 M pH = −log(5.1 × 10−4) = 3.29; pOH = 14.00 − pH = 10.71; [OH−] = 10−10.71 = 1.9 × 10−11 M 46. ⇌ HBz Initial Change Equil. Ka = C H+ ~0 + Bz- HBz = C6H5CO2H 0 C = [HBz]0 = concentration of HBz that dissolves to give saturated solution. x mol/L HBz dissociates to reach equilibrium −x → +x +x x x C−x x2 [H + ][Bz − ] = 6.4 × 10−5 = , where x = [H+] [HBz] C−x 6.4 × 10−5 = [ H + ]2 ; pH = 2.80; [H+] = 10−2.80 = 1.6 × 10−3 M + C − [H ] C − 1.6 × 10−3 = (1.6 × 10 −3 ) 2 = 4.0 × 10−2, C = (4.0 × 10−2) + (1.6 × 10−3) = 4.2 × 10−2 M 6.4 × 10 −5 The molar solubility of C6H5CO2H is 4.2 × 10−2 mol/L. 4.2 × 10 −2 mol C 6 H 5 CO 2 H 122.1 g C 6 H 5 CO 2 H × × 0.100 L = 0.51 g per 100. mL solution L mol C 6 H 5 CO 2 H 2 tablets × 47. [HC9H7O4] = HC9H7O4 Initial Change Equil. 0.325 g HC9 H 7 O 4 1 mol HC9 H 7 O 4 × tablet 180.15 g = 0.0152 M 0.237 L ⇌ H+ + C9H7O4− ~0 0 0.0152 M x mol/L HC9H7O4 dissociates to reach equilibrium !x → +x +x x x 0.0152 ! x 206 CHAPTER 7 Ka = 3.3 × 10 −4 = ACIDS AND BASES − [ H + ] [C 9 H 7 O 4 ] x2 x2 = ≈ , x = 2.2 × 10 −3 M [HC9 H 7 O 4 ] 0.0152 − x 0.0152 Assumption that 0.0152 – x ≈ 0.0152 fails the 5% rule: 2.2 × 10 −3 × 100 = 14% 0.0152 Using successive approximations or the quadratic equation gives an exact answer of x = 2.1 × 10 −3 M. [H+] = x = 2.1 × 10 −3 M; pH = !log(2.1 × 10 −3 ) = 2.68 48. HF and HOC6H5 are both weak acids with Ka values of 7.2 × 10−4 and 1.6 × 10−10, respectively. Because the Ka value for HF is much greater than the Ka value for HOC6H5, HF will be the dominant producer of H+ (we can ignore the amount of H+ produced from HOC6H5 because it will be insignificant). ⇌ HF Initial Change Equil. H+ + F− 1.0 M ~0 0 x mol/L HF dissociates to reach equilibrium −x → +x +x 1.0 − x x x Ka = 7.2 × 10−4 = x2 x2 [H + ][F − ] = ≈ 1 .0 − x 1.0 [ HF] x = [H+] = 2.7 × 10−2 M; pH = −log(2.7 × 10−2) = 1.57; assumptions good. Solving for [OC6H5−] using HOC6H5 ⇌ H+ + OC6H5− equilibrium: Ka = 1.6 × 10−10 = − − [H + ][OC 6 H 5 ] (2.7 × 10 −2 )[OC 6 H 5 ] = , [HOC 6 H 5 ] 1.0 [OC6H5−] = 5.9 × 10−9 M Note that this answer indicates that only 5.9 × 10−9 M HOC6H5 dissociates, which indicates that HF is truly the only significant producer of H+ in this solution. 49. a. HCl is a strong acid. It will produce 0.10 M H+. HOCl is a weak acid. Let's consider the equilibrium: HOCl Initial Change Equil. ⇌ H+ + OCl− Ka = 3.5 × 10−8 0.10 M 0.10 M 0 x mol/L HOCl dissociates to reach equilibrium −x → +x +x 0.10 − x 0.10 + x x CHAPTER 7 ACIDS AND BASES Ka = 3.5 × 10−8 = 207 [H + ][OCl − ] (0.10 + x)( x) = ≈ x, x = 3.5 × 10−8 M [HOCl] 0.10 − x Assumptions are great (x is 0.000035% of 0.10). We are really assuming that HCl is the only important source of H+, which it is. The [H+] contribution from HOCl, x, is negligible. Therefore, [H+] = 0.10 M; pH = 1.00. b. HNO3 is a strong acid, giving an initial concentration of H+ equal to 0.050 M. Consider the equilibrium: HC2H3O2 ⇌ H+ + C2H3O2− Ka = 1.8 × 10−5 0.50 M 0.050 M 0 x mol/L HC2H3O2 dissociates to reach equilibrium −x → +x +x 0.50 − x 0.050 + x x Initial Change Equil. − [H + ][C 2 H 3O 2 ] (0.050 + x) x (0.050) x = ≈ Ka = 1.8 × 10 = [HC 2 H 3O 2 ] (0.50 − x) 0.50 −5 x = 1.8 × 10−4; assumptions are good (well within the 5% rule). [H+] = 0.050 + x = 0.050 M and pH = 1.30 50. Let HA symbolize the weak acid. Set up the problem like a typical weak acid equilibrium problem. HA Initial Change Equil. ⇌ H+ A− + ~0 0 0.15 M x mol/L HA dissociates to reach equilibrium !x → +x +x 0.15 ! x x x If the acid is 3.0% dissociated, then x = [H+] is 3.0% of 0.15: x = 0.030 × (0.15 M) = 4.5 × 10 −3 M. Now that we know the value of x, we can solve for Ka. Ka = ( 4.5 × 10 −3 ) 2 x2 [H + ][A − ] = = = 1.4 × 10 −4 −3 0.15 − x [HA] 0.15 − ( 4.5 × 10 ) 51. HX Initial Change Equil. ⇌ H+ + X− I ~0 0 where I = [HX]0 x mol/L HX dissociates to reach equilibrium −x → +x +x I−x x x From the problem, x = 0.25(I) and I − x = 0.30 M. 208 CHAPTER 7 ACIDS AND BASES I − 0.25(I) = 0.30 M, I = 0.40 M and x = 0.25(0.40 M) = 0.10 M Ka = 52. (0.10) 2 x2 [H + ][X − ] = = = 0.033 I −x 0.30 [HX] In all parts of this problem, acetic acid (HC2H3O2) is the best weak acid present. We must solve a weak acid problem. HC2H3O2 a. Initial Change Equil. ⇌ H+ + C2H3O2− 0.50 M ~0 0 x mol/L HC2H3O2 dissociates to reach equilibrium −x → +x +x 0.50 − x x x Ka = 1.8 × 10−5 = − [H + ][C 2 H 3O 2 ] x2 x2 = = [HC 2 H 3O 2 ] 0.50 − x 0.50 x = [H+] = [C2H3O2−] = 3.0 × 10−3 M; assumptions good. Percent dissociation = [H + ] 3.0 × 10 −3 H 100 = H 100 = 0.60% [HC 2 H 3O 2 ]0 0.50 b. The setup for solutions b and c are similar to solution a except that the final equation is different because the new concentration of HC2H3O2 is different. Ka = 1.8 × 10−5 = x2 x2 ≈ 0.050 − x 0.050 x = [H+] = [C2H3O2−] = 9.5 × 10−4 M; assumptions good. Percent dissociation = c. Ka = 1.8 × 10−5 = 9.5 × 10 −4 × 100 = 1.9% 0.050 x2 x2 ≈ 0.0050 − x 0.0050 x = [H+] = [C2H3O2−] = 3.0 × 10−4 M; check assumptions. Assumption that x is negligible is borderline (6.0% error). We should solve exactly. Using the method of successive approximations (see Appendix 1 of the text): x2 x2 = 1.8 × 10−5 = , x = 2.9 × 10−4 0.0047 0.0050 − (3.0 × 10 − 4 ) Next trial also gives x = 2.9 × 10−4. CHAPTER 7 ACIDS AND BASES Percent dissociation = 209 2.9 × 10 −4 × 100 = 5.8% 5.0 × 10 −3 d. As we dilute a solution, all concentrations are decreased. Dilution will shift the equilibrium to the side with the greater number of particles. For example, suppose we double the volume of an equilibrium mixture of a weak acid by adding water; then: ⎛ [H + ]eq ⎞ ⎛ [X − ]eq ⎟⎜ ⎜ ⎜ 2 ⎟⎜ 2 ⎠⎝ Q= ⎝ ⎛ [HX]eq ⎞ ⎜⎜ ⎟⎟ ⎝ 2 ⎠ ⎞ ⎟ ⎟ ⎠ = 1 K a 2 Q < Ka, so the equilibrium shifts to the right or toward a greater percent dissociation. e. [H+] depends on the initial concentration of weak acid and on how much weak acid dissociates. For solutions a-c, the initial concentration of acid decreases more rapidly than the percent dissociation increases. Thus [H+] decreases. 53. pH = 2.77, [H+] = 10−2.77 = 1.7 × 10−3 M HOCN ⇌ Initial 0.0100 Equil. 0.0100 − x H+ ~0 x + OCN− 0 x x = [H+] = [OCN−] = 1.7 × 10−3 M; [HOCN] = 0.0100 − x = 0.0100 − 0.0017 = 0.0083 M Ka = 54. (1.7 × 10 −3 ) 2 [H + ][OCN − ] = = 3.5 × 10−4 −3 [HOCN] 8.3 × 10 HClO4 is a strong acid with [H+] = 0.040 M. This equals the [H+] in the trichloroacetic acid solution. Set up the problem using the Ka equilibrium reaction for CCl3CO2H. CCl3CO2H Initial Equil. ⇌ 0.050 M 0.050 − x H+ ~0 x + CCl3CO2− 0 x − 55. Ka = [H + ][CCl 3CO 2 ] x2 = ; from the problem, x = [H+] = 4.0 × 10−2 M. [CCl 3CO 2 H ] 0.050 − x Ka = (4.0 × 10 −2 ) 2 = 0.16 0.050 − ( 4.0 × 10 − 2 ) Major species: HCOOH and H2O; major source of H+: HCOOH 210 CHAPTER 7 HCOOH Initial Change Equil. ⇌ H+ ACIDS AND BASES + HCOO− C ~0 0 where C = [HCOOH]0 x mol/L HCOOH dissociates to reach equilibrium → +x +x −x C−x x x Ka = 1.8 × 10−4 = x2 [H + ][HCOO − ] = , where x = [H+] C−x [HCOOH] 1.8 × 10−4 = [ H + ]2 ; because pH = 2.70: [H+] = 10−2.70 = 2.0 × 10−3 M C − [H + ] 1.8 × 10−4 = ( 2.0 × 10 −3 ) 2 4.0 × 10 −6 −3 ) = , C − (2.0 × 10 , C = 2.4 × 10−2 M C − (2.0 × 10 −3 ) 1.8 × 10 − 4 A 0.024 M formic acid solution will have pH = 2.70. 56. Let HSac = saccharin and I = [HSac]0. HSac Initial Equil. ⇌ I I!x Ka = 2.0 × 10 −12 = 2.0 × 10 −12 = H+ ~0 x + Sac− Ka = 10−11.70 = 2.0 × 10 −12 0 x x2 ; x = [H+] = 10 −5.75 = 1.8 × 10 −6 M I−x (1.8 × 10 −6 ) 2 , I = 1.6 M = [HSac]0. I − (1.8 × 10 −6 ) 100.0 g HC7H4NSO3 × 1 mol 1L 1000 mL × × = 340 mL 183.19 g 1.6 mol L Solutions of Bases 57. NO3−: Kb << Kw because HNO3 is a strong acid. All conjugate bases of strong acids have no base strength in water. H2O: Kb = Kw = 1.0 × 10−14; NH3: Kb = 1.8 × 10−5; C5H5N: Kb =1.7 × 10−9 Base strength = NH3 > C5H5N > H2O > NO3− (As Kb increases, base strength increases.) 58. Excluding water, these are the conjugate acids of the bases in the preceding exercise. In general, the stronger the base, the weaker is the conjugate acid. Note: Even though NH4+ and C5H5NH+ are conjugate acids of weak bases, they are still weak acids with Ka values between Kw and 1. Prove this to yourself by calculating the Ka values for NH4+ and C5H5NH+ (Ka = Kw/Kb). Acid strength = HNO3 > C5H5NH+ > NH4+ > H2O CHAPTER 7 59. ACIDS AND BASES a. C6H5NH2 b. C6H5NH2 211 c. OH− d. CH3NH2 The base with the largest Kb value is the strongest base (K b , C6 H 5 NH 2 = 3.8 × 10 −10 , K b , CH 3 NH 2 = 4.4 × 10 −4 ). OH− is the strongest base possible in water. 60. a. HClO4 (a strong acid) b. C6H5NH3+ c. C6H5NH3+ The acid with the largest Ka value is the strongest acid. To calculate Ka values for C6H5NH3+ and CH3NH3+, use Ka = Kw/Kb , where Kb refers to the bases C6H5NH2 or CH3NH2. 61. NaOH(aq) → Na+(aq) + OH−(aq); NaOH is a strong base that completely dissociates into Na+ and OH−. The initial concentration of NaOH will equal the concentration of OH− donated by NaOH. a. [OH−] = 0.10 M; pOH = −log[OH−] = −log(0.10) = 1.00 pH = 14.00 − pOH = 14.00 − 1.00 = 13.00 Note that H2O is also present, but the amount of OH− produced by H2O will be insignificant compared to the 0.10 M OH− produced from the NaOH. b. The [OH−] concentration donated by the NaOH is 1.0 × 10−10 M. Water by itself donates 1.0 × 10−7 M. In this exercise, water is the major OH− contributor, and [OH−] = 1.0 × 10−7 M. pOH = −log(1.0 × 10−7) = 7.00; pH = 14.00 − 7.00 = 7.00 c. [OH−] = 2.0 M; pOH = −log(2.0) = −0.30; pH = 14.00 − (−0.30) = 14.30 62. a. Ca(OH)2 → Ca2+ + 2 OH−; Ca(OH)2 is a strong base and dissociates completely. [OH−] = 2(0.00040) = 8.0 × 10−4 M; pOH = −log[OH−] = 3.10; pH = 14.00 − pOH = 10.90 b. 25 g KOH 1 mol KOH × = 0.45 mol KOH/L L 56.1 g KOH KOH is a strong base, so [OH−] = 0.45 M; pOH = −log(0.45) = 0.35; pH = 13.65 c. 150.0 g NaOH 1 mol × = 3.750 M; NaOH is a strong base, so [OH−] = 3.750 M. L 40.00 g pOH = −log(3.750) = −0.5740 and pH = 14.0000 − (−0.5740) = 14.5740 Although we are justified in calculating the answer to four decimal places, in reality the pH can only be measured to ±0.01 pH units. 63. pH = 10.50; pOH = 14.00 − 10.50 = 3.50; [OH−] = 10−3.50 = 3.2 × 10−4 M 212 CHAPTER 7 ACIDS AND BASES Ba(OH)2(aq) → Ba2+(aq) + 2 OH−(aq); Ba(OH)2 donates 2 mol OH− per mol Ba(OH)2. [Ba(OH)2] = 3.2 × 10−4 M OH− × 1 M Ba (OH) 2 = 1.6 × 10−4 M Ba(OH)2 2 M OH − A 1.6 × 10−4 M Ba(OH)2 solution will produce a pH = 10.50 solution. 64. pOH = 14.00 – 11.56 = 2.44; [OH−] = [KOH] = 10 −2.44 = 3.6 × 10 −3 M 3.6 × 10 −3 mol KOH 56.1 g KOH × = 0.16 g KOH L mol KOH 0.8000 L × 65. Neutrally charged organic compounds containing at least one nitrogen atom generally behave as weak bases. The nitrogen atom has an unshared pair of electrons around it. This lone pair of electrons is used to form a bond to H+. 66. These are solutions of weak bases in water. In each case, we must solve an equilibrium weak base problem. a. (C2H5)3N + H2O Initial Change Equil. ⇌ (C2H5)3NH+ + OH− Kb = 4.0 × 10 −4 0.20 M 0 ~0 x mol/L of (C2H5)3N reacts with H2O to reach equilibrium !x → +x +x x x 0.20 ! x Kb = 4.0 × 10 −4 = [(C 2 H 5 ) 3 NH + ][OH − ] x2 x2 = ≈ , x = [OH−] = 8.9 × 10 −3 M [(C 2 H 5 ) 3 N ] 0.20 − x 0.20 Assumptions good (x is 4.5% of 0.20). [OH−] = 8.9 × 10 −3 M [H+] = Kw 1.0 × 10 −14 = = 1.1 × 10 −12 M; pH = 11.96 − −3 [OH ] 8.9 × 10 HONH2 + H2O b. Initial Equil. ⇌ HONH3+ 0.20 M 0.20 ! x Kb = 1.1 × 10 −8 = 0 x + OH− Kb = 1.1 × 10 −8 ~0 x x2 x2 ≈ , x = [OH−] = 4.7 × 10 −5 M; assumptions good. 0.20 − x 0.20 [H+] = 2.1 × 10 −10 M; pH = 9.68 67. This is a solution of a weak base in water. We must solve a weak base equilibrium problem. CHAPTER 7 ACIDS AND BASES C2H5NH2 Initial Change Equil. 213 ⇌ + H2O C2H5NH3+ + OH− Kb = 5.6 × 10−4 0.20 M 0 ~0 x mol/L C2H5NH2 reacts with H2O to reach equilibrium !x → +x +x 0.20 ! x x x + Kb = [C 2 H 5 NH 3 ][OH − ] x2 x2 = ≈ (assuming x << 0.20) [C 2 H 5 NH 2 ] 0.20 − x 0.20 1.1 × 10 −2 × 100 = 5.5% 0.20 The assumption fails the 5% rule. We must solve exactly using either the quadratic equation or the method of successive approximations (see Appendix 1 of the text). Using successive approximations and carrying extra significant figures: x = 1.1 × 10 −2 ; checking assumption: x2 x2 = = 5.6 × 10−4, x = 1.0 × 10 −2 M (consistent answer) 0.20 − 0.011 0.189 x = [OH−] = 1.0 × 10 −2 M; [H+] = ⇌ (C2H5)2NH + H2O 68. Initial Change Equil. Kw 1.0 × 10 −14 = 1.0 × 10 −12 M; pH = 12.00 = − −2 [OH ] 1.0 × 10 (C2H5)2NH2+ + OH− Kb = 1.3 × 10 −3 0.050 M 0 ~0 x mol/L (C2H5)2NH reacts with H2O to reach equilibrium !x → +x +x x x 0.050 ! x Kb = 1.3 × 10 −3 = + [(C 2 H 5 ) 2 NH 2 ][OH − ] x2 x2 = ≈ [(C 2 H 5 ) 2 NH] 0.050 − x 0.050 x = 8.1 × 10 −3 ; assumption is bad (x is 16% of 0.20). Using successive approximations: x2 1.3 × 10 −3 = , x = 7.4 × 10 −3 0.050 − 0.081 1.3 × 10 −3 = x2 , x = 7.4 × 10 −3 (consistent answer) 0.050 − 0.074 [OH−] = x = 7.4 × 10 −3 M; [H+] = Kw/[OH−] = 1.4 × 10 −12 M; pH = 11.85 69. 5.0 × 10 −3 g 1 mol × = 1.7 × 10−3 M = [codeine]0; let cod = codeine (C18H21NO3). 0.0100 L 299.4 g 214 CHAPTER 7 ACIDS AND BASES Solving the weak base equilibrium problem: cod + H2O Initial Change Equil. ⇌ codH+ OH− + Kb = 10−6.05 = 8.9 × 10−7 0 ~0 1.7 × 10−3 M x mol/L codeine reacts with H2O to reach equilibrium −x → +x +x x x 1.7 × 10−3 − x Kb = 8.9 × 10−7 = x2 x2 , x = 3.9 × 10−5 ; assumptions good. ≈ (1.7 × 10 −3 − x) 1.7 × 10 −3 [OH−] = 3.9 × 10−5 M; [H+] = Kw/[OH−] = 2.6 × 10−10 M; pH = −log[H+] = 9.59 70. Codeine = C18H21NO3; codeine sulfate = C36H44N2O10S The formula for codeine sulfate works out to (codeine H+)2SO42−, where codeine H+ = HC18H21NO3+. Two codeine molecules are protonated by H2SO4, forming the conjugate acid of codeine. The SO42− then acts as the counter ion to give a neutral compound. Codeine sulfate is an ionic compound that is more soluble in water than codeine, allowing more of the drug into the bloodstream. 71. To solve for percent ionization, just solve the weak base equilibrium problem. a. NH3 + H2O ⇌ NH4+ Initial 0.10 M Equil. 0.10 − x Kb = 1.8 × 10−5 = 0 x 1.8 × 10−5 = Kb = 1.8 × 10−5 ~0 x [OH − ] 1.3 × 10 −3 M × 100 = × 100 = 1.3% [ NH 3 ]0 0.10 M NH3 + H2O Initial Equil. OH− x2 x2 ≈ , x = [OH−] = 1.3 × 10−3 M; assumptions good. 0.10 − x 0.10 Percent ionization = b. + 0.010 M 0.010 − x ⇌ NH4+ + 0 x OH− ~0 x x2 x2 ≈ , x = [OH−] = 4.2 × 10−4 M; assumptions good. 0.010 − x 0.010 Percent ionization = 4.2 × 10 −4 × 100 = 4.2% 0.010 Note: For the same base, the percent ionization increases as the initial concentration of base decreases. CHAPTER 7 ACIDS AND BASES c. CH3NH2 + H2O 215 ⇌ CH3NH3+ 0.10 M 0.10 − x Initial Equil. + 0 x OH− Kb = 4.38 × 10−4 ~0 x x2 x2 ≈ , x = 6.6 × 10−3; assumption fails the 5% rule (x is 0.10 − x 0.10 6.6% of 0.10). Using successive approximations and carrying extra significant figures: 4.38 × 10−4 = x2 x2 = = 4.38 × 10−4, x = 6.4 × 10−3 0.10 − 0.0066 0.093 Percent ionization = 72. (consistent answer) 6.4 × 10 −3 × 100 = 6.4% 0.10 1.0 g quinine 1 mol quinine × = 1.6 × 10−3 M quinine; let Q = quinine = C20H24N2O2. 1.9000 L 324.4 g quinine Q Initial Change Equil. + H2O ⇌ QH+ + OH− Kb = 10−5.1 = 8 × 10−6 1.6 × 10−3 M 0 ~0 x mol/L quinine reacts with H2O to reach equilibrium → +x +x −x −3 1.6 × 10 − x x x Kb = 8 × 10−6 = [QH + ][OH − ] x2 x2 ≈ = [Q ] (1.6 × 10 −3 − x) 1.6 × 10 −3 x = 1 × 10−4; assumption fails 5% rule (x is 6% of 0.0016). Using successive approximations: x2 = 8 × 10−6, x = 1 × 10−4 M (consistent answer) −3 −4 (1.6 × 10 − 1 × 10 ) x = [OH−] = 1 × 10−4 M; pOH = 4.0; pH = 10.0 73. Using the Kb reaction to solve where PT = p-toluidine (CH3C6H4NH2): PT Initial Change Equil. + H2O ⇌ PTH+ + OH− 0.016 M 0 ~0 x mol/L of PT reacts with H2O to reach equilibrium −x → +x +x 0.016 − x x x 216 CHAPTER 7 Kb = ACIDS AND BASES x2 [PTH + ][OH − ] = 0.016 − x [PT ] Since pH = 8.60: pOH = 14.00 − 8.60 = 5.40 and [OH−] = x = 10−5.40 = 4.0 × 10−6 M Kb = (4.0 × 10 −6 ) 2 = 1.0 × 10−9 0.016 − ( 4.0 × 10 −6 ) 74. HONH2 + H2O Initial Equil. ⇌ I I–x Kb = 1.1 × 10 −8 = HONH3+ + OH− 0 x ~0 x Kb = 1.1 × 10 −8 I = [HONH2]0 x2 I−x From problem, pH = 10.00, so pOH = 4.00 and x = [OH−] = 1.0 × 10 −4 M. 1.1 × 10 −8 = (1.0 × 10 −4 ) 2 , I = 0.91 M I − (1.0 × 10 − 4 ) Mass HONH2 = 0.2500 L × 0.91 mol HONH 2 33.03 g HONH 2 × = 7.5 g HONH2 L mol HONH 2 Polyprotic Acids 75. 76. − H3C6H5O7(aq) ⇌ H2C6H5O7 (aq) + H (aq) [H 2 C 6 H 5 O 7 ][H + ] K a1 = [ H 3C 6 H 5 O 7 ] H2C6H5O7−(aq) ⇌ HC6H5O72−(aq) + H+(aq) Ka2 = HC6H5O72−(aq) ⇌ C6H5O73−(aq) + H+(aq) K a3 = − + 2− [HC 6 H 5 O 7 ][H + ] − [H 2 C 6 H 5O 7 ] 3− [C 6 H 5 O 7 ][H + ] 2− [HC 6 H 5 O 7 ] H2CO3 is a weak acid with K a1 = 4.3 × 10 −7 and K a 2 = 4.8 × 10 −11 . The [H+] concentration in solution will be determined from the K a1 reaction because K a1 >> K a 2 . Because K a1 << 1, the [H+] < 0.10 M; only a small percentage of the 0.10 M H2CO3 will dissociate into HCO3− and H+. So statement a best describes the 0.10 M H2CO3 solution. H2SO4 is a strong acid as well as a very good weak acid ( K a1 >> 1, K a 2 = 1.2 × 10 −2 ). All of the 0.10 M H2SO4 solution will dissociate into 0.10 M H+ and 0.10 M HSO4−. However, because HSO4− is a good weak acid due to the relatively large Ka value, some of the 0.10 M HSO4− will dissociate into some more H+ and SO42−. Therefore, the [H+] will be greater than 0.10 M but will not reach 0.20 M because only some of 0.10 M HSO4− will dissociate. Statement c is best for a 0.10 M H2SO4 solution. CHAPTER 7 77. ACIDS AND BASES 217 The reactions are: H3AsO4 ⇌ H+ + H2AsO4− K a1 = 5 × 10−3 H2AsO4− ⇌ H+ + HAsO42− K a 2 = 8 × 10−8 HAsO42− ⇌ H+ + AsO43− K a 3 = 6 × 10−10 We will deal with the reactions in order of importance, beginning with the largest Ka, K a1 . H3AsO4 ⇌ H+ 0.20 M 0.20 - x Initial Equil. 5 × 10−3 = ~0 x + H2AsO4− K a1 = 5 × 10−3 = − [H + ][H 2 AsO 4 ] [H 3 AsO 4 ] 0 x x2 x2 ≈ , x = 3 × 10−2 M; assumption fails the 5% rule. x − 0.20 0.20 Solving by the method of successive approximations: 5 × 10−3 = x2/(0.20 − 0.03), x = 3 × 10−2 (consistent answer) [H+] = [H2AsO4−] = 3 × 10−2 M; [H3AsO4] = 0.20 - 0.03 = 0.17 M 2− Because K a 2 = [H + ][HAsO 4 ] − [H 2 AsO 4 ] = 8 × 10−8 is much smaller than the K a1 value, very little of − H2AsO4 (and HAsO42−) dissociates compared to H3AsO4. Therefore, [H+] and [H2AsO4−] will not change significantly by the K a 2 reaction. Using the previously calculated concentrations of H+ and H2AsO4− to calculate the concentration of HAsO42−: 2− (3 × 10 −2 )[HAsO 4 ] , [HAsO42−] = 8 × 10−8 M 8 × 10 = −2 3 × 10 −8 The assumption that the K a 2 reaction does not change [H+] and [H2AsO4−] is good. We repeat the process using K a 3 to get [AsO43−]. K a 3 = 6 × 10−10 = 3− 3− [H + ][AsO 4 ] 2− [HAsO 4 ] = (3 × 10 −2 )[AsO 4 ] 8 × 10 −8 [AsO43−] = 1.6 × 10−15 ≈ 2 × 10−15 M; assumption good. 218 CHAPTER 7 ACIDS AND BASES So in 0.20 M analytical concentration of H3AsO4: [H3AsO4] = 0.17 M; [H+] = [H2AsO4−] = 3 × 10−2 M; [HAsO42−] = 8 × 10−8 M; [AsO43−] = 2 × 10−15 M; [OH−] = Kw/[H+] = 3 × 10−13 M 78. The relevant reactions are: H2CO3 ⇌ H+ + HCO3− K a1 = 4.3 × 10−7; HCO3− ⇌ H+ + CO32− K a 2 = 4.8 × 10−11 Initially, we deal only with the first reaction (since K a1 >> K a 2 ), and then let those results control values of concentrations in the second reaction. ⇌ H2CO3 Initial Equil. 0.010 M 0.010 − x K a1 = 4.3 × 10−7 = H+ + HCO3− ~0 x 0 x − [H + ][HCO 3 ] x2 x2 = ≈ [H 2 CO 3 ] 0.010 − x 0.010 x = 6.6 × 10−5 M = [H+] = [HCO3−]; assumptions good. HCO3− Initial Equil. 6.6 × 10−5 M 6.6 × 10−5 − y ⇌ H+ + 6.6 × 10−5 M 6.6 × 10−5 + y If y is small, then [H+] = [HCO3−] and K a 2 = 4.8 × 10−11 = CO32− 0 y 2− [H + ][CO 3 ] − [HCO 3 ] ≈ y. y = [CO32−] = 4.8 × 10−11 M; assumptions good. The amount of H+ from the second dissociation is 4.8 × 10−11 M or: 4.8 × 10 −11 × 100 = 7.3 × 10−5 % −5 6.6 × 10 This result justifies our treating the equilibria separately. If the second dissociation contributed a significant amount of H+, we would have to treat both equilibria simultaneously. The reaction that occurs when acid is added to a solution of HCO3− is: HCO3−(aq) + H+(aq) → H2CO3(aq) → H2O(l) + CO2(g) The bubbles are CO2(g) and are formed by the breakdown of unstable H2CO3 molecules. We should write H2O(l) + CO2(aq) or CO2(aq) for what we call carbonic acid. It is for convenience, however, that we write H2CO3(aq). CHAPTER 7 79. ACIDS AND BASES 219 For H2C6H6O6. K a1 = 7.9 × 10 −5 and K a 2 = 1.6 × 10 −12 . Because K a1 >> K a 2 , the amount of H+ produced by the K a 2 reaction will be negligible. 1 mol H 2 C 6 H 6 O 6 176.12 g = 0.0142 M 0.2000 L 0.500 g × [H2C6H6O6]0 = H2C6H6O6(aq) Initial Equil. ⇌ 0.0142 M 0.0142 ! x K a1 = 7.9 × 10 −5 = HC6H6O6−(aq) + H+(aq) 0 x K a1 = 7.9 × 10 −5 ~0 x x2 x2 ≈ , x = 1.1 × 10 −3 ; assumption fails the 5% rule. 0.0142 0.0142 − x Solving by the method of successive approximations: 7.9 × 10 −5 = x2 , x = 1.0 × 10 −3 M (consistent answer) 0.0142 − 1.1 × 10 −3 Because H+ produced by the K a 2 reaction will be negligible, [H+] = 1.0 × 10 −3 and pH = 3.00. 80. Because K a 2 for H2S is so small, we can ignore the H+ contribution from the K a 2 reaction. ⇌ H2S Initial Equil. 0.10 M 0.10 ! x K a1 = 1.0 × 10 −7 = H+ HS− ~0 x 0 x K a1 = 1.0 × 10 −7 x2 x2 ≈ , x = [H+] = 1.0 × 10 −4 ; assumptions good. 0.10 − x 0.10 pH = !log(1.0 × 10 −4 ) = 4.00 Use the K a 2 reaction to determine [S2−]. Initial Equil. HS− 1.0 × 10 −4 M 1.0 × 10 −4 ! x K a 2 = 1.0 × 10 −19 = ⇌ H+ + −4 1.0 × 10 M 1.0 × 10 −4 + x (1.0 × 10 −4 + x) x (1.0 × 10 −4 )x ≈ (1.0 × 10 − 4 − x) 1.0 × 10 − 4 x = [S2−] = 1.0 × 10 −19 M; assumptions good. S2− 0 x 220 81. CHAPTER 7 ACIDS AND BASES The dominant H+ producer is the strong acid H2SO4. A 2.0 M H2SO4 solution produces 2.0 M HSO4- and 2.0 M H+. However, HSO4− is a weak acid that could also add H+ to the solution. ⇌ HSO4− H+ SO42− + 2.0 M 2.0 M 0 x mol/L HSO4− dissociates to reach equilibrium −x → +x +x 2.0 − x 2.0 + x x Initial Change Equil. K a 2 = 1.2 × 10−2 = 2− [H + ][SO 4 ] − [HSO 4 ] = (2.0 + x) x 2.0( x) ≈ , x = 1.2 × 10−2 M 2.0 − x 2 .0 Because x is 0.60% of 2.0, the assumption is valid by the 5% rule. The amount of additional H+ from HSO4− is 1.2 × 10−2 M. The total amount of H+ present is: [H+] = 2.0 + (1.2 × 10−2) = 2.0 M; pH = −log(2.0) = -0.30 Note: In this problem H+ from HSO4− could have been ignored. However, this is not usually the case in more dilute solutions of H2SO4. 82. For H2SO4, the first dissociation occurs to completion. The hydrogen sulfate ion, HSO4−, is a weak acid with K a 2 = 1.2 × 10−2. We will consider this equilibrium for additional H+ production: HSO4− Initial Change Equil. ⇌ H+ + SO42− 0.0050 M 0.0050 M 0 x mol/L HSO4− dissociates to reach equilibrium −x → +x +x 0.0050 − x 0.0050 + x x K a 2 = 0.012 = (0.0050 + x) x ≈ x, x = 0.012; assumption is horrible (240% error). 0.0050 − x Using the quadratic formula: 6.0 × 10−5 − (0.012)x = x2 + (0.0050)x, x2 + (0.017)x − 6.0 × 10−5 = 0 x= − 0.017 ± (2.9 × 10 −4 + 2.4 × 10 −4 )1/ 2 − 0.017 ± 0.023 = , x = 3.0 × 10−3 M 2 2 [H+] = 0.0050 + x = 0.0050 + 0.0030 = 0.0080 M; pH = 2.10 Note: We had to consider both H2SO4 and HSO4− for H+ production in this problem. CHAPTER 7 ACIDS AND BASES 221 Acid-Base Properties of Salts 83. a. These are strong acids like HCl, HBr, HI, HNO3, H2SO4, or HClO4. b. These are salts of the conjugate acids of the bases in Table 7.3. These conjugate acids are all weak acids. NH4Cl, CH3NH3NO3, and C2H5NH3Br are three examples. Note that the anions used to form these salts are conjugate bases of strong acids; this is so because they have no acidic or basic properties in water (with the exception of HSO4−, which has weak acid properties). c. These are strong ases like LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2. d. These are salts of the conjugate bases of the neutrally charged weak acids in Table 7.2. The conjugate bases of weak acids are weak bases themselves. Three examples are NaClO2, KC2H3O2, and CaF2. The cations used to form these salts are Li+, Na+, K+, Rb+, Cs+, Ca2+, Sr2+, or Ba2+ because these cations have no acidic or basic properties in water. Notice that these are the cations of the strong bases you should memorize. e. There are two ways to make a neutral salt. The easiest way is to combine a conjugate base of a strong acid (except for HSO4−) with one of the cations from a strong base. These ions have no acidic/basic properties in water, so salts of these ions are neutral. Three examples are NaCl, KNO3, and SrI2. Another type of strong electrolyte that can produce neutral solutions are salts that contain an ion with weak acid properties combined with an ion of opposite charge having weak base properties. If the Ka for the weak acid ion is equal to the Kb for the weak base ion, then the salt will produce a neutral solution. The most common example of this type of salt is ammonium acetate (NH4C2H3O2). For this salt, Ka for NH4+ = Kb for C2H3O2− = 5.6 × 10 −10 . This salt at any concentration produces a neutral solution. 84. Ka × Kb = Kw, −log(Ka × Kb) = −log Kw −log Ka − log Kb = −log Kw, pKa + pKb = pKw = 14.00 (Kw = 1.0 × 10−14 at 25°C) 85. One difficult aspect of acid-base chemistry is recognizing what types of species are present in solution, that is, whether a species is a strong acid, strong base, weak acid, weak base, or a neutral species. Below are some ideas and generalizations to keep in mind that will help in recognizing types of species present. a. Memorize the following strong acids: HCl, HBr, HI, HNO3, HClO4, and H2SO4 b. Memorize the following strong bases: LiOH, NaOH, KOH, RbOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2 c. Weak acids have a Ka value of less than 1 but greater than Kw. Some weak acids are listed in Table 7.2 of the text. Weak bases have a Kb value of less than 1 but greater than Kw. Some weak bases are listed in Table 7.3 of the text. 222 CHAPTER 7 ACIDS AND BASES d. Conjugate bases of weak acids are weak bases, that is, all have a Kb value of less than 1 but greater than Kw. Some examples of these are the conjugate bases of the weak acids listed in Table 7.2 of the text. e. Conjugate acids of weak bases are weak acids, that is, all have a Ka value of less than 1 but greater than Kw. Some examples of these are the conjugate acids of the weak bases listed in Table 7.3 of the text. f. Alkali metal ions (Li+, Na+, K+, Rb+, Cs+) and some alkaline earth metal ions (Ca2+, Sr2+, Ba2+) have no acidic or basic properties in water. g. Conjugate bases of strong acids (Cl−, Br-, I−, NO3−, ClO4−, HSO4−) have no basic properties in water (Kb << Kw), and only HSO4- has any acidic properties in water. Let’s apply these ideas to this problem to see what types of species are present. a. HI: Strong acid; HF: weak acid (Ka = 7.2 × 10−4) NaF: F− is the conjugate base of the weak acid HF, so F− is a weak base. The Kb value for F− = Kw/Ka, HF = 1.4 × 10−11. Na+ has no acidic or basic properties. NaI: Neutral (pH = 7.0); Na+ and I− have no acidic/basic properties. In order of increasing pH, we place the compounds from most acidic (lowest pH) to most basic (highest pH). Increasing pH: HI < HF < NaI < NaF. b. NH4Br: NH4+ is a weak acid (Ka = 5.6 × 10−10), and Br- is a neutral species. HBr: Strong acid KBr: Neutral; K+ and Br- have no acidic/basic properties. NH3: Weak base, Kb = 1.8 × 10−5 Increasing pH: HBr < NH4Br < KBr < NH3 Most Most acidic basic c. C6H5NH3NO3: C6H5NH3+ is a weak acid (K a = K w /K b, C6 H 5 NH 2 = 1.0 × 10−14/3.8 × 10−10 = 2.6 × 10−5), and NO3− is a neutral species. NaNO3: Neutral; Na+ and NO3− have no acidic/basic properties. NaOH: Strong base HOC6H5: Weak acid (Ka = 1.6 × 10−10) KOC6H5: OC6H5− is a weak base (K b = K w /K a, HOC6 H 5 = 6.3 × 10−5), and K+ is a neutral species. C6H5NH2: Weak base (Kb = 3.8 × 10−10) HNO3: Strong acid CHAPTER 7 ACIDS AND BASES 223 This is a little more difficult than the previous parts of this problem because two weak acids and two weak bases are present. Between the weak acids, C6H5NH3+ is a stronger weak acid than HOC6H5 since the Ka value for C6H5NH3+ is larger than the Ka value for HOC6H5. Between the two weak bases, because the Kb value for OC6H5- is larger than the Kb value for C6H5NH2, OC6H5− is a stronger weak base than C6H5NH2. Increasing pH: HNO3 < C6H5NH3NO3 < HOC6H5 < NaNO3 < C6H5NH2 < KOC6H5 < NaOH Most acidic Most basic 86. Reference Table 7.6 of the text and the solution to Exercise 85 for some generalizations on acid-base properties of salts. The letters in parenthesis is(are) the generalization(s) listed in Exercise 85 that identifies that species. CaBr2: Neutral; Ca2+ and Br- have no acidic/basic properties (f and g). KNO2: NO2− is a weak base, Kb = (1.0 × 10−14)/(4.0 × 10−4) = 2.5 × 10−11 (c and d). Ignore K+ (f). HClO4: Strong acid (a) HNO2: Weak acid, Ka = 4.0 × 10−4 (c) HONH3ClO4: HONH3+ is a weak acid, Ka = K w /K b, HONH 2 = (1.0 × 10−14)/(1.1 × 10−8 ) = 9.1 × 10−7 (c and e). Ignore ClO4− (g). NH4NO2: NH4+ is a weak acid, Ka = 5.6 × 10−10 (c and e). NO2− is a weak base, Kb = 2.5 × 10−11 (c and d). Because the Ka value for NH4+ is a slightly larger than Kb for NO2−, the solution will be slightly acidic with a pH a little lower than 7.0. Using the information above (identity and the Ka or Kb values), the ordering is: Most acidic → most basic: HClO4 > HNO2 > HONH3ClO4 > NH4NO2 > CaBr2 > KNO2 87. Reference Table 7.6 of the text and the solution to Exercise 85 for some generalizations on acid-base properties of salts. a. KCl → K+ + Cl− neutral; K+ and Cl− have no effect on pH. b. C2H5NH3CN → C2H5NH3+ + CN− basic; C2H5NH3+ is a weak acid (K a = K w /K b,C 2 H 5 NH 2 = 1.0 × 10−14/5.6 × 10−4 = 1.8 × 10−11), and CN− is a weak base (Kb = Kw/Ka, HCN = 1.0 × 10−14/6.2 × 10−10 = 1.6 × 10−5). Because K b , CN − > K a , C 2 H 5 NH3+ , the solution of C2H5NH3CN will be basic. c. C5H5NHF → C5H5NH+ + F− acidic; C5H5NH+ is a weak acid (K a = K w /K b,C5 H 5 N = 5.9 × 10 −6 ) , and F− is a weak base (Kb = Kw/Ka, HF = 1.4 × 10−11). Because K a , C5 H5 NH + > K b, F− , the solution of C5H5NHF will be acidic. 224 CHAPTER 7 ACIDS AND BASES d. NH4C2H3O2 → NH4+ + C2H3O2− neutral; NH4+ is a weak acid (Ka = 5.6 × 10−10), and C2H3O2− is a weak base (K b = K w /K a, HC 2 H 3O 2 = 5.6 × 10 −10 ). Because K a , NH + = 4 K b, C 2 H 3O 2 − , the solution of NH4C2H3O2 will have pH = 7.00. e. NaHSO3 → Na+ + HSO3− acidic; Na+ has no acidic/basic properties. HSO3− is a weak acid (K a 2 = 1.0 × 10−7), and HSO3− is also a weak base (K b = K w /K a1 , H 2SO 3 = 1.0 × 10−14/1.5 × 10−2 = 6.7 × 10−13). HSO3− is a stronger acid than a base because Ka > Kb. Therefore, the solution is acidic. f. NaHCO3 → Na+ + HCO3− basic; ignore Na+; HCO3− is a weak acid (K a 2 = 4.8 × 10−11), and HCO3− is a weak base (Kb = K w /K a1 , H 2CO3 = 2.3 × 10−8) . HCO3− is a stronger base than an acid because Kb > Ka. Therefore, the solution is basic. 88. a. CH3NH3Cl → CH3NH3+ + Cl−: CH3NH3+ is a weak acid. Cl− is the conjugate base of a strong acid. Cl− has no basic (or acidic) properties. CH3NH3+ ⇌ CH3NH2 + H+ Ka = [CH 3 NH 2 ][H + ] + [CH 3 NH 3 ] = Kw 1.00 × 10 −14 = Kb 4.38 × 10 − 4 = 2.28 × 10 −11 CH3NH3+ ⇌ CH3NH2 + H+ 0 ~0 0.10 M x mol/L CH3NH3+ dissociates to reach equilibrium Change !x → +x +x Equil. 0.10 ! x x x 2 2 x x ≈ (assuming x << 0.10) Ka = 2.28 × 10 −11 = 0.10 − x 0.10 Initial x = [H+] = 1.5 × 10 −6 M; pH = 5.82; assumptions good. b. NaCN → Na+ + CN−: CN− is a weak base. Na+ has no acidic (or basic) properties. CN− + H2O Initial Change Equil. ⇌ HCN + OH− Kb = Kw 1.0 × 10 −14 = Ka 6.2 × 10 −10 0.050 M 0 ~0 Kb = 1.6 × 10 −5 x mol/L CN− reacts with H2O to reach equilibrium !x → +x +x 0.050 ! x x x Kb = 1.6 × 10 −5 = x2 [HCN][OH − ] x2 = ≈ 0.050 − x 0.050 [CN − ] x = [OH−] = 8.9 × 10 −4 M; pOH = 3.05; pH = 10.95; assumptions good. CHAPTER 7 89. ACIDS AND BASES 225 a. KNO2 → K+ + NO2−: NO2− is a weak base. Ignore K+. NO2− + H2O ⇌ HNO2 + OH− Initial 0.12 M Equil. 0.12 ! x 0 x [OH − ][HNO 2 ] Kb = 2.5 × 10 −11 = − [ NO 2 ] Kb = Kw 1.0 × 10 −14 = 2.5 × 10 −11 = −4 Ka 4.0 × 10 ~0 x x2 x2 ≈ 0.12 − x 0.12 = x = [OH−] = 1.7 × 10 −6 M; pOH = 5.77; pH = 8.23; assumptions good. b. NaOCl → Na+ + OCl−: OCl− is a weak base. Ignore Na+. OCl− + H2O ⇌ HOCl + OH− Initial 0.45 M Equil. 0.45 ! x Kb = 2.9 × 10 −7 = 0 x Kb = Kw 1.0 × 10 −14 = = 2.9 × 10 −7 Ka 3.5 × 10 −8 ~0 x x2 x2 [HOCl][OH − ] = ≈ 0.45 − x 0.45 [OCl − ] x = [OH−] = 3.6 × 10 −4 M; pOH = 3.44; pH = 10.56; assumptions good. c. NH4ClO4 → NH4+ + ClO4−: NH4+ is a weak acid. ClO4− is the conjugate base of a strong acid. ClO4− has no basic (or acidic) properties. −14 K NH4+ ⇌ NH3 + H+ Ka = w = 1.0 × 10 −5 = 5.6 × 10 −10 Kb 1.8 × 10 Initial 0.40 M Equil. 0.40 ! x Ka = 5.6 × 10 −10 = 0 x [ NH 3 ][H + ] + [ NH 4 ] ~0 x = x2 x2 ≈ 0.40 − x 0.40 x = [H+] = 1.5 × 10 −5 M; pH = 4.82; assumptions good. 90. Solution is acidic from HSO4− ⇌ H+ + SO42−. Solving the weak acid problem: HSO4− Initial Equil. 0.10 M 0.10 - x ⇌ H+ ~0 x + SO42− 0 x Ka = 1.2 × 10−2 226 CHAPTER 7 Ka = 1.2 × 10−2 = 2− [H + ][SO 4 ] − [HSO 4 ] = ACIDS AND BASES x2 x2 ≈ , x = 0.035 0.10 − x 0.10 Assumption is not good (35% error). Using successive approximations: x2 x2 ≈ = 1.2 × 10−2, x = 0.028 0.10 − x 0.10 − 0.035 x2 = 1.2 × 10−2, x = 0.029 M (consistent answer) 0.10 − 0.028 x = [H+] = 0.029 M; pH = 1.54 If we add Na2CO3 to a solution of NaHSO4, the base CO32− will react with the acid HSO4−. Depending on relative amounts, two reactions are possible. CO32−(aq) + HSO4−(aq) or ⇌ HCO3−(aq) + SO42−(aq) CO32− (aq) + 2 HSO4−(aq) ⇌ 2 SO42−(aq) + H2O(l) + CO2(g) 91. NaN3 → Na+ + N3−; Azide (N3−) is a weak base since it is the conjugate base of a weak acid. All conjugate bases of weak acids are weak bases (Kw < Kb < 1). Ignore Na+. N3− + H2O Initial Change Equil. Kb = ⇌ HN3 + OH− Kb = Kw 1.0 × 10 −14 = = 5.3 × 10−10 −5 Ka 1.9 × 10 0.010 M 0 ~0 x mol/L of N3 reacts with H2O to reach equilibrium −x → +x +x 0.010 − x x x [HN 3 ][OH − ] − [N3 ] , 5.3 × 10−10 = x = [OH−] = 2.3 × 10−6 M; [H+] = x2 x2 ≈ (assuming x << 0.010) 0.010 − x 0.010 1.0 × 10 −14 = 4.3 × 10−9 M; assumptions good. −6 2.3 × 10 [HN3] = [OH−] = 2.3 × 10−6 M; [Na+] = 0.010 M; [N3−] = 0.010 − 2.3 × 10−6 = 0.010 M 92. C2H5NH3Cl → C2H5NH3+ + Cl−; C2H5NH3+ is the conjugate acid of the weak base C2H5NH2 (Kb = 5.6 × 10−4). As is true for all conjugate acids of weak bases, C2H5NH3+ is a weak acid. Cl− has no basic (or acidic) properties. Ignore Cl−. Solving the weak acid problem: CHAPTER 7 ACIDS AND BASES C2H5NH3+ Initial Change Equil. ⇌ 227 C2H5NH2 H+ + Ka = Kw/(5.6 × 10−4) = 1.8 × 10−11 0 ~0 0.25 M x mol/L C2H5NH3+ dissociates to reach equilibrium −x → +x +x 0.25 − x x x Ka = 1.8 × 10−11 = [C 2 H 5 NH 2 ][H + ] + [C 2 H 5 NH 3 ] = x2 x2 ≈ (assuming x << 0.25) 0.25 − x 0.25 x = [H+] = 2.1 × 10−6 M; pH = 5.68; assumptions good. [C2H5NH2] = [H+] = 2.1 × 10−6 M; [C2H5NH3+] = 0.25 M; [Cl−] = 0.25 M [OH−] = Kw/[H+] = 4.8 × 10−9 M 93. All these salts contain Na+, which has no acidic/basic properties and a conjugate base of a weak acid (except for NaCl, where Cl− is a neutral species). All conjugate bases of weak acids are weak bases since Kb values for these species are between Kw and 1. To identify the species, we will use the data given to determine the Kb value for the weak conjugate base. From the Kb value and data in Table 7.2 of the text, we can identify the conjugate base present by calculating the Ka value for the weak acid. We will use A- as an abbreviation for the weak conjugate base. A− + H2O Initial Change Equil. Kb = ⇌ HA + OH− 0.100 mol/1.00 L 0 ~0 x mol/L A− reacts with H2O to reach equilibrium −x → +x +x x x 0.100 − x x2 [HA ][OH − ] = ; from the problem, pH = 8.07: 0.100 − x [A − ] pOH = 14.00 − 8.07 = 5.93; [OH−] = x = 10−5.93 = 1.2 × 10−6 M Kb = (1.2 × 10 −6 ) 2 = 1.4 × 10−11 = Kb value for the conjugate base of a weak acid. −6 0.100 − (1.2 × 10 ) The Ka value for the weak acid equals Kw/Kb: Ka = 1.0 × 10 −14 = 7.1 × 10−4 1.4 × 10 −11 From Table 7.2 of the text, this Ka value is closest to HF. Therefore, the unknown salt is NaF. 228 94. CHAPTER 7 ACIDS AND BASES BHCl → BH+ + Cl−; Cl− is the conjugate base of the strong acid HCl, so Cl− has no acidic/ basic properties. BH+ is a weak acid because it is the conjugate acid of a weak base B. Determining the Ka value for BH+: BH+ Initial Change Equil. Ka = ⇌ B + H+ 0 ~0 0.10 M x mol/L BH+ dissociates to reach equilibrium −x → +x +x x x 0.10 − x x2 [B][H + ] = ; from the problem, pH = 5.82: 0.10 − x [BH + ] [H+] = x = 10−5.82 = 1.5 × 10−6 M; Ka = (1.5 × 10 −6 ) 2 = 2.3 × 10−11 −6 0.10 − (1.5 × 10 ) Kb for the base B = Kw/Ka = (1.0 × 10−14)/(2.3 × 10−11) = 4.3 × 10−4. From Table 7.3 of the text, this Kb value is closest to CH3NH2, so the unknown salt is CH3NH3Cl. 95. Major species: Co(H2O)63+ (Ka = 1.0 × 10−5), Cl− (neutral), and H2O (Kw = 1.0 × 10−14); Co(H2O)63+ will determine the pH since it is a stronger acid than water. Solving the weak acid problem in the usual manner: ⇌ Co(H2O)63+ Initial Equil. Co(H2O)5(OH)2+ 0.10 M 0.10 - x Ka = 1.0 × 10−5 = + 0 x H+ Ka = 1.0 × 10−5 ~0 x x2 x2 ≈ , x = [H+] = 1.0 × 10−3 M 0.10 − x 0.10 pH = −log(1.0 × 10−3) = 3.00; assumptions good. 96. Major species present are H2O, C5H5NH+ [Ka = K w /K b, C5H 5 N = (1.0 × 10−14)/(1.7 × 10−9) = 5.9 × 10−6], and F− [Kb = K w /K a, HF = (1.0 × 10−14)/(7.2 × 10−4) = 1.4 × 10−11]. The reaction to consider is the best acid present (C5H5NH+) reacting with the best base present (F−). Let’s solve by first setting up an ICE table. C5H5NH+(aq) Initial Change Equil. 0.200 M −x 0.200 − x + F−(aq) ⇌ 0.200 M −x → 0.200 − x C5H5N(aq) + HF(aq) 0 +x x 0 +x x CHAPTER 7 ACIDS AND BASES K = K a, C H 5 K= 5 NH × + 1 229 = 5.9 × 10−6 × K a , HF 1 = 8.2 × 10−3 7.2 × 10 − 4 [C 5 H 5 N][HF] x2 −3 , 8.2 × 10 = ; taking the square root of both sides: [C 5 H 5 NH + ][F − ] (0.200 − x) 2 0.091 = x , x = 0.018 − (0.091)x, x = 0.016 M 0.200 − x From the setup to the problem, x = [C5H5N] = [HF] = 0.016 M, and 0.200 − x = 0.200 − 0.016 = 0.184 M = [C5H5NH+] = [F−]. To solve for the [H+], we can use either the Ka equilibrium for C5H5NH+ or the Ka equilibrium for HF. Using C5H5NH+ data: K a, C H 5 5 = 5.9 × 10−6 = NH + [C 5 H 5 N ][H + ] (0.016)[H + ] = , [H+] = 6.8 × 10−5 M + 0 . 184 [C 5 H 5 NH ] pH = −log(6.8 × 10−5) = 4.17 As one would expect, because the Ka for the weak acid is larger than the Kb for the weak base, a solution of this salt should be acidic. 97. Major species: NH4+, OCl−, and H2O; Ka for NH4+ = (1.0 × 10−14)/(1.8 × 10−5) = 5.6 × 10 −10 and Kb for OCl− = (1.0 × 10−14)/(3.5 × 10−8) = 2.9 × 10 −7 . Because OCl− is a better base than NH4+ is an acid, the solution will be basic. The dominant equilibrium is the best acid (NH4+) reacting with the best base (OCl−) present. NH4+ OCl− + 0.50 M –x 0.50 – x Initial Change Equil. K = K a , NH + × 4 K = 0.016 = 1 K a , HOCl 0.50 M –x → 0.50 – x NH3 0 +x x + HOCl 0 +x x = (5.6 × 10−10)/(3.5 × 10−8) = 0.016 [ NH 3 ][HOCl] + ⇌ − [ NH 4 ][OCl ] = x(x) (0.50 − x)(0.50 − x) x2 x = 0.016, = (0.016)1/2 = 0.13, x = 0.058 M 2 0 . 50 − x (0.50 − x) To solve for the H+, use any pertinent Ka or Kb value. Using Ka for NH4+: K a , NH 4 + = 5.6 × 10 −10 = [ NH 3 ][H + ] + [ NH 4 ] = (0.058)[H + ] , [H+] = 4.3 × 10 −9 M, pH = 8.37 0.50 − 0.058 230 98. CHAPTER 7 ACIDS AND BASES Major species: Na+, PO43− (a weak base), and H2O; the Kb value for PO43− is much larger than the Kb values for HPO42− and H2PO4−. We can ignore the contribution of OH− from the Kb reactions for HPO42− and H2PO4− . K b for PO 4 3− = Kw 1.0 × 10 −14 = = 0.021 K a3 4.8 × 10 −13 Note: Kb for HPO4− = K w /K a 2 = 1.6 × 10 −7 and Kb for H2PO4− = K w /K a1 = 1.3 × 10 −12. Indeed, Kb for PO43− >> Kb values for HPO4− and and H2PO4−. ⇌ PO43− + H2O Initial Equil. HPO42− + OH− 0.10 M 0.10 ! x Kb = 0.021 = 0 x Kb = 0.021 ~0 x x2 ; because Kb is so large, the 5% assumption will not hold. Solving 0.10 − x using the quadratic equation: x2 + (0.021)x – 0.0021 = 0, x = [OH−] = 3.7 × 10 −2 , pOH = 1.43, pH = 12.57 Solutions of Dilute Acids and Bases 99. HBrO Initial Change Equil. ⇌ H+ + BrO− Ka = 2 × 10−9 ~0 0 1.0 × 10−6 M x mol/L HBrO dissociates to reach equilibrium −x → +x +x −6 x x 1.0 × 10 − x Ka = 2 × 10−9 = x2 x2 ≈ , x = [H+] = 4 × 10−8 M; pH = 7.4 −6 −6 (1.0 × 10 − x) 1.0 × 10 Let’s check the assumptions. This answer is impossible! We can't add a small amount of an acid to a neutral solution and get a basic solution. The highest pH possible for an acid in water is 7.0. In the correct solution we would have to take into account the autoionization of water. 100. C6H5OH Initial 4.0 × 10−5 M Equil. 4.0 × 10−5 − x ⇌ C6H5O− 0 x + H+ ~0 x C6H5OH = phenol CHAPTER 7 Ka = ACIDS AND BASES 231 x2 x2 −10 , 1.6 × 10 ≈ , x = [H+] = 8.0 × 10−8 M −5 −5 (4.0 × 10 − x) 4.0 × 10 Check assumptions. The assumption that the H+ contribution from water is negligible is poor. Whenever the calculated pH is greater than 6.0 ([H+] < 1 × 10−6 M) for an acid solution, the H+ contribution from water should be considered. From Section 7.9 of the text, try [H+] = (Ka[HA]0 + Kw)1/2 . [H+] = [(1.6 × 10−10)(4.0 × 10−5) + (1.0 × 10−14)]1/2 = 1.3 × 10−7 M [ H + ]2 − K w = 5.3 × 10−8. Assumption [H + ] This equation will work if [HA]0 = 4.0 × 10−5 >> good. [H+] = 1.3 × 10−7 M; pH = 6.89 Note: If the assumption that ([H+]2 − Kw)/[H+] << Ka is bad, then the full equation derived in Section 7.9 of the text should be used. 101. HCN ⇌ Initial 5.0 × 10−4 M Equil. 5.0 × 10−4 − x Ka = H+ ~0 x + CN− Ka = 6.2 × 10−10 0 x x2 x2 = 6.2 × 10−10, x = 5.6 × 10−7; check assumptions. ≈ −4 −4 (5.0 × 10 − x) 5.0 × 10 The assumption that the H+ contribution from water is negligible is poor. Whenever the calculated pH is greater than 6.0 for a weak acid, the water contribution to [H+] must be considered. From Section 7.9 in text: if [ H + ]2 − K w << [HCN]0 = 5.0 × 10−4, then we can use [H+] = (Ka[HCN]o + Kw)1/2. + [H ] Using this formula: [H+] = [(6.2 × 10−10)(5.0 × 10−4) + (1.0 × 10−14)]1/2, [H+] = 5.7 × 10−7 M Checking assumptions: [ H + ]2 − K w = 5.5 × 10−7 << 5.0 × 10−4 + [H ] Assumptions good. pH = −log(5.7 × 10−7) = 6.24 102. We can't neglect the [H+] contribution from H2O since this is a very dilute solution of the strong acid. Following the strategy developed in Section 7.10 of the text, we first determine the charge balance equation and then manipulate this equation to get into one unknown. Charge balance: [H+] = [NO3−] + [OH−], [H+] = [NO3−] + Kw/[H+] [H+]2 − 1.0 × 10−14 = [H+](5.0 × 10−8), [H+]2 − (5.0 × 10−8)[H+] − 1.0 × 10−14 = 0 232 CHAPTER 7 ACIDS AND BASES Using the quadratic formula: [H+] = 1.3 × 10−7 M; pH = 6.89 103. We can't neglect the [H+] contribution from H2O since this is a very dilute solution of the strong acid. Following the strategy developed in Section 7.10 of the text, we first determine the charge balance equation and then manipulate this equation to get into one unknown. [Positive charge] = [negative charge] [H+] = [Cl−] + [OH−] = 7.0 × 10−7 + Kw Kw (because [Cl−] = 7.0 × 10−7 and [OH−] = ) + [H ] [H + ] [ H + ]2 − K w = 7.0 × 10−7, [H+]2 − (7.0 × 10−7)[H+] − 1.0 × 10−14 = 0 + [H ] Using the quadratic formula to solve: [H+] = − (−7.0 × 10 −7 ) ± [(−7.0 × 10 −7 ) 2 − 4(1)(−1.0 × 10 −14 )]1/ 2 2(1) [H+] = 7.1 × 10−7 M; pH = −log(7.1 × 10−7) = 6.15 104. Because this is a very dilute solution of NaOH, we must worry about the amount of OH− donated from the autoionization of water. NaOH → Na+ + OH− H2O ⇌ H+ + OH− Kw = [H+][OH−] = 1.0 × 10−14 This solution, like all solutions, must be charged balanced; that is, [positive charge] = [negative charge]. For this problem, the charge balance equation is: [Na+] + [H+] = [OH−], where [Na+] = 1.0 × 10−7 M and [H+] = Kw [OH − ] Substituting into the charge balance equation: 1.0 × 10−7 + 1.0 × 10 −14 = [OH−], [OH−]2 − (1.0 × 10−7)[OH−] − 1.0 × 10−14 = 0 − [OH ] Using the quadratic formula to solve: [OH−] = − (−1.0 × 10 −7 ) ± [(−1.0 × 10 −7 ) 2 − 4(1)(−1.0 × 10 −14 )]1/ 2 2(1) [OH−] = 1.6 × 10−7 M; pOH = −log(1.6 × 10−7) = 6.80; pH = 7.20 CHAPTER 7 ACIDS AND BASES 233 Additional Exercises 105. a. NH3 + H3O+ ⇌ NH4+ + H2O Keq = + Kb 1.8 × 10 −5 [ NH 4 ] 1 = = = = 1.8 × 109 + + −14 Kw 1.0 × 10 [ NH 3 ][H ] K a for NH 4 b. NO2− + H3O+ ⇌ H2O + HNO2 Keq = c. NH4+ + CH3CO2− ⇌ NH3 + CH3CO2H [HNO 2 ] − + [ NO 2 ][ H ] Keq = = 1 1 = = 2.5 × 103 Ka 4.0 × 10 − 4 [ NH 3 ][CH 3CO 2 H ] + − [ NH 4 ][CH 3CO 2 ] × [H + ] [H + ] + K a for NH 4 Kw = Keq = K a for CH 3CO 2 H (K b for NH 3 )(K a for CH 3CO 2 H) Keq = 1.0 × 10 −14 = 3.1 × 10−5 (1.8 × 10 −5 ) (1.8 × 10 −5 ) d. H3O+ + OH− ⇌ 2 H2O Keq = e. NH4+ + OH− ⇌ NH3 + H2O f. HNO2 + OH− ⇌ 1 = 1.0 × 1014 Kw Keq = 1 = 5.6 × 104 K b for NH 3 H2O + NO2− − K a for HNO 2 [ NO 2 ] 4.0 × 10 −4 [H + ] = = = 4.0 × 1010 × Keq = − + −14 Kw 1.0 × 10 [HNO 2 ][OH ] [H ] 106. a. In the lungs there is a lot of O2, and the equilibrium favors Hb(O2)4. In the cells there is a deficiency of O2, and the equilibrium favors HbH44+. b. CO2 is a weak acid, CO2 + H2O ⇌ HCO3− + H+. Removing CO2 essentially decreases H+. Hb(O2)4 is then favored, and O2 is not released by hemoglobin in the cells. Breathing into a paper bag increases CO2 in the blood, thus increasing [H+], which shifts the reaction left. c. CO2 builds up in the blood, and it becomes too acidic, driving the equilibrium to the left. Hemoglobin can't bind O2 as strongly in the lungs. Bicarbonate ion acts as a base in water and neutralizes the excess acidity. 234 CHAPTER 7 ACIDS AND BASES 107. The light bulb is bright because a strong electrolyte is present; that is, a solute is present that dissolves to produce a lot of ions in solution. The pH meter value of 4.6 indicates that a weak acid is present. (If a strong acid were present, the pH would be close to zero.) Of the possible substances, only HCl (strong acid), NaOH (strong base), and NH4Cl are strong electrolytes. Of these three substances, only NH4Cl contains a weak acid (the HCl solution would have a pH close to zero, and the NaOH solution would have a pH close to 14.0). NH4Cl dissociates into NH4+ and Cl- ions when dissolved in water. Cl− is the conjugate base of a strong acid, so it has no basic (or acidic properties) in water. NH4+, however, is the conjugate acid of the weak base NH3, so NH4+ is a weak acid and would produce a solution with a pH = 4.6 when the concentration is ~1.0 M. NH4Cl is the solute. 108. From the pH, C7H4ClO2− is a weak base. Use the weak base data to determine Kb for C7H4ClO2− (which we will abbreviate as CB−). CB− Initial Equil. + H2O ⇌ 0.20 M 0.20 - x HCB OH− + 0 x ~0 x Because pH = 8.65, pOH = 5.35 and [OH−] = 10−5.35 = 4.5 × 10−6 M = x. Kb = x2 [HCB][OH − ] (4.5 × 10 −6 ) 2 = = = 1.0 × 10−10 − −6 0.20 − x [CB ] 0.20 − (4.5 × 10 ) Because CB− is a weak base, HCB, chlorobenzoic acid, is a weak acid. Solving the weak acid problem: HCB Initial Equil. Ka = 0.20 M 0.20 − x ⇌ H+ + ~0 x CB− 0 x Kw x2 x2 1.0 × 10 −14 −4 , 1 . 0 10 = × = ≈ Kb 0.20 − x 0.20 1.0 × 10 −10 x = [H+] = 4.5 × 10−3 M; pH = 2.35; assumptions good. 109. CaO(s) + H2O(l) → Ca(OH)2(aq); Ca(OH)2(aq) → Ca2+(aq) + 2 OH−(aq) 0.25 g CaO × [OH−] = 1 mol CaO 1 mol Ca (OH) 2 2 mol OH − × × 56.08 g 1 mol CaO mol Ca (OH) 2 = 5.9 × 10 −3 M 1 .5 L pOH = !log(5.9 × 10 −3 ) = 2.23, pH = 14.00 – 2.23 = 11.77 110. 10.0 g NaOCN × 1 mol = 0.154 mol NaOCN 65.01 g CHAPTER 7 ACIDS AND BASES 10.0 g H2C2O4 × 235 1 mol = 0.111 mol H2C2O4 90.04 g Mol NaOCN 0.154 mol (actual) = = 1.39 Mol H 2 C 2 O 4 0.111 mol The balanced reaction requires a larger 2 : 1 mole ratio. Therefore, NaOCN in the numerator is limiting. Because there is a 2 : 2 mole correspondence between mol NaOCN reacted and mol HNCO produced, 1.54 mol HNCO will be produced. HNCO Initial Equil. ⇌ 0.154 mol/0.100 L 1.54 ! x Ka = 1.2 × 10 −4 = H+ NCO− + ~0 x Ka = 1.2 × 10 −4 0 x x2 x2 ≈ , x = [H+] = 1.4 × 10 −2 M 1.54 − x 1.54 pH = –log(1.4 × 10 −2 ) = 1.85; assumptions good. 111. 30.0 mg papH + Cl − 1000 mL 1g 1 mol papH + Cl − 1 mol papH + × × × × mL soln L 1000 mg 378.85 g mol papH + Cl − = 0.0792 M ⇌ papH+ Initial 0.0792 M Equil. 0.0792 ! x Ka = 2.5 × 10 −6 = pap + H+ 0 x Ka = Kw K b, pap = 2.1 × 10 −14 = 2.5 × 10 −6 8.33 × 10 −9 ~0 x x2 x2 ≈ , x = [H+] = 4.4 × 10 −4 M 0.0792 − x 0.0792 pH = -log(4.4 × 10 −4 ) = 3.36; assumptions good. 112. For this problem we will abbreviate CH2=CHCO2H as Hacr and CH2=CHCO2− as acr−. a. Hacr Initial Equil. 0.10 M 0.10 − x ⇌ H+ ~0 x + acr0 x 236 CHAPTER 7 Ka = ACIDS AND BASES x2 x2 , 5.6 × 10−5 ≈ , x = [H+] = 2.4 × 10−3 M; pH = 2.62 0.10 − x 0.10 Assumptions good. b. Percent dissociation = 2.4 × 10 −3 × 100 = 2.4% 0.10 c. For 0.010% dissociation: [acr−] = 1.0 × 10−4(0.10) = 1.0 × 10−5 M Ka = [ H + ] (1.0 × 10 −5 ) [H + ][acr − ] , 5.6 × 10−5 = , [H+] = 0.56 M [Hacr] 0.10 − (1.0 × 10 −5 ) d. acr− is a weak base and the major source of OH− in this solution. acr- + H2O ⇌ Hacr Initial 0.050 M Equil. 0.050 − x [OH − ][ Hacr] Kb = [acr − ] + Kb = ~0 x Kb = 1.8 × 10−10 0 x , 1.8 × 10−10 = Kw 1.0 × 10 −14 = Ka 5.6 × 10 −5 OH− x2 x2 ≈ 0.050 − x 0.050 x = [OH−] = 3.0 × 10−6 M; pOH = 5.52; pH = 8.48; assumptions good. 113. Fe(H2O)63+ + H2O a. Initial Equil. Ka = ⇌ Fe(H2O)5(OH)2+ 0 x 0.10 M 0.10 − x [H 3O + ][Fe( H 2 O) 5 (OH ) 2 + ] 3+ [Fe(H 2 O) 6 ] , 6.0 × 10−3 = + H3O+ ~0 x x2 x2 ≈ 0.10 − x 0.10 x = 2.4 × 10−2 M; assumption is poor (24% error). Using successive approximations: x2 = 6.0 × 10−3, x = 0.021 0.10 − 0.024 x2 x2 = 6.0 × 10−3, x = 0.022; = 6.0 × 10−3, x = 0.022 0.10 − 0.021 0.10 − 0.022 x = [H+] = 0.022 M; pH = 1.66 CHAPTER 7 ACIDS AND BASES [Fe(H 2 O) 5 (OH) 2 + ] b. 3+ [Fe( H 2 O) 6 ] = 237 0.0010 [H + ] (0.0010) ; Ka = 6.0 × 10−3 = 0.9990 0.9990 Solving: [H+] = 6.0 M; pH = −log(6.0) = −0.78 c. Because of the lower charge, Fe2+(aq) will not be as strong an acid as Fe3+(aq). A solution of iron(II) nitrate will be less acidic (have a higher pH) than a solution with the same concentration of iron(III) nitrate. 114. At a pH = 0.00, the [H+] = 10−0.00 = 1.0 M. Begin with 1.0 L × 2.0 mol/L NaOH = 2.0 mol OH−. We will need 2.0 mol HCl to neutralize the OH− plus an additional 1.0 mol excess to reduce to a pH of 0.00. We need 3.0 mol HCl total to achieve pH = 0.00. 115. 0.50 M HA, Ka = 1.0 × 10−3; 0.20 M HB, Ka = 1.0 × 10−10; 0.10 M HC, Ka = 1.0 × 10−12 Major source of H+ is HA because its Ka value is significantly larger than other Ka values. HA Initial Equil. Ka = ⇌ 0.50 M 0.50 − x H+ + ~0 x A− 0 x x2 x2 0.022 , 1.0 × 10−3 ≈ , x = 0.022 M = [H+], × 100 = 4.4% error 0.50 − x 0.50 0.50 Assumption good. Let's check out the assumption that only HA is an important source of H+. (0.022) [B − ] , [B−] = 9.1 × 10−10 M For HB: 1.0 × 10−10 = (0.20) At most, HB will produce an additional 9.1 × 10−10 M H+. Even less will be produced by HC. Thus our original assumption was good. [H+] = 0.022 M. 116. [HA]0 = 1.0 mol = 0.50 mol/L; solve using the Ka equilibrium reaction. 2 .0 L HA Initial Equil. Ka = 0.50 M 0.50 − x ⇌ H+ ~0 x + A− 0 x x2 [H + ][ A − ] = ; in this problem, [HA] = 0.45 M so: [HA] 0.50 − x [HA] = 0.45 M = 0.50 M − x, x = 0.05 M; Ka = 117. (0.05) 2 = 6 × 10−3 0.45 Since NH3 is so concentrated, we need to calculate the OH− contribution from the weak base NH3. 238 CHAPTER 7 ⇌ NH3 + H2O NH4+ 0 x Initial 15.0 M Equil. 15.0 − x Kb = 1.8 × 10−5 = Kb = 1.8 × 10−5 OH− + ACIDS AND BASES 0.0100 M (Assume no volume change.) 0.0100 + x x(0.0100 + x) x(0.0100) ≈ , x = 0.027; assumption is horrible 15.0 − x 15.0 (x is 270% of 0.0100). Using the quadratic formula: (1.8 × 10−5)(15.0 − x) = (0.0100)x + x2, x2 + (0.0100)x − 2.7 × 10−4 = 0 x = 1.2 × 10−2 M, [OH−] = (1.2 × 10−2) + 0.0100 = 0.022 M 118. [H+]0 = (1.0 × 10−2) + (1.0 × 10−2) = 2.0 × 10−2 M from strong acids HCl and H2SO4. HSO4− is a good weak acid (Ka = 0.012). However, HCN is a poor weak acid (Ka = 6.2 × 10−10) and can be ignored. Calculating the H+ contribution from HSO4-: HSO4− Initial Equil. Ka = ⇌ H+ SO42− + 0.020 M 0.020 + x 0.010 M 0.010 − x Ka = 0.012 0 x x(0.020 + x) x(0.020) , 0.012 ≈ , x = 0.0060; assumption poor (60% error). 0.010 − x 0.010 Using the quadratic formula: x2 + (0.032)x − 1.2 × 10−4 = 0, x = 3.4 × 10−3 M [H+] = 0.020 + x = 0.020 + (3.4 × 10−3) = 0.023 M; pH = 1.64 119. a. The initial concentrations are halved since equal volumes of the two solutions are mixed. HC2H3O2 Initial Equil. 0.100 M 0.100 − x Ka = 1.8 × 10−5 = ⇌ H+ + 5.00 × 10−4 M 5.00 × 10−4 + x C2H3O2− 0 x x(5.00 × 10 −4 + x) x(5.00 × 10 −4 ) ≈ 0.100 − x 0.100 x = 3.6 × 10−3; assumption is horrible. Using the quadratic formula: x2 + (5.18 × 10−4)x − 1.8 × 10−6 = 0 x = 1.1 × 10−3 M; [H+] = 5.00 × 10−4 + x = 1.6 × 10−3 M; pH = 2.80 b. x = [C2H3O2−] = 1.1 × 10−3 M 120. a. NH4(HCO3) → NH4+ + HCO3− CHAPTER 7 ACIDS AND BASES K a , NH + = 4 239 Kw 1.0 × 10 −14 1.0 × 10 −14 −10 = 5.6 × 10 ; K = = = 2.3 × 10−8 −5 −7 b , HCO 3 − K 1.8 × 10 4.3 × 10 a1 Solution is basic since HCO3− is a stronger base than NH4+ is as an acid. The acidic properties of HCO3− were ignored because K a 2 is very small (4.8 × 10−11). b. NaH2PO4 → Na+ + H2PO4−; ignore Na+. K 1.0 × 10 −14 = 1.3 × 10−12 K a , H PO − = 6.2 × 10−8; K b , H PO − = w = 2 2 4 2 4 K a1 7.5 × 10 −3 Solution is acidic since Ka > Kb. c. Na2HPO4 → 2 Na+ + HPO42−; ignore Na+. Ka 3, HPO 4 2− = 4.8 × 10−13; K b , HPO 4 2− = Kw 1.0 × 10 −14 = = 1.6 × 10−7 Ka2 6.2 × 10 −8 Solution is basic since Kb > Ka. d. NH4(H2PO4) → NH4+ + H2PO4− NH4+ is weak acid and H2PO4− is also acidic (see part b). Solution with both ions present will be acidic. e. NH4(HCO2) → NH4+ + HCO2−; from Appendix 5, K a , HCO 2 H = 1.8 × 10−4. K a , NH + = 5.6 × 10−10; K b , HCO − = 4 2 Kw 1.0 × 10 −14 = = 5.6 × 10−11 Ka 1.8 × 10 − 4 Solution is acidic since NH4+ is a stronger acid than HCO2− is a base. 121. Because the values of K a1 and K a 2 are fairly close to each other, we should consider the amount of H+ produced by the K a1 and K a 2 reactions. H3C6H5O7 ⇌ H2C6H5O7− 0 0.15 M x 0.15 - x 2 2 x x ≈ 8.4 × 10−4 = , x = 1.1 × 10−2; 0.15 − x 0.15 Initial Equil. + H+ K a1 = 8.4 × 10−4 ~0 x assumption fails the 5% rule. Solving more exactly using the method of successive approximations: 8.4 × 10−4 = x2 , x = 1.1 × 10−2 M (consistent answer) (0.15 − 1.1 × 10 − 2 ) Now let’s solve for the H+ contribution from the K a 2 reaction. 240 CHAPTER 7 H2C6H5O7− ⇌ HC6H5O72− 1.1 × 10−2 M 1.1 × 10−2 − x Initial Equil. 1.8 × 10−5 = H+ + 0 x ACIDS AND BASES K a 2 = 1.8 × 10−5 1.1 × 10−2 M 1.1 × 10−2 + x x (1.1 × 10 −2 + x) x (1.1 × 10 −2 ) ≈ , x = 1.8 × 10−5 M; assumption good −2 −2 (1.1 × 10 − x) 1.1 × 10 (0.2% error). At most, 1.8 × 10−5 M H+ will be added from the K a 2 reaction. [H+]total = (1.1 × 10−2) + (1.8 × 10−5) = 1.1 × 10−2 M Note that the H+ contribution from the K a 2 reaction was negligible compared to the H+ contribution from the K a1 reaction even though the two Ka values only differed by a factor of 50. Therefore, the H+ contribution from the K a 3 reaction will also be negligible since K a 3 < Ka2 . Solving: pH = −log(1.1 × 10−2) = 1.96 Challenge Problems 122. 1.000 L × 1.00 × 10 −4 mol HA = 1.00 × 10−4 mol HA L 25.0% dissociation gives: mol H+ = 0.250 × (1.00 × 10−4) = 2.50 × 10−5 mol mol A− = 0.250 × (1.00 × 10−4) = 2.50 × 10−5 mol mol HA− = 0.750 × (1.00 × 10−4) = 7.50 × 10−5 mol ⎛ 2.50 × 10 −5 ⎞ ⎛ 2.50 × 10 −5 ⎞ ⎟ ⎟⎜ ⎜ ⎟ ⎟⎜ V V [H + ][A − ] ⎜⎝ ⎠ ⎝ ⎠ −4 = 1.00 × 10 = Ka = −5 [ HA] ⎛ 7.50 × 10 ⎞ ⎜ ⎟ ⎜ ⎟ V ⎝ ⎠ 1.00 × 10−4 = (2.50 × 10 −5 ) 2 (2.50 × 10 −5 ) 2 , V = = 0.0833 L = 83.3 mL (7.50 × 10 −5 )(V) (1.00 × 10 − 4 )(7.50 × 10 −5 ) The volume goes from 1000. mL to 83.3 mL, so 917 mL of water evaporated. 123. a. HCO3− + HCO3− ⇌ H2CO3 + CO32− CHAPTER 7 ACIDS AND BASES 2− Keq = [H 2 CO 3 ][CO 3 ] − − [HCO 3 ][HCO 3 ] × 241 Ka2 [H + ] 4.8 × 10 −11 = = = 1.1 × 10−4 + −7 K [H ] 4.3 × 10 a1 b. [H2CO3] = [CO32−] since the reaction in part a is the principal equilibrium reaction. c. H2CO3 ⇌ 2 H+ + CO32− 2− Keq = [H + ]2 [CO 3 ] = K a1 × K a 2 [H 2 CO 3 ] Because [H2CO3] = [CO32−] from part b, [H+]2 = K a1 × K a 2 . [H+] = (K a1 × K a 2 )1/ 2 , or taking the −log of both sides: pH = pK a1 + pK a 2 2 d. [H+] = [(4.3 × 10−7) × (4.8 × 10−11)]1/2, [H+] = 4.5 × 10−9 M; pH = 8.35 124. a. NaHCO3(aq) → Na+(aq) + HCO3−(aq); NaHSO4 → Na+(aq) + HSO4−(aq) Na+ has no acidic (or basic) properties. HCO3− is a weak acid with Ka = 4.8 × 10−11. HCO3− is also the conjugate base of the weak acid H2CO3, which makes it a weak base. HCO3− is amphoteric; the dominant equilibrium of the best acid reacting with the best base present in a bicarbonate solution is: HCO3−(aq) + HCO3−(aq) ⇌ H2CO3(aq) + CO32−(aq) Because the best acid and best base present are the same species, adding more HCO3− adds both the acid and the base to the equilibrium at the same time. The H2CO3 and CO32− concentrations are increased by the same proportions as more HCO3− is added. The proportional increase is determined only by the Ka value for HCO3− and the Kb value for HCO3−. Thus bicarbonate solutions are concentration independent. For HSO4− solutions, the dominant equilibrium of the best acid reacting with the best base present is: HSO4−(aq) + H2O(l) ⇌ SO42−(aq) + H3O+(aq) This is just the Ka reaction for HSO4−. HSO4− is the conjugate base of the strong acid H2SO4, so HSO4− is a much worse base than water. Water is the best base present in bisulfate solutions. When more HSO4− is added, more H3O+ will be produced, resulting in a more acidic pH. The pH of HSO4− solutions does depend on the concentration of HSO4− present. b. The dominant equilibrium reaction is: HCO3−(aq) + HCO3−(aq) ⇌ H2CO3(aq) + CO32−(aq) 242 CHAPTER 7 ACIDS AND BASES From this reaction, the equilibrium concentrations of H2CO3 and CO32− must be equal to each other. If we add the K a1 reaction for H2CO3 to the K a 2 reaction for HCO3−, the result is: 2− [H + ]2 [CO 3 ] K = K a1 × K a 2 = H2CO3 ⇌ 2 H+ + CO32− [H 2 CO 3 ] 2− + 2 Because [H2CO3] = [CO3 ]: [H ] = K a1 × K a 2 [H+] = ( K a1 × K a 2 )1/2 or taking the −log of both sides: pH = pH = pK a1 + pK a 2 2 = Initial Equil. 0.010 M 0.010 − x 1.2 × 10 −2 = 2 − log(4.3 × 10 −7 ) − log(4.8 × 10 −11 ) 6.37 + 10.32 , pH = = 8.35 2 2 ⇌ HSO4− c. pK a1 + pK a 2 Ka = 1.2 × 10 −2 H+ + SO42− ~0 x 0 x x2 ; solving using the quadratic equation: 0.010 − x x = [H+] = 6.5 × 10−3 M; pH = 2.19 125. HC2H3O2 Initial 1.00 M Equil. 1.00 − x 1.8 × 10 −5 = ⇌ H+ + ~0 x C2H3O2− Ka = 1.8 × 10 −5 0 x x2 x2 ≈ , x = [H+] = 4.24 × 10 −3 M (using one extra sig. fig.) 1.00 − x 1.00 pH = !log(4.24 × 10 −3 ) = 2.37; assumptions good. We want to double the pH to 2(2.37) = 4.74 by addition of the strong base NaOH. As is true with all strong bases, they are great at accepting protons. In fact, they are so good that we can assume they accept protons 100% of the time. The best acid present will react the strong base. This is HC2H3O2. The initial reaction that occurs when the strong base is added is: HC2H3O2 + OH− → C2H3O2− + H2O Note that this reaction has the net effect of converting HC2H3O2 into its conjugate base, C2H3O2−. For a pH = 4.74, let’s calculate the ratio of [C2H3O2−]/[HC2H3O2] necessary to achieve this pH. CHAPTER 7 ACIDS AND BASES HC2H3O2 ⇌H + + C2H3O2 − 243 − [H + ][C 2 H 3 O 2 ] Ka = [ HC 2 H 3 O 2 ] When pH = 4.74, [H+] = 10 −4.74 = 1.8 × 10 −5 . Ka = 1.8 × 10 −5 = − − (1.8 × 10 −5 )[C 2 H 3O 2 ] [C 2 H 3O 2 ] , = 1.0 [HC 2 H 3O 2 ] [HC 2 H 3O 2 ] For a solution having pH = 4.74, we need to have equal concentrations (equal moles) of C2H3O2− and HC2H3O2. Therefore, we need to add an amount of NaOH that will convert onehalf of the HC2H3O2 into C2H3O2−. This amount is 0.50 M NaOH. HC2H3O2 + OH− → C2H3O2− + H2O 0.50 M 0 !0.50 → +0.50 0 0.50 M Before 1.00 M Change !0.50 After 0.50 M completion From the preceding stoichiometry problem, adding enough NaOH(s) to produce a 0.50 M OH− solution will convert one-half the HC2H3O2 into C2H3O2−; this results in a solution with pH = 4.74. Mass NaOH = 1.00 L × 126. dRT Molar mass = = P 0.50 mol NaOH 40.00 g NaOH × = 20. g NaOH L mol 5.11 g / L × 0.08206 L atm × 298 K K mol = 125 g/mol 1.00 atm 1 mol 125 g = 0.120 M; pH = 1.80, [H+] = 10−1.80 = 1.6 × 10−2 M 0.100 L 1.50 g × [HA]0 = HA ⇌ Initial 0.120 M Equil. 0.120 − x Ka = 127. H+ + A− ~0 x 0 x where x = [H+] = 1.6 × 10−2 M [H + ][A − ] (1.6 × 10 −2 ) 2 = = 2.5 × 10−3 [HA] 0.120 − 0.016 Major species: BH+, X−, and H2O; because BH+ is the best acid and X− is the best base in solution, the principal equilibrium is: 244 CHAPTER 7 Initial 0.100 M Equil. 0.100 − x K= K a , BH + K a , HX = ⇌ X− BH+ + B + 0.100 M 0.100 − x 0 x ACIDS AND BASES HX 0 x [B][HX ] , where [B] = [HX] and [BH+] = [X−] + − [BH ][X ] To solve for the Ka of HX, let’s use the equilibrium expression to derive a general expression that relates pH to the pKa for BH+ and to the pKa for HX. K a , BH + K a , HX = [HX ]2 [H + ][X − ] [HX] [H + ] ; K , = = a , HX [HX] K a , HX [ X − ]2 [X − ] ⎛ [H + ] [HX]2 ⎜ = = ⎜ K a , HX [ X − ]2 ⎝ 2 ⎞ ⎟ , [ H + ]2 = K × K a , HX a , BH + ⎟ K a , HX ⎠ pK a , BH + + pK a , HX Taking the −log of both sides: pH = 2 K a , BH + This is a general equation that applies to all BHX type salts. Solving the problem: Kb for B = 1.0 × 10−3; Ka for BH+ = pH = 8.00 = 128. 11.00 + pK a , HX 2 Kw = 1.0 × 10−11 Kb , pK a , HX = 5.00 and Ka for HX = 10−5.00 = 1.0 × 10−5 Major species: NH4+, C2O42−, and H2O; reacting the best acid with the best base: NH4+ + C2O42− ⇌ NH3 + HC2O4− K= K a , NH 4 K a , HC O 2 Initial Change Equil. K= 0.200 M −x 0.200 − x 0.100 M −x → 0.100 − x 0 +x x 0 +x x + 4 − K = 9.2 × 10−6 ( x)( x) = 9.2 × 10−6; solving: x = 4.3 × 10−4 M (0.200 − x)(0.100 − x) Use either Ka expression to solve for [H+]. K a 2 = 6.1 × 10−5 = 2− [H + ][C 2 O 4 ] − [HC 2 O 4 ] = [H + ](0.100 − 4.3 × 10 −4 ) , [H+] = 2.6 × 10−7 M; (4.3 × 10 − 4 ) pH = 6.59 We get the same answer using the Ka equilibrium for NH4+. CHAPTER 7 ACIDS AND BASES Ka = 5.6 × 10−10 = 129. [H + ][ NH 3 ] + [ NH 4 ] = 245 [H + ](4.3 × 10 −4 ) , [H+] = 2.6 × 10 −7 M; pH = 6.59 −4 (0.200 − 4.3 × 10 ) 0.0500 M HCO2H (HA), Ka = 1.77 × 10−4; 0.150 M CH3CH2CO2H (HB), Ka = 1.34 × 10−5 Because two comparable weak acids are present, each contributes to the total pH. Charge balance: [H+] = [A−] + [B−] + [OH−] = [A−] + [B−] + Kw/[H+] Mass balance for HA and HB: 0.0500 = [HA] + [A−] and 0.150 = [HB] + [B−] [H + ][A − ] [H + ][B − ] = 1.77 × 10 − 4 ; = 1.34 × 10−5 [HA] [HB] We have five equations and five unknowns. Manipulate the equations to solve. [H+] = [A−] + [B−] + Kw/[H+]; [H+]2 = [H+][A−] + [H+][B−] + Kw [H+][A−] = (1.77 × 10−4)[HA] = (1.77 × 10−4) (0.0500 − [A−]) If [A−] << 0.0500, then [H+][A−] ≈ (1.77 × 10−4) (0.0500) = 8.85 × 10−6. Similarly, assume [H+][B−] ≈ (1.34 × 10−5)(0.150) = 2.01 × 10−6. [H+]2 = 8.85 × 10−6 + 2.01 × 10−6 + 1.00 × 10−14, [H+] = 3.30 × 10−3 mol/L 8.85 × 10 −6 ≈ 2.68 × 10−3 Check assumptions: [H+][A−] ≈ 8.85 × 10−6, [A−] ≈ −3 3.30 × 10 Assumed 0.0500 − [A−] ≈ 0.0500. This assumption is borderline (2.68 × 10−3 is 5.4% of 0.0500). The HB assumption is good (0.4% error). Using successive approximations to refine the [H+][A−] value: [H+] = 3.22 × 10−3 M, pH = −log(3.22 × 10−3) = 2.492 Note: If we treat each acid separately: H+ from HA = 2.9 × 10−3 H+ from HB = 1.4 × 10−3 _________________________________________________ 4.3 × 10−3 M = [H+]total This assumes the acids did not suppress each other’s ionization. They do, and we expect the [H+] to be less than 4.3 × 10−3 M. We get such an answer. 130. HA Initial Equil. ⇌ C C − 1.00 × 10−4 H+ + ~0 1.00 × 10−4 A− 0 1.00 × 10−4 Ka = 1.00 × 10−6 C = [HA]0 for pH = 4.000 x = [H+] = 1.00 × 10−4 M 246 CHAPTER 7 Ka = ACIDS AND BASES (1.00 × 10 −4 ) 2 = 1.00 × 10−6; solving: C = 0.0101 M −4 (C − 1.00 × 10 ) The solution initially contains 50.0 × 10−3 L × 0.0101 mol/L = 5.05 × 10−4 mol HA. We then dilute to a total volume V in liters. The resulting pH = 5.000, so [H+] = 1.00 × 10−5. In the typical weak acid problem, x = [H+], so: ⇌ HA Initial Equil. Ka = H+ 5.05 × 10−4 mol/V (5.05 × 10−4/V) − (1.00 × 10−5) + ~0 1.00 × 10−5 A− 0 1.00 × 10−5 (1.00 × 10 −5 ) 2 = 1.00 × 10−6 (5.05 × 10 − 4 /V) − (1.00 × 10 −5 ) 1.00 × 10−4 = (5.05 × 10−4/V) − 1.00 × 10−5 V = 4.59 L; 50.0 mL are present initially, so we need to add 4540 mL of water. 131. HA Initial Change Equil. [HA]0 −x [HA]0 − x ⇌ H+ → ~0 +x x + A− Ka = 5.00 × 10−10 0 +x x From the problem: pH = 5.650, so [H+] = x = 10−5.650 = 2.24 × 10−6 M 5.00 × 10−10 = x2 (2.24 × 10 −6 ) 2 = , [HA]0 = 1.00 × 10−2 M [HA]0 − x ([HA]0 − 2.24 × 10 −6 ) After the water is added, the pH of the solution is between 6 and 7, so the water contribution to the [H+] must be considered. The general expression for a very dilute weak acid solution is: Ka = [ H + ]2 − K w [ H + ]2 − K w [HA]0 − [H + ] pH = 6.650; [H+] = 10−6.650 = 2.24 × 10−7 M Let V = volume of water added: 5.00 × 10−10 = (2.24 × 10 −7 ) 2 − (1.00 × 10 −14 ) ⎛ 0.0500 ⎞ (2.24 × 10 −7 ) 2 − (1.00 × 10 −14 ) ⎟⎟ − (1.00 × 10 − 2 )⎜⎜ 2.24 × 10 −7 ⎝ 0.0500 + V ⎠ Solving, V = 6.16 L of water were added. 132. Major species = Na+, HSO4−, NH3, and H2O; reaction: HSO4− + NH3 ⇌ SO42− + NH4+ CHAPTER 7 ACIDS AND BASES 2− K= + [SO 4 ][ NH 4 ] − [HSO 4 ][ NH 3 ] = K a , HSO K a , NH 4 4 − + 247 = 1.2 × 10 −2 = 2.1 × 107 5.6 × 10 −10 Because K is a large number, let the reaction go to completion, and then solve the back equilibrium problem. ⇌ HSO4− + NH3 Before After 0.10 M 0 0.10 M 0 SO42− + 0 0.10 M NH4+ 0 0.10 M Allow the reaction to attain equilibrium: Initial Change Equil. HSO4− + NH3 0 +x x 0 +x x ⇌ SO42− NH4+ + 0.10 M ← −x 0.10 − x 0.10 M −x 0.10 − x (0.10 − x) 2 (0.10) 2 = 2.1 × 107, x = 2.2 × 10−5 M; assumptions good. ≈ 2 2 x x [HSO4−] = 2.2 × 10−5 M; [SO42−] = 0.10 M [NH3] = 2.2 × 10−5 M; [NH4+] = 0.10 M Using one of the Ka equilibrium expressions to solve for [H+]: K a , HSO − = 4 133. [H + ] (0.10) = 1.2 × 10 − 2 , [H+] = 2.6 × 10−6 M; pH = 5.59 2.2 × 10 −5 0.135 mol CO 2 = 5.40 × 10−2 mol CO2/L = 5.40 × 10−2 M H2CO3; 0.105 M CO32− 2.50 L The best acid (H2CO3) reacts with the best base present (CO32−) for the principal equilibrium. H2CO3 + CO32− → 2 HCO3− K= K a1 , H 2CO 3 K a 2 , H 2CO3 = 4.3 × 10 −7 = 9.0 × 103 4.8 × 10 −11 Because K >> 1, assume all CO2 (H2CO3) is converted into HCO3−; that is, 5.40 × 10−2 mol/L CO32− is converted into HCO3−. [HCO3−] = 2(5.40 × 10−2) = 0.108 M; [CO32−] = 0.105 - 0.0540 = 0.051 M Note: If we solve for the [H2CO3] using these concentrations, we get [H2CO3] = 2.5 × 10−5 M; our assumption that the reaction goes to completion is good (2.5 × 10−5 is 0.05% of 0.051). Whenever K >> 1, always assume the reaction goes to completion. 248 CHAPTER 7 ACIDS AND BASES To solve for the [H+] in equilibrium with HCO3− and CO32−, use the Ka expression for HCO3−. HCO3− ⇌ H+ + CO32− 4.8 × 10−11 = K a 2 = 4.8 × 10−11 2− [H + ][CO 3 ] − [HCO 3 ] = [H + ] (0.051) 0.108 [H+] = 1.0 × 10−10; pH = 10.00; assumptions good. 134. H2O ⇌ H+ + OH−; Kw = [H+][OH−]; B + H2O Kb = ⇌ HB+ + OH− [HB+ ][OH − ] [B] Charge balance: [H+] + [HB+] = [OH−]; material balance: [B]o = [B] + [HB+] So: [OH−] = [H+] + [HB+] [OH−] = Kw Kw + [HB+] or: [HB+] = [OH−] − − [OH ] [OH − ] [B] = [B]0 − [HB+] ⎛ Kw [B] = [B]0 − ⎜⎜ [OH − ] − [OH − ] ⎝ ⎛ Kw ⎞ ⎟ [OH − ] ⎜⎜ [OH − ] − − ⎟ [OH ] ⎠ Kb = ⎝ ⎛ Kw ⎞ ⎟ [B]0 − ⎜⎜ [OH − ] − [OH − ] ⎟⎠ ⎝ Assuming [B]0 >> Kb ≈ ⎞ ⎟ ⎟ ⎠ = [OH − ]2 − K w [OH − ]2 − K w [B]0 − [OH − ] [OH − ]2 − K w , then: [OH − ] [OH − ]2 − K w [OH − ]2 − 1.0 × 10 −14 , 6.1 × 10−11 = [B]0 2.0 × 10 −5 [OH−] = 1.1 × 10−7; pOH = 6.96; pH = 7.04 (assumption good) 135. Major species: H2O, Na+, and NO2−; NO2− is a weak base. NO2− + H2O ⇌ HNO2 + OH−; CHAPTER 7 ACIDS AND BASES 249 Because this is a very dilute solution of a weak base, the OH− contribution from H2O must be considered. The weak base equations for dilute solutions are analogous to the weak acid equations derived in Section 7.9 of the text. For A− type bases (A− + H2O Kb = ⇌ HA + OH−), the general equation is: [OH − ]2 − K w [OH − ]2 − K w [A − ]0 − [OH − ] When [A−]0 >> [OH − ]2 − K w [OH − ]2 − K w , then K = b [OH − ] [A − ]0 and: [OH−] = (Kb[A−]0 + Kw)1/2 1/ 2 ⎛ 1.0 × 10 −14 ⎞ × (6.0 × 10 − 4 ) + (1.0 × 10 −14 ) ⎟⎟ Try: [OH ] = ⎜⎜ −4 ⎝ 4.0 × 10 ⎠ − Checking assumption: 6.0 × 10−4 >> = 1.6 × 10−7 M (1.6 × 10 −7 ) 2 − (1.0 × 10 −14 ) = 9.8 × 10−8 −7 1.6 × 10 Assumption good. [OH−] = 1.6 × 10−7 M; pOH = 6.80; pH = 7.20 136. Major species: H+, HSO4−, and H2O (water is important!) Charge balance: [H+] = [OH−] + [HSO4−] + 2[SO42−] [HSO4−]0 = [SO42−] + [HSO4−] = 1.00 × 10−7 M (from the 1.00 × 10−7 M H2SO4) Kw = [H+][OH−] = 1.0 × 10−14 Material balance: 2− Ka = 1.2 × 10−2 = [H+] = [H + ][SO 4 ] − [HSO 4 ] ; [HSO4−] = (1.00 × 10−7) − [SO42−] Kw Kw + (1.00 × 10−7) − [SO42−] + 2[SO42−], [SO42−] = [H+] − − (1.00 × 10−7) + [H ] [H + ] ⎞ ⎛ Kw [H + ]⎜⎜ [H + ] − − (1.00 × 10 −7 ) ⎟⎟ + [H ] [H ][SO 4 ] ⎠ ⎝ 1.2 H 10−2 = = − K [HSO 4 ] w (1.00 × 10 −7 ) − [H + ] + + (1.00 × 10 −7 ) [H + ] + 2− This is a complicated expression to solve. Because this is such a dilute solution of H2SO4 (1.00 × 10−7 M ), the Ka equilibrium expression for HSO4− dictates that [SO42−] >> [HSO4−]. Let’s assume that [SO42−] = 1.00 × 10−7 M (assume most of the HSO4− dissociates): 250 CHAPTER 7 [H+] = ACIDS AND BASES Kw Kw + (1.00 × 10−7) + [SO42−] = + 2.00 × 10−7 + [H ] [H + ] Solving: [H+] = 2.4 × 10−7 M; pH = 6.62 2− Assumption good: [SO 4 ] − [HSO 4 ] = Ka 1.2 × 10 −2 = = 5.0 × 104. + −7 [H ] 2.4 × 10 We do have mostly SO42− at equilibrium. 137. Major species: H2O, NH3, H+, and Cl−; The H+ from the strong acid will react with the best base present (NH3). Because strong acids are great at donating protons, the reaction between H+ and NH3 essentially goes to completion, that is, until one or both of the reactants runs out. The reaction is: NH3 + H+ → NH4+ Because equal volumes of 1.0 × 10−4 M NH3 and 1.0 × 10−4 M H+ are mixed, both reactants are in stoichiometric amounts, and both reactants will run out at the same time. After reaction, only NH4+ and Cl− remain. Cl− has no basic properties since it is the conjugate base of a strong acid. Therefore, the only species with acid-base properties is NH4+, a weak acid. The initial concentration of NH4+ will be exactly one-half of 1.0 × 10−4 M since equal volumes of NH3 and HCl were mixed. Now we must solve the weak acid problem involving 5.0 × 10−5 M NH4+. K ⇌ H+ + NH3 Ka = w = 5.6 × 10−10 NH4+ Kb −5 Initial 5.0 × 10 M ~0 0 x x Equil. 5.0 × 10−5 − x Ka = x2 x2 ≈ = 5.6 × 10−10, x = 1.7 × 10−7 M; check assumptions. (5.0 × 10 −5 − x) 5.0 × 10 −5 We cannot neglect [H+] that comes from H2O. As discussed in Section 7.9 of the text, assume 5.0 × 10−5 >> ([H+]2 − Kw)/[H+]. If this is the case, then: [H+] = (Ka[HA]0 + Kw)1/2 = 1.9 × 10−7 M; checking assumption: [ H + ]2 − K w = 1.4 × 10-7 << 5.0 × 10−5 (assumption good) [H + ] So: [H+] = 1.9 × 10−7 M; pH = 6.72 138. Ca(OH)2 (s) → Ca2+(aq) + 2 OH−(aq) This is a very dilute solution of Ca(OH)2, so we can't ignore the OH− contribution from H2O. From the dissociation of Ca(OH)2 alone, 2[Ca2+] = [OH−]. Including the H2O autoionization into H+ and OH−, the overall charge balance is: CHAPTER 7 ACIDS AND BASES 251 2[Ca2+] + [H+] = [OH−] 2(3.0 × 10−7 M) + Kw/[OH−] = [OH−], [OH−]2 = (6.0 × 10−7)[OH−] + Kw [OH−]2 − (6.0 × 10−7)[OH−] − 1.0 × 10−14 = 0; using quadratic formula: [OH−] = 6.2 × 10−7 M Marathon Problems 139. To determine the pH of solution A, the Ka value for HX must be determined. Use solution B to determine Kb for X−, which can then be used to calculate Ka for HX (Ka = Kw/Kb). Solution B: X− Initial Change Equil. + H2O 0.0500 M −x 0.0500 − x ⇌ HX → 0 +x x + OH− Kb = [HX][OH − ] [X − ] ~0 +x x Kb = x2 ; from the problem, pH = 10.02, so pOH = 3.98 and [OH−] = x = 10−3.98 0.0500 − x Kb = (10 −3.98 ) 2 = 2.2 × 10−7 −3.98 0.0500 − 10 Solution A: Ka, HX = K w /K b, X − = (1.0 × 10−14)/(2.2 × 10−7) = 4.5 × 10−8 ⇌ H+ 0.100 M −x → 0.100 − x ~0 +x x HX Initial Change Equil. Ka = 4.5 × 10−8 = + X− Ka = 4.5 × 10−8 = [H + ][X − ] [HX] 0 +x x x2 x2 ≈ , x = [H+] = 6.7 × 10−5 M 0.100 − x 0.100 Assumptions good (x is 0.067% of 0.100); pH = 4.17 Solution C: Major species: H2O, HX (Ka = 4.5 × 10−8), Na+, and OH−; the OH− from the strong base is exceptional at accepting protons. OH− will react with the best acid present (HX), and we can assume that OH− will react to completion with HX, that is, until one (or both) of the reactants runs out. Because we have added one volume of substance to another, we have diluted both solutions from their initial concentrations. What hasn’t changed is the moles of each reactant. So let’s work with moles of each reactant initially. 252 CHAPTER 7 ACIDS AND BASES 0.100 mol HX = 5.00 × 10−3 mol HX L Mol HX = 0.0500 L × Mol OH− = 0.0150 L × 0.250 mol NaOH 1 mol OH − × = 3.75 × 10−3 mol OH− L mol NaOH Now lets determine what is remaining in solution after OH− reacts completely with HX. Note that OH− is the limiting reagent. HX OH− + Before 5.00 × 10−3 mol −3.75 × 10−3 Change After 1.25 × 10−3 mol completion → 3.75 × 10−3 mol −3.75 × 10−3 0 → X− + 0 +3.75 × 10−3 3.75 × 10−3 mol H2O − +3.75 × 10−3 − After reaction, the solution contains HX, X−, Na+ and H2O. The Na+ (like most +1 metal ions) has no effect on the pH of water. However, HX is a weak acid and its conjugate base, X−, is a weak base. Since both Ka and Kb reactions refer to these species, we could use either reaction to solve for the pH; we will use the Kb reaction. To solve the equilibrium problem using the Kb reaction, we need to convert to concentration units since Kb is in concentration units of mol/L. [HX] = 1.25 × 10 −3 mol 3.75 × 10 −3 mol = 0.0192 M; [X−] = = 0.0577 M (0.0500 + 0.0150) L 0.0650 L [OH−] = 0 (We reacted all of it to completion.) X− Initial Change Equil. + H2O ⇌ HX + 0.0577 M 0.0192 M x mol/L of X− reacts to reach equilibrium −x → +x 0.0577 − x 0.0192 + x Kb = 2.2 × 10−7 = x = [OH−] = (0.0192 + x)x (0.0192) x ≈ 0.0577 − x 0.0577 OH− Kb = 2.2 × 10−7 0 +x x (assuming x is << 0.0192) (2.2 × 10 −7 )(0.0577) = 6.6 × 10−7 M; assumptions great (x is 0.0034% of 0.0192 0.0192). [OH−] = 6.6 × 10−7 M, pOH = 6.18, pH = 14.00 = 6.18 = 7.82 = pH of solution C The combination is 4-17-7-82. 140. a. Strongest acid from group I = HCl; weakest base (smallest Kb) from group II = NaNO2 0.20 M HCl + 0.20 M NaNO2; major species = H+, Cl−, Na+, NO2−, and H2O CHAPTER 7 ACIDS AND BASES 253 Let the H+ react to completion with the NO2−; then solve the back equilibrium problem. + NO2− H+ Before After 0.10 M 0 → HNO2 0.10 M 0 HNO2 0 0.10 M (Molarities are halved due to dilution.) Ka = 4.0 × 10−4 ⇌ H+ + NO2− → 0 +x x 0 +x x Initial 0.10 M Change −x Equil. 0.10 − x x2 = 4.0 × 10−4; solving, x = [H+] = 6.1 × 10−3 M; pH = 2.21 0.10 − x b. Weakest acid from group I = (C2H5)3NHCl; best base from group II = KOI; The dominant equilibrium will be the best base reacting with the best acid. OI− + (C2H5)3NH+ Initial Equil. K= 0.10 M 0.10 − x K a , (C 2 H 5 ) 3 NH K a , HOI + ⇌ 0.10 M 0.10 − x = HOI + (C2H5)3N 0 x 0 x 1.0 × 10 −14 1 × = 1.25 (carrying extra sig. fig.) −4 4.0 × 10 2.0 × 10 −11 x2 x = 1.25, = 1.12, x = 0.053 M 2 0.10 − x (0.10 − x) So: [HOI] = 0.053 M and [OI−] = 0.10 – x = 0.047 M; using the Ka equilibrium constant for HOI to solve for [H+]: 2.0 × 10−11 = [H + ](0.047) , [H+] = 2.3 × 10−11 M; pH = 10.64 (0.053) c. Ka for (C2H5)3NH+ = Kb for NO2− = 1.0 × 10 −14 = 2.5 × 10−11 4.0 × 10 − 4 1.0 × 10 −14 = 2.5 × 10−11 4.0 × 10 − 4 Because Ka = Kb, mixing (C2H5)3NHCl with NaNO2 will result in a solution with pH = 7.00. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA Buffers 15. A buffer solution is one that resists a change in its pH when either hydroxide ions or protons (H+) are added. Any solution that contains a weak acid and its conjugate base or a weak base and its conjugate acid is classified as a buffer. The pH of a buffer depends on the [base]/[acid] ratio. When H+ is added to a buffer, the weak base component of the buffer reacts with the H+ and forms the acid component of the buffer. Even though the concentrations of the acid and base components of the buffer change some, the ratio of [base]/[acid] does not change that much. This translates into a pH that doesn’t change much. When OH− is added to a buffer, the weak acid component is converted into the base component of the buffer. Again, the [base]/[acid] ratio does not change a lot (unless a large quantity of OH− is added), so the pH does not change much. H+(aq) + CO32−(aq) → HCO3−(aq); OH−(aq) + HCO3−(aq) → H2O(l) + CO32−(aq) 16. When [HA] = [A−] (or [BH+] = [B]) for a buffer, the pH of the solution is equal to the pKa value for the acid component of the buffer (pH = pKa because [H+] = Ka). A best buffer has equal concentrations of the acid and base components so it is equally efficient at absorbing H+ and OH−. For a pH = 4.00 buffer, we would choose the acid component having a Ka close to 10 −4.00 = 1.0 × 10 −4 (pH = pKa for a best buffer). For a pH = 10.00 buffer, we would want the acid component of the buffer to have a Ka close to 10 −10.00 = 1.0 × 10 −10 . Of course, we can have a buffer solution made from a weak base and its conjugate acid. For a pH = 10.00 buffer, our conjugate acid should have Ka ≈ 1.0 × 10 −10 , which translates into a Kb value of the base close to 1.0 × 10 −4 (Kb = Kw/Ka for conjugate acid-base pairs). The capacity of a buffer is a measure of how much strong acid or strong base the buffer can neutralize. All the buffers listed have the same pH (= pKa = 4.74) because they all have a 1 : 1 concentration ratio between the weak acid and the conjugate base. The 1.0 M buffer has the greatest capacity; the 0.01 M buffer the least capacity. In general, the larger the concentrations of weak acid and conjugate base, the greater is the buffer capacity, that is, the more strong acid or strong base that can be neutralized with little pH change. 17. pH = pKa + log ⎛ [base] ⎞ [base] [base] ⎟⎟ < 0. ; when [acid] > [base], then < 1 and log ⎜⎜ [acid] [acid] ⎝ [acid] ⎠ From the Henderson-Hasselbalch equation, if the log term is negative, then pH < pKa. When one has more acid than base in a buffer, the pH will be on the acidic side of the pKa value; 254 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 255 that is, the pH is at a value lower than the pKa value. When one has more base than acid in a buffer ([conjugate base] > [weak acid]), then the log term in the Henderson-Hasselbalch equation is positive, resulting in pH > pKa. When one has more base than acid in a buffer, the pH is on the basic side of the pKa value; that is, the pH is at a value greater than the pKa value. The other scenario you can run across in a buffer is when [acid] = [base]. Here, the log term is equal to zero, and pH = pKa. 18. NH3 + H2O ⇌ NH4 + OH − + + [ NH 4 ][OH − ] Kb = ; taking the −log of the Kb expression: [ NH 3 ] −log Kb = −log[OH−] − log + + [ NH 4 ] [ NH 4 ] , −log[OH−] = −log Kb + log [ NH 3 ] [ NH 3 ] + pOH = pKb + log 19. [acid] [ NH 4 ] or pOH = pKb + log [ NH 3 ] [base] a. This is a weak acid problem. Let HC3H5O2 = HOPr and C3H5O2− = OPr−. HOPr(aq) Initial Change Equil. ⇌ H+(aq) + OPr−(aq) Ka = 1.3 × 10−5 ~0 0 0.100 M x mol/L HOPr dissociates to reach equilibrium −x → +x +x 0.100 − x x x Ka = 1.3 × 10−5 = [H + ][OPr − ] x2 x2 = ≈ [HOPr] 0.100 − x 0.100 x = [H+] = 1.1 × 10−3 M; pH = 2.96; assumptions good by the 5% rule. b. This is a weak base problem. OPr−(aq) + H2O(l) Initial Change Equil. ⇌ HOPr(aq) + OH−(aq) Kb = Kw = 7.7 × 10−10 Ka 0.100 M 0 ~0 − x mol/L OPr reacts with H2O to reach equilibrium −x → +x +x 0.100 − x x x Kb = 7.7 × 10−10 = [HOPr][OH − ] x2 x2 = ≈ 0.100 − x 0.100 [OPr − ] x = [OH−] = 8.8 × 10−6 M; pOH = 5.06; pH = 8.94; assumptions good. c. Pure H2O, [H+] = [OH−] = 1.0 × 10−7 M; pH = 7.00 256 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA d. This solution contains a weak acid and its conjugate base. This is a buffer solution. We will solve for the pH through the weak acid equilibrium reaction. HOPr(aq) Initial Change Equil. 1.3 × 10−5 = ⇌ H+(aq) + OPr−(aq) Ka = 1.3 × 10−5 ~0 0.100 M 0.100 M x mol/L HOPr dissociates to reach equilibrium −x → +x +x 0.100 − x x 0.100 + x (0.100 + x)( x) (0.100)( x) = x = [H+] ≈ 0.100 − x 0.100 [H+] = 1.3 × 10−5 M; pH = 4.89; assumptions good. Alternately, we can use the Henderson-Hasselbalch equation to calculate the pH of buffer solutions. pH = pKa + log [base] ⎛ 0.100 ⎞ 5 = pKa + log ⎜ ⎟ = pKa = −log(1.3 × 10− ) = 4.89 0 . 100 [acid] ⎝ ⎠ The Henderson-Hasselbalch equation will be valid when an assumption of the type 0.1 + x ≈ 0.1 that we just made in this problem is valid. From a practical standpoint, this will almost always be true for useful buffer solutions. If the assumption is not valid, the solution will have such a low buffering capacity it will not be of any use to control the pH. Note: The Henderson-Hasselbalch equation can only be used to solve for the pH of buffer solutions. 20. a. We have a weak acid (HOPr = HC3H5O2) and a strong acid (HCl) present. The amount of H+ donated by the weak acid will be negligible. To prove it lets consider the weak acid equilibrium reaction: HOPr Initial Change Equil. ⇌ H+ + OPr− Ka = 1.3 × 10−5 0.100 M 0.020 M 0 x mol/L HOPr dissociates to reach equilibrium −x → +x +x 0.100 − x 0.020 + x x [H+] = 0.020 + x ≈ 0.020 M; pH = 1.70; assumption good (x = 6.5 × 10−5 is << 0.020). Note: The H+ contribution from the weak acid HOPr was negligible. The pH of the solution can be determined by only considering the amount of strong acid present. b. Added H+ reacts completely with the best base present, OPr−. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA OPr− Before Change After + 0.100 M −0.020 0.080 H+ → 257 HOPr 0.020 M −0.020 → 0 0 +0.020 0.020 M Reacts completely After reaction, a weak acid, HOPr , and its conjugate base, OPr−, are present. This is a buffer solution. Using the Henderson-Hasselbalch equation where pKa = −log (1.3 × 10−5) = 4.89: pH = pKa + log [base] (0.080) = 4.89 + log = 5.49; assumptions good. [acid] (0.020) c. This is a strong acid problem. [H+] = 0.020 M; pH = 1.70 d. Added H+ reacts completely with the best base present, OPr−. OPr− Before Change After H+ → HOPr 0.020 M −0.020 0 → 0.100 M +0.020 Reacts completely 0.120 + 0.100 M −0.020 0.080 A buffer solution results (weak acid + conjugate base). Using the HendersonHasselbalch equation: pH = pKa + log 21. (0.080) [base] = 4.71; assumptions good. = 4.89 + log (0.120) [acid] a. OH− will react completely with the best acid present, HOPr. HOPr Before Change After + 0.100 M −0.020 0.080 OH− 0.020 M −0.020 0 → OPr- → 0 +0.020 0.020 + H2O Reacts completely A buffer solution results after the reaction. Using the Henderson-Hasselbalch equation: pH = pKa + log [base] (0.020) = 4.89 + log = 4.29; assumptions good. [acid] (0.080) b. We have a weak base and a strong base present at the same time. The amount of OHadded by the weak base will be negligible. To prove it, let’s consider the weak base equilibrium: 258 CHAPTER 8 OPr− Initial Change Equil. + H2O ⇌ APPLICATIONS OF AQUEOUS EQUILIBRIA HOPr + OH− Kb = 7.7 × 10−10 0.100 M 0 0.020 M − x mol/L OPr reacts with H2O to reach equilibrium −x → +x +x 0.100 − x x 0.020 + x [OH−] = 0.020 + x ≈ 0.020 M; pOH = 1.70; pH = 12.30; assumption good. Note: The OH− contribution from the weak base OPr− was negligible (x = 3.9 × 10−9 M as compared to 0.020 M OH- from the strong base). The pH can be determined by only considering the amount of strong base present. c. This is a strong base in water. [OH−] = 0.020 M; pOH = 1.70; pH = 12.30 d. OH− will react completely with HOPr, the best acid present. HOPr Before Change After + 0.100 M −0.020 0.080 OH− OPr− → 0.020 M −0.020 → 0 0.100 M +0.020 0.120 + H2O Reacts completely Using the Henderson-Hasselbalch equation to solve for the pH of the resulting buffer solution: pH = pKa + log 22. [base] (0.120) = 4.89 + log = 5.07; assumptions good. [acid] (0.080) Consider all the results to Exercises 19, 20, and 21: Solution a b c d Initial pH 2.96 8.94 7.00 4.89 After Added H+ 1.70 5.49 1.70 4.71 After Added OH− 4.29 12.30 12.30 5.07 The solution in Exercise 19d is a buffer; it contains both a weak acid (HC3H5O2) and a weak base (C3H5O2−). Solution d shows the greatest resistance to changes in pH when either a strong acid or a strong base is added, which is the primary property of buffers. 23. Major species: HF, F−, K+, and H2O. K+ has no acidic or basic properties. This is a solution containing a weak acid and its conjugate base. This is a buffer solution. One appropriate equilibrium reaction you can use is the Ka reaction of HF, which contains both HF and F−. However, you could also use the Kb reaction for F− and come up with the same answer. Alternately, you could use the Henderson-Hasselblach equation to solve for the pH. For this problem, we will use the Ka reaction and set up an ICE table to solve for the pH. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA ⇌ HF Initial Change Equil. F− + 259 H+ 0.60 M 1.00 M ~0 x mol/L HF dissociates to reach equilibrium !x → +x +x 1.00 + x x 0.60 ! x Ka = 7.2 × 10 −4 = [F − ][ H + ] (1.00 + x)( x) (1.00)( x) = ≈ (assuming x << 0.60) 0.60 [ HF] 0.60 − x x = [H+] = 0.60 × (7.2 × 10−4) = 4.3 × 10 −4 M; assumptions good. pH = −log(4.3 × 10−4) = 3.37 24. Major species: HONH2 (Kb = 1.1 × 10 −8 ), HONH3+, Cl−, and H2O; Cl− has no acidic/basic properties. We have a weak base and its conjugate acid present at the same time in solution. We have a buffer solution. To solve for the pH of a buffer, one can set up an ICE table using the Ka reaction for HONH3+, or set up an ICE table using the Kb reaction for HONH2, or use the Henderson-Hasselbalch equation. Using the Henderson-Hasselbalch equation: pH = pKa + log ⎛ 1.0 × 10 −14 ⎞ [HONH 2 ] [base] ⎟ + log = − log ⎜⎜ −8 ⎟ + [acid] [HONH 3 ] ⎝ 1.1 × 10 ⎠ ⎛ 0.100 ⎞ pH = −log(9.1 × 10−7) + log ⎜ ⎟ = 6.04 + 0.00 = 6.04 ⎝ 0.100 ⎠ Note that pH = pKa for a buffer solution when [weak base] = [conjugate acid]. 25. Major species after NaOH added: HF, F−, K+, Na+, OH−, and H2O. The OH− from the strong base will react with the best acid present (HF). Any reaction involving a strong base is assumed to go to completion. Because all species present are in the same volume of solution, we can use molarity units to do the stoichiometry part of the problem (instead of moles). The stoichiometry problem is: OH− Before Change After + HF 0.10 mol/1.00 L 0.60 M !0.10 M !0.10 M 0 0.50 → F− → 1.00 M +0.10 M 1.10 + H2O Reacts completely After all the OH− reacts, we are left with a solution containing a weak acid (HF) and its conjugate base (F−). This is what we call a buffer problem. We will solve this buffer problem using the Ka equilibrium reaction. One could also use the Kb equilibrium reaction or use the Henderson-Hasselbalch equation to solve for the pH. 260 CHAPTER 8 ⇌ HF Initial Change Equil. APPLICATIONS OF AQUEOUS EQUILIBRIA F− + H+ 0.50 M 1.10 M ~0 x mol/L HF dissociates to reach equilibrium !x → +x +x 1.10 + x x 0.50 ! x Ka = 7.2 × 10 −4 = (1.10 + x)( x) (1.10)( x) ≈ , x = [H+] = 3.3 × 10 −4 M; pH = 3.48; 0.50 0.50 − x assumptions good. Note: The added NaOH to this buffer solution changes the pH only from 3.37 to 3.48. If the NaOH were added to 1.0 L of pure water, the pH would change from 7.00 to 13.00. Major species after HCl added: HF, F−, H+, K+, Cl−, and H2O; the added H+ from the strong acid will react completely with the best base present (F−). H+ Before 0.20 mol 1.00 L Change After !0.20 M 0 F− + → 1.00 M HF 0.60 M !0.20 M → 0.80 +0.20 M 0.80 Reacts completely After all the H+ has reacted, we have a buffer solution (a solution containing a weak acid and its conjugate base). Solving the buffer problem: HF Initial Equil. ⇌ F− 0.80 M 0.80 ! x Ka = 7.2 × 10 −4 = + 0.80 M 0.80 + x H+ 0 x (0.80 + x)( x) (0.80)( x) ≈ , x = [H+] = 7.2 × 10 −4 M; pH = 3.14; 0.80 0.80 − x assumptions good. Note: The added HCl to this buffer solution changes the pH only from 3.37 to 3.14. If the HCl were added to 1.0 L of pure water, the pH would change from 7.00 to 0.70. 26. Major species: H2O, Cl−, Na+, HONH2, HONH3+, and OH−; the added strong base dominates the initial reaction mixture. Let the OH− react completely with the best acid present (HONH3+). HONH3+ Before Change After 0.100 M !0.020 0.080 + OH− → 0.020 M !0.020 → 0 HONH2 + H2O 0.100 M +0.020 0.120 A buffer solution results. Using the Henderson-Hasselbalch equation: Reacts completely CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA pH = pKa + log ⎛ Kw [base ] ; pK a = − log ⎜ ⎜ K b , HONH [acid ] 2 ⎝ pH = 6.04 + log [HONH 2 ] + [HONH 3 ] = 6.04 + log 261 −14 ⎞ ⎛ ⎞ ⎟ = − log ⎜ 1.0 × 10 ⎟ = 6.04 ⎜ 1.1 × 10 −8 ⎟ ⎟ ⎝ ⎠ ⎠ (0.120) = 6.04 + 0.18 = 6.22 (0.080) Major species: H2O, Cl−, HONH2, HONH3+, and H+; the added strong acid dominates the initial reaction mixture. Let the H+ react completely with HONH2, the best base present. HONH2 Before Change After + 0.100 M !0.020 0.080 H+ → HONH3+ 0.020 M !0.020 0 → 0.100 M +0.020 Reacts completely 0.120 A buffer solution results after reaction. Using the Henderson-Hasselbalch equation: pH = 6.04 + log [HONH 2 ] + [HONH 3 ] = 6.04 + log (0.080) = 6.04 ! 0.18 = 5.86 (0.120) 1 mol HC 7 H 5 O 2 122.12 g = 0.880 M 0.2000 L 21.5 g HC 7 H 5 O 2 × 27. [HC7H5O2] = 1 mol NaC 7 H 5 O 2 × 144.10 g − [C7H5O2 ] = 0.2000 L We have a buffer solution since we have both a weak the same time. One can use the Ka reaction or the Kb reaction for the acid component of the buffer. 37.7 g NaC 7 H 5 O 2 × HC7H5O2 Initial Change Equil. ⇌ H+ + − 1 mol C 7 H 5 O 2 mol NaC 7 H 5 O 2 = 1.31 M acid and its conjugate base present at reaction to solve. We will use the Ka C7H5O2− ~0 1.31 M 0.880 M x mol/L of HC7H5O2 dissociates to reach equilibrium !x → +x +x x 1.31 + x 0.880 – x Ka = 6.4 × 10 −5 = x(1.31 + x) x(1.31) ≈ , x = [H+] = 4.3 × 10 −5 M 0.880 0.880 − x pH = !log(4.3 × 10 −5 ) = 4.37; assumptions good. Alternatively, we can use the Henderson-Hasselbalch equation to calculate the pH of buffer solutions. 262 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA − pH = pKa + log [C H O ] [base] = pKa + log 7 5 2 [HC 7 H 5 O 2 ] [acid] ⎛ 1.31 ⎞ pH = !log(6.4 × 10 −5 ) + log ⎜ ⎟ = 4.19 + 0.173 = 4.36 ⎝ 0.880 ⎠ Within round-off error, this is the same answer we calculated solving the equilibrium problem using the Ka reaction. The Henderson-Hasselbalch equation will be valid when an assumption of the type 1.31 + x ≈ 1.31 that we just made in this problem is valid. From a practical standpoint, this will almost always be true for useful buffer solutions. If the assumption is not valid, the solution will have such a low buffering capacity that it will be of no use to control the pH. Note: The Henderson-Hasselbalch equation can only be used to solve for the pH of buffer solutions. 28. 50.0 g NH4Cl × 1 mol NH 4 Cl = 0.935 mol NH4Cl added to 1.00 L; [NH4+] = 0.935 M 53.49 g NH 4 Cl Using the Henderson Hasselbalch equation to solve for the pH of this buffer solution: pH = pKa + log 29. ⎛ 0.75 ⎞ = !log(5.6 × 10 −10 ) + log ⎜ ⎟ = 9.25 ! 0.096 = 9.15 [ NH 4 ] ⎝ 0.935 ⎠ [ NH 3 ] + C5H5NH+ ⇌ H+ + C5H5N Ka = Kw 1.0 × 10 −14 = = 5.9 × 10−6 Kb 1.7 × 10 −9 pKa = −log(5.9 × 10−6) = 5.23 We will use the Henderson-Hasselbalch equation to calculate the concentration ratio necessary for each buffer. pH = pKa + log [C 5 H 5 N ] [base] , pH = 5.23 + log [acid] [C 5 H 5 NH + ] a. 4.50 = 5.23 + log log [C 5 H 5 N ] [C 5 H 5 NH + ] [C 5 H 5 N ] = −0.73 [C 5 H 5 NH + ] [C 5 H 5 N ] = 10−0.73 = 0.19 + [C 5 H 5 NH ] b. 5.00 = 5.23 + log log [C 5 H 5 N ] [C 5 H 5 NH + ] [C 5 H 5 N ] = −0.23 [C 5 H 5 NH + ] [C 5 H 5 N ] = 10−0.23 = 0.59 + [C 5 H 5 NH ] CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA c. 5.23 = 5.23 + log [C 5 H 5 N ] [C 5 H 5 NH + ] d. 5.50 = 5.23 + log [C 5 H 5 N ] = 100.0 = 1.0 [C 5 H 5 NH + ] 30. 263 [C 5 H 5 N ] [C 5 H 5 NH + ] [C 5 H 5 N ] = 100.27 = 1.9 [C 5 H 5 NH + ] Added OH− converts HC2H3O2 into C2H3O2−: HC2H3O2 + OH− → C2H3O2− + H2O From this reaction, the moles of C2H3O2− produced equal the moles of OH− added. Also, the total concentration of acetic acid plus acetate ion must equal 2.0 M (assuming no volume change on addition of NaOH). Summarizing for each solution: [C2H3O2−] + [HC2H3O] = 2.0 M and [C2H3O2−] produced = [OH−] added − a. pH = pKa + log − [C 2 H 3 O 2 ] [C H O ] ; for pH = pKa, log 2 3 2 =0 [HC 2 H 3O 2 ] [HC 2 H 3O 2 ] − Therefore, [C 2 H 3 O 2 ] = 1.0 and [C2H3O2−] = [HC2H3O2]. [HC 2 H 3O 2 ] Because [C2H3O2−] + [HC2H3O2] = 2.0 M: [C2H3O2−] = [HC2H3O2] = 1.0 M = [OH−] added To produce a 1.0 M C2H3O2− solution, we need to add 1.0 mol of NaOH to 1.0 L of the 2.0 M HC2H3O2 solution. The resulting solution will have pH = pKa = 4.74. − b. 4.00 = 4.74 + log − [C 2 H 3 O 2 ] [C 2 H 3 O 2 ] , = 10−0.74 = 0.18 [HC 2 H 3O 2 ] [HC 2 H 3O 2 ] [C2H3O2−] = 0.18[HC2H3O2] or [HC2H3O2] = 5.6[C2H3O2−] Because [C2H3O2−] + [HC2H3O2] = 2.0 M: [C2H3O2−] + 5.6[C2H3O2−] = 2.0 M, [C2H3O2−] = 2 .0 = 0.30 M = [OH−] added 6.6 We need to add 0.30 mol of NaOH to 1.0 L of 2.0 M HC2H3O2 solution to produce 0.30 M C2H3O2−. The resulting solution will have pH = 4.00. − c. 5.00 = 4.74 + log − [C 2 H 3 O 2 ] [C 2 H 3 O 2 ] , = 100.26 = 1.8 [HC 2 H 3O 2 ] [HC 2 H 3O 2 ] 1.8[HC2H3O2] = [C2H3O2−] or [HC2H3O2] = 0.56[C2H3O2−] 1.56[C2H3O2−] = 2.0 M, [C2H3O2−] = 1.3 M = [OH−] added We need to add 1.3 mol of NaOH to 1.0 L of 2.0 M HC2H3O2 to produce a solution with pH = 5.00. 264 31. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA When H+ is added, it converts C2H3O2− into HC2H3O2: C2H3O2− + H+ → HC2H3O2. From this reaction, the moles of HC2H3O2 produced must equal the moles of H+ added and the total concentration of acetate ion + acetic acid must equal 1.0 M (assuming no volume change). Summarizing for each solution: [C2H3O2−] + [HC2H3O2] = 1.0 M and [HC2H3O2] = [H+] added − a. pH = pKa + log [C 2 H 3 O 2 ] ; for pH = pKa, [C2H3O2−] = [HC2H3O2]. [HC 2 H 3O 2 ] For this to be true, [C2H3O2−] = [HC2H3O2] = 0.50 M = [H+] added, which means that 0.50 mol of HCl must be added to 1.0 L of the initial solution to produce a solution with pH = pKa. − − [C H O ] [C 2 H 3 O 2 ] b. 4.20 = 4.74 + log 2 3 2 , = 10 −0.54 = 0.29 [HC 2 H 3O 2 ] [HC 2 H 3O 2 ] [C2H3O2−] = 0.29[HC2H3O2]; 0.29[HC2H3O2] + [HC2H3O2] = 1.0 M [HC2H3O2] = 0.78 M = [H+] added 0.78 mol of HCl must be added to produce a solution with pH = 4.20. − − [C H O ] [C 2 H 3 O 2 ] = 100.26 = 1.8 c. 5.00 = 4.74 + log 2 3 2 , [HC 2 H 3O 2 ] [HC 2 H 3O 2 ] [C2H3O2−] = 1.8[HC2H3O2]; 1.8[HC2H3O2] + [HC2H3O2] = 1.0 M [HC2H3O2] = 0.36 M = [H+] added 0.36 mol of HCl must be added to produce a solution with pH = 5.00. − 32. pH = pKa + log − [C 2 H 3 O 2 ] [C H O ] , 4.00 = !log(1.8 × 10 −5 ) + log 2 3 2 [HC 2 H 3O 2 ] [HC 2 H 3O 2 ] − [C 2 H 3 O 2 ] = 0.18; this is also equal to the mole ratio between C2H3O2− and HC2H3O2. [HC 2 H 3O 2 ] Let x = volume of 1.00 M HC2H3O2 and y = volume of 1.00 M NaC2H3O2 x + y = 1.00 L, x = 1.00 – y x(1.00 mol/L) = mol HC2H3O2; y(1.00 mol/L) = mol NaC2H3O2 = mol C2H3O2− y y = 0.18 or = 0.18; solving: y = 0.15 L, so x = 1.00 ! 0.15 = 0.85 L. x 1.00 − y We need 850 mL of 1.00 M HC2H3O2 and 150 mL of 1.00 M NaC2H3O2 to produce a buffer solution at pH = 4.00. Thus: CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 265 − 33. pH = pKa + log [C 2 H 3 O 2 ] ; pKa = −log(1.8 × 10−5) = 4.74 [HC 2 H 3O 2 ] Because the buffer components, C2H3O2− and HC2H3O2, are both in the same volume of water, the concentration ratio of [C2H3O2−]/[HC2H3O2] will equal the mole ratio of mol C2H3O2−/mol HC2H3O2. − 5.00 = 4.74 + log mol C 2 H 3O 2 0.200 mol ; mol HC2H3O2 = 0.5000 L × = 0.100 mol mol HC 2 H 3O 2 L − 0.26 = log mol C 2 H 3O 2 , 0.100 mol mol C 2 H 3O 2 0.100 mol − Mass NaC2H3O2 = 0.18 mol NaC2H3O2 × 34. [H+] added = a. 82.03 g = 15 g NaC2H3O2 mol 0.010 mol = 0.040 M; the added H+ reacts completely with NH3 to form NH4+. 0.2500 L H+ → NH4+ 0.040 M −0.040 0 → 0.15 M +0.040 0.19 + NH3 Before Change After = 100.26 = 1.8, mol C2H3O2− = 0.18 mol 0.050 M −0.040 0.010 Reacts completely A buffer solution still exists after H+ reacts completely. Using the HendersonHasselbalch equation: pH = pKa + log b. NH3 Before Change After 0.50 M −0.040 0.46 ⎛ 0.010 ⎞ = −log(5.6 × 10−10) + log ⎜ ⎟ = 9.25 + (−1.28) = 7.97 [ NH 4 ] ⎝ 0.19 ⎠ [ NH 3 ] + + H+ → 0.040 M −0.040 → 0 NH4+ 1.50 M +0.040 1.54 A buffer solution still exists. pH = pKa + log Reacts completely ⎛ 0.46 ⎞ , 9.25 + log ⎜ ⎟ = 8.73 [ NH 4 ] ⎝ 1.54 ⎠ [ NH 3 ] + Note: The two buffers differ in their capacity and not their initial pH (both buffers had an initial pH = 8.77). Solution b has the greatest capacity since it has the largest concentrations of weak acid and conjugate base. Buffers with greater capacities will be able to absorb more added H+ or OH−. 35. a. pKb for C6H5NH2 = −log(3.8 × 10−10) = 9.42; pKa for C6H5NH3+ = 14.00 − 9.42 = 4.58 pH = pKa + log [C 6 H 5 NH 2 ] + [C 6 H 5 NH 3 ] , 4.20 = 4.58 + log 0.50 M + [C 6 H 5 NH 3 ] 266 CHAPTER 8 −0.38 = log 0.50 M + [C 6 H 5 NH 3 ] b. 4.0 g NaOH × 36. , [C6H5NH3+] = [C6H5NH3Cl] = 1.2 M 1 mol NaOH 1 mol OH − 0.10 mol × = 0.10 mol OH−; [OH−] = = 0.10 M 40.00 g mol NaOH 1 .0 L C6H5NH3+ OH− + 1.2 M −0.10 1.1 Before Change After APPLICATIONS OF AQUEOUS EQUILIBRIA → 0.10 M −0.10 → 0 C6H5NH2 + H2O 0.50 M +0.10 0.60 ⎛ 0.60 ⎞ A buffer solution exists. pH = 4.58 + log ⎜ ⎟ = 4.32 ⎝ 1.1 ⎠ − − [ NO 2 ] [ NO 2 ] , 3.55 = −log(4.0 × 10−4) + log pH = pKa + log [HNO 2 ] [HNO 2 ] − 3.55 = 3.40 + log − [ NO 2 ] [ NO 2 ] , = 100.15 = 1.4 [HNO 2 ] [HNO 2 ] Let x = volume (L) HNO2 solution needed, then 1.00 − x = volume of NaNO2 solution needed to form this buffer solution. − [ NO 2 ] = 1.4 = [HNO 2 ] 0.50 mol NaNO 2 0.50 − (0.50) x L = 0.50 mol HNO 2 (0.50) x x× L (1.00 − x) × (0.70)x = 0.50 − (0.50)x , (1.20)x = 0.50, x = 0.42 L We need 0.42 L of 0.50 M HNO2 and 1.00 − 0.42 = 0.58 L of 0.50 M NaNO2 to form a pH = 3.55 buffer solution. − 37. a. pH = pKa + log − [HCO 3 ] [HCO 3 ] [base] = 6.37 + log , 7.40 = −log(4.3 × 10−7) + log [H 2 CO 3 ] [H 2 CO 3 ] [acid] − [HCO 3 ] [H 2 CO 3 ] [CO 2 ] 1 = 0.091 = 101.03 = 11; = = − − [H 2 CO 3 ] 11 [HCO 3 ] [HCO 3 ] −8 b. 7.15 = −log(6.2 × 10 ) + log 2− [HPO 4 ] − [H 2 PO 4 ] = 10−0.06 = 0.9, 2− [ HPO 4 ] − [H 2 PO 4 ] − [H 2 PO 4 ] 2− [ HPO 4 ] = 2− , 7.15 = 7.21 + log 1 = 1.1 ≈ 1 0 .9 [HPO 4 ] − [H 2 PO 4 ] CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 267 c. A best buffer has approximately equal concentrations of weak acid and conjugate base, so pH ≈ pKa for a best buffer. The pKa value for a H3PO4/H2PO4− buffer is −log(7.5 × 10−3) = 2.12. A pH of 7.15 is too high for a H3PO4/H2PO4− buffer to be effective. At this high of pH, there would be so little H3PO4 present that we could hardly consider it a buffer; this solution would not be effective in resisting pH changes, especially when strong base is added. 38. The reaction OH− + CH3NH3+ → CH3NH2 + H2O goes to completion for solutions a, c, and d (no reaction occurs between the species in solution b because both species are bases). After the OH− reacts completely, there must be both CH3NH3+ and CH3NH2 in solution for it to be a buffer. The important components of each solution (after the OH− reacts completely) is(are): a. 0.05 M CH3NH2 (no CH3NH3+ remains, no buffer) b. 0.05 M OH− and 0.1 M CH3NH2 (two bases present, no buffer) c. 0.05 M OH− and 0.05 M CH3NH2 (too much OH− added, no CH3NH3+ remains, no buffer) d. 0.05 M CH3NH2 and 0.05 M CH3NH3+ (a buffer solution results) Only the combination in mixture d results in a buffer. Note that the concentrations are halved from the initial values. This is so because equal volumes of two solutions were added together, which halves the concentrations. 39. a. No; a solution of a strong acid (HNO3) and its conjugate base (NO3−) is not generally considered a buffer solution. b. No; two acids are present (HNO3 and HF), so it is not a buffer solution. c. H+ reacts completely with F−. Since equal volumes are mixed, the initial concentrations in the mixture are 0.10 M HNO3 and 0.20 M NaF. H+ Before Change After 0.10 M !0.10 0 + F− 0.20 M !0.10 0.10 → HF → 0 +0.10 0.10 Reacts completely After H+ reacts completely, a buffer solution results; that is, a weak acid (HF) and its conjugate base (F−) are both present in solution in large quantities. d. No; a strong acid (HNO3) and a strong base (NaOH) do not form buffer solutions. They will neutralize each other to form H2O. 40. Because we have added two solutions together, the concentration of each reagent has changed. What hasn’t changed is the moles or millimoles of each reagent. Let’s determine the millimoles of each reagent present by multiplying the volume in milliters by the molarity in units of mmol/mL. 268 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 100.0 mL × 0.100 M = 10.0 mmol NaF; 100.0 mL × 0.025 M = 2.5 mmol HCl H+ + F− → HF; 2.5 mmol H+ converts 2.5 mmol F− into 2.5 mmol HF. After the reaction, a buffer solution results containing 2.5 mmol HF and (10.0 − 2.5 =) 7.5 mmol F− in 200.0 mL of solution. pH = pKa + log 41. ⎛ 7.5 mmol / 200.0 mL ⎞ [F− ] ⎟⎟ = 3.62; assumptions good. = 3.14 + log ⎜⎜ [HF] ⎝ 2.5 mmol / 200.0 mL ⎠ A best buffer has large and equal quantities of weak acid and conjugate base. Because [acid] [base] = pKa + 0 = pKa (pH ≈ pKa for a best = [base] for a best buffer, pH = pKa + log [acid] buffer). The best acid choice for a pH = 7.00 buffer would be the weak acid with a pKa close to 7.0 or Ka ≈ 1 × 10−7. HOCl is the best choice in Table 7.2 (Ka = 3.5 × 10−8; pKa = 7.46). To make this buffer, we need to calculate the [base]/[acid] ratio. 7.00 = 7.46 + log [base] [OCl − ] = 10−0.46 = 0.35 , [acid] [HOCl] Any OCl−/HOCl buffer in a concentration ratio of 0.35 : 1 will have a pH = 7.00. One possibility is [NaOCl] = 0.35 M and [HOCl] = 1.0 M. 42. For a pH = 5.00 buffer, we want an acid with a pKa close to 5.00. For a conjugate acid-base pair, 14.00 = pKa + pKb. So for a pH = 5.00 buffer, we want the base to have a pKb close to (14.0 − 5.0 =) 9.0 or a Kb close to 1 × 10−9. The best choice in Table 7.3 is pyridine (C5H5N) with Kb = 1.7 × 10−9. pH = pKa + log K 1.0 × 10 −14 [base] ; Ka = w = = 5.9 × 10−6 −9 Kb [acid] 1.7 × 10 5.00 = −log(5.9 × 10-6) + log [C 5 H 5 N ] [base] , = 10−0.23 = 0.59 + [acid] [C 5 H 5 NH ] There are many possibilities to make this buffer. One possibility is a solution of [C5H5N] = 0.59 M and [C5H5NHCl] = 1.0 M. The pH of this solution will be 5.00 because the base to acid concentration ratio is 0.59 : 1. 43. To solve for [KOCl], we need to use the equation derived in Section 8.3 of the text on the exact treatment of buffered solutions. The equation is: ⎛ [ H + ]2 − K w [H + ]⎜⎜ [A − ]0 + [H + ] ⎝ Ka = [ H + ]2 − K w [ HA]0 − [H + ] ⎞ ⎟ ⎟ ⎠ CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 269 Because pH = 7.20, [H+] = 10−7.20 = 6.3 × 10−8 M. ⎛ (6.3 × 10 −8 ) 2 − (1.0 × 10 −14 ) 6.3 × 10 −8 ⎜⎜ [OCl − ] + 6.3 × 10 −8 ⎝ Ka = 3.5 × 10−8 = (6.3 × 10 −8 ) 2 − (1.0 × 10 −14 ) 1.0 × 10 −6 − 6.3 × 10 −8 3.5 × 10−8 = 6.3 × 10 −8 ([OCl − ] − 9.57 × 10 −8 ) (1.0 × 10 −6 ) + (9.57 × 10 −8 ) ⎞ ⎟ ⎟ ⎠ (Carrying extra sig. figs.) 3.83 × 10−14 = 6.3 × 10-8([OCl−] − 9.57 × 10−8), [OCl−] = [KOCl] = 7.0 × 10−7 M 44. B + H2O ⇌ BH+ + OH− Kb = [BH + ][OH − ] [B] The equation for the exact treatment of B/BHCl type buffers would be analogous to the equation for HA/NaA type buffers. The equation is: ⎛ [OH − ]2 − K w [OH − ]⎜⎜ [BH + ]0 + [OH − ] ⎝ Kb = [OH − ]2 − K w [B]0 − [OH − ] ⎞ ⎟ ⎟ ⎠ Solving the buffer problem using the regular procedures: HONH2 + H2O ⇌ HONH3+ + OH− Kb = 1.1 × 10−8 1.0 × 10−5 M ~0 1.0 × 10−4 M x mol/L of HONH2 reacts with H2O to reach equilibrium +x +x Change −x Equil. 1.0 × 10−4 − x 1.0 × 10−5 + x x Initial Kb = 1.1 × 10−8 = + [HONH 3 ][OH − ] (1.0 × 10 −5 + x) x (1.0 × 10 −5 ) x = ≈ [HONH 2 ] (1.0 × 10 − 4 − x) 1.0 × 10 − 4 (Assuming x << 1.0 × 10−5.) x = [OH−] = 1.1 × 10−7 M; assumption that x << 1.0 × 10−5 is good (x is 1.1% of 1.0 × 10−5). In the regular procedure to solve the buffer problem, the problem reduced down to the expression: + Kb = [HONH 3 ] 0 [OH − ] [HONH 2 ] 0 270 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA This expression holds if x is negligible as compared to [HONH3+]0 and [HONH2]0 as it was in this problem. Now we want to know if we need to worry about the contribution of OH− from water. From the equation for the exact treatment of buffers, if ([OH−]2 − Kw) / [OH−] is much less than [HONH3+]0 and [HONH2]0, then the exact equation reduces to: + Kb = [OH − ][HONH 3 ]0 [ HONH 2 ]0 This is the same expression we ended up with to solve the problem using the regular procedures. Checking the neglected term using the [OH−] calculated above: [OH − ]2 − K w (1.1 × 10 −7 ) 2 − (1.0 × 10 −14 ) = = 1.9 × 10−8 [OH − ] 1.1 × 10 −7 This is indeed much smaller than [HONH3+]0 and [HONH2]0 (1.9 × 10−8 is 0.19% of 1.0 × 10−5). So for this problem we would calculate the same [OH−] using the exact equation as we calculated using the regular procedures. In general, we only need to use the exact equation when the buffering materials have a concentration of 10−6 M or less. 45. Using regular procedures, pH = pKa = −log(1.6 × 10−7) = 6.80 since [A−]0 = [HA]0 in this buffer solution. However, the pH is very close to that of neutral water, so maybe we need to consider the H+ contribution from water. Another problem with this answer is that x (= [H+]) is not small as compared with [HA]0 and [A−]0 , which was assumed when solving using the regular procedures. Because the concentrations of the buffer components are less than 10−6 M, let us use the expression for the exact treatment of buffers to solve. ⎛ [ H + ]2 − K w [H + ]⎜⎜ [ A − ]0 + [H + ] ⎝ −7 Ka = 1.6 × 10 = [ H + ]2 − K w [HA]0 − [H + ] ⎞ ⎟ ⎟ ⎠ = ⎛ [H + ]2 − (1.0 × 10 −14 ) [H + ]⎜⎜ 5.0 × 10 −7 + [H + ] ⎝ [ H + ]2 − (1.0 × 10 −14 ) 5.0 × 10 −7 − [H + ] ⎞ ⎟ ⎟ ⎠ Solving exactly requires solving a cubic equation. Instead, we will use the method of successive approximations where our initial guess for [H+] = 1.6 × 10−7 M (the value obtained using the regular procedures). ⎛ (1.6 × 10 −7 ) 2 − (1.0 × 10 −14 ) [H + ]⎜⎜ 5.0 × 10 −7 + 1.6 × 10 −7 ⎝ 1.6 × 10−7 = (1.6 × 10 −7 ) 2 − (1.0 × 10 −14 ) 5.0 × 10 −7 − 1.6 × 10 −7 ⎞ ⎟ ⎟ ⎠ , [H+] = 1.1 × 10−7 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 271 We continue the process using 1.1 × 10−7 as our estimate for [H+]. This gives [H+] = 1.5 × 10−7. We continue the process until we get a self consistent answer. After three more iterations, we converge on [H+] = 1.3 × 10−7 M. Solving for the pH: pH = −log(1.3 × 10−7) = 6.89 Note that if we were to solve this problem exactly (using the quadratic formula) while ignoring the H+ contribution from water, the answer comes out to [H+] = 1.0 × 10−7 M. We get a significantly different answer when we consider the H+ contribution from H2O. Acid-Base Titrations 46. (f) all points after stoichiometric point (a) and (e) pH (c) (d) (b) mL base HA + OH− → A− + H2O; Added OH− from the strong base converts the weak acid HA into its conjugate base A−. Initially before any OH− is added (point d), HA is the dominant species present. After OH− is added, both HA and A− are present, and a buffer solution results (region b). At the equivalence point (points a and e), exactly enough OH- has been added to convert all the weak acid HA into its conjugate base A−. Past the equivalence point (region f), excess OH− is present. For the answer to part b, we included almost the entire buffer region. The maximum buffer region (or the region which is the best buffer solution) is around the halfway point to equivalence (point c). At this point, enough OH− has been added to convert exactly one-half of the weak acid present initially into its conjugate base, so [HA] = [A−] and pH = pKa. A best buffer has about equal concentrations of weak acid and conjugate base present. 272 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 47. (d) (c) (b) pH (a) and (e) (f) all points after stoichiometric point mL acid B + H+ → BH+; added H+ from the strong acid converts the weak base B into its conjugate acid BH+. Initially, before any H+ is added (point d), B is the dominant species present. After H+ is added, both B and BH+ are present, and a buffered solution results (region b). At the equivalence point (points a and e), exactly enough H+ has been added to convert all the weak base present initially into its conjugate acid BH+. Past the equivalence point (region f), excess H+ is present. For the answer to b, we included almost the entire buffer region. The maximum buffer region is around the halfway point to equivalence (point c), where [B] = [BH+]. Here, pH = pKa , which is a characteristic of a best buffer. 48. Let’s review the strong acid-strong base titration using the example (case study) covered in Section 8.5 of the text. The example used was the titration of 50.0 mL of 0.200 M HNO3 titrated by 0.100 M NaOH. See Fig. 8.1 for the titration curve. Here are the important points. a. Initially, before any strong base has been added. Major species: H+, NO3−, and H2O. To determine the pH, determine the [H+] in solution after the strong acid has completely dissociated, as we always do for strong acid problems. b. After some strong base has been added, up to the equilivance point. For our example, this is from just after 0.00 mL NaOH added up to just before 100.0 mL NaOH added. Major species before any reaction: H+, NO3−, Na+, OH−, and H2O. Na+ and NO3− have no acidic or basic properties. In this region, the OH− from the strong base reacts with some of the H+ from the strong acid to produce water (H+ + OH− → H2O). As is always the case when something strong reacts, we assume the reaction goes to completion. Major species after reaction: H+, NO3−, Na+, and H2O: To determine the pH of the solution, we first determine how much of the H+ is neutralized by the OH−. Then we determine the excess [H+] and take the –log of this quantity to determine pH. From 0.1 to 99.9 mL NaOH added, the excess H+ from the strong acid determines the pH. c. The equivalence point (100.0 mL NaOH added). Major species before reaction: H+, NO3−, Na+, OH−, and H2O. Here, we have added just enough OH− to neutralize all of the H+ from the strong acid (moles OH− added = moles H+ present). After the stoichiometry reaction (H+ + OH− → H2O), both H+ and OH− have run out (this is the definition of the equivalence point). Major species after reaction: Na+, NO3−, and H2O. All we have in CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 273 solution are some ions with no acidic or basic properties (NO3− and Na+ in H2O). The pH = 7.00 at the equivalence point of a strong acid-strong base titration. d. Past the equivalence point (volume of NaOH added > 100.0 mL). Major species before reaction H+, NO3−, Na+, OH−, and H2O. After the stoichiometry reaction goes to completion (H+ + OH− → H2O), we have excess OH− present. Major species after reaction: OH−, Na+, NO3−, and H2O. We determine the excess [OH−] and convert this into the pH. After the equivalence point, the excess OH− from the strong base determines the pH. See Fig. 8.2 for a titration curve of a strong base by a strong acid. The stoichiometry problem is still the same, H+ + OH− → H2O, but what is in excess after this reaction goes to completion is the reverse of the strong acid-strong base titration. The pH up to just before the equivalence point is determined by the excess OH− present. At the equivalence point, pH = 7.00 because we have added just enough H+ from the strong acid to react with all the OH− from the strong base (moles of base present = moles of acid added). Past the equivalence point, the pH is determined by the excess H+ present. As can be seen from Figs. 8.1 and 8.2, both strong by strong titrations have pH = 7.00 at the equivalence point, but the curves are the reverse of each other before and after the equivalence point. 49. Titration i is a strong acid titrated by a strong base. The pH is very acidic until just before the equivalence point; at the equivalence point, pH = 7.00; and past the equivalence the pH is very basic. Titration ii is a strong base titrated by a strong acid. Here the pH is very basic until just before the equivalence point; at the equivalence point, pH = 7.00; and past the equivalence point, the pH is very acidic. Titration iii is a weak base titrated by a strong acid. The pH starts out basic because a weak base is present. However, the pH will not be as basic as in titration ii, where a strong base is titrated. The pH drops as HCl is added; then at the halfway point to equivalence, pH = pKa. Because Kb = 4.4 × 10 −4 for CH3NH2, CH3NH3+ has Ka = Kw/Kb = 2.3 × 10 −11 and pKa = 10.64. So, at the halfway point to equivalence for this weak base-strong acid titration, pH = 10.64. The pH continues to drop as HCl is added; then at the equivalence point the pH is acidic (pH < 7.00) because the only important major species present is a weak acid (the conjugate acid of the weak base). Past the equivalence point the pH becomes more acidic as excess HCl is added. Titration iv is a weak acid titrated by a strong base. The pH starts off acidic, but not nearly as acidic as the strong acid titration (i). The pH increases as NaOH is added; then, at the halfway point to equivalence, pH = pKa for HF = !log(7.2 × 10 −4 ) = 3.14. The pH continues to increase past the halfway point; then at the equivalence point, the pH is basic (pH > 7.0) because the only important major species present is a weak base (the conjugate base of the weak acid). Past the equivalence point, the pH becomes more basic as excess NaOH is added. a. All require the same volume of titrant to reach the equivalence point. At the equivalence point for all these titrations, moles acid = moles base (MAVA = MBVB). Because all the molarities and volumes are the same in the titrations, the volume of titrant will be the same (50.0 mL titrant added to reach equivalence point). 274 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA b. Increasing initial pH: i < iv < iii < ii; the strong acid titration has the lowest pH, the weak acid titration is next, followed by the weak base titration, with the strong base titration having the highest pH. c. i < iv < iii < ii; the strong acid titration has the lowest pH at the halfway point to equivalence, and the strong base titration has the highest halfway point pH. For the weak acid titration, pH = pKa = 3.14, and for the weak base titration, pH = pKa = 10.64. d. Equivalence point pH: iii < ii = i < iv; the strong-by-strong titrations have pH = 7.00 at the equivalence point. The weak base titration has an acidic pH at the equivalence point, and a weak acid titration has a basic equivalence point pH. The only different answer when the weak acid and weak base are changed would be for part c. This is for the halfway point to equivalence, where pH = pKa. HOC6H5; Ka = 1.6 × 10 −10 , pKa = !log(1.6 × 10 −10 ) = 9.80 C5H5NH+, Ka = Kw K b , C5 H 5 N = 1.0 × 10 −14 = 5.9 × 10 −6 , pKa = 5.23 1.7 × 10 −9 From the pKa values, the correct ordering at the halfway point to equivalence would be i < iii < iv < ii. Note that for the weak base-strong acid titration using C5H5N, the pH is acidic at the halfway point to equivalence, whereas the weak acid-strong base titration using HOC6H5 is basic at the halfway point to equivalence. This is fine; this will always happen when the weak base titrated has a Kb < 1 × 10 −7 (so Ka of the conjugate acid is greater than 1 × 10−7) and when the weak acid titrated has a Ka < 1 × 10 −7 (so Kb of the conjugate base is greater than 1 × 10−7). 50. HA + OH− → A− + H2O; it takes 25.0 mL of 0.100 M NaOH to reach the equivalence point where mmol HA = mmol OH− = 25.0 mL(0.100 M) = 2.50 mmol. At the equivalence point, some HCl is added. The H+ from the strong acid reacts to completion with the best base present, A−. H+ Before Change After + 13.0 mL × 0.100 M !1.3 mmol 0 A− → 2.5 mmol !1.3 mmol 1.2 mmol HA 0 +1.3 mmol 1.3 mmol A buffer solution is present after the H+ has reacted completely. pH = pKa + log ⎛ 1.2 mmol / VT [A − ] , 4.7 = pKa + log ⎜⎜ [HA] ⎝ 1.3 mmol / VT ⎞ ⎟⎟ ⎠ Because the log term will be negative [log(1.2/1.3) = !0.035)], the pKa value of the acid must be greater than 4.7. CHAPTER 8 51. APPLICATIONS OF AQUEOUS EQUILIBRIA 275 a. Because all acids are the same initial concentration, the pH curve with the highest pH at 0 mL of NaOH added will correspond to the titration of the weakest acid. This is curve f. b. The pH curve with the lowest pH at 0 mL of NaOH added will correspond to the titration of the strongest acid. This is pH curve a. The best point to look at to differentiate a strong acid from a weak acid titration (if initial concentrations are not known) is the equivalence point pH. If the pH = 7.00, the acid titrated is a strong acid; if the pH is greater than 7.00, the acid titrated is a weak acid. c. For a weak acid-strong base titration, the pH at the halfway point to equivalence is equal to the pKa value. The pH curve, which represents the titration of an acid with Ka = 1.0 × 10−6, will have a pH = −log(1 × 10−6) = 6.0 at the halfway point. The equivalence point, from the plots, occurs at 50 mL NaOH added, so the halfway point is 25 mL. Plot d has a pH ≈ 6.0 at 25 mL of NaOH added, so the acid titrated in this pH curve (plot d) has Ka ≈ 1 × 10−6. 52. The three key points to emphasize in your sketch are the initial pH, the pH at the halfway point to equivalence, and the pH at the equivalence point. For all the weak bases titrated, pH = pKa at the halfway point to equivalence (50.0 mL HCl added) because [weak base] = [conjugate acid] at this point. Here, the weak base with Kb = 1 × 10 −5 has a conjugate acid with Ka = 1 × 10−9, so pH = 9.0 at the halfway point. The weak base with Kb = 1 × 10 −10 has a pH = 4.0 at the halfway point to equivalence. For the initial pH, the strong base has the highest pH (most basic), whereas the weakest base has the lowest pH (least basic). At the equivalence point (100.0 mL HCl added), the strong base titration has pH = 7.0. The weak bases titrated have acidic pH’s because the conjugate acids of the weak bases titrated are the major species present. The weakest base has the strongest conjugate acid so its pH will be lowest (most acidic) at the equivalence point. strong base pH 7.0 Kb = 10-5 Kb = 10-10 Volume HCl added (mL) 53. This is a strong acid (HClO4) titrated by a strong base (KOH). Added OH− from the strong base will react completely with the H+ present from the strong acid to produce H2O. 276 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA a. Only strong acid present. [H+] = 0.200 M; pH = 0.699 0.100 mmol OH − = 1.00 mmol OHb. mmol OH− added = 10.0 mL × mL mmol H+ present = 40.0 mL × 0.200 mmol H + = 8.0 mmol H+ mL Note: The units millimoles are usually easier numbers to work with. The units for molarity are moles per liter but are also equal to millimoles per milliliter. H+ Before Change After + 8.00 mmol −1.00 mmol 7.00 mmol OH− → H2O 1.00 mmol −1.00 mmol 0 Reacts completely The excess H+ determines the pH. [H+]excess = 7.00 mmol H + = 0.140 M 40.0 mL + 10.0 mL pH = −log(0.140) = 0.854 c. mmol OH− added = 40.0 mL × 0.100 M = 4.00 mmol OH− H+ Before After [H+]excess = + 8.00 mmol 4.00 mmol OH− → H2O 4.00 mmol 0 4.00 mmol = 0.0500 M; pH = 1.301 (40.0 + 40.0) mL d. mmol OH- added = 80.0 mL × 0.100 M = 8.00 mmol OH−; this is the equivalence point because we have added just enough OH- to react with all the acid present. For a strong acid-strong base titration, pH = 7.00 at the equivalence point because only neutral species are present (K+, ClO4−, H2O). e. mmol OH− added = 100.0 mL × 0.100 M = 10.0 mmol OH− H+ Before After 8.00 mmol 0 + OH− → H2O 10.0 mmol 2.0 mmol Past the equivalence point, the pH is determined by the excess OH- present. [OH−]excess = 54. 2.0 mmol = 0.014 M; pOH = 1.85; pH = 12.15 (40.0 + 100.0) mL This is a strong base, Ba(OH)2, titrated by a strong acid, HCl. The added strong acid will neutralize the OH− from the strong base. As is always the case when a strong acid and/or strong base reacts, the reaction is assumed to go to completion. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 277 a. Only a strong base is present, but it breaks up into two moles of OH− ions for every mole of Ba(OH)2. [OH−] = 2 × 0.100 M = 0.200 M; pOH = 0.699; pH = 13.301 b. mmol OH− present = 80.0 mL × 0.100 mmol Ba (OH) 2 2 mmol OH − × mL mmol Ba (OH) 2 = 16.0 mmol OH− mmol H+ added = 20.0 mL × OH− Before Change After + 16.0 mmol −8.00 mmol 8.0 mmol [OH−]excess = 0.400 mmol H + = 8.00 mmol H+ mL H+ → H2O 8.00 mmol −8.00 mmol 0 Reacts completely 8.0 mmol OH − = 0.080 M; pOH = 1.10; pH = 12.90 80.0 mL + 20.0 mL c. mmol H+ added = 30.0 mL × 0.400 M = 12.0 mmol H+ OH− + H+ → H2O Before After 16.0 mmol 4.0 mmol 12.0 mmol 0 [OH−]excess = 4.0 mmol OH − = 0.036 M; pOH = 1.44; pH = 12.56 (80.0 + 30.0) mL d. mmol H+ added = 40.0 mL × 0.400 M = 16.0 mmol H+; this is the equivalence point. Because the H+ will exactly neutralize the OH− from the strong base, all we have in solution is Ba2+, Cl−, and H2O. All are neutral species, so pH = 7.00. e. mmol H+ added = 80.0 mL × 0.400 M = 32.0 mmol H+ OH− Before After [H+]excess = 55. 16.0 mmol 0 + H+ → H2O 32.0 mmol 16.0 mmol 16.0 mmol H + = 0.100 M; pH = 1.000 (80.0 + 80.0) mL This is a weak acid (HC2H3O2) titrated by a strong base (KOH). a. Only weak acid is present. Solving the weak acid problem: 278 CHAPTER 8 HC2H3O2 Initial Change Equil. APPLICATIONS OF AQUEOUS EQUILIBRIA ⇌ H+ + C2H3O2− 0.200 M ~0 0 x mol/L HC2H3O2 dissociates to reach equilibrium −x → +x +x x x 0.200 − x Ka = 1.8 × 10−5 = x2 x2 ≈ , x = [H+] = 1.9 × 10−3 M 0.200 − x 0.200 pH = 2.72; assumptions good. b. The added OH− will react completely with the best acid present, HC2H3O2. mmol HC2H3O2 present = 100.0 mL × mmol OH− added = 50.0 mL × HC2H3O2 Before Change After + 20.0 mmol −5.00 mmol 15.0 mmol 0.200 mmol HC 2 H 3O 2 = 20.0 mmol HC2H3O2 mL 0.100 mmol OH − = 5.00 mmol OH− mL OH− → 5.00 mmol −5.00 mmol → 0 C2H3O2− + H2O 0 +5.00 mmol 5.00 mmol Reacts completely After reaction of all the strong base, we have a buffer solution containing a weak acid (HC2H3O2) and its conjugate base (C2H3O2−). We will use the Henderson-Hasselbalch equation to solve for the pH. pH = pKa + log − ⎛ 5.00 mmol/VT [C 2 H 3 O 2 ] = −log (1.8 × 10−5) + log ⎜⎜ [HC 2 H 3O 2 ] ⎝ 15.0 mmol/VT ⎞ ⎟⎟, where VT = ⎠ total volume ⎛ 5.00 ⎞ pH = 4.74 + log ⎜ ⎟ = 4.74 + (-0.477) = 4.26 ⎝ 15.0 ⎠ Note that the total volume cancels in the Henderson-Hasselbalch equation. For the [base]/[acid] term, the mole ratio equals the concentration ratio because the components of the buffer are always in the same volume of solution. c. mmol OH− added = 100.0 mL × (0.100 mmol OH−/mL) = 10.0 mmol OH−; the same amount (20.0 mmol) of HC2H3O2 is present as before (it doesn’t change). As before, let the OH− react to completion, then see what is remaining in solution after this reaction. HC2H3O2 Before After 20.0 mmol 10.0 mmol + OH− 10.0 mmol 0 → C2H3O2− + H2O 0 10.0 mmol CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 279 A buffer solution results after reaction. Because [C2H3O2−] = [HC2H3O2] = 10.0 mmol/total volume, pH = pKa. This is always true at the halfway point to equivalence for a weak acid-strong base titration, pH = pKa. pH = −log(1.8 × 10−5) = 4.74 d. mmol OH− added = 150.0 mL × 0.100 M = 15.0 mmol OH−. Added OH− reacts completely with the weak acid. HC2H3O2 Before After + 20.0 mmol 5.0 mmol OH− → 15.0 mmol 0 C2H3O2− + H2O 0 15.0 mmol We have a buffer solution after all the OH− reacts to completion. Using the HendersonHasselbalch equation: pH = 4.74 + log − ⎛ 15.0 mmol [C 2 H 3 O 2 ] = 4.74 + log ⎜⎜ [HC 2 H 3O 2 ] ⎝ 5.0 mmol ⎞ ⎟⎟ ⎠ pH = 4.74 + 0.48 = 5.22 e. mmol OH− added = 200.00 mL × 0.100 M = 20.0 mmol OH−; as before, let the added OH− react to completion with the weak acid; then see what is in solution after this reaction. HC2H3O2 Before After 20.0 mmol 0 + OH− 20.0 mmol 0 → C2H3O2− + H2O 0 20.0 mmol This is the equivalence point. Enough OH− has been added to exactly neutralize all the weak acid present initially. All that remains that affects the pH at the equivalence point is the conjugate base of the weak acid (C2H3O2−). This is a weak base equilibrium problem. K 1.0 × 10 −14 C2H3O2− + H2O ⇌ HC2H3O2 + OH− Kb = w = Kb 1.8 × 10 −5 Initial Change Equil. 20.0 mmol/300.0 mL 0 0 Kb = 5.6 × 10−9 x mol/L C2H3O2 reacts with H2O to reach equilibrium −x → +x +x 0.0667 − x x x Kb = 5.6 × 10−10 = x2 x2 ≈ , x = [OH−] = 6.1 × 10−6 M 0.0667 − x 0.0667 pOH = 5.21; pH = 8.79; assumptions good. 280 CHAPTER 8 f. APPLICATIONS OF AQUEOUS EQUILIBRIA mmol OH− added = 250.0 mL × 0.100 M = 25.0 mmol OH− HC2H3O2 Before After OH− + 20.0 mmol 0 C2H3O2− + → 25.0 mmol 5.0 mmol H2O 0 20.0 mmol After the titration reaction, we have a solution containing excess OH− and a weak base C2H3O2−. When a strong base and a weak base are both present, assume that the amount of OH− added from the weak base will be minimal; that is, the pH past the equivalence point is determined by the amount of excess strong base. [OH−]excess = 56. 5.0 mmol = 0.014 M; pOH = 1.85; pH = 12.15 100.0 mL + 250.0 mL This is a weak base (H2NNH2) titrated by a strong acid (HNO3). To calculate the pH at the various points, let the strong acid react completely with the weak base present; then see what is in solution. a. Only a weak base is present. Solve the weak base equilibrium problem. H2NNH2 + H2O Initial Equil. ⇌ H2NNH3+ + OH− 0 x 0.100 M 0.100 - x Kb = 3.0 × 10−6 = ~0 x x2 x2 ≈ , x = [OH−] = 5.5 × 10−4 M 0.100 − x 0.100 pOH = 3.26; pH = 10.74; assumptions good. b. mmol H2NNH2 present = 100.0 mL × mmol H+ added = 20.0 mL × H2NNH2 Before Change After 10.0 mmol −4.00 mmol 6.0 mmol + 0.100 mmol H 2 NNH 2 = 10.0 mmol H2NNH2 mL 0.200 mmol H + = 4.00 mmol H+ mL H+ 4.00 mmol −4.00 mmol 0 → H2NNH3+ → 0 +4.00 mmol 4.00 mmol Reacts completely A buffer solution results after the titration reaction. Solving using the HendersonHasselbalch equation: pH = pKa + log K 1.0 × 10 −14 [base] = 3.3 × 10−9 ; Ka = w = Kb [acid] 3.0 × 10 −6 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA ⎛ 6.0 mmol / VT pH = −log(3.3 × 10−9) + log ⎜⎜ ⎝ 4.00 mmol / VT 281 ⎞ ⎟⎟, where VT = total volume, which ⎠ cancels. pH = 8.48 + log(1.5) = 8.48 + 0.18 = 8.66 c. mmol H+ added = 25.0 mL × 0.200 M = 5.00 mmol H+ H2NNH2 Before After H+ + 10.0 mmol 5.0 mmol H2NNH3+ → 5.00 mmol 0 0 5.00 mmol This is the halfway point to equivalence where [H2NNH3+] = [H2NNH2]. At this point, pH = pKa (which is characteristic of the halfway point for any weak base-strong acid titration). pH = −log(3.3 × 10−9) = 8.48 d. mmol H+ added = 40.0 mL × 0.200 M = 8.00 mmol H+ H2NNH2 Before After H+ + 10.0 mmol 2.0 mmol H2NNH3+ → 8.00 mmol 0 0 8.00 mmol A buffer solution results. pH = pKa + log ⎛ 2.0 mmol / VT [base] = 8.48 + log ⎜⎜ [acid] ⎝ 8.00 mmol / VT ⎞ ⎟⎟ = 8.48 + (-0.60) = 7.88 ⎠ e. mmol H+ added = 50.0 mL × 0.200 M = 10.0 mmol H+ H2NNH2 Before After H+ + 10.0 mmol 0 → 10.0 mmol 0 H2NNH3+ 0 10.0 mmol As is always the case in a weak base-strong acid titration, the pH at the equivalence point is acidic because only a weak acid (H2NNH3+) is present. Solving the weak acid equilibrium problem: H2NNH3+ Initial Equil. ⇌ 10.0 mmol/150.0 mL 0.0667 − x Ka = 3.3 × 10−9 = H+ 0 x + H2NNH2 0 x x2 x2 ≈ , x = [H+] = 1.5 × 10−5 M 0.0667 − x 0.0667 pH = 4.82; assumptions good. 282 CHAPTER 8 f. APPLICATIONS OF AQUEOUS EQUILIBRIA mmol H+ added = 100.0 mL × 0.200 M = 20.0 mmol H+ H2NNH2 Before After H+ + 10.0 mmol 0 → 20.0 mmol 10.0 mmol H2NNH3+ 0 10.0 mmol Two acids are present past the equivalence point, but the excess H+ will determine the pH of the solution since H2NNH3+ is a weak acid. [H+]excess = 57. 10.0 mmol 100.0 mL + 100.0 mL = 0.0500 M; pH = 1.301 We will do sample calculations for the various parts of the titration. All results are summarized in Table 8.1 at the end of Exercise 60. At the beginning of the titration, only the weak acid HC3H5O3 is present. Let HLac = HC3H5O3 and Lac− = C3H5O3−. HLac Initial Change Equil. 1.4 × 10−4 = ⇌ H+ + Lac− Ka = 10−3.86 = 1.4 × 10−4 0.100 M ~0 0 x mol/L HLac dissociates to reach equilibrium −x → +x +x x x 0.100 − x x2 x2 ≈ , x = [H+] = 3.7 × 10−3 M; pH = 2.43; assumptions good. 0.100 − x 0.100 Up to the stoichiometric point, we calculate the pH using the Henderson-Hasselbalch equation. This is the buffer region. For example, at 4.0 mL of NaOH added: initial mmol HLac present = 25.0 mL × mmol OH− added = 4.0 mL × 0.100 mmol = 2.50 mmol HLac mL 0.100 mmol = 0.40 mmol OH− mL Note: The units millimoles are usually easier numbers to work with. The units for molarity are moles per liter but are also equal to millimoles per milliliter. The 0.40 mmol of added OH− converts 0.40 mmol HLac to 0.40 mmol Lac− according to the equation: HLac + OH− → Lac− + H2O Reacts completely. mmol HLac remaining = 2.50 − 0.40 = 2.10 mmol; mmol Lac− produced = 0.40 mmol We have a buffer solution. Using the Henderson-Hasselbalch equation where pKa = 3.86: CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA pH = pKa + log [Lac − ] (0.40) = 3.86 + log [HLac] (2.10) 283 (Total volume cancels, so we can use use the ratio of moles or millimoles.) pH = 3.86 − 0.72 = 3.14 Other points in the buffer region are calculated in a similar fashion. Perform a stoichiometry problem first, followed by a buffer problem. The buffer region includes all points up to 24.9 mL OH− added. At the stoichiometric point (25.0 mL OH− added), we have added enough OH−to convert all of the HLac (2.50 mmol) into its conjugate base (Lac−). All that is present is a weak base. To determine the pH, we perform a weak base calculation. [Lac−]0 = 2.50 mmol = 0.0500 M 25.0 mL + 25.0 mL Lac− + H2O Initial Change Equil. Kb = ⇌ HLac + OH− Kb = 1.0 × 10 −14 = 7.1 × 10−11 1.4 × 10 − 4 0.0500 M 0 0 x mol/L Lac− reacts with H2O to reach equilibrium −x → +x +x 0.0500 − x x x x2 x2 ≈ = 7.1 × 10−11 0.0500 − x 0.0500 x = [OH−] = 1.9 × 10−6 M; pOH = 5.72; pH = 8.28; assumptions good. Past the stoichiometric point, we have added more than 2.50 mmol of NaOH. The pH will be determined by the excess OH- ion present. An example of this calculation follows. At 25.1 mL: OH− added = 25.1 mL × 0.100 mmol = 2.51 mmol OH− mL 2.50 mmol OH− neutralizes all the weak acid present. The remainder is excess OH−. Excess OH− = 2.51 − 2.50 = 0.01 mmol OH− [OH−]excess = 0.01 mmol = 2 × 10−4 M; pOH = 3.7; pH = 10.3 (25.0 + 25.1) mL All results are listed in Table 8.1 at the end of the solution to Exercise 60. 58. Results for all points are summarized in Table 8.1 at the end of the solution to Exercise 60. At the beginning of the titration, we have a weak acid problem: 284 CHAPTER 8 ⇌ HOPr Initial Change Equil. Ka = APPLICATIONS OF AQUEOUS EQUILIBRIA H+ + OPr- HOPr = HC3H5O2 OPr- = C3H5O2− 0.100 M ~0 0 x mol/L HOPr acid dissociates to reach equilibrium −x → +x +x 0.100 − x x x x2 x2 [H + ][OPr − ] = 1.3 × 10−5 = ≈ 0.100 − x 0.100 [HOPr] x = [H+] = 1.1 × 10−3 M; pH = 2.96; assumptions good. The buffer region is from 4.0 to 24.9 mL of OH− added. We will do a sample calculation at 24.0 mL OH− added. Initial mmol HOPr present = 25.0 mL × mmol OH− added = 24.0 mL × 0.100 mmol = 2.50 mmol HOPr mL 0.100 mmol = 2.40 mmol OH− mL The added strong base converts HOPr into OPr−. HOPr Before Change After + 2.50 mmol −2.40 0.10 mmol OH− → 2.40 mmol −2.40 → 0 OPr- + 0 +2.40 2.40 mmol H2O Reacts completely A buffer solution results. Using the Henderson-Hasselbalch equation where pKa = −log(1.3 × 10−5) = 4.89: pH = pKa + log [base] [OPr − ] = 4.89 + log [acid] [HOPr] ⎛ 2.40 ⎞ pH = 4.89 + log ⎜ ⎟ = 4.89 + 1.38 = 6.27 (Volume cancels, so we can use the ⎝ 0.10 ⎠ millimole ratio in the log term.) All points in the buffer region 4.0 mL to 24.9 mL are calculated this way. See Table 8.1 at the end of Exercise 60 for all the results. At the stoichiometric point, only a weak base (OPr−) is present: CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA OPr- Initial Change Equil. Kb = + ⇌ H2O 285 OH− + HOPr 2.50 mmol 0 0 = 0.0500 M 50.0 mL x mol/L OPr− reacts with H2O to reach equilibrium −x → +x +x 0.0500 − x x x [OH − ][HOPr] K w x2 x2 −10 = = 7.7 H 10 = ≈ Ka 0.0500 − x 0.0500 [OPr − ] x = 6.2 × 10−6 M = [OH−], pOH = 5.21, pH = 8.79; assumptions good. Beyond the stoichiometric point, the pH is determined by the excess strong base added. The results are the same as those in Exercise 57 (see Table 8.1). For example at 26.0 mL NaOH added: [OH−] = 59. 2.60 mmol − 2.50 mmol = 2.0 × 10−3 M; pOH = 2.70; pH = 11.30 (25.0 + 26.0) mL At beginning of the titration, only the weak base NH3 is present. As always, solve for the pH using the Kb reaction for NH3. NH3 + H2O Initial Equil. Kb = ⇌ NH4+ 0.100 M 0.100 − x 0 x + OH− Kb = 1.8 × 10−5 ~0 x x2 x2 ≈ = 1.8 × 10−5 0.100 − x 0.100 x = [OH−] = 1.3 × 10−3 M; pOH = 2.89; pH = 11.11; assumptions good. In the buffer region (4.0 − 24.9 mL), we can use the Henderson-Hasselbalch equation: Ka = [ NH 3 ] 1.0 × 10 −14 = 5.6 × 10−10; pKa = 9.25; pH = 9.25 + log −5 + 1.8 × 10 [ NH 4 ] We must determine the amounts of NH3 and NH4+ present after the added H+ reacts completely with the NH3. For example, after 8.0 mL HCl added: initial mmol NH3 present = 25.0 mL × 0.100 mmol = 2.50 mmol NH3 mL 286 CHAPTER 8 mmol H+ added = 8.0 mL × APPLICATIONS OF AQUEOUS EQUILIBRIA 0.100 mmol = 0.80 mmol H+ mL Added H+ reacts with NH3 to completion: NH3 + H+ → NH4+ mmol NH3 remaining = 2.50 − 0.80 = 1.70 mmol; mmol NH4+ produced = 0.80 mmol pH = 9.25 + log 1.70 = 9.58 0.80 (Mole ratios can be used since the total volume cancels.) Other points in the buffer region are calculated in similar fashion. Results are summarized in Table 8.1 at the end of Exercise 60. At the stoichiometric point (25.0 mL H+ added), just enough HCl has been added to convert all the weak base (NH3) into its conjugate acid (NH4+). Perform a weak acid calculation. [NH4+]0 = 2.50 mmol/50.0 mL = 0.0500 M ⇌ NH4+ Initial Equil. H+ + 0.0500 M 0.0500 - x 5.6 × 10−10 = 0 x NH3 Ka = 5.6 × 10−10 0 x x2 x2 ≈ , x = [H+] = 5.3 × 10−6 M; pH = 5.28; assumptions 0.0500 − x 0.0500 good. Beyond the stoichiometric point, the pH is determined by the excess H+. For example, at 28.0 mL of H+ added: H+ added = 28.0 mL × 0.100 mmol = 2.80 mmol H+ mL Excess H+ = 2.80 mmol − 2.50 mmol = 0.30 mmol excess H+ [H+]excess = 0.30 mmol = 5.7 × 10−3 M; pH = 2.24 (25.0 + 28.0) mL All results are summarized in Table 8.1. 60. Initially, a weak base problem: py Initial Equil. Kb = + 0.100 M 0.100 − x H2O ⇌ Hpy+ + 0 x OH− py is pyridine. ~0 x [Hpy + ][OH − ] x2 x2 = ≈ ≈ 1.7 × 10−9 [py] 0.100 − x 0.100 x = [OH−] = 1.3 × 10−5 M; pOH = 4.89; pH = 9.11; assumptions good. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 287 Buffer region (4.0 − 24.5 mL): Added H+ reacts completely with py: py + H+ → Hpy+. Determine the moles (or millimoles) of py and Hpy+ after reaction, then use the HendersonHasselbalch equation to solve for the pH. Ka = Kw 1.0 × 10 −14 [py] = = 5.9 × 10−6; pKa = 5.23; pH = 5.23 + log −9 Kb 1.7 × 10 [Hpy + ] Results in the buffer region are summarized in Table 8.1, which follows this problem. See Exercise 59 for a similar sample calculation. At the stoichiometric point (25.0 mL H+ added), this is a weak acid problem since just enough H+ has been added to convert all the weak base into its conjugate acid. The initial concentration of [Hpy+] = 0.0500 M. Hpy+ Initial Equil. 5.9 × 10−6 = 0.0500 M 0.0500 − x ⇌ py H+ + 0 x Ka = 5.9 × 10−6 0 x x2 x2 ≈ , x = [H+] = 5.4 × 10−4 M; pH = 3.27; asumptions 0.0500 − x 0.0500 good. Beyond the equivalence point, the pH determination is made by calculating the concentration of excess H+. See Exercise 8.59 for an example. All results are summarized in Table 8.1. Table 8.1 Summary of pH Results for Exercises 57 − 60 (Graph follows) Titrant mL Exercise 57 Exercise 58 Exercise 59 0.0 4.0 8.0 12.5 20.0 24.0 24.5 24.9 25.0 25.1 26.0 28.0 30.0 2.43 3.14 3.53 3.86 4.46 5.24 5.6 6.3 8.28 10.3 11.29 11.75 11.96 2.96 4.17 4.56 4.89 5.49 6.27 6.6 7.3 8.79 10.3 11.30 11.75 11.96 11.11 9.97 9.58 9.25 8.65 7.87 7.6 6.9 5.28 3.7 2.71 2.24 2.04 Exercise 60___ 9.11 5.95 5.56 5.23 4.63 3.85 3.5 3.27 2.71 2.25 2.04 288 61. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA a. This is a weak acid-strong base titration. At the halfway point to equivalence, [weak acid] = [conjugate base], so pH = pKa (always for a weak acid-strong base titration). pH = −log(6.4 × 10−5) = 4.19 mmol HC7H5O2 present = 100.0 mL × 0.10 M = 10. mmol HC7H5O2. For the equivalence point, 10. mmol of OH− must be added. The volume of OH− added to reach the equivalence point is: 10. mmol OH− H 1 mL 0.10 mmol OH − = 1.0 × 102 mL OH− At the equivalence point, 10. mmol of HC7H5O2 is neutralized by 10. mmol of OH− to produce 10. mmol of C7H5O2−. This is a weak base. The total volume of the solution is 100.0 mL + 1.0 × 102 mL = 2.0 × 102 mL. Solving the weak base equilibrium problem: C7H5O2− + H2O ⇌ Initial 10. mmol/2.0 × 102 mL Equil. 0.050 − x Kb = 1.6 × 10−10 = HC7H5O2 + OH− Kb = 0 x 1.0 × 10 −14 = 1.6 × 10−10 6.4 × 10 −5 0 x x2 x2 ≈ , x = [OH−] = 2.8 × 10−6 M 0.050 − x 0.050 pOH = 5.55; pH = 8.45; assumptions good. b. At the halfway point to equivalence for a weak base-strong acid titration, pH = pKa because [weak base] = [conjugate acid]. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA Ka = 289 Kw 1.0 × 10 −14 = = 1.8 × 10−11; pH = pKa = −log(1.8 × 10−11) = 10.74 −4 Kb 5.6 × 10 For the equivalence point (mmol acid added = mmol base present): mmol C2H5NH2 present = 100.0 mL × 0.10 M = 10. mmol C2H5NH2 mL H+ added = 10. mmol H+ × 1 mL 0.20 mmol H + = 50. mL H+ The strong acid added completely converts the weak base into its conjugate acid. Therefore, at the equivalence point, [C2H5NH3+]0 = 10. mmol/(100.0 + 50.) mL = 0.067 M. Solving the weak acid equilibrium problem: C2H5NH3+ ⇌ H+ + Initial 0.067 M Equil. 0.067 − x Ka = 1.8 × 10−11 = C2H5NH2 0 x 0 x x2 x2 ≈ , x = [H+] = 1.1 × 10−6 M 0.067 − x 0.067 pH = 5.96; assumptions good. c. In a strong acid-strong base titration, the halfway point has no special significance other than that exactly one-half of the original amount of acid present has been neutralized. mmol H+ present = 100.0 mL × 0.50 M = 50. mmol H+ mL OH− added = 25 mmol OH− × H+ + 1 mL = 1.0 × 102 mL OH− 0.25 mmol OH− → H2O Before After 50. mmol 25 mmol 25 mmol 0 [H+]excess = 25 mmol = 0.13 M; pH = 0.89 (100.0 + 1.0 × 10 2 ) mL At the equivalence point of a strong acid-strong base titration, only neutral species are present (Na+, Cl−, and H2O), so the pH = 7.00. 62. 75.0 mL × 0.10 mmol 0.10 mmol = 7.5 mmol HA; 30.0 mL × = 3.0 mmol OH− added mL mL The added strong base reacts to completion with the weak acid to form the conjugate base of the weak acid and H2O. 290 CHAPTER 8 HA Before After OH− + 7.5 mmol 4.5 mmol APPLICATIONS OF AQUEOUS EQUILIBRIA → A− 3.0 mmol 0 + H2O 0 3.0 mmol A buffer results after the OH− reacts to completion. Using the Henderson-Hasselbalch equation: ⎛ 3.0 mmol / 105.0 mmol ⎞ [A − ] ⎟⎟ pH = pKa + log , 5.50 = pKa + log ⎜⎜ [HA] ⎝ 4.5 mmol / 105.0 mmol ⎠ pKa = 5.50 ! log(3.0/4.5) = 5.50 – (!0.18) = 5.68; Ka = 10 −5.68 = 2.1 × 10 −6 63. a. 1.00 L × 0.100 mol/L = 0.100 mol HCl added to reach stoichiometric point. The 10.00-g sample must have contained 0.100 mol of NaA. 10.00 g = 100. g/mol 0.100 mol b. 500.0 mL of HCl added represents the halfway point to equivalence. Thus pH = pKa = 5.00 and Ka = 1.0 × 10−5. At the equivalence point, enough H+ has been added to convert all the A− present initially into HA. The concentration of HA at the equivalence point is: [HA]0 = 0.100 mol = 0.0909 M 1.10 L HA Initial Equil. 0.0909 M 0.0909 - x Ka = 1.0 × 10−5 = ⇌ H+ 0 x + A− Ka = 1.0 × 10−5 0 x x2 x2 ≈ 0.0909 − x 0.0909 x = 9.5 × 10−4 M = [H+]; pH = 3.02; assumptions good. 64. An acid-base indicator marks the end point of a titration by changing color. Acid-base indicators are weak acids themselves. We abbreviate the acid form of an indicator as HIn and the conjugate base form as In−. The reason there is a color change with indicators is that the HIn form has one color associated with it, whereas the In− form has a different color associated with it. Which form dominates in solution and dictates the color is determined by the pH of the solution. The related quilibrium is HIn ⇌ H+ + In−. In a very acidic solution, there are lots of H+ ions present, which drives the indicator equilibrium to the left. The HIn form dominates, and the color of the solution is the color due to the HIn form. In a very basic solution, H+ has been removed from solution. This drives the indicator equilibrium to the right, and the In− form dominates. In very basic solutions, the solution takes on the color of the In− form. In between very acidic and very basic solutions, there is a range of pH values where the solution has significant amounts of both the HIn and In− forms present. This is CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 291 where the color change occurs, and we want this pH to be close to the stoichiometric point of the titration. The pH at which the color change occurs is determined by the Ka of the indicator. Equivalence point: when enough titrant has been added to react exactly with the substance in the solution being titrated. Endpoint: when the indicator changes color. We want the indicator to tell us when we have reached the equivalence point. We can detect the endpoint visually and assume that it is the equivalence point for doing stoichiometric calculations. They don’t have to be as close as 0.01 pH units since, at the equivalence point, the pH is changing very rapidly with added titrant. The range over which an indicator changes color only needs to be close to the pH of the equivalence point. The two forms of an indicator are different colors. The HIn form has one color and the In− form has another color. To see only one color, that form must be in an approximately tenfold excess or greater over the other form. When the ratio of the two forms is less than 10, both colors are present. To go from [HIn]/[In−] = 10 to [HIn]/[In−] = 0.1 requires a change of 2 pH units (a 100-fold decrease in [H+]) as the indicator changes from the HIn color to the In− color. From Figure 8.8, thymol blue has three colors associated with it: orange, yellow, and blue. In order for this to happen, thymol blue must be a diprotic acid. The H2In form has the orange color, the HIn− form has the yellow color, and the In2− form has the blue color associated with it. Thymol blue cannot be monoprotic; monoprotic indicators only have two colors associated with them (either the HIn color or the In− color). 65. HIn ⇌ In− + H+ Ka = [In − ][H + ] = 1.0 × 10 −9 [HIn] a. In a very acid solution, the HIn form dominates, so the solution will be yellow. b. The color change occurs when the concentration of the more dominant form is approximately ten times as great as the less dominant form of the indicator. 10 [HIn] ⎛1⎞ = ; Ka = 1.0 × 10 −9 = ⎜ ⎟ [H+], [H+] = 1 × 10 −8 M; pH = 8.0 at color − 1 [In ] ⎝ 10 ⎠ change c. This is way past the equivalence point (100.0 mL OH− added), so the solution is very basic and the In− form of the indicator dominates. The solution will be blue. 66. HIn ⇌ In− + H+ Ka = [In − ][H + ] = 10 −3.00 = 1.0 × 10−3 [HIn] At 7.00% conversion of HIn into In−, [In−]/[HIn] = 7.00/93.00. Ka = 1.0 × 10 −3 = [ In − ] 7.00 × [H + ] = × [H + ], [H+] = 1.3 × 10 −2 M, pH = 1.89 [HIn] 93.00 The color of the base form will start to show when the pH is increased to 1.89. 292 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 67. pH > 5 for bromcresol green to be blue. pH < 8 for thymol blue to be yellow. The pH is between 5 and 8. 68. a. yellow b. green (Both yellow and blue forms are present.) c. yellow d. 69. 70. 71. 72. blue When choosing an indicator, we want the color change of the indicator to occur approximately at the pH of the equivalence point. Since the pH generally changes very rapidly at the equivalence point, we don’t have to be exact. This is especially true for strong acid-strong base titrations. The following are some indicators where the color change occurs at about the pH of the equivalence point. Exercise pH at Eq. Pt. Indicator 53 55 7.00 8.79 bromthymol blue or phenol red o-cresolphthalein or phenolphthalein Exercise pH at Eq. Pt. Indicator 54 56 7.00 4.82 bromthymol blue or phenol red bromcresol green Exercise pH at Eq. Pt. Indicator 57 59 8.28 5.28 o-cresolphthalein or phenolphthalein bromcresol green Exercise pH at Eq. Pt. Indicator 58 60 8.79 3.27 o-cresolphthalein or phenolphthalein 2,4-dinitrophenol The titration in Exercise 60 will be very difficult to mark the equivalence point. The pH break at the equivalence point is too small. 73. The color of the indicator will change over the approximate range of pH = pKa ± 1 = 5.3 ± 1. Therefore, the useful pH range of methyl red where it changes color would be about 4.3 (red) to 6.3 (yellow). Note that at pH < 4.3, the HIn form of the indicator dominates, and the color of the solution is the color of HIn (red). At pH > 6.3, the In- form of the indicator dominates, and the color of the solution is the color of In− (yellow). In titrating a weak acid with base, we start off with an acidic solution with pH < 4.3, so the color would change from red to reddish orange at pH ≈ 4.3. In titrating a weak base with acid, the color change would be from yellow to yellowish orange at pH ≈ 6.3. Only a weak base-strong acid titration would have an acidic pH at the equivalence point, so only in this type of titration would the color change of methyl red indicate the approximate endpoint. 74. For bromcresol green, the resulting green color indicates that both HIn and In− are present in significant amounts. This occurs when pH ≈ pKa of the indicator. From results of the CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 293 bromcresol green indicator, pH ≈ 5.0. Note that the results of the first two indicators are inconclusive. HX ⇌ H+ + X−; from the typical weak acid setup: [H+] = [X−] ≈ 1 × 10−5 M and [HX] ≈ 0.01 M Ka = 75. [H + ][X − ] (1 × 10 −5 ) 2 = = 1 × 10−8 [HX] 0.01 100.0 mL × 0.0500 M = 5.00 mmol H3X initially a. Because K a1 >> K a 2 >> K a 3 , pH initially is determined by H3X equilibrium reaction. ⇌ H3X Initial Equil. 0.0500 M 0.0500 − x K a1 = 1.0 × 10−3 = H+ + H2X− ~0 x 0 x x2 x2 ≈ , x = 7.1 × 10−3; assumption poor. 0.0500 − x 0.0500 Using the quadratic formula: x2 + (1.0 × 10−3)x − 5.0 × 10−5 = 0, x = 6.6 × 10−3 M = [H+]; pH = 2.18 b. 1.00 mmol OH− added converts H3X into H2X−. After this reaction goes to completion, 4.00 mmol H3X and 1.00 mmol H2X− are in a total volume of 110.0 mL. Solving the buffer problem: ⇌ H3X Initial Equil. 0.0364 M 0.0364 − x K a1 = 1.0 × 10−3 = H+ ~0 x + H2X− 0.00909 M 0.00909 + x x(0.00909 + x) ; assumption that x is small does not work here. 0.0364 − x Using the quadratic formula and carrying extra sig. figs. x2 + (1.01 × 10−2)x − 3.64 × 10−5 = 0, x = 2.8 × 10−3 M = [H+]; pH = 2.55 c. 2.50 mmol OH− added results in 2.50 mmol H3X and 2.50 mmol H2X− after OH− reacts completely with H3X. This is the first halfway point to equivalence. pH = p K a1 = 3.00; assumptions good (5% error). d. 5.00 mmol OH− added results in 5.00 mmol H2X− after OH− reacts completely with H3X. This is the first stoichiometric point. 294 CHAPTER 8 pH = pK a1 + pK a 2 3.00 + 7.00 = 5.00 2 = 2 APPLICATIONS OF AQUEOUS EQUILIBRIA e. 6.00 mmol OH−added results in 4.00 mmol H2X− and 1.00 mmol HX2− after OH− reacts completely with H3X and then reacts completely with H2X−. Using the H2X− ⇌ H+ + HX2− reaction: pH = pK a 2 + log f. [HX 2 − ] = 7.00 − log(1.00/4.00) = 6.40; assumptions good. [H 2 X − ] 7.50 mmol KOH added results in 2.50 mmol H2X− and 2.50 mmol HX2− after OH− reacts completely. This is the second halfway point to equivalence. pH = pK a 2 = 7.00; assumptions good. g. 10.0 mmol OH− added results in 5.0 mmol HX2− after OH− reacts completely. This is the second stoichiometric point. pH = pK a 2 + pK a 3 2 = 7.00 + 12.00 = 9.50 2 h. 12.5 mmol OH− added results in 2.5 mmol HX2− and 2.5 mmol X3− after OH− reacts completely with H3X first, then H2X−, and finally HX2-. This is the third halfway point to equivalence. Usually pH = pK a 3 but normal assumptions don't hold. We must solve for the pH exactly. [X3−-] = [HX2−] = 2.5 mmol/225.0 mL = 1.1 × 10−2 M X3− + H2O Initial Equil. ⇌ + 0.011 M 0.011 + x 0.011 M 0.011 − x Kb = 1.0 × 10−2 = HX2− OH− Kb = Kw = 1.0 × 10−2 K a3 0 x x(0.011 + x) ; using the quadratic formula: 0.011 − x x2 + (2.1 × 10−2)x − 1.1 × 10−4 = 0, x = 4.3 × 10−3 M = OH−; pH = 11.63 i. 15.0 mmol OH− added results in 5.0 mmol X3− after OH− reacts completely. This is the third stoichiometric point. X3− + H2O ⇌ HX2− + OH− Initial 5.0 mmol = 0.020 M 250.0 mL 0 0 Equil. 0.020 − x x x Kb = Kw = 1.0 × 10−2 K a3 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA Kb = 295 x2 x2 , 1.0 × 10−2 ≈ , x = 1.4 × 10−2; assumption poor. 0.020 − x 0.020 Using the quadratic formula: x2 + (1.0 × 10−2)x − 2.0 × 10−4 = 0 x = [OH−] = 1.0 × 10−2 M; pH = 12.00 j. 20.0 mmol OH− added results in 5.0 mmol X3− and 5.0 mmol OH− excess after OH− reacts completely. Because Kb for X3− is fairly large for a weak base, we have to worry about the OH− contribution from X3−. [X3−] = [OH−] = 5.0 mmol = 1.7 × 10−2 M 300.0 mL X3− Initial Equil. + H2O ⇌ OH− 1.7 × 10−2 M 1.7 × 10−2 − x Kb = 1.0 × 10−2 = + 1.7 × 10−2 M 1.7 × 10−2 + x HX2− 0 x (1.7 × 10 −2 + x) x (1.7 × 10 − 2 − x) Using the quadratic formula: x2 + (2.7 × 10−2)x − 1.7 × 10−4 = 0, x = 5.3 × 10−3 M [OH−] = (1.7 × 10−2) + x = (1.7 × 10−2) + (5.3 × 10−3) = 2.2 × 10−2 M; pH = 12.34 76. a. Because K a1 >> K a 2 >> K a13 , the initial pH is determined by H3A. Consider only the first dissociation. ⇌ H3A Initial Equil. K a1 = 0.100 M 0.100 − x H+ ~0 x + H2A− 0 x [H + ][H 2 A − ] x2 x2 = = 1.5 × 10−4 ≈ , x = 3.9 × 10−3 [H 3 A] 0.100 − x 0.100 [H+] = 3.9 × 10−3 M; pH = 2.41; assumptions good. b. 10.0 mL × 1.00 M = 10.0 mmol NaOH. Began with 100.0 mL × 0.100 M = 10.0 mmol H3A. Added OH− converts H3A into H2A−. This takes us to the first stoichiometric point where the amphoteric H2A− is the major species present. pH = pK a1 + pK a 2 2 = 3.82 + 7.52 = 5.67 2 296 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA c. 25.0 mL × 1.00 M = 25.0 mmol NaOH added. After OH− reacts completely, the mixture contains 5.0 mmol HA2− and 5.0 mmol A3−. K a3 = [H + ][A 3− ] ; because [A3−] = [HA2−], [H+] = K a 3 ; pH = pK a 3 = 11.30 2− [HA ] This is the third halfway point to equivalence; assumptions good. 77. 0.200 g = 1.212 × 10−3 mol = 1.212 mmol H3A (carrying extra sig. figs.) 165.0 g/mol a. (10.50 mL)(0.0500 M) = 0.525 mmol OH− added; H3A + OH− → H2A− + H2O; 1.212 − 0.525 = 0.687 mmol H3A remains after OH− reacts completely and 0.525 mmol H2A− formed. Solving the buffer problem using the K a1 reaction gives: ⎞ ⎛ 0.525 (10 −3.73 )⎜ + 10 −3.73 ⎟ ⎠ = 1.5 × 10−4 ( pK = 3.82) ⎝ 60.50 K a1 = a1 0.687 − 10 −3.73 60.50 First stoichiometric point: pH = pKa1 + pKa 2 = 5.19 = 2 3.82 + pK a 2 2 pK a 2 = 6.56 ( K a 2 = 2.8 × 10−7) Second stoichiometric point: pH = pKa 2 + pKa 3 2 , 8.00 = 6.56 + pK a 3 2 −10 pK a 3 = 9.44 ( K a 3 = 3.6 × 10 ) b. 1.212 mmol H3A = (0.0500 M OH − )(VOH − ) , VOH − = 24.2 mL; 24.2 mL of OH− are necessary to reach the first stoichiometric point. It will require 60.5 mL to reach the third halfway point to equivalence, where pH = pK a 3 = 9.44. The pH at 59.0 mL of NaOH added should be a little lower than 9.44. c. 59.0 mL of 0.0500 M OH− = 2.95 mmol OH− added H3A + OH− → Before 1.212 mmol 2.95 mmol After 0 H2A− 1.74 + OH− Before 1.212 1.74 After 0.53 0 HA2− + OH− Before 1.212 0.53 After 0.68 mmol 0 H2A− + H2O 0 1.212 → HA2− + H2O 0 1.212 → A3− + H2O 0 0.53 mmol CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 297 Use the K a 3 reaction to solve for the [H+], making the normal assumptions. ⎛ 0.53 mmol ⎞ + ⎜⎜ ⎟ [H ] 109 mL ⎟⎠ ⎝ −10 , [H+] = 4.6 × 10−10 M; pH = 9.34 K a 3 = 3.6 × 10 = ⎛ 0.68 mmol ⎞ ⎜⎜ ⎟⎟ ⎝ 109 mL ⎠ Assumptions good. 78. 50.0 mL × 0.10 mmol H 2 A = 5.0 mmol H2A initially mL To reach the first equivalence point, 5.0 mmol OH− must be added. This occurs after addition of 50.0 mL of 0.10 M NaOH. At the first equivalence point for a diprotic acid, pH = ( pK a1 + pK a 2 )/2 = 8.00. Addition of 25.0 mL of 0.10 M NaOH will be the first halfway point to equivalence, where [H2A] = [HA−] and pH = pK a1 = 6.70. Solving for the Ka values: pK a1 = 6.70, K a1 = 10−6.70 = 2.0 × 10−7 pK a1 + pK a 2 2 79. 6.70 + pK a 2 = 8.00, 2 = 8.00, pK a 2 = 9.30, K a 2 = 10−9.30 = 5.0 × 10−10 a. Na+ is present in all solutions. The added H+ from HCl reacts completely with CO32− to convert it into HCO3−. After all CO32− is reacted (after point C, the first equivalence point), H+ then reacts completely with the next best base present, HCO3−. Point E represents the second equivalence point. The major species present at the various points after H+ reacts completely follow. A. CO32−, H2O, Na+ B. CO32−, HCO3−, H2O, Cl− , Na+ C. HCO3−, H2O, Cl−, Na+ D. HCO3−, CO2 (H2CO3), H2O, Cl−, Na+ E. CO2 (H2CO3), H2O, Cl−, Na+ F. H+ (excess), CO2 (H2CO3), H2O, Cl−, Na+ b. Point A (initially): CO32− + H2O ⇌ HCO3− + OH− Initial 0.100 M 0 ~0 Equil. 0.100 − x x x Kb = 2.1 × 10−4 = − [HCO 3 ][OH − ] 2− [CO 3 ] = K b , CO 2− = 3 Kw 1.0 × 10 −14 = Ka2 4.8 × 10 −11 Kb = 2.1 × 10−4 x2 x2 ≈ 0.100 − x 0.100 x = 4.6 × 10−3 M = [OH−]; pH = 11.66; assumptions good. 298 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA Point B: The first halfway point where [CO32−] = [HCO3−]. pH = pK a 2 = −log(4.8 × 10−11) = 10.32; assumptions good. First equivalence point (25.00 mL of 0.100 M HCl added). The amphoteric Point C: HCO3− is the major acid-base species present. pH = pH = pK a1 + pK a 2 2 ; pK a1 = −log(4.3 × 10−7) = 6.37 6.37 + 10.32 = 8.35 2 Point D: The second halfway point where [HCO3−] = [H2CO3]. pH = pK a1 = 6.37; assumptions good. Point E: This is the second equivalence point, where all of the CO32− present initially has been converted into H2CO3 by the added strong acid. 50.0 mL HCl added. [H2CO3] = 2.50 mmol/75.0 mL = 0.0333 M H2CO3 ⇌ H+ + HCO3− Initial 0.0333 M Equil. 0.0333 − x 0 x K a1 = 4.3 × 10−7 = x2 x2 ≈ 0.0333 − x 0.0333 K a1 = 4.3 × 10−7 0 x x = [H+] = 1.2 × 10−4 M; pH = 3.92; assumptions good. 80. a. HA− ⇌ H+ + A2- Ka = 1 × 10−8; when [HA−] = [A2−], pH = pK a 2 = 8.00. The titration reaction is A2− + H+ → HA− (goes to completion). Begin with 100.0 mL × 0.200 mmol/mL = 20.0 mmol A2−. We need to convert 10.0 mmol A2− into HA- by adding 10.0 mmol H+. This will produce a solution where [HA−] = [A2−] and pH = pK a 2 = 8.00. 10.0 mmol = 1.00 mmol/mL × V, V = 10.0 mL HCl b. At the second stoichiometric point, all A2− is converted into H2A. This requires 40.0 mmol HCl, which is 40.0 mL of 1.00 M HCl. [H2A]0 = 20.0 mmol = 0.143 M; because K a1 >> K a 2 , H2A is the major source of H+. 140.0 mL CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA H2A Initial Equil. K a1 = ⇌ H+ 0.143 M 0.143 − x + 299 HA− 0 x 0 x x2 x2 , 1.0 H 10−3 ≈ , x = 0.012 M; check assumptions: 0.143 − x 0.143 0.012 × 100 = 8.4%; can't neglect x. Using successive approximations: 0.143 x = 0.0115 (carrying extra sig. figs.); [H+] = 0.0115 M and pH = 1.94 Solubility Equilibria 81. In our setups, s = solubility in mol/L. Because solids do not appear in the Ksp expression, we do not need to worry about their initial or equilibrium amounts. Ag3PO4(s) a. Initial Change Equil. ⇌ 3 Ag+(aq) PO43−(aq) + 0 0 s mol/L of Ag3PO4(s) dissolves to reach equilibrium −s → +3s +s 3s s Ksp = 1.8 × 10−18 = [Ag+]3[PO43−] = (3s)3(s) = 27s4 27s4 = 1.8 × 10−18, s = (6.7 × 10−20)1/4 = 1.6 × 10−5 mol/L = molar solubility 1.6 × 10 −5 mol Ag 3 PO 4 418.7 g Ag 3 PO 4 × = 6.7 × 10−3 g/L L mol Ag 3 PO 4 b. CaCO3(s) Initial Equil. ⇌ s = solubility (mol/L) Ca2+(aq) 0 s + CO32−(aq) 0 s Ksp = 8.7 × 10−9 = [Ca2+][CO32−] = s2, s = 9.3 × 10−5 mol/L 9.3 × 10 −5 mol 100.1 g × = 9.3 × 10−3 g/L L mol 300 CHAPTER 8 c. ⇌ Hg2Cl2(s) Initial Equil. APPLICATIONS OF AQUEOUS EQUILIBRIA Hg22+(aq) s = solubility (mol/L) 2 Cl−(aq) + 0 s 0 2s Ksp = 1.1 × 10−18 = [Hg22+][Cl−]2 = (s)(2s)2 = 4s3, s = 6.5 × 10−7 mol/L 6.5 × 10 −7 mol 472.1 g × = 3.1 × 10−4 g/L L mol 82. a. PbI2(s) Initial Equil. ⇌ Pb2+(aq) s = solubility (mol/L) 2 I−(aq) + 0 s 0 2s Ksp = 1.4 × 10−8 = [Pb2+][I−]2 = s(2s)2 = 4s3 s = (1.4 × 10−8/4)1/3 = 1.5 × 10−3 mol/L = molar solubility b. CdCO3(s) Initial Equil. ⇌ s = solubility (mol/L) Cd2+(aq) + 0 s CO32−(aq) 0 s Ksp = 5.2 × 10−12 = [Cd2+][CO32−] = s2, s = 2.3 × 10−6 mol/L c. Sr3(PO4)2(s) Initial Equil. ⇌ s = solubility (mol/L) 3 Sr2+(aq) + 2 PO43−(aq) 0 3s 0 2s Ksp = 1 × 10−31 = [Sr2+]3[PO43−]2 = (3s)3(2s)2 = 108s5, s = 2 × 10−7 mol/L 83. In our setup, s = solubility of the ionic solid in mol/L. This is defined as the maximum amount of a salt that can dissolve. Because solids do not appear in the Ksp expression, we do not need to worry about their initial and equilibrium amounts. a. CaC2O4(s) Initial Change Equil. ⇌ Ca2+(aq) + C2O42−(aq) 0 0 s mol/L of CaC2O4(s) dissolves to reach equilibrium !s → +s +s s s From the problem, s = 6.1 × 10 −3 g 1 mol CaC 2 O 4 × = 4.8 × 10 −5 mol/L. L 128.10 g Ksp = [Ca2+][C2O42−] = (s)(s) = s2, Ksp = (4.8 × 10 −5 )2 = 2.3 × 10 −9 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA b. ⇌ BiI3(s) Initial Change Equil. Bi3+(aq) 301 3 I-(aq) + 0 0 s mol/L of BiI3(s) dissolves to reach equilibrium !s → +s +3s s 3s Ksp = [Bi3+][I−]3 = (s)(3s)3 = 27s4, Ksp = 27(1.32 × 10 −5 )4 = 8.20 × 10 −19 PbBr2(s) 84. Initial Change Equil. ⇌ Pb2+(aq) + 2 Br-(aq) 0 0 s mol/L of PbBr2(s) dissolves to reach equilibrium !s → +s +2s s 2s From the problem, s = [Pb2+] = 2.14 × 10 −2 M. So: Ksp = [Pb2+][Br−]2 = s(2s)2 = 4s3, Ksp = 4(2.14 × 10 −2 )3 = 3.92 × 10 −5 ⇌ Ag2C2O4(s) 85. Initial Equil. s = solubility (mol/L) 2 Ag+(aq) 0 2s + C2O42−(aq) 0 s From problem, [Ag+] = 2s = 2.2 × 10 −4 M, s = 1.1 × 10 −4 M Ksp = [Ag+]2[C2O42−] = (2s)2(s) = 4s3 = 4(1.1 × 10 −4 )3 = 5.3 × 10 −12 M2X3(s) 86. Initial Change Equil. ⇌ 2 M3+(aq) + 3 X2−(aq) Ksp = [M3+]2[X2−]3 s = solubility (mol/L) 0 0 s mol/L of M2X3(s) dissolves to reach equilibrium !s +2s +3s 2s 3s Ksp = (2s)2(3s)3 = 108s5 ; s = 3.60 × 10 −7 g 1 mol M 2 X 3 × = 1.25 × 10 −9 mol/L L 288 g Ksp = 108(1.25 × 10 −9 )5 = 3.30 × 10 −43 87. a. Because both solids dissolve to produce three ions in solution, we can compare values of Ksp to determine relative solubility. Because the Ksp for CaF2 is the smallest, CaF2(s) has the smallest molar solubility. 302 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA b. We must calculate molar solubilities because each salt yields a different number of ions when it dissolves. Ca3(PO4)2(s) Initial Equil. ⇌ s = solubility (mol/L) 3 Ca2+(aq) + 2 PO43−(aq) 0 3s Ksp = 1.3 × 10−32 0 2s Ksp = [Ca2+]3[PO43−]2 = (3s)3(2s)2 = 108s5, s = (1.3 × 10−32/108)1/5 = 1.6 × 10−7 mol/L ⇌ FePO4(s) Initial Equil. s = solubility (mol/L) Fe3+(aq) + 0 s PO43−(aq) Ksp = 1.0 × 10−22 0 s Ksp = [Fe3+][PO43−] = s2, s = 1.0 × 10 −22 = 1.0 × 10−11 mol/L FePO4 has the smallest molar solubility. 88. Ag2SO4(s) a. Initial Equil. ⇌ 2 Ag+(aq) + s = solubility (mol/L) 0 2s SO42−(aq) 0 s Ksp = 1.2 × 10−5 = [Ag+]2[SO42−] = (2s)2s = 4s3, s = 1.4 × 10−2 mol/L b. ⇌ Ag2SO4(s) Initial Equil. 2 Ag+(aq) + s = solubility (mol/L) 0.10 M 0.10 + 2s SO42−(aq) 0 s Ksp = 1.2 × 10−5 = (0.10 + 2s)2(s) ≈ (0.10)2(s), s = 1.2 × 10−3 mol/L; assumption good. ⇌ Ag2SO4(s) c. Initial Equil 2 Ag+(aq) + s = solubility (mol/L) SO42−(aq) 0.20 M 0.20 + s 0 2s 1.2 × 10−5 = (2s)2(0.20 + s) ≈ 4s2(0.20), s = 3.9 × 10−3 mol/L; assumption good. Note: Comparing the solubilities in parts b and c to part a illustrates that the solubility of a salt decreases when a common ion is present. 89. Fe(OH)3(s) a. ⇌ Fe3+ + 3 OH− pH = 7.0, [OH−] = 1 × 10−7 M 0 1 × 10−7 M s mol/L of Fe(OH)3(s) dissolves to reach equilibrium = molar solubility → +s +3s Change −s s 1 × 10−7 + 3s Equil. Initial CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 303 Ksp = 4 × 10−38 = [Fe3+][OH−]3 = (s)(1 × 10−7 + 3s)3 ≈ s(1 × 10−7)3 s = 4 × 10−17 mol/L; assumption good (3s << 1 × 10−7). Fe(OH)3(s) b. Initial Change Equil. ⇌ Fe3+ 3 OH− + pH = 5.0, OH− = 1 × 10−9 M 0 1 × 10−9 M (Buffered) s mol/L dissolves to reach equilibrium −s → +s ⎯ (Assume no pH change in buffer) s 1 × 10−9 Ksp = 4 × 10−38 = [Fe3+][OH−]3 = (s)(1 × 10−9)3, s = 4 × 10−11 mol/L = molar solubility c. ⇌ Fe(OH)3(s) Initial Change Equil. Fe3+ + 3 OH− pH = 11.0, [OH− ] = 1 × 10−3 M 0 0.001 M s mol/L dissolves to reach equilibrium −s → +s − s 0.001 (Buffered) (Assume no pH change) Ksp = 4 × 10−38 = [Fe3+][OH− ]3 = (s)(0.001)3, s = 4 × 10−29 mol/L = molar solubility Note: As [OH− ] increases, solubility decreases. This is the common ion effect. Ce(IO3)3(s) 90. Initial Equil. ⇌ s = solubility (mol/L) Ce3+(aq) + 3 IO3−(aq) 0.20 M 0.20 + 3s 0 s Ksp = [Ce3+][IO3−]3 = s(0.20 + 3s)3 From the problem, s = 4.4 × 10−8 mol/L; solving for Ksp: Ksp = (4.4 × 10−8) × [0.20 + 3(4.4 × 10−8)]3 = 3.5 × 10−10 91. ZnS(s) Initial Equil. ⇌ s = solubility (mol/L) Zn2+ + 0.050 M 0.050 + s S2− Ksp = [Zn2+][S2−] 0 s Ksp = 2.5 × 10 −22 = (0.050 + s)(s) ≈ (0.050)s, s = 5.0 × 10 −21 mol/L; assumption good. Mass ZnS that dissolves = 0.3000 L × 5.0 × 10 −21 mol ZnS 97.45 g ZnS × = 1.5 × 10 −19 g L mol 304 92. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA For 99% of the Mg2+ to be removed, we need, at equilibrium, [Mg2+] = 0.01(0.052 M). Using the Ksp equilibrium constant, calculate the [OH−] required to reach this reduced [Mg2+]. Mg(OH)2(s) ⇌ Mg2+(aq) + 2 OH−(aq) Ksp = 8.9 × 10 −12 8.9 × 10 −12 = [Mg2+][OH−]2 = [0.01(0.052 M)] [OH−]2, [OH−] = 1.3 × 10 −4 M (extra sig. fig.) pOH = !log(1.3 × 10 −4 ) = 3.89; pH = 10.11; at a pH = 10.1, 99% of the Mg2+ in seawater will be removed as Mg(OH)2(s). 93. If the anion in the salt can act as a base in water, then the solubility of the salt will increase as the solution becomes more acidic. Added H+ will react with the base, forming the conjugate acid. As the basic anion is removed, more of the salt will dissolve to replenish the basic anion. The salts with basic anions are Ag3PO4, CaCO3, CdCO3 and Sr3(PO4)2. Hg2Cl2 and PbI2 do not have any pH dependence because Cl− and I− are terrible bases (the conjugate bases of a strong acids). + + excess H 2− Ag3PO4(s) + H (aq) → 3 Ag (aq) + HPO4 (aq) + 2+ + 2+ CaCO3(s) + H → Ca + HCO3 − excess H excess H 3 Ag+(aq) + H3PO4(aq) + Ca2+ + H2CO3 [H2O(l) + CO2(g)] + Cd2+ + H2CO3 [H2O(l) + CO2(g)] CdCO3(s) + H → Cd + HCO3− Sr3(PO4)2(s) + 2 H+ → 3 Sr2+ + 2 HPO42− 94. + excess H + 3 Sr2+ + 2 H3PO4 The relevant reactions are: 2 Ag+(aq) + CrO42−(aq) ⇌ Ag2CrO4(s); (red) Ag+(aq) + Cl−(aq) ⇌ AgCl(s) (white) In the first reaction, solid Ag2CrO4 is formed, leaving some Ag+(aq) and CrO42−(aq) in solution. As Cl−(aq) is added, the AgCl solid forms, lowering the concentration of Ag+(aq). Thus equilibrium is shifted to the left (first reaction above), and the red solid disappears. As more CrO42−(aq) goes into solution, the solution turns yellow. This means that Ag2CrO4(s) is more soluble than AgCl(s), which we can verify using Ksp values. The yellow color is due to the presence of CrO42−(aq). Ag2CrO4(s) Initial Change Equil. ⇌ 2 Ag+(aq) + CrO42−(aq) 0 +2s 2s 0 +s s (2s)2(s) = Ksp = 9.0 × 10−12, s = 1.3 × 10−4 M CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA AgCl(s) ⇌ 305 Ag+(aq) + Cl−(aq) Initial Change Equil. 0 +s s 0 +s s s2 = Ksp = 1.6 × 10−10, s = 1.3 × 10−5 M Thus Ag2CrO4(s) is more soluble than AgCl(s). 95. a. AgF b. Pb(OH)2 c. Sr(NO2)2 d. Ni(CN)2 All these salts have anions that are bases. The anions of the other choices are conjugate bases of strong acids. They have no basic properties in water and, therefore, do not have solubilities that depend on pH. 96. 0.020 mmol mL = 7.5 × 10−3 M 200. mL 75.0 mL × 2+ [Ba ]0 = [SO42−]0 = 0.040 mmol mL = 2.5 × 10−2 M 200. mL 125 mL × Q = [Ba2+]0[SO42−]0 = (7.5 × 10−3)(2.5 × 10−2) = 1.9 × 10−4 > Ksp (1.5 × 10−9) A precipitate of BaSO4(s) will form. BaSO4(s) ⇌ Ba2+ + SO42− 0.025 M 0.0075 M Let 0.0075 mol/L Ba2+ react with SO42− to completion because Ksp << 1. Change ← −0.0075 −0.0075 Reacts completely After 0 0.0175 New initial (carry extra sig. fig.) s mol/L BaSO4 dissolves to reach equilibrium Change −s → +s +s s 0.0175 + s Equil. Before Ksp = 1.5 × 10−9 = [Ba2+][SO42−] = (s)(0.0175 + s) ≈ s(0.0175) s = 8.6 × 10−8 mol/L; [Ba2+] = 8.6 × 10−8 M; [SO42−] = 0.018 M; assumption good. 97. [BaBr2]0 = 0.150 L(1.0 × 10 −4 mol / L) = 6.0 × 10−5 M 0.250 L 306 CHAPTER 8 [K2C2O4]0 = APPLICATIONS OF AQUEOUS EQUILIBRIA 0.100 L(6.0 × 10 −4 mol / L) = 2.4 × 10−4 M 0.250 L Q = [Ba2+]0[C2O42−]0 = (6.0 × 10−5)(2.4 × 10−4) = 1.5 × 10−8 M Because Q < Ksp, BaC2O4(s) will not precipitate. The final concentration of ions will be: [Ba2+] = 6.0 × 10−5 M, [Br-] = 1.2 × 10−4 M [K+] = 4.8 × 10−4 M, [C2O42−] = 2.4 × 10−4 M 98. 50.0 mL × 0.10 M = 5.0 mmol Pb2+; 50.0 mL × 1.0 M = 50. mmol Cl−. For this solution, Q > Ksp, so PbCl2 precipitates. Assume precipitation of PbCl2(s) is complete. 5.0 mmol Pb2+ requires 10. mmol of Cl− for complete precipitation, which leaves 40. mmol Cl− in excess. Now let some of the PbCl2(s) re-dissolve to establish equilibrium ⇌ PbCl2(s) Pb2+(aq) + 2 Cl−(aq) Initial 0 40.0 mmol/100.0 mL s mol/L of PbCl2(s) dissolves to reach equilibrium Equil. s 0.40 + 2s Ksp = [Pb2+][Cl−]2, 1.6 × 10−5 = s(0.40 + 2s)2 ≈ s(0.40)2 s = 1.0 × 10−4 mol/L; assumption good. At equilibrium: [Pb2+] = s = 1.0 × 10−4 mol/L and [Cl−] = 0.40 + 2s, 0.40 + 2(1.0 × 10−4) = 0.40 M 99. Ag3PO4(s) ⇌ 3 Ag+(aq) + PO43−(aq); when Q is greater than Ksp, precipitation will occur. We will calculate the [Ag+]0 necessary for Q = Ksp. Any [Ag+]0 greater than this calculated number will cause precipitation of Ag3PO4(s). In this problem, [PO43−]0 = [Na3PO4]o = 1.0 × 10−5 M. Ksp = 1.8 × 10−18; Q = 1.8 × 10−18 = [Ag+]03[PO43−]0 = [Ag+]03(1.0 × 10−5 M) 1/ 3 ⎛ 1.8 × 10 −18 ⎞ ⎟ [Ag+]0 = ⎜⎜ −5 ⎟ ⎝ 1.0 × 10 ⎠ , [Ag+]0 = 5.6 × 10−5 M When [Ag+]0 = [AgNO3]0 is greater than 5.6 × 10−5 M, Ag3PO4(s) will precipitate. 100. From Table 8.5, Ksp for NiCO3 = 1.4 × 10−7 and Ksp for CuCO3 = 2.5 × 10−10. From the Ksp values, CuCO3 will precipitate first because it has the smaller Ksp value and will be least soluble. For CuCO3(s), precipitation begins when: CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA [CO32−] = K sp , CuCO3 [Cu 2 + ] = 307 2.5 × 10 −10 = 1.0 × 10−9 M CO32− 0.25 M For NiCO3(s) to precipitate: [CO32−] = K sp , NiCO3 [ Ni 2 + ] = 1.4 × 10 −7 = 5.6 × 10−7 M CO32− 0.25 M Determining the [Cu2+] when NiCO3(s) begins to precipitate: [Cu2+] = K sp , CuCO 3 = 2− [CO 3 ] 2.5 × 10 −10 = 4.5 × 10−4 M Cu2+ 5.6 × 10 −7 M For successful separation, 1% Cu2+ or less of the initial amount of Cu2+ (0.25 M) must be present before NiCO3(s) begins to precipitate. The percent of Cu2+ present when NiCO3(s) begins to precipitate is: 4.5 × 10 −4 M × 100 = 0.18% Cu2+ 0.25 M Because less than 1% Cu2+ remains of the initial amount, the metals can be separated through slow addition of Na2CO3(aq). 101. + a. 2+ 2+ Ag , Mg , Cu NaCl(aq) AgCl(s) 2+ Mg , Cu2 + - NH3(aq) - contains OH Mg(OH) 2(s) 2+ Cu(NH3) 4 (aq) H2S(aq) CuS(s) 308 CHAPTER 8 2+ b. 2+ Pb , C a , Fe APPLICATIONS OF AQUEOUS EQUILIBRIA 2+ NaCl(aq) 2+ Ca , F e PbCl2 (s) 2+ Na2SO 4 (aq) or H 2SO 4(aq) CaSO 4(s) Fe 2+ H2 S(aq) - make basic FeS(s) - - - Cl , Br , I c. AgNO3(aq) AgCl(s), AgBr(s), AgI(s) NH3(aq) + AgBr(s) + AgI(s) Ag(NH3)2 (aq) + - Cl (aq) Na2S2O3 AgI(s) d. - Ag(S2O3)23 (aq) + + - Br (aq) + Pb2 , Bi3 Na2SO4(aq) or H2SO4(aq) PbSO4(s) + Bi3 H2S(aq) - make basic Bi2S3(s) CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 309 Complex Ion Equilibria 102. K1 = 7.9 × 103 Mn2+ + C2O42− ⇌ MnC2O4 2− 2MnC2O4 + C2O4 ⇌ Mn(C2O4)2 K2 = 7.9 × 101 ____________________________________________________________ Mn2+(aq) + 2 C2O42−(aq) ⇌ Mn(C2O4)22−(aq) Kf = K1K2 = 6.2 × 105 103. Hg2+(aq) + 2 I−(aq) → HgI2(s), orange ppt; HgI2(s) + 2 I−(aq) → HgI42−(aq) Soluble complex ion 104. Ag+(aq) + Cl−(aq) ⇌ AgCl(s), white ppt.; AgCl(s) + 2 NH3(aq) ⇌ Ag(NH3)2+(aq) + Cl−(aq) Ag(NH3)2+(aq) + Br−(aq) ⇌ AgBr(s) + 2 NH3(aq), pale yellow ppt. = AgBr(s) AgBr(s) + 2 S2O32−(aq) ⇌ Ag(S2O3)23−(aq) + Br−(aq) Ag(S2O3)23−(aq) + I−(aq) ⇌ AgI(s) + 2 S2O32−(aq), yellow ppt. = AgI(s) The least soluble salt (smallest Ksp value) must be AgI because it forms in the presence of Cl−and Br−. The most soluble salt (largest Ksp value) must be AgCl because it forms initially but never re-forms. The order of Ksp values is Ksp (AgCl) > Ksp (AgBr) > Ksp (AgI). The order of formation constants is Kf [Ag(S2O3)23−] > Kf [Ag(NH3)2+] because addition of S2O32− causes the AgBr(s) precipitate to dissolve, but the presence of NH3 was unable to prevent AgBr(s) from forming. This assumes concentrations are about equal. 105. 65 g KI 1 mol KI × = 0.78 M KI 0.500 L 166.0 g KI The formation constant for HgI42− is an extremely large number. Because of this, we will let the Hg2+ and I− ions present initially react to completion and then solve an equilibrium problem to determine the Hg2+ concentration. Hg2+ Before Change After Change Equil. + ⇌ 4 I− HgI42− K = 1.0 × 1030 0.010 M 0.78 M 0 −0.010 −0.040 → +0.010 Reacts completely (K large) 0 0.74 0.010 New initial x mol/L HgI42− dissociates to reach equilibrium +x +4x ← −x x 0.74 + 4x 0.010 − x 2− K = 1.0 × 1030 = 1.0 × 1030 ≈ [HgI 4 ] 2+ − 4 [Hg ][I ] = (0.010 − x) ; making usual assumptions: ( x)(0.74 + 4 x) 4 (0.010) , x = [Hg2+] = 3.3 × 10−32 M ; assumptions good. 4 ( x)(0.74) 310 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA Note: 3.3 × 10−32 mol/L corresponds to one Hg2+ ion per 5 × 107 L. It is very reasonable to approach the equilibrium in two steps. The reaction does essentially go to completion. Fe3+ 106. Initial Change Equil. ⇌ 6 CN− + Fe(CN)63− K = 1 × 1042 0 2.0 M 0.090 mol/0.60 L = 0.15 M 3− x mol/L Fe(CN)6 dissociates to reach equilibrium +6x ← −x +x x 2.0 + 6x 0.15 − x K = 1 × 1042 = 3− [Fe(CN ) 6 ] (0.15 − x) 0.15 = , 1 × 1042 ≈ 6 3+ − 6 ( x)(2.0 + 6 x) x(2.0) 6 [Fe ][CN ] x = [Fe3+] = 2 × 10−45 M; [Fe(CN)63−] = 0.15 M − x = 0.15 M; assumptions good. 107. Ag+ + NH3 ⇌ AgNH3+ K1 = 2.1 × 103 + + AgNH3 + NH3 ⇌ Ag(NH3)2 K2 = 8.2 × 103 __________________________________________________ K = K1K2 = 1.7 × 107 Ag+ + 2 NH3 ⇌ Ag(NH3)2+ The initial concentrations are halved because equal volumes of the two solutions are mixed. Let the reaction go to completion since K is large; then solve an equilibrium problem. Ag+ Before After Equil. + 2 NH3 0.20 M 0 x K = 1.7 × 107 = 2.0 M 1.6 1.6 + 2x ⇌ Ag(NH3)2+ 0 0.20 0.20 ! x + [Ag( NH 3 ) 2 ] 0.20 − x 0.20 = ≈ , x = 4.6 × 10 −9 M; + 2 2 [Ag ][ NH 3 ] x (1.6 + 2 x) x (1.6) 2 assumptions good. [Ag+] = x = 4.6 × 10 −9 M; [NH3] = 1.6 M; [Ag(NH3)2+] = 0.20 M Use either the K1 or K2 equilibrium expression to calculate [AgNH3+]. AgNH3+ + NH3 8.2 × 103 = ⇌ Ag(NH3)2+ + [ Ag( NH 3 ) 2 ] + [AgNH 3 ][ NH 3 ] = K2 = 8.2 × 103 0.20 + [AgNH 3 ](1.6) , [AgNH3+] = 1.5 × 10 −5 M CHAPTER 8 108. APPLICATIONS OF AQUEOUS EQUILIBRIA 311 AgCl(s) ⇌ Ag+ + Cl− Ksp = 1.6 × 10 −10 Ag+ + 2 NH3 ⇌ Ag(NH3)2+ Kf = 1.7 × 107 ________________________________________________________________ K = Ksp × Kf = 2.7 × 10 −3 AgCl(s) + 2 NH3 ⇌ Ag(NH3)2+(aq) + Cl−(aq) ⇌ AgCl(s) + 2NH3 Ag(NH3)2+ + Cl− 1.0 M 0 0 s mol/L of AgCl(s) dissolves to reach equilibrium = molar solubility s s 1.0 – 2s Initial Equil. K = 2.7 × 10 −3 + [Ag( NH 3 ) 2 ][Cl − ] s2 = = ; taking the square root: [ NH 3 ]2 (1.0 − 2s ) 2 s = (2.7 × 10 −3 )1/2 = 5.2 × 10 −2 , s = 4.7 × 10 −2 mol/L 1 .0 − 2 s In pure water, the solubility of AgCl(s) is (1.6 × 10 −10 )1/2 = 1.3 × 10 −5 mol/L. Notice how the presence of NH3 increases the solubility of AgCl(s) by over a factor of 3500. 109. a. Cu(OH)2 ⇌ Cu2+ + 2 OH− Ksp = 1.6 × 10−19 Cu2+ + 4 NH3 ⇌ Cu(NH3)42+ Kf = 1.0 × 1013 ____________________________________________________________________ K = KspKf = 1.6 × 10−6 Cu(OH)2(s) + 4 NH3(aq) ⇌ Cu(NH3)42+(aq) + 2 OH−(aq) b. Cu(OH)2(s) + 4 NH3 ⇌ Cu(NH3)42+ + 2 OH− K = 1.6 × 10-6 5.0 M 0 0.0095 M s mol/L Cu(OH)2 dissolves to reach equilibrium Equil. 5.0 − 4s s 0.0095 + 2s Initial K = 1.6 × 10−6 = 2+ [Cu ( NH 3 ) 4 ][OH − ]2 s (0.0095 + 2s) 2 = [ NH 3 ]4 (5.0 − 4s) 4 If s is small: 1.6 × 10−6 = s (0.0095) 2 , s = 11. mol/L (5.0 ) 4 Assumptions are not good. We will solve the problem by successive approximations. scalc = 1.6 × 10 −6 (5.0 − 4 sguess ) 4 (0.0095 + 2 sguess ) 2 ; the results from six trials are: 312 CHAPTER 8 sguess: scalc: APPLICATIONS OF AQUEOUS EQUILIBRIA 0.10, 0.050, 0.060, 0.055, 0.056 1.6 × 10−2, 0.071, 0.049, 0.058, 0.056 Thus the solubility of Cu(OH)2 is 0.056 mol/L in 5.0 M NH3. 110. a. CuCl(s) Initial Equil. ⇌ Cu+ s = solubility (mol/L) + 0 s Cl− 0 s Ksp = 1.2 × 10−6 = [Cu+][Cl−] = s2, s = 1.1 × 10−3 mol/L b. Cu+ forms the complex ion CuCl2- in the presence of Cl-. We will consider both the Ksp reaction and the complex ion reaction at the same time. CuCl(s) ⇌ Cu+(aq) + Cl−(aq) Ksp = 1.2 × 10−6 + (aq) + 2 Cl−(aq) ⇌ CuCl2−(aq) Kf = 8.7 × 104 Cu ___________________________________________________________________________________________ CuCl(s) + Cl−(aq) ⇌ CuCl2−(aq) CuCl(s) + Cl− Initial Equil. ⇌ K = Ksp × Kf = 0.10 CuCl2− 0.10 M 0.10 − s − K = 0.10 = [CuCl 2 ] − [Cl ] = 0 s where s = solubility of CuCl(s) in mol/L s , 1.0 × 10−2 − (0.10)s = s 0.10 − s (1.10)s = 1.0 × 10−2, s = 9.1 × 10−3 mol/L 111. AgBr(s) ⇌ Ag+ + Br− Ksp = 5.0 × 10−13 Kf = 2.9 × 1013 Ag+ + 2 S2O32- ⇌ Ag(S2O3)23− _____________________________________________________ AgBr(s) + 2 S2O32− ⇌ Ag(S2O3)23− + Br− K = Ksp × Kf = 14.5 AgBr(s) Initial Change Equil. K= + 2 S2O32− ⇌ Ag(S2O3)23− + 0.500 M 0 s mol/L AgBr(s) dissolves to reach equilibrium −s −2s → +s 0.500 − 2s s s2 = 14.5; taking the square root of both sides: (0.500 − 2s ) 2 Br− 0 +s s (Carry extra sig. figs.) CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 313 s = 3.81, s = 1.91 − (7.62)s, s = 0.222 mol/L 0.500 − 2 s 1.00 L × 112. 0.222 mol AgBr 187.8 g AgBr × = 41.7 g AgBr = 42 g AgBr L mol AgBr a. AgI(s) ⇌ Ag+(aq) + Initial s = solubility (mol/L) Equil. 0 s I−(aq) Ksp = [Ag+][I−] = 1.5 × 10−16 0 s Ksp = 1.5 × 10−16 = s2, s = 1.2 × 10−8 mol/L b. AgI(s) Ag+ + 2 NH3 AgI(s) + 2 NH3(aq) AgI(s) Initial Equil. ⇌ ⇌ Ag+ + I− Ag(NH3)2+ Ksp = 1.5 × 10−16 Kf = 1.7 × 107 ⇌ Ag(NH3)2+(aq) + I−(aq) K = Ksp × Kf = 2.6 × 10−9 + 2 NH3 ⇌ Ag(NH3)2+ + I− 3.0 M 0 0 s mol/L of AgI(s) dissolves to reach equilibrium = molar solubility 3.0 − 2s s s + K= [Ag( NH 3 ) 2 ][I − ] s2 s2 −9 = ≈ , 2.6 × 10 , s = 1.5 × 10−4 mol/L 2 2 2 [ NH 3 ] (3.0 − 2 s) (3.0) Assumption good. c. The presence of NH3 increases the solubility of AgI. Added NH3 removes Ag+ from solution by forming the complex ion, Ag(NH3)2+. As Ag+ is removed, more AgI(s) will dissolve to replenish the Ag+ concentration. 113. Test tube 1: Added Cl− reacts with Ag+ to form a silver chloride precipitate. The net ionic equation is Ag+(aq) + Cl-(aq) → AgCl(s). Test tube 2: Added NH3 reacts with Ag+ ions to form a soluble complex ion, Ag(NH3)2+. As this complex ion forms, Ag+ is removed from the solution, which causes the AgCl(s) to dissolve. When enough NH3 is added, all the silver chloride precipitate will dissolve. The equation is AgCl(s) + 2 NH3(aq) → Ag(NH3)2+(aq) + Cl−(aq). Test tube 3: Added H+ reacts with the weak base, NH3, to form NH4+. As NH3 is removed from the Ag(NH3)2+ complex ion, Ag+ ions are released to solution and can then react with Cl− to re-form AgCl(s). The equations are Ag(NH3)2+(aq) + 2 H+(aq) → Ag+(aq) + 2 NH4+(aq), and Ag+(aq) + Cl−(aq) → AgCl(s). 314 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA Additional Exercises 114. 1.0 mL × 1.0 mmol = 1.0 mmol Cd2+ added to the ammonia solution mL Thus [Cd2+]0 = 1.0 × 10−3 mol/L. We will first calculate the equilibrium Cd2+ concentration using the complex ion equilibrium, and then determine if this Cd2+ concentration is large enough to cause precipitation of Cd(OH)2(s). Cd2+ Before Change After Change Equil. + 4 NH3 ⇌ Cd(NH3)42+ Kf = 1.0 × 107 1.0 × 10−3 M 5.0 M 0 −1.0 × 10−3 −4.0 × 10−3 → +1.0 × 10−3 Reacts completely 0 4.996 ≈ 5.0 1.0 × 10−3 New initial x mol/L Cd(NH3)42+ dissociates to reach equilibrium +x +4x ← −x x 5.0 + 4x 0.0010 − x Kf = 1.0 × 107 = (0.010 − x) (0.010) ≈ 4 ( x)(5.0 + 4 x) ( x)(5.0) 4 x = [Cd2+] = 1.6 × 10−13 M; assumptions good. This is the maximum [Cd2+] possible. Now we will determine if Cd(OH)2(s) forms at this concentration of Cd2+. In 5.0 M NH3 we can calculate the pH: ⇌ NH3 + H2O Initial 5.0 M Equil. 5.0 − y Kb = 1.8 × 10−5 = NH4+ + OH− 0 y Kb = 1.8 × 10−5 ~0 y + y2 y2 [ NH 4 ][OH − ] ≈ , y = [OH−] = 9.5 × 10−3 M; assumptions = [ NH 3 ] 5.0 − y 5 .0 good. We now calculate the value of the solubility quotient, Q: Q = [Cd2+][OH−]2 = (1.6 × 10−13)(9.5 × 10−3)2 Q = 1.4 × 10−17 < Ksp (5.9 × 10−15); 115. therefore, no precipitate forms. a. The optimum pH for a buffer is when pH = pKa. At this pH a buffer will have equal neutralization capacity for both added acid and base. As shown next, because the pKa for TRISH+ is 8.1, the optimal buffer pH is about 8.1. Kb = 1.19 × 10−6; Ka = Kw/Kb = 8.40 × 10−9; pKa = −log(8.40 × 10−9) = 8.076 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA b. pH = pKa + log 315 [TRIS] [TRIS] , 7.00 = 8.076 + log + [TRISH ] [TRISH + ] [TRIS] = 10-1.08 = 0.083 (at pH = 7.00) + [TRISH ] 9.00 = 8.076 + log c. [TRIS] [TRIS] , = 100.92 = 8.3 (at pH = 9.00) + [TRISH ] [TRISH + ] 50.0 g TRIS 1 mol × = 0.206 M = 0.21 M = [TRIS] 2.0 L 121.14 g 65.0 g TRISHCl 1 mol × = 0.206 M = 0.21 M = [TRISHCl] = [TRISH+] 2 .0 L 157.60 g pH = pKa + log [TRIS] (0.21) = 8.076 + log = 8.08 + (0.21) [TRISH ] The amount of H+ added from HCl is: (0.50 × 10−3 L) × 12 mol/L = 6.0 × 10−3 mol H+ The H+ from HCl will convert TRIS into TRISH+. The reaction is: TRIS Before Change After 0.21 M −0.030 0.18 H+ + → 6.0 × 10 −3 = 0.030 M 0.2005 −0.030 → 0 TRISH+ 0.21 M +0.030 0.24 Reacts completely Now use the Henderson-Hasselbalch equation to solve this buffer problem. ⎛ 0.18 ⎞ pH = 8.076 + log ⎜ ⎟ = 7.95 ⎝ 0.24 ⎠ 116. i. This is the result when you have a salt that breaks up into two ions. Examples of these salts (but not all) include AgCl, SrSO4, BaCrO4, and ZnCO3. ii. This is the result when you have a salt that breaks up into three ions, either two cations and one anion or one cation and two anions. Some examples are SrF2, Hg2I2, and Ag2SO4. iii. This is the result when you have a salt that breaks up into four ions, either three cations and one anion (Ag3PO4) or one cation and three anions (ignoring the hydroxides, there are no examples of this type of salt in Table 8.5). iv. This is the result when you have a salt that breaks up into five ions, either three cations and two anions [Sr3(PO4)2] or two cations and three anions (no examples of this type of salt are in Table 8.5). 316 117. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA A best buffer is when pH ≈ pKa; these solutions have about equal concentrations of weak acid and conjugate base. Therefore, choose combinations that yield a buffer where pH ≈ pKa; that is, look for acids whose pKa is closest to the pH. a. Potassium fluoride + HCl will yield a buffer consisting of HF (pKa = 3.14) and F−. b. Benzoic acid + NaOH will yield a buffer consisting of benzoic acid (pKa = 4.19) and benzoate anion. c. Sodium acetate + acetic acid (pKa = 4.74) is the best choice for pH = 5.0 buffer since acetic acid has a pKa value closest to 5.0. d. HOCl and NaOH: This is the best choice to produce a conjugate acid-base pair with pH = 7.0. This mixture would yield a buffer consisting of HOCl (pKa = 7.46) and OCl−. Actually, the best choice for a pH = 7.0 buffer is an equimolar mixture of ammonium chloride and sodium acetate. NH4+ is a weak acid (Ka = 5.6 × 10−10), and C2H3O2− is a weak base (Kb = 5.6 × 10−10). A mixture of the two will give a buffer at pH = 7.0 because the weak acid and weak base are the same strengths (Ka for NH4+ = Kb for C2H3O2−). NH4C2H3O2 is commercially available, and its solutions are used for pH = 7.0 buffers. e. Ammonium chloride + NaOH will yield a buffer consisting of NH4+ (pKa = 9.26) and NH3. 118. a. pH = pKa = −log(6.4 × 10−5) = 4.19 since [HBz] = [Bz-], where HBz = C6H5CO2H and [Bz-] = C6H5CO2−. b. [Bz−] will increase to 0.120 M and [HBz] will decrease to 0.080 M after OH− reacts completely with HBz. The Henderson-Hasselbalch equation is derived from the Ka dissociation reaction. pH = pKa + log (0.120) [Bz − ] , pH = 4.19 + log = 4.37; assumptions good. (0.080) [HBz] Bz− c. + H2O Initial 0.120 M Equil. 0.120 − x Kb = ⇌ HBz + 0.080 M 0.080 + x OH− 0 x Kw 1.0 × 10 −14 (0.080 + x)( x) (0.080)( x) = ≈ = −5 Ka (0.120 − x) 0.120 6.4 × 10 x = [OH−] = 2.34 × 10−10 M (carrying extra sig. fig.); assumptions good. pOH = 9.63; pH = 4.37 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 317 d. We get the same answer. Both equilibria involve the two major species, benzoic acid and benzoate anion. Both equilibria must hold true. Kb is related to Ka by Kw and [OH−] is related to [H+] by Kw, so all constants are interrelated. 119. NaOH added = 50.0 mL × 0.500 mmol = 25.0 mmol NaOH mL NaOH left unreacted = 31.92 mL HCl × 0.289 mmol 1 mmol NaOH × = 9.22 mmol NaOH mL mmol HCl NaOH reacted with aspirin = 25.0 − 9.22 = 15.8 mmol NaOH 15.8 mmol NaOH × Purity = 1 mmol aspirin 180.2 mg × = 1420 mg = 1.42 g aspirin 2 mmol NaOH mmol 1.42 g × 100 = 99.5% 1.427 g Here, a strong base is titrated by a strong acid. The equivalence point will be at pH = 7.0. Bromthymol blue would be the best indicator since it changes color at pH ≈ 7 (from base color to acid color). See Fig. 8.8 of the text. 120. At 4.0 mL NaOH added: 2.43 − 3.14 ΔpH = = 0.18 ΔmL 0 − 4 .0 The other points are calculated in a similar fashion. The results are summarized and plotted below. As can be seen from the plot, the advantage of this approach is that it is much easier to accurately determine the location of the equivalence point. 121. mL pH 0 4.0 8.0 12.5 20.0 24.0 24.5 24.9 25.0 25.1 26.0 28.0 30.0 2.43 3.14 3.53 3.86 4.46 5.24 5.6 6.3 8.28 10.3 11.29 11.75 11.96 |ΔpH/ΔmL| − 0.18 0.098 0.073 0.080 0.20 0.7 2 20 20 1 0.23 0.11 At the equivalence point, P2− is the major species. P2− is a weak base in water because it is the conjugate base of a weak acid. 318 CHAPTER 8 P2− ⇌ H2O HP− 0 .5 g 1 mol × = 0.024 M 0.1 L 204.2 g 0.024 − x Initial Equil. Kb = + APPLICATIONS OF AQUEOUS EQUILIBRIA OH− + 0 ~0 x x (carry extra sig. fig.) Kw x2 x2 [HP − ][OH − ] 1.0 × 10 −14 −9 = H 10 = , 3.2 ≈ = Ka 0.024 − x 0.024 P 2− 10 −5.51 x = [OH−] = 8.8 × 10−6 M; pOH = 5.1; pH = 8.9; assumptions good. Phenolphthalein would be the best indicator for this titration because it changes color at pH ≈ 9 (from acid color to base color). Cr3+ 122. Before Change After Change Equil. 1.0 × 1023 ≈ a. CrEDTA− + 2 H+ (Buffer) Reacts completely New initial (Buffer) (0.0010 − x) (1.0 × 10 −6 ) 2 [CrEDTA − ][H + ] 2 = (x)(0.049 + x) [Cr 3+ ][H 2 EDTA 2− ] (0.0010) (1.0 × 10 −12 ) , x = [Cr3+] = 2.0 × 10−37 M; assumptions good. x(0.049) Pb(OH)2(s) Initial Equil. ⇌ 0.0010 M 0.050 M 0 1.0 × 10−6 M −0.0010 −0.0010 → +0.0010 No change 0 0.049 0.0010 1.0 × 10−6 x mol/L CrEDTA− dissociates to reach equilibrium +x +x ← −x − x 0.049 + x 0.0010 − x 1.0 × 10−6 Kf = 1.0 × 1023 = 123. + H2EDTA2− ⇌ s = solubility (mol/L) Pb2+ + 2 OH− 1.0 × 10−7 M from water 1.0 × 10−7 + 2s 0 s Ksp = 1.2 × 10−15 = [Pb2+][OH−]2 = s(1.0 × 10−7 + 2s)2 ≈ s(2s2) = 4s3 s = [Pb2+] = 6.7 × 10−6 M; assumption is good by the 5% rule. b. Pb(OH)2(s) ⇌ Pb2+ + 2 OH− 0 0.10 M pH = 13.00, [OH−] = 0.10 M s mol/L Pb(OH)2(s) dissolves to reach equilibrium Equil. s 0.10 (Buffered solution) Initial 1.2 × 10−15 = (s)(0.10)2, s = [Pb2+] = 1.2 × 10−13 M CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 319 c. We need to calculate the Pb2+ concentration in equilibrium with EDTA4-. Since K is large for the formation of PbEDTA2−, let the reaction go to completion, and then solve an equilibrium problem to get the Pb2+ concentration. Pb2+ Before Change After Equil. + EDTA4− ⇌ PbEDTA2− K = 1.1 × 1018 0.010 M 0.050 M 0 0.010 mol/L Pb2+ reacts completely (large K) −0.010 −0.010 → +0.010 Reacts completely 0 0.040 0.010 New initial x mol/L PbEDTA2− dissociates to reach equilibrium x 0.040 + x 0.010 − x 1.1 × 1018 = (0.010 − x) 0.010 ≈ , x = [Pb2+] = 2.3 × 10−19 M; assumptions good. ( x)(0.040 + x) x(0.040) Now calculate the solubility quotient for Pb(OH)2 to see if precipitation occurs. The concentration of OH− is 0.10 M since we have a solution buffered at pH = 13.00. Q = [Pb2+]0[OH−]02 = (2.3 × 10−19)(0.10)2 = 2.3 × 10−21 < Ksp (1.2 × 10−15) Pb(OH)2(s) will not form since Q is less than Ksp. Ba(OH)2(s) 124. ⇌ Ba2+(aq) + 2 OH−(aq) Ksp = [Ba2+][OH−]2 = 5.0 × 10 −3 Initial s = solubility (mol/L) Equil. 0 s ~0 2s Ksp = 5.0 × 10 −3 = s(2s)2 = 4s3, s = 0.11 mol/L; assumption good. [OH−] = 2s = 2(0.11) = 0.22 mol/L; pOH = 0.66, pH = 13.34 Sr(OH)2(s) ⇌ Equil. Sr2+(aq) + 2 OH−(aq) s Ksp = [Sr2+][OH−]2 = 3.2 × 10 −4 2s Ksp = 3.2 × 10 −4 = 4s3, s = 0.043 mol/L; asssumption good. [OH−] = 2(0.043) = 0.086 M; pOH = 1.07, pH = 12.93 Ca(OH)2(s) Equil. ⇌ Ca2+(aq) + 2 OH−(aq) s Ksp = [Ca2+][OH−]2 = 1.3 × 10 −6 2s Ksp = 1.3 × 10 −6 = 4s3, s = 6.9 × 10 −3 mol/L; assumption good. [OH−] = 2(6.9 × 10 −3 ) = 1.4 × 10 −2 mol/L; pOH = 1.85, pH = 12.15 320 125. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA HC2H3O2 ⇌ H+ + C2H3O2−; let C0 = initial concentration of HC2H3O2 From normal weak acid setup: Ka = 1.8 × 10−5 = [H+] = 10−2.68 = 2.1 × 10−3 M; 1.8 × 10−5 = − [H + ][C 2 H 3O 2 ] [ H + ]2 = [ HC 2 H 3O 2 ] C 0 − [H + ] ( 2.1 × 10 −3 ) 2 , C0 = 0.25 M C 0 − ( 2.1 × 10 −3 ) 25.0 mL × 0.25 mmol/mL = 6.3 mmol HC2H3O2 Need 6.3 mmol KOH = VKOH × 0.0975 mmol/mL, VKOH = 65 mL 126. In the final solution: [H+] = 10−2.15 = 7.1 × 10−3 M Beginning mmol HCl = 500.0 mL × 0.200 mmol/mL = 100. mmol HCl Amount of HCl that reacts with NaOH = 1.50 × 10−2 mmol/mL × V 7.1 × 10 −3 mmol final mmol H + 100. − 0.0150 V = = mL total volume 500.0 + V 3.6 + (7.1 × 10−3)V = 100. − (1.50 × 10−2)V, (2.21 × 10−2)V = 100. − 3.6 V = 4.36 × 103 mL = 4.36 L = 4.4 L NaOH 127. 0.400 mol/L × VNH3 = mol NH3 = mol NH4+ after reaction with HCl at the equivalence point. At the equivalence point: [NH4+]0 = NH4+ Initial Equil. Ka = 0.267 M 0.267 − x ⇌ + 0.400 × VNH3 mol NH 4 = = 0.267 M total volume 1.50 × VNH3 H+ + 0 x NH3 0 x Kw x2 x2 1.0 × 10 −14 −10 ≈ = = , 5.6 × 10 0.267 − x 0.267 Kb 1.8 × 10 −5 x = [H+] = 1.2 × 10−5 M; pH = 4.92; assumptions good. 128. 50.0 mL × 0.100 M = 5.00 mmol NaOH initially At pH = 10.50, pOH = 3.50, [OH−] = 10−3.50 = 3.2 × 10−4 M mmol OH− remaining = 3.2 × 10−4 mmol/mL × 73.75 mL = 2.4 × 10−2 mmol CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 321 mmol OH− that reacted = 5.00 - 0.024 = 4.98 mmol Because the weak acid is monoprotic, 23.75 mL of the weak acid solution contains 4.98 mmol HA. [HA]0 = 129. 4.98 mmol = 0.210 M 23.75 mL HA + OH− → A− + H2O, where HA = acetylsalicylic acid mmol HA present = 27.36 mL OH− × Molar mass of HA = 0.5106 mmol OH − 1 mmol HA × = 13.97 mmol HA mL OH − mmol OH − 2.51 g HA = 180. g/mol 13.97 × 10 −3 mol HA To determine the Ka value, use the pH data. After complete neutralization of acetylsalicylic acid by OH−, we have 13.97 mmol of A- produced from the neutralization reaction. A− will react completely with the added H+ and re-form acetylsalicylic acid HA. mmol H+ added = 15.44 mL × A− Before Change After + 13.97 mmol −6.985 6.985 mmol 0.4524 mmol H + = 6.985 mmol H+ mL H+ → 6.985 mmol −6.985 → 0 HA 0 +6.985 6.985 mmol Reacts completely We have back titrated this solution to the halfway point to equivalence, where pH = pKa (assuming HA is a weak acid). This is true because after H+ reacts completely, equal milliliters of HA and A− are present, which only occurs at the halfway point to equivalence. Assuming acetylsalicylic acid is a weak acid, then pH = pKa = 3.48. Ka = 10−3.48 = 3.3 × 10−4. 130. [X−]0 = 5.00 M and [Cu+]0 = 1.0 × 10−3 M since equal volumes of each reagent are mixed. Because the K values are large, assume that the reaction goes completely to CuX32−; then solve an equilibrium problem. Cu+ Before After Equil. K= 1.0 × 10−3 M 0 x + 3 X− ⇌ 5.00 M 5.00 − 3(10−3) = 5.00 5.00 + 3x CuX32− K = K1K2K3 = 1.0 × 109 0 1.0 × 10−3 1.0 × 10−3 − x 1.0 × 10−3 (1.0 × 10 −3 − x) 9 , 1.0 × 10 ≈ , x = [Cu+] = 8.0 × 10−15 M; x(5.00)3 x(5.00 + 3 x) 3 assumptions good. 322 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA [CuX32−] = 1.0 × 10−3 − 8.0 × 10−15 = 1.0 × 10−3 M 2− K3 = [CuX3 ] − − [CuX2 ][X ] , 1.0 × 103 = (1.0 × 10−3 ) − [CuX2 ](5.00) , [CuX2−] = 2.0 × 10−7 M Summarizing: [CuX32−] = 1.0 × 10−3 M [CuX2−] = 2.0 × 10−7 M [Cu2+] = 8.0 × 10−15 M (answer a) (answer b) (answer c) 131. K a 3 is so small (4.8 × 10−13) that a break is not seen at the third stoichiometric point. 132. We will see only the first stoichiometric point in the titration of salicylic acid because K a 2 is so small. For adipic acid the Ka values are fairly close to each other. Both protons will be titrated almost simultaneously, giving us only one break. The stoichiometric points will occur when 1 mol of H+ is added per mole of salicylic acid present and when 2 mol of H+ is added per mole of adipic acid present. Thus the 25.00-mL volume corresponded to the titration of salicylic acid, and the 50.00-mL volume corresponded to the titration of adipic acid. Challenge Problems 133. mmol HC3H5O2 present initially = 45.0 mL × mmol C3H5O2− present initially = 55.0 mL × 0.750 mmol = 33.8 mmol HC3H5O2 mL 0.700 mmol = 38.5 mmol C3H5O2− mL The initial pH of the buffer is: 38.5 mmol − [C 3 H 5 O 2 ] 38.5 100.0 mL = 4.89 + log = −log(1.3 × 10−5) + log = 4.95 pH = pKa + log 33.8 mmol [HC3 H 5 O 2 ] 33.8 100.0 mL Note: Because the buffer components are in the same volume of solution, we can use the mole (or millimole) ratio in the Henderson-Hasselbalch equation to solve for pH instead of using the concentration ratio of [C3H5O2−]/[HC3H5O2]. The total volume always cancels for buffer solutions. When NaOH is added, the pH will increase, and the added OH− will convert HC3H5O2 into C3H5O2−. The pH after addition of OH− increases by 2.5%, so the resulting pH is: 4.95 + 0.025(4.95) = 5.07 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 323 At this pH, a buffer solution still exists, and the millimole ratio between C3H5O2− and HC3H5O2 is: − pH = pKa + log − mmol C3 H 5 O 2 mmol C3 H 5 O 2 , 5.07 = 4.89 + log mmol HC3 H 5 O 2 mmol HC3 H 5 O 2 − mmol C3 H 5 O 2 = 100.18 = 1.5 mmol HC3 H 5 O 2 Let x = mmol OH− added to increase pH to 5.07. Because OH− will essentially react to completion with HC3H5O2, the setup to the problem using millimoles is: HC3H5O2 Before Change After + 33.8 mmol −x 33.8 − x OH− → C3H5O2− x mmol −x 0 → 38.5 mmol +x Reacts completely 38.5 + x − mmol C3 H 5 O 2 38.5 + x = 1.5 = , 1.5(33.8 − x) = 38.5 + x, x = 4.9 mmol OH− added mmol HC3 H 5 O 2 33.8 − x The volume of NaOH necessary to raise the pH by 2.5% is: 4.9 mmol NaOH × 1 mL = 49 mL 0.10 mmol NaOH 49 mL of 0.10 M NaOH must be added to increase the pH by 2.5%. 134. Major species: PO43−, OH−, H+, 5.00 mmol 5.00 mmol 15.0 mmol CN−, Na+, 7.50 mmol K+, Cl−, H2O PO43− and CN− are weak bases. PO43− + H2O ⇌ HPO42− + OH− Kb = Kw/ K a 3 = 2.1 × 10−2 CN− + H2O ⇌ HCN + OH− Kb = Kw/Ka = 1.6 × 10−5 One of the keys to this problem is to recognize that pK a 2 for H3PO4 = 7.21 [−log(6.2 × 10−8) = 7.21]. The K a 2 reaction for H3PO4 is: H2PO4− ⇌ HPO42− + H+ The pH of the final solution will equal 7.21 when we have a buffer solution with [H2PO4− ] = [HPO42−]. Let’s see what is in solution after we let the best acid and best base react. In each of the following reactions, something strong is reacting, so we assume the reactions go to completion. The first reaction to run to completion is H+ + OH− → H2O. 324 CHAPTER 8 H+ + Before 15.0 mmol After 10.0 mmol OH− APPLICATIONS OF AQUEOUS EQUILIBRIA → H2O 5.00 mmol 0 The next best base present is PO43−. H+ + Before 10.0 mmol After 5.0 mmol PO43− → 5.00 mmol 0 HPO42− 0 5.00 mmol The next best base present is CN−. H+ + Before 5.0 mmol After 0 CN− → 7.50 mmol 2.5 mmol HCN 0 5.0 mmol We need to add 2.5 mmol H+ to convert all the CN− into HCN; then all that remains is 5.00 mmol HPO42− and 7.5 mmol HCN (a very weak acid with Ka = 6.2 × 10−10). From here, we would need to add another 2.5 mmol H+ in order to convert one-half the HPO42− present into its conjugate acid so that [HPO42−] = [H2PO4−] and pH = pK a 2 = 7.21. Adding 5.0 mmol H+ to the original solution: H+ + Before 5.0 mmol After 2.5 mmol H+ Before 2.5 mmol After 0 CN− → 2.5 mmol 0 + HPO42− 5.00 mmol 2.5 mmol HCN 5.0 mmol 7.5 mmol → H2PO4− 0 2.5 mmol After 5.0 mmol H+ (HNO3) is added to the original mixture, we are left with [HPO42−] = [H2PO4−] so that pH = pK a 2 = 7.21. Note that HCN, with Ka = 6.2 × 10−10, is too weak of an acid to interfere with the H2PO4−/ HPO42− buffer. Volume HNO3 = 5.0 mmol HNO3 × 135. 1 mL = 50. mL HNO3 0.100 mmol HNO3 a. Best acid will react with the best base present, so the dominate equilibrium is: NH4+ + X− ⇌ NH3 + HX Keq = [ NH3 ][HX] + [ NH4 ][X − ] = K a , NH + 4 K a , HX CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 325 Because initially [NH4+]0 = [X−]0 and [NH3]0 = [HX]0 = 0, at equilibrium [NH4+] = [X−] and [NH3] = [HX]. Therefore: Keq = K a , NH + 4 K a , HX [HX]2 [X − ]2 = The Ka expression for HX is: K a , HX = [H + ] [H + ][X − ] [HX] = , K a , HX [HX] [X − ] Substituting into the Keq expression: Keq = Rearranging: [H+]2 = K a , NH pH = 4 + K a , NH + 4 K a , HX [HX]2 ⎛ [H + ] = − 2 =⎜ ⎜ K a , HX [X ] ⎝ ⎞ ⎟ ⎟ ⎠ 2 × K a , HX , or taking the −log of both sides: pK a , NH + + pK a , HX 4 2 b. Ammonium formate = NH4(HCO2) K a , NH 4 + = pH = 1.0 × 10 −14 1.8 × 10 −5 = 5.6 × 10−10, pKa = 9.25; K a , HCO 2 H = 1.8 × 10−4, pKa = 3.74 pK a , NH4 + + pK a , HCO2 H 2 = 9.25 + 3.74 = 6.50 2 Ammonium acetate = NH4(C2H3O2); K a , HC 2 H 3O 2 = 1.8 × 10−5; pKa = 4.74 pH = 9.25 + 4.74 = 7.00 2 Ammonium bicarbonate = NH4(HCO3); K a , H 2 CO3 = 4.3 × 10−7; pKa = 6.37 pH = 9.25 + 6.37 = 7.81 2 c. NH4+(aq) + OH−(aq) → NH3(aq) + H2O(l); C2H3O2−(aq) + H+(aq) → HC2H3O2(aq) 136. a. Major species: H+, HSO4−, H2C6H6O6, and H2O; HSO4− is the best acid, with H2O as the best base. 326 CHAPTER 8 HSO4− Initial Change Equil. ⇌ 0.050 M −x 0.050 − x APPLICATIONS OF AQUEOUS EQUILIBRIA H+ + 0.050 M +x 0.050 + x SO42− 0 +x x (0.050 + x)( x) = 1.2 × 10−2; we must use the quadratic equation. (0.050 − x) x = 8.5 × 10−3 M, [H+] = 0.050 + (8.5 × 10−3) = 5.85 × 10−2 M, pH = 1.23 HSO4−, 5.0 mmol b. Major species: H+, 5.0 mmol H2C6H6O6, OH−, Na+, H2O 20. mmol 10. mmol React OH− to completion. React the best base with the best acid. H+ + OH− → H2O Before 5.0 After 0 10. 5 – – HSO4− + OH− → H2O + SO42− Before 5.0 After 0 5 0 – – 0 5.0 mmol After we have let OH− react to completion, the best acid remaining is H2C6H6O6, and the best base remaining is SO42−. React these two together. Initial 20./200. Change –x Equil. 0.10 – x K= K a1 , H 2 C 6 H 6 O 6 K a 2 , H 2SO 4 ⇌ SO42− H2C6H6O6 + 5.0 mmol/200. mL –x 0.025 – x = 6.6 × 10−3; HC6H6O6− + HSO4− 0 +x x 0 +x x x2 = 6.6 × 10−3 (0.10 − x)(0.025 − x) Using the quadratic equation, x = 3.7 × 10−3 M. Use either K a1 for H2C6H6O6 or K a 2 for H2SO4 to calculate [H+]. − For example, 7.9 × 10−5 = [H + ][HC6 H 6O6 ] [H + ] (0.0037) = [H 2 C6 H 6 O 6 ] (0.10 − 0.0037) [H+] = 2.1 × 10−3 M, pH = 2.68 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA c. Major species: H+, HSO4−, 5.0 mmol 5.0 mmol H2C6H6O6, OH−, Na+, H2O 20. mmol 30. mmol React OH− to completion first. React the best acid with the best base. H+ + OH− → H2O Before 5.0 After 0 30. 25 – – HSO4− + OH− → Before 5.0 After 0 25 20. – – H2C6H6O6 + OH− Before 20. After 0 H2O + SO42− 0 5.0 ⇌ HC6H6O6− + H2O 20. 0 0 20. – – After we let all of the OH− react completely, the major species are: HC6H6O6−, 20. mmol SO42−, H2O, Na+ 5.0 mmol HC6H6O6− is the best acid as well as the best base present (amphoteric species). ⇌ H2C6H6O6 + C6H6O62− d. Major species: H+, HSO4−, H2C6H6O6, 5.0 mmol 5.0 mmol 20. mmol OH−, Na+, H2O 50. mmol Dominant reaction: HC6H6O6− + HC6H6O6− pH = pK a 1 + pK a 2 2 = 4.10 + 11.80 = 7.95 2 React OH− first to completion. React the best acid with the best base. H+ + OH− → Before 5.0 After 0 50. 45 H2O – – HSO4− + OH− → H2O Before 5.0 After 0 45 40. – – + SO42− 0 5.0 H2C6H6O6 + OH− → H2O + HC6H6O6− Before After 20. 0 40. 20. – – 0 20. 327 328 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA HC6H6O6− + OH− → C6H6O62− + H2O Before After 20. 0 20. 0 0 20. mmol After all the OH− reacts completely, we have a solution of the weak base C6H6O62− (SO42− is a much weaker base than C6H6O62−, so we can ignore it). Solving the weak base problem. C6H6O62− Initial Change Equil. Kb = + ⇌ H2O 20. mmol/600. mL −x 0.033 − x HC6H6O6− + 0 +x x OH− 0 +x x Kw x2 = 6.3 × 10−3 = 0.033 − x Ka 2 Using the quadratic equation: x = [OH−] = 1.2 × 10−2 M; pOH = 1.92, pH = 12.08 137. a. CuBr(s) ⇌ Cu+ + Br− Ksp = 1.0 × 10−5 Kf = 1.0 × 1011 Cu+ + 3 CN− ⇌ Cu(CN)32− ________________________________________________ CuBr(s) + 3 CN ⇌ Cu(CN)32− + Br− K = 1.0 × 106 Because K is large, assume that enough CuBr(s) dissolves to completely use up the 1.0 M CN−; then solve the back equilibrium problem to determine the equilibrium concentrations. CuBr(s) + 3 CN− Before Change After ⇌ Cu(CN)32− + Br− x 1.0 M 0 0 x mol/L of CuBr(s) dissolves to react completely with 1.0 M CN− −x −3x → +x +x 0 1.0 − 3x x x For reaction to go to completion, 1.0 − 3x = 0 and x = 0.33 mol/L. Now solve the back equilibrium problem. CuBr(s) + 3 CN− Initial Change Equil. ⇌ Cu(CN)32− + Br− 0 0.33 M 0.33 M Let y mol/L of Cu(CN)32− react to reach equilibrium. +3y ← −y −y 3y 0.33 − y 0.33 − y K = 1.0 × 106 = (0.33 − y ) 2 (0.33) 2 ≈ , y = 1.6 × 10−3 M; assumptions good. (3y) 3 27 y 3 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 329 Of the initial 1.0 M CN−, only 3(1.6 × 10−3) = 4.8 × 10−3 M is present at equilibrium. Indeed, enough CuBr(s) did dissolve to essentially remove the initial 1.0 M CN−. This amount, 0.33 mol/L, is the solubility of CuBr(s) in 1.0 M NaCN. b. [Br−] = 0.33 − y = 0.33 − 1.6 × 10−3 = 0.33 M c. [CN−] = 3y = 3(1.6 × 10−3) = 4.8 × 10−3 M 138. Major species: H+, HSO4−, Ba2+, NO3−, and H2O; Ba2+ will react with the SO42− produced from Ka reaction for HSO4−. ⇌ 2− SO4 ⇌ HSO4− H+ + SO42− K a 2 = 1.2 × 10−2 Ba2+ + BaSO4(s) K = 1/Ksp = 1/(1.5 × 10−9) = 6.7 × 108 _________________________________________________________________ Ba2+ + HSO4− ⇌ H+ + BaSO4(s) Koverall = (1.2 × 10−2) × (6.7 × 108) = 8.0 × 106 Because Koverall is so large, the reaction essentially goes to completion. Because H2SO4 is a strong acid, [HSO4−]0 = [H+]0 = 0.10 M. Ba2+ + HSO4− Before Change After Change Equil. 0.30 M −0.10 0.20 +x 0.20 + x K = 8.0 × 106 = ⇌ 0.10 M −0.10 → 0 +x ← x H+ + BaSO4(s) 0.10 M +0.10 0.20 M −x 0.20 − x New initial 0.20 − x 0.20 ≈ , x = 1.3 × 10−7 M; assumptions good. (0.20 + x) x 0.20( x) [H+] = 0.20 − 1.3 × 10−7 = 0.20 M; pH = 0.70 [Ba2+] = 0.20 + 1.3 × 10−7 = 0.20 M From the initial reaction essentially going to completion, 1.0 L(0.10 mol HSO4−/L) = 0.10 mol HSO4− reacted; this will produce 0.10 mol BaSO4(s). Only 1.3 × 10−7 mol of this dissolves to reach equilibrium, so 0.10 mol BaSO4(s) is produced. 0.10 mol BaSO4 × 139. a. SrF2(s) 233.4 g BaSO 4 = 23 g BaSO4 produced mol ⇌ Sr2+(aq) + 2 F−(aq) Initial 0 0 s mol/L SrF2 dissolves to reach equilibrium Equil. s 2s [Sr2+][F−]2 = Ksp = 7.9 × 10−10 = 4s3, s = 5.8 × 10−4 mol/L in pure water 330 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA b. Greater, because some of the F− would react with water: F− + H2O ⇌ HF + OH− Kb = Kw = 1.4 × 10−11 K a , HF This lowers the concentration of F−, forcing more SrF2 to dissolve. c. SrF2(s) ⇌ Sr2+ + 2 F− Ksp = 7.9 × 10−10 = [Sr2+][F−]2 Let s = solubility = [Sr2+]; then 2s = total F− concentration. Since F− is a weak base, some of the F− is converted into HF. Therefore: total F− concentration = 2s = [F−] + [HF] HF ⇌ H+ + F− Ka = 7.2 × 10−4 = 7.2 × 10−2 = [H + ][F − ] 1.0 × 10 −2 [F − ] = (since pH = 2.00 buffer) [HF] [HF] [F − ] , [HF] = 14[F−]; Solving: [HF] [Sr2+] = s; 2s = [F−] + [HF] = [F−] + 14[F−], 2s = 15[F−], [F−] = 2s/15 Ksp = 7.9 × 10 140. −10 2 ⎛ 2s ⎞ = [Sr ][F ] = (s) ⎜ ⎟ , s = 3.5 × 10−3 mol/L in pH = 2.00 solution ⎝ 15 ⎠ 2+ − 2 Ksp = [Ni2+][S2−] = 3 × 10 −21 K a1 = 1.0 × 10-7 H2S ⇌ H+ + HS− HS− ⇌ H+ + S2K a 2 = 1 × 10 −19 _________________________________________ [H + ] 2 [S 2 − ] H2S ⇌ 2 H+ + S2− K = K a1 × K a 2 = 1 × 10 −26 = [H 2 S] Because K is very small, only a tiny fraction of the H2S will react. At equilibrium, [H2S] = 0.10 M and [H+] = 1 × 10 −3 . [S2−] = K[H 2S] (1 × 10 −26 )(0.10) = = 1 × 10 −21 M [ H + ]2 (1 × 10 −3 ) 2 NiS(s) ⇌ Ni2+(aq) + S2−(aq) Ksp = 3.0 × 10 −21 Precipitation of NiS will occur when Q > Ksp. We will calculate [Ni2+] for Q = Ksp. Q = Ksp = [Ni2+][S2−] = 3.0 × 10 −21 , [Ni2+] = 3.0 × 10 −21 = 3M 1 × 10 − 21 CHAPTER 8 141. APPLICATIONS OF AQUEOUS EQUILIBRIA 331 a. Al(OH)3(s) ⇌ Al3+ + 3 OH−; Al(OH)3(s) + OH- ⇌ Al(OH)4− S = solubility = total Al3+ concentration = [Al3+] + [Al(OH)4−] [Al3+] = K sp − 3 [OH ] = Ksp × [H + ]3 Kw 3 , because [OH−]3 = (Kw/[H+])3 − Kw KK w [Al(OH) 4 ] = K; [OH−] = ; [Al(OH)4−] = K[OH−] = + − [H ] [H + ] [OH ] S = [Al3+] + [Al(OH)4−] = [H+]3Ksp/Kw3 + KKw/[H+] b. Ksp = 2 × 10−32; Kw = 1.0 × 10−14; K = 40.0 S= [H + ]3 (2 × 10 −32 ) 40.0 (1.0 × 10 −14 ) 4.0 × 10 −13 + 3 10 + = [H ] (2 × 10 ) + (1.0 × 10 −14 ) 3 [H + ] [H + ] pH solubility (S, mol/L) log S _________________________________________________ 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 2 × 10−2 2 × 10−5 4.2 × 10−7 4.0 × 10−6 4.0 × 10−5 4.0 × 10−4 4.0 × 10−3 4.0 × 10−2 4.0 × 10−1 −1.7 −4.7 −6.38 −5.40 −4.40 −3.40 −2.40 −1.40 −0.40 As expected, the solubility of Al(OH)3(s) is increased by very acidic solutions and by very basic solutions. 142. Solubility in pure water: CaC2O4(s) ⇌ Initial s = solubility (mol/L) Equil. Ca2+ + C2O42− 0 s Ksp = 2 × 10−9 0 s Ksp = s2 = 2 × 10−9, s = solubility = 4.47 × 10−5 = 4 × 10−5 mol/L Solubility in 1.0 M H+: 332 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA CaC2O4(s) C2O42− + H+ HC2O4− + H+ ⇌ ⇌ ⇌ Ca2+ + C2O42− HC2O4− H2C2O4 Ksp = 2 × 10−9 K = 1/ K a 2 = 1.6 × 104 K = 1/ K a1 = 15 CaC2O4(s) + 2 H+ ⇌ Ca2+ + H2C2O4 Koverall = 5 × 10−4 ___________________________________________________________________________________ Initial 0.10 M 0 0 s mol/L of CaC2O4(s) dissolves to reach equilibrium Equil. 0.10 – 2s s s 5 × 10−4 = s2 s , = (5 × 10 − 4 )1/ 2 , s = 2 × 10 −3 mol / L 2 0.10 − 2 s (0.10 − 2s ) Solubility in 0.10 M H + 2 × 10 −3 = = 50 Solubility in pure water 4 × 10 −5 CaC2O4(s) is 50 times more soluble in 0.10 M H+ than in pure water. This increase in solubility is due to the weak base properties of C2O42−. 143. For HOCl, Ka = 3.5 × 10−8 and pKa = −log(3.5 × 10−8) = 7.46. This will be a buffer solution because the pH is close to the pKa value. pH = pKa + log [OCl − ] [OCl − ] [OCl − ] , 8.00 = 7.46 + log , = 100.54 = 3.5 [HOCl] [HOCl] [HOCl] 1.00 L × 0.0500 M = 0.0500 mol HOCl initially. Added OH− converts HOCl into OCl−. The total moles of OCl− and HOCl must equal 0.0500 mol. Solving where n = moles: nOCl − + nHOCl = 0.0500 and nOCl − = (3.5) nHOCl (4.5)nHOCl = 0.0500, nHOCl = 0.011 mol; nOCl − = 0.039 mol Need to add 0.039 mol NaOH to produce 0.039 mol OCl−. 0.039 mol = V × 0.0100 M, V = 3.9 L NaOH Note: Normal buffer assumptions hold. 144. a. HA ⇌ H+ + A− Ka = 5.0 × 10−10; [HA]0 = 1.00 × 10−4 M Because this is a dilute solution of a very weak acid, H2O cannot be ignored as a source of H+. From Section 7.9 of text, try: [H+] = (Ka[HA]0 + Kw)1/2 = 2.4 × 10−7 M; pH = 6.62 Check assumption: CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 333 [ H + ]2 − K w = 2.0 × 10−7 << 1.0 × 10−-4; assumption good. pH = 6.62 + [H ] b. 100.0 mL × (1.00 × 10−4 mmol/mL) = 1.00 × 10−2 mmol HA 5.00 mL × (1.00 × 10−3 mmol/mL) = 5.00 × 10−3 mmol NaOH added; let OH− react completely with HA. After reaction, 5.0 × 10−3 mmol HA and 5.00 × 10−3 mmol A− are in 105.0 mL. [A−]0 = [HA]0 = 5.00 × 10−3 mmol/105.0 mL = 4.76 × 10−5 M A− + H2O Initial Equil. 4.76 × 10−5 M 4.76 × 10−5 − x Kb = 2.0 × 10−5 = ⇌ HA + OH− 4.76 × 10−5 M 4.76 × 10−5 + x Kb = Kw/Ka = 2.0 × 10−5 0 x (4.76 × 10 −5 + x) x ; x will not be small compared to 4.76 × 10−5. (4.76 × 10 −5 − x) Using the quadratic formula and carrying extra sig. figs.: x2 + (6.76 × 10−5)x − 9.52 × 10−10 = 0 x = [OH−] = 1.2 × 10−5 M; pOH = 4.92; pH = 9.08 c. At the stoichiometric point, all the HA is converted into A−. A− Initial Equil. + H2O ⇌ 0.0100 mmol = 9.09 × 10−5 M 110.0 mL 9.09 × 10−5 − x Kb = 2.0 × 10−5 = HA + OH− 0 0 x x x2 x2 ≈ , x = 4.3 × 10−5; assumption poor. −5 −5 (9.09 × 10 − x) 9.09 × 10 Using the quadratic formula and carrying extra sig. figs.: x2 + (2.0 × 10−5)x − (1.82 × 10−9) = 0, x = 3.4 × 10−5 M = [OH−] pOH = 4.47; pH = 9.53 Assumption to ignore H2O contribution to OH− is good. 145. Kb = 2.0 × 10−5 a. 200.0 mL × 0.250 mmol Na3PO4/mL = 50.0 mmol Na3PO4 135.0 mL × 1.000 mmol HCl/mL = 135.0 mmol HCl 100.0 mL × 0.100 mmol NaCN/mL = 10.0 mmol NaCN 334 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA Let H+ from the HCl react to completion with the bases in solution. In general, react the strongest base first and so on. Here, 110.0 mmol of HCl reacts to convert all CN− to HCN and all PO43− to H2PO4−. At this point 10.0 mmol HCN, 50.0 mmol H2PO4−, and 25.0 mmol HCl are in solution. The remaining HCl reacts completely with H2PO4−, converting 25.0 mmol to H3PO4. The final solution contains 25.0 mmol H3PO4, (50.0 − 25.0 =) 25.0 mmol H2PO4−, and 10.0 mmol HCN. HCN (Ka = 6.2 × 10−10) is a much weaker acid than either H3PO4 ( K a1 = 7.5 × 10−3) or H2PO4− ( K a 2 = 6.2 × 10−8), so ignore it. We have a buffer solution. Principal equilibrium reaction is: H3PO4 Initial Equil. ⇌ 25.0 mmol/435.0 mL 0.0575 − x K a1 = 7.5 × 10−3 = H+ 0 x + H2PO4− K a1 = 7.5 × 10−3 25.0/435.0 0.0575 + x x(0.0575 + x) ; normal assumptions don't hold here. 0.0575 − x Using the quadratic formula and carrying extra sig. figs.: x2 + (0.0650)x − 4.31 × 10−4 = 0, x = 0.0061 M = [H+]; pH = 2.21 b. [HCN] = 146. 10.0 mmol = 2.30 × 10−2 M; HCN dissociation will be minimal. 435.0 mL 50.0 mL × 0.100 M = 5.00 mmol H2SO4; 30.0 mL × 0.100 M = 3.00 mmol HOCl 25.0 mL × 0.200 M = 5.00 mmol NaOH; 10.0 mL × 0.150 M = 1.50 mmol KOH 25.0 mL × 0.100 M = 2.50 mmol Ba(OH)2 = 5.00 mmol OH−; we've added 11.50 mmol OH− total. Let OH− react completely with the best acid present (H2SO4). 10.00 mmol OH− + 5.00 mmol H2SO4 → 10.00 mmol H2O + 5.00 mmol SO42− OH− still remains after reacting completely with H2SO4. OH− will then react with the next best acid (HOCl). The remaining 1.50 mmol OH− will convert 1.50 mmol HOCl into 1.50 mmol OCl−, resulting in a solution with 1.50 mmol OCl− and (3.00 − 1.50 =) 1.50 mmol HOCl. The major species at this point are HOCl, OCl−, SO42−, and H2O plus cations that don't affect pH. SO42− is an extremely weak base (Kb = 8.3 × 10−13). We have a buffer solution composed of HOCl and OCl−. Because [HOCl] = [OCl−]: [H+] = Ka = 3.5 × 10−8 M; pH = 7.46; assumptions good. 147. Major species PO43−, H+, HSO4−, H2O, and Na+; let the best base (PO43−) react with the best acid (H+). Assume the reaction goes to completion because H+ is reacting. Note that the concentrations are halved when equal volumes of the two reagents are mixed. CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA PO43− + Before 0.25 M After 0.20 M → HPO42− 0.050 M 0 0 0.050 M H+ 335 Major species: PO43−, HPO42−, HSO4−, H2O, and Na+; react the best base (PO43−) with the best acid (HSO4−). Because K for this reaction is very large, assume the reaction goes to completion. K a , HSO − 4 PO43− + HSO4− → HPO42− + SO42− K= = 2.5 × 1010 K a , HPO 2− 4 Before 0.20 M After 0.15 M 0.050 M 0 0.050 M 0.100 M 0 0.050 M Major species: PO43−, HPO42−, SO42− (a very weak base with Kb = 8.3 × 10−13), H2O, and Na+; because the best base present (PO43−) and best acid present (HPO42−) are conjugate acidbase pairs, a buffer solution exists. Because Kb for PO43− is a relatively large value (Kb = K w /K a, HPO 2− = 0.021), the usual assumptions that the amount of base that reacts to each 4 equilibrium is negligible compared with the initial concentration of base will not hold. Solving using the Kb reaction for PO43−: PO43− + Initial Change Equil. H2O 0.15 M −x 0.15 − x Kb = 0.021 = ⇌ HPO42− → 0.100 M +x 0.100 + x + OH− Kb = 0.021 0 +x x (0.100 + x)( x) ; using quadratic equation: 0.15 − x x = [OH−] = 0.022 M; pOH = 1.66; pH = 12.34 148. a. At the third halfway point, pH = pK a 3 = −log(4.8 × 10−13) = 12.32. b. At third equivalence point, the reaction is: PO43− + H2O Initial Change Equil. 10. mmol 400. mL −x 0.025 − x – ⇌ HPO42− + OH− 0 0 +x x +x x Kb = Kw 1.0 × 10 −14 = K a3 4.8 × 10 −13 Kb = 2.1 × 10−2 336 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA x2 = 2.1 × 10−2; using the quadratic equation: 0.025 − x x = 1.5 × 10−2 M = [OH−]; pOH = 1.82; pH = 12.18 c. The pH at the third halfway point must be more acidic (lower pH) than the pH at the third equivalence point. Therefore, the pH at the third halfway point cannot equal 12.32. In part a we assumed that x was negligible: HPO42− + H2O Initial Change Equil. ⇌ 5.0 mmol/350. mL −x 0.014 − x PO43− + H3O+ 5.0/350. +x 0.014 + x (0.014 + x)( x) (0.014)( x) ≈ , (0.014 − x) (0.014) K a 3 = 4.8 × 10−13 0 +x x x = 4.8 × 10−13 M This looks fine, but this is a situation where we must use the Kb reaction for the weak base PO43− to solve for the pH. The [OH−] in solution is not negligible compared to 0.014 M, so the usual assumptions don’t hold here. The usual buffer assumptions don’t hold in very acidic or very basic solutions. In this very basic solution, we must use the Kb reaction and the quadratic equation: ⇌ PO43− + H2O HPO42− + OH− PO43− + H2O d. Initial Change Equil. 0.014 M −x 0.014 − x – – – ⇌ HPO42− + OH− 0.014 M +x 0.014 + x Kb = 2.1 × 10−2 0 +x x (0.14 + x)( x) = 2.1 × 10−2; using the quadratic equation: (0.14 − x) x = [OH−] = 7.0 × 10−3 M; pOH = 2.15; pH = 11.85 This pH answer makes more sense because it is below the pH at the third equivalence point calculated in part b of this problem (pH = 12.18). 149. H3A: pK a1 = 3.00, pK a 2 = 7.30, pK a 3 = 11.70 The pH at the second stoichiometric point is: pH = pK a 2 + pK a 3 2 = 7.30 + 11.70 = 9.50 2 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 337 Thus to reach a pH of 9.50, we must go to the second stoichiometric point. 100.0 mL × 0.0500 M = 5.00 mmol H3A initially. To reach the second stoichiometric point, we need 10.0 mmol OH− = 1.00 mmol/mL × VNaOH. Solving for VNaOH: VNaOH = 10.0 mL (to reach pH = 9.50) pH = 4.00 is between the first halfway point to equivalence (pH = pK a1 = 3.00) and the first pK a1 + pK a 2 stoichiometric point (pH = = 5.15). 2 This is the buffer region controlled by H3A ⇌ H2A− + H+. pH = pK a1 + log [H 2 A − ] [H A − ] [H 2 A − ] , 4.00 = 3.00 + log 2 , = 10. [H 3 A] [H 3 A] [H 3 A] Because both species are in the same volume, the mole ratio also equals 10. Let n = mmol: nH 2A − nH 3 A = 10. and nH 2A − + nH 3A = 5.00 mmol (mole balance) 11nH 3A = 5.00, n H 3A = 0.45 mmol; nH 2A − = 4.55 mmol We need to add 4.55 mmol OH− to get 4.55 mmol H2A− from the original H3A present. 4.55 mmol = 1.00 mmol/mL × VNaOH, VNaOH = 4.55 mL of NaOH (to reach pH = 4.00) Note: Normal buffer assumptions are good. 150. 100.0 mL × 0.100 M = 10.0 mmol H3A initially a. 100.0 mL × 0.0500 mmol/mL = 5.00 mmol OH− added This is the first halfway point to equivalence, where [H3A] = [H2A−] and pH = pK a1 . pH = −log(5.0 × 10− 4) = 3.30; assumptions good. b. Since pK a 2 = 8.00, a buffer mixture of H2A− and HA2− can produce a pH = 8.67 solution. 8.67 = 8.00 + log [HA 2 − ] [HA 2 − ] , = 10+0.67 = 4.7 [H 2 A − ] [H 2 A − ] Both species are in the same volume, so the mole ratio also equals 4.7. Let n = mmol: nHA 2− nH 2A − = 4.7, nHA 2− = (4.7)nH 2A − ; nHA 2− + nH 2A − = 10.0 mmol (mole balance) 338 CHAPTER 8 (5.7)nH 2A − = 10.0 mmol, nH APPLICATIONS OF AQUEOUS EQUILIBRIA 2A − = 1.8 mmol; nHA 2− = 8.2 mmol To reach this point, we must add a total of 18.2 mmol NaOH. 10.0 mmol OH− converts all of the 10.0 mmol H3A into H2A-. The next 8.2 mmol OH− converts 8.2 mmol H2Ainto 8.2 mmol HA2− , leaving 1.8 mmol H2A−. 18.2 mmol = 0.0500 M × V, V = 364 mL NaOH Note: Normal buffer assumptions are good. 151. a. V1 corresponds to the titration reaction of CO32− + H+ → HCO3−; V2 corresponds to the titration reaction of HCO3− + H+ → H2CO3. Here, there are two sources of HCO3−: NaHCO3 and the titration of Na2CO3, so V2 > V1. b. V1 corresponds to two titration reactions: OH− + H+ → H2O and CO32− + H+ → HCO3−. V2 corresponds to just one titration reaction: HCO3− + H+ → H2CO3. Here, V1 > V2 due to the presence of OH− , which is titrated in the V1 region. c. 0.100 mmol HCl/mL × 18.9 mL = 1.89 mmol H+; Because the first stoichiometric point only involves the titration of Na2CO3 by H+, 1.89 mmol of CO32− has been converted into HCO3−. The sample contains 1.89 mmol Na2CO3 × 105.99 mg/mmol = 2.00 × 102 mg = 0.200 g Na2CO3. The second stoichiometric point involves the titration of HCO3− by H+. 0.100 mmol H + × 36.7 mL = 3.67 mmol H+ = 3.67 mmol HCO3− mL 1.89 mmol NaHCO3 came from the first stoichiometric point of the Na2CO3 titration. 3.67 − 1.89 = 1.78 mmol HCO3− came from NaHCO3 in the original mixture. 1.78 mmol NaHCO3 × 84.01 mg NaHCO3/mmol = 1.50 × 102 mg NaHCO3 = 0.150 g NaHCO3 Mass % Na2CO3 = Mass % NaHCO3 = 0.200 g × 100 = 57.1% Na2CO3 (0.200 + 0.150) g 0.150 g × 100 = 42.9% NaHCO3 0.350 g CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA 339 Marathon Problem 152. a. Because K a1 >> K a 2 , the amount of H+ contributed by the K a 2 reaction will be negligible. The [H+] donated by the K a1 reaction is 10−2.06 = 8.7 × 10−3 M H+. ⇌ H2A Initial Equil. H+ [H2A]0 [H2A]0 − x K a1 = 5.90 × 10−2 = ~0 x HA− K a1 = 5.90 × 10−2 0 x [H2A]0 = initial concentration x2 (8.7 × 10 −3 ) 2 , [H2A]0 = 1.0 × 10−2 M = [H 2 A ]0 − x [H 2 A]0 − 8.7 × 10 −3 Mol H2A present initially = 0.250 L × Molar mass H2A = + 1.0 × 10 −2 mol H 2 A = 2.5 × 10−3 mol H2A L 0.225 g H 2 A = 90. g/mol 2.5 × 10 −3 mol H 2 A b. H2A + 2 OH− → A2− + H2O; at the second equivalence point, the added OH− has converted all the H2A into A2−, so A2− is the major species present that determines the pH. The millimoles of A2− present at the equivalence point equal the millimoles of H2A present initially (2.5 mmol), and the millimoles of OH− added to reach the second equivalence point are 2(2.5 mmol) = 5.0 mmol OH−added. The only information we need now in order to calculate the K a 2 value is the volume of Ca(OH)2 added in order to reach the second equivalent point. We will use the Ksp value for Ca(OH)2 to help solve for the volume of Ca(OH)2 added. Ca(OH)2(s) Initial Equil. ⇌ + 2 OH− Ca2+ s = solubility (mol/L) 0 s Ksp = 1.3 × 10−6 = [Ca2+][OH−]2 ~0 2s Ksp = 1.3 × 10−6 = (s)(2s)2 = 4s3, s = 6.9 × 10−3 M Ca(OH)2; assumptions good. The volume of Ca(OH)2 required to deliver 5.0 mmol OH− (the amount of OH− necessary to reach the second equivalence point) is: 5.0 mmol OH− × 1 mmol Ca (OH) 2 2 mmol OH − × 1 mL 6.9 × 10 −3 mmol Ca (OH) 2 = 362 mL = 360 mL Ca(OH)2 At the second equivalence point, the total volume of solution is: 250. mL + 360 mL = 610 mL 340 CHAPTER 8 APPLICATIONS OF AQUEOUS EQUILIBRIA Now we can solve for K a 2 using the pH data at the second equivalence point. Because the only species present that has any effect on pH is the weak base A2−, the setup to the problem requires the Kb reaction for A2−. −14 K A2− + H2O ⇌ HA− + OH− Kb = w = 1.0 × 10 Ka2 Ka2 Initial Equil. Kb = 2.5 mmol 610 mmol 4.1 × 10−3 M − x 0 0 x x 1.0 × 10 −14 x2 = Ka2 4.1 × 10 −3 − x From the problem: pH = 7.96, so [OH−] = 10−6.04 = 9.1 × 10−7 M = x Kb = 1.0 × 10 −14 (9.1 × 10 −7 ) 2 = = 2.0 × 10−10; K a 2 = 5.0 × 10−5 Ka2 (4.1 × 10 −3 ) − (9.1 × 10 −7 ) Note: The amount of OH− donated by the weak base HA− will be negligible because the Kb value for A2− is more than a 1000 times the Kb value for HA−. In addition, because the pH is less than 8.0 at the second equivalence point, the amount of OH− added by H2O may need to be considered. Using the equation derived in Exercise 7.135, we get the same K a 2 value as calculated above by ignoring the OH− contribution from H2O. CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY The Nature of Energy 15. Ball A: PE = mgz = 2.00 kg × 9.81 m 196 kg m 2 × 10.0 m = = 196 J s2 s2 At point I: All this energy is transferred to ball B. All of B's energy is kinetic energy at this point. Etotal = KE = 196 J. At point II, the sum of the total energy will equal 196 J. At point II: PE = mgz = 4.00 kg × 9.81 m × 3.00 m = 118 J s2 KE = Etotal − PE = 196 J − 118 J = 78 J 16. Plot a represents an exothermic reaction. In an exothermic process, the bonds in the product molecules are stronger (on average) than those in the reactant molecules. The net result is that the quantity of energy Δ(PE) is transferred to the surroundings as heat when reactants are converted to products. For an endothermic process (plot b), energy flows into the system as heat to increase the potential energy of the system. In an endothermic process, the products have higher potential energy (weaker bonds on average) than the reactants. 17. Path-dependent functions for a trip from Chicago to Denver are those quantities that depend on the route taken. One can fly directly from Chicago to Denver or one could fly from Chicago to Atlanta to Los Angeles and then to Denver. Some path-dependent quantities are miles traveled, fuel consumption of the airplane, time traveling, airplane snacks eaten, etc. State functions are path independent; they only depend on the initial and final states. Some state functions for an airplane trip from Chicago to Denver would be longitude change, latitude change, elevation change, and overall time zone change. 18. Only when there is a volume change can PV work be done. In pathway I (steps 1 + 2), only the first step does PV work (step 2 has a constant volume of 30.0 L). In pathway II (steps 3 + 4), only step 4 does PV work (step 3 has a constant volume of 10.0 L). 341 342 CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY Pathway I: w = !PΔV = !2.00 atm(30.0 L ! 10.0 L) = -40.0 L atm × 101.3 J L atm = !4.05 × 103 J Pathway II: w = !PΔV = !1.00 atm(30.0 L ! 10.0 L) = !20.0 L atm × 101.3 J L atm = !2.03 × 103 J Note: The sign is minus (!) because the system is doing work on the surroundings (an expansion). We get different values of work for the two pathways; both pathways have the same initial and final states. Because w depends on the pathway, work cannot be a state function. 19. Step 1: ΔE1 = q + w = 72 J + 35 J = 107 J; step 2: ΔE2 = 35 J − 72 J = −37 J ΔEoverall = ΔE1 + ΔE2 = 107 J − 37 J = 70. J 20. a. ΔE = q + w = !23 J + 100. J = 77 J b. w = !PΔV = !1.90 atm(2.80 L ! 8.30 L) = 10.5 L atm × 101.3 J = 1060 J L atm ΔE = q + w = 350. J + 1060 = 1410 J c. w = !PΔV = !1.00 atm(29.1 L ! 11.2 L) = !17.9 L atm × 101.3 J = !1810 J L atm ΔE = q + w = 1037 J ! 1810 J = !770 J 20.8 J × 39.1 mol × (38.0 − 0.0)°C = 30,900 J C mol = 30.9 kJ 101.3 J = −12,400 J = −12.4 kJ w = −PΔV = −1.00 atm × (998 L − 876 L) = −122 L atm × L atm ΔE = q + w = 30.9 kJ + (−12.4 kJ) = 18.5 kJ 21. q = molar heat capacity × mol × ΔT = 22. In this problem, q = w = −950. J. −950. J × 1 L atm = −9.38 L atm of work done by the gases 101.3 J w = −PΔV, −9.38 L atm = 23. o − 650. atm × (Vf − 0.040 L), Vf − 0.040 = 11.0 L, Vf = 11.0 L 760 H2O(g) → H2O(l); ΔE = q + w; q = −40.66 kJ; w = −PΔV CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY Volume of 1 mol H2O(l) = 1.000 mol H2O(l) × 343 18.02 g 1 cm 3 × = 18.1 cm3 = 18.1 mL mol 0.996 g w = −PΔV = −1.00 atm × (0.0181 L − 30.6 L) = 30.6 L atm × 101.3 J = 3.10 × 103 J L atm = 3.10 kJ ΔE = q + w = −40.66 kJ + 3.10 kJ = −37.56 kJ 24. ΔE = q + w, !102.5 J = 52.5 J + w, w = !155.0 J × 1 L atm = !1.530 L atm 101.3 J w = !PΔV, !1.530 L atm = !0.500 atm × ΔV, ΔV = 3.06 L ΔV = Vf – Vi, 3.06 L = 58.0 L ! Vi, Vi = 54.9 L = initial volume Properties of Enthalpy 25. ΔH = ΔE + PΔV at constant P; from the definition of enthalpy, the difference between ΔH and ΔE at constant P is the quantity PΔV. Thus when a system at constant P can do pressurevolume work, then ΔH ≠ ΔE. When the system cannot do PV work, then ΔH = ΔE at constant pressure. An important way to differentiate ΔH from ΔE is to concentrate on q, the heat flow; the heat flow by a system at constant pressure equals ΔH, and the heat flow by a system at constant volume equals ΔE. 26. a. The combustion of gasoline releases heat, so this is an exothermic process. b. H2O(g) → H2O(l); heat is released when water vaper condenses, so this is an exothermic process. c. To convert a solid to a gas, heat must be absorbed, so this is an endothermic process. d. Heat must be added (absorbed) in order to break a bond, so this is an endothermic process. 27. One should try to cool the reaction mixture or provide some means of removing heat since the reaction is very exothermic (heat is released). The H2SO4(aq) will get very hot and possibly boil unless cooling is provided. 28. This is an endothermic reaction, so heat must be absorbed in order to convert reactants into products. The high-temperature environment of internal combustion engines provides the heat. 29. 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) ΔH = -1652 kJ; note that 1652 kJ of heat is released when 4 mol Fe reacts with 3 mol O2 to produce 2 mol Fe2O3. 344 CHAPTER 9 a. 4.00 mol Fe × ENERGY, ENTHALPY, AND THERMOCHEMISTRY − 1652 kJ = !1650 kJ; 1650 kJ of heat released 4 mol Fe b. 1.00 mol Fe2O3 × − 1652 kJ = !826 kJ; 826 kJ of heat released 2 mol Fe 2 O 3 c. 1.00 g Fe × 1 mol Fe − 1652 kJ × = !7.39 kJ; 7.39 kJ of heat released 55.85 g 4 mol Fe d. 10.0 g Fe × 1 mol O 2 1 mol Fe = 0.179 mol Fe; 2.00 g O2 × = 0.0625 mol O2 55.85 g 32.00 g 0.179 mol Fe/0.0625 mol O2 = 2.86; the balanced equation requires a 4 mol Fe/3 mol O2 = 1.33 mole ratio. O2 is limiting since the actual mole Fe/mole O2 ratio is greater than the required mole ratio. 0.0625 mol O2 × 30. − 1652 kJ = !34.4 kJ; 34.4 kJ of heat released 3 mol O 2 From Example 9.1, q = 1.3 × 108 J. Because the heat transfer process is only 60.% 100. J = 2.2 × 108 J efficient, the total energy required is: 1.3 × 108 J × 60. J Mass C3H8 = 2.2 × 108 J × 1 mol C3 H 8 44.09 g C3 H 8 × = 4.4 × 103 g C3H8 3 mol C3 H 8 2221 × 10 J 31. When a liquid is converted into gas, there is an increase in volume. The 2.5 kJ/mol quantity is the work done by the vaporization process in pushing back the atmosphere. 32. ΔH = ΔE + PΔV; from this equation, ΔH > ΔE when ΔV > 0, ΔH < ΔE when ΔV < 0, and ΔH = ΔE when ΔV = 0. Concentrate on the moles of gaseous products versus the moles of gaseous reactants to predict ΔV for a reaction. a. There are 2 moles of gaseous reactants converting to 2 moles of gaseous products, so ΔV = 0. For this reaction, ΔH = ΔE. b. There are 4 moles of gaseous reactants converting to 2 moles of gaseous products, so ΔV < 0 and ΔH < ΔE. c. There are 9 moles of gaseous reactants converting to 10 moles of gaseous products, so ΔV > 0 and ΔH > ΔE. The Thermodynamics of Ideal Gases 33. Consider the constant volume process first. CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY n = 1.00 × 103 g × 1 mol 44.60 J 44.60 J = 33.3 mol C2H6; Cv = = o 30.07 g K mol C mol ΔE = nCvΔT = (33.3 mol)(44.60 J °C−1 mol−1)(75.0 − 25.0°C) = 74,300 J = 74.3 kJ ΔE = q + w; since ΔV = 0, w = 0; ΔE = qv = 74.3 kJ ΔH = ΔE + ΔPV = ΔE + nRΔT ΔH = 74.3 kJ + (33.3 mol)(8.3145 J K−1 mol−1)(50.0 K)(1 kJ/1000 J) ΔH = 74.3 kJ + 13.8 kJ = 88.1 kJ Now consider the constant pressure process. qp = ΔH = nCpΔT = (33.3 mol)(52.92 J K−1 mol−1)(50.0 K) qp = 88,100 J = 88.1 kJ = ΔH w = −PΔV = −nRΔT = −(33.3 mol)(8.3145 J K−1 mol−1)(50.0 K) = −13,800 J = −13.8 kJ ΔE = q + w = 88.1 kJ − 13.8 kJ = 74.3 kJ Summary q ΔE ΔH w 34. 88.0 g N2O × Constant V 74.3 kJ 74.3 kJ 88.1 kJ 0 Constant P 88.1 kJ 74.3 kJ 88.1 kJ −13.8 kJ 1 mol N 2 O = 2.00 mol N2O 44.02 g N 2 O At constant pressure, qp = ΔH. ΔH = nCpΔT = (2.00 mol)(38.70 J °C−1 mol−1)(55°C − 165°C) ΔH = −8510 J = −8.51 kJ = qp w = −PΔV = −nRΔT = −(2.00 mol)(8.3145 J K−1 mol−1)(−110. K) = 1830 J = 1.83 kJ ΔE = q + w = −8.51 kJ + 1.83 kJ = −6.68 kJ 35. Pathway I: Step 1: (5.00 mol, 3.00 atm, 15.0 L) → (5.00 mol, 3.00 atm, 55.0 L) w = −PΔV = −(3.00 atm)(55.0 − 15.0 L) = −120. L atm 345 346 CHAPTER 9 w = −120. L atm H ENERGY, ENTHALPY, AND THERMOCHEMISTRY 101.3 J 1 kJ × = −12.2 kJ L atm 1000 J ΔH = qp = nCpΔT = nCp × C Δ (PV ) Δ (PV ) = p ; Δ(PV) = (P2V2 − P1V1) nR R For an ideal monatomic gas: Cp = ΔH = qp = 5 5 ⎛ 5 ⎞ Δ( PV ) R; ΔH = ⎜ R ⎟ = Δ (PV ) 2 2 ⎝2 ⎠ R 5 5 Δ(PV) = (165 − 45.0) L atm = 300. L atm 2 2 ΔH = qp = 300. L atm × 101.3 J 1 kJ × = 30.4 kJ L atm 1000 J ΔE = q + w = 30.4 kJ − 12.2 kJ = 18.2 kJ Step 2: (5.00 mol, 3.00 atm, 55.0 L) → (5.00 mol, 6.00 atm, 20.0 L) ⎛ 3 ⎞ ⎛ Δ (PV ) ⎞ 3 ΔE = nCvΔT = n ⎜ R ⎟ ⎜ ⎟ = ΔPV ⎝ 2 ⎠ ⎝ nR ⎠ 2 ΔE = 3 (120. − 165) L atm = −67.5 L atm (Carry an extra significant figure.) 2 ΔE = −67.5 L atm × 101.3 J 1 kJ × = −6.8 kJ L atm 1000 J ⎛ 5 ⎞ ⎛ Δ (PV ) ⎞ 5 ΔH = nCpΔT = n ⎜ R ⎟ ⎜ ⎟ = ΔPV ⎝ 2 ⎠ ⎝ nR ⎠ 2 ΔH = 5 (−45 L atm) = −113 L atm (Carry an extra significant figure.) 2 ΔH = −113 L atm × 101.3 J 1 kJ × = −11.4 = −11 kJ L atm 1000 J w = −PextΔV = −(6.00 atm)(20.0 − 55.0) L = 210. L atm w = 210. L atm × 101.3 J 1 kJ × = 21.3 kJ L atm 1000 J ΔE = q + w, −6.8 kJ = q + 21.3 kJ, q = −28.1 kJ CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY Summary: Path I q w ΔE ΔH Step 1 Step 2 30.4 kJ −12.2 kJ 18.2 kJ 30.4 kJ −28.1 kJ 21.3 kJ −6.8 kJ −11 kJ Total 2.3 kJ 9.1 kJ 11.4 kJ 19 kJ Pathway II: Step 3: (5.00 mol, 3.00 atm, 15.0 L) → (5.00 mol, 6.00 atm, 15.0 L) ΔE = qv = 3 5 Δ(PV) = (90.0 − 45.0) L atm = 67.5 L atm 2 2 ΔE = qv = 67.5 L atm × 101.3 J 1 kJ × = 6.84 kJ L atm 1000 J w = −PΔV = 0 because ΔV = 0 ΔH = ΔE + Δ(PV) = 67.5 L atm + 45.0 L atm = 112.5 L atm = 11.40 kJ Step 4: (5.00 mol, 6.00 atm, 15.0 L) → (5.00 mol, 6.00 atm, 20.0 L) ⎛ 5 ⎞ ⎛ Δ (PV ) ⎞ 5 ΔH = qp = nCpΔT = ⎜ R ⎟ ⎜ ⎟ = ΔPV ⎝ 2 ⎠ ⎝ nR ⎠ 2 ΔH = 5 (120. - 90.0) L atm = 75 L atm 2 ΔH = qp = 75 L atm × 101.3 J 1 kJ × = 7.6 kJ L atm 1000 J w = −PΔV = − (6.00 atm)(20.0 − 15.0) L = −30. L atm w = −30. L atm × 101.3 J 1 kJ × = −3.0 kJ L atm 1000 J ΔE = q + w = 7.6 kJ − 3.0 kJ = 4.6 kJ Summary: Path II q w ΔE ΔH Step 3 Step 4 6.84 kJ 0 6.84 kJ 11.40 kJ 7.6 kJ −3.0 kJ 4.6 kJ 7.6 kJ Total 14.4 kJ −3.0 kJ 11.4 kJ 19.0 kJ 347 348 CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY State functions are independent of the particular pathway taken between two states; path functions are dependent on the particular pathway. In this problem, the overall values of ΔH and ΔE for the two pathways are the same; hence ΔH and ΔE are state functions. The overall values of q and w for the two pathways are different; hence q and w are path functions. 36. For a monoatomic gas, Cv = (3/2)R and Cp = (5/2)R. Step 1: (2.00 mol, 10.0 atm, 10.0 L) → (2.00 mol, 10.0 atm, 5.0 L) ⎛ 5 ⎞ ⎛ Δ (PV ) ⎞ 5 ΔH = qp = nCpΔT = n ⎜ R ⎟ ⎜ ⎟ = ΔPV ⎝ 2 ⎠ ⎝ nR ⎠ 2 ΔH = qp = 5 (50. − 100.) = −125 L atm × 101.3 J L−1 atm−1 = −12.7 kJ 2 (We will carry all calculations to 0.1 kJ.) w = −PΔV = −(10.0 atm)(5.0 − 10.0) L = 50. L atm = 5.1 kJ ΔE = q + w = −12.7 + 5.1 = −7.6 kJ Step 2: (2.00 mol, 10.0 atm, 5.0 L) → (2.00 mol, 20.0 atm, 5.0 L) ΔE = qv = nCvΔT = 3 3 Δ(PV) = (100 − 50.) = 75 L atm = 7.6 kJ; w = 0 since ΔV = 0 2 2 ΔH = ΔE + Δ(PV) = 75 L atm + 50. L atm = 125 L atm = 12.7 kJ Step 3: (2.00 mol, 20.0 atm, 5.0 L) → (2.00 mol, 20.0 atm, 25.0 L) ΔH = qp = 5 5 Δ(PV) = (500. − 100) = 1.0 × 103 L atm = 101.3 kJ 2 2 w = −PΔV = −(20.0 atm)(25.0 − 5.0) L = −400. L atm = −40.5 kJ ΔE = q + w = 101.3 − 40.5 = 60.8 kJ Summary: q w ΔE ΔH Step 1 Step 2 Step 3 Total −12.7 kJ 5.1 kJ −7.6 kJ −12.7 kJ 7.6 kJ 0 7.6 kJ 12.7 kJ 101.3 kJ −40.5 kJ 60.8 kJ 101.3 kJ 96.2 kJ −35.4 kJ 60.8 kJ 101.3 kJ CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY 349 Calorimetry and Heat Capacity 37. In calorimetry, heat flow is determined into or out of the surroundings. Because ΔEuniv = 0 by the first law of thermodynamics, ΔEsys = !ΔEsurr; what happens to the surroundings is the exact opposite of what happens to the system. To determine heat flow, we need to know the heat capacity of the surroundings, the mass of the surroundings that accepts/donates the heat, and the change in temperature. If we know these quantities, qsurr can be calculated and then equated to qsys (!qsurr = qsys). For an endothermic reaction, the surroundings (the calorimeter contents) donates heat to the system. This is accompanied by a decrease in temperature of the surroundings. For an exothermic reaction, the system donates heat to the surroundings (the calorimeter), so temperature increases. qP = ΔH; qV = ΔE; a coffee-cup calorimeter is at constant (atmospheric) pressure. The heat released or gained at constant pressure is ΔH. A bomb calorimeter is at constant volume. The heat released or gained at constant volume is ΔE. 38. a. s = specific heat capacity = Energy = s × m × ΔT = b. Molar heat capacity = c. 1250 J = 39. 0.24 J 0.24 J = since ΔT(K) = ΔT(°C). o Kg Cg 0.24 J × 150.0 g × (298 K − 273 K) = 9.0 × 102 J Kg 0.24 J 107.9 g Ag × = o mol Ag Cg o 26 J C mol 0.24 J 1250 × m × (15.2°C − 12.0°C), m = = 1.6 × 103 g Ag o 0.24 × 3.2 Cg Specific heat capacity is defined as the amount of heat necessary to raise the temperature of one gram of substance by one degree Celsius. Therefore, H2O(l) with the largest heat capacity value requires the largest amount of heat for this process. The amount of heat for H2O(l) is: energy = s × m × ΔT = 4.18 J × 25.0 g × (37.0°C − 15.0°C) = 2.30 × 103 J o Cg The largest temperature change when a certain amount of energy is added to a certain mass of substance will occur for the substance with the smallest specific heat capacity. This is Hg(l), and the temperature change for this process is: 1000 J 10.7 kJ × energy kJ = 140°C ΔT = = 0.14 J s×m × 550. g o Cg 40. Heat gain by water = heat loss by metal = s × m × ΔT, where s = specific heat capacity. 350 CHAPTER 9 Heat gain = ENERGY, ENTHALPY, AND THERMOCHEMISTRY 4.18 J × 150.0 g × (18.3°C - 15.0°C) = 2100 J o Cg A common error in calorimetry problems is sign errors. Keeping all quantities positive helps to eliminate sign errors. Heat loss = 2100 J = s × 150.0 g × (75.0°C − 18.3°C), s = 41. 2100 J 56.7 C × 150.0 g = 0.25 J °C−1 g−1 o Heat loss by hot water = heat gain by cold water; keeping all quantities positive to avoid sign errors: 4.18 J 4.18 J × mhot × (55.0°C ! 37.0°C) = o × 90.0 g × (37.0 °C ! 22.0°C) o Cg Cg mhot = 42. 90.0 g × 15.0 o C = 75.0 g hot water needed 18.0 o C Heat loss by Al + heat loss by Fe = heat gain by water; keeping all quantities positive to avoid sign error: 0.89 J 0.45 J × 5.00 g Al × (100.0°C − Tf) + o × 10.00 g Fe × (100.0 − Tf) o Cg Cg 4.18 J = o × 97.3 g H2O × (Tf − 22.0°C) Cg 4.5(100.0 − Tf) + 4.5(100.0 − Tf) = 407(Tf − 22.0), 450 − (4.5)Tf + 450 − (4.5)Tf = 407Tf − 8950 416Tf = 9850, Tf = 23.7°C 43. 50.0 × 10−3 L × 0.100 mol/L = 5.00 × 10−3 mol of both AgNO3 and HCl are reacted. Thus 5.00 × 10−3 mol of AgCl will be produced because there is a 1 : 1 mole ratio between reactants. Heat lost by chemicals = heat gained by solution Heat gain = 4.18 J × 100.0 g × (23.40 − 22.60)°C = 330 J o Cg Heat loss = 330 J; this is the heat evolved (exothermic reaction) when 5.00 × 10−3 mol of AgCl is produced. So q = −330 J and ΔH (heat per mol AgCl formed) is negative with a value of: ΔH = − 330 J 1 kJ × = −66 kJ/mol −3 5.00 × 10 mol 1000 J CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY 351 Note: Sign errors are common with calorimetry problems. However, the correct sign for ΔH can be determined easily from the ΔT data; i.e., if ΔT of the solution increases, then the reaction is exothermic because heat was released, and if ΔT of the solution decreases, then the reaction is endothermic because the reaction absorbed heat from the water. For calorimetry problems, keep all quantities positive until the end of the calculation and then decide the sign for ΔH. This will help eliminate sign errors. 44. NH4NO3(s) → NH4+(aq) + NO3−(aq) ΔH = ?; mass of solution = 75.0 g + 1.60 g = 76.6 g Heat lost by solution = heat gained as NH4NO3 dissolves. To help eliminate sign errors, we will keep all quantities positive (q and ΔT) and then deduce the correct sign for ΔH at the end of the problem. Here, because temperature decreases as NH4NO3 dissolves, heat is absorbed as NH4NO3 dissolves, so this is an endothermic process (ΔH is positive). Heat lost by solution = 4.18 J × 76.6 g × (25.00 − 23.34)°C = 532 J = heat gained as o Cg NH NO dissolves 4 ΔH = 45. 80.05 g NH 4 NO 3 532 J 1 kJ × × = 26.6 kJ/mol NH4NO3 dissolving 1.60 g NH 4 NO 3 mol NH 4 NO 3 1000 J Because ΔH is exothermic, the temperature of the solution will increase as CaCl2(s) dissolves. Keeping all quantities positive: heat loss as CaCl2 dissolves = 11.0 g CaCl2 × heat gained by solution = 8.08 × 103 J = Tf − 25.0°C = 46. 3 1 mol CaCl 2 81.5 kJ × = 8.08 kJ 110.98 g CaCl 2 mol CaCl 2 4.18 J × (125 + 11.0) g × (Tf − 25.0°C) o Cg 8.08 × 103 = 14.2°C, Tf = 14.2°C + 25.0°C = 39.2°C 4.18 × 136 0.100 L × 0.500 mol HCl = 5.00 × 10−2 mol HCl L 0.300 L × 0.100 mol Ba (OH) 2 = 3.00 × 10−2 mol Ba(OH)2 L To react with all the HCl present, 5.00 × 10−2/2 = 2.50 × 10−2 mol Ba(OH)2 is required. Because 0.0300 mol Ba(OH)2 is present, HCl is the limiting reactant. 118 kJ = 2.95 kJ of heat is evolved by reaction 2 mol HCl 4.18 J × 400.0 g × ΔT Heat gained by solution = 2.95 × 103 J = o Cg 5.00 × 10−2 mol HCl × ΔT = 1.76°C = Tf − Ti = Tf − 25.0°C, Tf = 26.8°C 352 47. CHAPTER 9 Heat gain by calorimeter = ENERGY, ENTHALPY, AND THERMOCHEMISTRY 1.56 kJ × 3.2°C = 5.0 kJ = heat loss by quinone o C Heat loss = 5.0 kJ, which is the heat evolved (exothermic reaction) by the combustion of 0.1964 g of quinone. Because we are at constant volume, qv = ΔE. ΔEcomb = 48. − 5.0 kJ = −25 kJ/g; 0.1964 g ΔEcomb = − 25 kJ 108.09 g × = −2700 kJ/mol g mol a. Heat gain by calorimeter = heat loss by CH4 = 6.79 g CH4 × Heat capacity of calorimeter = 340. kJ = 31.5 kJ/°C 10.8 o C b. Heat loss by C2H2 = heat gain by calorimeter = 16.9°C × ΔEcomb = 49 1 mol CH 4 802 kJ × 16.04 g mol = 340. kJ 31. 5 kJ = 532 kJ o C − 532 kJ 26.04 g × = !1.10 × 103 kJ/mol 12.6 g C 2 H 2 mol C 2 H 2 a. C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l) b. A bomb calorimeter is at constant volume, so heat released = qv = ΔE: ΔE = − 24.00 kJ 342.30 g = −5630 kJ/mol C12H22O11 × 1.46 g mol c. ΔH = ΔE + Δ(PV) = ΔE + Δ(nRT) = ΔE + ΔnRT, where Δn = moles of gaseous products – moles of gaseous reactants. For this reaction, Δn = 12 −12 = 0, so ΔH = ΔE = −5630 kJ/mol. 50. A(l) → A(g) ΔHvap = 30.7 kJ w = −PΔV = −ΔnRT, where Δn = nproducts − nreactants = 1 − 0 = 1 w = −(1 mol)(8.3145 J K−1 mol−1)(80. + 273 K) = −2940 J; because pressure is constant: ΔE = qp + w = ΔH + w = 30.7 kJ + (−2.94 kJ) = 27.8 kJ Hess’s Law 51. ΔH = 2(−394 kJ) 2 C + 2 O2 → 2 CO2 ΔH = −286 kJ H2 + 1/2 O2 → H2O ΔH = − (−1300.kJ) 2 CO2 + H2O → C2H2 + 5/2 O2 ________________________________________________ ΔH = 226 kJ 2 C(s) + H2(g) → C2H2(g) CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY 353 Note: The enthalpy change for a reaction that is reversed is the negative quantity of the enthalpy change for the original reaction. If the coefficients in a balanced reaction are multiplied by an integer, then the value of ΔH is multiplied by the same integer. 52. 53. ΔH = 1/2 (167.4 kJ) ClF + 1/2 O2 → 1/2 Cl2O + 1/2 F2O ΔH = !1/2 (341.4 kJ) 1/2 Cl2O + 3/2 F2O → ClF3 + O2 ΔH = 1/2 (!43.4 kJ) F2 + 1/2 O2 → F2O ____________________________________________________________ ΔH = !108.7 kJ ClF(g) + F2(g) → ClF3(g) CaC2 → CaO + H2O → 2 CO2 + H2O → Ca + 1/2 O2 → 2 C + 2 O2 → Ca + 2 C Ca(OH)2 C2H2 + 5/2 O2 CaO 2 CO2 ΔH = !(!62.8 kJ) ΔH = !653.1 kJ ΔH = !(!1300. kJ) ΔH = !635.5 kJ ΔH = 2(!393.5 kJ) ______________________________________________________________________________________________ CaC2(s) + 2 H2O(l) → Ca(OH)2(aq) + C2H2(g) 54. ΔH = !713 kJ To avoid fractions, let's first calculate ΔH for the reaction: 6 FeO(s) + 6 CO(g) → 6 Fe(s) + 6 CO2(g) ΔH = −2(18 kJ) 6 FeO + 2 CO2 → 2 Fe3O4 + 2 CO 2 Fe3O4 + CO2 → 3 Fe2O3 + CO ΔH = − (−39 kJ) ΔH = 3(-23 kJ) 3 Fe2O3 + 9 CO → 6 Fe + 9 CO2 ____________________________________________________ 6 FeO(s) + 6 CO(g) → 6 Fe(s) + 6 CO2(g) ΔH = −66 kJ 55. − 66 kJ = −11 kJ 6 So for: FeO(s) + CO(g) → Fe(s) + CO2(g) ΔH = C4H4(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l) ΔHcomb = −2341 kJ C4H8(g) + 6 O2(g) → 4 CO2(g) + 4 H2O(l) ΔHcomb = −2755 kJ H2(g) + 1/2 O2(g) → H2O(l) ΔHcomb = −286 kJ By convention, H2O(l) is produced when enthalpies of combustion are given, and because per mole quantities are given, the combustion reaction refers to 1 mole of that quantity reacting with O2(g). Using Hess’s Law to solve: ΔH1 = −2341 kJ C4H4(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l) 4 CO2(g) + 4 H2O(l) → C4H8(g) + 6 O2(g) ΔH2 = − (−2755 kJ) ΔH3 = 2(−286 kJ) 2 H2(g) + O2(g) → 2 H2O(l) ___________________________________________________________________ ΔH = ΔH1 + ΔH2 + ΔH3 = −158 kJ C4H4(g) + 2 H2(g) → C4H8(g) 354 CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY 56. NO + O3 → NO2 + O2 ΔH = −199 kJ ΔH = −1/2 (−427 kJ) 3/2 O2 → O3 ΔH = −1/2 (495 kJ) O → 1/2 O2 ______________________________________________________ NO(g) + O(g) → NO2(g) ΔH = −233 kJ 57. ΔH = 177.4 kJ C6H4(OH)2 → C6H4O2 + H2 H2O2 → H2 + O2 ΔH = −(−191.2 kJ) ΔH = 2(−241.8 kJ) 2 H2 + O2 → 2 H2O(g) ΔH = 2(−43.8 kJ) 2 H2O(g) → 2 H2O(l) ________________________________________________________________ C6H4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2 H2O(l) ΔH = −202.6 kJ 58. P4O10 → P4 + 5 O2 ΔH = −(−2967.3 kJ) ΔH = 10(−285.7 kJ) 10 PCl3 + 5 O2 → 10 Cl3PO ΔH = −6(−84.2 kJ) 6 PCl5 → 6 PCl3 + 6 Cl2 P4 + 6 Cl2 → 4 PCl3 ΔH = −1225.6 kJ _______________________________________________________ P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g) ΔH = −610.1 kJ 59. ΔH = −4(46 kJ) 2 N2(g) + 6 H2(g) → 4 NH3(g) ΔH = −3(-484 kJ) 6 H2O(g) → 6 H2(g) + 3 O2(g) __________________________________________________ 2 N2(g) + 6 H2O(g) → 3 O2(g) + 4 NH3(g) ΔH = 1268 kJ No, because the reaction is very endothermic (requires a lot of heat to react), it would not be a practical way of making ammonia because of the high energy costs required. Standard Enthalpies of Formation 60. The change in enthalpy that accompanies the formation of one mole of a compound from its elements, with all substances in their standard states, is the standard enthalpy of formation for a compound. The reactions that refer to ΔH of are: Na(s) + 1/2 Cl2(g) → NaCl(s); H2(g) + 1/2 O2(g) → H2O(l) 6 C(graphite, s) + 6 H2(g) + 3 O2(g) → C6H12O6(s); Pb(s) + S(s) + 2 O2(g) → PbSO4(s) 61. In general: ΔH° = ∑ n p ΔH of , products − ∑ n r ΔH of, reactants , and all elements in their standard state have ΔH of = 0 by definition. a. The balanced equation is: 2 NH3(g) + 3 O2(g) + 2 CH4(g) → 2 HCN(g) + 6 H2O(g) CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY 355 ΔH° = (2 mol HCN × ΔH of , HCN + 6 mol H2O(g) H ΔH of , H 2O ) − (2 mol NH3 × ΔH of , NH 3 + 2 mol CH4 × ΔH of , CH 4 ) ΔH° = [2(135.1) + 6(−242)] − [2(−46) + 2(−75)] = −940. kJ b. Ca3(PO4)2(s) + 3 H2SO4(l) → 3 CaSO4(s) + 2 H3PO4(l) ⎡ ⎛ − 1433 kJ ⎞ ⎛ − 1267 kJ ⎞⎤ ΔH° = ⎢3 mol CaSO 4 (s)⎜ ⎟ + 2 mol H 3 PO 4 (l)⎜ ⎟⎥ ⎝ mol ⎠ ⎝ mol ⎠⎦ ⎣ ⎡ ⎛ − 814 kJ ⎞⎤ ⎛ − 4126 kJ ⎞ − ⎢1 mol Ca 3 (PO 4 ) 2 (s)⎜ ⎟⎥ ⎟ + 3 mol H 2SO 4 (l)⎜ ⎝ mol ⎠⎦ ⎝ mol ⎠ ⎣ ΔH° = −6833 kJ − (-6568 kJ) = −265 kJ c. NH3(g) + HCl(g) → NH4Cl(s) ΔH° = (1 mol NH4Cl × ΔH of , NH 4 Cl ) − (1 mol NH3 × ΔH of , NH 3 + 1 mol HCl × ΔH of , HCl ) ⎡ ⎡ ⎛ − 314 kJ ⎞⎤ ⎛ − 46 kJ ⎞ ⎛ − 92 kJ ⎞⎤ ΔH° = ⎢1 mol ⎜ ⎟⎥ − ⎢1 mol ⎜ ⎟ + 1 mol ⎜ ⎟⎥ ⎝ mol ⎠⎦ ⎝ mol ⎠ ⎝ mol ⎠⎦ ⎣ ⎣ ΔH° = −314 kJ + 138 kJ = −176 kJ d. The balanced equation is: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) ⎡ ⎡ ⎛ − 242 kJ ⎞ ⎤ ⎛ − 278 kJ ⎞⎤ ⎛ − 393.5 kJ ⎞ ΔH° = ⎢2 mol ⎜ ⎟ ⎥ − ⎢1 mol ⎜ ⎟⎥ ⎟ + 3 mol ⎜ ⎝ mol ⎠ ⎦ ⎝ mol ⎠⎦ ⎝ mol ⎠ ⎣ ⎣ ΔH° = −1513 kJ − (−278 kJ) = −1235 kJ e. SiCl4(l) + 2 H2O(l) → SiO2(s) + 4 HCl(aq) Because HCl(aq) is H+(aq) + Cl−(aq), ΔH of = 0 - 167 = -167 kJ/mol. ⎡ ⎛ − 167 kJ ⎞ ⎛ − 911 kJ ⎞ ⎤ ΔH° = ⎢4 mol ⎜ ⎟ + 1 mol ⎜ ⎟⎥ ⎝ mol ⎠ ⎝ mol ⎠ ⎦ ⎣ ⎡ ⎛ − 687 kJ ⎞ ⎛ − 286 kJ ⎞ ⎤ − ⎢1 mol ⎜ ⎟ + 2 mol ⎜ ⎟⎥ ⎝ mol ⎠ ⎝ mol ⎠ ⎦ ⎣ ΔH° = −1579 kJ − (−1259 kJ) = −320. kJ 356 CHAPTER 9 f. ENERGY, ENTHALPY, AND THERMOCHEMISTRY MgO(s) + H2O(l) → Mg(OH)2(s) ⎡ ⎛ − 925 kJ ⎞ ⎤ ⎡ ⎛ − 286 kJ ⎞ ⎤ ⎛ − 602 kJ ⎞ ΔH° = ⎢1 mol ⎜ ⎟ ⎥ − ⎢1 mol ⎜ ⎟⎥ ⎟ + 1 mol ⎜ ⎝ mol ⎠ ⎦ ⎣ ⎝ mol ⎠ ⎦ ⎝ mol ⎠ ⎣ ΔH° = −925 kJ − (−888 kJ) = −37 kJ 62. a. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g); ΔH° = ∑ n p ΔH of , products − ∑ n r ΔH of, reactants ⎡ ⎛ − 46 kJ ⎞⎤ ⎛ − 242 kJ ⎞ ⎤ ⎡ ⎛ 90. kJ ⎞ ΔH° = ⎢4 mol ⎜ ⎟⎥ = −908 kJ ⎟ ⎥ − ⎢4 mol ⎜ ⎟ + 6 mol ⎜ ⎝ mol ⎠⎦ ⎝ mol ⎠ ⎦ ⎣ ⎝ mol ⎠ ⎣ 2 NO(g) + O2(g) → 2 NO2(g) ⎡ ⎛ 34 kJ ⎞ ⎤ ⎡ ⎛ 90. kJ ⎞⎤ ΔH° = ⎢2 mol ⎜ ⎟ ⎥ − ⎢2 mol ⎜ ⎟⎥ = −112 kJ ⎝ mol ⎠ ⎦ ⎣ ⎝ mol ⎠⎦ ⎣ 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) ⎡ ⎛ − 207 kJ ⎞ ⎛ 90. kJ ⎞⎤ ΔH° = ⎢2 mol ⎜ ⎟ + 1 mol ⎜ ⎟⎥ ⎝ mol ⎠ ⎝ mol ⎠⎦ ⎣ ⎡ ⎛ 34 kJ ⎞ ⎛ − 286 kJ ⎞ ⎤ − ⎢3 mol ⎜ ⎟ + 1 mol ⎜ ⎟ ⎥ = −140. kJ ⎝ mol ⎠ ⎝ mol ⎠ ⎦ ⎣ Note: All ΔH of values are assumed ± 1 kJ. b. 12 NH3(g) + 15 O2(g) → 12 NO(g) + 18 H2O(g) 12 NO(g) + 6 O2(g) → 12 NO2(g) 12 NO2(g) + 4 H2O(l) → 8 HNO3(aq) + 4 NO(g) 4 H2O(g) → 4 H2O(l) _________________________________________________ 12 NH3(g) + 21 O2(g) → 8 HNO3(aq) + 4 NO(g) + 14 H2O(g) The overall reaction is exothermic since each step is exothermic. 63. ⎛ − 416 kJ ⎞ 4 Na(s) + O2(g) → 2 Na2O(s), ΔH° = 2 mol ⎜ ⎟ = −832 kJ ⎝ mol ⎠ 2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g) ⎡ ⎛ − 470. kJ ⎞ ⎤ ⎡ ⎛ − 286 kJ ⎞⎤ ΔH° = ⎢2 mol ⎜ ⎟ ⎥ − ⎢2 mol ⎜ ⎟⎥ = −368 kJ ⎝ mol ⎠ ⎦ ⎣ ⎝ mol ⎠⎦ ⎣ CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY 357 2Na(s) + CO2(g) → Na2O(s) + CO(g) ⎡ ⎛ − 393.5 kJ ⎞⎤ ⎛ − 110.5 kJ ⎞⎤ ⎡ ⎛ − 416 kJ ⎞ ΔH° = ⎢1 mol ⎜ ⎟⎥ = −133 kJ ⎟⎥ − ⎢1 mol ⎜ ⎟ + 1 mol ⎜ ⎝ mol ⎠⎦ ⎝ mol ⎠⎦ ⎣ ⎝ mol ⎠ ⎣ In reactions 2 and 3, sodium metal reacts with the "extinguishing agent." Both reactions are exothermic and each reaction produces a flammable gas, H2 and CO, respectively. 64. 3 Al(s) + 3 NH4ClO4(s) → Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g) ⎡ ⎛ − 1676 kJ ⎞⎤ ⎛ − 704 kJ ⎞ ⎛ − 242 kJ ⎞ ⎛ 90. kJ ⎞ ΔH° = ⎢6 mol ⎜ ⎟⎥ ⎟ + 1 mol ⎜ ⎟ + 3 mol ⎜ ⎟ + 1 mol ⎜ ⎝ mol ⎠⎦ ⎝ mol ⎠ ⎝ mol ⎠ ⎝ mol ⎠ ⎣ ⎡ ⎛ − 295 kJ ⎞ ⎤ − ⎢3 mol ⎜ ⎟ ⎥ = −2677 kJ ⎝ mol ⎠ ⎦ ⎣ 65. 5 N2O4(l) + 4 N2H3CH3(l) → 12 H2O(g) + 9 N2(g) + 4 CO2(g) ⎡ ⎛ − 393.5 kJ ⎞⎤ ⎛ − 242 kJ ⎞ ΔH° = ⎢12 mol ⎜ ⎟⎥ ⎟ + 4 mol ⎜ mol ⎝ mol ⎠⎦ ⎝ ⎠ ⎣ ⎡ ⎛ − 20. kJ ⎞ ⎛ 54 kJ ⎞ ⎤ − ⎢5 mol ⎜ ⎟ + 4 mol ⎜ ⎟ ⎥ = −4594 kJ ⎝ mol ⎠ ⎝ mol ⎠ ⎦ ⎣ 66. For Exercise 64, a mixture of 3 mol Al and 3 mol NH4ClO4 yields 2677 kJ of energy. The mass of the stoichiometric reactant mixture is: 26.98 g ⎞ ⎛ 117.49 g ⎞ ⎛ ⎜ 3 mol × ⎟ + ⎜ 3 mol × ⎟ = 433.41 g mol ⎠ ⎝ mol ⎠ ⎝ For 1.000 kg of fuel: 1.000 × 103 g × − 2677 kJ = −6177 kJ 433.41 g In Exercise 65, we get 4594 kJ of energy from 5 mol of N2O4 and 4 mol of N2H3CH3. The 92.02 g ⎞ ⎛ 46.08 g ⎞ ⎛ mass is: ⎜ 5 mol × ⎟ + ⎜ 4 mol × ⎟ = 644.42 kJ mol ⎠ ⎝ mol ⎠ ⎝ For 1.000 kg of fuel: 1.000 × 103 g × − 4594 kJ = −7129 kJ 644.42 g Thus we get more energy per kilogram from the N2O4/N2H3CH3 mixture. 358 67. CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY a. ΔH° = 3 mol(227 kJ/mol) − 1 mol(49 kJ/mol) = 632 kJ b. Because 3 C2H2(g) is higher in energy than C6H6(l), acetylene will release more energy per gram when burned in air. 68. 69. a. C2H4(g) + O3(g) → CH3CHO(g) + O2(g) ΔH° = −166 kJ − (143 kJ + 52 kJ) = −361 kJ b. O3(g) + NO(g) → NO2(g) + O2(g) ΔH° = 34 kJ − (90. kJ + 143 kJ) = −199 kJ c. SO3(g) + H2O(l) → H2SO4(aq) ΔH° = −909 kJ − [−396 kJ + (-286 kJ)] = −227 kJ d. 2 NO(g) + O2(g) → 2 NO2(g) ΔH° = 2(34) kJ − 2(90.) kJ = −112 kJ 2 ClF3(g) + 2 NH3(g) → N2(g) + 6 HF(g) + Cl2(g) ΔH° = (6 ΔH of , HF ) − (2 ΔH of , ClF3 + 2ΔH of , NH 3 ΔH° = −1196 kJ ) ⎛ − 271 kJ ⎞ ⎛ − 46 kJ ⎞ o −1196 kJ = 6 mol ⎜ ⎟ − 2 ΔH f , ClF3 − 2 mol ⎜ ⎟ ⎝ mol ⎠ ⎝ mol ⎠ −1196 kJ = −1626 kJ − 2 ΔH fo, ClF3 + 92 kJ, ΔH fo, ClF3 = 70. C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l) (−1626 + 92 + 1196) kJ − 169 kJ = 2 mol mol ΔH° = −1411 kJ ΔH° = −1411.1 kJ = 2(−393.5) kJ + 2(−285.8) kJ − ΔH fo, C 2 H 4 −1411.1 kJ = −1358.6 kJ − ΔH fo, C 2 H 4 , ΔH fo, C 2 H 4 = 52.5 kJ/mol Energy Consumption and Sources 71. Mass of H2O = 1.00 gal × 3.785 L 1000 mL 1.00 g × × = 3790 g H2O gal L mL Energy required (theoretical) = s × m × ΔT = 4.18 J × 3790 g × 10.0 °C = 1.58 × 105 J o Cg For an actual (80.0% efficient) process, more than this quantity of energy is needed since heat is always lost in any transfer of energy. The energy required is: 1.58 × 105 J × 100 . J = 1.98 × 105 J 80 .0 J Mass of C2H2 = 1.98 × 105 J × 1 mol C 2 H 2 26.04 g C 2 H 2 × = 3.97 g C2H2 3 mol C 2 H 2 1300. × 10 J CHAPTER 9 72. ENERGY, ENTHALPY, AND THERMOCHEMISTRY 359 C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) ΔH° = [2(−393.5 kJ) + 3(−286 kJ)] − (−278 kJ) = −1367 kJ/mol ethanol − 1367 kJ 1 mol × = −29.67 kJ/g mol 46.068 g 73. CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l) ΔH° = [!393.5 kJ + 2(!286 kJ)] ! (!239 kJ) = !727 kJ/mol CH3OH 1 mol − 727 kJ × = !22.7 kJ/g versus !29.67 kJ/g for ethanol mol 32.04 g Ethanol has a higher fuel value than methanol. 74. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ΔH° = [3(−393.5 kJ) + 4(−286 kJ)] − (−104 kJ) = −2221 kJ/mol C3H8 − 2221 kJ 1 mol − 50.37 kJ = versus −47.7 kJ/g for octane (Example 9.8) × mol 44.096 g g The two fuel values are very close. An advantage of propane is that it burns more cleanly. The boiling point of propane is -42°C. Thus it is more difficult to store propane, and there are extra safety hazards associated with using high-pressure compressed-gas tanks. Additional Exercises 75. ΔEoverall = ΔEstep 1 + ΔEstep 2; this is a cyclic process, which means that the overall initial state and final state are the same. Because ΔE is a state function, ΔEoverall = 0 and ΔEstep 1 = −ΔEstep 2. ΔEstep 1 = q + w = 45 J + (−10. J) = 35 J ΔEstep 2 = −ΔEstep 1 = −35 J = q + w, −35 J = −60. J + w, w = 25 J 76. H = E + PV, ΔH = ΔE + Δ(PV), ΔE = ΔH − Δ(PV) Assume that H2O(g) is ideal. Δ(PV) = Δ(nRT) = RTΔn at constant T We go from n = 0 to n = 1, thus Δn = 1. [H2O(l) is a liquid.] ΔE = ΔH − (8.3145 J K−1 mol−1)(298 K)(1 mol) = ΔH − 2480 J ΔH° = −242 kJ/mol − (−286 kJ/mol) = 44 kJ/mol ΔE = 44 kJ − 2.48 kJ, ΔE = 41.52 kJ = 42 kJ 360 CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY 77. H2(g) + 1/2 O2(g) → H2O(l) ΔH° = ΔH of , H 2O ( l) = −285.8 kJ H2O(l) → H2(g) + 1/2 O2(g) ΔH° = 285.8 kJ ΔE° = ΔH° − PΔV = ΔH° − ΔnRT ⎛ 1 kJ ⎞ ⎟⎟ ΔE° = 285.8 kJ − (1.50 − 0 mol)(8.3145 J K−1 mol−1)(298 K) ⎜⎜ ⎝ 1000 J ⎠ ΔE° = 285.8 kJ − 3.72 kJ = 282.1 kJ 78. 400 kcal × 4.18 kJ = 1.67 × 103 kJ ≈ 2 × 103 kJ kcal ⎛ 2.54 cm 1m ⎞ 1 kg ⎞ 9.81 m ⎛ ⎟⎟ × ⎟ = 160 J ≈ 200 J PE = mgz = ⎜⎜180 lb × × ⎜⎜ 8 in × × 2 m 100 cm ⎟⎠ 2.205 lb ⎠ s ⎝ ⎝ 200 J of energy is needed to climb one step. The total number of steps to climb are: 2 × 106 J × 79. 1 step = 1 × 104 steps 200 J The specific heat of water is 4.18 J °C−1 g−1, which is equal to 4.18 kJ °C−1 kg−1 We have 1.00 kg of H2O, so: 1.00 kg × 4.18 kJ = 4.18 kJ/°C o C kg This is the portion of the heat capacity that can be attributed to H2O. Total heat capacity = Ccal + C H 2O , 80. Ccal = 10.84 − 4.18 = 6.66 kJ/°C Heat released = 1.056 g × 26.42 kJ/g = 27.90 kJ = heat gain by water and calorimeter ⎛ 4.18 kJ ⎞ ⎛ 6.66 kJ ⎞ × 0.987 kg × ΔT ⎟⎟ + ⎜ o × ΔT ⎟ Heat gain = 27.90 kJ = ⎜⎜ o C ⎠ ⎝ C kg ⎠ ⎝ 27.90 = (4.13 + 6.66)ΔT = (10.79)ΔT, ΔT = 2.586°C 2.586°C = Tf − 23.32°C, Tf = 25.91°C 81. Na2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2 NaNO3(aq) ΔH = ? CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY 1.00 L × 361 2.00 mol 0.750 mol = 2.00 mol Na2SO4; 2.00 L × = 1.50 mol Ba(NO3)2 L L The balanced equation requires a 1 : 1 mole ratio between Na2SO4 and Ba(NO3)2. Because we have fewer moles of Ba(NO3)2 present, it is limiting and 1.50 mol BaSO4 will be produced [there is a 1 : 1 mole ratio between Ba(NO3)2 and BaSO4]. Heat gain by solution = heat loss by reaction 1000 mL 2.00 g × = 6.00 × 103 g 1L mL 6.37 J Heat gain by solution = o × 6.00 × 103 g × (42.0 ! 30.0)EC = 4.59 × 105 J Cg Mass of solution = 3.00 L × Because the solution gained heat, the reaction is exothermic; q = !4.59 × 105 J for the reaction. ΔH = − 4.59 × 105 J = !3.06 × 105 J/mol = !306 kJ/mol 1.50 mol BaSO 4 82. |qsurr| = |qsolution + qcal|; we normally assume that qcal is zero (no heat gain/loss by the calorimeter). However, if the calorimeter has a nonzero heat capacity, then some of the heat absorbed by the endothermic reaction came from the calorimeter. If we ignore qcal, then qsurr is too small, giving a calculated ΔH value that is less positive (smaller) than it should be. 83. HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) ΔH = !56 kJ 0.2000 L × 0.400 mol HCl = 8.00 × 10 −2 mol HCl L 0.1500 L × 0.500 mol NaOH = 7.50 × 10 −2 mol NaOH L Because the balanced reaction requires a 1 : 1 mole ratio between HCl and NaOH, and because fewer moles of NaOH are actually present as compared with HCl, NaOH is the limiting reagent. 7.50 × 10 −2 mol NaOH × 84. − 56 kJ = !4.2 kJ; 4.2 kJ of heat is released. mol NaOH a. 2 SO2(g) + O2(g) → 2 SO3(g); w = −PΔV; because the volume of the piston apparatus decreased as reactants were converted to products (ΔV < 0), w is positive (w > 0). b. COCl2(g) → CO(g) + Cl2(g); because the volume increased (ΔV > 0), w is negative (w < 0). c. N2(g) + O2(g) → 2 NO(g); because the volume did not change (ΔV = 0), no PV work is done (w = 0). 362 CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY In order to predict the sign of w for a reaction, compare the coefficients of all the product gases in the balanced equation to the coefficients of all the reactant gases. When a balanced reaction has more moles of product gases than moles of reactant gases (as in b), the reaction will expand in volume (ΔV positive), and the system does work on the surroundings. When a balanced reaction has a decrease in the moles of gas from reactants to products (as in a), the reaction will contract in volume (ΔV negative), and the surroundings will do compression work on the system. When there is no change in the moles of gas from reactants to products (as in c), ΔV = 0 and w = 0. 85. w = −PΔV; Δn = moles of gaseous products − moles of gaseous reactants. Only gases can do PV work (we ignore solids and liquids). When a balanced reaction has more moles of product gases than moles of reactant gases (Δn positive), the reaction will expand in volume (ΔV positive), and the system will do work on the surroundings. For example, in reaction c, Δn = 2 − 0 = 2 moles, and this reaction would do expansion work against the surroundings. When a balanced reaction has a decrease in the moles of gas from reactants to products (Δn negative), the reaction will contract in volume (ΔV negative), and the surroundings will do compression work on the system, e.g., reaction a, where Δn = 0 − 1 = −1. When there is no change in the moles of gas from reactants to products, ΔV = 0 and w = 0, e.g., reaction b, where Δn = 2 − 2 = 0. When ΔV > 0 (Δn > 0), then w < 0, and the system does work on the surroundings (c and e). When ΔV < 0 (Δn < 0), then w > 0, and the surroundings do work on the system (a and d). When ΔV = 0 (Δn = 0), then w = 0 (b). 86. The specific heat capacities are: 0.89 J °C−1 g−1 (Al) and 0.45 J °C−1 g−1 (Fe). Al would be the better choice. It has a higher heat capacity and a lower density than Fe. Using Al, the same amount of heat could be dissipated by a smaller mass, keeping the mass of the amplifier down. 87. a. aluminum oxide = Al2O3; 2 Al(s) + 3/2 O2(g) → Al2O3(s) b. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) c. Ba(OH)2(aq) + 2 HCl(aq) → 2 H2O(l) + BaCl2(aq) d. 2 C(graphite, s) + 3/2 H2(g) + 1/2 Cl2(g) → C2H3Cl(g) e. C6H6(l) + 15/2 O2(g) → 6 CO2(g) + 3 H2O(l) Note: ΔHcomb values assume 1 mole of compound combusted. f. NH4Br(s) → NH4+(aq) + Br-(aq) Challenge Problems 88. If the gas is monoatomic: Cv = 3 5 R = 12.47 J K−1 mol−1 and Cp = R = 20.79 J K−1 mol−1 2 2 CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY 363 If the gas is behaving ideally, then Cp − Cv = R = 8.3145 J K−1 mol−1. At constant volume: qv = 2079 J = nCvΔT Cv = 2079 J 2079 J = = 20.79 J K−1 mol−1 nΔT (1 mol)(400.0 − 300.0 K ) Because Cv ≠ 3/2 R = 12.47 J, the gas is not a monoatomic gas. At constant pressure: qp = nCpΔT qp = ΔE − w = 1305 J − (−150. J) = 1455 J (Gas expansion, so system does work on surroundings.) Cp = qp nΔT = 1455 J = 29.1 J K−1 mol−1 (1 mol)(600.0 − 550.0 K ) Cp − Cv = 29.1 − 20.79 = 8.3 J K−1 mol−1 = R The gas is behaving ideally since Cp − Cv = R. 89. H2O(s) → H2O(l) ΔH = ΔHfus; for 1 mol of supercooled water at −15.0°C (or 258.2 K), ΔHfus, 258.2 K = 10.9 kJ/2.00 mol = 5.45 kJ/mol. Using Hess’s law and the equation ΔH = nCpΔT: H2O(s, 273.2 K) → H2O(s, 258.2 K) H2O(s, 258.2 K) → H2O(l, 258.2 K) H2O(l, 258.2 K) → H2O(l, 273.2 K) ΔH1 = 1 mol(37.5 J K−1 mol−1)(−15.0 K) = −563 J = −0.563 kJ ΔH2 = 1 mol(5.45 kJ/mol) = 5.45 kJ ΔH3 = 1 mol(75.3 J K−1 mol−1)(15.0 K) = 1130 J = 1.13 kJ ______________________________________________________________________________________________________________ H2O(s, 273. 2 K) → H2O(l, 273.2 K) ΔHfus, 273.2 = ΔH1 + ΔH2 + ΔH3 ΔHfus, 273.2 = −0.563 kJ + 5.45 kJ + 1.13 kJ, ΔHfus, 273.2 = 6.02 kJ/mol 90. Energy needed = 20. × 103 g C12 H 22 O11 1 mol C12 H 22 O11 5640 kJ × × = 3.3 × 105 kJ/h h 342.3 g C12 H 22 O11 mol Energy from sun = 1.0 kW/m2 = 1000 W/m2 = 10,000 m2 × 1000 J 1.0 kJ = s m2 s m2 1.0 kJ 60 s 60 min × × = 3.6 × 107 kJ/h 2 min h sm Percent efficiency = 3.3 × 105 kJ energy used per hour × 100 = × 100 = 0.92% total energy per hour 3.6 × 10 7 kJ 364 91. CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY Molar heat capacity of H2O(l) = 4.184 J K−1 g−1(18.015 g/mol) = 75.37 J K−1 mol−1 Molar heat capacity of H2O(g) = 2.02 J K−1 g−1(18.015 g/mol) = 36.4 J K−1 mol−1 Using Hess’s law and the equation ΔH = nCpΔT: H2O(l, 298.2 K) → H2O(l, 373.2 K) ΔH1 = 1 mol(75.37 J K−1 mol−1)(75.0 K)(1 kJ/1000 J) = 5.65 kJ H2O(l, 373.2 K) → H2O(g, 373.2 K) ΔH2 = 1 mol(40.66 kJ/mol) = 40.66 kJ H2O(g, 373.2 K) → H2O(g, 298.2 K) ΔH3 = 1 mol(36.4 J K−1 mol−1)(−75.0 K)(1 kJ/1000 J) = −2.73 kJ __________________________________________________________________________ H2O(l, 298.2 K) → H2O(g, 298.2 K) ΔH vap, 298.2 K = ΔH1 + ΔH2 + ΔH3 = 43.58 kJ/mol Using ΔH of values in Appendix 4 (which are determined at 25° C): ΔH°vap = −242 kJ − (−286 kJ) = 44 kJ To two significant figures, the two calculated ΔHvap values agree (as they should). 92. a. Using Hess's law and the equation ΔH = nCpΔT: ΔH1 = -83.3 kJ CH3Cl(248°C) + H2(248°C) → CH4(248°C) + HCl(248°C) CH3Cl(25°C) → CH3Cl(248°C) ΔH2 = 1 mol(48.5 J °C−1 mol−1)(223°C) = 10,800 J = 10.8 kJ ΔH3 = 1(28.9)(223)(1 kJ/1000 J) = 6.44 kJ H2(25°C) → H2(248°C) CH4(248°C) → CH4(25°C) ΔH4 = 1(41.3)(−223)(1/1000) = −9.21 kJ HCl(248°C) → HCl(25°C) ΔH5 = 1(29.1)(−223)(1/1000) = −6.49 kJ ____________________________________________________________________________ CH3Cl(25°C) + H2(25°C) → CH4(25°C) + HCl(25°C) ΔH° = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5 ΔH° = −83.3 kJ + 10.8 kJ + 6.44 kJ − 9.21 kJ − 6.49 kJ = −81.8 kJ b. ΔH° = (ΔH of , CH 4 + ΔH of , HCl ) − ( ΔH of , CH 3Cl + ΔH of , H 2 ) −81.8 kJ = (−75 kJ − 92 kJ) − (ΔH of , CH 3Cl + 0), ΔH of , CH 3Cl = −85 kJ/mol 93. Energy used in 8.0 hours = 40. kWh = Energy from the sun in 8.0 hours = 40. kJ h 3600 s × = 1.4 × 105 kJ s h 10. kJ 60 s 60 min × × × 8.0 h = 2.9 × 104 kJ/m2 2 min h sm Only 13% of the sunlight is converted into electricity: 0.13 × (2.9 × 104 kJ/m2) × area = 1.4 × 105 kJ, area = 37 m2 CHAPTER 9 94. ENERGY, ENTHALPY, AND THERMOCHEMISTRY 365 a. The gas will flow out of the 4.0 L bulb into the 20.0-L bulb. Eventually, the gas will be evenly dispersed throughout the two bulbs. The pressure will be the same inside both bulbs. The moles of gas (n) and the temperature (T) are constant. However, as the gas expands from 4.0 L to 24.0 L, the pressure will decrease. Pinitial = Pi = nRT 2.4 mol(0.08206 L atm K −1 mol −1 )(305 K ) = = 15 atm Vi 4.0 L Pfinal = Pf = Pi Vi 15 atm(4.0 L) = = 2.5 atm Vf 24.0 L b. Assuming the gas behaves ideally, then ΔE = nCvΔT and ΔH = nCpΔT. Because ΔT = 0, ΔE = 0 and ΔH = 0. From ΔE = q + w = 0, q = −w. Here we have an expansion of a gas where w = −PexΔV. The gas is expanding against a vacuum (Pex = 0), so w = 0. We call this a free expansion of a gas. Work can only occur in the expansion of a gas when the gas expands against a certain nonzero external pressure. Because w = 0, q = −w = 0. For a free expansion of an ideal gas at a constant temperature, ΔE = 0, ΔH = 0, q = 0, and w = 0. If the expansion occurs against some nonzero constant external pressure, then ΔE = 0, ΔH = 0, q = −w = PexΔV. c. The driving force is the natural tendency of processes to spontaneously proceed toward states that have the highest probability of existing. In this problem the gas mixed evenly throughout both bulbs is the most probable (likely) state to occur. This driving force is related to entropy, which will be discussed in detail in Chapter 10. 95. For an isothermal (constant T) process involving the expansion or compression of a gas, ΔE = nCvΔT = 0 (ΔH is also zero). Because ΔE = q + w = 0, q = −w = −(−PexΔV). So if the expansion or compression occurs against some nonzero external pressure, w ≠ 0 and q ≠ 0. Instead, q = −w = PexΔV (if the gas expands or contracts against some constant pressure). 96. There are five parts to this problem. We need to calculate: 1. q required to heat H2O(s) from !30.EC to 0EC; use the specific heat capacity of H2O(s) 2. q required to convert 1 mol H2O(s) at 0EC into 1 mol H2O(l) at 0EC; use ΔHfusion 3. q required to heat H2O(l) from 0EC to 100.EC; use the specific heat capacity of H2O(l) 4. q required to convert 1 mol H2O(l) at 100.EC into 1 mol H2O(g) at 100.EC; use ΔHvaporization 5. q required to heat H2O(g) from 100.EC to 140.EC; use the specific heat capacity of H2O(g) We will sum up the heat required for all five parts, and this will be the total amount of heat required to convert 1.00 mol of H2O(s) at !30.EC to H2O(g) at 140.EC. 366 CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY q1 = 2.03 J EC−1 g−1 × 18.02 g × [0 – (!30.)]EC = 1.1 × 103 J q2 = 1.00 mol × 6.01 × 103 J/mol = 6.01 × 103 J q3 = 4.18 J °C−1 g−1 × 18.02 g × (100. – 0)EC = 7.53 × 103 J q4 = 1.00 mol × 40.7 × 104 J/mol = 4.07 × 104 J q5 = 2.02 J °C−1 g−1 × 18.02 g × (140. – 100.)EC = 1.5 × 103 J qtotal = q1 + q2 + q3 + q4 + q5 = 5.68 × 104 J = 56.8 kJ Marathon Problems 97. X → CO2(g) + H2O(l) + O2(g) + A(g) ΔH = −1893 kJ/mol (unbalanced) To determine X, we must determine the moles of X reacted, the identity of A, and the moles of A produced. For the reaction at constant P (ΔH = q): − q H 2O = q rxn = -4.184 J °C-1 g-1(1.000 × 104 g)(29.52 - 25.00 °C)(1 kJ/1000 J) qrxn = −189.1 kJ (carrying extra significant figsures) Because ΔH = −1893 kJ/mol for the decomposition reaction, and because only -189.1 kJ of heat was released for this reaction, 189.1 kJ × (1 mol X/1893 kJ) = 0.100 mol X were reacted. Molar mass of X = 22.7 g X = 227 g/mol 0.100 mol X From the problem, 0.100 mol X produced 0.300 mol CO2, 0.250 mol H2O, and 0.025 mol O2. Therefore, 1.00 mol X contains 3.00 mol CO2, 2.50 mol H2O, and 0.25 mol O2. ⎛ 44.0 g ⎞ ⎛ 18.0 g ⎞ 1.00 mol X = 227 g = 3.00 mol CO2 ⎜ ⎟ + 2.50 mol H2O ⎜ ⎟ ⎝ mol ⎠ ⎝ mol ⎠ ⎛ 32.0 g ⎞ + 0.25 mol O2 ⎜ ⎟ + (mass of A) ⎝ mol ⎠ Mass of A in 1.00 mol X = 227 g − 132 g − 45.0 g − 8.0 g = 42 g A To determine A, we need the moles of A produced. The total moles of gas produced can be determined from the gas law data provided in the problem. Because H2O(l) is a product, we need to subtract PH 2O from the total pressure. ntotal = PV ; Ptotal = Pgases + PH 2O ; Pgases = 778 torr − 31 torr = 747 torr RT CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY 367 ⎛ 1L ⎞ ⎟ = 12.0 L V = height × area; area = πr2; V = (59.8 cm)(π)(8.00 cm)2 ⎜⎜ 3 ⎟ ⎝ 1000 cm ⎠ T = 273.15 + 29.52 = 302.67 K ⎛ 1 atm ⎞ ⎟ (12.0 L) 747 torr⎜⎜ 760 torr ⎟⎠ PV ⎝ ntotal = = = 0.475 mol = mol CO2 + mol O2 + mol A 0.08206 L atm RT (302.67 K ) K mol Mol A = 0.475 mol total − 0.300 mol CO2 − 0.025 mol O2 = 0.150 mol A Because 0.100 mol X reacted, 1.00 mol X would contain 1.50 mol A, which from a previous calculation represents 42 g A. Molar mass of A = 42 g A = 28 g/mol 1.50 mol A Because A is a gaseous element, the only element that is a gas and has this molar mass is N2(g). Thus A = N2(g). a. Now we can determine the formula of X. X → 3 CO2(g) + 2.5 H2O(l) + 0.25 O2(g) + 1.5 N2(g). For a balanced reaction, X = C3H5N3O9, which, for your information, is nitroglycerine. ⎛ 1 atm ⎞ ⎟⎟ (12.0 L − 0) = −12.3 L atm b. w = −PΔV = −778 torr ⎜⎜ ⎝ 760 torr ⎠ ⎛ 8.3145 J K −1 mol −1 −12.3 L atm ⎜⎜ −1 −1 ⎝ 0.08206 L atm K mol ⎞ ⎟ = −1250 J = −1.25 kJ, w = −1.25 kJ ⎟ ⎠ c. ΔE = q + w, where q = ΔH since at constant pressure. For 1 mol of X decomposed: w = −1.25 kJ/0.100 mol = −12.5 kJ/mol ΔE = ΔH + w = −1893 kJ/mol + (−12.5 kJ/mol) = −1906 kJ/mol ΔH of for C3H5N3O9 can be estimated from standard enthalpies of formation data and assuming ΔH rxn = ΔH orxn . For the balanced reaction given in part a, where ΔH orxn = −1893 kJ: −1893 kJ = (3ΔH of , CO 2 + 2.5ΔH of , H 2O + 0.25ΔH of , O 2 + 1.5ΔH of , N 2 ) − (ΔH of , C3H 5 N 3O9 ) −1893 kJ = [3(−393.5) kJ + 2.5(−286) kJ + 0 + 0] − ΔH of , C3H 5 N 3O9 ΔH of , C3H 5 N 3O9 = −2.5 kJ/mol = −3 kJ/mol 368 98. CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY ⎛ 2 x + y/2 ⎞ CxHy + ⎜ ⎟ Ο2 → x CO2 + y/2 H2O 2 ⎝ ⎠ [x(!393.5) + y/2 (!242)] ! ΔH oC x H y = !2044.5, !(393.5)x ! 121y ! ΔH C x H y = !2044.5 dgas = P • MM , where MM = average molar mass of CO2/H2O mixture RT 0.751 g/L = 1.00 atm × MM , MM of CO2/H2O mixture = 29.1 g/mol 0.08206 L atm × 473 K K mol Let a = mol CO2 and 1.00 ! a = mol H2O (assuming 1.00 total moles of mixture) (44.01)a + (1.00 ! a) × 18.02 = 29.1; solving: a = 0.426 mol CO2 ; mol H2O = 0.574 mol y 0.574 y 2 = , 2.69 = , y = (2.69)x Thus: 0.426 x x For whole numbers, multiply by three, which gives y = 8, x = 3. Note that y = 16, x = 6 is possible, along with other combinations. Because the hydrocarbon has a lower density than Kr, the molar mass of CxHy must be less than the molar mass of Kr (83.80 g/mol). Only C3H8 works. !2044.5 = !393.5(3) ! 121(8) ! ΔH oC3H8 , ΔH oC3H8 = !104 kJ/mol CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY Spontaneity and Entropy 12. a. A spontaneous process is one that occurs without any outside intervention. b. Entropy is how energy is distributed among energy levels in the “particles” that constitute a given system. Entropy is closely associated with probability, where the most probable arrangement (state) is the highest entropy state. c. Positional probability is a type of probability that depends on the number of arrangements in space that yield a particular state. d. The system is the portion of the universe in which we are interested. e. The surroundings are everything else in the universe besides the system. f. 13. The universe is everything; universe = system + surroundings. We draw all the possible arrangements of the two particles in the three levels. 2 kJ 1 kJ 0 kJ Total E = x xx x x x 0 kJ 1 kJ 2 kJ xx x x xx __ __ 2 kJ 3 kJ 4 kJ B A B A A B A B 1 kJ 1 kJ 2 kJ 2 kJ The most likely total energy is 2 kJ. 14. 2 kJ 1 kJ 0 kJ AB Etotal = 0 kJ AB AB 2 kJ 4 kJ B A A B 3 kJ 3 kJ The most likely total energy is 2 kJ. 15. Processes a, b, d, and g are spontaneous. Processes c, e, and f require an external source of energy in order to occur since they are nonspontaneous. 369 370 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 16. Of the three phases, gases have the greatest positional probability (greatest entropy), followed by liquids, with solids having the smallest positional disorder (smallest entropy). Thus a, b, c, e, and g involve an increase in entropy. All lead to an increase in positional probability. 17. a. Positional probability increases; there is a greater volume accessible to the randomly moving gas molecules, which increases disorder. b. The positional probability doesn't change. There is no change in volume and thus no change in the numbers of positions of the molecules. c. Positional probability decreases because the volume decreases (P and V are inversely related). 18. a. H2 at 100°C and 0.5 atm; higher temperature and lower pressure mean greater volume and hence greater positional probability. b. N2; N2 at STP has the greater volume. c. H2O(l) has greater positional probability than H2O(s). 19. There are more ways to roll a seven. We can consider all the possible throws by constructing a table. One die 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 Sum of the two dice There are six ways to get a seven, more than any other number. The seven is not favored by energy; rather, it is favored by probability. To change the probability, we would have to expend energy (do work). 20. Arrangement I: S = k ln Ω; Ω = 1; S = k ln 1 = 0 Arrangement II: Ω = 4; S = k ln 4 = (1.38 × 10−23 J/K) ln 4, S = 1.91 × 10−23 J/K Arrangement III: Ω = 6; S = k ln 6 = 2.47 × 10−23 J/K CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 371 Energy, Enthalpy, and Entropy Changes Involving Ideal Gases and Physical Changes 21. 1.00 × 103 g C2H6 × 1 mol = 33.3 mol 30.07 g qv = ΔE = nCvΔT = 33.3 mol(44.60 J K−1 mol−1)(48.4 K) = 7.19 × 104 J = 71.9 kJ At constant volume, 71.9 kJ of energy is required, and ΔE = 71.9 kJ. At constant pressure (assuming ethane acts as an ideal gas): Cp = Cv + R = 44.60 + 8.31 = 52.91 J K−1 mol−1 Energy required = qp = ΔH = nCpΔT = 33.3 mol(52.91 J K−1 mol−1)(48.4 K) = 8.53 × 104 J = 85.3 kJ For the constant pressure process, ΔE = 71.9 kJ, as calculated previously (ΔE is unchanged). 22. It takes nCpΔT amount of energy to carry out this process. The internal energy of the system increases by nCvΔT. So the fraction that goes into raising the internal energy is: nC v ΔT C 20.8 = v = = 0.715 nC p ΔT Cp 29.1 The remainder of the energy (PΔV = nRΔT) goes into expanding the gas against the constant pressure. 100.0 g N2 × 1 mol = 3.570 mol 28.014 g qv =ΔE = nCvΔT = 3.570 mol(20.8 J K−1 mol−1)(60.0 K) = 4.46 × 103 J = 4.46 kJ 23. Heat gain by He = heat loss by N2; because ΔT in °C = ΔT in K, the units on the heat capacities also could be J °C−1 mol−1. (0.400 mol)(12.5 J °C−1 mol−1)(Tf - 20.0 °C) = (0.600 mol)(20.7 J °C−1 mol−1)(100.0 °C − Tf) (5.00)Tf − 100. = 1240 − (12.4)Tf, Tf = 24. 1340 = 77.0 °C 17.4 The volumes for each step are: a. P1 = 5.00 atm, n = 1.00 mol, T = 350. K; V1 = b. P2 = 2.24 atm, V2 = nRT = 12.8 L P2 nRT = 5.74 L P1 c. P3 = 1.00 atm, V3 = 28.7 L 372 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY The process can be carried out in the following steps: (P1, V1) → (P2, V1) w = −PΔV = 0 (constant volume process) (P2, V1) → (P2, V2) w = −(2.24 atm)(12.8 − 5.74 L) = −16 L atm (P2, V2) → (P3, V2) w = 0 (P3, V2) → (P3, V3) w = −(1.00 atm)(28.7 − 12.8 L) = −15.9 L atm wtotal = −16 − 15.9 = −32 L atm; −32 L atm × 101.3 J = −3200 J = total work L atm wrev = −nRT ln(P1/P2) = −(1.00 mol)(8.3145 J K−1 mol−1)(350. K) ln(5.00/1.00) = -4680 J 25. P1V1 = P2V2 since n and T are constant; P2 = P1V1 (5.0)(1.0) = = 2.5 atm V2 (2.0) Gas expands isothermally against no pressure, so ΔE = 0, w = 0, and q = 0. ΔE = 0, so qrev = −wrev = nRT ln(V2/V1); T = PV = 61 K nR qrev = (1.0 mol)(8.3145 J K−1 mol−1)(61 K) ln(2.0/1.0) = 350 J 26. ⎛V ⎞ ⎛ 20.0 L ⎞ ⎟⎟ = −1720 J a. wrev = −nRT ln ⎜⎜ 2 ⎟⎟ = −(1.00 mol)(8.3145 J K−1 mol−1)(298 K) ln ⎜⎜ ⎝ 10.0 L ⎠ ⎝ V1 ⎠ For isothermal expansion: ΔE = 0, so qrev = -wrev = 1720 J b. w = −PΔV = −1.23 atm(20.0 L − 10.0 L) = −12.3 L atm −12.3 L atm × 101.3 J L−1 atm−1 = −1250 J ΔE = 0 for isothermal expansion, so q = 1250 J. 27. a. qv =ΔE = nCvΔT = (1.000 mol)(28.95 J K−1 mol−1)(350.0 − 298.0 K) qv = 1.51 × 103 J = 1.51 kJ qp = ΔH = nCpΔT = (1.000)(37.27)(350.0 − 298.0) = 1.94 × 103 J = 1.94 kJ b. ΔS = S350 − S298 = nCp ln(T2/T1) S350 − 213.64 J/K = (1.000 mol)(37.27 J K−1 mol−1) ln(350.0/298.0) S350 = 213.64 J/K + 5.994 J/K = 219.63 J/K = molar entropy at 350.0 K and 1.000 atm CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 373 c. ΔS = nR ln(V2/V1), V = nRT/P, ΔS = nR ln(P1/P2) = S(350, 1.174) − S(350, 1.000) ΔS = S(350, 1.174) − 219.63 J/K = (1.000 mol)(8.3145 J K−1 mol−1) ln(1.000 atm/1.174 atm) ΔS = −1.334 J/K = S(350, 28. 1.174) − 219.63, S(350, 1.174) = 218.30 J K−1 mol−1 a. He(g, 0.100 mol, 25°C, 1.00 atm) → He(g, 0.100 mol, 25°C, 5.00 L) ΔS = nR ln(V2/V1) = Sfinal − Sinitial; Si = 0.100 mol(126.1 J K−1 mol−1) = 12.6 J/K nRT = V1 = P1 0.100 mol × 0.08206 L atm × 298 K K mol = 2.45 L 1.00 atm Sf − 12.6 J/K = (0.100 mol)(8.3145 J K−1 mol−1) ln(5.00 L/2.45 L) Sf = 12.6 J/K + 0.593 J/K = 13.2 J/K b. He(3.00 mol, 25°C, 1.00 atm) → He(3.00 mol, 25°C, 3000.0 L) ΔS = nR ln(V2/V1) = Sfinal − Sinitial; Si = 3.00 mol(126.1 J K−1 mol−1) = 378 J/K nRT = V1 = P1 3.00 mol × 0.08206 L atm × 298 K K mol = 73.4 L 1.00 atm ⎛ 3000.0 L ⎞ ⎟⎟ = 92.6 J/K Sf − 378 J/K = (3.00 mol)(8.3145 J K−1 mol−1) ln ⎜⎜ ⎝ 73.4 L ⎠ Sf = 378 + 92.6 = 471 J/K 29. For A(l, 125°C) → A(l, 75°C): ΔS = nCp ln(T2/T1) = 1.00 mol(75.0 J K−1 mol−1) ln(348 K/398 K) = −10.1 J/K For A(l, 75°C) → A(g, 155°C): ΔS = 75.0 J K−1 mol−1 For A(g, 155°C) → A(g,125°C): ΔS = nCp ln(T2/T1) = 1.00 mol(29.0 J K−1 mol−1) ln(398 K/428 K) = −2.11 J/K The sum of the three step gives A(l, 125°C) → A(g, 125°C). ΔS for this process is the sum of ΔS for each of the three steps. ΔS = −10.1 + 75.0 − 2.11 = 62.8 J/K For a phase change, ΔS = ΔH/T. At 125°C: ΔHvap = TΔS = 398 K(62.8 J/K) = 2.50 × 104 J 374 30. CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY Calculate q by breaking up the total process into several steps, and use the formula q = nCΔT or the ΔH values to calculate q for each step. q = (1.000 mol)(37.5 J K−1 mol−1)(30.0 K) + 6010 J + (1.000 mol)(75.3 J K−1 mol−1)(100.0 K) + 40,700 J + (1.000 mol)( 36.4 J K−1 mol−1)(40.0 K) q = 1130 J + 6010 J + 7530 J + 40,700 J + 1460 J = 56,800 J = 56.8 kJ At constant pressure: qp = ΔH = 56.8 kJ 1.000 mol × 0.08206 L atm × 372.2 K K mol = 30.6 L 1.00 atm At 100.0°C: V = nRT = P At 140.0°C: V = nR (413.2 K ) = 33.9 L P Work is only done when vaporization occurs and when the vapor expands as T is increased from 100.0°C to 140.0°C. w = −PΔV; wtotal = −1.00 atm(30.6 L − 0.018 L) − 1.00 atm(33.9 - 30.6 L) w = −30.6 L atm − 3.3 L atm = −33.9 L atm; −33.9 L atm × 101.3 J = −3430 J = −3.43 kJ L atm ΔE = q + w = 56.8 kJ − 3.43 kJ = 53.4 kJ Calculate ΔS by breaking up the total process into several steps and using ΔS = nCp ln(T2/T1) or ΔS = ΔH/T (for phase changes) to calculate ΔS for each step. ⎛ 273.2 K ⎞ 6010 J ⎟⎟ + ΔS = (1.000 mol)(37.5 J K−1 mol−1) ln ⎜⎜ ⎝ 243.2 K ⎠ 273.2 K ⎛ 373.2 K ⎞ 40,700 J ⎟⎟ + + (1.000 mol)(75.3 J K−1 mol−1) ln ⎜⎜ 373.2 K ⎝ 273.2 K ⎠ ⎛ 413.2 K ⎞ ⎟⎟ + (1.000 mol)(36.4 J K−1 mol−1) ln ⎜⎜ ⎝ 373.2 K ⎠ ΔS = 4.36 J/K + 22.0 J/K + 23.5 J/K + 109 J/K + 3.71 J/K = 163 J/K Summary: q = 56.8 kJ; ΔH = 56.8 kJ; w = −3.43 kJ; ΔE = 53.4 kJ; ΔS = 163 J/K 31. Calculate the final temperature by equating heat loss to heat gain. Keep all quantities positive to avoid sign errors. CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 375 (3.00 mol)(75.3 J °C−1 mol−1)(Tf − 0°C) = (1.00 mol)(75.3 J °C−1 mol−1)(100.°C − Tf) Solving: Tf = 25°C = 298 K Now we can calculate ΔS for the various changes using ΔS = nCp ln (T2/T1). Heat 3 mol H2O: ΔS1 = (3.00 mol)(75.3 J K−1 mol−1) ln(298 K/273 K) = 19.8 J/K Cool 1 mol H2O: ΔS2 = (1.00 mol)(75.3 J K−1 mol−1) ln(298/373) = −16.9 J/K ΔStotal = ΔSheat + ΔScool = 19.8 - 16.9 = 2.9 J/K 32. 18.02 g ice = 1.000 mol ice; 54.05 g H2O = 3.000 mol H2O Heat gained by the ice: (1.000 mol)(37.5 J °C−1 mol−1)(10.0°C) + 6.01 × 103 J + (1.000 mol)(75.3 J °C−1 mol−1)(Tf − 0.0) Heat lost by H2O(l) = 3.000 mol(75.3 J °C−1 mol−1)(100.0°C − Tf) Heat gain by ice = heat lost by H2O; 375 J + 6010 J + (75.3)Tf = 22,600 J − 226Tf Solving: Tf = 53.8°C Now calculate ΔS for the various changes using ΔS = nCp ln(T2/T1) or, for a phase change, ΔS = ΔH/T. ⎛ 273.2 K ⎞ 6010 J ⎟⎟ + ΔSice = (1.000 mol)(37.5 J K−1 mol−1) ln ⎜⎜ ⎝ 263.2 K ⎠ 273.2 K ⎛ 327.0 ⎞ + (1.000 mol)(75.3 J K−1 mol−1) ln ⎜ ⎟ ⎝ 273.2 ⎠ ΔSice = 1.40 J/K + 22.0 J/K + 13.5 J/K = 36.9 J/K ΔSwater = (3.000 mol)(75.3 J K−1 mol−1) ln(327.0/373.2) = −29.9 J/K ΔStotal = ΔSice + ΔSwater = 36.9 − 29.9 = 7.0 J/K Entropy and the Second Law of Thermodynamics: Free Energy 33. Living organisms need an external source of energy to carry out these processes. Green plants use the energy from sunlight to produce glucose from carbon dioxide and water by photosynthesis. In the human body the energy released from the metabolism of glucose helps drive the synthesis of proteins. For all processes combined, ΔSuniv must be greater than zero (second law). 376 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 34. It appears that the sum of the two processes has no net change. This is not so. By the second law of thermodynamics, ΔSuniv must have increased even though it looks as if we have gone through a cyclic process. 35. ΔSsurr is primarily determined by heat flow. This heat flow into or out of the surroundings comes from the heat flow out of or into the system. In an exothermic process (ΔH < 0), heat flows into the surroundings from the system. The heat flow into the surroundings increases the random motions in the surroundings and increases the entropy of the surroundings (ΔSsurr > 0). This is a favorable driving force for spontaneity. In an endothermic reaction (ΔH > 0), heat is transferred from the surroundings into the system. This heat flow out of the surroundings decreases the random motions in the surroundings and decreases the entropy of the surroundings (ΔSsurr < 0). This is unfavorable. The magnitude of ΔSsurr also depends on the temperature. The relationship is inverse; at low temperatures, a specific amount of heat exchange makes a larger percent change in the surroundings than the same amount of heat flow at a higher temperature. The negative sign in the ΔSsurr = !ΔH/T equation is necessary to get the signs correct. For an exothermic reaction where ΔH is negative, this increases ΔSsurr, so the negative sign converts the negative ΔH value into a positive quantity. For an endothermic process where ΔH is positive, the sign of ΔSsurr is negative, and the negative sign converts the positive ΔH value into a negative quantity. 36. a. To boil a liquid requires heat. Hence this is an endothermic process. All endothermic processes decrease the entropy of the surroundings (ΔSsurr is negative). b. This is an exothermic process. Heat is released when gas molecules slow down enough to form the solid. In exothermic processes, the entropy of the surroundings increases (ΔSsurr is positive). 37. 38. a. ΔSsurr = − ΔH − (−2221 kJ ) = = 7.45 kJ/K = 7.45 × 103 J/K T 298 K b. ΔSsurr = − ΔH − 112 kJ = = −0.376 kJ/K = −376 J/K T 298 K a. Cgraphite(s); graphite has the larger positional probability. The diamond form is a very ordered structure. b. C2H5OH(g); the gaseous state has a much larger volume associated with it and, in turn, a much greater positional probability. c. CO2(g); the gaseous state has a much larger volume associated with it and, in turn, a much greater positional probability. d. N2O(g); more complicated molecule with more parts, greater positional probability. e. HCl(g); larger molecule, more parts (electrons), greater positional probability. 39. a. Decrease in positional probability; ΔS°(−) CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 377 b. Increase in positional probability; ΔS°(+) c. Decrease in positional probability (Δn < 0); ΔS°(−) d. Decrease in positional probability (Δn < 0); ΔS°(−) e. HCl(g) has a greater positional probability due to the huge volume of gas; ΔS°(−). f. Increase in positional probability; ΔS°(+) For c, d, and e, concentrate on the gaseous products and reactants. When there are more gaseous product molecules than gaseous reactant molecules (Δn > 0), then ΔS° will be positive. When Δn is negative, then ΔS° is negative. 40. a. 2 H2S(g) + SO2(g) → 3 Srhombic(s) + 2 H2O(g); because there are more molecules of reactant gases as compared to product molecules of gas (Δn = 2 − 3 < 0), ΔS° will be negative as positional probability decreases when the moles of gas decrease. ΔS° = ∑ n p S oproducts − ∑ n r S oreactants ΔS° = [3 mol Srhombic(s)(32 J K−1 mol−1) + 2 mol H2O(g)(189 J K−1 mol−1)] − [2 mol H2S(g)(206 J K−1 mol−1) + 1 mol SO2(g)(248 J K−1 mol−1)] ΔS° = 474 J/K − 660. J/K = −186 J/K b. 2 SO3(g) → 2 SO2(g) + O2(g); Because Δn of gases is positive (Δn = 3 − 2), ΔS° will be positive as positional probability increases when the moles of gas increase. ΔS = 2 mol(248 J K−1 mol−1) + 1 mol(205 J K−1 mol−1) − [2 mol(257 J K−1 mol−1)] = 187 J/K c. Fe2O3(s) + 3 H2(g) → 2 Fe(s) + 3 H2O(g); because Δn of gases = 0 (Δn = 3 − 3), we can’t easily predict if ΔS° will be positive of negative. ΔS = 2 mol(27 J K−1 mol−1) + 3 mol(189 J K−1 mol−1) − [1 mol(90. J K−1 mol−1) + 3 mol(131 J K−1 mol−1)] ΔS = 138 J/K 41. C2H2(g) + 4 F2(g) → 2 CF4(g) + H2(g); ΔS° = 2SoCF4 + SoH 2 − [ SoC 2 H 2 + 4SoF2 ] − 358 J/K = (2 mol)SoCF4 + 131 J/K − [201 J/K + 4(203 J/K)], SoCF4 = 262 J K−1 mol−1 42. From Appendix 4, SE = 198 J K−1 mol−1 for CO(g), and SE = 27 J K−1 mol−1 for Fe(s). Let S ol = SE for Fe(CO)5(l) and S og = SE for Fe(CO)5(g). 378 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY ΔSE = !677 J/K = 1 mol(S ol ) – [1 mol(27 J K−1 mol−1) + 5 mol(198 J K−1 mol−1)] S ol = 340. J K−1 mol−1 ΔSE = 107 J/K = 1 mol(S og ) – 1 mol(340. J K−1 mol−1) S og = SE for Fe(CO)5(g) = 447 J K−1 mol−1 43. −144 J/K = (2 mol) SoAlBr3 − [2(28 J/K) + 3(152 J/K)], SoAlBr3 = 184 J K−1 mol−1 44. At the boiling point, ΔG = 0, so ΔH = TΔS. ΔS = ΔH 27.5 kJ / mol = = 8.93 × 10−2 kJ K−1 mol−1 = 89.3 J K−1 mol−1 T ( 273 + 35) K ΔH 58.51 × 103 J / mol = = 629.7 K ΔS 92.92 J K −1 mol −1 45. At the boiling point, ΔG = 0 so ΔH = TΔS. T = 46. C2H5OH(l) → C2H5OH(g); at the boiling point, ΔG = 0 and ΔSuniv = 0. For the vaporization process, ΔS is a positive value, whereas ΔH is a negative value. To calculate ΔSsys, we will determine ΔSsurr from ΔH and the temperature; then ΔSsys = !ΔSsurr for a system at equilibrium. ΔSsurr = 38.7 × 103 J / mol − ΔH = = !110. J K−1 mol−1 T 351 K ΔSsys = !ΔSsurr = !(!110.) = 110. J K−1 mol−1 47. a. NH3(s) → NH3(l); ΔG = ΔH − TΔS = 5650 J/mol − 200. K(28.9 J K−1 mol−1) ΔG = 5650 J/mol − 5780 J/mol = −130 J/mol Yes, NH3 will melt because ΔG < 0 at this temperature. b. At the melting point, ΔG = 0, so T = 48. ΔH 5650 J / mol = = 196 K. ΔS 28.9 J K −1 mol −1 a. Srhombic → Smonoclinic; this phase transition is spontaneous (ΔG < 0) at temperatures above 95°C. ΔG = ΔH − TΔS; for ΔG to be negative only above a certain temperature, then ΔH is positive and ΔS is positive (see Table 10.6 of text). b. Because ΔS is positive, Srhombic is the more ordered crystalline structure (has the smaller positional probability associated with it). CHAPTER 10 49. SPONTANEITY, ENTROPY, AND FREE ENERGY 379 Solid I → solid II; equilibrium occurs when ΔG = 0. ΔG = ΔH ! TΔS, ΔH = TΔS, T = ΔH/ΔS = − 743.1 J / mol = 43.7 K = !229.5°C − 17.0 J K −l mol −1 Free Energy and Chemical Reactions 50. a. + 2 O2(g) → CO2(g) + 2 H2O(g) CH4(g) _____________________________________________________________________ ΔH of −75 kJ/mol 0 −393.5 −242 ΔG of −51 kJ/mol 0 −394 −229 (Appendix 4 data) S° 186 J K-1 mol-1 205 214 189 _____________________________________________________________________ ΔH° = ∑ n p ΔH of , products − ∑ n r ΔH of, reactants ; ΔSo = ∑ n pSoproducts − ∑ n r Soreactants ΔH° = 2 mol(−242 kJ/mol) + 1 mol(−393.5 kJ/mol) − [1 mol(−75 kJ/mol)] = −803 kJ ΔS° = 2 mol(189 J K−1 mol−1) + 1 mol(214 J K−1 mol−1) − [1 mol(186 J K−1 mol−1) + 2 mol(205 J K−1 mol−1)] = −4 J/K There are two ways to get ΔG°. We can use ΔG° = ΔH° − TΔS° (be careful of units): ΔG° = ΔH° − TΔS° = −803 × 103 J − (298 K)(−4 J/K) = −8.018 × 105 J = −802 kJ or we can use ΔG of values, where ΔG° = ∑ n p ΔG of , products − ∑ n r ΔG of, reactants : ΔG° = 2 mol(−229 kJ/mol) + 1 mol(−394 kJ/mol) − [1 mol(−51 kJ/mol)] ΔG° = −801 kJ (Answers are the same within round-off error.) b. 6 CO2(g) + 6 H2O(l) → C6H12O6(s) + 6 O2(g) _______________________________________________________ ΔH of −393.5 kJ/mol −286 −1275 0 S° 214 J K−1 mol−1 70. 212 205 ______________________________________________________ ΔH° = −1275 − [6(−286) + 6(−393.5)] = 2802 kJ ΔS° = 6(205) + 212 − [6(214) + 6(70.)] = −262 J/K ΔG° = 2802 kJ − (298 K)( −0.262 kJ/K) = 2880. kJ 380 CHAPTER 10 c. SPONTANEITY, ENTROPY, AND FREE ENERGY P4O10(s) + 6 H2O(l) → 4 H3PO4(s) ______________________________________________ ΔH of −2984 −286 −1279 (kJ/mol) S° 229 70. 110. (J K−1 mol−1) _______________________________________________ ΔH° = 4 mol(−1279 kJ/mol) − [1 mol(−2984 kJ/mol) + 6 mol(−286 kJ/mol)] = −416 kJ ΔS° = 4(110.) − [229 + 6(70.)] = −209 J/K ΔG° = ΔH° − TΔS° = −416 kJ − (298 K)(−0.209 kJ/K) = −354 kJ d. HCl(g) + NH3(g) → NH4Cl(s) ___________________________________________________________ ΔH of −92 −46 −314 (kJ/mol) S° 187 193 96 (J K−1 mol−1) ___________________________________________________________ ΔH° = −314 − (−92 − 46) = −176 kJ; ΔS° = 96 − (187 + 193) = −284 J/K ΔG° = ΔH° − TΔS° = −176 kJ − (298 K)(−0.284 kJ/K) = −91 kJ 51. −5490. kJ = 8(−394 kJ) + 10(−237 kJ) − 2 ΔG of , C 4 H10 , ΔG of , C 4 H10 = −16 kJ/mol 52. Of the functions ΔGE, ΔHE, and ΔSE, ΔGE has the greatest dependence on temperature. The temperature is usually assumed to be 25EC. However, if other temperatures are used in a reaction, we can estimate ΔG° at that different temperature by assuming ΔH° and ΔS° are temperature independent (which is not always the best assumption). We calculate ΔHE and ΔS° values for a reaction using Appendix 4 data, and then use the different temperature in the ΔG° = ΔHE ! TΔS° equation to determine (estimate) ΔG° at that different temperature. 53. ΔG° = -58.03 kJ − (298 K)(−0.1766 kJ/K) = −5.40 kJ ΔG° = 0 = ΔH° − TΔS°, T = ΔH o − 58.03 kJ = = 328.6 K o − 0 .1766 kJ / K ΔS ΔG° is negative below 328.6 K, where the favorable ΔH° term dominates. 54. a. CH2 O CH2(g) + HCN(g) CH2 CHCN(g) + H2O(l) CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 381 ΔH° = 185.0 − 286 − (−53 + 135.1) = −183 kJ; ΔS° = 274 + 70. − (242 + 202) = −100. J/K ΔG° = ΔH° − TΔS° = −183 kJ − 298 K(−0.100 kJ/K) = −153 kJ b. HC≡CH(g) + HCN(g) → CH2=CHCN(g) ΔH° = 185.0 − (135.1 + 227) = −177 kJ; ΔS° = 274 − (202 + 201) = −129 J/K o T = 70.°C = 343 K; ΔG 343 = ΔH° − TΔS° = −177 kJ − 343 K(−0.129 kJ/K) o ΔG 343 = −177 kJ + 44 kJ = −133 kJ c. 4 CH2=CHCH3(g) + 6 NO(g) → 4 CH2=CHCN(g) + 6 H2O(g) + N2(g) ΔH° = 6(−242) + 4(185.0) − [4(20.9) + 6(90.)] = −1336 kJ ΔS° = 192 + 6(189) + 4(274) − [6(211) + 4(266.9)] = 88 J/K o T = 700.°C = 973 K; ΔG 973 = ΔH° − TΔS° = −1336 kJ − 973 K(0.088 kJ/K) o ΔG 973 = −1336 kJ − 86 kJ = −1422 kJ 55. a. ΔG° = 2(−270. kJ) − 2(−502 kJ) = 464 kJ b. Because ΔG° is positive, this reaction is not spontaneous at standard conditions at 298 K. c. ΔG° = ΔH° − TΔS°, ΔH° = ΔG° + TΔS° = 464 kJ + 298 K(0.179 kJ/K) = 517 kJ We need to solve for the temperature when ΔG° = 0: ΔG° = 0 = ΔH° − TΔS°, T = ΔH o 517 kJ = = 2890 K o 0.179 kJ/K ΔS This reaction will be spontaneous at standard conditions (ΔG° < 0) when T > 2890 K. At these temperatures, the favorable entropy term will dominate. 56. C2H4(g) + H2O(g) → CH3CH2OH(l) ΔH° = −278 − (52 − 242) = −88 kJ; ΔS° = 161 − (219 + 189) = −247 J/K When ΔG° = 0, ΔH° = TΔS°, so T = ΔH o − 88 × 103 J = = 360 K. − 247 J/K ΔSo 382 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY At standard concentrations, ΔG = ΔG°, so the reaction will be spontaneous when ΔG° < 0. Since the signs of ΔH° and ΔS° are both negative, this reaction will be spontaneous at temperatures below 360 K (where the favorable ΔH° term will dominate). C2H6(g) + H2O(g) → CH3CH2OH(l) + H2(g) ΔH° = −278 − (−84.7 − 242) = 49 kJ; ΔS° = 131 + 161 − (229.5 + 189) = −127 J/K This reaction can never be spontaneous at standard conditions because of the signs of ΔH° and ΔS°. Thus the reaction C2H4(g) + H2O(g) → C2H5OH(l) would be preferred at standard conditions. 57. CH4(g) + CO2(g) → CH3CO2H(l) ΔH° = −484 − [−75 + (−393.5)] = −16 kJ; ΔS° = 160. − (186 + 214) = −240. J/K ΔG° = ΔH° − TΔS° = −16 kJ − (298 K)(−0.240 kJ/K) = 56 kJ At standard concentrations, where ΔG = ΔG°, this reaction is spontaneous only at temperatures below T = ΔH°/ΔS° = 67 K (where the favorable ΔH° term will dominate, giving a negative ΔG° value). This is not practical. Substances will be in condensed phases and rates will be very slow at this extremely low temperature. CH3OH(g) + CO(g) → CH3CO2H(l) ΔH° = −484 − [−110.5 + (−201)] = −173 kJ; ΔS° = 160. − (198 + 240.) = −278 J/K ΔG° = −173 kJ − (298 K)(−0.278 kJ/K) = −90. kJ This reaction also has a favorable enthalpy and an unfavorable entropy term. This reaction is spontaneous at temperatures below T = ΔH°/ΔS° = 622 K (assuming standard concentrations). The reaction of CH3OH and CO will be preferred at standard conditions. It is spontaneous at high enough temperatures that the rates of reaction should be reasonable. 58. 6 C(s) + 6 O2(g) → 6 CO2(g) ΔG° = 6(−394 kJ) ΔG° = 3(−237 kJ) 3 H2(g) + 3/2 O2(g) → 3 H2O(l) 6 CO2(g) + 3 H2O(l) → C6H6(l) + 15/2 O2(g) ΔG° = −1/2 (−6399 kJ) ____________________________________________________________________ ΔG° = 125 kJ 6 C(s) + 3 H2(g) → C6H6(l) 59. Enthalpy is not favorable, so ΔS must provide the driving force for the change. Thus ΔS is positive. There is an increase in positional probability, so the original enzyme has the more ordered structure (has the smaller positional probability). 60. a. When a bond is formed, energy is released, so ΔH is negative. Because there are more reactant molecules of gas than product molecules of gas (Δn < 0), ΔS will be negative. CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 383 b. ΔG = ΔH − TΔS; for this reaction to be spontaneous (ΔG < 0), the favorable enthalpy term must dominate. The reaction will be spontaneous at low temperatures, where the ΔH term dominates. 61. Because there are more product gas molecules than reactant gas molecules (Δn > 0), ΔS will be positive. From the signs of ΔH and ΔS, this reaction is spontaneous at all temperatures. It will cost money to heat the reaction mixture. Because there is no thermodynamic reason to do this, the purpose of the elevated temperature must be to increase the rate of the reaction, i.e., kinetic reasons. Free Energy: Pressure Dependence and Equilibrium 62. All reactions want to minimize their free energy. This is the driving force for any process. As long as ΔG is a negative, the process occurs. The equilibrium position represents the lowest total free energy available to any particular reaction system. Once equilibrium is reached, the system cannot minimize its free energy anymore. Converting from reactants to products or products to reactants will increase the total free energy of the system, which reactions do not want to do. 63. At constant temperature and pressure, the sign of ΔG (positive or negative) tells us which reaction is spontaneous (the forward or reverse reaction). If ΔG < 0, then the forward reaction is spontaneous and if ΔG > 0, then the reverse reaction is spontaneous. If ΔG = 0, then the reaction is at equilibrium (neither the forward nor reverse reaction is spontaneous). ΔG° gives the equilibrium position by deter-mining K for a reaction utilizing the equation ΔG° = !RT ln K. ΔG° can only be used to predict spontaneity when all reactants and products are present at standard pressures of 1 atm and/or standard concentrations of 1 M. 64. ΔG° = ∑ n p ΔG of , products − ∑ n r ΔG of, reactants , ΔG° = 2(−371) − [2(−300.)] = −142 kJ 2 ⎛ PSO 3 ⎜ ΔG = ΔG° + RT ln Q = −142 kJ + RT ln 2 ⎜ PSO × PO 2 2 ⎝ At 10.0 atm: ΔG = −142 kJ + 65. ⎞ ⎟ ; note: ΔG = ΔG° when all gases are ⎟ ⎠ at 1.00 atm. ⎛ (10.0) 2 ⎞ 8.3145 J K −1 mol −1 ⎟ = −148 kJ (298 K ) ln⎜⎜ 2 ⎟ 1000 J / kJ ⎝ (10.0) (10.0) ⎠ ΔG = ΔG° + RT ln Q = ΔG° + RT ln PN 2O 4 2 PNO 2 ΔG° = 1 mol(98 kJ/mol) − 2 mol(52 kJ/mol) = −6 kJ a. These are standard conditions, so ΔG = ΔG° since Q = 1 and ln Q = 0. Because ΔG° is negative, the forward reaction is spontaneous. The reaction shifts right to reach equilibrium. 384 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY b. ΔG = −6 × 103 J + 8.3145 J K−1 mol−1(298 K) ln 0.50 (0.21) 2 ΔG = −6 × 103 J + 6.0 × 103 J = 0 Since ΔG = 0, this reaction is at equilibrium (no shift). c. ΔG = −6 × 103 J + 8.3145 J K−1 mol−1(298 K) ln 1.6 (0.29) 2 ΔG = −6 × 103 J + 7.3 × 103 J = 1.3 × 103 J = 1 × 103 J Since ΔG is positive, the reverse reaction is spontaneous, so the reaction shifts to the left to reach equilibrium. 66. ΔG° = 3(0) + 2(−229) − [2(−34) + 1(−300.)] = −90. kJ ΔG = ΔG° + RT ln PH2 2O PH2 2S × PSO 2 = −90. kJ + ⎤ (8.3145)(298) ⎡ (0.030) 2 kJ ⎢ln ⎥ −4 2 1000 ⎣ (1.0 × 10 ) (0.010) ⎦ ΔG = −90. kJ + 39.7 kJ = −50. kJ 67. ΔH° = 2 ΔH fo, NH 3 = 2(−46) = −92 kJ; ΔG° = 2ΔG of , NH 3 = 2(−17) = −34 kJ ΔS° = 2(193 J/K) − [192 J/K + 3(131 J/K)] = −199 J/K; ΔG° = −RT ln K K= ⎛ ⎞ − ΔG o − (−34,000 J ) ⎟ = e13.72 = 9.1 × 105 = exp⎜⎜ −1 −1 ⎟ RT ( 8 . 3145 J K mol )( 298 K ) ⎝ ⎠ Note: When determining exponents, we will round off after the calculation is complete. This helps eliminate excessive round off error. a. ΔG = ΔG° + RT ln 2 PNH 3 PN 2 × PH3 2 = −34 kJ + (8.3145 J K −1 mol −1 )(298 K ) (50.) 2 ln 1000 J / kJ (200.)(200.) 3 ΔG = −34 kJ − 33 kJ = −67 kJ b. ΔG = −34 kJ + (8.3145 J K −1 mol −1 )(298 K ) (200.) 2 ln 1000 J / kJ ( 200.)(600.) 3 ΔG = −34 kJ − 34.4 kJ = −68 kJ c. Assume ΔH° and ΔS° are temperature independent. CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 385 o o ΔG100 = ΔH° − TΔS°, ΔG100 = −92 kJ − (100. K)(−0.199 kJ/K) = −72 kJ o ΔG100 = ΔG100 + RT ln Q = −72 kJ + (10.) 2 (8.3145 J K −1 mol −1 )(100. K ) ln 1000 J / kJ (50.)(200.) 3 ΔG100 = −72 kJ − 13 kJ = −85 kJ d. ΔG o700 = −92 kJ − (700. K)(−0.199 kJ/K) = 47 kJ ΔG700 = 47 kJ + (10.) 2 (8.3145 J K −1 mol −1 )(700. K ) ln 1000 J / kJ (50.)(200.) 3 ΔG700 = 47 kJ − 88 kJ = −41 kJ 68. a. ΔG° = −RT ln K = b. − 8.3145 J (298 K) ln(1.00 × 10−14) = 7.99 × 104 J = 79.9 kJ/mol K mol o ΔG 313 = −RT ln K = − 8.3145 J (313 K) ln(2.92 × 10−14) = 8.11 × 104 J = 81.1 kJ/mol K mol 69. ΔGE = !RT ln K. To determine K at a temperature other than 25EC, one needs to know ΔGE at that temperature. We assume ΔHE and ΔSE are temperature-independent and use the equation ΔGE = ΔHE - TΔSE to estimate ΔGE at the different temperature. For K = 1, we want ΔGE = 0, which occurs when ΔHE = TΔSE. Again, assume ΔHE and ΔSE are temperature independent, and then solve for T (= ΔHE/ΔSE). At this temperature, K = 1 because ΔGE = 0. This only works for reactions where the signs of ΔHE and ΔSE are the same (either both positive or both negative). When the signs are opposite, K will always be greater than one (when ΔHE is negative and ΔSE is positive) or K will always be less than one (when ΔHE is positive and ΔSE is negative). When the signs of ΔHE and ΔSE are opposite, K can never equal one. 70. a. ΔH of (kJ/mol) NH3(g) O2(g) NO(g) H2O(g) NO2(g) HNO3(l) H2O(l) S° (J K−1 mol−1) −46 0 90. −242 34 −174 −286 193 205 211 189 240. 156 70. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) ΔH° = 6(−242) + 4(90.) − [4(−46)] = −908 kJ 386 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY ΔS° = 4(211) + 6(189) − [4(193) + 5(205)] = 181 J/K ΔG° = −908 kJ − 298 K(0.181 kJ/K) = −962 kJ ΔG° = −RT ln K, ln K = − ΔG o − (−962 × 103 J ) = = 388 RT 8.3145 J K −1 mol −1 × 298 K ln K = 2.303 log K, log K = 168, K = 10168 (an extremely large value) 2 NO(g) + O2(g) → 2 NO2(g) ΔH° = 2(34) − [2(90.)] = −112 kJ; ΔS° = 2(240.) − [2(211) + (205)] = −147 J/K ΔG° = −112 kJ − (298 K)(−0.147 kJ/K) = −68 kJ K = exp ⎤ ⎡ − ΔG o − (−68,000 J ) 27.44 = exp ⎢ = 8.3 × 1011 ⎥ = e −1 −1 RT 8 . 3145 J K mol ( 298 K ) ⎦ ⎣ Note: When determining exponents, we will round off after the calculation is complete. 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g) ΔH° = 2(−174) + (90.) − [3(34) + (−286)] = −74 kJ ΔS° = 2(156) + (211) − [3(240.) + (70.)] = −267 J/K ΔG° = −74 kJ − (298 K)(−0.267 kJ/K) = 6 kJ K = exp ⎤ ⎡ − ΔG o − 6000 J −2.4 −2 = exp ⎢ ⎥ = e = 9 × 10 −1 −1 RT ⎣ 8.3145 J K mol (298 K ) ⎦ b. ΔG° = −RT ln K; T = 825 + 273 = 1098 K; We must determine ΔG° at 1098 K. o ΔG1098 = ΔH° − TΔS° = −908 kJ − (1098 K)(0.181 kJ/K) = −1107 kJ K = exp o ⎡ ⎤ − ΔG1098 − (−1.107 × 10 6 J ) 121.258 = exp ⎢ = 4.589 × 1052 ⎥ = e −1 −1 RT ⎣ 8.3145 J K mol (1098 K ) ⎦ c. There is no thermodynamic reason for the elevated temperature since ΔH° is negative and ΔS° is positive. Thus the purpose for the high temperature must be to increase the rate of the reaction. 71. K= 2 PNF 3 PN2 2 × PF32 = (0.48) 2 = 4.4 × 104 0.021(0.063) 3 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 387 o ΔG 800 = −RT ln K = −8.3145 J K−1 mol−1(800. K) ln(4.4 × 104) = −7.1 × 104 J/mol = −71 kJ/mol 72. 2 SO2(g) + O2(g) → 2 SO3(g); ΔG° = 2(−371 kJ) − [2(−300. kJ)] = −142 kJ − ΔG o − ( −142,000 J ) = = 57.311 RT 8.3145 J K −1 mol −1 (298 K ) ΔG° = −RT ln K, ln K = K = e57.311 = 7.76 × 1024 K = 7.76 × 1024 = 2 PSO 3 2 PSO × PO 2 2 = ( 2.0) 2 , PSO 2 = 1.0 × 10−12 atm 2 PSO 2 × (0.50) From the negative value of ΔG°, this reaction is spontaneous at standard conditions. Because there are more molecules of reactant gases than product gases, ΔS° will be negative (unfavorable). Therefore, this reaction must be exothermic (ΔH° < 0). When ΔH° and ΔS° are both negative, the reaction will be spontaneous at relatively low temperatures, where the favorable ΔH° term dominates. 73. Because the partial pressure of C(g) decreased, the net change that occurs for this reaction to reach equilibrium is for products to convert to reactants. A(g) Initial Change Equil. + 0.100 atm +x 0.100 + x 2 B(g) ⇌ C(g) 0.100 atm +2x 0.100 + 2x ← 0.100 atm !x 0.100 ! x From the problem, PC = 0.040 atm = 0.100 – x, x = 0.060 atm. The equilibrium partial pressures are: PA = 0.100 + x = 0.100 + 0.060 = 0.160 atm, PB = 0.100 + 2(0.60) = 0.220 atm, and PC = 0.040 atm. K= 0.040 = 5.2 0.160(0.220) 2 ΔGE = !RT ln K = !8.3145 J K−1 mol−1(298 K) ln(5.2) = !4.1 × 103 J/mol = !4.1 kJ/mol 74. ⎛ ⎞ − (−30,500 J ) ⎟ = 2.22 × 105 a. ΔG° = −RT ln K, K = exp ⎜⎜ −1 −1 ⎟ ⎝ 8.3145 J K mol × 298 K ⎠ b. C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) ΔG° = 6 mol(−394 kJ/mol) + 6 mol(−237 kJ/mol) − 1 mol(−911 kJ/mol) = −2875 kJ 2875 kJ 1 mol ATP × = 94.3 mol ATP; 94.3 molecules ATP/molecule glucose mol glucose 30.5 kJ 388 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY This is an overstatement. The assumption that all the free energy goes into this reaction is false. Actually, only 38 moles of ATP are produced by metabolism of 1 mole of glucose. 75. HgbO2 → Hgb + O2 ΔG° = − (−70 kJ) Hgb + CO → HgbCO ΔG° = −80 kJ _________________________________________ HgbO2 + CO → HgbCO + O2 ΔG° = −10 kJ ⎛ − ΔG o ΔG° = −RT ln K, K = exp ⎜⎜ ⎝ RT 76. a. ln K = ⎞ ⎛ ⎞ − (−10 × 103 J ) ⎟ = exp⎜ ⎟ ⎟ ⎜ (8.3145 J K −1 mol −1 )(298 K ) ⎟ = 60 ⎠ ⎝ ⎠ − ΔG o − 14,000 J = = −5.65, K = e−5.65 = 3.5 × 10−3 −1 −1 RT (8.3145 J K mol )(298 K ) ΔG° = 14 kJ Glutamic acid + NH3 → Glutamine + H2O ATP + H2O → ADP + H2PO4− ΔG° = −30.5 kJ ___________________________________________________________________________ Glutamic acid + ATP + NH3 → Glutamine + ADP + H2PO4− ΔG° = 14 − 30.5 = −17 kJ b. ln K = 77. − ΔG o − (−17,000 J ) = = 6.86, K = e6.86 = 9.5 × 102 −1 −1 RT 8.3145 J K mol (298 K ) At 25.0°C: ΔG° = ΔH° − TΔS° = −58.03 × 103 J/mol − (298.2 K)(−176.6 J K−1 mol−1) = −5.37 × 103 J/mol ΔG° = −RT ln K, ln K = − ΔG o − (−5.37 × 103 J / mol) = = 2.166; K = e2.166 = 8.72 RT (8.3145 J K −1 mol −1 )(298.2 K ) At 100.0°C: ΔG° = −58.03 × 103 J/mol − (373.2 K)(−176.6 J K−1 mol−1) = 7.88 × 103 J/mol ln K = 78. − (7.88 × 10 3 J / mol) = −2.540, K = e−2.540 = 0.0789 (8.3145 J K −1 mol −1 )(373.2 K ) − ΔH o ⎛ 1 ⎞ ΔSo is in the form of a straight line equation (y = mx + ⎜ ⎟+ R ⎝T⎠ R b). A graph of ln K vs. 1/T will yield a straight line with slope = m = −ΔH°/R and a y The equation ln K = intercept = b = ΔS°/R. From the plot: slope = Δy 0 − 40. = = −1.3 × 104 K Δx 3.0 × 10 −3 K −1 − 0 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 389 −1.3 × 104 K = −ΔH°/R, ΔH° = 1.3 × 104 K × 8.3145 J K−1 mol−1 = 1.1 × 105 J/mol y intercept = 40. = ΔS°/R, ΔS° = 40. × 8.3145 J K−1 mol−1 = 330 J K−1 mol−1 As seen here, when ΔH° is positive, the slope of the ln K vs. 1/T plot is negative. When ΔH° is negative, as in an exothermic process, the slope of the ln K vs. 1/T plot will be positive (slope = −ΔH°/R). 79. A graph of ln K vs. 1/T will yield a straight line with slope equal to -ΔH°/R and y intercept equal to ΔS°/R (see Exercise 10.78). a. Temp (°C) 0 25 35 40. 50. T(K) 273 298 308 313 323 1000/T (K−1) 3.66 3.36 3.25 3.19 3.10 Kw ln Kw 1.14 × 10−15 1.00 × 10−14 2.09 × 10−14 2.92 × 10−14 5.47 × 10−14 −34.408 −32.236 −31.499 −31.165 −30.537 ⎛1⎞ The straight-line equation (from a calculator) is: ln K = −6.91 × 103 ⎜ ⎟ − 9.09 ⎝T⎠ o − ΔH Slope = −6.91 × 103 K = R ΔH° = − (−6.91 × 103 K × 8.3145 J K−1 mol−1) = 5.75 × 104 J/mol = 57.5 kJ/mol ΔSo y intercept = −9.09 = , ΔS° = −9.09 × 8.3145 J K−1 mol−1 = −75.6 J K−1 mol−1 R b. From part a, ΔH° = 57.5 kJ/mol and ΔS° = −75.6 J K−1 mol−1. Assuming that ΔH° and ΔS° are temperature independent: 390 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY ΔG° = 57,500 J/mol − 647 K(−75.6 J K−1 mol−1) = 106,400 J/mol = 106.4 kJ/mol 80. The ln K vs. 1/T plot gives a straight line with slope = !ΔHE/R and y intercept = ΔSE/R. 1.352 × 104 K = !ΔHE/R, ΔHE = !(8.3145 J K−1 mol−1)(1.352 × 104 K) ΔHE = !1.124 × 105 J/mol = !112.4 kJ/mol !14.51 = ΔSE/R, ΔSE = (!14.51)(8.3145 J K−1 mol−1) = !120.6 J K−1 mol−1 Note that the signs for ΔH° and ΔS° make sense. When a bond forms, ΔH° < 0 and ΔS° < 0. 81. From Exercise 10.78, ln K = − ΔH o ΔSo + , R = 8.3145 J K−1 mol−1 RT R For two sets of K and T: − ΔH o ⎛ 1 ⎞ ΔSo − ΔH o ⎛ 1 ⎞ ΔSo ⎜ ⎟+ ⎜⎜ ⎟⎟ + ln K1 = ; ln K 2 = R ⎝ T1 ⎠ R R ⎜⎝ T2 ⎟⎠ R Subtracting the first expression from the second: ln K2 − ln K1 = ΔH o ⎛ 1 1 ⎞ K ΔH o ⎛ 1 1 ⎞ ⎜⎜ − ⎟⎟ ⎜⎜ − ⎟⎟ or ln 2 = R ⎝ T1 T2 ⎠ K1 R ⎝ T1 T2 ⎠ ⎛ 3.25 × 10 −2 ⎞ ⎛ 1 ΔH o 1 ⎞ ⎟= ⎜ ⎟ − ln⎜⎜ −1 −1 ⎜ ⎟ 8.84 8.3145 J K mol ⎝ 298 K 348 K ⎟⎠ ⎠ ⎝ −5.61 = (5.8 ×10−5 mol/J)(ΔH°), ΔH° = −9.7 × 104 J/mol For K = 8.84 at T = 25°C: ln(8.84) = ΔSo − (−9.7 × 10 4 J / mol) + (8.3145 J K −1 mol −1 )(298 K ) 8.3145 J K −1 mol −1 ΔSo = −37, ΔS° = −310 J K−1 mol−1 8.3145 We get the same value for ΔS° using K = 3.25 × 10−2 at T = 348 K data. ΔG° = −RT ln K. When K = 1.00 then ΔG° = 0 since ln(1.00) = 0. ΔG° = 0 = ΔH° − TΔS°. Assuming that ΔH° and ΔS° do not depend on temperature: ΔH° = TΔS°, T = ΔH o − 9.7 × 10 4 J / mol = = 310 K ΔSo − 310 J K −1 mol −1 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 391 Adiabatic Processes 82. Two equations derived in Section 10.14 for reversible adiabatic processes are T1V1γ-1 = T2V2γ−1 and P1V1γ = P2V2γ where γ = Cp/Cv. For a monoatomic ideal gas, Cp = (5/2)R and Cv = (3/2)R, so γ = 5/3 and γ − 1 = 2/3. ⎛V ⎞ T2 = T1 ⎜⎜ 1 ⎟⎟ ⎝ V2 ⎠ γ −1 γ ⎛ 5.00 L ⎞ ⎟⎟ = 298 K⎜⎜ ⎝ 12.5 L ⎠ 2/3 ⎛V ⎞ ⎛ 5.00 L ⎞ ⎟⎟ P2 = P1 ⎜⎜ 1 ⎟⎟ = 1.00 atm⎜⎜ ⎝ 12.5 L ⎠ ⎝ V2 ⎠ = 162 K 5/3 = 0.217 atm Because we have an adiabatic process, q = 0 and ΔE = w = nCvΔT. We need to calculate n. n= PV 1.00 atm × 5.00 L = = 0.204 mol RT 0.08206 L atm K −1 mol −1 × 298 K ΔE = w = 0.204 mol(3/2)(8.3145 J K−1 mol−1)(162 K − 298 K) = −346 J 83. For reversible adaiabatic process, P1V1( = P2V2(, where ( = Cp/Cv. For an ideal gas, Cp = Cv + R. Cp = Cv + R = 20.5 J K−1 mol−1 + 8.3145 J K−1 mol−1 = 28.8 J K−1 mol−1 V2( = V1 = Cp P1V1γ 28.8 , where ( = = = 1.40 Cv P2 20.5 nRT1 1.75 mol(0.08206 L atm K −1 mol−1 ) (294 K ) = = 28.1 L 1.50 atm P1 V21.40 = 1.50 atm (28.1 L)1.40 = 35.6, V2 = (35.6)1/1.40 = 12.8 L 4.50 atm To calculate w = ΔE = nCVΔT, we need T2 in order to calculate ΔT. T2 = 4.50 atm(12.8 L) P2 V2 = = 401 K nR 1.75 mol (0.08206 L atm K −1 mol −1 ) Note: we also could have used the formula T2/T1 = (V1/V2)(−1 to calculate T2. w = ΔE = nCVΔT = 1.75 mol(20.5 J K−1 mol−1)(401 K – 294 K) = 3.84 × 103 J 392 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY Note that for the adiabatic reversible compression, T increases, so ΔE increases. As work is added during the compression, the internal energy of the system increases (ΔE = w). 84. To calculate ΔE and ΔH, we need to determine the molar heat capacity of the ideal gas. As derived in Section 10.14 of the text, for a reversible adiabatic change (q = 0): ⎛ T2 ⎞ ⎜⎜ ⎟⎟ ⎝ T1 ⎠ Cv R ⎛T ⎞ ⎛V ⎞ ⎛ V1 ⎞ ⎟⎟ ; C v ln⎜⎜ 2 ⎟⎟ = R ln⎜⎜ 1 ⎟⎟, ⎝ T1 ⎠ ⎝ V2 ⎠ ⎝ V2 ⎠ = ⎜⎜ Since V2 = 2V1: ⎛V ⎞ ln⎜⎜ 1 ⎟⎟ V Cv = ⎝ 2⎠ R ⎛T ⎞ ln⎜⎜ 2 ⎟⎟ ⎝ T1 ⎠ Cv ln(1/2) = = 3.24 R ln(239 K/296 K) Cv = (3.24)(8.3145 J K−-1 mol−1) = 26.9 J K−1 mol−1 ΔE = nCvΔT = 1.50 mol(26.9 J K−1 mol−1)(239 K − 296 K) = −2,300 J = −2.3 kJ ΔH = ΔE + nRΔT = −2300 J + 1.50 mol(8.3145 J K−1 mol−1)(239 K − 296 K) ΔH = −3.0 × 103 J = −3.0 kJ 85. a. Isothermal reversible expansion: ΔT = 0 so ΔE = 0 w = −nRT ln(V2/V1) = −nRT ln(P1/P2) w = −1.00 mol(8.3145 J K−-1 mol−1)(300. K) ln(5.00/1.00) = −4.01 × 103 J Since ΔE = 0, q = −w = 4.01 × 103 J. b. Isothermal irreversible expansion against a constant external pressure of 1.00 atm, so ΔE = 0, and q = −w. w = −PexΔV; V1 = Vf = 1.00 mol (0.08206 L atm K −1 mol −1 ) 300. K = 4.92 L 5.00 atm 1.00 mol (0.08206)(300. K ) = 24.6 L 1.00 atm w = −(1.00 atm)(24.6 L – 4.92 L) × 101.3 J L−1 atm−1 = −1.99 × 103 J q = −w = 1.99 × 103 J As expected, we get more work from the reversible process. CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 393 c. Adiabatic reversible expansion: q = 0 so ΔE = w, ΔE = nCvΔT; we need Cv and ΔT to calculate ΔE and, in turn, w. In order to calculate T2, we need to determine V2. P1V1( = P2V2(, where ( = Cp/Cv. From part b, V1 = 4.92 L. For a gas behaving ideally: Cp = Cv + R, Cv = Cp – R = 37.1 J K−1 mol−1 – 8.3145 J K−1 mol−1 = 28.8 J K−1 mol−1 Cp = CV P V1.29 37.1 = 1.29; V21.29 = 1 1 28.8 P2 V21.29 = T2 = 5.00 atm (4.92 L)1.29 = 39.0, V2 = 17.1 L 1.00 atm P2 V2 1.00 atm (17.1 L) = = 208 K nR 1.00 mol(0.08206 L atm K −1 mol −1 ) ΔE = w = nCvΔT = 1.00 mol(28.8 J K−1 mol−1)(208 K – 300. K) = −2.65 × 103 J As expected, we do not get as much work in the adiabatic reversible process as when the gas expands isothermally and reversibly. 86. a. Isothermal reversible expansion: ΔSuniv = 0 (reversible process); ΔT = 0, so ΔE = 0, and qrev = – wrev = nRT ln(P1/P2). ΔSsys = q rev nRT ln(P1 / P2 ) = = nR ln(P1/P2) T T ΔSsys = 1.00 mol(8.3145 J K−1 mol−1) ln(5.00/1.00) = 13.4 J/K ΔSuniv = 0 = ΔSsys + ΔSsurr, ΔSsurr = ΔSsys = −13.4 J/K b. Isothermal irreversible expansion against a constant pressure: ΔSsys = 13.4 J/K (a state function); ΔE = 0 = q + w, q = −w = PexΔV Vi = 1.00 mol(0.08206 L atm K −1 mol −1 ) (300. K ) = 4.92 L 5.00 atm Vf = 1.00 mol(0.08206) (300. K ) = 24.6 L 1.00 atm q = PexΔV = 1.00 atm(24.6 L – 4.92 L) × 101.3 J L−1 atm−1 = 1.99 × 103 J 394 CHAPTER 10 ΔSsurr = SPONTANEITY, ENTROPY, AND FREE ENERGY − q actual − 1.99 × 103 J = = −6.63 J/K T 300. K ΔSuniv = ΔSsys + ΔSsurr = 13.4 J/K – 6.63 J/K = 6.77 J/K c. Adiabatic reversible expansion: ΔSuniv = 0 (reversible); qrev = 0 so ΔSsys = qrev/T = 0. − q actual − q rev = = 0. Or one could say that because ΔSuniv = 0, ΔSsys = −ΔSsurr, T T and since ΔSsys = 0, ΔSsurr must equal zero. ΔSsurr = Additional Exercises 87. When an ionic solid dissolves, positional probability increases, so ΔSsys is positive. Since temperature increased as the solid dissolved, this is an exothermic process, and ΔSsurr is positive (ΔSsurr = −ΔH/T). Since the solid did dissolve, the dissolving process is spontaneous, so ΔSuniv is positive (as it must be when ΔSsys and .ΔSsurr are both positive). 88. As any process occurs, ΔSuniv will increase; ΔSuniv cannot decrease. Time also goes in one direction, just as ΔSuniv goes in one direction. 89. wmax = ΔG; when ΔG is negative, the magnitude of ΔG is equal to the maximum possible useful work obtainable from the process (at constant T and P). When ΔG is positive, the magnitude of ΔG is equal to the minimum amount of work that must be expended to make the process spontaneous. Due to waste energy (heat) in any real process, the amount of useful work obtainable from a spontaneous process is always less than wmax, and for a nonspontaneous reaction, an amount of work greater than wmax must be applied to make the process spontaneous. 90. The introduction of mistakes is an effect of entropy. The purpose of redundant information is to provide a control to check the "correctness" of the transmitted information. 91. HF(aq) ⇌ H+(aq) + F−(aq); ΔG = ΔG° + RT ln [H + ][F − ] [ HF] ΔG°= −RT ln K = −(8.3145 J K−1 mol−1)(298 K) ln(7.2 × 10−4) = 1.8 × 104 J/mol a. The concentrations are all at standard conditions, so ΔG = ΔG = 1.8 × 104 J/mol (since Q = 1.0 and ln Q = 0). Because ΔG° is positive, the reaction shifts left to reach equilibrium. b. ΔG = 1.8 × 104 J/mol + (8.3145 J K−1 mol−1)(298 K) ln (2.7 × 10 −2 ) 2 0.98 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 395 ΔG = 1.8 × 104 J/mol − 1.8 × 104 J/mol = 0 Because ΔG = 0, the reaction is at equilibrium (no shift). c. ΔG = 1.8 × 104 + 8.3145(298) ln (1.0 × 10 −5 ) 2 = −1.1 × 104 J/mol; shifts right 1.0 × 10 −5 d. ΔG = 1.8 × 104 + 8.3145(298) ln 7.2 × 10 −4 (0.27) = 1.8 × 104 − 1.8 × 104 = 0; 0.27 at equilibrium e. ΔG = 1.8 × 104 + 8.3145(298) ln 92. 1.0 × 10 −3 (0.67) = 2 × 103 J/mol; shifts left 0.52 ⎛ [K + ] K+(blood) ⇌ K+(muscle) ΔG° = 0; ΔG = RT ln ⎜⎜ + m ⎝ [ K ]b ΔG = ⎞ ⎟ ; ΔG = wmax ⎟ ⎠ 8.3145 J ⎛ 0.15 ⎞ 3 (310. K) ln ⎜ ⎟ , ΔG = 8.8 × 10 J/mol = 8.8 kJ/mol K mol ⎝ 0.0050 ⎠ At least 8.8 kJ of work must be applied to transport 1 mol K+. Other ions will have to be transported in order to maintain electroneutrality. Either anions must be transported into the cells, or cations (Na+) in the cell must be transported to the blood. The latter is what happens: [Na+] in blood is greater than [Na+] in cells as a result of this pumping. 8.8 kJ 1 mol ATP × = 0.29 mol ATP + 30.5 kJ mol K 93. ΔS° will be negative because 2 moles of gaseous reactants forms 1 mole of gaseous product. For ΔG° to be negative, ΔH must be negative (exothermic). For this sign combination of ΔH° and ΔS°, K decreases as T increases because ΔG° becomes more positive (ΔG° = −RT ln K). Therefore, the ratio of the partial pressure of PCl5 (a product) to the partial pressure of PCl3 (a reactant) will decrease when T is raised. 94. At equilibrium: PH 2 ⎛ ⎞ ⎛ 0.08206 L atm ⎞ 1.10 × 1013 molecules ⎜ ⎟⎜ ⎟⎟ (298 K ) 23 ⎜ mol K 6.022 × 10 molecules / mol ⎟⎠ ⎜⎝ nRT ⎠ ⎝ = = = V 1.00 L 4.47 × 10−10 atm The pressure of H2 decreased from 1.00 atm to 4.47 × 10−10 atm. Essentially all the H2 and Br2 has reacted. Therefore, PHBr = 2.00 atm since there is a 2:1 mole ratio between HBr and 396 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY H2 in the balanced equation. Since we began with equal moles of H2 and Br2, then we will have equal moles of H2 and Br2 at equilibrium. Therefore, PH 2 = PBr2 = 4.47 × 10−10 atm. K= 2 PHBr (2.00) 2 = = 2.00 × 1019; assumptions good. PH 2 × PBr2 ( 4.47 × 10 −10 ) 2 ΔG° = −RT ln K = −(8.3145 J K−1 mol−1)(298 K) ln(2.00 × 1019) = −1.10 × 105 J/mol ΔS° = 95. ΔH o − ΔG o − 103,800 J − (−1.10 × 105 J ) = = 20 J/K T 298 K Using Le Chatelier's principle: A decrease in pressure (volume increases) will favor the side with the greater number of particles. Thus 2 I(g) will be favored at low pressure. Looking at ΔG: ΔG = ΔG° + RT ln (PI2 / PI 2 ); ln(PI2 / PI 2 ) > 0 when PI = PI 2 = 10 atm, and ΔG is positive (not spontaneous). But at PI = PI 2 = 0.10 atm, the logarithm term is negative. If |RT ln Q| > ΔG°, then ΔG becomes negative, and the reaction is spontaneous. 96. Ba(NO3)2(s) ⇌ Ba2+(aq) + 2 NO3−(aq) K = Ksp; ΔG° = −561 + 2(−109) − (−797) = 18 kJ ΔG° = −RT ln Ksp, ln Ksp = − ΔG o − 18,000 J = = −7.26 RT 8.3145 J K −1 mol −1 (298 K ) Ksp = e−7.26 = 7.0 × 10−4 97. NaCl(s) ⇌ Na+(aq) + Cl−(aq) K = Ksp = [Na+][Cl−] ΔGE = [(!262 kJ) + (!131 kJ)] – (!384 kJ) = !9 kJ = !9000 J ⎡ ⎤ − (−9000 J ) ΔGE = !RT ln Ksp, Ksp = exp ⎢ ⎥ = 38 = 40 −1 −1 ⎣ 8.3145 J K mol × 298 K ⎦ NaCl(s) ⇌ Na+(aq) + Cl−(aq) Initial s = solubility (mol/L) 0 Equil. s Ksp = 40 0 s Ksp = 40 = s(s), s = (40)1/2 = 6.3 = 6 M = [Cl−] 98. ΔGE = ΔHE ! TΔSE = !28.0 × 103 J – 298 K(!175 J/K) = 24,200 J ΔGE = !RT ln K, ln K = K = e −9.767 = 5.73 × 10 −5 − ΔG o − 24,200 J = = !9.767 RT 8.3145 J K −1 mol −1 × 298 K CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY B Initial Change Equil. + 0.125 M !x 0.125 ! x Kb = 5.73 × 10 −5 = H2O ⇌ BH+ 0 +x x + OH− 397 K = Kb = 5.73 × 10 −5 ~0 +x x x2 x2 [BH + ][OH − ] = ≈ , x = [OH−] = 2.68 × 10 −3 M 0.125 − x 0.125 [B] pOH = !log(2.68 × 10 −3 ) = 2.572; pH = 14.000 – 2.572 = 11.428; assumptions good. 99. 100. ΔS is more favorable (less negative) for reaction 2 than for reaction 1, resulting in K2 > K1. In reaction 1, seven particles in solution are forming one particle in solution. In reaction 2, four particles are forming one, which results in a smaller decrease in positional probability than for reaction 1. ΔGE = !RT ln K = ΔHE ! TΔSE; HX(aq) ⇌ H+(aq) + X−(aq) Ka reaction; the value of Ka for HF is less than one, while the other hydrogen halide acids have Ka > 1. In terms of ΔGE, HF must have a positive ΔG orxn value, while the other H-X acids have ΔGrxn < 0. The reason for the sign change in the Ka value between HF versus HCl, HBr, and HI is entropy. ΔS for the dissociation of HF is very large and negative. There is a high degree of ordering that occurs as the water molecules associate (hydrogen bond) with the small F− ions. The entropy of hydration strongly opposes HF dissociating in water, so much so that it overwhelms the favorable hydration energy, making HF a weak acid. 101. Note that these substances are not in the solid state but are in the aqueous state; water molecules are also present. There is an apparent increase in ordering (decrease in positional probability) when these ions are placed in water as compared to the separated state. The hydrating water molecules must be in a highly ordered arrangement when surrounding these anions. 102. ΔS = ΔH vap q rev 8.20 × 103 J / mol = ; for methane: ΔS = = 73.2 J K−1 mol−1 T T 112 K For hexane: ΔS = Vmet = 28.9 × 103 J / mol = 84.5 J K−1 mol−1 342 K nRT 1.00 mol(0.08206)(112 K ) nRT = = 9.19 L; Vhex = = R(342 K) = 28.1 L P 1.00 atm P ΔShex − ΔSmet = 84.5 − 73.2 = 11.3 J K−1 mol−1; R ln(Vhex/Vmet) = 9.29 J K−1 mol−1 As the molar volume of a gas increases, ΔSvap also increases. In the case of hexane and methane, the difference in molar volume accounts for 82% of the difference in the entropies. 103. S = k ln Ω; S has units of J K−1 mol−1, and k has units of J/K (k = 1.38 × 10−23 J/K). 398 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY To make units match: S (J K−1 mol−1) = NAk ln Ω when NA = Avogadro's number 189 J K−1 mol−1 = 8.31 J K−1 mol−1 ln Ωg −(70. J K−1 mol−1 = 8.31 J K−1 mol−1 ln Ωl) __________________________________________________ Subtracting: 119 J K−1 mol−1 = 8.31 J K−1 mol−1(ln Ωg − ln Ωl) 14.3 = ln(Ωg/Ωl), 104. Ωg Ω1 = e14.3 = 1.6 × 106 H2O(l, 298 K) → H2O(g, V = 1000. L/mol); Break process into two steps: Step 1: H2O(l, 298 K) → H2O(g, 298 K, V = nR (298) = 24.5 L) 1.00 atm ΔS = SoH 2O ( g ) − SoH 2O ( l ) = 189 J/K − 70. J/K = 119 J/K Step 2: H2O(g, 298 K, 24.5 L) → H2O(g, 298 K, 1000. L) ΔS = nR ln ⎛ 1000. L ⎞ V2 ⎟⎟ = 30.8 J/K = (1.00 mol)(8.3145 J K−1 mol−1) ln ⎜⎜ V1 ⎝ 24.5 L ⎠ ΔStotal = 119 + 30.8 = 150. J/K ΔG = 44.02 × 103 J − 298 K(150. J/K) = −700 J; spontaneous For H2O(l, 298 K) → H2O(g, 298 K, V = 100. L/mol): ⎛ 100. L ⎞ ⎟⎟ = 131 J/K ΔS = 119 J/K + (8.3145 J/K) ln ⎜⎜ ⎝ 24.5 L ⎠ ΔG = 44.02 × 103 J − 298 K(131 J/K) = 5.0 × 103 J; not spontaneous 105. Isothermal: ΔH = 0 (assume ideal gas) ⎛V ⎞ ⎛ 1.00 L ⎞ ⎟⎟ = −38.3 J/K ΔS = nR ln ⎜⎜ 2 ⎟⎟ = (1.00 mol)(8.3145 J K−1 mol−1) ln ⎜⎜ ⎝ 100.0 L ⎠ ⎝ V1 ⎠ ΔG = ΔH − TΔS = 0 − (300. K)(−38.3 J/K) = +11,500 J = 11.5 kJ 106. a. Free expansion w = 0, since Pext = 0; ΔE = nCvΔT, since ΔT = 0, ΔE = 0 ΔE = q + w, q = 0; ΔH = nCpΔT, since ΔT = 0, ΔH = 0 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 399 ⎛V ⎞ ⎛ 40.0 L ⎞ ⎟⎟ = 2.39 J/K ΔS = nR ln ⎜⎜ 2 ⎟⎟ = (1.00 mol)(8.3145 J K−1 mol−1) ln ⎜⎜ ⎝ 30.0 L ⎠ ⎝ V1 ⎠ ΔG = ΔH − TΔS = 0 − 300. K(2.39 J/K) = −717 J b. Reversible expansion ΔE = 0; ΔH = 0; ΔS = 2.39 J/K; ΔG = −717 J; these are state functions. ⎛V ⎞ ⎛ 40.0 L ⎞ ⎟⎟ = −718 J wrev = −nRT ln ⎜⎜ 2 ⎟⎟ = −(1.00 mol)(8.3145 J K−1 mol−1)(300. K) ln ⎜⎜ ⎝ 30.0 L ⎠ ⎝ V1 ⎠ ΔE = 0 = q + w, qrev = -wrev = 718 J Summary: (a) Free expansion q w ΔE ΔH ΔS ΔG 107. b) Reversible 0 0 0 0 2.39 J/K −717 J 718 J −718 J 0 0 2.39 J/K −717 J a. Isothermal: ΔE = 0 and ΔH = 0 if gas is ideal. ΔS = nR ln(P1/P2) = (1.00 mol)(8.3145 J K−1 mol−1) ln(5.00 atm/2.00 atm) = 7.62 J/K T= PV 5.00 atm × 5.00 L = = 305 K nR 1.00 mol × 0.08206 L atm K −1 mol −1 ΔG = ΔH − TΔS = 0 − (305 K)(7.62 J/K) = −2320 J w = −PΔV = −(2.00 atm)ΔV, where Vf = nRT = 12.5 L 2.00 atm w = −2.00 atm (12.5 − 5.00 L) × (101.3 J L−1 atm−1) = −1500 J ΔE = 0 = q + w, q = 1500 J b. Second law, ΔSuniv > 0 for spontaneous processes: ΔSuniv = ΔSsys + ΔSsurr = ΔSsys − ΔSuniv = 7.62 J/K − q actual T 1500 J = 7.62 − 4.9 = 2.7 J/K; thus the process is spontaneous. 305 K 400 108. CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY Isothermal: ΔE = 0, ΔH = 0; Vf = nRT/5.00 atm = 2.00 L ΔS = nR ln(P1/P2) = (1.00 mol)(8.3145 J K−1 mol−1) ln(1.50/5.00) = −10.0 J/K w = −5.00 atm(2.00 L − 6.67 L)(101.3 J L−1 atm−1) = 2370 J ΔE = 0 = q + w, q = −2370 J; ΔSsurr = 2370 J −q = = 19.4 J/K T 122 K ΔSuniv = ΔSsys + ΔSsurr = −10.0 J/K + 19.4 J/K = 9.4 J/K ΔG = ΔH − TΔS = 0 − (122 K)(−10.0 J/K) = 1220 J Challenge Problems 109. a. V1 = nRT1 2.00 mol × 0.08206 L atm K −1 mol −1 × 298 K = = 24.5 L P1 2.00 atm For an adiabatic reversible process, P1V1γ = P2V2γ and T1V1γ−-1 = T2V2γ−1, where γ = Cp/Cv. Because argon is a monoatomic gas, Cp = (5/2)R and Cv = (3/2)R, so γ = 5/3. V2γ = γ P1V1 2.00 atm (24.5 L) 5 / 3 = = 413, V2 = (413)3/5 = 37.1 L P2 1.00 atm We can either use the ideal gas law or the T1Vγ−1 = T2V2γ−1 equation to calculate the final temperature. Using the ideal gas law: T2 = P2 V2 1.00 atm × 37.1 L = = 226 K nR 2.00 mol × 0.08206 L atm K −1 mol −1 b. For an adiabatic process (q = 0), ΔE = w = nCvΔT. For an expansion against a fixed external pressure, w = −PΔV. From the ideal gas equation (see part a), V1 = 24.5 L. w = −PΔV = nCvΔT ⎛ 101.3 J ⎞ ⎛ 8.3145 J ⎞ ⎟⎟ = 2.00 mol(3/2) ⎜⎜ ⎟⎟(T2 − 298 K ) 1.00 atm(V2 – 24.5 L) ⎜⎜ ⎝ L atm ⎠ ⎝ mol K ⎠ We will ignore units from here. Note that both sides of the equation are in units of J. (−101)V2 + 2480 = (24.9)T2 − 7430 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY To solve for T2, we need to find an expression for V2.. Using the ideal gas equation: V2 = nRT2 2.00 mol (0.08206)T2 = = (0.164)T2; substituting: P2 1.00 −101(0.164 T2) + 2480 = (24.9)T2 − 7430, (41.5)T2 = 9910, T2 = 239 K 110. For the processes to be spontaneous, ΔSuniv > 0, and ΔSuniv = ΔS + ΔSsurr. ΔS = q rev q −q −q , ΔSsurr = , ΔSuniv = rev T T T Since the processes are isothermal, ΔH = 0, ΔE = 0, q = −w; n = 1.0 mol, V1 = 5.0 L, P1 = 5.0 atm, giving T = 305 K from the ideal gas equation. a. If P2 = 2.0 atm, then V2 = 12.5 L. ⎛V ⎞ ⎛ 12.5 ⎞ qrev = nRT ln ⎜⎜ 2 ⎟⎟ = 1.0(8.3145)(305) ln ⎜ ⎟ = 2320 J ⎝ 5.0 ⎠ ⎝ V1 ⎠ w = −P(ΔV) = −2.0 atm(12.5 L − 5.0 L) = −15 L atm = −1520 J q = 1520 J; ΔSuniv = 2320 − 1520 = 2.6 J/K 305 Because ΔSuniv > 0, the process is spontaneous. ⎛V ⎞ ⎛ 5.0 ⎞ b. qrev = nRT ln ⎜⎜ 2 ⎟⎟ = (1.0)(8.3145)(305) ln ⎜ ⎟ = −2320 J ⎝ 12.5 ⎠ ⎝ V1 ⎠ w = −P(ΔV) = 37.5 L atm = 3800 J, q = −3800 J ΔSuniv = − 2320 − ( −3800) = 4.9 J/K (ΔSuniv > 0) 305 q rev , so TΔS = qrev. T For the expansion, ΔG = −2320 J; for the compression, ΔG = +2320 J. c. ΔG = ΔH − TΔS; ΔH = 0 for both; ΔS = Because pressure is not constant, ΔG cannot be used to predict spontaneity. 401 402 111. CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY q rev ; isothermal: ΔT = 0 so ΔE = 0 and q = −w. T q 855 J Reversible expansion: qrev = 855 J; ΔS = rev = = 2.87 J/K T 298 K For the compression, we go back to the initial state for the overall process (expansion then compression). Because the initial and overall final state are the same, all state functions like ΔS must equal zero. a. ΔS = ΔSoverall = 0 = ΔSexp + ΔScomp, ΔScomp = −ΔSexp = −2.87 J/K Note: Even though the compression step is not a reversible process, it still must have qrev = −855 J. ΔScomp = q rev − 855 J = = −2.87 J/K T 298 K b. ΔSuniv, overall = ΔSsys, overall + ΔSsurr, overall ΔSsys, overall = 2.87 J/K – 2.87 J/K = 0 ΔSsurr, exp = − q actual − 855 J = = – 2.87 J/K T 298 K From the problem, the compression step is isothermal, so q = −w. The work done on the system is 2(855) J, so q = −2(855) J. ΔSsurr, comp = − q actual − [−2(855) J ] = = 5.74 J/K T 298 K ΔSsurr, overall = –2.87 J/K + 5.74 J/K = 2.87 J/K ΔSuniv, overall = 0 + 2.87 J/K = 2.87 J/K 112. It is true that ΔS is related to qrev and ΔSsurr is related to qactual: ΔS = − q actual q rev ; ΔSsurr = T T However, it is not true that the magnitude of qrev is always greater than the magnitude of qactual. Reference Table 10.3 and specifically the compression experiments. For the isothermal compression of an ideal gas, the magnitude of qactual is always greater than qrev. Only when the compression is done reversibly does qactual = qrev. Also consider a sample of ice and water at 0°C and 1 atm. Here this system is at equilibrium, so qrev = qactual (ΔS = ΔSsurr). One cannot say that |ΔS| is always larger than |ΔSsurr| because |qrev| is not always greater than |qactual|. CHAPTER 10 113. SPONTANEITY, ENTROPY, AND FREE ENERGY 403 The system is the 1.00-L sample of water. The process is the cooling of the water from 90.°C to 25°C. The surroundings are the room (at 25°C) and everything else. 1.00 × 103 mL × 1.00 g 1 mol × = 55.5 mol H2O mL 18.02 g ΔS = nCp ln(T2/T1) = 55.5 mol(75.3 J K−1 mol−1) ln(298/363) = −825 J/K ΔSsurr = − q actual ; qactual = nCpΔT = 55.5 mol(75.3 J K−1 mol−1)(−65 K) = −2.72 × 105 J T ΔSsurr = − q actual − (−2.72 × 105 J ) = = 913 J/K 298 K T ΔSuniv = ΔS + ΔSsurr = −825 J/K + 913 J/K = 88 J/K Not too surprising, this is a spontaneous process due to the favorable ΔSsurr term. 114. a. Vessel 1: At 0°C, this system is at equilibrium, so ΔSuniv = 0 and ΔS = ΔSsurr. Because the vessel is perfectly insulated, q = 0 so ΔSsurr = 0 = ΔSsys. b. Vessel 2: The presence of salt in water lowers the freezing point of water to a temperature below 0°C. In vessel 2 the conversion of ice into water will be spontaneous at 0°C, so ΔSuniv > 0. Because the vessel is perfectly insulated, ΔSsurr = 0. Therefore, ΔSsys must be positive (ΔS > 0) in order for ΔSuniv to be positive. 115. Kp = PCO 2 ; to prevent Ag2CO3 from decomposing, PCO 2 should be greater than Kp. From Exercise 10.78, ln K = ln K2 ΔH o = K1 R ⎛1 1 ⎜⎜ − T2 ⎝ T1 ΔSo − ΔH o + . For two conditions of K and T, the equation is: RT R ⎞ ⎟⎟ ⎠ Let T1 = 25°C = 298 K, K1 = 6.23 × 10−3 torr; T2 = 110.°C = 383 K, K2 = ? ln K2 79.14 × 103 J / mol = 6.23 × 10 −3 torr 8.3145 J K −1 mol −1 ln K2 = 7.1, 6.23 × 10 −3 ⎛ 1 1 ⎜⎜ − 383 K ⎝ 298 K ⎞ ⎟⎟ ⎠ K2 = e7.1 = 1.2 × 103, K2 = 7.5 torr −3 6.23 × 10 To prevent decomposition of Ag2CO3, the partial pressure of CO2 should be greater than 7.5 torr. 404 CHAPTER 10 116. ⇌ HX(aq) Initial Equil. 0.10 M 0.10 – x SPONTANEITY, ENTROPY, AND FREE ENERGY H+(aq) ~0 x + X−(aq) Ka = [H + ][X − ] [HX] 0 x From problem: x = [H+] = 10 −5.83 = 1.5 × 10 −6 ; Ka = (1.5 × 10 −6 ) 2 = 2.3 × 10 −11 0.10 − 1.5 × 10 −6 ΔGE = !RT ln K = !8.3145 J K−1 mol−1(298 K) ln(2.3 × 10 −11 ) = 6.1 × 104 J/mol = 61 kJ/mol 117. 3 O2(g) ⇌ ln K = − ΔG o − 326 × 103 J = = −131.573, K = e−131.573 = 7.22 × 10−58 −1 −1 RT (8.3145 J K mol ) (298 K ) 2 O3(g); ΔH° = 2(143) = 286 kJ; ΔG° = 2(163) = 326 kJ We need the value of K at 230. K. From Exercise 10.78: ln K = − ΔH o ΔSo + RT R For two sets of K and T: ln K2 ΔH o ⎛ 1 1 ⎞ ⎜⎜ ⎟ − = K1 R ⎝ T1 T2 ⎟⎠ Let K2 = 7.22 × 10−58, T2 = 298; K1 = K230, T1 = 230. K; ΔH° = 286 × 103 J ln 7.22 × 10 −58 286 × 103 ⎛ 1 1 ⎞ = − ⎜ ⎟ = 34.13 K 230 8.3145 ⎝ 230. 298 ⎠ 7.22 × 10 −58 = e34.13 = 6.6 × 1014, K230 = 1.1 × 10−72 K 230 K230 = 1.1 × 10−72 = PO2 3 PO3 2 = PO2 3 (1.0 × 10 −3 atm) 3 , PO3 = 3.3 × 10−41 atm The volume occupied by one molecule of ozone is: V= nRT (1 / 6.022 × 10 23 mol)(0.8206 L atm K −1 mol−1 )(230. K ) = , V = 9.5 × 1017 L − 41 P (3.3 × 10 atm) Equilibrium is probably not maintained under these conditions. When only two ozone molecules are in a volume of 9.5 × 1017 L, the reaction is not at equilibrium. Under these conditions, Q > K and the reaction shifts left. But with only two ozone molecules in this huge volume, it is extremely unlikely that they will collide with each other. At these conditions, the concentration of ozone is not large enough to maintain equilibrium. CHAPTER 10 118. SPONTANEITY, ENTROPY, AND FREE ENERGY 405 ΔH = nCpΔT = 5/2 R(ΔT) = 5/2 (8.3145 J K−1 mol−1)(300. − 200. K) = 2080 J ΔE = nCvΔT =3/2 R(ΔT) = 3/2 (8.3145 J K−1 mol−1)(300. − 200. K) = 1250 J w = −P(ΔV) = −(nR)(ΔT) = 1.00 mol(8.3145 J K−1 mol−1)(100. K) = 831 J q = nCPΔT = 5/2 RΔT = ΔH = 2080 J ⎛T ⎞ ⎛ 5 ⎞ ⎛ 300. ⎞ ΔS = nC p ln ⎜⎜ 2 ⎟⎟ = (1.00) ⎜ R ⎟ ln ⎜ ⎟ = 8.43 J/K ⎝ 2 ⎠ ⎝ 200. ⎠ ⎝ T1 ⎠ ΔS due to ΔV = nR ln ⎛T ⎞ V2 = nR ln ⎜⎜ 2 ⎟⎟ = 3.37 J/K V1 ⎝ T1 ⎠ ⎛V ⎞ ⎛3 ⎞ ⎛T ⎞ ΔS due to ΔT = nC V ln ⎜⎜ 2 ⎟⎟ = n ⎜ R ⎟ ln ⎜⎜ 2 ⎟⎟ = 5.06 J/K ⎝ 2 ⎠ ⎝ T1 ⎠ ⎝ V1 ⎠ G = H − TS, ΔG = ΔH − Δ(TS) = ΔH − (T2S2 − T1S1) Since ΔS = 8.43 J/K, S°1 = 8.00 J/K, so So2 = 16.43 J/K. ΔG = 2080 J − [(300.)(16.43) − (200.)(8.00)] = −1250 J 119. a. ΔG° = G oB − G oA = 11,718 − 8996 = 2722 J ⎛ − ΔG o ⎞ ⎡ ⎤ − 2722 J ⎟ = exp ⎢ K = exp⎜⎜ ⎥ = 0.333 −1 −1 ⎟ ⎣ (8.3145 J K mol )(298 K ) ⎦ ⎝ RT ⎠ b. Since Q = 1.00 > K, reaction shifts left. Let x = atm of B(g) that reacts to reach equilibrium. A(g) Initial Equil. K= 1.00 atm 1.00 + x ⇌ B(g) K = PB/PA 1.00 atm 1.00 − x 1.00 − x = 0.333, 1.00 − x = 0.333 + (0.333)x, x = 0.50 atm 1.00 + x PB = 1.00 − 0.50 = 0.50 atm; PA = 1.00 + 0.50 = 1.50 atm c. ΔG = ΔG° + RT ln Q = ΔG° + RT ln(PB/PA) ΔG = 2722 J + (8.3145)(298) ln(0.50/1.50) = 2722 J − 2722 J = 0 (carrying extra sig. figs.) 406 120. CHAPTER 10 The liquid water will evaporate at first, and eventually, an equilibrium will be reached (physical equilibrium). • • • • • • 121. SPONTANEITY, ENTROPY, AND FREE ENERGY Since evaporation is an endothermic process, ΔH is positive. Since H2O(g) has a greater positional probability, ΔS is positive. Since we don’t know the relative magnitudes of ΔH and ΔS, we cannot identify the sign for ΔG (since it is not a constant pressure problem, ΔG does not tell us about spontaneity). The water will become cooler (the higher-energy water molecules leave); thus ΔTwater will be negative. Since the vessel is insulated, q = 0, so ΔSsurr = 0. Because the process occurs, it is spontaneous, so ΔSuniv is positive. Step 1: ΔE = 0 and ΔH = 0 since ΔT = 0 w = −PΔV = −(9.87 × 10−3 atm)ΔV; V = nRT , R = 0.08206 L atm K−1 mol−1 P ⎛ 1 1⎞ ΔV = Vf − Vi = nRT ⎜⎜ − ⎟⎟ ⎝ Pf Pi ⎠ ⎛ ⎞ 1 1 ⎟ ΔV = 1.00 mol(0.08206)(298 K) ⎜⎜ − −3 −2 2.45 × 10 atm ⎟⎠ ⎝ 9.87 × 10 atm ΔV = 1480 L (we will carry all values to three sig. figs.) w = −(9.87 × 10−3 atm)(1480 L) = −14.6 L atm(101.3 J L−1 atm−1) = −1480 J ⎛P ⎞ ΔE = q + w = 0, q = −w = +1480 J; ΔS = nR ln ⎜⎜ 1 ⎟⎟ ⎝ P2 ⎠ ⎛ 2.45 × 10 −2 atm ⎞ ⎟ , ΔS = 7.56 J/K ΔS = 1.00 mol(8.3145 J K−1 mol−1) ln ⎜⎜ −3 ⎟ ⎝ 9.87 × 10 atm ⎠ ΔG = ΔH − TΔS = 0 − 298 K(7.56 J/K) = −2250 J Step 2: ΔE = 0, ΔH = 0 ⎛ ⎞ nRT nRT 101.3 J ⎟L × w = − (4.93 × 10 −3 atm)⎜⎜ − = −1240 J −3 −3 ⎟ L atm 9.87 × 10 atm ⎠ ⎝ 4.93 × 10 ⎛ 9.87 × 10 −3 atm ⎞ ⎟ = 5.77 J/K q = −w = 1240 J; ΔS = nR ln ⎜⎜ −3 ⎟ ⎝ 4.93 × 10 atm ⎠ ΔG = 0 − 298K(5.77 J/K) = −1720 J CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 407 Step 3: ΔE = 0, ΔH = 0 ⎛ ⎞ nRT nRT 101.3 J ⎟L × w = −(2.45 × 10−3 atm) ⎜⎜ − = −1250 J −3 −3 ⎟ L atm 4.93 × 10 ⎠ ⎝ 2.45 × 10 ⎛ 4.93 × 10 −3 atm ⎞ ⎟ = 5.81 J/K q = −w = 1250 J; ΔS = nR ln ⎜⎜ −3 ⎟ ⎝ 2.45 × 10 atm ⎠ ΔG = 0 − 298 K(5.81 J/K) = −1730 J q 122. w ΔE ΔS ΔH ΔG Step 1 Step 2 Step 3 1480 J 1240 J 1250 J −1480 J −1240 J −1250 J 0 0 0 7.56 J/K 5.77 J/K 5.81 J/K 0 0 0 −2250 J −1720 J −1730 J Total 3970 J −3970 J 0 19.14 J/K 0 −5.70 × 103 J a. Isothermal: ΔE = 0, ΔH = 0; PV = nRT, R = 0.08206 L atm K−1 mol−1 ⎞ ⎛ nRT nRT 101.3 J ⎟L × = −2230 J w = −PΔV = − 2.45 × 10 −3atm⎜⎜ − −3 −2 ⎟ L atm 2.45 × 10 ⎠ ⎝ 2.45 × 10 q = −w = 2230 J ⎛ 2.45 × 10 −2 atm ⎞ ⎛P ⎞ ⎟ = 19.1 J/K ΔS = nR ln ⎜⎜ 1 ⎟⎟ = (1.00 mol)(8.3145 J K−1 mol−-1) ln ⎜⎜ −3 ⎟ ⎝ P2 ⎠ ⎝ 2.45 × 10 atm ⎠ ΔG = ΔH − TΔS = 0 − (298 K)(19.1 J/K) = −5.69 × 103 J = −5.69 kJ b. ΔE = 0; ΔH = 0; ΔS = 19.1 J/K; ΔG = −5.69 × 103 J; Same as in part a since these are all state functions. ΔS = q rev , qrev = TΔS = 298 K(19.1 J/K) = 5.69 × 103 J = 5.69 kJ T ΔE = 0 = q + w, wrev = −qrev = −5.69 × 103 J = −5.69 kJ c. ΔE = 0; ΔH = 0; ΔS = −19.1 J/K; ΔG = 5690 J a.) (The signs are opposite those in part ⎛ ⎞ nRT nRT 101.3 J ⎟L × − = 22,300 J = 22.3 kJ w = −2.45 × 10−2 atm ⎜⎜ −2 −3 ⎟ L atm 2.45 × 10 ⎠ ⎝ 2.45 × 10 ΔE = q + w = 0, q = −22.3 kJ 408 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY d. 2.45 x 10-2 P (atm) (a) (c) (b) 2.45 x 10-3 V1 e. ΔSsurr = V2 − q actual − 2230 J ; ΔSsurr, a = = −7.48 J/K T 298 K ΔSsurr, b = −ΔS = −19.1 J/K (reversible process); ΔSsurr, c = 123. 22,300 J = 74.8 J/K 298 K a. ΔG° = 2 mol(−394 kJ/mol) − 2 mol(−137 kJ/mol) = −514 kJ ⎛ − ΔG o ⎞ ⎛ ⎞ − (−514,000 J ) ⎟ = exp⎜ ⎟ = 1.24 × 1090 K = exp ⎜⎜ −1 −1 ⎜ ⎟ ⎟ ⎝ (8.3145 J K mol ) (298 K ) ⎠ ⎝ RT ⎠ b. ΔS° = 2(214 J/K) − [2(198 J/K) + 205 J/K] = −173 J/K 2 CO(1.00 atm) + O2(1.00 atm) → 2 CO2(1.00 atm) ΔS° = −173 J/K 2 CO2(1.00 atm) → 2 CO2(10.0 atm) ΔS = nR ln(P1/P2) = −38.3 J/K ΔS = nR ln(P1/P2) = 38.3 J/K 2 CO(10.0 atm) → 2 CO(1.00 atm) O2(10.0 atm) → O2(1.00 atm) ΔS = nR ln(P1/P2) = 19.1 J/K __________________________________________________________________________ ΔS = −173 + 19.1 = −154 J/K 2 CO(10.0 atm) + O2(10.0 atm) → CO2(10.0 atm) 124. O2(g) → 2 SO3(g) 2 SO2(g) + _________________________________________ ΔH of −297 kJ/mol 0 −396 205 257 S° 248 J K−1 mol−1 __________________________________________ ΔH o298 = 2(−396) − 2(−297) = −198 kJ; ΔSo298 = 2(257) − [205 + 2(248)] = −187 J/K Set up a thermochemical cycle to convert to T = 227°C = 500. K. 2 SO2(g, 227°C) → 2 SO2(g, 25°C) ΔH1 = nCpΔT O2(g, 227°C) → O2(g, 25°C) ΔH2 = nCpΔT 2 SO2(g, 25°C) + O2(g, 25°C) → 2 SO3(g, 25°C) ΔH3 = ΔH o298 = −198 kJ ΔH4 = nCpΔT 2 SO3(g, 25°C) → 2 SO3(g, 227°C) ___________________________________________________________________________ o ΔH 500 = ΔH1 + ΔH2 + ΔH3 + ΔH4 2 SO2(g, 227°C) + O2(g, 227°C) → 2 SO3(g, 227°C) CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY o ΔH 500 = 2 mol × 409 39.9 J K −1 mol −1 29.4 × (−202 K) + 1 × × (−202) − 198 kJ + 1000 J / kJ 1000 2× 50.7 × 202 1000 o ΔH 500 = −16.1 kJ − 5.94 kJ − 198 kJ + 20.5 kJ = −199.5 kJ = −200. kJ For the same cycle but using the equation ΔS = nCp ln (T2/T1) and ΔSo298 : o ΔS500 = 2(39.9) ln(298 K/500. K) + 1(29.4) ln(298/500.) − 187 + 2(50.7) ln(500./298) o ΔS500 = −41.3 J/K − 15.2 J/K − 187 J/K + 52.5 J/K = −191 J/K 125. T(°C) T(K) −200. -180. -160. -140. -100. -60. -30. -10. 0 73 93 113 133 173 213 243 263 273 Cp (J K−1 mol−1) 12 15 17 19 24 29 33 36 37 Cp/T (J K−2 mol−1) 0.16 0.16 0.15 0.14 0.14 0.14 0.14 0.14 0.14 Total area of Cp/T versus T plot = ΔS = I + II + III (See following plot.) ΔS = (0.16 J K−2 mol−1)(20. K) + (0.14 J K−2 mol−1)(180. K) + 1/2 (0.02 J K−2 mol−1)(40. K) ΔS = 3.2 + 25 + 0.4 = 29 J K−1 mol−1 126. We can set up three equations in three unknowns: 28.7262 = a + 300.0 b + (300.0)2 c 29.2937 = a + 400.0 b + (400.0)2 c 29.8545 = a + 500.0 b + (500.0)2 c 410 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY These can be solved by several methods. One way involves setting up a matrix and solving with a calculator, such as: ⎛1 300.0 90,000 ⎞ ⎛ a ⎞ ⎛ 28.7262 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜1 400.0 160,000 ⎟ ⎜ b ⎟ = ⎜ 29.2937 ⎟ ⎜1 500.0 250,000 ⎟ ⎜ c ⎟ ⎜ 29.8545 ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ The solution is: a = 26.98; b = 5.91 × 10−3; c = −3.4 × 10−7 At 900. K: Cp = 26.98 + 5.91 × 10−3(900.) − 3.4 × 10−7(900.)2 = 32.02 J K−1 mol−1 T2 ΔS = n ∫ T1 T2 ΔS = a C p dT T dT ∫ T +b T1 T2 (a + bT + cT 2 ) dT , n = 1.00 mol T T1 = n∫ T2 ∫ dT + c T1 T2 ⎛ T2 ⎞ c(T22 − T12 ) ⎜ ⎟ T dT a ln b ( T T ) + − + = 2 1 ∫ ⎜T ⎟ 2 ⎝ 1⎠ T1 Solving using T2 = 900. K and T1 = 100. K: ΔS = 59.3 + 4.73 − 0.14 = 63.9 J/K 127. To calculate ΔSsys at 10.0°C, we need a place to start. From the data in the problem, we can calculate ΔSsys at the melting point (5.5°C). For a phase change, ΔSsys = qrev/T = ΔH/T, where ΔH is determined at the melting point (5.5°C). Solving for ΔH at 5.5°C (using a thermochemical cycle): C6H6(l, 25.0°C) → C6H6(s, 25.0°C) ΔH = −10.04 kJ ΔH = nCpΔT/1000 = 2.59 kJ C6H6(l, 5.5°C) → C6H6(l, 25.0°C) C6H6(s, 25.0°C) → C6H6(s, 5.5°C) ΔH = nCpΔT/1000 = −1.96 kJ __________________________________________________________ C6H6(l, 5.5°C) → C6H6(s, 5.5°C) At the melting point, ΔSsys = ΔH = −9.41 kJ − 9.41 × 103 J ΔH = = −33.8 J/K. T 278.7 K For the phase change at 10.0°C (283.2 K): ΔS = −33.8 J/K C6H6(l, 278.7 K) → C6H6(s, 278.7 K) ΔS = nCp ln(T2/T1) = −2.130 J/K C6H6(l, 283.2 K) → C6H6(l, 278.7 K) C6H6(s, 278.7 K) → C6H6(s, 283.2 K) ΔS = nCp ln(T2/T1) = 1.608 J/K ___________________________________________________________________ C6H6(l, 283.2 K) → C6H6(s, 283.2 K) ΔSsys = −34.3 J/K To calculate ΔSsurr, we need ΔH at 10.0°C (ΔSsurr = − ΔH ). T CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 411 C6H6(l, 25.0°C) → C6H6(s, 25.0°C) ΔH = −10.04 kJ ΔH = nCpΔT/1000 = 2.00 kJ C6H6(l, 10.0°C) → C6H6(l, 25.0°C) C6H6(s, 25.0°C) → C6H6(s, 10.0°C) ΔH = nCpΔT/1000 = −1.51 kJ _____________________________________________________________ C6H6(l, 10.0°C) → C6H6(s, 10.0°C) ΔSsurr = ΔH = −9.55 kJ − ΔH − (−9.55 × 103 J ) = = 33.7 J/K T 283.2 K Marathon Problems 128. a. ΔS° will be negative since there is a decrease in the number of moles of gas. b. Since ΔS° is negative, ΔH° must be negative for the reaction to be spontaneous at some temperatures. Therefore, ΔSsurr is positive. c. Ni(s) + 4 CO(g) ⇌ Ni(CO)4(g) ΔH° = −607 − 4(−110.5) = −165 kJ; ΔS° = 417 − [4(198) + (30.)] = −405 J/K d. ΔG° = 0 = ΔH° − TΔS°; T = ΔH o − 165 × 103 J = = 407 K or 134°C − 405 J / K ΔSo e. T = 50.°C + 273 = 323 K o ΔG 323 = −165 kJ − (323 K)(−0.405 kJ/K) = −34 kJ ln K = f. − ΔG o − (−34,000 J ) = = 12.66, K = e12.66 = 3.1 × 105 −1 −1 RT (8.3145 J K mol ) (323 K ) T = 227°C + 273 = 500. K o ΔG 500 = −165 kJ − (500. K)(−0.405 kJ/K) = 38 kJ ln K = − 38,000 = −9.14, K = e−9.14 = 1.1 × 10−4 (8.3145) (500.) g. The temperature change causes the value of the equilibrium constant to change from a large value (greater than 1) favoring formation of Ni(CO)4 to a small value (less than 1) favoring the decomposition of Ni(CO)4 into pure Ni and CO. This is exactly what is wanted in order to purify a nickel sample. h. Ni(CO)4(l) ⇌ Ni(CO)4(g) Kp = PNi ( CO ) 4 412 CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY At 42°C (the boiling point): ΔG° = 0 = ΔH° − TΔS° ΔS° = ΔH o 29.0 × 103 J = = 92.1 J/K T 315 K o = ΔH° − TΔS° = 29.0 × 103 J − 425 K(92.1 J/K) = −10,100 J At 152°C: ΔG152 ΔG° = −RT ln Kp, ln Kp = 10,100 J = 2.86, Kp = e2.86 = 17 atm 8.3145 (425 K ) A maximum pressure of 17 atm can be attained before Ni(CO)4(g) will liquify. 129. a. ΔE = 0, ΔH = 0. Because ΔE = 0 = q + w, q = −w. b. w = −(1.00 atm)(2.00 L)(101.3 J L−1 atm−1) = −203 J; ΔE = ΔH = 0, q = 203 J c. w = −[(1.33 atm)(1.00 L) + (1.00 atm)(1.00 L)] × 101.3 J = −236 J L atm q = 236 J; ΔE = ΔH = 0 ⎛V ⎞ ⎡ ⎛ V ⎞⎤ 101.3 J = −281 J d. w = −nRT ln⎜⎜ 2 ⎟⎟ = ⎢− PV ln⎜⎜ 2 ⎟⎟⎥ ⎝ V1 ⎠ ⎢⎣ ⎝ V1 ⎠⎦⎥ L atm q = 281 J; ΔE = ΔH = 0 ⎛ 101.3 J ⎞ ⎟⎟ = 405 J e. w = −(2.00 atm)(−2.00 L) × ⎜⎜ ⎝ L atm ⎠ q = −405 J; ΔE = ΔH = 0 f. g. ⎛ 101.3 J ⎞ ⎟⎟ = 337 J w = −[(1.33 atm)(−1.00 L) + (2.00 atm)(−1.00 L)] × ⎜⎜ ⎝ L atm ⎠ q = −337 J; ΔE = ΔH = 0 ⎛ V ⎞⎤ 101.3 J ⎛V ⎞ ⎡ w = − nRT ln⎜⎜ 2 ⎟⎟ = ⎢− PV ln⎜⎜ 2 ⎟⎟⎥ = 281 J ⎝ V1 ⎠ ⎢⎣ ⎝ V1 ⎠⎥⎦ L atm q = −281 J; ΔE = ΔH = 0 Note: Overall work for the one-step process = −203 + 405 = 202 J, (q = −202 J). Overall work for the two-process = −236 + 337 = 101 J, (q = −101 J). Overall work for the reversible process = w = −281 + 281 = 0, (q = 0). Thus in an overall reversible process (expansion + compression), the system and surroundings are unchanged. In an irreversible process, this is not the case. CHAPTER 11 ELECTROCHEMISTRY Galvanic Cells, Cell Potentials, and Standard Reduction Potentials 15. Electrochemistry is the study of the interchange of chemical and electrical energy. A redox (oxidation-reduction) reaction is a reaction in which one or more electrons are transferred. In a galvanic cell, a spontaneous redox reaction occurs that produces an electric current. In an electrolytic cell, electricity is used to force a nonspontaneous redox reaction to occur. 16. Magnesium is an alkaline earth metal; Mg will oxidize to Mg2+. The oxidation state of hydrogen in HCl is +1. To be reduced, the oxidation state of H must decrease. The obvious choice for the hydrogen product is H2(g), where hydrogen has a zero oxidation state. The balanced reaction is Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g). Mg goes from the 0 to the +2 oxidation state by losing two electrons. Each H atom goes from the +1 to the 0 oxidation state by gaining one electron. Since there are two H atoms in the balanced equation, a total of two electrons are gained by the H atoms. Hence two electrons are transferred in the balanced reaction. When the electrons are transferred directly from Mg to H+, no work is obtained. In order to harness this reaction to do useful work, we must control the flow of electrons through a wire. This is accomplished by making a galvanic cell that separates the reduction reaction from the oxidation reaction in order to control the flow of electrons through a wire to produce a voltage. 17. A typical galvanic cell diagram is: e e Salt bridge Anode (oxidation) Cations Anions 413 Cathode (reduction) 414 CHAPTER 11 ELECTROCHEMISTRY The diagram for all cells will look like this. The contents of each half-cell will be identified for each reaction, with all concentrations at 1.0 M and partial pressures at 1.0 atm. Note that cations always flow into the cathode compartment and anions always flow into the anode compartment. This is required to keep each compartment electrically neutral. a. Reference Table 11.1 for standard reduction potentials. Remember that E ocell = E°(cathode) − E°(anode); in the Solutions Guide, we will represent E°(cathode) as E oc and represent −E°(anode) as − E oa . Also remember that standard potentials are not multiplied by the integer used to obtain the overall balanced equation. (Cl2 + 2 e− → 2 Cl−) × 3 E oc = 1.36 V − E oa = −1.33 V 7 H2O + 2 Cr3+ → Cr2O72− + 14 H+ + 6 e− ________________________________________________________________________ 7 H2O(l) + 2 Cr3+(aq) + 3 Cl2(g) → Cr2O72− (aq) + 6 Cl− (aq) + 14 H+(aq) E ocell = 0.03 V The contents of each compartment are: Cathode: Pt electrode; Cl2 bubbled into solution, Cl− in solution Anode: Pt electrode; Cr3+, H+, and Cr2O72− in solution We need a nonreactive metal to use as the electrode in each case since all the reactants and products are in solution. Pt is the most common choice. Another possibility is graphite. b. Cu2+ + 2 e− → Cu E oc = 0.34 V − E oa = 2.37 V Mg → Mg2+ + 2 e− ______________________________________________________________ E ocell = 2.71 V Cu2+(aq) + Mg(s) → Cu(s) + Mg2+(aq) Cathode: Cu electrode; Cu2+ in solution Anode: Mg electrode; Mg2+ in solution c. 5 e− + 6 H+ + IO3− → 1/2 I2 + 3 H2O E oc = 1.20 V − E oa = −0.77 V (Fe2+ → Fe3+ + e−) × 5 ___________________________________________________________________ 6 H+ + IO3− + 5 Fe2+ → 5 Fe3+ + 1/2 I2 + 3 H2O E ocell = 0.43 V or 12 H+(aq) + 2 IO3−(aq) + 10 Fe2+(aq) → 10 Fe3+(aq) + I2(s) + 6 H2O(l) Cathode: Pt electrode; IO3−, I2, and H2SO4 (H+ source) in solution Anode: Pt electrode; Fe2+ and Fe3+ in solution Note: I2(s) would make a poor electrode since it sublimes. E ocell = 0.43 V CHAPTER 11 d. ELECTROCHEMISTRY 415 (Ag+ + e− → Ag) × 2 E oc = 0.80 V − E oa = 0.76 V Zn → Zn2+ + 2 e− _________________________________________________ Zn(s) + 2 Ag+(aq) → 2 Ag(s) + Zn2+(aq) E ocell = 1.56 V Cathode: Ag electrode; Ag+ in solution Anode: Zn electrode; Zn2+ in solution 18. a. (5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O) × 2 E oc = 1.51 V − E oa = −0.54 V (2 I− → I2 + 2 e−) × 5 _______________________________________________________________________ 16 H+(aq) + 2 MnO4−(aq) + 10 I−(aq) → 5 I2(aq) + 2 Mn2+(aq) + 8 H2O(l) E ocell = 0.97 V This reaction is spontaneous at standard conditions because E ocell > 0. b. (5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O) × 2 E oc = 1.51 V − E oa = !2.87 V (2 F− → F2 + 2 e−) × 5 ________________________________________________________________________ 16 H+(aq) + 2 MnO4−(aq) + 10 F−(aq) → 5 F2(aq) + 2 Mn2+(aq) + 8 H2O(l) E ocell = −1.36 V This reaction is not spontaneous at standard conditions because E ocell < 0. c. H2 → 2H+ + 2 e− E oc = 0.00 V − E oa = !2.23 V H2 + 2 e− → 2H− ___________________________________________ E ocell = !2.23 V; not spontaneous 2H2(g) → 2H+(aq) + 2 H−(aq) d. Au3+ + 3 e− → Au E oc = 1.50 V − E oa = −0.80 V (Ag → Ag+ + e−) × 3 _________________________________________________________________ Au3+(aq) + 3 Ag(s) → Au(s) + 3 Ag+(aq) E ocell = 0.70 V; spontaneous 19. Reference Exercise 11.17 for a typical galvanic cell design. The contents of each half-cell compartment are identified below, with all solute concentrations at 1.0 M and all gases at 1.0 atm. For each pair of half-reactions, the half-reaction with the largest standard reduction potential will be the cathode reaction, and the half-reaction with the smallest reduction potential will be reversed to become the anode reaction. Only this combination gives a spontaneous overall reaction, i.e., a reaction with a positive overall standard cell potential. 416 CHAPTER 11 a. Cl2 + 2 e− → 2 Cl− ELECTROCHEMISTRY E oc = 1.36 V − E oa = −1.09 V 2 Br- → Br2 + 2 e− _________________________________________________ Cl2(g) + 2 Br-(aq) → Br2(aq) + 2 Cl−(aq) E ocell = 0.27 V The contents of each compartment are: Cathode: Pt electrode; Cl2(g) bubbled in, Cl- in solution Anode: Pt electrode; Br2 and Br- in solution b. (2 e− + 2 H+ + IO4− → IO3− + H2O) × 5 E oc = 1.60 V − E oa = −1.51 V (4 H2O + Mn2+ → MnO4− + 8 H+ + 5 e−) × 2 ________________________________________________________________________ 10 H+ + 5 IO4− + 8 H2O + 2 Mn2+ → 5 IO3− + 5 H2O + 2 MnO4− + 16 H+ E ocell = 0.09 V This simplifies to: 3 H2O(l) + 5 IO4−(aq) + 2 Mn2+(aq) → 5 IO3−(aq) + 2 MnO4−(aq) + 6 H+(aq) E ocell = 0.09 V Cathode: Pt electrode; IO4−, IO3−, and H2SO4 (as a source of H+) in solution Anode: Pt electrode; Mn2+, MnO4−, and H2SO4 in solution c. H2O2 + 2 H+ + 2 e− → 2 H2O E oc = 1.78 V − E oa = −0.68 V H2O2 → O2 + 2 H+ + 2 e− _______________________________________________________________________ 2 H2O2(aq) → 2 H2O(l) + O2(g) E ocell = 1.10 V Cathode: Pt electrode; H2O2 and H+ in solution Anode: Pt electrode; O2(g) bubbled in, H2O2 and H+ in solution d. (Fe3+ + 3 e− → Fe) × 2 E oc = −0.036 V − E oa = 1.18 V (Mn → Mn2+ + 2 e−) × 3 ________________________________________________________________________ 2 Fe3+(aq) + 3 Mn(s) → 2 Fe(s) + 3 Mn2+(aq) E ocell = 1.14 V Cathode: Fe electrode; Fe3+ in solution; Anode: Mn electrode; Mn2+ in solution 20. Locate the pertinent half-reactions in Table 11.1, and then figure which combination will give a positive standard cell potential. In all cases the anode compartment contains the species CHAPTER 11 ELECTROCHEMISTRY 417 with the smallest standard reduction potential. In part a, the copper compartment is the anode, and in part b, the cadmium compartment is the anode. Au3+ + 3 e− → Au a. E oc = 1.50 V – E oa = −0.16 V (Cu+ → Cu2+ + e−) × 3 __________________________________________________________ Au3+(aq) + 3 Cu+(aq) → Au(s) + 3 Cu2+(aq) E ocell = 1.34 V (VO2+ + 2 H+ + e− → VO2+ + H2O) × 2 b. E oc = 1.00 V – E oa = 0.40 V Cd → Cd2+ + 2e_____________________________________________________________________ 2 VO2+(aq) + 4 H+(aq) + Cd(s) → 2 VO2+(aq) + 2 H2O(l) + Cd2+(aq) E ocell = 1.40 V 21. 22. In standard line notation, the anode is listed first and the cathode is listed last. A double line separates the two compartments. By convention, the electrodes are on the ends, with all solutes and gases toward the middle. A single line is used to indicate a phase change. We also included all concentrations. 19a. Pt | Br− (1.0 M), Br2 (1.0 M) || Cl2 (1.0 atm) | Cl− (1.0 M) | Pt 19b. Pt | Mn2+ (1.0 M), MnO4− (1.0 M), H+ (1.0 M) | | IO4− (1.0 M), IO3− (1.0 M), H+ (1.0 M) | Pt 19c. Pt | H2O2 (1.0 M), H+ (1.0 M) | O2 (1.0 atm) | | H2O2 (1.0 M), H+ (1.0 M) | Pt 19d. Mn | Mn2+ (1.0 M) | | Fe3+ (1.0 M) | Fe The reduction half-reaction for the SCE is: Hg2Cl2 + 2 e− → 2 Hg + 2 Cl− ESCE = 0.242 V For a spontaneous reaction to occur, Ecell must be positive. Using the standard reduction potentials in Table 11.1 and the given SCE potential, deduce which combination will produce a positive overall cell potential. a. Cu2+ + 2 e− → Cu E° = 0.34 V Ecell = 0.34 - 0.242 = 0.10 V; SCE is the anode. b. Fe3+ + e− → Fe2+ E° = 0.77 V Ecell = 0.77 - 0.242 = 0.53 V; SCE is the anode. c. AgCl + e− → Ag + Cl− E° = 0.22 V Ecell = 0.242 − 0.22 = 0.02 V; SCE is the cathode. 418 CHAPTER 11 d. Al3+ + 3 e− → Al ELECTROCHEMISTRY E° = −1.66 V Ecell = 0.242 + 1.66 = 1.90 V; SCE is the cathode. e. Ni2++ 2 e− → Ni E° = −0.23 V Ecell = 0.242 + 0.23 = 0.47 V; SCE is the cathode. 23. a. 2 H+ + 2 e− → H2 E° = 0.00 V; Cu → Cu2+ + 2 e− −E° = -0.34 V E ocell = −0.34 V; no, H+ cannot oxidize Cu to Cu2+ at standard conditions ( E ocell < 0). b. Fe3+ + e− → Fe2+ E° = 0.77 V; 2 I− → I2 + 2 e− −E° = -0.54 V E ocell = 0.77 − 0.54 = 0.23 V; yes, Fe3+ can oxidize I− to I2. c. H2 → 2 H+ + 2 e− −E° = 0.00 V; Ag+ + e− → Ag E° = 0.80 V E ocell = 0.80 V; yes, H2 can reduce Ag+ to Ag at standard conditions ( E ocell > 0). d. Fe2+ → Fe3+ + e− −E° = -0.77 V; Cr3+ + e− → Cr2+ E ocell = −0.50 − 0.77 = −1.27 V; conditions. 24. K+ < H2O −2.92 −0.83 < Cd2+ −0.40 < I2 < AuCl4− 0.54 0.99 < IO3− 1.20 Good reducing agents are easily oxidized. The reducing agents are on the right side of the reduction half-reactions listed in Table 11.1. The best reducing agents have the most negative standard reduction potentials (E°); i.e., the best reducing agents have the most positive !E° value. The ordering from worst to best reducing agents is: !E°(V) 26. no, Fe2+ cannot reduce Cr3+ to Cr2+ at standard Good oxidizing agents are easily reduced. Oxidizing agents are on the left side of the reduction half-reactions listed in Table 11.1. We look for the largest, most positive standard reduction potentials to correspond to the best oxidizing agents. The ordering from worst to best oxidizing agents is: E°(V) 25. E° = −0.50 V F− < H2O !2.92 −1.23 Cl2 + 2 e− → 2 Cl− Pb2+ + 2 e− → Pb Na+ + e- → Na < I2 < Cu+ −1.20 −0.16 E° = 1.36 V E° = −0.13 V E° = −2.71 V < H− < 2.23 Ag+ + e− → Ag Zn2+ + 2 e− → Zn K 2.92 E° = 0.80 V E° = −0.76 V a. Oxidizing agents (species reduced) are on the left side of the above reduction halfreactions. Of the species available, Ag+ would be the best oxidizing agent since it has the most positive E° value. CHAPTER 11 ELECTROCHEMISTRY 419 b. Reducing agents (species oxidized) are on the right side of the reduction half-reactions. Of the species available, Zn would be the best reducing agent since it has the most positive −E° value. c. SO42− + 4 H+ + 2 e− → H2SO3 + H2O E oc = 0.20 V; SO42− can oxidize Pb and Zn at standard conditions. When SO42- is coupled with these reagents, E ocell is positive. d. Al → Al3+ + 3 e− − E oa = 1.66 V; Al can oxidize Ag+ and Zn2+ at standard conditions since E ocell > 0. 27. a. 2 Br- → Br2 + 2 e− − E oa = −1.09 V; 2 Cl− → Cl2 + 2 e− − E oa = −1.36 V; E oc > 1.09 V to oxidize Br−; E oc < 1.36 V to not oxidize Cl−; Cr2O72−, O2, MnO2, and IO3− are all possible because when all these oxidizing agents are coupled with Br−, they give E ocell > 0, and when coupled with Cl−, they give E ocell < 0 (assuming standard conditions). b. Mn → Mn2+ + 2 e− − E oa = 1.18 V; Ni → Ni2+ + 2 e− − E oa = 0.23 V; any oxidizing agent with −0.23 V > E oc > −1.18 V will work. PbSO4, Cd2+, Fe2+, Cr3+, Zn2+, and H2O will be able to oxidize Mn but not oxidize Ni (assuming standard conditions). 28. a. Cu2+ + 2 e− → Cu E oc = 0.34 V; Cu2+ + e− → Cu+ E oc = 0.16 V; to reduce Cu2+ to Cu but not reduce Cu2+ to Cu+, the reducing agent must have a − E oa value between −0.34 and −0.16 V (so E ocell is positive only for the Cu2+-to-Cu reduction). The reducing agents (species oxidized) are on the right side of the half-reactions in Table 11.1. The reagents at standard conditions, which have a − E oa value between −0.34 and −0.16 V, are Ag (in 1.0 M Cl−) and H2SO3. b. Br2 + 2 e− → 2 Br− E oc = 1.09 V; I2 + 2 e− → 2 I− E oc = 0.54 V; from Table 11.1, VO2+, Au (in 1.0 M Cl−), NO, ClO2−, Hg22+, Ag, Hg, Fe2+, H2O2, and MnO4− are all capable at standard conditions of reducing Br2 to Br− but not reducing I2 to I−. When these reagents are coupled with Br2, E ocell > 0, and when coupled with I2, E ocell < 0. 29. ClO− + H2O + 2 e− → 2 OH− + Cl− E oc = 0.90 V − E oa = 0.10 V 2 NH3 + 2 OH− → N2H4 + 2 H2O + 2 e− __________________________________________________________ ClO−(aq) + 2 NH3(aq) → Cl−(aq) + N2H4(aq) + H2O(l) E ocell = 1.00 V Because E ocell is positive for this reaction, at standard conditions ClO− can spontaneously oxidize NH3 to the somewhat toxic N2H4. 30. Consider the strongest oxidizing agent combined with the strongest reducing agent from Table 11.1: 420 CHAPTER 11 ELECTROCHEMISTRY F2 + 2 e− → 2 F− E oc = 2.87 V − E oa = 3.05 V (Li → Li+ + e−) × 2 _______________________________________________ F2(g) + 2 Li(s) → 2 Li+(aq) + 2 F−(aq) E ocell = 5.92 V The claim is impossible. The strongest oxidizing agent and reducing agent when combined only give E ocell of about 6 V. Cell Potential, Free Energy, and Equilibrium 31. An extensive property is one that depends directly on the amount of substance. The free energy change for a reaction depends on whether 1 mole of product is produced or 2 moles of product are produced or 1 million moles of product are produced. This is not the case for cell potentials, which do not depend on the amount of substance. The equation that relates ΔG to E is ΔG = !nFE. It is the n term that converts the intensive property E into the extensive property ΔG. n is the number of moles of electrons transferred in the balanced reaction that ΔG is associated with. 32. Fe2+ + 2 e− → Fe E° = !0.44 V = !0.44 J/C ΔG° = !nFE° = !(2 mol e−)(96,485 C/mol e−)(!0.44 J/C)(1 kJ/1000 J) = 85 kJ 85 kJ = 0 ! (ΔG of , Fe2+ + 0) , ΔG of , Fe2+ = !85 kJ We can get ΔG of , Fe3+ two ways. Consider: Fe3+ + e− → Fe2+ E° = 0.77 V ΔG° = !(1 mol e)(96,485 C/mol e-)(0.77 J/C) = !74,300 J = !74 kJ ΔG° = 74 kJ Fe2+ → Fe3+ + e− 2+ − ΔG° = !85 kJ Fe → Fe + 2 e ___________________________________________________ ΔG° = !11 kJ, ΔG of , Fe3+ = !11 kJ/mol Fe → Fe3+ + 3 e− Or consider: Fe3+ + 3 e− → Fe E° = !0.036 V ΔG° = !(3 mol e−)(96,485 C/mol e−)(!0.036 J/C) = 10,400 J ≈ 10. kJ 10. kJ = 0 ! (ΔG of , Fe3+ + 0) , ΔG of , Fe3+ = !10. kJ/mol; Round-off error explains the 1-kJ discrepancy. 33. Because the cells are at standard conditions, wmax = ΔG = ΔG° = − nFE ocell . See Exercise 11.20 for the balanced overall equations and E ocell . 20a. wmax = −(3 mol e−)(96,485 C/mol e−)(1.34 J/C) = −3.88 × 105 J = −388 kJ 20b. wmax = −(2 mol e−)(96,485 C/mol e−)(1.40 J/C) = −2.70 × 105 J = −270. kJ CHAPTER 11 34. ELECTROCHEMISTRY 421 a. Possible reaction: I2(s) + 2 Cl− (aq) → 2 I−(aq) + Cl2(g) E ocell = 0.54 V − 1.36 V = −0.82 V; this reaction is not spontaneous at standard conditions since E ocell < 0. No reaction occurs. b. Possible reaction: Cl2(g) + 2 I−(aq) → I2(s) + 2 Cl−(aq) E ocell = 0.82 V; this reaction is spontaneous at standard conditions since E ocell > 0. The reaction will occur. Cl2(g) + 2 I−(aq) → I2(s) + 2 Cl−(aq) E ocell = 0.82 V = 0.82 J/C ΔG° = − nFE ocell = −(2 mol e−)(96,485 C/mol e−)(0.82 J/C) = −1.6 × 105 J = −160 kJ E° = nE o 2(0.82) 0.0591 log K, log K = = = 27.75, K = 1027.75 = 5.6 × 1027 n 0.0591 0.0591 Note: When determining exponents, we will round off to the correct number of significant figures after the calculation is complete in order to help eliminate excessive round-off errors. c. Possible reaction: 2 Ag(s) + Cu2+(aq) → Cu(s) + 2 Ag+(aq) reaction occurs. E ocell = −0.46 V; no d. Fe2+ can be oxidized or reduced. The other species present are H+, SO42−, H2O, and O2 from air. Only O2 in the presence of H+ has a large enough standard reduction potential to oxidize Fe2+ to Fe3+ (resulting in E ocell > 0). All other combinations, including the possible reduction of Fe2+, give negative cell potentials. The spontaneous reaction is: 4 Fe2+(aq) + 4 H+(aq) + O2(g) → 4 Fe3+(aq) + 2 H2O(l) E ocell = 1.23 - 0.77 = 0.46 V ΔG° = − nFE ocell = −(4 mol e−)(96,485 C/mol e−)(0.46 J/C)(1 kJ/1000 J) = −180 kJ log K = 35. 4(0.46) = 31.13, K = 1.3 × 1031 0.0591 Reference Exercise 11.19 for the balanced reactions and standard cell potentials. balanced reactions are necessary to determine n, the moles of electrons transferred. 19a. Cl2(aq) + 2 Br-(aq) → Br2(aq) + 2 Cl−(aq) The E ocell = 0.27 V = 0.27 J/C, n = 2 mol e− ΔG° = − nFE ocell = −(2 mol e−)(96,485 C/mol e−)(0.27 J/C) = −5.2 × 104 J = −52 kJ E ocell = 0.0591 nE o 2(0.27) log K, log K = = = 9.14, K = 109.14 = 1.4 × 109 n 0.0591 0.0591 422 CHAPTER 11 ELECTROCHEMISTRY Note: When determining exponents, we will round off to the correct number of significant figures after the calculation is complete in order to help eliminate excessive round-off errors. 19b. ΔG° = −(10 mol e−)(96,485 C/mol e−)(0.09 J/C) = −9 × 104 J = −90 kJ log K = 19c. ΔG° = −(2 mol e−)(96,485 C/mol e−)(1.10 J/C) = −2.12 × 105 J = −212 kJ log K = 19d. a. 2(1.10) = 37.225, K = 1.68 × 1037 0.0591 ΔG° = −(6 mol e−)(96,485 C/mol e−)(1.14 J/C) = −6.60 × 105 J = −660. kJ log K = 36. 10(0.09) = 15.2, K = 1015.2 = 2 × 1015 0.0591 6(1.14) = 115.736, K = 5.45 × 10115 0.0591 Cl2 + 2 e− → 2 Cl− E oc = 1.36 V − E oa = −0.954 V (ClO2− → ClO2 + e−) × 2 ____________________________________________________________ 2 ClO2−(aq) + Cl2(g) → 2 ClO2(aq) + 2 Cl−(aq) E ocell = 0.41 V = 0.41 J/C ΔG° = − nFE ocell = −(2 mol e−)(96,485 C/mol e−)(0.41 J/C) = −7.91 × 104 J = −79 kJ ΔG° = −RT ln K, so K = exp(−ΔG°/RT) K = exp[(7.9 × 104 J)/(8.3145 J K−1 mol−1)(298 K)] = 7.0 × 1013 or log K = b. 37. a. nE o 2(0.41) = = 13.87, K = 1013.87 = 7.4 × 1013 0.0591 0.0591 (H2O + ClO2 → ClO3− + 2 H+ + e−) × 5 5 e + 4 H+ + ClO2 → Cl− + 2 H2O ________________________________________________ 3 H2O(l) + 6 ClO2(g) → 5 ClO3−(aq) + Cl−(aq) + 6 H+(aq) − (4 H+ + NO3− + 3 e− → NO + 2 H2O) × 2 E oc = 0.96 V − E oa = 1.18 V (Mn → Mn2+ + 2 e−) × 3 _______________________________________________________________________ 3 Mn(s) + 8 H+(aq) + 2 NO3−(aq) → 2 NO(g) + 4 H2O(l) + 3 Mn2+(aq) E ocell = 2.14 V CHAPTER 11 ELECTROCHEMISTRY 423 5 × (2 e− + 2 H+ + IO4− → IO3− + H2O) E oc = 1.60 V − E oa = −1.51 V 2 × (Mn2+ + 4 H2O → MnO4− + 8 H+ + 5 e−) ________________________________________________________________________ 5 IO4−(aq) + 2 Mn2+(aq) + 3 H2O(l) → 5 IO3−(aq) + 2 MnO4−(aq) + 6 H+(aq) E ocell = 0.09 V b. Nitric acid oxidation (see part a for E ocell ): ΔG° = − nFE ocell = −(6 mol e−)(96,485 C/mol e−)(2.14 J/C) = −1.24 × 106 J = −1240 kJ o nE 6(2.14) log K = = = 217, K ≈ 10217 0.0591 0.0591 Periodate oxidation (see part a for E ocell ): ΔG° = −(10 mol e−)(96,485 C/mol e−)(0.09 J/C)(1 kJ/1000 J) = −90 kJ log K = 38. 10(0.09) = 15.2, K = 1015.2 = 2 × 1015 0.0591 2 H2(g) + O2(g) → 2 H2O(l); oxygen goes from the zero oxidation state to the -2 oxidation state in H2O. Because two moles of O appear in the balanced reaction, n = 2(2) = 4 mol electrons transferred. a. E ocell = 0.0591 0.0591 log K = log(1.28 × 1083), E ocell = 1.23 V n 4 ΔG° = − nFE ocell = −(4 mol e−)(96,485 C/mol e−)(1.23 J/C) = −4.75 × 105 J = −475 kJ b. Because moles of gas decrease as reactants are converted into products, ΔS° will be negative (unfavorable). Because the value of ΔG° is negative, ΔH° must be negative (ΔG° = ΔH° − TΔS°). c. ΔG = wmax = ΔH − TΔS. Because ΔS is negative, as T increases, ΔG becomes more positive (closer to zero). Therefore, wmax will decrease as T increases. 39. ΔG° = !nFE° = ΔH° ! TΔS°, E° = TΔSo ΔH o − nF nF If we graph E° versus T, we should get a straight line (y = mx + b). The slope (m) of the line is equal to ΔS°/nF, and the y intercept is equal to -ΔH°/nF. From the equation above, E° will have a small temperature dependence when ΔS° is close to zero. 40. CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l) ΔG° = ∑ n p ΔG of , products − ∑ n r ΔG of , reactants = 2(−237) + (−394) − (−166) = −702 kJ 424 CHAPTER 11 ELECTROCHEMISTRY The balanced half-reactions are: H2O + CH3OH → CO2 + 6 H+ + 6 e− and O2 + 4 H+ + 4 e− → 2 H2O For 3/2 mol O2, 6 moles of electrons will be transferred (n = 6). ΔG° = −nFE°, E° = − ΔG o − (−702,000 J ) = = 1.21 J/C = 1.21 V nF (6 mol e − ) (96,485 C / mol e − ) For this reaction: ΔS° = 2(70.) + 214 − [127 + 3/2 (205)] = −81 J/K From Exercise 11.39, E° = TΔSo ΔH o − . nF nF Because ΔS° is negative, E° will decrease with an increase in temperature. 41. a. Cu+ + e− → Cu E oc = 0.52 V Cu+ → Cu2+ + e− − E oa = !0.16 V ____________________________________________________________ E ocell = 0.36 V; spontaneous 2 Cu+(aq) → Cu2+(aq) + Cu(s) ΔG° = − nFE ocell = !(1 mol e−)(96,485 C/mol e−)(0.36 J/C) = !34,700 J = -35 kJ E ocell = nE o 1(0.36) 0.0591 log K, log K = = = 6.09, K = 106.09 = 1.2 × 106 n 0.0591 0.0591 b. Fe2+ + 2 e− → Fe E oc = !0.44 V (Fe2+ → Fe3+ + e−) × 2 − E oa = !0.77 V _____________________________________________________________ E ocell = !1.21 V; not spontaneous 3 Fe2+(aq) → 2 Fe3+(aq) + Fe(s) c. HClO2 + 2 H+ + 2 e− → HClO + H2O E oc = 1.65 V HClO2 + H2O → ClO3− + 3 H+ + 2 e− − E oa = !1.21 V ______________________________________________________________________ 2 HClO2(aq) → ClO3−(aq) + H+(aq) + HClO(aq) E ocell = 0.44 V; spontaneous ΔG° = − nFE ocell = !(2 mol e−)(96,485 C/mol e−)(0.44 J/C) = !84,900 J = !85 kJ log K = 42. nE o 2(0.44) = = 14.89, K = 7.8 × 1014 0.0591 0.0591 NO3− is a spectator ion. The reaction that occurs is Ag+ reacting with Zn. CHAPTER 11 ELECTROCHEMISTRY 425 (Ag+ + e− → Ag) × 2 E oc = 0.80 V 2+ − ! E oa = 0.76 V Zn → Zn + 2 e _________________________________________ 2 Ag+ + Zn → 2 Ag + Zn2+ E ocell = 1.56 V log K = 43. Al3+ + 3 e− → Al E oc = −1.66 V − E oa = 2.07 V Al + 6 F− → AlF63− + 3 e− ___________________________________________________ Al3+(aq) + 6 F−(aq) → AlF63−(aq) K=? E ocell = 0.41 V log K = 44. nE 2(1.56) = , K = 1052.792 = 6.19 × 1052 0.0591 0.0591 nE o 3(0.41) = = 20.81, K = 1020.81 = 6.5 × 1020 0.0591 0.0591 The Ksp reaction is FeS(s) ⇌ Fe2+(aq) + S2−(aq); K = Ksp. Manipulate the given equations so that when they are added together, we get the Ksp reaction. Then we can use the value of E ocell for the reaction to determine Ksp. FeS + 2 e− → Fe + S2− E oc = !1.01 V Fe → Fe2+ + 2 e− − E oa = 0.44 V ____________________________________________ E ocell = !0.57 V Fe(s) → Fe2+(aq) + S2−(aq) log Ksp = 45. nE o 2(−0.57) = = !19.29, Ksp = 10 −19.29 = 5.1 × 10 −20 0.0591 0.0591 CuI + e− → Cu + I− E oCuI = ? Cu → Cu+ + e− − E oa = −0.52 V _________________________________________________ CuI(s) → Cu+(aq) + I−(aq) E ocell = E oCuI − 0.52 V For this overall reaction, K = Ksp = 1.1 × 10−12: E ocell = 0.0591 0.0591 log Ksp = log(1.1 × 10−12) = −0.71 V n 1 E ocell = −0.71 V = E oCuI − 0.52, E oCuI = −0.19 V 426 CHAPTER 11 ELECTROCHEMISTRY Galvanic Cells: Concentration Dependence 46. Cr2O72− + 14 H+ + 6 e− → 2 Cr3+ + 7 H2O E oc = 1.33 V 3+ − (Al → Al + 3 e ) × 2 − E oa = 1.66 V _________________________________________________________ E ocell = 2.99 V Cr2O72− + 14 H+ + 2 Al → 2 Al3+ + 2 Cr3+ + 7 H2O E = EE ! 0.0591 0.0591 [Al3+ ]2 [Cr 3+ ]2 log Q , E = 2.99 V ! log 2− n 6 [Cr2 O 7 ][H + ]14 3.01 = 2.99 ! 0.0591 (0.30) 2 (0.15) 2 , log n (0.55)[H + ]14 ⎛ 3.7 × 10 −3 ⎞ − 6(0.02) ⎟ = log ⎜⎜ + 14 ⎟ 0.0591 ⎝ [H ] ⎠ 3.7 × 10 −3 = 10 −2.0 = 0.01, [H+]14 = 0.37, [H+] = 0.93 = 0.9 M, pH = -log(0.9) = 0.05 [H + ]14 47. Au3+ + 3 e− → Au a. E oc = 1.50 V (Tl → Tl+ + e−) × 3 − E oa = 0.34 V _________________________________________________ Au3+(aq) + 3 Tl(s) → Au(s) + 3 Tl+(aq) E ocell = 1.84 V b. ΔG° = − nFE ocell = −(3 mol e−)(96,485 C/mol e−)(1.84 J/C) = −5.33 × 105 J = −533 kJ log K = nE o 3(1.84) = = 93.401, K = 1093.401 = 2.52 × 1093 0.0591 0.0591 c. At 25°C, Ecell = E ocell − Ecell = 1.84 V − [Tl + ]3 0.0591 log Q, where Q = . n [Au 3+ ] 0.0591 0.0591 [Tl + ]3 (1.0 × 10 −4 ) 3 log = 1.84 − log 3 3 [Au 3+ ] 1.0 × 10 − 2 Ecell = 1.84 − (−0.20) = 2.04 V 48. (Cr2+ → Cr3+ + e−) × 2 Co + 2 e− → Co _____________________________________ 2 Cr2+(aq) + Co2+(aq) → 2 Cr3+(aq) + Co(s) 2+ E ocell = 0.0591 0.0591 log K = log(2.79 × 107) = 0.220 V n 2 CHAPTER 11 ELECTROCHEMISTRY E = E° − 427 0.0591 [Cr 3+ ]2 ( 2 .0 ) 2 0.0591 log = 0.220 V − log = 0.151 V n 2 [Cr 2+ ]2 [Co 2 + ] (0.30) 2 (0.20) ΔG = −nFE = −(2 mol e−)(96,485 C/mol e−)(0.151 J/C) = −2.91 × 104 J = −29.1 kJ 49. (Pb2+ + 2 e− → Pb) × 3 E oc = −0.13 V (Al → Al3+ + 3 e−) × 2 − E oa = 1.66 V ________________________________________________________ 3 Pb2+(aq) + 2 Al(s) → 3 Pb(s) + 2 Al3+(aq) E ocell = 1.53 V From the balanced reaction, when the [Al3+] has increased by 0.60 mol/L (Al3+ is a product in the spontaneous reaction), then the Pb2+ concentration has decreased by 3/2 (0.60 mol/L) = 0.90 M. Ecell = 1.53 V − [Al3+ ]2 (1.60) 2 0.0591 0.0591 = 1.53 − log log 6 6 [Pb 2 + ]3 (0.10) 3 Ecell = 1.53 V − 0.034 V = 1.50 V 50. a. E = E° − RT 0.0591 ln Q, or at 25°C, E = E° − log Q. nF n For Cu2+(aq) + 2 e− → Cu(s) E = 0.34 V − E° = 0.34 V; E = E° − 0.0591 log(1/[Cu2+]) n 0.0591 log(1/0.10) = 0.34 V − 0.030 V = 0.31 V 2 b. E = 0.34 − 0.0591 log(1/2.0) = 0.34 V − (−8.9 × 10−3 V) = 0.35 V 2 c. E = 0.34 − 0.0591 log(1/1.0 × 10−4) = 0.34 − 0.12 = 0.22 V 2 d. 5 e− + 8 H+(aq) + MnO4− (aq) → Mn2+(aq) + 4 H2O(l) E = E° − E° = 1.51 V [Mn 2 + ] 0.0591 0.0591 (0.010) log = 1.51 V − log − + 8 5 5 (0.10) (1.0 × 10 −3 )8 [MnO 4 ][H ] E = 1.51 − e. E = 1.51 − 0.0591 (23) = 1.51 V − 0.27 V = 1.24 V 5 ⎡ (0.010) ⎤ 0.0591 log ⎢ = 1.51 − 0.083 = 1.43 V 8⎥ 5 ⎣ (0.10) (0.10) ⎦ 428 51. CHAPTER 11 ELECTROCHEMISTRY a. n = 2 for this reaction (lead goes from Pb → Pb2+ in PbSO4). E = E ocell − ⎛ ⎞ 1 1 0.0591 ⎟ = 2.04 V − 0.0591 log log ⎜ + 2 − 2 ⎟ 2 ⎜ 2 2 (4.5) (4.5) 2 ⎝ [H ] [HSO 4 ] ⎠ 2.04 V − (−0.077 V) = 2.12 V b. We can calculate ΔG° from ΔG° = ΔH° − TΔS° and then E° from ΔG° = −nFE°, or we can use the equation derived in Exercise 11.39. E c. o − 20 = TΔS o − ΔH o (253 K )(263.5 J / K ) + 315.9 × 103 J = = 1.98 J/C = 1.98 V nF (2 mol e − )(96,485 C / mol e − ) E − 20 = E o− 20 − RT 1 RT ln Q = 1.98 V − ln + 2 − nF nF [H ] [ HSO 4 ]2 E−20 = 1.98 V − (8.3145 J K −1 mol −1 )(253 K ) 1 ln = 1.98 V − (−0.066 V) − − 2 (2 mol e )(96,485 C / mol e ) (4.5) (4.5) 2 = 2.05 V d. As the temperature decreases, the cell potential decreases. Also, oil becomes more viscous at lower temperatures, which adds to the difficulty of starting an engine on a cold day. The combination of these two factors results in batteries failing more often on cold days than on warm days. 52. a. Ag2CrO4(s) + 2 e− → 2 Ag(s) + CrO42−(aq) Hg2Cl2 + 2 e− → 2 Hg + 2 Cl− E° = 0.446 V ESCE = 0.242 V SCE will be the oxidation half-reaction with Ecell = 0.446 - 0.242 = 0.204 V. ΔG = −nFEcell = −2(96,485)(0.204)J = −3.94 × 104 J = −39.4 kJ b. In SCE, we assume all concentrations are constant. Therefore, only CrO42− appears in the Q expression, and it will appear in the numerator since CrO42− is produced in the reduction half-reaction. To calculate Ecell at nonstandard CrO42- concentrations, we use the following equation. o Ecell = Ecell − c. Ecell = 0.204 − 0.0591 0.0591 log[CrO42−] = 0.204 V − log[CrO42−] 2 2 0.0591 log(1.00 × 10−5) = 0.204 V − (−0.148 V) = 0.352 V 2 d. 0.504 V = 0.204 V − (0.0591/2) log[CrO42−] log[CrO42−] = −10.152, [CrO42−] = 10−10.152 = 7.05 × 10−11 M CHAPTER 11 e. ELECTROCHEMISTRY 429 Eoc = 0.446 V Ag2CrO4 + 2 e− → 2 Ag + CrO42− o (Ag → Ag+ + e−) × 2 − Ea = -0.80 V _________________________________________________________________ Ag2CrO4(s) → 2 Ag+(aq) + CrO4−(aq) Eocell = 53. Eocell = −0.35 V; K = Ksp = ? 0.0591 (−0.35 V ) ( 2) = −11.84, Ksp = 10−11.84 = 1.4 × 10−12 log Ksp, log Ksp = n 0.0591 Concentration cell: a galvanic cell in which both compartments contain the same components but at different concentrations. All concentration cells have E ocell = 0 because both compartments contain the same contents. The driving force for the cell is the different ion concentrations at the anode and cathode. The cell produces a voltage as long as the ion concentrations are different. Equilibrium for a concentration cell is reached (E = 0) when the ion concentrations in the two compartments are equal. The net reaction in a concentration cell is: Ma+(cathode, x M) → Ma+(anode, y M) E ocell = 0 and the Nernst equation is: E = EE − 0.0591 [M a + (anode)] 0.0591 log a + , log Q = − n a [M (cathode)] where a is the number of electrons transferred. To register a potential (E > 0), the log Q term must be a negative value. This occurs when Ma+(cathode) > Ma+(anode). The higher ion concentration is always at the cathode, and the lower ion concentration is always at the anode. The magnitude of the cell potential depends on the magnitude of the differences in ion concentrations between the anode and cathode. The larger the difference in ion concentrations, the more negative is the log Q term, and the more positive is the cell potential. Thus, as the difference in ion concentrations between the anode and cathode compartments increases, the cell potential increases. This can be accomplished by decreasing the ion concentration at the anode and/or by increasing the ion concentration at the cathode. 54. When NaCl is added to the anode compartment, Ag+ reacts with Cl− to form AgCl(s). Adding Cl− lowers the Ag+ concentration, which causes an increase in the cell potential. To determine Ksp for AgCl (Ksp = [Ag+][Cl−]), we must know the equilibrium Ag+ and Cl− concentrations. Here, [Cl−] is given, and we use the Nernst equation to calculate the [Ag+] at the anode. 55. As is the case for all concentration cells, Ecell = 0, and the smaller ion concentration is always in the anode compartment. The general Nernst equation for the Ni | Ni2+ (x M) | | Ni2+(y M) | Ni concentration cell is: o 430 CHAPTER 11 ELECTROCHEMISTRY [ Ni 2+ ]anode 0.0591 0.0591 log Q = log n 2 [ Ni 2 + ]cathode o Ecell = Ecell − o a. Because both compartments are at standard conditions ([Ni2+] = 1.0 M), Ecell = Ecell = 0 V. No electron flow occurs. b. Cathode = 2.0 M Ni2+; anode = 1.0 M Ni2+; electron flow is always from the anode to the cathode, so electrons flow to the right in the diagram. Ecell = [ Ni 2+ ]anode − 0.0591 − 0.0591 1 .0 log = = 8.9 × 10−3 V log 2+ 2 2 2 .0 [ Ni ]cathode c. Cathode = 1.0 M Ni2+; anode = 0.10 M Ni2+; electrons flow to the left in the diagram. Ecell = − 0.0591 0.10 log = 0.030 V 2 1.0 d. Cathode = 1.0 M Ni2+; anode = 4.0 × 10−5 M Ni2+; electrons flow to the left in the diagram. Ecell = 4.0 × 10 −5 − 0.0591 = 0.13 V log 2 1.0 e. Since both concentrations are equal, log(2.5/2.5) = log(1.0) = 0, and Ecell = 0. No electron flow occurs. 56. Cathode: Eoc = 0.80 V M2+ + 2e− → M(s) o − Ea = −0.80 V Anode: M(s) → M2+ + 2e− ______________________________________________________ M2+(cathode) → M2+(anode) Ecell = 0.44 V = 0.00 V − log [M2+]anode = − Eocell = 0.00 V [M 2+ ]anode [M 2 + ]anode 0.0591 0.0591 , 0.44 = − log log 2 2 1 .0 [M 2+ ]cathode 2(0.44) = −14.89, [M2+]anode = 1.3 × 10−15 M 0.0591 Because we started with equal numbers of moles of SO42− and M2+, [M2+] = [SO42−] at equilibrium. Ksp = [M2+][SO42−] = (1.3 × 10−15)2 = 1.7 × 10−30 57. Cu2+(aq) + H2(g) → 2 H+(aq) + Cu(s) Eocell = 0.34 V − 0.00V = 0.34 V and n = 2 CHAPTER 11 ELECTROCHEMISTRY 431 o Since PH 2 = 1.0 atm and [H+] = 1.0 M: Ecell = Ecell − a. Ecell = 0.34 V − 1 0.0591 log 2 [Cu 2+ ] 0.0591 1 log = 0.34 V − 0.11 V = 0.23 V 2 2.5 × 10 − 4 b. Use the Ksp expression to calculate the Cu2+ concentration in the cell. Cu(OH)2(s) ⇌ Cu2+(aq) + 2 OH−(aq) Ksp = 1.6 × 10−19 = [Cu2+][OH−]2 From the problem, [OH−] = 0.10 M, so: [Cu2+] = Ecell = Eocell − 1.6 × 10 −19 = 1.6 × 10−17 M 2 (0.10) 1 0.0591 1 0.0591 log = 0.34 − = 0.34 − 0.50 log 2+ 2 2 [Cu ] 1.6 × 10 −17 = −0.16 V Because Ecell < 0, the forward reaction is not spontaneous, but the reverse reaction is spontaneous. The Cu electrode becomes the anode, and Ecell = 0.16 V for the reverse reaction. The cell reaction is 2 H+(aq) + Cu(s) → Cu2+(aq) + H2(g). c. 0.195 V = 0.34 V − 0.0591 1 1 log , log = 4.91, [Cu2+] = 10−4.91 2+ 2 [Cu ] [Cu 2+ ] = 1.2 × 10−5 M Note: When determining exponents, we will carry extra significant figures. o o d. Ecell = Ecell − (0.0591/2) log (1/[Cu2+]) = Ecell + 0.0296 log [Cu2+]; this equation is in the form of a straight-line equation, y = mx + b. A graph of Ecell versus log[Cu2+] will yield a straight line with slope equal to 0.0296 V or 29.6 mV. 58. o 3 Ni2+(aq) + 2 Al(s) → 2 Al3+(aq) + 3 Ni(s) Ecell = −0.23 V + 1.66 V = 1.43 V; n = 6 Ecell = Eocell − 0.0591 [Al3+ ]2 0.0591 [Al3+ ]2 log log , 1.82 V = 1.43 V − n 6 [ Ni 2 + ]3 (1.0) 3 log [Al3+]2 = −39.59, [Al3+]2 = 10−39.59, [Al3+] = 1.6 × 10−20 M Al(OH)3(s) ⇌ Al3+(aq) + 3 OH−(aq) Ksp = [Al3+] [OH−]3; from the problem, [OH−] = 1.0 × 10−4 M. Ksp = (1.6 × 10−20)(1.0 × 10−4)3 = 1.6 × 10−32 59. a. Ag+(x M, anode) → Ag+(0.10 M, cathode); for the silver concentration cell, EE = 0.00 (as is always the case for concentration cells) and n = 1. 432 CHAPTER 11 E = 0.76 V = 0.00 ! 0.76 = -0.0591 log [Ag + ]anode 0.0591 log 1 [Ag + ]cathode [Ag + ]anode [Ag + ]anode = 10 −12.86 , [Ag+]anode = 1.4 × 10 −14 M , 0.10 0.10 b. Ag+(aq) + 2 S2O32−(aq) ⇌ Ag(S2O3)23−(aq) 3− K= ELECTROCHEMISTRY [Ag(S2 O 3 ) 2 ] 2− [Ag + ][S2 O 3 ]2 = 1.0 × 10 −3 = 2.9 × 1013 1.4 × 10 −14 (0.050) 2 Electrolysis 60. a. Al3+ + 3 e− → Al; 3 mol e− is needed to produce 1 mol Al from Al3+. 1.0 × 103 g × b. 1.0 g Ni × 1 mol Al 3 mol e − 96,485 C 1s × × × = 1.07 × 105 s − 27.0 g mol Al 100.0 C mol e = 3.0 × 101 hours 1 mol 2 mol e − 96,485 C 1s × × × = 33 s − 58.7 g mol Ni 100.0 C mol e c. 5.0 mol Ag × 61. 15 A = 1 mol e − 96,485 C 1s × × = 4.8 × 103 s = 1.3 hours − mol Ag 100.0 C mol e 15 C 60 s 60 min × × = 5.4 × 104 C of charge passed in 1 hour s min mol a. 5.4 × 104 C × 1 mol e − 1 mol Co 58.9 g × × = 16 g Co 96,485 C 2 mol e − mol b. 5.4 × 104 C × 1 mol e − 1 mol Hf 178.5 g × × = 25 g Hf 96,485 C 4 mol e − mol c. 2 I− → I2 + 2e−; 5.4 × 104 C × 1 mol e − 1 mol I 2 253.8 g I 2 × × = 71 g I2 − 96,485 C 2 mol e mol I 2 d. Cr is in the +6 oxidation state in CrO3. Six moles of e− are needed to produce 1 mol Cr from molten CrO3. 5.4 × 104 C × 1 mol e − 1 mol Cr 52.0 g Cr × × = 4.9 g Cr 96,485 C 6 mol e − mol Cr CHAPTER 11 62. ELECTROCHEMISTRY 2.30 min × 433 2.00 C 1 mol e − 1 mol Ag 60 s = 138 s; 138 s × × × = 2.86 × 10−3 mol Ag − min s 96,485 C mol e [Ag+] = 2.86 × 10−3 mol Ag+/0.250 L = 1.14 × 10−2 M 63. The oxidation state of bismuth in BiO+ is +3 because oxygen has a !2 oxidation state in this ion. Therefore, 3 moles of electrons are required to reduce the bismuth in BiO+ to Bi(s). 10.0 g Bi × 1 mol Bi 3 mol e − 96,485 C 1s × × × = 554 s = 9.23 min − 209.0 g Bi mol Bi 25.0 C mol e 64. When molten salts are electrolyzed, there is only one species present that can be oxidized (the anion in simple salts), and there is only one species that can be reduced (the cation in simple salts). When H2O is present, as is the case when aqueous solutions are electrolyzed, we must consider the oxidation and reduction of water as potential reactions that can occur. When water is present, more reactions can take place, making predictions more difficult. 65. First determine the species present, and then reference Table 11.1 to help you identify each species as a possible oxidizing agent (species reduced) or as a possible reducing agent (species oxidized). Of all the possible oxidizing agents, the species that will be reduced at the cathode will have the most positive E oc value; the species that will be oxidized at the anode will be the reducing agent with the most positive − E oa value. a. Species present: Ni2+ and Br-; Ni2+ can be reduced to Ni, and Br− can be oxidized to Br2 (from Table 11.1). The reactions are: Cathode: Ni2+ + 2e− → Ni Anode: 2 Br- → Br2 + 2 e− E oc = −0.23 V − E oa = −1.09 V b. Species present: Al3+ and F−; Al3+ can be reduced, and F− can be oxidized. The reactions are: Cathode: Al3+ + 3 e− → Al E oc = −1.66 V 2 F− → F2 + 2 e− − E oa = −2.87 V Anode: c. Species present: Mn2+ and I-; Mn2+ can be reduced, and I− can be oxidized. The reactions are: Cathode: Mn2+ + 2 e− → Mn Anode: 2 I− → I2 + 2 e− E oc = −1.18 V − E oa = −0.54 V d. For aqueous solutions, we must now consider H2O as a possible oxidizing agent and a possible reducing agent. Species present: Ni2+, Br−, and H2O. Possible cathode reactions are: 434 CHAPTER 11 Ni2+ + 2e− → Ni ELECTROCHEMISTRY E oc = −0.23 V 2 H2O + 2 e− → H2 + 2 OH− E oc = −0.83 V Because it is easier to reduce Ni2+ than H2O (assuming standard conditions), Ni2+ will be reduced by the above cathode reaction. Possible anode reactions are: 2 Br- → Br2 + 2 e− − E oa = −1.09 V 2 H2O → O2 + 4 H+ + 4 e− − E oa = −1.23 V Because Br− is easier to oxidize than H2O (assuming standard conditions), Br− will be oxidized by the above anode reaction. e. Species present: Al3+, F−, and H2O; Al3+ and H2O can be reduced. The reduction potentials are E oc = −1.66 V for Al3+ and E oc = −0.83 V for H2O (assuming standard conditions). H2O should be reduced at the cathode (2 H2O + 2 e− → H2 + 2 OH−). F− and H2O can be oxidized. The oxidation potentials are − E oa = −2.87 V for F− and − E oa = −1.23 V for H2O (assuming standard conditions). From the potentials, we would predict H2O to be oxidized at the anode (2 H2O → O2 + 4 H+ + 4 e−). f. Species present: Mn2+, I−, and H2O; Mn2+ and H2O can be reduced. The possible cathode reactions are: Mn2+ + 2 e− → Mn E oc = −1.18 V 2 H2O + 2 e− → H2 + 2 OH− E oc = −0.83 V Reduction of H2O should occur at the cathode because E oc for H2O is most positive. I− and H2O can be oxidized. The possible anode reactions are: 2 I− → I2 + 2 e− 2 H2O → O2 + 4 H+ + 4 e− − E oa = −0.54 V − E oa = −1.23 V Oxidation of I− will occur at the anode because − E oa for I− is most positive. 66. a. Species present: Na+, SO42−, and H2O. From the potentials, H2O is the most easily oxidized and the most easily reduced species present. The reactions are: Cathode: 2 H2O + 2 e− → H2(g) + 2 OH−; Anode: 2 H2O → O2(g) + 4 H+ + 4 e− CHAPTER 11 ELECTROCHEMISTRY 435 b. When water is electrolyzed, a significantly higher voltage than predicted is necessary to produce the chemical change (called overvoltage). This higher voltage is probably great enough to cause some SO42− to be oxidized instead of H2O. Thus the volume of O2 generated would be less than expected, and the measured volume ratio would be greater than 2:1. 67. Au3+ + 3 e− → Au Ag+ + e- → Ag Ni2+ + 2 e− → Ni Cd2+ + 2 e− → Cd E° = 1.50 V E° = 0.80 V E° = −0.23 V E° = −0.40 V 2 H2O + 2e− → H2 + 2 OH− E° = −0.83 V Au(s) will plate out first since it has the most positive reduction potential, followed by Ag(s), which is followed by Ni(s), and finally, Cd(s) will plate out last since it has the most negative reduction potential of the metals listed. 68. Mol e− = 50.0 min × 60 s 2.50 C 1 mol e − × × = 7.77 × 10−2 mol e− min s 96,485 C Mol Ru = 2.618 g Ru × 1 mol Ru = 2.590 × 10−2 mol Ru 101.1 g Ru Mol e − 7.77 × 10 −2 mol e − = = 3.00 Mol Ru 2.590 × 10 − 2 mol Ru The charge on the ruthenium ions is +3 (Ru3+ + 3 e− → Ru). 69. To begin plating out Pd: Ec = 0.62 V − 0.0591 [Cl − ]4 0.0591 (1.0) 4 log 0 . 62 log = − 2− 2 2 0.020 [PdCl 4 ] Ec = 0.62 V − 0.050 V = 0.57 V When 99% of Pd has plated out, [PdCl4−] = Ec = 0.62 − 1 (0.020) = 0.00020 M. 100 0.0591 (1.0) 4 log = 0.62 V − 0.11V = 0.51 V 2 2.0 × 10 − 4 To begin Pt plating: Ec = 0.73 V − 0.0591 (1.0) 4 log = 0.73 − 0.050 = 0.68 V 2 0.020 When 99% of Pt is plated: Ec = 0.73 − To begin Ir plating: Ec = 0.77 V − 0.0591 (1.0) 4 log = 0.73 − 0.11 = 0.62 V 2 2.0 × 10 − 4 0.0591 (1.0) 4 log = 0.77 − 0.033 = 0.74 V 3 0.020 436 CHAPTER 11 When 99% of Ir is plated: Ec = 0.77 − ELECTROCHEMISTRY 0.0591 (1.0) 4 log = 0.77 − 0.073 = 0.70 V 3 2.0 × 10 − 4 Yes, since the range of potentials for plating out each metal do not overlap, we should be able to separate the three metals. The exact potential to apply depends on the oxidation reaction. The order of plating will be Ir(s) first, followed by Pt(s), and finally, Pd(s) as the potential is gradually increased. 70. 74.1 s × 2.00 C 1 mol e − 1 mol M = 5.12 × 10 −4 mol M, where M = unknown metal × × s 96,485 C 3 mol e − Molar mass = 71. 0.107 g M 209 g = ; the element is bismuth. −4 mol 5.12 × 10 mol M Alkaline earth metals form +2 ions, so 2 mol of e− are transferred to form the metal M. Mol M = 748 s × 5.00 C 1 mol e − 1 mol M = 1.94 × 10−2 mol M × × s 96,485 C 2 mol e − Molar mass of M = 72. 0.471 g M = 24.3 g/mol; MgCl2 was electrolyzed. 1.94 × 10 − 2 mol M 150. × 103 g C 6 H 8 N 2 1 mol C 6 H 8 N 2 1h 1 min 2 mol e − 96,485 C × × × × × h 60 min 60 s 108.14 g C 6 H 8 N 2 mol C 6 H 8 N 2 mol e − = 7.44 × 104 C/s or a current of 7.44 × 104 A 73. F2 is produced at the anode: 2 F− → F2 + 2 e− 2.00 h × 60 min 60 s 10.0 C 1 mol e − × × × = 0.746 mol e− h min s 96,485 C 0.746 mol e− × V= 1 mol F2 nRT = 0.373 mol F2; PV = nRT, V = − P 2 mol e (0.373 mol)(0.08206 L atm K −1 mol −1 )(298 K ) = 9.12 L F2 1.00 atm K is produced at the cathode: K+ + e− → K 0.746 mol e− × 1 mol K 39.10 g K = 29.2 g K × mol K mol e − CHAPTER 11 ELECTROCHEMISTRY 74. 15000 J h 60 s 60 min × × = 5.4 × 107 J or 5.4 × 104 kJ s min h 15 kWh = To melt 1.0 kg Al requires: 1.0 × 103 g Al × 437 (Hall process) 1 mol Al 10.7 kJ × = 4.0 × 102 kJ 26.98 g mol Al It is feasible to recycle Al by melting the metal because, in theory, it takes less than 1% of the energy required to produce the same amount of Al by the Hall process. 75. In the electrolysis of aqueous sodium chloride, H2O is reduced in preference to Na+, and Cl− is oxidized in preference to H2O. The anode reaction is 2 Cl− → Cl2 + 2 e−, and the cathode reaction is 2 H2O + 2 e− → H2 + 2 OH−. The overall reaction is 2 H2O(l) + 2 Cl− (aq) → Cl2(g) + H2(g) + 2 OH− (aq). From the 1 : 1 mole ratio between Cl2 and H2 in the overall balanced reaction, if 6.00 L of H2(g) is produced, then 6.00 L of Cl2(g) also will be produced since moles and volume of gas are directly proportional at constant T and P (see Chapter 5 of text). 76. The half-reactions for the electrolysis of water are: (2 e− + 2 H2O → H2 + 2 OH−) × 2 2 H2O → 4 H+ + O2 + 4 e− __________________________ 2 H2O(l) → 2 H2(g) + O2(g) Note: 4 H+ + 4 OH− → 4 H2O, and n = 4 for this reaction as it is written. 15.0 min × 2 mol H 2 60 s 25.0 C 1 mol e − × × × = 1.17 × 10−2 mol H2 min s 96,485 C 4 mol e − At STP, 1 mole of an ideal gas occupies a volume of 22.42 L (see Chapter 5 of the text). 1.17 × 10−2 mol H2 × 22.42 L = 0.262 L = 262 mL H2 mol H 2 1.17 × 10−2 mol H2 × 1 mol O 2 22.42 mol L × = 0.131 L = 131 mL O2 2 mol H 2 mol O 2 Additional Exercises 77. (CO + O2− → CO2 + 2 e−) × 2 O2 + 4 e− → 2 O2− __________________________ 2 CO + O2 → 2 CO2 ΔG = −nFE, E = − ΔG − (−380 × 10 −3 J ) = = 0.98 V nF (4 mol e − )(96,485 C / mol e − ) 438 78. CHAPTER 11 ELECTROCHEMISTRY As a battery discharges, Ecell decreases, eventually reaching zero. A charged battery is not at equilibrium. At equilibrium, Ecell = 0 and ΔG = 0. We get no work out of an equilibrium system. A battery is useful to us because it can do work as it approaches equilibrium. Both fuel cells and batteries are galvanic cells that produce cell potentials to do useful work. However, fuel cells, unlike batteries, have the reactants continuously supplied and can produce a current indefinitely. The overall reaction in the hydrogen-oxygen fuel cell is 2 H2(g) + O2(g) → 2 H2O(l). The half-reactions are: 4 e− + O2 + 2 H2O → 4 OH− cathode 2 H2 + 4 OH− → 4 H2O + 4 e− anode Utilizing the standard potentials in Table 11.1, E ocell = 0.40 V + 0.83 V = 1.23 V for the hydrogen-oxygen fuel cell. As with all fuel cells, the H2(g) and O2(g) reactants are continuously supplied. See Figure 11.16 for a schematic of this fuel cell. 79. For C2H5OH, H has a +1 oxidation state, and O has a !2 oxidiation state. This dictates a !2 oxidation state for C. For CO2, O has a !2 oxidiation state, so carbon has a +4 oxidiation state. Six moles of electrons are transferred per mole of carbon oxidized (C goes from !2 → +4). Two moles of carbon are in the balanced reaction, so n = 12. wmax = !1320 kJ = ΔG = !nFE, !1320 × 103 J = !nFE = !(12 mol e−)(96,485 C/mol e−)E E = 1.14 J/C = 1.14 V 80. O2 + 2 H2O + 4 e− → 4 OH− E oc = 0.40 V ! E oa = 0.83 V (H2 + 2 OH− → 2 H2O + 2 e−) × 2 ________________________________________________________ 2 H2(g) + O2(g) → 2 H2O(l) E ocell = 1.23 V = 1.23 J/C Because standard conditions are assumed, wmax = ΔG° for 2 mol H2O produced. ΔG° = − nFE ocell = !(4 mol e−)(96,485 C/mol e−)(1.23 J/C) = !475,000 J = !475 kJ For 1.00 × 103 g H2O produced, wmax is: 1.00 × 103 g H2O × 1 mol H 2 O − 475 kJ × = !13,200 kJ = wmax 18.02 g H 2 O 2 mol H 2 O The work done can be no larger than the free energy change. The best that could happen is that all the free energy released would go into doing work, but this does not occur in any real process because there is always waste energy in a real process. Fuel cells are more efficient in converting chemical energy into electrical energy; they are also less massive. The major CHAPTER 11 ELECTROCHEMISTRY 439 disadvantage is that they are expensive. In addition, H2(g) and O2(g) are an explosive mixture if ignited, much more so than fossil fuels. 81. Cadmium goes from the zero oxidation state to the +2 oxidation state in Cd(OH)2. Because one mole of Cd appears in the balanced reaction, n = 2 mol electrons transferred. At standard conditions: wmax = ΔG° = −nFE°, wmax = −(2 mol e−)(96,485 C/mol e−)(1.10 J/C) = −2.12 × 105 J = −212 kJ 82. The corrosion of a metal can be viewed as the process of returning metals to their natural state. The natural state of metals is to have positive oxidation numbers. This corrosion is the oxidation of a pure metal (oxidation number = 0) into its ions. For corrosion of iron to take place, you must have: a. exposed iron surface–a reactant b. O2(g)–a reactant c. H2O(l)–a reactant, but also provides a medium for ion flow (it provides the salt bridge) d. ions–needed to complete the salt bridge Because water is a reactant and acts as a salt bridge for corrosion, cars do not rust in dry-air climates, whereas corrosion is a big problem in humid climates. Salting roads in the winter also increases the severity of corrosion. The dissolution of the salt into ions on the surface of a metal increases the conductivity of the aqueous solution and accelerates the corrosion process. 83. a. Paint: covers the metal surface so that no contact occurs between the metal and air. This only works as long as the painted surface is not scratched. b. Durable oxide coatings: covers the metal surface so that no contact occurs between the metal and air. c. Galvanizing: coating steel with zinc; Zn forms an effective oxide coating over steel; also, zinc is more easily oxidized than the iron in the steel. d. Sacrificial metal: attaching a more easily oxidized metal to an iron surface; the more active metal is preferentially oxidized instead of iron. e. Alloying: adding chromium and nickel to steel; the added Cr and Ni form oxide coatings on the steel surface. f. Cathodic protection: a more easily oxidized metal is placed in electrical contact with the metal we are trying to protect. It is oxidized in preference to the protected metal. The protected metal becomes the cathode electrode, thus cathodic protection. 440 84. CHAPTER 11 ELECTROCHEMISTRY Metals corrode because they oxidize easily. Referencing Table 11.1, most metals are associated with negative standard reduction potentials. This means the reverse reactions, the oxidation half-reactions, have positive oxidation potentials, indicating that they oxidize fairly easily. Another key point is that the reduction of O2 (which is a reactant in corrosion processes) has a more positive E ored than most of the metals (for O2, E ored = 0.40 V). This means that when O2 is coupled with most metals, the reaction will be spontaneous because E ocell > 0, so corrosion occurs. The noble metals (Ag, Au, and Pt) all have standard reduction potentials greater than that of O2. Therefore, O2 is not capable of oxidizing these metals at standard conditions. Note: The standard reduction potential for Pt → Pt2+ + 2 e− is not in Table 11.1. As expected, its reduction potential is greater than that of O2 (E oPt = 1.19 V). 85. Zn → Zn2+ + 2 e− − E oa = 0.76 V; Fe → Fe2+ + 2 e− − E oa = 0.44 V It is easier to oxidize Zn than Fe, so the Zn will be oxidized, protecting the iron of the Monitor's hull. 86. (Al3+ + 3 e− → Al) × 2 E oc = −1.66 V − E oa = ? (M → M2+ + 2 e−) × 3 _________________________________________________________________ 3 M(s) + 2 Al3+(aq) → 2 Al(s) + 3 M2+(aq) E ocell = − E oa − 1.66 V ΔG° = − nFE ocell , −411 × 103 J = −(6 mol e−)(96,485 C/mol e−)( E ocell ), E ocell = 0.71 V E ocell = − E oa − 1.66 V = 0.71 V, − E oa = 2.37 or E oc = −2.37 From Table 11.1, the reduction potential for Mg2+ + 2 e− → Mg is −2.37 V, which fits the data. Hence the metal is magnesium. 87. The potential oxidizing agents are NO3− and H+. Hydrogen ion cannot oxidize Pt under either condition. Nitrate cannot oxidize Pt unless there is Cl− in the solution. Aqua regia has both Cl− and NO3-. The overall reaction is: (NO3− + 4 H+ + 3 e− → NO + 2 H2O) × 2 E oc = 0.96 V − E oa = −0.755 V (4 Cl− + Pt → PtCl42− + 2 e−) × 3 _________________________________________________________________________________ 12 Cl−(aq) + 3 Pt(s) + 2 NO3−(aq) + 8 H+(aq) → 3 PtCl42-(aq) + 2 NO(g) + 4 H2O(l) E ocell = 0.21 V 88. a. 3 e− + 4 H+ + NO3− → NO + 2 H2O E° = 0.96 V Nitric acid can oxidize Co to Co2+ ( E ocell > 0), but is not strong enough to oxidize Co to Co3+ ( E ocell < 0). Co2+ is the primary product assuming standard conditions. CHAPTER 11 ELECTROCHEMISTRY 441 b. Concentrated nitric acid is about 16 mol/L. [H+] = [NO3−] = 16 M; assume PNO = 1 atm. E = 0.96 V − P 0.0591 0.0591 1 log + 4 NO − = 0.96 − log 3 3 (16) 5 [H ] [ NO 3 ] E = 0.96 − (−0.12) = 1.08 V No, concentrated nitric acid still will only be able to oxidize Co to Co2+. 89. a. Ecell = Eref + 0.05916 pH, 0.480 V = 0.250 V + 0.05916 pH pH = 0.480 − 0.250 = 3.888; uncertainty = ±1 mV = ± 0.001 V 0.05916 pHmax = 0.481 − 0.250 = 3.905; 0.05916 pHmin = 0.479 − 0.250 = 3.871 0.05916 Thus if the uncertainty in potential is ±0.001 V, then the uncertainty in pH is ±0.017, or about ±0.02 pH units. For this measurement, [H+] = 10−3.888 = 1.29 × 10−4 M. For an error of +1 mV, [H+] = 10−3.905 = 1.24 × 10−4 M. For an error of −1 mV, [H+] = 10-3.871 = 1.35 × 10−4 M. So the uncertainty in [H+] is ±0.06 × 10−4 M = ±6 × 10−6 M. b. From part a, we will be within ±0.02 pH units if we measure the potential to the nearest ±0.001 V (±1 mV). 90. a. ΔG° = ∑ n p ΔG of , products − ∑ n r ΔG or, reactants = 2(−480.) + 3(86) − [3(−40.)] = −582 kJ From oxidation numbers, n = 6. ΔG° = −nFE°, E° = log K = b. − ΔG o − (−582,000 J ) = = 1.01 V nF 6(96,485) C nE o 6(1.01) = = 102.538, K = 10102.538 = 3.45 × 10102 0.0591 0.0591 3 × (2 e− + Ag2S → 2 Ag + S2−) E oAg 2S = ? − E oa = 1.66 V 2 × (Al → Al3+ + 3 e−) ___________________________________________________________ 3 Ag2S(s) + 2 Al(s) → 6 Ag(s) + 3 S2−(aq) + 2 Al3+(aq) E ocell = 1.01 V E ocell = E oAg 2S + 1.66, E oAg 2S = 1.01 − 1.66 = −0.65 V 442 91. CHAPTER 11 ELECTROCHEMISTRY 2 Ag+(aq) + Cu(s) → Cu2+(aq) + 2 Ag(s) E ocell = 0.80 - 0.34 V = 0.46 V; A galvanic cell produces a voltage as the forward reaction occurs. Any stress that increases the tendency of the forward reaction to occur will increase the cell potential, whereas a stress that decreases the tendency of the forward reaction to occur will decrease the cell potential. a. Added Cu2+ (a product ion) will decrease the tendency of the forward reaction to occur, which will decrease the cell potential. b. Added NH3 removes Cu2+ in the form of Cu(NH3)42+. Because a product ion is removed, this will increase the tendency of the forward reaction to occur, which will increase the cell potential. c. Added Cl− removes Ag+ in the form of AgCl(s). Because a reactant ion is removed, this will decrease the tendency of the forward reaction to occur, which will decrease the cell potential. d. Q1 = Q2 = [Cu 2 + ]0 ; as the volume of solution is doubled, each concentration is halved. [Ag + ]02 1/2 [Cu 2 + ]0 2[Cu 2+ ]0 = = 2Ql (1/2 [ Ag + ]0 ) 2 [Ag + ]02 The reaction quotient is doubled because the concentrations are halved. Because reactions are spontaneous when Q < K, and because Q increases when the solution volume doubles, the reaction is closer to equilibrium, which will decrease the cell potential. e. Because Ag(s) is not a reactant in this spontaneous reaction, and because solids do not appear in the reaction quotient expressions, replacing the silver electrode with a platinum electrode will have no effect on the cell potential. 92. a. e e Salt bridge Cathode Anode Pt 1.0 M Fe2+ 1.0 M Fe3+ Au 1.0 M Au3+ CHAPTER 11 ELECTROCHEMISTRY b. Au3+(aq) + 3 Fe2+(aq) → 3 Fe3+(aq) + Au(s) E cell = E ocell − 443 E ocell = 1.50 − 0.77 = 0.73 V 0.0591 [Fe3+ ]3 0.0591 log log Q = 0.73 V − 3 n [Au 3+ ][Fe 2 + ]3 Since [Fe3+] = [Fe2+] = 1.0 M: 0.31 V = 0.73 V − 0.0591 1 log 3 [Au 3+ ] 3(−0.42) 1 = − log , log [Au3+] = −21.32, [Au3+] = 10−21.32 = 4.8 × 10−22 M 0.0591 [Au 3+ ] Au3+ + 4 Cl− ⇌ AuCl4-; because the equilibrium Au3+ concentration is so small, assume [AuCl4−] ≈ [Au3+]0 ≈ 1.0 M; i.e., assume K is large so that the reaction essentially goes to completion. K= [AuCl −4 ] 1.0 = = 2.1 × 1025; asumption good (K is large). 3+ − 4 − 22 4 [Au ][Cl ] (4.8 × 10 )(0.10) Challenge Problems 93. Chromium(III) nitrate [Cr(NO3)3] has chromium in the +3 oxidation state. 1 mol Cr 3 mol e− 96, 485 C × × = 6.40 × 103 C of charge 1.15 g Cr × − mol Cr 52.00 g mol e For the Os cell, 6.40 × 103 C of charge also was passed. 1 mol e − 1 mol Os 3 = 0.0166 mol Os; 6.40 × 10 C × = 0.0663 mol e− 3.15 g Os × 96,485 C 190.2 g Mol e − 0.0663 = = 3.99 ≈ 4 Mol Os 0.0166 This salt is composed of Os4+ and NO3− ions. The compound is Os(NO3)4, osmium(IV) nitrate. For the third cell, identify X by determining its molar mass. Two moles of electrons are transferred when X2+ is reduced to X. Molar mass = 2.11 g X = 63.6 g/mol 1 mol e − 1 mol X 3 6.40 × 10 C × × 96,485 C 2 mol e − This is copper (Cu). 444 94. CHAPTER 11 Eocell = 0.400 V – 0.240 V = 0.160 V; E = EE ! 0.180 = 0.160 ! ELECTROCHEMISTRY 0.0591 log Q n 0.0591 0.120 log(9.32 × 10 −3 ), 0.020 = , n=6 n n Six moles of electrons are transferred in the overall balanced reaction. We now have to figure out how to get 6 mol e− into the overall balanced equation. The two possibilities are to have ion charges of +1 and +6 or +2 and +3; only these two combinations yield a 6 when common multiples are determined when adding the reduction half-reaction to the oxidation half-reaction. Because N forms +2 charged ions, M must form for +3 charged ions. The overall cell reaction can now be determined. (M3+ + 3 e− → M) × 2 E oc = 0.400 V ! E oa = !0.240 V (N → N2+ + 2 e−) × 3 _____________________________________________ 2 M3+ + 3 N → 3 N2+ + 2 M E ocell = 0.160 V Q = 9.32 × 10 −3 = [ N 2+ ]0 3 [M 3+ ]0 2 = (0.10) 3 , [M3+] = 0.33 M 3+ 2 [M ] wmax = ΔG = !nFE = !6(96,485)(0.180) = !1.04 × 105 J = !104 kJ The maximum amount of work this cell could produce is 104 kJ. 95. a. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) E ocell = 1.10 V Ecell = 1.10 V − 0.0591 [ Zn 2 + ] log 2 [Cu 2+ ] Ecell = 1.10 V − 0.0591 0.10 = 1.10 V − (!0.041 V) = 1.14 V log 2 2.50 b. 10.0 h × 60 min 60 s 10.0 C 1 mol e − 1 mol Cu × × × × = 1.87 mol Cu produced h min s 96,485 C 2 mol e − The Cu2+ concentration decreases by 1.87 mol/L, and the Zn2+ concentration will increase by 1.87 mol/L. [Cu2+] = 2.50 − 1.87 = 0.63 M; [Zn2+] = 0.10 + 1.87 = 1.97 M Ecell = 1.10 V − 0.0591 1.97 = 1.10 V − 0.015 V = 1.09 V log 2 0.63 CHAPTER 11 ELECTROCHEMISTRY c. 1.87 mol Zn consumed × 445 65.38 g Zn = 122 g Zn mol Zn Mass of electrode = 200. − 122 = 78 g Zn 1.87 mol Cu formed × 63.55 g Cu = 119 g Cu mol Cu Mass of electrode = 200. + 119 = 319 g Cu d. Three things could possibly cause this battery to go dead: 1. All the Zn is consumed. 2. All the Cu2+ is consumed. 3. Equilibrium is reached (Ecell = 0). We began with 2.50 mol Cu2+ and 200. g Zn × 1 mol Zn/65.38 g Zn = 3.06 mol Zn. Cu2+ is the limiting reagent and will run out first. To react all the Cu2+ requires: 2.50 mol Cu2+ × 2 mol e − 96,485 C 1s 1h × × × = 13.4 h 2+ − 10.0 C 3600 s mol Cu mol e For equilibrium to be reached: E = 0 = 1.10 V − 0.0591 [ Zn 2+ ] log 2 [Cu 2+ ] [ Zn 2 + ] = K = 102(1.10)/0.0591 = 1.68 × 1037 2+ [Cu ] This is such a large equilibrium constant that virtually all the Cu2+ must react to reach equilibrium. So the battery will go dead in 13.4 hours. 96. a. Emeas = Eref − 0.05916 log[F−], 0.4462 = 0.2420 − 0.05916 log[F−] log[F−] = −3.4517, [F−] = 3.534 × 10−4 M b. pH = 9.00; pOH = 5.00; [OH−] = 1.0 × 10−5 M 0.4462 = 0.2420 − 0.05916 log[[F−] + 10.0(1.0 × 10−5)] log([F−] + 1.0 × 10−4) = −3.452, [F−] + 1.0 × 10−4 = 3.532 × 10−4, [F−] = 2.5 × 10−4 M True value is 2.5 × 10−4, and by ignoring the [OH−], we would say [F−] was 3.5 × 10−4, so: Percent error = 1.0 × 10 −4 2.5 × 10 − 4 × 100 = 40.% 446 CHAPTER 11 c. [F−] = 2.5 × 10−4 M; [OH−] = ELECTROCHEMISTRY [F − ] 2.5 × 10 −4 = 50. = k[OH − ] 10.0[OH − ] 2.5 × 10 −4 = 5.0 × 10−7 M; pOH = 6.30; pH = 7.70 10.0 × 50. d. HF ⇌ H+ + F− Ka = [H + ][F − ] = 7.2 × 10−4; if 99% is F−, then [F−]/[HF] = 99. [HF] 99[H+] = 7.2 × 10−4, [H+] = 7.3 × 10−6 M; pH = 5.14 e. The buffer controls the pH so that there is little HF present and there is little response to OH−. Typically, a buffer of pH = 6.0 is used. 97. 3 × (e− + 2 H+ + NO3− → NO2 + H2O) a. E oc = 0.775 V − E oa = −0.957 V 2 H2O + NO → NO3− + 4 H+ + 3 e− _______________________________________________________________________ 2 H+(aq) + 2 NO3−(aq) + NO(g) → 3 NO2(g) + H2O(l) E ocell = −0.182 V K = ? log K = Eo 3(−0.182) = = −9.239, K = 10−9.239 = 5.77 × 10−10 0.0591 0.0591 b. Let C = concentration of HNO3 = [H+] = [NO3−]. 5.77 × 10−10 = 3 PNO 2 PNO × [ H + ]2 × [ NO 3− ]2 = 3 PNO 2 PNO × C 4 If 0.20% NO2 by moles and Ptotal = 1.00 atm: PNO 2 = 0.20 mol NO 2 × 1.00 atm = 2.0 × 10−3 atm; PNO = 1.00 − 0.0020 = 1.00 atm 100. mol total 5.77 × 10−10 = 98. (2.0 × 10 −3 ) 3 , C = 1.9 M HNO3 (1.00)C 4 2 Ag+(aq) + Ni(s) → Ni2+(aq) + Ag(s); the cell is dead at equilibrium (E = 0). E ocell = 0.80 V + 0.23 V = 1.03 V 0 = 1.03 V − 0.0591 log K, K = 7.18 × 1034 2 K is very large. Let the forward reaction go to completion. CHAPTER 11 ELECTROCHEMISTRY 2 Ag+ + Ni → Ni2+ + 2 Ag Before 1.0 M After 0 M 447 K = [Ni2+]/[Ag+]2 = 7.18 × 10 34 1.0 M 1.5 M Now solve the back-equilibrium problem. 2 Ag+ + Ni Initial 0 Change +2x Equil. 2x K = 7.18 × 10 ⇌ ← 34 = 1.5 M −x 1.5 − x 1 .5 − x 1.5 ≈ ; solving, x = 2.3 × 10−18 M. Assumptions good. 2 (2 x) (2 x) 2 −18 [Ag+] = 2x = 4.6 × 10 99. Ni2+ + 2 Ag 2 H+ + 2 e− → H2 M; [Ni2+] = 1.5 − 2.3 × 10−18 = 1.5 M E oc = 0.000 V − E oa = −(−0.440V) Fe → Fe2+ + 2e− _________________________________________________ 2 H+(aq) + Fe(s) → H2(g) + Fe3+(aq) E ocell = 0.440 V Ecell = E ocell PH 2 × [Fe3+ ] 0.0591 − log Q, where n = 2 and Q = n [ H + ]2 To determine Ka for the weak acid, first use the electrochemical data to determine the H+ concentration in the half-cell containing the weak acid. 0.333 V = 0.440 V − 0.0591 1.00 atm(1.00 × 10 −3 M ) log 2 [ H + ]2 0.107(2) 1.0 × 10 −3 1.0 × 10 −3 , = log = 103.621 = 4.18 × 103, [H+] = 4.89 × 10−4 M + 2 + 2 0.0591 [H ] [H ] Now we can solve for the Ka value of the weak acid HA through the normal setup for a weak acid problem. [H + ][A − ] HA(aq) ⇌ H+(aq) + A−(aq) Ka = [HA] Initial 1.00 M ~0 0 Equil. 1.00 − x x x Ka = (4.89 × 10 −4 ) 2 x2 , where x = [H+] = 4.89 × 10-4 M, Ka = = 2.39 × 10−7 −4 1.00 − x 1.00 − 4.89 × 10 448 CHAPTER 11 ELECTROCHEMISTRY 100. e e Salt bridge Cathode Anode Cu Cu 1.00 M Cu2+ 1.0 × 10-4 M Cu2+ a. E° = 0 (concentration cell), E = 0 − ⎛ 1.0 × 10 −4 ⎞ 0.0591 ⎟ , E = 0.12 V log ⎜⎜ ⎟ 2 ⎝ 1.00 ⎠ b. Cu2+(aq) + 4 NH3(aq) ⇌ Cu(NH3)42+(aq); because [Cu2+] << [NH3], [NH3]0 = 2.0 M = [NH3]eq. Also, Koverall = K1•K2•K3•K4 = 1.0 × 1013, so the reaction lies far to the right. Let the reaction go to completion. Cu2+ Before After + 4 NH3 1.0 × 10−4 M 0 2.0 M 2.0 → Cu(NH3)42+ 0 1.0 × 10−4 Now allow the reaction to get to equilibrium. Cu2+ Initial Equil. 0 x + 4 NH3 → 2.0 M 2.0 + 4x Cu(NH3)42+ 1.0 × 10−4 M 1.0 × 10−4 − x (1.0 × 10 −4 − x) 1.0 × 10 −4 = 1.0 × 1013, x = [Cu2+] = 6.3 × 10−19 M ≈ 4 4 x(2.0 + 4 x) x ( 2 .0 ) Thus E = 0 − 101. a. ⎛ 6.3 × 10 −19 ⎞ 0.0591 ⎟ , E = 0.54 V log ⎜⎜ ⎟ 2 ⎠ ⎝ 1.00 (Ag+ + e− → Ag) × 2 E oc = 0.80 V − E oa = −0.34 V Cu → Cu2+ + 2 e− _________________________________________________ 2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+(aq) E ocell = 0.46 V CHAPTER 11 ELECTROCHEMISTRY Ecell = E ocell − 449 [Cu 2 + ] 0.0591 log Q, where n = 2 and Q = . n [Ag + ]2 To calculate Ecell, we need to use the Ksp data to determine [Ag+]. AgCl(s) ⇌ Ag+(aq) + Cl−(aq) Initial s = solubility (mol/L) Equil. 0 s Ksp = 1.6 × 10-10 = [Ag+][Cl−] 0 s Ksp = 1.6 × 10−10 = s2, s = [Ag+] = 1.3 × 10−5 mol/L Ecell = 0.46 V − 0.0591 2 .0 log = 0.46 V − 0.30 = 0.16 V 2 (1.3 × 10 −5 ) 2 b. Cu2+(aq) + 4 NH3(aq) ⇌ Cu(NH4)42+(aq) K = 1.0 × 1013 = [Cu ( NH 3 ) 24+ ] [Cu 2+ ][ NH 3 ]4 Because K is very large for the formation of Cu(NH3)42+, the forward reaction is dominant. At equilibrium, essentially all the 2.0 M Cu2+ will react to form 2.0 M Cu(NH3)42+. This reaction requires 8.0 M NH3 to react with all the Cu2+ in the balanced equation. Therefore, the moles of NH3 added to 1.0-L solution will be larger than 8.0 mol since some NH3 must be present at equilibrium. In order to calculate how much NH3 is present at equilibrium, we need to use the electrochemical data to determine the Cu2+ concentration. Ecell = E ocell − log 0.0591 [Cu 2 + ] 0.0591 log log Q, 0.52 V = 0.46 V − n 2 (1.3 × 10 −5 ) 2 [Cu 2+ ] [Cu 2+ ] − 0.06(2) = = − 2.03, = 10−2.03 = 9.3 × 10−3 0.0591 (1.3 × 10 −5 ) 2 (1.3 × 10 −5 ) 2 [Cu2+] = 1.6 × 10−12 = 2 × 10−12 M (We carried extra significant figures in the calculation.) Note: Our assumption that the 2.0 M Cu2+ essentially reacts to completion is excellent as only 2 × 10−12 M Cu2+ remains after this reaction. Now we can solve for the equilibrium [NH3]. K = 1.0 × 1013 = [Cu ( NH 3 ) 24+ ] ( 2.0) = , [NH3] = 0.6 M 2+ 4 [Cu ][ NH 3 ] ( 2 × 10 −12 ) [ NH 3 ]4 Because 1.0 L of solution is present, 0.6 mol NH3 remains at equilibrium. The total moles of NH3 added is 0.6 mol plus the 8.0 mol NH3 necessary to form 2.0 M Cu(NH3)42+. Therefore, 8.0 + 0.6 = 8.6 mol NH3 was added. 450 CHAPTER 11 (Ag+ + e− → Ag) × 2 102. ELECTROCHEMISTRY E oc = 0.80 V − E oa = −(−0.13) Pb → Pb2+ + 2 e− ______________________________________________ 2 Ag+(aq) + Pb(s) → 2 Ag(s) + Pb2+(aq) E ocell = 0.93 V E = E° − log 0.0591 [ Pb 2+ ] 0.0591 (1.8) log log , 0.83 V = 0.93 V − + 2 n 2 [Ag ] [Ag + ]2 (1.8) 0.10(2) (1.8) = = 3.38, = 103.38, [Ag+] = 0.027 M + 2 + 2 0.0591 [Ag ] [Ag ] ⇌ Ag2SO4(s) Initial s = solubility (mol/L) Equil. 2 Ag+(aq) + SO42−(aq) 0 2s Ksp = [Ag+]2[SO42-] 0 s From the problem: 2s = 0.027 M, s = 0.027/2 Ksp = (2s)2(s) = (0.027)2(0.027/2) = 9.8 × 10−6 103. E oc = 0.80 V (Ag+ + e− → Ag) × 2 − E oa = 0.40 V Cd → Cd2+ + 2 e− _________________________________________________ E ocell = 1.20 V 2 Ag+(aq) + Cd(s) → Cd2+(aq) + 2 Ag(s) Overall complex ion reaction: Ag+(aq) + 2 NH3(aq) → Ag(NH3)2+(aq) K = K1K2 = 2.1 × 103(8.2 × 103) = 1.7 × 107 Because K is large, we will let the reaction go to completion, and then solve the backequilibrium problem. Before After Change Equil. Ag+ 1.00 M 0 x x + 2 NH3 15.0 M 13.0 +2x 13.0 + 2x ⇌ Ag(NH3)2+ 0 1.00 ← −x 1.00 − x K = 1.7 × 107 New initial + K= [Ag( NH 3 ) 2 ] 1.00 − x 1.00 ; 1.7 × 10 7 = ≈ + 2 2 [Ag ][ NH 3 ] x(13.0 + 2 x) x(13.0) 2 Solving: x = 3.5 × 10−10 M = [Ag+]; assumptions good. E = E° − ⎡ ⎤ 0.0591 [Cd 2 + ] 0.0591 1.0 log ⎢ log = 1.20 V − + 2 −10 2 ⎥ 2 2 [Ag ] ⎣ (3.5 × 10 ) ⎦ E = 1.20 − 0.56 = 0.64 V CHAPTER 11 104. ELECTROCHEMISTRY 451 Standard reduction potentials can only be manipulated and added together when electrons in the reduction half-reaction exactly cancel with the electrons in the oxidation half-reaction. We will solve this problem by applying the equation ΔGE = !nFEE to the half-reactions. ΔG of for e− = 0. M3+ + 3 e− → M ΔGE = !nFEE = !3(96,485)(0.10) = −2.9 × 104 J Because M and e− have ΔG of = 0: −2.9 × 104 J = ! ΔG of , M 3+ , ΔG of , M 3+ = 2.9 × 104 J M2+ + 2 e− → M ΔGE = !nFEE = !2(96,485)(0.50) = −9.6 × 104 J −9.6 × 104 J = ! ΔG of , M 2+ , ΔG of , M 2+ = 9.6 × 104 J M3+ + e− → M2+ EE = 105. ΔGE = 9.6 × 104 J – (2.9 × 104 J) = 6.7 × 104 J − ΔG o − (6.7 × 10 4 ) = = −0.69 V; M3+ + e− → M2+ nF (1)(96,485) E° = −0.69 V a. From Table 11.1: 2 H2O + 2 e− → H2 + 2 OH− E° = -0.83 V E ocell = E oH 2 O − E oZr = !0.83 V + 2.36 V = 1.53 V Yes, the reduction of H2O to H2 by Zr is spontaneous at standard conditions since E ocell > 0. b. (2 H2O + 2 e− → H2 + 2 OH−) × 2 Zr + 4 OH− → ZrO2•H2O + H2O + 4 e− ______________________________________ 3 H2O(l) + Zr(s) → 2 H2(g) + ZrO2•H2O(s) c. ΔG° = −nFE° = −(4 mol e−)(96,485 C/mol e−)(1.53 J/C) = −5.90 × 105 J = −590. kJ E = E° − E° = 0.0591 log Q; at equilibrium, E = 0 and Q = K. n 0.0591 4(1.53) log K, log K = = 104, K ≈ 10104 n 0.0591 d. 1.00 × 103 kg Zr × 1000 g 1 mol Zr 2 mol H 2 × × = 2.19 × 104 mol H2 kg 91.22 g Zr mol Zr 452 CHAPTER 11 2.19 × 104 mol H2 × V= ELECTROCHEMISTRY 2.016 g H 2 = 4.42 × 104 g H2 mol H 2 nRT (2.19 × 10 4 mol)(0.08206 L atm K −1 mol −1 )(1273 K ) = = 2.3 × 106 L H2 P 1.0 atm e. Probably yes; less radioactivity overall was released by venting the H2 than what would have been released if the H2 had exploded inside the reactor (as happened at Chernobyl). Neither alternative is pleasant, but venting the radioactive hydrogen is the less unpleasant of the two alternatives. Marathon Problems 106. a. Cu2+ + 2 e− → Cu E oc = 0.34 V − E oa = 1.20 V V → V2+ + 2 e− _________________________________________________ E ocell = 1.54 V Cu2+(aq) + V(s) → Cu(s) + V2+(aq) Ecell = E ocell − [V 2+ ] [V 2+ ] 0.0591 log Q, where n = 2 and Q = = . n [Cu 2+ ] 1.00 M To determine Ecell, we must know the initial [V2+], which can be determined from the stoichiometric point data. At the stoichiometric point, moles H2EDTA2− added = moles V2+ present initially. 0.0800 mol H 2 EDTA 2− 1 mol V 2 + × Mol V2+ present initially = 0.5000 L × L mol H 2 EDTA 2− = 0.0400 mol V2+ [V2+]0 = 0.0400 mol V 2+ = 0.0400 M 1.00 L Ecell = 1.54 V − 0.0591 0.0400 log = 1.54 V − (−0.0413) = 1.58 V 2 1.00 b. Use the electrochemical data to solve for the equilibrium [V2+]. Ecell = E ocell − 0.0591 [V 2+ ] log , n [Cu 2 + ] 1.98 V = 1.54 V − 0.0591 [V 2+ ] log 2 1.00 M [V2+] = 10−(0.44)(2)/0.0591 = 1.3 × 10−15 M H2EDTA2−(aq) + V2+(aq) ⇌ VEDTA2−(aq) + 2 H+(aq) K= [ VEDTA 2− ][H + ]2 [H 2 EDTA 2 − ][V 2+ ] In this titration reaction, equal moles of V2+ and H2EDTA2− are reacted at the stoichiometric point. Therefore, equal moles of both reactants must be present at CHAPTER 11 ELECTROCHEMISTRY 453 equilibrium, so [H2EDTA2−] = [V2+] = 1.3 × 10−15 M. In addition, because [V2+] at equilibrium is very small compared to the initial 0.0400 M concentration, the reaction essentially goes to completion. The moles of VEDTA2− produced will equal the moles of V2+ reacted (= 0.0400 mol). At equilibrium, [VEDTA2−] = 0.0400 mol/(1.00 L + 0.5000 L) = 0.0267 M. Finally, because we have a buffer solution, the pH is assumed not to change, so [H+] = 10−10.00 = 1.0 × 10−10 M. Calculating K for the reaction: K= [ VEDTA 2− ][H + ]2 (0.0267)(1.0 × 10 −10 ) 2 = = 1.6 × 108 2− 2+ −15 −15 [H 2 EDTA ][V ] (1.3 × 10 )(1.3 × 10 ) c. At the halfway point, 250.0 mL of H2EDTA2− has been added to 1.00 L of 0.0400 M V2+. Exactly one-half the 0.0400 mol of V2+ present initially has been converted into VEDTA2−. Therefore, 0.0200 mol of V2+ remains in 1.00 + 0.2500 = 1.25 L solution. Ecell = 1.54 V − 0.0591 [V 2+ ] 0.0591 (0.0200 / 1.25) log = 1.54 − log 2+ 2 1.00 2 [Cu ] Ecell = 1.54 − (−0.0531) = 1.59 V 107. Begin by choosing any reduction potential as 0.00 V. For example, let’s assume B2+ + 2 e− → B E° = 0.00 V From the data, when B/B2+ and E/E2+ are together as a cell, EE = 0.81 V. E2+ + 2 e− → E must have a potential of !0.81 V or 0.81 V since E may be involved in either the reduction or the oxidation half-reaction. We will arbitrarily choose E to have a potential of !0.81 V. Setting the reduction potential at !0.81 for E and 0.00 for B, we get the following table of potentials. B2+ + 2 e− → B 0.00 V E2+ + 2 e− → E !0.81 V D2+ + 2 e− → D 0.19 V C2+ + 2 e− → C !0.94 V A2+ + 2 e− → A !0.53 V From largest to smallest: D2+ + 2 e− → D 0.19 V B2+ + 2 e− → B 0.00 V A2+ + 2 e− → A !0.53 V E2+ + 2 e− → E !0.81 V C2+ + 2 e− → C !0.94 V 454 CHAPTER 11 ELECTROCHEMISTRY A2+ + 2 e− → A is in the middle. Let’s call this 0.00 V. We get: D2+ + 2 e− → D 0.72 V B2+ + 2 e− → B 0.53 V A2+ + 2 e− → A 0.00 V E2+ + 2 e− → E !0.28 V C2+ + 2 e− → C !0.41 V Of course, since the reduction potential of E could have been assumed to 0.81 V instead of !0.81 V, we can also get: C2+ + 2 e− → C 0.41 V E2+ + 2 e− → E 0.28 V 2+ -− 0.00 V 2+ − B +2e →B !0.53 V D2+ + 2 e− → D !0.72 V A +2e →A One way to determine which table is correct is to add metal C to a solution with D2+ and metal D to a solution with C2+. If D comes out of solution, the first table is correct. If C comes out of solution, the second table is correct. CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY Light and Matter 21. ν = c 3.00 × 108 m / s = = 3.0 × 1010 s−1 λ 1.0 × 10 − 2 m E = hν = 6.63 × 10−34 J s × 3.0 × 1010 s−1 = 2.0 × 10−23 J/photon 2.0 × 10 −23 J 6.02 × 10 23 photons = = 12 J/mol photon mol 22. The wavelength is the distance between consecutive wave peaks. Wave a shows 4 wavelengths and wave b shows 8 wavelengths. Wave a: λ = 1.6 × 10 −3 m = 4.0 × 10−4 m 4 Wave b: λ = 1.6 × 10 −3 m = 2.0 × 10−4 m 8 Wave a has the longer wavelength. Because frequency and photon energy are both inversely proportional to wavelength, wave b will have the higher frequency and larger photon energy since it has the shorter wavelength. ν = c 3.00 × 108 m/s = = 1.5 × 1012 s−1 −4 λ 2.0 × 10 m E = hc 6.63 × 10 −34 J s × 3.00 × 108 m/s = = 9.9 × 10−22 J −4 λ 2.0 × 10 m Because both waves are examples of electromagnetic radiation, both waves travel at the same speed, c, the speed of light. From Figure 12.3 of the text, both of these waves represent infrared electromagnetic radiation. 455 456 23. CHAPTER 12 a. λ = QUANTUM MECHANICS AND ATOMIC THEORY c 3.00 × 108 m / s = = 5.0 × 10 −6 m 13 −1 ν 6.0 × 10 s b. From Figure 12.3, this is infrared EMR. c. E = hν = 6.63 × 10 −34 J s × 6.0 × 1013 s −1 = 4.0 × 10 −20 J/photon 4.0 × 10 −20 J 6.022 × 10 23 photons × = 2.4 × 104 J/mol photon mol d. Frequency and photon energy are directly related (E = hv). Because 5.4 × 1013 s −1 EMR has a lower frequency than 6.0 × 1013 s −1 EMR, the 5.4 × 1013 s −1 EMR will have less energetic photons. 24. E = hν = hc 6.63 × 10 −34 J s × 3.00 × 108 m / s = = 8.0 × 10−18 J/photon 1m λ 25 nm × 1 × 109 nm 8.0 × 10 −18 J 6.02 × 10 23 photons = 4.8 × 106 J/mol × photon mol 25. The energy needed to remove a single electron is: 279.7 kJ 1 mol × = 4.645 × 10−22 kJ = 4.645 × 10−19 J mol 6.0221 × 10 23 E = 26. hc hc 6.6261 × 10 −34 J s × 2.9979 × 108 m / s = = 4.277 × 10−7 m = 427.7 nm , λ = λ E 4.645 × 10 −19 J Referencing Figure 12.3 of the text, 2.12 × 10−10 m electromagnetic radiation is X rays. λ = c 2.9979 × 108 m / s = = 2.799 m ν 107.1 × 10 6 s −1 From the wavelength calculated above, 107.1 MHz electromagnetic radiation is FM radiowaves. λ = hc 6.626 × 10 −34 J s × 2.998 × 108 m / s = = 5.00 × 10−7 m E 3.97 × 10 −19 J The 3.97 × 10−19 J/photon electromagnetic radiation is visible (green) light. CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 457 The photon energy and frequency order will be the exact opposite of the wavelength ordering because E and ν are both inversely related to λ. From the previously calculated wavelengths, the order of photon energy and frequency is: FM radiowaves < visible (green) light < X rays longest λ shortest λ lowest ν highest ν smallest E largest E 27. The energy to remove a single electron is: 208.4 kJ 1 mol × = 3.461 × 10−22 kJ = 3.461 × 10−19 J = Ew 23 mol 6.022 × 10 Energy of 254-nm light is: E = hc (6.626 × 10 −34 J s)(2.998 × 108 m / s) = = 7.82 × 10−19 J −9 λ 254 × 10 m Ephoton = EK + Ew, EK = 7.82 × 10−19 J − 3.461 × 10−19 J = 4.36 × 10−19 J = maximum KE 28. Planck’s discovery that heated bodies give off only certain frequencies of light and Einstein’s study of the photoelectric effect support the quantum theory of light. The wave-particle duality is summed up by saying all matter exhibits both particulate and wave properties. Electromagnetic radiation, which was thought to be a pure waveform, transmits energy as if it has particulate properties. Conversely, electrons, which were thought to be particles, have a wavelength associated with them. This is true for all matter. Some evidence supporting wave properties of matter are: 1. Electrons can be diffracted like light. 2. The electron microscope uses electrons in a fashion similar to the way in which light is used in a light microscope. However, wave properties of matter are only important for small particles with a tiny mass, e.g., electrons. The wave properties of larger particles are not significant. 29. The photoelectric effect refers to the phenomenon in which electrons are emitted from the surface of a metal when light strikes it. The light must have a certain minimum frequency (energy) in order to remove electrons from the surface of a metal. Light having a frequency below the minimum results in no electrons being emitted, whereas light at or higher than the minimum frequency does cause electrons to be emitted. For light having a frequency higher than the minimum frequency, the excess energy is transferred into kinetic energy for the emitted electron. Albert Einstein explained the photoelectric effect by applying quantum theory. 30. a. 10.% of speed of light = 0.10 × 3.00 × 108 m/s = 3.0 × 107 m/s 458 CHAPTER 12 λ= QUANTUM MECHANICS AND ATOMIC THEORY h 6.63 × 10 −34 J s = 2.4 × 10 −11 m = 2.4 × 10 −2 nm , λ = − 31 7 mv 9.11 × 10 kg × 3.0 × 10 m / s Note: For units to come out, the mass must be in kg since 1 J = 1 kg m2/s2. b. λ = h 6.63 × 10 −34 J s = = 3.4 × 10 −34 m = 3.4 × 10 −25 nm mv 0.055 kg × 35 m/s This number is so small that it is insignificant. We cannot detect a wavelength this small. The meaning of this number is that we do not have to worry about the wave properties of large objects. 31. 32. a. λ= h 6.626 × 10 −34 J s = = 1.32 × 10 −13 m − 27 8 mv 1.675 × 10 kg × (0.0100 × 2.998 × 10 m/s) b. λ= h h 6.626 × 10 −34 J s = = 5.3 × 103 m/s , v = −12 − 27 mv λm 75 × 10 m × 1.675 × 10 kg λ= h h , v= ; for λ = 1.0 × 102 nm = 1.0 × 10 −7 m: λm mv v = 6.63 × 10 −34 J s = 7.3 × 103 m/s −31 −7 9.11 × 10 kg × 1.0 × 10 m For λ = 1.0 nm = 1.0 × 10 −9 m: v = 33. m= 6.63 × 10 −34 J s = 7.3 × 105 m/s 9.11 × 10 −31 kg × 1.0 × 10 −9 m h 6.626 × 10 −34 kg m 2 / s = = 6.68 × 10 − 26 kg / atom −15 8 λv 3.31 × 10 m × (0.0100 × 2.998 × 10 m/s) 6.68 × 10 −26 kg 6.022 × 10 23 atoms 1000 g × × = 40.2 g/mol atom mol kg The element is calcium (Ca). Hydrogen Atom: The Bohr Model 34. The Bohr model assumes that the electron in hydrogen can orbit the nucleus at specific distances from the nucleus. Each orbit has a specific energy associated with it. Therefore, the electron in hydrogen can only have specific energies; not all energies are allowed. The term quantized refers to the allowed energy levels for the electron in hydrogen. CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 459 The great success of the Bohr model is that it could explain the hydrogen emission spectrum. The electron in H moves about the allowed energy levels by absorbing or emitting certain photons of energy. The photon energies absorbed or emitted must be exactly equal to the energy difference between any two allowed energy levels. Because not all energies are allowed in hydrogen (energy is quantized), not all energies of EMR are absorbed/emitted. The Bohr model predicted the exact wavelengths of light that would be emitted for a hydrogen atom. Although the Bohr model has great success for hydrogen and other one-electron ions, it does not explain emission spectra for elements/ions having more than one electron. The fundamental flaw is that we cannot know the exact motion of an electron as it moves about the nucleus; therefore, well-defined circular orbits are not appropriate. 35. For the H atom (Z = 1): En = -2.178 × 10−18 J/n2; for a spectral transition, ΔE = Ef − Ei: ⎛ 1 1 ⎞ ΔE = −2.178 × 10−18 J ⎜⎜ 2 − 2 ⎟⎟ ni ⎠ ⎝ nf where ni and nf are the levels of the initial and final states, respectively. A positive value of ΔE always corresponds to an absorption of light, and a negative value of ΔE always corresponds to an emission of light. ⎛ 1 1⎞ a. ΔE = −2.178 × 10−18 J ⎜ 2 − 2 ⎟ = −2.178 × 10−18 J ⎜2 3 ⎟⎠ ⎝ ⎛1 1⎞ ⎜ − ⎟ ⎝4 9⎠ ΔE = −2.178 × 10−18 J × (0.2500 - 0.1111) = −3.025 × 10−19 J The photon of light must have precisely this energy (3.025 × 10−19 J). |ΔE| = Ephoton = hν = 6.6261 × 10 −34 J s × 2.9979 × 108 m / s hc hc , λ = = λ ΔE 3.025 × 10 −19 J = 6.567 × 10−7 m = 656.7 nm From Figure 12.3, this is visible electromagnetic radiation (red light). ⎛ 1 1 ⎞ b. ΔE = −2.178 × 10−18 J ⎜ 2 − 2 ⎟ = −4.084 × 10−19 J ⎜2 4 ⎟⎠ ⎝ λ = 6.6261 × 10 −34 J s × 2.9979 × 108 m / s hc = 4.864 × 10−7 m = 486.4 nm = −19 ΔE 4.084 × 10 J This is visible electromagnetic radiation (green-blue light). 460 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY ⎛1 1 ⎞ c. ΔE = −2.178 × 10−18 J ⎜ 2 − 2 ⎟ = −1.634 × 10−18 J ⎜1 2 ⎟⎠ ⎝ 6.6261 × 10 −34 J s × 2.9979 × 108 m / s = 1.216 × 10−7 m = 121.6 nm λ = 1.634 × 10 −18 J This is ultraviolet electromagnetic radiation. ⎛ 1 1 ⎞ d. ΔE = −2.178 × 10−18 J ⎜ 2 − 2 ⎟ = −1.059 × 10−19 J ⎜3 4 ⎟⎠ ⎝ λ= 6.6261 × 10 −34 J s × 2.9979 × 108 m / s hc = 1.876 × 10−6 m or 1876 nm = −19 ΔE 1.059 × 10 J This is infrared electromagnetic radiation. 36. Ionization from n = 1 corresponds to the transition ni = 1 → nf = ∞ where E∞ = 0. ⎛1⎞ ΔE = E∞ − E1 = −E1 = RH ⎜ 2 ⎟ = RH , ΔE = 2.178 × 10−18 J = Ephoton ⎜1 ⎟ ⎝ ⎠ λ = hc (6.6261 × 10 −34 J s)(2.9979 × 108 m / s) = = 9.120 × 10−8 m = 91.20 nm −18 E 2.178 × 10 J ⎛ 1⎞ ⎛1⎞ To ionize from n = 3, ΔE = 0 − E3 = RH ⎜ 2 ⎟ = 2.178 × 10−18 J ⎜ ⎟ ⎜3 ⎟ ⎝9⎠ ⎝ ⎠ ΔE = Ephoton = 2.420 × 10−19 J; λ = 37. hc = 8.208 × 10−7 m = 820.8 nm E a. False; it takes less energy to ionize an electron from n = 3 than from the ground state. b. True c. False; the energy difference between n = 3 and n = 2 is smaller than the energy difference to n = 2 electronic transition than for the n = 3 to n = 1 transition. E and λ are inversely proportional to each other (E = hc/λ). d. True e. False; the ground state in hydrogen is n = 1 and all other allowed energy states are called excited states; n = 2 is the first excited state, and n = 3 is the second excited state. CHAPTER 12 38. QUANTUM MECHANICS AND ATOMIC THEORY ⎛ 1 1 ⎞ ΔE = −2.178 × 10−18 J ⎜⎜ 2 − 2 ⎟⎟ = −2.178 × 10−18 J ni ⎠ ⎝ nf λ = 461 ⎛ 1 1 ⎞⎟ ⎜ = 2.091 × 10−18 J = Ephoton − 2 ⎟ ⎜ 52 1 ⎠ ⎝ hc 6.6261 × 10 −34 J s × 2.9979 × 108 m / s = = 9.500 × 10−8 m = 95.00 nm E 2.091 × 10 −18 J Because wavelength and energy are inversely related, visible light (λ ≈ 400−700 nm) is not energetic enough to excite an electron in hydrogen from n = 1 to n = 5. ⎛ 1 1 ⎞ ΔE = −2.178 × 10−18 J ⎜ 2 − 2 ⎟ = 4.840 × 10−19 J ⎜6 2 ⎟⎠ ⎝ λ = hc 6.6261 × 10 −34 J s × 2.9979 × 108 m / s = = 4.104 × 10−7 m = 410.4 nm E 4.840 × 10 −19 J Visible light with λ = 410.4 nm will excite an electron from the n = 2 to the n = 6 energy level. 39. |ΔE| = Ephoton = hc 6.6261 × 10 −34 J s × 2.9979 × 108 m / s = = 5.001 × 10−19 J λ 397.2 × 10 −9 m ΔE = −5.001 × 10−19 J because we have an emission. ⎛ 1 1 ⎞ −5.001 × 10−19 J = E2 − En = −2.178 × 10−18 J ⎜ 2 − 2 ⎟ ⎜2 n ⎟⎠ ⎝ 1 1 1 0.2296 = − 2 , 2 = 0.0204, n = 7 4 n n 40. |ΔE| = Ephoton = hν = 6.626 × 10 −34 J s × 6.90 × 1014 s −1 = 4.57 × 10 −19 J ΔE = !4.57 × 10 −19 J because we have an emission. 1 ⎞ ⎛ 1 !4.57 × 10 −19 J = En – E5 = !2.178 × 10 −18 J ⎜ 2 − 2 ⎟ 5 ⎠ ⎝n 1 1 1 − = 0.210, = 0.250, n2 = 4, n = 2 2 2 25 n n The electronic transition is from n = 5 to n = 2. 41. ⎛ 1 ⎞ ΔE = E4 ! En = -En = 2.178 × 10 −18 J ⎜ 2 ⎟ ⎝n ⎠ 462 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY hc 6.626 × 10 −34 J s × 2.9979 × 108 m / s = 1.36 × 10 −19 J = −9 λ 1460 × 10 m Ephoton = ⎛ 1 ⎞ Ephoton = ΔE = 1.36 × 10 −19 J = 2.178 × 10 −18 ⎜ 2 ⎟, n2 = 16.0, n = 4 ⎝n ⎠ 42. For one electron species, En = − R H Z2 , where RH = 2.178 × 10−18 J, and Z = atomic number n2 (nuclear charge). The electronic transition is n = 1 → n = ∞ (E∞ = 0). This is called the ionization energy (IE). Because E∞ = 0, the IE is given by the energy of state n = 1 (ΔE = E∞ − E1 = −E1 = RHZ2/l2 = RHZ2). a. H: Z = 1; IE = 2.178 × 10−18 J (l)2 = 2.178 × 10−18 J/atom IE = 2.178 × 10 −18 J 1 kJ 6.0221 × 10 23 atoms = 1311.6 kJ/mol ≈ 1312 kJ/mol × × atom 1000 J mol For any one-electron species, the ionization energy per mole is given as: IE = 1311.6 kJ 2 (Z ) mol (We will carry an extra significant figure.) We get this by combining IE = 2.178 × 10−18 J (Z2) and: 2.178 × 10 −18 J 6.0221 × 10 23 atoms 1 kJ 1311.6 kJ × × = atom mol 1000 J mol b. He+: Z = 2; IE = 1311.6 kJ/mol × 22 = 5246 kJ/mol c. Li2+: Z = 3; IE = 1311.6 kJ/mol × 32 = 1.180 × 104 kJ/mol d. C5+: Z = 6; IE = 1311.6 kJ/mol × 62 = 4.722 × 104 kJ/mol (Assume n = 1 for all.) e. Fe25+: Z = 26; IE = 1311.6 kJ/mol × (26)2 = 8.866 × 105 kJ/mol 43. Ephoton = hc 6.6261 × 10 −34 J s × 2.9979 × 108 m/s = = 7.839 × 10 −19 J −9 λ 253.4 × 10 m ΔE = 7.839 × 10 −19 J The general energy equation for one-electron ions is En = !2.178 × 10 −18 J (Z2)/n2, where Z = atomic number. ⎛ 1 1 ΔE = !2.178 × 10 −18 J (Z)2 ⎜⎜ 2 − 2 ni ⎝ nf ⎞ ⎟ , Z = 4 for Be3+ ⎟ ⎠ CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 463 2 ⎛ 1 1 ⎞ ⎟ ΔE = !7.839 × 10 −19 J = !2.178 × 10 −18 (4)2 ⎜⎜ 2 − ⎟ n 5 ⎝ f ⎠ 7.839 × 10 −19 1 1 1 + = 2 , 2 = 0.06249, nf = 4 −18 25 nf nf 2.178 × 10 × 16 This emission line corresponds to the n = 5 → n = 4 electronic transition. Wave Mechanics and Particle in a Box 44. Units of ΔE × Δt = J × s, the same as the units of Planck's constant. Linear momentum p is equal to mass times velocity, p = mv. Units of ΔpΔx = kg × 45. kg m 2 kg m 2 m ×m= = ×s=J×s s s s2 a. Δp = mΔv = 9.11 × 10−31 kg × 0.100 m/s = ΔpΔx ≥ b. Δx = 9.11 × 10 −32 kg m s h 6.626 × 10 −34 J s h , Δx = = = 5.79 × 10−4 m −32 4π Δp 4π 4 × 3.142 × (9.11 × 10 kg m / s) h 6.626 × 10 −34 J s = = 3.64 × 10−33 m 4 π Δp 4 × 3.142 × 0.145 kg × 0.100 m / s c. The diameter of an H atom is roughly 1.0 × 10−8 cm. The uncertainty in position is much larger than the size of the atom. d. The uncertainty is insignificant compared to the size of a baseball. 46. 47. En = h2 24h 2 n2h 2 2 2 ; ΔE = E − E = (5 − 1 ) = 5 1 8mL2 8mL2 8mL2 ΔE = 24(6.626 × 10 −34 J s) 2 = 9.04 × 10−16 J 8(9.109 × 10 −31 kg)( 40.0 × 10 −12 m) 2 ΔE = hc (6.626 × 10 −34 J s)( 2.998 × 108 m / s) hc , λ = = = 2.20 × 10−10 m = 0.220 nm ΔE λ 9.04 × 10 −16 J En = n2h 2 9h 2 4h 2 5h 2 ; ΔE = E − = − E = 3 2 8mL2 8mL2 8mL2 8mL2 464 CHAPTER 12 ΔE = hc (6.626 × 10 −34 J s)( 2.998 × 108 m / s) = = 2.46 × 10−20 J −9 λ 8080 × 10 m ΔE = 2.46 × 10−20 J = 48. ΔE = En − E1 = ΔE = QUANTUM MECHANICS AND ATOMIC THEORY 5(6.626 × 10 −34 J s) 2 5h 2 = , L = 3.50 × 10−9 m = 3.50 nm 8(9.109 × 10 −31 kg) L2 8mL2 n2h 2 12 h 2 − 8mL2 8mL2 hc (6.6261 × 10 −34 J s)( 2.9979 × 108 m / s) = = 1.446 × 10−20 J λ 1.374 × 10 −5 m ΔE = 1.446 × 10−20 J = n 2 (6.626 × 10 −34 J s) 2 8(9.109 × 10 −31 kg) (10.0 × 10 −9 m) 2 − 1.446 × 10−20 = (6.02 × 10−22)n2 − 6.02 × 10−22, n2 = 49. En = n2h 2 ; 8mL2 (6.626 × 10 −34 J s) 2 8(9.109 × 10 −31 kg) (10.0 × 10 −9 m) 2 1.506 × 10 −20 = 25.0, n = 5 6.02 × 10 − 22 as L increases, En will decrease, and the spacing between energy levels will also decrease. 50. Total area = 1; area of one hump = 1/3 Shaded area = 1/6 = probability of finding the electron between x = 0 and x = L/6 in a one-dimensional box with n = 3. 51. En = n2h 2 , n = 1 for ground state; from equation, as L increases, En decreases. 8mL2 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY Using numbers: 10−6 m box: E1 = 465 h2 h2 (1 × 1012 m−2); 10−10 m box: E1 = (1 × 1020 m−2) 8m 8m As expected, the electron in the 1 × 10−6 m box has the lowest ground state energy. Orbitals and Quantum Numbers 52. Quantum numbers give the allowed solutions to Schrödinger equation. Each solution is an allowed energy level called a wave function or an orbital. Each wave function solution is described by three quantum numbers, n, R, and mR. The physical significance of the quantum numbers are: n: Gives the energy (it completely specifies the energy only for the H atom or ions with one electron) and the relative size of the orbitals. ℓ: Gives the type (shape) of orbital. mℓ: Gives information about the direction in which the orbital is pointing. The specific rules for assigning values to the quantum numbers n, ℓ, and mℓ are covered in Section 12.9. In Section 12.10, the spin quantum number ms is discussed. Since we cannot locate electrons, we cannot see if they are spinning. The spin is a convenient model. It refers to the ability of the two electrons that can occupy any specific orbital to produce two differently oriented magnetic moments. 53. The 2p orbitals differ from each other in the direction in which they point in space. The 2p and 3p orbitals differ from each other in their size, energy, and number of nodes. A nodal surface in an atomic orbital is a surface in which the probability of finding an electron is zero. The 1p, 1d, 2d, 1f, 2f, and 3f orbitals are not allowed solutions to the Schrödinger equation. For n = 1, R ≠ 1, 2, 3, etc., so 1p, 1d, and 1f orbitals are forbidden. For n = 2, R ≠ 2, 3, 4, etc., so 2d and 2f orbitals are forbidden. For n = 3, R ≠ 3, 4, 5, etc., so 3f orbitals are forbidden. The penetrating term refers to the fact that there is a higher probability of finding a 4s electron closer to the nucleus than a 3d electron. This leads to a lower energy for the 4s orbital relative to the 3d orbitals in polyelectronic atoms and ions. 54. b. For ℓ = 3, mℓ can range from -3 to +3; thus +4 is not allowed. c. n cannot equal zero. d. ℓ cannot be a negative number. The quantum numbers in part a are allowed. 55. a. For n = 3, ℓ = 3 is not possible. d. ms cannot equal !1. e. ℓ cannot be a negative number. f. For ℓ = 1, mℓ cannot equal 2. The quantum numbers in parts b and c are allowed. 466 56. CHAPTER 12 5p: three orbitals QUANTUM MECHANICS AND ATOMIC THEORY 3d z 2 : one orbital 4d: five orbitals n = 5: ℓ = 0 (1 orbital), ℓ = 1 (3 orbitals), ℓ = 2 (5 orbitals), ℓ = 3 (7 orbitals), ℓ = 4 (9 orbitals); total for n = 5 is 25 orbitals. n = 4: ℓ = 0 (1), ℓ = 1 (3), ℓ = 2 (5), ℓ = 3 (7); total for n = 4 is 16 orbitals. 57. 1p, 0 electrons (ℓ ≠ 1 when n = 1); 6d x 2 − y 2 , 2 electrons (specifies one atomic orbital); 4f, 14 electrons (7 orbitals have 4f designation); 7py, 2 electrons (specifies one atomic orbital); 2s, 2 electrons (specifies one atomic orbital); n = 3, 18 electrons (3s, 3p, and 3d orbitals are possible; there are one 3s orbital, three 3p orbitals, and five 3d orbitals). 58. ψ2 gives the probability of finding the electron at that point. 59. The diagrams of the orbitals in the text give only 90% probabilities of where the electron may reside. We can never be 100% certain of the location of the electrons due to Heisenburg’s uncertainty principle. 60. 2p 3p ψ2 ψ2 r r 61. For r = ao and θ = 0° (Z = 1 for H): ψ 2pz ⎛ ⎞ 1 1 ⎜ ⎟ = 1/ 2 ⎜ −11 ⎟ 4(2 π) ⎝ 5.29 × 10 ⎠ 3/ 2 (1) e−1/2 cos 0 = 1.57 × 1014; ψ2 = 2.46 × 1028 For r = ao and θ = 90°: ψ 2 p z = 0 because cos 90° = 0; ψ2 = 0; the xy plane is a node for the 2pz atomic orbital. 62. A node occurs when ψ = 0. ψ300 = 0 when 27 − 18σ + 2σ2 = 0. Solving using the quadratic formula: σ = 18 ± (18) 2 − 4(2)(27) 4 = 18 ± 108 4 σ = 7.10 or σ = 1.90; because σ = r/ao, the nodes occur at r = (7.10)ao = 3.76 × 10−10 m and at r = (1.90)ao = 1.01 × 10−10 m, where r is the distance from the nucleus. CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 467 Polyelectronic Atoms 63. Valence electrons are the electrons in the outermost principal quantum level of an atom (those electrons in the highest n value orbitals). The electrons in the lower n value orbitals are all inner core or just core electrons. The key is that the outermost electrons are the valence electrons. When atoms interact with each other, it will be the outermost electrons that are involved in these interactions. In addition, how tightly the nucleus holds these outermost electrons determines atomic size, ionization energy, and other properties of atoms. Elements in the same group have similar valence electron configurations and, as a result, have similar chemical properties. 64. The widths of the various blocks in the periodic table are determined by the number of electrons that can occupy the specific orbital(s). In the s block, we have one orbital (R = 0, mR = 0) that can hold two electrons; the s block is two elements wide. For the f block, there are 7 degenerate f orbitals (R = 3, mR = !3, !2, !1, 0, 1, 2, 3), so the f block is 14 elements wide. The g block corresponds to R = 4. The number of degenerate g orbitals is 9. This comes from the 9 possible mR values when R = 4 (mR = !4, !3, !2, !1, 0, 1, 2, 3, 4). With 9 orbitals, each orbital holding two electrons, the g block would be 18 elements wide. The h block has R = 5, mR = !5, !4, !3, !2, !1, 0, 1, 2, 3, 4, 5. With 11 degenerate h orbitals, the h block would be 22 elements wide. 65. a. n = 4: ℓ can be 0, 1, 2, or 3. Thus we have s (2 e−), p (6 e−), d (10 e−) and f (14 e−) orbitals present. Total number of electrons to fill these orbitals is 32. b. n = 5, mℓ = +1: for n = 5, ℓ = 0, 1, 2, 3, 4; for ℓ = 1, 2, 3, 4, all can have mℓ = +1. Four distinct orbitals, thus 8 electrons. c. n = 5, ms = +1/2: for n = 5, ℓ = 0, 1, 2, 3, 4. Number of orbitals = 1, 3, 5, 7, 9 for each value of ℓ, respectively. There are 25 orbitals with n = 5. They can hold 50 electrons, and 25 of these electrons can have ms = +1/2. d. n = 3, ℓ = 2: these quantum numbers define a set of 3d orbitals. There are 5 degenerate 3d orbitals that can hold a total of 10 electrons. e. n = 2, ℓ = 1: these define a set of 2p orbitals. There are 3 degenerate 2p orbitals that can hold a total of 6 electrons. f. It is impossible for n = 0. Thus no electrons can have this set of quantum numbers. g. The four quantum numbers completely specify a single electron. h. n = 3: 3s, 3p, and 3d orbitals all have n = 3. These orbitals can hold 18 electrons, and 9 of these electrons can have ms = +1/2. i. n = 2, ℓ = 2: this combination is not possible (ℓ ≠ 2 for n = 2). Zero electrons in an atom can have these quantum numbers. j. n = 1, ℓ = 0, mℓ = 0: these define a 1s orbital that can hold 2 electrons. 468 66. CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY Si: 1s22s22p63s23p2 or [Ne]3s23p2; Ga: 1s22s22p63s23p64s23d104p1 or [Ar]4s23d104p1 As: [Ar]4s23d104p3; Ge: [Ar]4s23d104p2; Al: [Ne]3s23p1; Cd: [Kr]5s24d10 S: [Ne]3s23p4; Se: [Ar]4s23d104p4 67. The following are complete electron configurations. Noble gas shorthand notation could also be used. Sc: 1s22s22p63s2 3p64s23d1; Fe: 1s22s22p63s2 3p64s23d6 P: 1s22s22p63s2 3p3; Cs: 1s22s22p63s2 3p64s23d104p65s24d105p66s1 Eu: 1s22s22p63s2 3p64s23d104p65s24d105p66s24f65d1* Pt: 1s22s22p63s2 3p64s23d104p65s24d105p66s24f145d8* Xe: 1s22s22p63s2 3p64s23d104p65s24d105p6; Br: 1s22s22p63s2 3p64s23d104p5 *Note: These electron configurations were written down using only the periodic table. Actual electron configurations are: Eu: [Xe]6s24f7 and Pt: [Xe]6s14f145d9 68. Cl: ls22s22p63s23p5 or [Ne]3s23p5 As:122s22p63s23p64s23d104p3 or [Ar]4s23d104p3 Sr: 1s22s22p63s23p64s23d104p65s2 or [Kr]5s2 W: [Xe]6s24f145d4 Pb: [Xe]6s24f145d106p2 Cf: [Rn]7s25f10* *Note: Predicting electron configurations for lanthanide and actinide elements is difficult since they have 0, 1, or 2 electrons in d orbitals. This is the actual Cf configuration. 69. Exceptions: Cr, Cu, Nb, Mo, Tc, Ru, Rh, Pd, Ag, Pt, and Au; Tc, Ru, Rh, Pd, and Pt do not correspond to the supposed extra stability of half-filled and filled subshells. 70. a. Both In and I have one unpaired 5p electron, but only the nonmetal I would be expected to form a covalent compound with the nonmetal F. One would predict an ionic compound to form between the metal In and the nonmetal F. I: [Kr]5s24d105p5 ↑↓ ↑↓ ↑ 5p b. From the periodic table, this will be element 120. Element 120: [Rn]7s25f146d107p68s2 c. Rn: [Xe]6s24f145d106p6; note that the next discovered noble gas will also have 4f electrons (as well as 5f electrons). d. This is chromium, which is an exception to the predicted filling order. Cr has 6 unpaired electrons, and the next most is 5 unpaired electrons for Mn. CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY Cr: [Ar]4s13d5 ↑ 4s 71. 469 ↑ ↑ ↑ ↑ ↑ 3d The two exceptions are Cr and Cu. Cr: 1s22s22p63s23p64s13p5; Cr has 6 unpaired electrons. 1s 2s 2p 3p or or 4s 3s 4s 3d 3d Cu: 1s22s22p63s23p64s13d10; Cu has 1 unpaired electron. 1s 2s 2p 3s 3p or 4s 72. 4s 3d Hg: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d10 a. From the electron configuration for Hg, we have 3s2, 3p6, and 3d10 electrons; 18 total electrons with n = 3. b. 3d10, 4d10, 5d10; 30 electrons are in d atomic orbitals. c. 2p6, 3p6, 4p6, 5p6; each set of np orbitals contain one pz atomic orbital. Because we have 4 sets of filled np orbitals and two electrons can occupy the pz orbital, there are 4(2) = 8 electrons in pz atomic orbitals. d. All the electrons are paired in Hg, so one-half of the electrons are spin-up (ms = +1/2) and the other half are spin-down (ms = !1/2); 40 electrons have spin-up. 73. Element 115, Uup, is in Group 5A under Bi (bismuth): Uup: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s25f146d107p3 a. 5s2, 5p6, 5d10, and 5f14; 32 electrons have n = 5 as one of their quantum numbers b. ℓ = 3 are f orbitals. 4f14 and 5f14 are the f orbitals used. They are all filled so 28 electrons have ℓ = 3. 470 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY c. p, d, and f orbitals all have one of the degenerate orbitals with mℓ = 1. There are 6 orbitals with mℓ = 1 for the various p orbitals used; there are 4 orbitals with mℓ =1 for the various d orbitals used; and there are 2 orbitals with mℓ = 1 for the various f orbitals used. We have a total of 6 + 4 + 2 = 12 orbitals with mℓ = 1. Eleven of these orbitals are filled with 2 electrons, and the 7p orbitals are only half-filled. The number of electrons with mℓ = 1 is 11 × (2 e−) + 1 × (1 e−) = 23 electrons. d. The first 112 electrons are all paired; one-half of these electrons (56 e−) will have ms = !1/2. The 3 electrons in the 7p orbitals singly occupy each of the three degenerate 7p orbitals; the three electrons are spin parallel, so the 7p electrons either have ms = +1/2 or ms = !1/2. Therefore, either 56 electrons have ms = !1/2 or 59 electrons have ms = !1/2. 74. 75. Ti : [Ar]4s23d2 n ℓ mℓ ms 4s 4 0 0 +1/2 4s 4 0 0 !1/2 3d 3 2 !2 +1/2 3d 3 2 !1 +1/2 Only one of 10 possible combinations of mℓ and ms for the first d electron. For the ground state, the second d electron should be in a different orbital with spin parallel; 4 possibilities. We get the number of unpaired electrons by examining the incompletely filled subshells. The paramagnetic substances have unpaired electrons, and the ones with no unpaired electrons are not paramagnetic (they are called diamagnetic). Li: 1s22s1 ↑ ; paramagnetic with 1 unpaired electron. 2s N: 1s22s22p3 ↑ ↑ ↑ ; paramagnetic with 3 unpaired electrons. 2p Ni: [Ar]4s23d8 ↑↓ ↑↓ ↑↓ ↑ ↑ ; paramagnetic with 2 unpaired electrons. 3d Te: [Kr]5s24d105p4 ↑↓ ↑ ↑ ; paramagnetic with 2 unpaired electrons. 5p Ba: [Xe]6s2 ↑↓ ; not paramagnetic because no unpaired electrons are present. 6s Hg: [Xe]6s24f145d10 ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ; not paramagnetic because no unpaired electrons. 5d 76. None of the s block elements have 2 unpaired electrons. In the p block, the elements with either ns2np2 or ns2np4 valence electron configurations have 2 unpaired electrons. For elements 1-36, these are elements C, Si, and Ge (with ns2np2) and elements O, S, and Se (with ns2np4). For the d block, the elements with configurations nd2 or nd8 have two unpaired CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 471 electrons. For elements 1-36, these are Ti (3d2) and Ni (3d8). A total of 8 elements from the first 36 elements have two unpaired electrons in the ground state. 77. The s block elements with ns1 for a valence electron configuration have one unpaired electrons. These are elements H, Li, Na, and K for the first 36 elements. The p block elements with ns2np1 or ns2np5 valence electron configurations have one unpaired electron. These are elements B, Al, and Ga (ns2np1) and elements F, Cl, and Br (ns2np5) for the first 36 elements. In the d block, Sc ([Ar]4s23d1) and Cu ([Ar]4s13d10) each have one unpaired electron. A total of 12 elements from the first 36 elements have one unpaired electron in the ground state. 78. O: 1s22s22px22py2 (↑↓ ↑↓ ); there are no unpaired electrons in this oxygen atom. This configuration would be an excited state, and in going to the more stable ground state (↑↓ ↑ ↑ ), energy would be released. 79. We get the number of unpaired electrons by examining the incompletely filled subshells. O: [He]2s22p4 2p4: ↑↓ ↑ ↑ Two unpaired e− O+: [He]2s22p3 2p3: ↑ Three unpaired e− O-: [He]2s22p5 2p5: ↑↓ ↑↓ ↑ Os: [Xe]6s24f145d6 5d6: ↑↓ ↑ Zr: [Kr]5s24d2 4d2: ↑ ↑ One unpaired e− ↑ ↑ ↑ Four unpaired e− Two unpaired e− ↑ S: [Ne]3s23p4 3p4: ↑↓ ↑ ↑ Two unpaired e− F: [He]2s22p5 2p5: ↑↓ ↑↓ ↑ One unpaired e− 3p6: ↑↓ ↑↓ ↑↓ Zero unpaired e− Ar: [Ne]3s23p6 80. ↑ a. Excited state of boron b. Ground state of neon B ground state: 1s22s22p1 c. Excited state of fluorine F ground state: 1s22s22p5 d. Excited state of iron Fe ground state: [Ar]4s23d6 The Periodic Table and Periodic Properties 81. Ionization energy: P(g) → P+(g) + e−; electron affinity: P(g) + e− → P−(g) 82. Across a period, the positive charge from the nucleus increases as protons are added. The number of electrons also increase, but these outer electrons do not completely shield the increasing nuclear charge from each other. The general result is that the outer electrons are 472 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY more strongly bound as one goes across a period, which results in larger ionization energies (and smaller size). Aluminum is out of order because the electrons in the filled 3s orbital shield some of the nuclear charge from the 3p electron. Hence the 3p electron is less tightly bound than a 3s electron, resulting in a lower ionization energy for aluminum as compared with magnesium. The ionization energy of sulfur is lower than phosphorus because of the extra electronelectron repulsions in the doubly occupied sulfur 3p orbital. These added repulsions, which are not present in phosphorus, make it slightly easier to remove an electron from sulfur as compared to phosphorus. 83. As successive electrons are removed, the net positive charge on the resulting ion increases. This increase in positive charge binds the remaining electrons more firmly, and the ionization energy increases. The electron configuration for Si is 1s22s22p63s23p2. There is a large jump in ionization energy when going from the removal of valence electrons to the removal of core electrons. For silicon, this occurs when the fifth electron is removed since we go from the valence electrons in n = 3 to the core electrons in n = 2. There should be another big jump when the thirteenth electron is removed, i.e., when a 1s electron is removed. 84. Both trends are a function of how tightly the outermost electrons are held by the positive charge in the nucleus. An atom in which the outermost electrons are held tightly will have a small radius and a large ionization energy. Conversely, an atom in which the outermost electrons are held weakly will have a large radius and a small ionization energy. The trends of radius and ionization energy should be opposite of each other. Electron affinity is the energy change associated with the addition of an electron to a gaseous atom. Ionization energy is the energy it takes to remove an electron from a gaseous atom. Because electrons are always attracted to the positive charge of the nucleus, energy will always have to be added to break the attraction and remove the electron from a neutral charged atom. Ionization energies are always endothermic for neutral charged atoms. Adding an electron is more complicated. The added electron will be attracted to the nucleus; this attraction results in energy being released. However, the added electron will encounter the other electrons, which results in electron-electron repulsions; energy must be added to overcome these repulsions. Which of the two opposing factors dominates determines whether the overall electron affinity for an element is exothermic or endothermic. 85. Size (radius) decreases left to right across the periodic table, and size increases from top to bottom of the periodic table. a. S < Se < Te b. Br < Ni < K c. F < Si < Ba d. Be < Na < Rb e. Ne < Se < Sr f. O < P < Fe All follow the general radius trend. 86. The ionization energy trend is the opposite of the radius trend; ionization energy (IE), in general, increases left to right across the periodic table and decreases from top to bottom of the periodic table. CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY a. Te < Se < S b. K < Ni < Br c. Ba < Si < F d. Rb < Na < Be e. Sr < Se < Ne f. 473 Fe < P < O All follow the general ionization energy (IE) trend. 87. a. Ba b. K c. O; in general, Group 6A elements have a lower ionization energy than neighboring Group 5A elements. This is an exception to the general ionization energy trend across the periodic table. d. S2−; this ion has the most electrons compared to the other sulfur species present. S2− has the largest number of electron-electron repulsions, which leads to S2− having the largest size and smallest ionization energy. e. Cs; this follows the general ionization energy trend. 88. a. He b. Cl c. Element 117 is the next halogen to be discovered (under At), element 119 is the next alkali metal to be discovered (under Fr), and element 120 is the next alkaline earth metal to be discovered (under Ra). From the general radius trend, the halogen (element 117) will be the smallest. d. Si e. Na+; this ion has the fewest electrons compared to the other sodium species present. Na+ has the smallest number of electron-electron repulsions, which makes it the smallest ion with the largest ionization energy. 89. As: [Ar]4s23d104p3; Se: [Ar]4s23d104p4; the general ionization energy trend predicts that Se should have a higher ionization energy than As. Se is an exception to the general ionization energy trend. There are extra electron-electron repulsions in Se because two electrons are in the same 4p orbital, resulting in a lower ionization energy for Se than predicted. 90. Expected order from IE trend: Be < B < C < N < O B and O are exceptions to the general IE trend. The IE of O is lower because of the extra electron-electron repulsions present when two electrons are paired in the same orbital. This makes it slightly easier to remove an electron from O compared to N. B is an exception because of the smaller penetrating ability of the 2p electron in B compared to the 2s electrons in Be. The smaller penetrating ability makes it slightly easier to remove an electron from B compared to Be. The correct IE ordering, taking into account the two exceptions, is: B < Be < C < O < N. 91. Size also decreases going across a period. Sc and Ti along with Y and Zr are adjacent elements. There are 14 elements (the lanthanides) between La and Hf, making Hf considerably smaller. 474 92. CHAPTER 12 a. Sg: [Rn]7s25f146d4 QUANTUM MECHANICS AND ATOMIC THEORY b. W c. Similar to Cr, SgO3, Sg2O3, SgO42−, and Sg2O72− are some likely possibilities. 93. a. Uus will have 117 electrons. [Rn]7s25f146d107p5 b. It will be in the halogen family and will be most similar to astatine (At). c. Like the other halogens: NaUus, Mg(Uus)2, C(Uus)4, O(Uus)2 d. Like the other halogens: UusO−, UusO2−, UusO3−, UusO4− 94. The electron affinity trend is very erratic. In general, EA becomes more positive in going down a group, and EA becomes more negative from left to right across a period (with many exceptions). a. I < Br < F < Cl; Cl is most exothermic (F is an exception). b. N < O < F, F is most exothermic. 95. Electron-electron repulsions become important when we try to add electrons to an atom. From the standpoint of electron-electron repulsions, larger atoms would have more favorable (more exothermic) electron affinities. Considering only electron-nucleus attractions, smaller atoms would be expected to have the more favorable (more exothermic) EA values. These two factors are the opposite of each other. Thus the overall variation in EA is not as great as ionization energy, in which attractions to the nucleus dominate. 96. O; the electron-electron repulsions will be much more severe for O− + e− → O2− than for O + e− → O−. 97. Electron-electron repulsions are much greater in O− than in S− because the electron goes into a smaller 2p orbital versus the larger 3p orbital in sulfur. This results in a more favorable (more exothermic) EA for sulfur. 98. Al (−44 kJ/mol), Si (−120), P (−74), S (-200.4), Cl (−348.7); based on the increasing nuclear charge, we would expect the EA to become more exothermic as we go from left to right in the period. Phosphorus is out of line. The reaction for the EA of P is: P(g) + e− → P−(g) [Ne]3s23p3 [Ne]3s23p4 The additional electron in P− will have to go into an orbital that already has one electron. There will be greater repulsions between the paired electrons in P−, causing the EA of P to be less favorable than predicted based solely on attractions to the nucleus. 99. a. The electron affinity of Mg2+ is ΔH for Mg2+(g) + e− → Mg+(g); this is just the reverse of the second ionization energy for Mg. EA(Mg2+) = −IE2(Mg) = −1445 kJ/mol (Table 12.6) CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 475 b. EA of Al+ is ΔH for Al+(g) + e− → Al(g); EA(Al+) = −IE1(Al) = −580 kJ/mol (Table 12.6) c. IE of Cl−is ΔH for Cl−(g) → Cl(g) + e−; IE(Cl−) = −EA(Cl) = +348.7 kJ/mol (Table 12.8) 100. d. Cl(g) → Cl+(g) + e− ΔH = IE1(Cl) = 1255 kJ/mol (Table 12.6) e. Cl+(g) + e− → Cl(g) ΔH = −IE1(Cl) = −1255 kJ/mol = EA(Cl+) a. More favorable EA: C, Br, K, and Cl; the electron affinity trend is very erratic. N, Ar, and Mg have positive EA values (unfavorable) due to their electron configurations (see text for detailed explanation). F has a more positive EA value than expected from its position in the periodic table. b. Higher IE: N, Ar, Mg, and F (follows the IE trend) c. Larger size: C , Br, K, and Cl (follows the radius trend) The Alkali Metals 101. Yes; the ionization energy general trend is to decrease down a group, and the atomic radius trend is to increase down a group. The data in Table 12.9 confirm both of these general trends. 102. It should be potassium peroxide (K2O2) because K+ ions are stable in ionic compounds. K2+ ions are not stable; the second ionization energy of K is very large compared to the first. 103. a. 6 Li(s) + N2(g) → 2 Li3N(s) b. 2 Rb(s) + S(s) → Rb2S(s) c. 2 Cs(s) + 2 H2O(l) → 2 CsOH(aq) + H2(g) d. 2 Na(s) + Cl2(g) → 2 NaCl(s) 104. ν= c 2.9979 × 108 m/s = = 6.582 × 1014 s−1 λ 455.5 × 10 −9 m E = hν = (6.626 × 10−34 J s)(6.582 × 1014 s−1) = 4.361 × 10−19 J 105. For 589.0 nm: ν = c 2.9979 × 108 m / s = = 5.090 × 1014 s −1 λ 589.0 × 10 −9 m E = hν = 6.6261 × 10 −34 J s × 5.090 × 1014 s −1 = 3.373 × 10 −19 J For 589.6 nm: ν = c/λ = 5.085 × 1014 s −1 ; E = hν = 3.369 × 10 −19 J 476 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY The energies in kJ/mol are: 106. 107. 3.373 × 10 −19 J × 1 kJ 6.0221 × 10 23 × = 203.1 kJ/mol 1000 J mol 3.369 × 10 −19 J × 1 kJ 6.0221 × 10 23 × = 202.9 kJ/mol 1000 J mol a. Li3N; lithium nitride b. NaBr; sodium bromide c. K2S; potassium sulfide d. Li3P; lithium phosphide e. RbH; rubidium hydride f. sodium hydride NaH; It should be element 119 with ground state electron configuration: [Rn]7s25f146d107p68s1 Additional Exercises 108. Each element has a characteristic spectrum. Thus the presence of the characteristic spectral lines of an element confirms its presence in any particular sample. 109. a. n 110. The valence electrons are strongly attracted to the nucleus for elements with large ionization energies. One would expect these species to readily accept another electron and have very exothermic electron affinities. The noble gases are an exception. The noble gases have a large IE but have an endothermic EA. Noble gases have a stable arrangement of electrons. Adding an electron disrupts this stable arrangement, resulting in unfavorable electron affinities. 111. Size decreases from left to right and increases going down the periodic table. Thus going one element right and one element down would result in a similar size for the two elements diagonal to each other. The ionization energies will be similar for the diagonal elements since the periodic trends also oppose each other. Electron affinities are harder to predict, but atoms with similar sizes and ionization energies should also have similar electron affinities. 112. Energy to make water boil = s × m × ΔT = Ephoton = b. n and ℓ 4.18 J × 50.0 g × 75.0°C = 1.57 × 104 J. o Cg hc 6.626 × 10 −34 J s × 2.998 × 108 m/s = = 2.04 × 10−24 J −2 λ 9.75 × 10 m 1.57 × 104 J × 1s = 20.9 s; 750. J 1.57 × 104 J × 1 photon = 7.70 × 1027 photons − 24 2.04 × 10 J CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 477 1000 m 1s × = 200 s (about 3 minutes) km 3.00 × 108 m 113. 60 × 106 km × 114. E= 310. kJ 1 mol × = 5.15 × 10−22 kJ = 5.15 × 10−19 J mol 6.022 × 10 23 λ= hc 6.626 × 10 −34 J s × (2.998 × 108 m / s) = = 3.86 × 10−7 m = 386 nm −19 E 5.15 × 10 J λ= 100 cm hc 6.626 × 10 −34 J s × 2.998 × 108 m / s = = 5.53 × 10 −7 m × −19 E m 3.59 × 10 J 115. = 5.53 × 10 −5 cm From the spectrum, λ = 5.53 × 10 −5 cm is greenish yellow light. ⎛ 1 1 ΔE = !RH ⎜ 2 − 2 ⎜n ni ⎝ f λ= ⎞ ⎟ = !2.178 × 10 −18 J ⎛⎜ 1 − 1 ⎞⎟ = -4.840 × 10 −19 J ⎟ 62 ⎠ ⎝2 ⎠ 2 116. 100 cm 6.6261 × 10 −34 J s × 2.9979 × 108 m / s hc = = 4.104 × 10 −7 m × −19 ΔE m 4.840 × 10 J = 4.104 × 10 −5 cm From the spectrum, λ = 4.104 × 10 −5 cm is violet light, so the n = 6 to n = 2 visible spectrum line is violet. 117. When the p and d orbital functions are evaluated at various points in space, the results sometimes have positive values and sometimes have negative values. The term phase is often associated with the + and ! signs. For example, a sine wave has alternating positive and negative phases. This is analogous to the positive and negative values (phases) in the p and d orbitals. 118. The general ionization energy trend says that ionization energy increases going left to right across the periodic table. However, one of the exceptions to this trend occurs between Groups 2A and 3A. Between these two groups, Group 3A elements usually have a lower ionization energy than Group 2A elements. Therefore, Al should have the lowest first ionization energy value, followed by Mg, with Si having the largest ionization energy. Looking at the values for the first ionization energy in the graph, the green plot is Al, the blue plot is Mg, and the red plot is Si. Mg (the blue plot) is the element with the huge jump between I2 and I3. Mg has two valence electrons, so the third electron removed is an inner core electron. Inner core electrons are always much more difficult to remove than valence electrons since they are closer to the nucleus, on average, than the valence electrons. 478 119. CHAPTER 12 a. QUANTUM MECHANICS AND ATOMIC THEORY Na(g) → Na+(g) + e− Cl(g) + e− → Cl−(g) IE1 = 495 kJ EA = !348.7 kJ ______________________________________________________________________________ Na(g) + Cl(g) → Na+(g) + Cl−(g) b. Mg(g) → Mg+(g) + e− F(g) + e−→ F-(g) ΔH = 146 kJ IE1 = 735 kJ EA = !327.8 kJ _____________________________________________________________________________ Mg(g) + F(g) → Mg+(g) + F−(g) c. ΔH = 407 kJ Mg+(g) → Mg2+(g) + e− F(g) + e− → F−(g) IE2 = 1445 kJ EA = !327.8 kJ _______________________________________________________________________ Mg+(g) + F(g) → Mg2+(g) + F−(g) ΔH = 1117 kJ d. From parts b and c, we get: Mg(g) + F(g) → Mg+(g) + F−(g) Mg+(g) + F(g) → Mg2+(g) + F−(g) ΔH = 407 kJ ΔH = 1117 kJ Mg(g) + 2 F(g) → Mg2+(g) + 2 F−(g) ΔH = 1524 kJ ______________________________________________________________________ 120. a. Se3+(g) → Se4+(g) + e− b. S−(g) + e− → S2−(g) d. Mg(g) → Mg+(g) + e− e. Mg(s) → Mg+(s) + e− c. Fe3+(g) + e− → Fe2+(g) 121. Valence electrons are easier to remove than inner core electrons. The large difference in energy between I2 and I3 indicates that this element has two valence electrons. This element is most likely an alkaline earth metal since alkaline earth metal elements all have two valence electrons. 122. All oxygen family elements have ns2np4 valence electron configurations, so this nonmetal is from the oxygen family. a. 2 + 4 = 6 valence electrons. b. O, S, Se, and Te are the nonmetals from the oxygen family (Po is a metal). c. Because oxygen family nonmetals form -2 charged ions in ionic compounds, K2X would be the predicted formula, where X is the unknown nonmetal. d. From the size trend, this element would have a smaller radius than barium. e. From the ionization energy trend, this element would have a smaller ionization energy than fluorine. 123. a. The 4+ ion contains 20 electrons. Thus the electrically neutral atom will contain 24 electrons. The atomic number is 24 which identifies it as chromium. CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 479 b. The ground state electron configuration of the ion must be: 1s22s22p63s23p64s03d2; there are 6 electrons in s orbitals. c. 12 d. 2 e. This is the isotope f. 3.01 × 1023 atoms × 50 24 Cr . There are 26 neutrons in the nucleus. 1 mol 49.9 g × = 24.9 g mol 6.022 × 10 23 atoms g. 1s22s22p63s23p64s13d5 is the ground state electron configuration for Cr. Cr is an exception to the normal filling order. 124. a. As we remove succeeding electrons, the electron being removed is closer to the nucleus, and there are fewer electrons left repelling it. The remaining electrons are more strongly attracted to the nucleus, and it takes more energy to remove these electrons. b. Al: 1s22s22p63s23p1; for I4, we begin removing an electron with n = 2. For I3, we remove an electron with n = 3 (the last valence electron). In going from n = 3 to n = 2 there is a big jump in ionization energy because the n = 2 electrons are much closer to the nucleus on average than the n = 3 electrons. Since the n = 2 electrons are closer to the nucleus, they are held more tightly and require a much larger amount of energy to remove compared to the n = 3 electrons. In general, valence electrons are much easier to remove than inner core electrons. c. Al4+; The electron affinity for Al4+ is ΔH for the reaction: Al4+(g) + e− →Al3+(g) ΔH = −I4 = −11,600 kJ/mol d. The greater the number of electrons, the greater the size. Size trend: Al4+ < Al3+ < Al2+ < Al+ < Al 125. a. Each orbital could hold 4 electrons. b. The first period corresponds to n = 1, which can only have 1s orbitals. The 1s orbital could hold 4 electrons; hence the first period would have four elements. The second period corresponds to n = 2, which has 2s and 2p orbitals. These four orbitals can each hold four electrons. A total of 16 elements would be in the second period. c. 20 126. d. 28 None of the noble gases and no subatomic particles had been discovered when Mendeleev published his periodic table. Thus there was no element out of place in terms of reactivity. There was no reason to predict an entire family of elements. Mendeleev ordered his table by mass; he had no way of knowing there were gaps in atomic numbers (they hadn't been invented yet). 480 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 127. At x = 0, the value of the square of the wave function must be zero. The particle must be inside the box. For ψ = A cos(Lx), at x = 0, cos(0) = 1 and ψ2 = A2. This violates the boundary condition. 128. Because two electrons can occupy each energy level, the n = 1 and n = 2 energy levels are filled and the first excited state is n = 3. En = ΔE = λ = 129. n2h 2 9h 2 4h 2 5h 2 ; ΔE = E − E = − = 3 2 8mL2 8mL2 8mL2 8mL2 5(6.626 × 10 −34 J s) 2 = 9.47 × 10−19 J −31 −10 2 8(9.109 × 10 kg) (5.64 × 10 m) hc 6.626 × 10 −34 (2.998 × 108 m/s) = = 2.10 × 10−7 m = 210. nm ΔE 9.47 × 10 −19 J a. Because wavelength is inversely proportional to energy, the spectral line to the right of B (at a larger wavelength) represents the lowest possible energy transition; this is n = 4 to n = 3. The B line represents the next lowest energy transition, which is n = 5 to n = 3, and the A line corresponds to the n = 6 to n = 3 electronic transition. b. Because this spectrum is for a one-electron ion, En = −2.178 × 10−18 J (Z2/n2). To determine ΔE and, in turn, the wavelength of spectral line A, we must determine Z, the atomic number of the one electron species. Use spectral line B data to determine Z. ⎛ Z2 Z2 ⎞ ΔE5 → 3 = −2.178 × 10−18 J ⎜⎜ 2 − 2 ⎟⎟ = −2.178 × 10−18 5 ⎠ ⎝3 E = ⎛ 16Z 2 ⎞ ⎜ ⎟ ⎜ 9 × 25 ⎟ ⎝ ⎠ hc 6.6261 × 10 −34 J s(2.9979 × 108 m / s) = = 1.394 × 10−18 J −9 λ 142.5 × 10 m Because an emission occurs, ΔE5 → 3 = −1.394 × 10−18 J. ⎛ 16 Z 2 ⎞ ⎟ , Z2 = 9.001, Z = 3; the ion is Li2+. ΔE = −1.394 × 10−18 J = −2.178 × 10−18 J ⎜⎜ ⎟ 9 × 25 ⎝ ⎠ Solving for the wavelength of line A: 1 ⎞ ⎛1 ΔE6 → 3 = −2.178 × 10−18(3)2 ⎜ 2 − 2 ⎟ = −1.634 × 10−18 J 6 ⎠ ⎝3 λ = 6.6261 × 10 −34 J s(2.9979 × 108 m / s) hc = = 1.216 × 10−7 m = 121.6 nm −18 ΔE 1.634 × 10 J CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 481 Challenge Problems 130. λ = h ; v rms = mv For one atom, R = 2.31 × 10−11 m = Molar mass = 3RT ; λ = m h m 3RT m = h 3RTm 8.3145 J 1 mol = 1.381 × 10−23 J K−1 atom−1 × K mol 6.022 × 10 23 atoms 6.626 × 10 −34 J s m 3(1.381 × 10 − 23 )(373 K ) , m = 5.32 × 10−26 kg = 5.32 × 10−23 g 5.32 × 10 −23 g 6.022 × 10 23 atoms = 32.0 g/mol × atom mol The atom is sulfur (S). 131. For one-electron species, En = − R H Z 2 /n 2 . IE is for the n = 1 → n = ∞ transition. So: IE = E∞ − E1 = −E1 = R H Z 2 /n 2 = RHZ2 4.72 × 10 4 kJ 1 mol 1000 J × × = 2.178 × 10−18 J (Z2); solving: Z = 6 23 mol kJ 6.022 × 10 Element 6 is carbon (X = carbon), and the charge for a one-electron carbon ion is 5+ (m = 5). The one-electron ion is C5+. 132. 1 ⎞ ⎛ 1 For hydrogen: ΔE = !2.178 × 10−18 J ⎜ 2 − 2 ⎟ = !4.574 × 10 −19 J 5 ⎠ ⎝2 For a similar blue light emission, He+ will need about the same ΔE value. For He+: En = !2.178 × 10 −18 J (Z2/n2), where Z = 2: ⎛ 22 22 ⎞ ΔE = !4.574 × 10 −19 J = !2.178 × 10−18 J ⎜⎜ 2 − 2 ⎟⎟ 4 ⎠ ⎝ nf 0.2100 = 4 4 4 − , 0.4600 = 2 , 2 16 nf nf nf = 2.949 The transition from n = 4 to n = 3 for He+ should emit similar-colored blue light as the n = 5 to n = 2 hydrogen transition; both these transitions correspond to very nearly the same energy change. 482 133. CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY a. Because the energy levels Exy are inversely proportional to L2, the nx = 2, ny = 1 energy level will be lower in energy than the nx = 1, ny = 2 energy level since Lx > Ly. The first three energy levels Exy in order of increasing energy are: E11 < E21 < E12 The quantum numbers are: Ground state (E11) First excited state (E21) Second excited state (E12) → nx = 1, ny = 1 → nx = 2, ny = 1 → nx = 1, ny = 2 b. E21 → E12 is the transition. Exy = E12 = ⎤ 1.76 × 1017 h 2 h2 ⎡ 12 22 + ⎢ ⎥ = 8m ⎣ (8.00 × 10 −9 m) 2 (5.00 × 10 −9 m) 2 ⎦ 8m E21 = ⎤ 1.03 × 1017 h 2 h2 ⎡ 22 12 + ⎢ ⎥ = 8m ⎣ (8.00 × 10 −9 m) 2 (5.00 × 10 −9 m) 2 ⎦ 8m ΔE = E12 − E21 = ΔE = λ = 134. 135. ny2 ⎞ h 2 ⎛⎜ nx2 ⎟ + 8m ⎜⎝ L2x L2y ⎟⎠ 1.76 × 1017 h 2 1.03 × 1017 h 2 7.3 × 1016 h 2 = − 8m 8m 8m (7.3 × 1016 m −2 ) (6.626 × 10 −34 J s) 2 = 4.4 × 10−21 J 8(9.11 × 10 −31 kg ) hc 6.626 × 10 −34 J s(2.998 × 108 m/s) = = 4.5 × 10−5 m − 21 ΔE 4.4 × 10 J The ratios for Mg, Si, P, Cl, and Ar are about the same. However, the ratios for Na, Al, and S are higher. For Na, the second IE is extremely high since the electron is taken from n = 2 (the first electron is taken from n = 3). For Al, the first electron requires a bit less energy than expected by the trend due to the fact it is a 3p electron. For S, the first electron requires a bit less energy than expected by the trend due to electrons being paired in one of the p orbitals. 1 ⎛Z⎞ ⎜ ⎟ ψ1s = π ⎜⎝ a 0 ⎟⎠ 3/ 2 1 ⎛ 1 ⎞ ⎜ ⎟ ψ1s = π ⎜⎝ a 0 ⎟⎠ 3/ 2 e−σ; Z = 1 for H, σ = ⎛−r⎞ exp⎜⎜ ⎟⎟ ⎝ a0 ⎠ Zr −r = , a0 = 5.29 × 10−11 m a0 a0 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 483 3 2 Probability is proportional to ψ : ψ 1 2s ⎛ − 2r ⎞ 1⎛ 1 ⎞ ⎟⎟ = ⎜⎜ ⎟⎟ exp⎜⎜ π ⎝ a0 ⎠ ⎝ a0 ⎠ (units of ψ2 = m−3) 3 a. ψ 1 2s ⎡ − 2 ( 0) ⎤ 1⎛ 1 ⎞ 30 −3 (at nucleus) = ⎜⎜ ⎟⎟ exp ⎢ ⎥ = 2.15 × 10 m π ⎝ a0 ⎠ a ⎣ 0 ⎦ If we assume this probability is constant throughout the 1 × 10−3 pm3 volume, then the total probability p is ψ12s × V. 1.0 × 10−3 pm3 = (1.0 × 10−3 pm) × (1 × 10−12 m/pm)3 = 1.0 × 10−39 m3 Total probability = p = (2.15 × 1030 m−3) × (1.0 × 10−39 m3) = 2.2 × 10−9 b. For an electron that is 1.0 × 10−11 m from the nucleus: 3 ψ 12s = ⎡ − 2(1.0 × 10 −11 ) ⎤ ⎞ 1⎛ 1 ⎜ ⎟ exp = 1.5 × 1030 m−3 ⎢ −11 ⎥ π ⎜⎝ 5.29 × 10 −11 ⎟⎠ × ( 5 . 29 10 ) ⎣ ⎦ V = 1.0 × 10−39 m3; p = ψ 12s × V = 1.5 × 10−9 ⎡ − 2(53 × 10 −12 ) ⎤ c. ψ 12s = 2.15 × 1030 m−3 exp ⎢ = 2.9 × 1029; V = 1.0 × 10−39 m3 −11 ⎥ × ( 5 . 29 10 ) ⎣ ⎦ p = ψ 12s × V = 2.9 × 10−10 d. V = 4 π [(10.05 × 10−12 m)3 − (9.95 × 10−12 m)3] = 1.3 × 10−34 m3 3 We shall evaluate ψ 12s at the middle of the shell, r = 10.00 pm, and assume ψ 12s is constant from r = 9.95 to 10.05 pm. The concentric spheres are assumed centered about the nucleus. ⎡ − 2(10.0 × 10 −12 m) ⎤ 30 −3 ψ 12s = 2.15 × 1030 m−3 exp ⎢ ⎥ = 1.47 × 10 m −11 ⎣ (5.29 × 10 m) ⎦ p = (1.47 × 1030 m−3) (1.3 × 10−34 m3) = 1.9 × 10−4 e. V = 4 π [(52.95 × 10−12 m)3 − (52.85 × 10−12 m)3] = 4 × 10−33 m3 3 Evaluate ψ 12s at r = 52.90 pm: ψ 12s = 2.15 × 1030 m−3 (e−2) = 2.91 × 1029 m−3; p = 1 × 10−3 136. a. Lx = Ly = Lz; Exyz = h 2 ( nx2 + ny2 + nz2 ) 8mL2 484 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY E111 = 3h 2 h2 6h 2 2 2 2 ; E (1 = + 1 + 2 ) = 112 8mL2 8mL2 8mL2 E122 = h2 9 h2 2 2 2 (1 + 2 + 2 ) = 8mL2 8mL2 b. E111: only a single state; E112: triple degenerate, either nx, ny or nz can equal 2; E122: triple degenerate, either nx, ny or nz can equal 1; E222: single state Cubic Box E222 ____ E122 These are no longer degenerate. _____ _____ E112 These are no longer degenerate. _____ _____ E111 137. Exyz = Rectangular box _____ h 2 ( nx2 + ny2 + nz2 ) 8mL2 , where L = Lx = Ly = Lz. The first four energy levels will be filled with the 8 electrons. The first four energy levels are: E111 = h 2 + 12 + 12 8mL2 E211 = E121 = E112 = = 3h 2 8mL2 6h 2 8mL2 (These three energy levels are degenerate.) The next energy levels correspond to the first excited state. The energy for these levels are: E221 = E212 = E122 = 9h 2 8mL2 (These three energy levels are degenerate.) The electronic transition in question is from one of the degenerate E211, E121, or E112 levels to one of the degenerate E221, E212, or E122 levels. CHAPTER 12 ΔE = 9h 2 6h 2 3h 2 − = 8mL2 8mL2 8mL2 ΔE = 3(6.626 × 10 −34 J s) 2 = 8.03 × 10−20 J −31 −9 2 8(9.109 × 10 kg )(1.50 × 10 m) λ = 138. QUANTUM MECHANICS AND ATOMIC THEORY hc (6.626 × 10 −34 J s)(2.998 × 108 m / s) = = 2.47 × 10−6 m = 2470 nm ΔE 8.03 × 10 − 20 J a. 1st period: p = 1, q = 1, r = 0, s = ±1/2 (2 elements) 2nd period: p = 2, q = 1, r = 0, s = ±1/2 (2 elements) 3rd period: p = 3, q = 1, r = 0, s = ±1/2 (2 elements) p = 3, q = 3, r = -2, s = ±1/2 (2 elements) p = 3, q = 3, r = 0, s = ±1/2 (2 elements) p = 3, q = 3, r = +2, s = ±1/2 (2 elements) 4th period: p = 4; q and r values are the same as with p = 3 (8 total elements) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 b. Elements 2, 4, 12, and 20 all have filled shells and will be least reactive. c. Draw similarities to the modern periodic table. XY could be X+Y−, X2+Y2− or X3+Y3−. Possible ions for each are: X+ could be elements 1, 3, 5, or 13; Y- could be 11 or 19. X2+ could be 6 or 14; Y2− could be 10 or 18. X3+ could be 7 or 15; Y3− could be 9 or 17. Note: X4+ and Y4− ions probably won’t form. 485 486 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY XY2 will be X2+(Y−)2; see preceding information for possible ions. X2Y will be (X+)2Y2−; see preceding information for possible ions. XY3 will be X3+(Y−)3; see preceding information for possible ions. X2Y3 will be (X3+)2(Y2−)3; see preceding information for possible ions. d. From a, we can see that eight electrons can have p = 3. e. p = 4, q = 3, r = 2, s = ±1/2 (2 electrons) f. p = 4, q = 3, r = −2 , s = ±1/2 (2) p = 4, q = 3, r = 0, s = ±1/2 (2) p = 4, q = 3, r = +2, s = ±1/2 (2) A total of 6 electrons can have p = 4 and q = 3. g. p = 3, q = 0, r = 0; this is not allowed; q must be odd. Zero electrons can have these quantum numbers. h. p = 5, q = 1, r = 0 p = 5, q = 3, r = −2, 0, +2 p = 5, q = 5, r = -4, -2, 0, +2, +4 i. p = 6, q = 1, r = 0, s = ±1/2 (2 electrons) p = 6, q = 3, r = −2, 0, +2; s = ±1/2 (6) p = 6, q = 5, r = −4, −2, 0, +2, +4; s = ±1/2 (10) Eighteen electrons can have p = 6. 139. a. Assuming the Bohr model applies to the 1s electron, E1s = −RHZ2/n2 = −RHZ2eff, where n = 1. IE = E∞ − E1s = 0 – E1s = RHZ2eff 2.462 × 10 6 kJ 1 mol 1000 J × × = 2.178 × 10−18 J (Zeff)2 23 mol kJ 6.0221 × 10 Solving: Zeff = 43.33 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 487 b. Silver is element 47, so Z = 47 for silver. Our calculated Zeff value is slightly less than 47. Electrons in other orbitals can penetrate the 1s orbital. Thus a 1s electron can be slightly shielded from the nucleus, giving a Zeff close to but less than Z. 140. The third IE refers to the following process: E2+(g) → E3+(g) + e− ΔH = IE3. The electron configurations for the 2+ charged ions of Na to Ar are: Na2+: Mg2+: 1s22s22p5 1s22s22p6 Al2+: Si2+: P2+: S2+: Cl2+: Ar2+: [Ne]3s1 [Ne]3s2 [Ne]3s23p1 [Ne]3s23p2 [Ne]3s23p3 [Ne]3s23p4 IE3 for sodium and magnesium should be extremely large compared with the others because n = 2 electrons are much more difficult to remove than n = 3 electrons. Between Na2+ and Mg2+, one would expect to have the same trend as seen with IE1(F) versus IE1(Ne); these neutral atoms have identical electron configurations to Na2+ and Mg2+. Therefore, the 1s22s22p5 ion (Na2+) should have a lower ionization energy than the 1s22s22p6 ion (Mg2+). The remaining 2+ ions (Al2+ to Ar2+) should follow the same trend as the neutral atoms having the same electron configurations. The general IE trend predicts an increase from [Ne]3s1 to [Ne]3s23p4. The exceptions occur between [Ne]3s2 and [Ne]3s23p1 and between [Ne]3s23p3 and [Ne]3s23p4. [Ne]3s23p1 is out of order because of the small penetrating ability of the 3p electron as compared with the 3s electrons. [Ne]3s23p4 is out of order because of the extra electron-electron repulsions present when two electrons are paired in the same orbital. Therefore, the correct ordering for Al2+ to Ar2+ should be Al2+ < P2+ < Si2+ < S2+ < Ar2+ < Cl2+, where P2+ and Ar2+ are out of line for the same reasons that Al and S are out of line in the general ionization energy trend for neutral atoms. IE Na2+ Mg2+ Al2+ Si2+ P2+ S2+ Cl2+ Ar2+ 488 CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY Note: The actual numbers in Table 12.6 support most of this plot. No IE3 is given for Na2+, so you cannot check this. The only deviation from our discussion is IE3 for Ar2+ which is greater than IE3 for Cl2+ instead of less than. Marathon Problem 141. a. Let λ = wavelength corresponding to the energy difference between the excited state, n = ?, and the ground state, n = 1. Use the information in part a to first solve for the energy difference ΔE1 → n, and then solve for the value of n. From the problem, λ = (λradio/3.00 × 107). ΔE1 → n = λradio = hc (3.00 × 10 7 ) hc hc = , λ = radio ΔE1 → n λ (λ radio / 3.00 × 10 7 ) c ν radio = c ; equating the two λradio expressions gives: 97.1 × 10 6 s −1 c hc (3.00 × 10 7 ) , ΔE1 → n = h(3.00 × 107)(97.1 × 106) = ΔE1 → n 97.1 × 10 6 ΔE1 → n = 6.626 × 10−34 J s(3.00 × 107)(97.1 × 106 s−1) = 1.93 × 10−18 J Now we can solve for the n value of the excited state. ⎛ 1 1⎞ ΔE1 → n = 1.93 × 10−18 J = −2.178 × 10−18 ⎜ 2 − 2 ⎟ ⎜n 1 ⎟⎠ ⎝ 1 − 1.93 × 10 −18 + 2.178 × 10 −18 = = 0.11, n = 3 = energy level of the excited n2 2.178 × 10 −18 state b. From de Broglie’s equation: λ = h 6.626 × 10 −34 J s = = 1.28 × 10−6 m mv 9.109 × 10 −31 kg (570. m / s) Let n = V = principal quantum number of the valence shell of element X. The electronic transition in question will be from n = V to n = 3 (as determined in part a). ΔEn →3 ⎛ 1 1 ⎞ = −2.178 × 10−18 J ⎜ 2 − 2 ⎟ ⎜3 n ⎟⎠ ⎝ |ΔEn → 3| = hc 6.626 × 10 −34 J s(2.998 × 108 m / s) = = 1.55 × 10−19 J λ 1.28 × 10 −6 m CHAPTER 12 QUANTUM MECHANICS AND ATOMIC THEORY 489 1 ⎞ ⎛1 ΔEn → 3 = −1.55 × 10−19 J = −2.178 × 10−18 J ⎜ − 2 ⎟ n ⎠ ⎝9 1 = n2 ⎛1⎞ − 1.55 × 10 −19 + 2.178 × 10 −18 ⎜ ⎟ ⎝ 9 ⎠ = 0.040, n = 5 −18 2.178 × 10 Thus V = 5 = the principal quantum number for the valence shell of element X; i.e., element X is in the fifth period (row) of the periodic table (element X = Rb − Xe). c. For n = 2, we can have 2s and 2p orbitals. None of the 2s orbitals have mℓ = −1, and only one of the 2p orbitals has mℓ = −1. In this one 2p atomic orbital, only one electron can have ms = −1/2. Thus, only one unpaired electron exists in the ground state for element X. From period 5 elements, X could be Rb, Y, Ag, In, or I because all of these elements only have one unpaired electron in the ground state. ⎛ Z2 ⎞ d. For a one-electron ion, En = −2.178 × 10−18 J ⎜ 2 ⎟ , where Z = atomic number. ⎜n ⎟ ⎝ ⎠ For He+, Z = 2. The ground state (n = 1) energy for hydrogen is −2.178 × 10−18 J. Equating the two energy level values: ⎛ 22 ⎞ −2.178 × 10−18 J ⎜⎜ 2 ⎟⎟ = −2.178 × 10−18 J, n = 2 ⎝n ⎠ Thus the azimuthal quantum number (ℓ) for the subshell of X that contains the unpaired electron is 2, which means the unpaired electron is in the d subshell. Although Y and Ag are both d-block elements, only Y has one unpaired electron in the d block. Silver is an exception to the normal filling order; Ag has the unpaired electron in the 5s orbital. The ground-state electron configurations are: Y: [Kr] 5s24d1 and Ag: [Kr] 5s14d10 Element X is yttrium (Y). CHAPTER 13 BONDING: GENERAL CONCEPTS Chemical Bonds and Electronegativity 11. Electronegativity is the ability of an atom in a molecule to attract electrons to itself. Electronegativity is a bonding term. Electron affinity is the energy change when an electron is added to a substance. Electron affinity deals with isolated atoms in the gas phase. A covalent bond is a sharing of electron pair(s) in a bond between two atoms. An ionic bond is a complete transfer of electrons from one atom to another to form ions. The electrostatic attraction of the oppositely charged ions is the ionic bond. A pure covalent bond is an equal sharing of shared electron pair(s) in a bond. A polar covalent bond is an unequal sharing. Ionic bonds form when there is a large difference in electronegativity between the two atoms bonding together. This usually occurs when a metal with a small electronegativity is bonded to a nonmetal having a large electronegativity. A pure covalent bond forms between atoms having identical or nearly identical eletronegativities. A polar covalent bond forms when there is an intermediate electronegativity difference. In general, nonmetals bond together by forming covalent bonds, either pure covalent or polar covalent. Ionic bonds form due to the strong electrostatic attraction between two oppositely charged ions. Covalent bonds form because the shared electrons in the bond are attracted to two different nuclei, unlike the isolated atoms where electrons are only attracted to one nuclei. The attraction to another nuclei overrides the added electron-electron repulsions. 12. a. There are two attractions of the form (+1)(−1) , where r = 1 × 10−10 m = 0.1 nm. r ⎡ (+1)(−1) ⎤ −18 −18 V = 2 × (2.31 × 10−19 J nm) ⎢ ⎥ = −4.62 × 10 J = −5 × 10 J ⎣ 0.1 nm ⎦ b. There are 4 attractions of +1 and −1 charges at a distance of 0.1 nm from each other. The two negative charges and the two positive charges repel each other across the diagonal of the square. This is at a distance of 2 × 0.1 nm. 490 CHAPTER 13 BONDING: GENERAL CONCEPTS ⎡ (+1)(−1) ⎤ −19 V = 4 × (2.31 × 10−19) ⎢ ⎥ + 2.31 × 10 ⎣ 0 .1 ⎦ 491 ⎡ (+1)(+1) ⎤ ⎢ ⎥ ⎢⎣ 2 (0.1) ⎥⎦ ⎡ (−1)(−1) ⎤ + 2.31 × 10−19 ⎢ ⎥ ⎢⎣ 2 (0.1) ⎥⎦ V = −9.24 × 10−18 J + 1.63 × 10−18 J + 1.63 × 10−18 J = −5.98 × 10−18 J = −6 × 10−18 J Note: There is a greater net attraction in arrangement b than in a. 13. Using the periodic table, we expect the general trend for electronegativity to be: 1. Increase as we go from left to right across a period 2. Decrease as we go down a group 14. a. C < N < O b. Se < S < Cl c. Sn < Ge < Si d. Tl < Ge < S e. Rb < K < Na f. The most polar bond will have the greatest difference in electronegativity between the two atoms. From positions in the periodic table, we would predict: a. Ge−F 15. Ga < B < O b. P−Cl c. S−F d. Ti−Cl e. Sn−H f. Tl−Br The general trends in electronegativity used in Exercises 13.13 and 13.14 are only rules of thumb. In this exercise we use experimental values of electronegativities and can begin to see several exceptions. The order of EN using Figure 13.3 is: a. C (2.6) < N (3.0) < O (3.4) same as predicted b. Se (2.6) = S (2.6) < Cl (3.2) different c. Si (1.9) < Ge (2.0) = Sn (2.0) different d. Tl (2.0) = Ge (2.0) < S (2.6) different e. Rb (0.8) = K (0.8) < Na (0.9) different f. Ga (1.8) < B (2.0) < O (3.4) same Most polar bonds using actual EN values: 16. a. Si−F (Ge−F predicted) b. P−Cl (same as predicted) c. S−F (same as predicted) d. Ti−Cl (same as predicted) e. C−H (Sn−H predicted) f. Al−Br (Tl−Br predicted) Electronegativity values increase from left to right across the periodic table. The order of electronegativities for the atoms from smallest to largest electronegativity will be H = P < C < N < O < F. The most polar bond will be F‒H since it will have the largest difference in 492 CHAPTER 13 BONDING: GENERAL CONCEPTS electronegativities, and the least polar bond will be P‒H since it will have the smallest difference in electronegativities (ΔEN = 0). The order of the bonds in decreasing polarity will be F‒H > O‒H > N‒H > C‒H > P‒H. 17. Ionic character is proportional to the difference in electronegativity values between the two elements forming the bond. Using the trend in electronegativity, the order will be: Br‒Br < N‒O < C‒F < Ca‒O < K‒F least most ionic character ionic character Note that Br‒Br, N‒O and C‒F bonds are all covalent bonds since the elements are all nonmetals. The Ca‒O and K‒F bonds are ionic, as is generally the case when a metal forms a bond with a nonmetal. (IE − EA) 18. F Cl Br I (IE − EA)/502 2006 kJ/mol 1604 1463 1302 4.0 3.2 2.9 2.6 EN (text) 2006/502 = 4.0 4.0 3.2 3.0 2.7 The values calculated from IE and EA show the same trend as (and agree fairly closely) with the values given in the text. Ionic Compounds 19. Anions are larger than the neutral atom, and cations are smaller than the neutral atom. For anions, the added electrons increase the electron-electron repulsions. To counteract this, the size of the electron cloud increases, placing the electrons further apart from one another. For cations, as electrons are removed, there are fewer electron-electron repulsions, and the electron cloud can be pulled closer to the nucleus. Isoelectronic: same number of electrons. Two variables, the number of protons and the number of electrons, determine the size of an ion. Keeping the number of electrons constant, we only have to consider the number of protons to predict trends in size. The ion with the most protons attracts the same number of electrons most strongly, resulting in a smaller size. 20. All of these ions have 18 e−; the smallest ion (Sc3+) has the most protons attracting the 18 e−, and the largest ion has the fewest protons (S2−). The order in terms of increasing size is Sc3+ < Ca2+ < K+ < Cl− < S2−. In terms of the atom size indicated in the question: K+ Ca2+ Sc3+ S2- Cl- CHAPTER 13 21. BONDING: GENERAL CONCEPTS 493 c. O2− > O− > O a. Cu > Cu+ > Cu2+ b. Pt2+ > Pd2+ > Ni2+ d. La3+ > Eu3+ > Gd3+ > Yb3+ e. Te2− > I− > Cs+ > Ba2+ > La3+ For answer a, as electrons are removed from an atom, size decreases. Answers b and d follow the radius trend. For answer c, as electrons are added to an atom, size increases. Answer e follows the trend for an isoelectronic series, i.e., the smallest ion has the most protons. 22. 23. a. Mg2+: 1s22s22p6 Sn2+: [Kr]5s24d10 K+: 1s22s22p63s23p6 Al3+: 1s22s22p6 Tl+: [Xe]6s24f145d10 As3+: [Ar]4s23d10 b. N3−, O2− and F−: 1s22s22p6 Te2-: [Kr]5s24d105p6 c. Be2+: 1s2 Rb+: [Ar]4s23d104p6 Ba2+: [Kr]5s24d105p6 Se2−: [Ar]4s23d104p6 I−: [Kr]5s24d105p6 a. Cs2S is composed of Cs+ and S2−. Cs+ has the same electron configuration as Xe, and S2− has the same configuration as Ar. b. SrF2; Sr2+ has the Kr electron configuration, and F− has the Ne configuration. c. Ca3N2; Ca2+ has the Ar electron configuration, and N3− has the Ne configuration. d. AlBr3; Al3+ has the Ne electron configuration, and Br− has the Kr configuration. 24. a. Sc3+ b. Te2− c. Ce4+ and Ti4+ d. Ba2+ All of these have the number of electrons of a noble gas. 25. Se2−, Br-, Rb+, Sr2+, Y3+, and Zr4+ are some ions that are isoelectronic with Kr (36 electrons). In terms of size, the ion with the most protons will hold the electrons tightest and will be the smallest. The size trend is: Zr4+ < Y3+ < Sr2+ < Rb+ < Br- < Se2− smallest largest 26. Lattice energy is proportional to Q1Q2/r, where Q is the charge of the ions and r is the distance between the ions. In general, charge effects on lattice energy are greater than size effects. a. LiF; Li+ is smaller than Cs+. b. NaBr; Br- is smaller than I−. c. BaO; O2− has a greater charge than Cl-. d. CaSO4; Ca2+ has a greater charge than Na+. 494 CHAPTER 13 e. K2O; O2− has a greater charge than F−. 27. f. BONDING: GENERAL CONCEPTS Li2O; The ions are smaller in Li2O. a. Al3+ and S2− are the expected ions. The formula of the compound would be Al2S3 (aluminum sulfide). b. K+ and N3−; K3N, potassium nitride c. Mg2+ and Cl−; MgCl2, magnesium chloride d. Cs+ and Br−; CsBr, cesium bromide 28. Ionic solids can be characterized as being held together by strong omnidirectional forces. i. For electrical conductivity, charged species must be free to move. In ionic solids the charged ions are held rigidly in place. Once the forces are disrupted (melting or dissolution), the ions can move about (conduct). ii. Melting and boiling disrupts the attractions of the ions for each other. If the forces are strong, it will take a lot of energy (high temperature) to accomplish this. iii. If we try to bend a piece of material, the atoms/ions must slide across each other. For an ionic solid, the following might happen: − + − + − + − + − + − + − + − + − + − + − − + − + − + − Strong repulsion Strong attraction Just as the layers begin to slide, there will be very strong repulsions causing the solid to snap across a fairly clean plane. These properties and their correlation to chemical forces will be discussed in detail in Chapter 16. K(s) → K(g) 29. + ΔH = 64 kJ (sublimation) − K(g) → K (g) + e 1/2 Cl2(g) → Cl(g) − − ΔH = 419 kJ (ionization energy) ΔH = 239/2 kJ (bond energy) Cl(g) + e → Cl (g) ΔH = !349 kJ (electron affinity) − ΔH = !690. kJ (lattice energy) K (g) + Cl (g) → KCl(s) __________________________________________________________ ΔH °f = !437 kJ/mol K(s) + 1/2 Cl2(g) → KCl(s) + CHAPTER 13 30. BONDING: GENERAL CONCEPTS Mg(s) → Mg(g) Mg(g) → Mg+(g) + e− Mg+(g) → Mg2+(g) + e− F2(g) → 2 F(g) 2 F(g) + 2 e− → 2 F−(g) Mg2+(g) + 2 F−(g) → MgF2(s) ΔH = 150. kJ ΔH = 735 kJ ΔH = 1445 kJ ΔH = 154 kJ ΔH = 2(-328) kJ ΔH = −2913 kJ 495 (sublimation) (IE1) (IE2) (BE) (EA) (LE) ______________________________________________________________________________________________ Mg(s) + F2(g) → MgF2(s) 31. ΔH of = −1085 kJ/mol Use Figure 13.11 as a template for this problem. Li(s) → Li(g) → 1/2 I2(g) → I(g) + e− → + Li (g) + I−(g) → Li(g) Li+(g) + e− I(g) I−(g) LiI(s) ΔHsub = ? ΔH = 520. kJ ΔH = 151/2 kJ ΔH = !295 kJ ΔH = !753 kJ ________________________________________________________________________ Li(s) + 1/2 I2(g) → LiI(s) ΔH = !272 kJ ΔHsub + 520. + 151/2 ! 295 ! 753 = !272, ΔHsub = 181 kJ 32. Two other factors that must be considered are the ionization energy needed to produce more positively charged ions and the electron affinity needed to produce more negatively charged ions. The favorable lattice energy more than compensates for the unfavorable ionization energy of the metal and for the unfavorable electron affinity of the nonmetal, as long as electrons are added to or removed from the valence shell. Once the valence shell is full, the ionization energy required to remove another electron is extremely unfavorable; the same is true for electron affinity when an electron is added to a higher n shell. These two quantities are so unfavorable after the valence shell is complete that they overshadow the favorable lattice energy, and the higher-charged ionic compounds do not form. 33. a. From the data given, less energy is required to produce Mg+(g) + O−(g) than to produce Mg2+(g) + O2−(g). However, the lattice energy for Mg2+O2− will be much more exothermic than for Mg+O− (due to the greater charges in Mg2+O2−). The favorable lattice energy term will dominate and Mg2+O2− forms. b. Mg+ and O− both have unpaired electrons. In Mg2+ and O2− there are no unpaired electrons. Hence Mg+O− would be paramagnetic; Mg2+O2− would be diamagnetic. Paramagnetism can be detected by measuring the mass of a sample in the presence and absence of a magnetic field. The apparent mass of a paramagnetic substance will be larger in a magnetic field because of the force between the unpaired electrons and the field. 496 34. CHAPTER 13 BONDING: GENERAL CONCEPTS Let us look at the complete cycle for Na2S. 2 Na(s) → 2 Na(g) 2 Na(g) → 2 Na+(g) + 2 e− S(s) → S(g) S(g) + e− → S−(g) S−(g) + e− → S2−(g) + 2 Na (g) + S2−(g) → Na2S(s) 2ΔHsub, Na = 2(109) kJ 2IE = 2(495) kJ ΔHsub, S = 277 kJ EA1 = !200. kJ EA2 = ? LE = !2203 kJ _______________________________________________________________________________________ ΔH °f = !365 kJ 2 Na(s) + S(s) → Na2S(s) ΔH °f = 2ΔH sub , Na + 2IE + ΔHsub, S + EA1 + EA2 + LE, !365 = !918 + EA2, EA2 = 553 kJ For each salt: ΔH °f = 2ΔHsub, M + 2IE + 277 ! 200. + LE + EA2 K2S: !381 = 2(90.) + 2(419) + 277 ! 200. ! 2052 + EA2, EA2 = 576 kJ Rb2S: !361 = 2(82) + 2(409) + 277 ! 200. ! 1949 + EA2, EA2 = 529 kJ Cs2S: !360. = 2(78) + 2(382) + 277 ! 200. ! 1850. + EA2, EA2 = 493 kJ We get values from 493 to 576 kJ. The mean value is: 540 ±50 kJ. 35. 553 + 576 + 529 + 493 = 538 kJ. We can represent the results as EA2 = 4 Ca2+ has a greater charge than Na+, and Se2− is smaller than Te2−. The effect of charge on the lattice energy is greater than the effect of size. We expect the trend from most exothermic to least exothermic to be: CaSe > CaTe > Na2Se > Na2Te (−2862) 36. (−2721) (−2130) (−2095 kJ/mol) Lattice energy is proportional to the charge of the cation times the charge of the anion, Q1Q2. Compound Q1Q2 Lattice Energy FeCl2 (+2)(−1) = −2 −2631 kJ/mol FeCl3 (+3)(−1) = −3 −5339 kJ/mol Fe2O3 (+3)(−2) = −6 −14,744 kJ/mol Bond Energies 37. This is what we observe. a. H H + Cl Cl 2H Cl CHAPTER 13 BONDING: GENERAL CONCEPTS Bonds broken: 497 Bonds formed: 1 H‒H (432 kJ/mol) 1 Cl‒Cl (239 kJ/mol) 2 H‒Cl (427 kJ/mol) ΔH = ΣDbroken ! ΣDformed, ΔH = 432 kJ + 239 kJ ! 2(427) kJ = !183 kJ b. N+ 3 H N 2H H N H H Bonds broken: Bonds formed: 6 N‒H (391 kJ/mol) 1 N ≡ N (941 kJ/mol) 3 H‒H (432 kJ/mol) ΔH = 941 kJ + 3(432) kJ ‒ 6(391) kJ = ‒109 kJ c. Sometimes some of the bonds remain the same between reactants and products. To save time, only break and form bonds that are involved in the reaction. H N+2 H C H H H H C N H H Bonds broken: Bonds formed: 1 C≡N (891 kJ/mol) 2 H−H (432 kJ/mol) 1 C−N (305 kJ/mol) 2 C−H (413 kJ/mol) 2 N−H (391 kJ/mol) ΔH = 891 kJ + 2(432 kJ) − [305 kJ + 2(413 kJ) + 2(391 kJ)] = −158 kJ H H N d. +2 F N H F 4 H F+ N N H Bonds broken: 1 N−N (160. kJ/mol) 4 N−H (391 kJ/mol) 2 F−F (154 kJ/mol) Bonds formed: 4 H−F (565 kJ/mol) 1 N≡N (941 kJ/mol) ΔH = 160. kJ + 4(391 kJ) + 2(154 kJ) − [4(565 kJ) + 941 kJ] = −1169 kJ 38. a. ΔH = 2ΔH fo, HCl = 2 mol(−92 kJ/mol) = −184 kJ (−183 kJ from bond energies) 498 CHAPTER 13 BONDING: GENERAL CONCEPTS b. ΔH = 2ΔH of , NH 3 = 2 mol(-46 kJ/mol) = −92 kJ (−109 kJ from bond energies) Comparing the values for each reaction, bond energies seem to give a reasonably good estimate for the enthalpy change of a reaction. The estimate is especially good for gas phase reactions. H H 39. H C N C H C C N H H Bonds broken: 1 C‒N (305 kJ/mol) Bonds formed: 1 C‒C (347 kJ/mol) ΔH = ΣDbroken ! ΣDformed, ΔH = 305 ! 347 = !42 kJ Note: Sometimes some of the bonds remain the same between reactants and products. To save time, only break and form bonds that are involved in the reaction. 40. H H H C C H H O H+3O Bonds broken: O 2O C O +3 H O H Bonds formed: 5 C−H (413 kJ/mol) 1 C−C (347 kJ/mol) 1 C−O (358 kJ/mol) 1 O−H (467 kJ/mol) 3 O=O (495 kJ/mol) 2 × 2 C=O (799 kJ/mol) 3 × 2 O−H (467 kJ/mol) ΔH = 5(413 kJ) + 347 kJ + 358 kJ + 467 kJ + 3(495 kJ) – [4(799 kJ) + 6(467 kJ)] = −1276 kJ 41. H−C/C−H + 5/2 O=O → 2 O=C=O + H−O−H Bonds broken: 2 C−H (413 kJ/mol) 1 C/C (839 kJ/mol) 5/2 O = O (495 kJ/mol) Bonds formed: 2 × 2 C=O (799 kJ/mol) 2 O−H (467 kJ/mol) ΔH = 2(413 kJ) + 839 kJ + 5/2 (495 kJ) – [4(799 kJ) + 2(467 kJ)] = !1228 kJ 42. Let x = bond energy for A2, and then 2x = bond energy for AB. ΔH = !285 kJ = x + 432 kJ – [2(2x)], 3x = 717, x = 239 kJ/mol = bond energy for A2 CHAPTER 13 BONDING: GENERAL CONCEPTS 499 43. H 4 N H O + 5 N C H O H N O H 12 H N O H + 9 N N + 4 O C O H Bonds broken: Bonds formed: 9 N‒N (160. kJ/mol) 4 N‒C (305 kJ/mol) 12 C‒H (413 kJ/mol) 12 N‒H (391 kJ/mol) 10 N=O (607 kJ/mol) 10 N‒O (201 kJ/mol) 24 O‒H (467 kJ/mol) 9 N≡N (941 kJ/mol) 8 C=O (799 kJ/mol) ΔH = 9(160.) + 4(305) + 12(413) + 12(391) + 10(607) + 10(201) − [24(467) + 9(941) + 8(799)] ΔH = 20,388 kJ − 26,069 kJ = −5681 kJ 44. a. I. H H * O H C H + C * H O Bonds broken (*): H * C N H C N C * C H H H Bonds formed (*): 1 C‒O (358 kJ) 1 C‒H (413 kJ) 1 O‒H (467 kJ) 1 C‒C (347 kJ) ΔHI = 358 kJ + 413 kJ − (467 kJ + 347 kJ) = −43 kJ II. OH H H * C * C * H H H C N H C H * + C C N H * O H O 500 CHAPTER 13 Bonds broken (*): BONDING: GENERAL CONCEPTS Bonds formed (*): 1 C‒O (358 kJ/mol) 1 C‒H (413 kJ/mol) 1 C‒C (347 kJ/mol) 1 H‒O (467 kJ/mol) 1 C=C (614 kJ/mol) ΔHII = 358 kJ + 413 kJ + 347 kJ − [467 kJ + 614 kJ] = +37 kJ ΔHoverall = ΔHI + ΔHII = −43 kJ + 37 kJ = −6 kJ b. H H H C 4 C C H H H 4 + 6 NO H C C + 6 H C H Bonds broken: N O H + N N H Bonds formed: 4 × 3 C‒H (413 kJ/mol) 4 C≡N (891 kJ/mol) 6 N=O (630. kJ/mol) 6 × 2 H‒O (467 kJ/mol) 1 N≡N (941 kJ/mol) ΔH = 12(413) + 6(630.) − [4(891) + 12(467) + 941] = −1373 kJ c. H H 2 C H H H C C H + 2 H N H H + 3 O2 C 2 H C + 6 C H Bonds broken: N H O H H Bonds formed: 2 × 3 C‒H (413 kJ/mol) 2 × 3 N‒H (391 kJ/mol) 3 O=O (495 kJ/mol) 2 C≡N (891 kJ/mol) 6 × 2 O‒H (467 kJ/mol) ΔH = 6(413) + 6(391) + 3(495) − [2(891) + 12(467)] = −1077 kJ 45. Because both reactions are highly exothermic, the high temperature is not needed to provide energy. It must be necessary for some other reason. The reason is to increase the speed of the reaction. This will be discussed in Chapter 15 on kinetics. 46. H H C O H + C O H H H O C C O H H CHAPTER 13 BONDING: GENERAL CONCEPTS Bonds broken: 501 Bonds formed: 1 C‒C (347 kJ/mol) 1 C=O (745 kJ/mol) 1 C‒O (358 kJ/mol) 1 C≡O (1072 kJ/mol) 1 C‒O (358 kJ/mol) ΔH = 1072 + 358 − (347 + 745 + 358) = −20. kJ CH3OH(g) + CO(g) → CH3COOH(l) ΔH° = −484 kJ − [(−201 kJ) + (−110.5 kJ)] = −173 kJ Using bond energies, ΔH = −20. kJ. For this reaction, bond energies give a much poorer estimate for ΔH as compared with gas phase reactions. The major reason for the large discrepancy is that not all species are gases in this exercise. Bond energies do not account for the energy changes that occur when liquids and solids form instead of gases. These energy changes are due to intermolecular forces and will be discussed in Chapter 16. 47. a. HF(g) → H(g) + F(g) H(g) → H+(g) + e− F(g) + e− → F−(g) ΔH = 565 kJ ΔH = 1312 kJ ΔH = −327.8 kJ ___________________________________________________________________ HF(g) → H+(g) + F−(g) b. HCl(g) → H(g) + Cl(g) H(g) → H+(g) + e− Cl(g) + e− → Cl−(g) ΔH = 1549 kJ ΔH = 427 kJ ΔH = 1312 kJ ΔH = −348.7 kJ ____________________________________________________________________ HCl(g) → H+(g) + Cl−(g) c. HI(g) → H(g) + I(g) H(g) → H+(g) + e− I(g) + e− → I−(g) ΔH = 1390. kJ ΔH = 295 kJ ΔH = 1312 kJ ΔH = −295.2 kJ ___________________________________________________________________ d. HI(g) → H+(g) + I−(g) ΔH = 1312 kJ H2O(g) → OH(g) + H(g) H(g) → H+(g) + e− OH(g) + e− → OH−(g) ΔH = 467 kJ ΔH = 1312 kJ ΔH = −180. kJ _____________________________________________________________________ H2O(g) → H+(g) + OH−(g) 48. ΔH = 1599 kJ a. Using SF4 data: SF4(g) → S(g) + 4 F(g) ΔH° = 4DSF = 278.8 kJ + 4(79.0 kJ) − (−775 kJ) = 1370. kJ DSF = 1370. kJ = 342.5 kJ/mol 4 mol SF bonds 502 CHAPTER 13 BONDING: GENERAL CONCEPTS Using SF6 data: SF6(g) → S(g) + 6 F(g) ΔH° = 6DSF = 278.8 kJ + 6(79.0 kJ) − (−1209 kJ) = 1962 kJ DSF = 1962 kJ = 327.0 kJ/mol 6 b. The S‒F bond energy in Table 13.6 is 327 kJ/mol. The value in the table was based on the S‒F bond in SF6. c. S(g) and F(g) are not the most stable form of the element at 25°C and 1 atm. The most stable forms are S8(s) and F2(g); ΔH of = 0 for these two species. 49. NH3(g) → N(g) + 3 H(g) ΔH° = 3DNH = 472.7 kJ + 3(216.0 kJ) − (−46.1 kJ) = 1166.8 kJ DNH = 1166. 8 kJ = 388.93 kJ/mol 3 mol NH bonds Dcalc = 389 kJ/mol compared with 391 kJ/mol in the table. There is good agreement. 50. ΔH of for H(g) is ΔH° for the reaction: 1/2 H2(g) → H(g); ΔH of for H(g) equals one-half the H‒H bond energy. Lewis Structures and Resonance 51. Drawing Lewis structures is mostly trial and error. However, the first two steps are always the same. These steps are (1) count the valence electrons available in the molecule/ion, and (2) attach all atoms to each other with single bonds (called the skeletal structure). Unless noted otherwise, the atom listed first is assumed to be the atom in the middle, called the central atom, and all other atoms in the formula are attached to this atom. The most notable exceptions to the rule are formulas that begin with H, e.g., H2O, H2CO, etc. Hydrogen can never be a central atom since this would require H to have more than two electrons. In these compounds, the atom listed second is assumed to be the central atom. After counting valence electrons and drawing the skeletal structure, the rest is trial and error. We place the remaining electrons around the various atoms in an attempt to satisfy the octet rule (or duet rule for H). Keep in mind that practice makes perfect. After practicing, you can (and will) become very adept at drawing Lewis structures. CHAPTER 13 BONDING: GENERAL CONCEPTS a. HCN has 1 + 4 + 5 = 10 valence electrons. H C N H C 503 b. PH3 has 5 + 3(1) = 8 valence electrons. H N P H H H Skeletal structure Skeletal structure uses 4 e−; 6 e− remain c. Cl C d. NH4+ has 5 + 4(1) ! 1 = 8 valence electrons. H H Cl Cl Cl C H Cl N Lewis structure e. H2CO has 2(1) + 4 + 6 = 12 valence electrons. + H H Cl Skeletal structure Lewis structure Skeletal structures uses 6 e−; 2 e− remain CHCl3 has 4 + 1 + 3(7) = 26 valence electrons. H H H Skeletal structure Lewis structure P Note: Subtract valence electrons for positive charged ions. Lewis structure f. SeF2 has 6 + 2(7) = 20 valence electrons. O F C H i. C O HBr has 1 + 7 = 8 valence electrons. H F H g. CO2 has 4 + 2(6) = 16 valence electrons. O Se Br h. O2 has 2(6) = 12 valence electrons. O O 504 52. CHAPTER 13 BONDING: GENERAL CONCEPTS a. POCl3 has 5 + 6 + 3(7) = 32 valence electrons. O Cl O P Cl Cl P Cl Cl Cl Skeletal structure Lewis structure Note: This structure uses all 32 e− while satisfying the octet rule for all atoms. This is a valid Lewis structure. SO42− has 6 + 4(6) + 2 = 32 valence electrons. 2- O O S Note: A negatively charged ion will have additional electrons to those that come from the valence shell of the atoms. O O XeO4, 8 + 4(6) = 32 e− PO43−, 5 + 4(6) + 3 = 32 e− 3- O O O Xe O O O P O O ClO4− has 7 + 4(6) + 1 = 32 valence electrons. - O O Cl O O Note: All these species have the same number of atoms and the same number of valence electrons. They also have the same Lewis structure. CHAPTER 13 BONDING: GENERAL CONCEPTS SO32−, 6 + 3(6) + 2 = 26 e− b. NF3 has 5 + 3(7) = 26 valence electrons. F N F F N F 505 2- F O F Skeletal structure S O O Lewis structure ClO3−, 7 + 3(6) + 1 = 26 e− PO33−, 5 + 3(6) + 3 = 26 e− 3O P O O O Cl O O Note: Species with the same number of atoms and valence electrons have similar Lewis structures. c. ClO2− has 7 + 2(6) + 1 = 20 valence electrons. O Cl O O Cl Skeletal structure - Lewis structure PCl2−, 5 + 2(7) + 1 = 20 e− SCl2, 6 + 2(7) = 20 e− Cl O S Cl Cl P Cl - Note: Species with the same number of atoms and valence electrons have similar Lewis structures. 53. Molecules/ions that have the same number of valence electrons and the same number of atoms will have similar Lewis structures. 54. a. NO2− has 5 + 2(6) + 1 = 18 valence electrons. The skeletal structure is: O ‒ N ‒ O To get an octet about the nitrogen and only use 18 e− , we must form a double bond to one of the oxygen atoms. O N O O N O 506 CHAPTER 13 BONDING: GENERAL CONCEPTS Because there is no reason to have the double bond to a particular oxygen atom, we can draw two resonance structures. Each Lewis structure uses the correct number of electrons and satisfies the octet rules, so each is a valid Lewis structure. Resonance structures occur when you have multiple bonds that can be in various positions. We say the actual structure is an average of these two resonance structures. NO3− has 5 + 3(6) + 1 = 24 valence electrons. We can draw three resonance structures for NO3−, with the double bond rotating between the three oxygen atoms. O O N O N O O O N O O O N2O4 has 2(5) + 4(6) = 34 valence electrons. We can draw four resonance structures for N2O4. O O N O N O N O O O N N O O O N O O O N O N O O b. OCN− has 6 + 4 + 5 + 1 = 16 valence electrons. We can draw three resonance structures for OCN−. O C N O C N O C N SCN− has 6 + 4 + 5 + 1 = 16 valence electrons. Three resonance structures can be drawn. S C N S C N S C N N3− has 3(5) + 1 = 16 valence electrons. As with OCN- and SCN-, three different resonance structures can be drawn. N N N N N N N N N CHAPTER 13 55. BONDING: GENERAL CONCEPTS 507 Ozone: O3 has 3(6) = 18 valence electrons. Two resonance structures can be drawn. O O O O O O Sulfur dioxide: SO2 has 6 + 2(6) = 18 valence electrons. Two resonance structures are possible. O S O O S O Sulfur trioxide: SO3 has 6 + 3(6) = 24 valence electrons. Three resonance structures are possible. O O O S S S O 56. O O O O O PAN (H3C2NO5) has 3(1) + 2(4) + 5 + 5(6) = 46 valence electrons. H H O C C This is the skeletal structure with complete octets about oxygen atoms (46 electrons used). O O O N O H This structure has used all 46 electrons, but there are only six electrons around one of the carbon atoms and the nitrogen atom. Two unshared pairs must become shared: i.e., we form two double bonds. H H O C C O O O N O H H H O C C H 57. H H O C C O O O H N O O O O (last form not important) N O CH3NCO has 4 + 3(1) + 5 + 4 + 6 = 22 valence electrons. The order of the elements in the formula give the skeletal structure. 508 CHAPTER 13 H H BONDING: GENERAL CONCEPTS H C N C O H H C H N C O H H C N C O H 58. Resonance occurs when more than one valid Lewis structure can be drawn for a particular molecule. A common characteristic of resonance structures is a multiple bond(s) that moves from one position to another. We say the electrons in the multiple bond(s) are delocalized in the molecule. This helps us rationalize why the bonds in a molecule that exhibit resonance are all equivalent in length and strength. Any one of the resonance structures indicates different types of bonds within that molecule. This is not correct, hence none of the individual resonance structures are correct. We think of the actual structure as an average of all the resonance structures; again, this helps explain the equivalent bonds within the molecule that experiment tells us we have. 59. Benzene has 6(4) + 6(1) = 30 valence electrons. Two resonance structures can be drawn for benzene. The actual structure of benzene is an average of these two resonance structures; i.e., all carbon-carbon bonds are equivalent with a bond length and bond strength somewhere between a single and a double bond. H H H C C C H H C H C C C C H C C C H H 60. H C H H We will use a hexagon to represent the six-membered carbon ring, and we will omit the four hydrogen atoms and the three lone pairs of electrons on each chlorine. If no resonance exists, we could draw four different molecules: Cl Cl Cl Cl Cl Cl Cl Cl If the double bonds in the benzene ring exhibit resonance, then we can draw only three different dichlorobenzenes. The circle in the hexagon in the following illustrations represent the delocalization of the three double bonds in the benzene ring (see Exercise 13.59). CHAPTER 13 BONDING: GENERAL CONCEPTS Cl Cl 509 Cl Cl Cl Cl With resonance, all carbon-carbon bonds are equivalent. We can’t distinguish between a single and double bond between adjacent carbons that have a chlorine attached. That only three isomers are observed supports the concept of resonance. Borazine (B3N3H6) has 3(3) + 3(5) + 6(1) = 30 valence electrons. The possible resonance structures are similar to those of benzene in Exercise 13.59. 61. H H H H B N B N H H N N B B H H B N B N H H H H 62. CH3 H CH3 B N H N B H CH3 B N H H H3C N B H H B N H H H H N B B N H H H B N N B CH3 CH3 B N B N CH3 H B N CH3 There are four different dimethylborazines. The circles in these structures represent the ability of borazine to form resonance structures (see Exercise 13.61), and CH3 is shorthand for three hydrogen atoms singly bonded to a carbon atom. There would be five structures if there were no resonance; all the structures drawn above plus an additional one related to the first Lewis structure above (see following illustration). H 510 CHAPTER 13 BONDING: GENERAL CONCEPTS CH3 CH3 B CH3 B and N 63. N CH3 Statements a and c are true. For statement a, XeF2 has 22 valence electrons and it is impossible to satisfy the octet rule for all atoms with this number of electrons. The best Lewis structure is: F Xe F For statement c, NO+ has 10 valence electrons, whereas NO− has 12 valence electrons. The Lewis structures are: N O + N O Because a triple bond is stronger than a double bond, NO+ has a stronger bond. For statement b, SF4 has five electron pairs around the sulfur in the best Lewis structure; it is an exception to the octet rule. Because OF4 has the same number of valence electrons as SF4, OF4 would also have to be an exception to the octet rule. However, Row 2 elements such as O never have more than 8 electrons around them, so OF4 does not exist. For statement d, two resonance structures can be drawn for ozone: O O O O O O When resonance structures can be drawn, the actual bond lengths and strengths are all equal to each other. Even though each Lewis structure implies the two O−O bonds are different, this is not the case in real life. In real life, both of the O−O bonds are equivalent. When resonance structures can be drawn, you can think of the bonding as an average of all of the resonance structures. 64. a. NO2, 5 + 2(6) = 17 e− O N O N2O4, 2(5) + 4(6) = 34 e− O Plus others O N N O O Plus other resonance structures b. BH3, 3 + 3(1) = 6 e− NH3, 5 + 3(1) = 8 e− H N H B H H H H CHAPTER 13 BONDING: GENERAL CONCEPTS 511 BH3NH3, 6 + 8 = 14 e− H H H B N H H H In reaction a, NO2 has an odd number of electrons, so it is impossible to satisfy the octet rule. By dimerizing to form N2O4, the odd electron on two NO2 molecules can pair up, giving a species whose Lewis structure can satisfy the octet rule. In general, odd-electron species are very reactive. In reaction b, BH3 is electron-deficient. Boron has only six electrons around it. By forming BH3NH3, the boron atom satisfies the octet rule by accepting a lone pair of electrons from NH3 to form a fourth bond. 65. PF5, 5 +5(7) = 40 valence electrons SF4, 6 + 4(7) = 34 e− F F F F S P F F F F F ClF3, 7 + 3(7) = 28 e− Br3−, 3(7) + 1 = 22 e− F Cl F Br Br Br F Row 3 and heavier nonmetals can have more than 8 electrons around them when they have to. Row 3 and heavier elements have empty d orbitals that are close in energy to valence s and p orbitals. These empty d orbitals can accept extra electrons. For example, P in PF5 has its five valence electrons in the 3s and 3p orbitals. These s and p orbitals have room for three more electrons, and if it has to, P can use the empty 3d orbitals for any electrons above 8. 66. SF6, 6 + 6(7) = 48 e− ClF5, 7 + 5(7) = 42 e− F F F S F F F F F F Cl F F 512 CHAPTER 13 BONDING: GENERAL CONCEPTS XeF4, 8 + 4(7) = 36 e− F F Xe F 67. F CO32− has 4 + 3(6) + 2 = 24 valence electrons. 2- O C O 2- O C C O O 2- O O O O Three resonance structures can be drawn for CO32−. The actual structure for CO32− is an average of these three resonance structures. That is, the three C‒O bond lengths are all equivalent, with a length somewhere between a single and a double bond. The actual bond length of 136 pm is consistent with this resonance view of CO32−. 68. The Lewis structures for the various species are below: C CO (10 e−): CO2 (16 e−): O CO32 (24 e -): O C Triple bond between C and O. Double bond between C and O. O 2- O C O 2- O C O O 2- O C O O O Average of 1 1/3 bond between C and O H CH3OH (14 e -): H C O H Single bond between C and O H As the number of bonds increases between two atoms, bond strength increases and bond length decreases. With this in mind, then: Longest → shortest C‒O bond: CH3OH > CO32− > CO2 > CO Weakest → strongest C‒O bond: CH3OH < CO32− < CO2 < CO CHAPTER 13 69. BONDING: GENERAL CONCEPTS N2 (10 e -): N N N2F4 (38 e -) : F N N F F N2F2 (24 e -): F N 513 Triple bond between N and N. N Single bond between N and N. F Double bond between N and N. F As the number of bonds increase between two atoms, bond strength increases and bond length decreases. From the Lewis structure, the shortest to longest N-N bonds is N2 < N2F2 < N2F4. Formal Charge 70. The nitrogen-nitrogen bond length of 112 pm is between a double (120 pm) and a triple (110 pm) bond. The nitrogen-oxygen bond length of 119 pm is between a single (147 pm) and a double bond (115 pm). The third resonance structure shown below doesn’t appear to be as important as the other two since there is no evidence from bond lengths for a nitrogen-oxygen triple bond or a nitrogen-nitrogen single bond as in the third resonance form. We can adequately describe the structure of N2O using the resonance forms: N N O N N O Assigning formal charges for all three resonance forms: N N O N N O N N O -1 +1 0 0 +1 -1 -2 +1 +1 For: , FC = 5 - 4 - 1/2(4) = -1 N N , FC = 5 - 1/2(8) = +1 , Same for , FC = 5 - 6 - 1/2(2) = -2 ; N O , FC = 6 - 4 - 1/2(4) = 0 ; O , FC = 6 - 2 - 1/2(6) = +1 N O N and , FC = 5 - 2 - 1/2(6) = 0 , FC = 6 - 6 - 1/2(2) = -1 N 514 CHAPTER 13 BONDING: GENERAL CONCEPTS We should eliminate N‒N≡O since it has a formal charge of +1 on the most electronegative element (O). This is consistent with the observation that the N‒N bond is between a double and triple bond, and that the N‒O bond is between a single and double bond. 71. See Exercise 13.52a for the Lewis structures of POCl3, SO42−, ClO4− and PO43−. All of these compounds/ions have similar Lewis structures to those of SO2Cl2 and XeO4 shown below. a. POCl3: P, FC = 5 − 1/2(8) = +1 b. SO42−: S, FC = 6 − 1/2(8) = +2 c. ClO4−: Cl, FC = 7 − 1/2(8) = +3 d. PO43−: P, FC = 5 − 1/2(8) = +1 e. SO2Cl2, 6 + 2(6) + 2(7) = 32 e− f. XeO4, 8 + 4(6) = 32 e− O Cl S O Cl O O g. ClO3−, 7 + 3(6) + 1 = 26 e− Cl O O S, FC = 6 − 1/2(8) = +2 O Xe Xe, FC = 8 − 1/2(8) = +4 h. NO43−, 5 + 4(6) + 3 = 32 e− 3- O O O N O O O Cl, FC = 7 − 2 − 1/2(6) = +2 72. N, FC = 5 − 1/2(8) = +1 For SO42−, ClO4−, PO4− , and ClO3−, only one of the possible resonance structures is drawn. a. Must have five bonds to P to minimize formal charge of P. The best choice is to form a double bond to O since this will give O a formal charge of zero and single bonds to Cl for the same reason. b. Must form six bonds to S to minimize formal charge of S. O Cl P Cl Cl P, FC = 0 2- O O S O O S, FC = 0 CHAPTER 13 BONDING: GENERAL CONCEPTS c. Must form seven bonds to Cl to minimize formal charge. d. Must form five bonds to P to to minimize formal charge. - O O 515 Cl O Cl, FC = 0 O 3- O P O P, FC = 0 O O e. f. O Cl S O Cl S, FC = 0 O O Xe O Xe, FC = 0 O g. O Cl O Cl, FC = 0 O h. We can’t . The following structure has a zero formal charge for N: 3- O O N O O but N does not expand its octet. We wouldn’t expect this resonance form to exist. 73. SCl, 6 + 7 = 13; the formula could be SCl (13 valence electrons), S2Cl2 (26 valence electrons), S3Cl3 (39 valence electrons), etc. For a formal charge of zero on S, we will need each sulfur in the Lewis structure to have two bonds to it and two lone pairs [FC = 6 – 4 – 1/2(4) = 0]. Cl will need one bond and three lone pairs for a formal charge of zero [FC = 7 – 6 – 1/2(2) = 0]. Since chlorine wants only one bond to it, it will not be a central atom here. With this in mind, only S2Cl2 can have a Lewis structure with a formal charge of zero on all atoms. The structure is: Cl 74. S S Cl OCN− has 6 + 4 + 5 + 1 = 16 valence electrons. Formal charge O C N O C N O C N 0 0 -1 -1 0 0 +1 0 -2 516 CHAPTER 13 BONDING: GENERAL CONCEPTS Only the first two resonance structures should be important. The third places a positive formal charge on the most electronegative atom in the ion and a -2 formal charge on N. CNO−: Formal charge C N O C N O C N O -2 +1 0 -1 +1 -1 -3 +1 +1 All the resonance structures for fulminate (CNO−) involve greater formal charges than in cyanate (OCN−), making fulminate more reactive (less stable). Molecular Structure and Polarity 75. The first step always is to draw a valid Lewis structure when predicting molecular structure. When resonance is possible, only one of the possible resonance structures is necessary to predict the correct structure because all resonance structures give the same structure. The Lewis structures are in Exercises 13.51, 13.52 and 13.54. The structures and bond angles for each follow. 13.51 a. HCN: linear, 180° b. PH3: trigonal pyramid, <109.5° c. CHCl3: tetrahedral, 109.5° d. NH4+: tetrahedral, 109.5° e. H2CO: trigonal planar, 120° f. g. CO2: linear, 180° h and i. O2 and HBr are both linear, but there is no bond angle in either. SeF2: V-shaped or bent, <109.5° Note: PH3 and SeF2 both have lone pairs of electrons on the central atom, which result in bond angles that are something less than predicted from a tetrahedral arrangement (109.5°). However, we cannot predict the exact number. For these cases we will just insert a less than sign to indicate this phenomenon. 13.52 a. All are tetrahedral; 109.5° b. All are trigonal pyramid; <109.5° c. All are V-shaped; <109.5° 13.54 a. NO2−: V-shaped, ≈ 120°; NO3−: trigonal planar, 120° N2O4: trigonal planar, 120° about both N atoms b. OCN−, SCN−, and N3− are all linear with 180° bond angles. CHAPTER 13 76. BONDING: GENERAL CONCEPTS 517 a. SeO3, 6 + 3(6) = 24 e− O 120o Se O 120o O 120o O Se O O Se O O O SeO3 has a trigonal planar molecular structure with all bond angles equal to 120°. Note that any one of the resonance structures could be used to predict molecular structure and bond angles. b. SeO2, 6 + 2(6) = 18 e− Se Se O O o ≈ 120 O O SeO2 has a V-shaped molecular structure. We would expect the bond angle to be approximately 120° as expected for trigonal planar geometry. Note: Both SeO3 and SeO2 structures have three effective pairs of electrons about the central atom. All the structures are based on a trigonal planar geometry, but only SeO3 is described as having a trigonal planar structure. Molecular structure always describes the relative positions of the atoms. c. PCl3 has 5 + 3(7) = 26 valence electrons. Cl P Cl d. SCl2 has 6 + 2(7) = 20 valence electrons Cl Cl Trigonal pyramid; all angles are <109.5°. S Cl V-shaped; angle is <109.5°. e. SiF4 has 4 + 4(7) = 32 valence electrons. F F Si F F Tetrahedral; all angles are 109.5°. Note: In PCl3, SCl2, and SiF4, there are four pairs of electrons about the central atom in each case. All the structures are based on a tetrahedral geometry, but only SiF4 has a tetrahedral structure. We consider only the relative positions of the atoms when describing the molecular structure. 518 77. CHAPTER 13 BONDING: GENERAL CONCEPTS From the Lewis structures (see Exercise 13.65), Br3− would have a linear molecular structure, ClF3 would have a T-shaped molecular structure, and SF4 would have a see-saw molecular structure. For example, consider ClF3 (28 valence electrons): The central Cl atom is surrounded by five electron pairs, which requires a trigonal bipyramid geometry. Since there are three bonded atoms and two lone pairs of electrons about Cl, we describe the molecular structure of ClF3 as T-shaped with predicted bond angles of about 90°. The actual bond angles will be slightly less than 90° due to the stronger repulsive effect of the lone pair electrons as compared to the bonding electrons. F Cl F F 78. From the Lewis structures (see Exercise 13.66), XeF4 would have a square planar molecular structure and ClF5 would have a square pyramid molecular structure. 79. a. XeCl2 has 8 + 2(7) = 22 valence electrons. Cl Xe Cl 180o There are five pairs of electrons about the central Xe atom. The structure will be based on a trigonal bipyramid geometry. The most stable arrangement of the atoms in XeCl2 is a linear molecular structure with a 180° bond angle. b. ICl3 has 7 + 3(7) = 28 valence electrons. Cl ≈ 90 o I Cl Cl T-shaped; The ClICl angles are ≈ 90°. Since the lone pairs will take up more space, the ClICl bond angles will probably be slightly less than 90°. ≈ 90 o c. TeF4 has 6 + 4(7) = 34 valence electrons. F ≈ 120 o F d. PCl5 has 5 + 5(7) = 40 valence electrons. 90 o F Te F 120 o Cl Cl Cl P Cl Cl ≈ 90 o See-saw or teeter-totter or distorted tetrahedron Trigonal bipyramid CHAPTER 13 BONDING: GENERAL CONCEPTS 519 All the species in this exercise have five pairs of electrons around the central atom. All the structures are based on a trigonal bipyramid geometry, but only in PCl5 are all the pairs bonding pairs. Thus PCl5 is the only one we describe the molecular structure as trigonal bipyramid. Still, we had to begin with the trigonal bipyramid geometry to get to the structures (and bond angles) of the others. 80. a. ICl5 , 7 + 5(7) = 42 e- b. XeCl4 , 8 + 4(7) = 36 e- ≈ 90o 90o Cl Cl Cl I 90o Cl Cl 90o Cl Cl Square pyramid, ≈ 90° bond angles Xe Cl 90o Cl Square planar, 90° bond angles c. SeCl6 has 6 + 6(7) = 48 valence electrons. Cl Cl Cl Se Cl Cl Octahedral, 90° bond angles Cl Note: All these species have six pairs of electrons around the central atom. All three structures are based on the octahedron, but only SeCl6 has an octahedral molecular structure. 81. Let us consider the molecules with three pairs of electrons around the central atom first; these molecules are SeO3 and SeO2, and both have a trigonal planar arrangement of electron pairs. Both these molecules have polar bonds, but only SeO2 has an overall net dipole moment. The net effect of the three bond dipoles from the three polar Se‒O bonds in SeO3 will be to cancel each other out when summed together. Hence SeO3 is nonpolar since the overall molecule has no resulting dipole moment. In SeO2, the two Se‒O bond dipoles do not cancel when summed together; hence SeO2 has a net dipole moment (is polar). Since O is more electronegative than Se, the negative end of the dipole moment is between the two O atoms, and the positive end is around the Se atom. The arrow in the following illustration represents the overall dipole moment in SeO2. Note that to predict polarity for SeO2, either of the two resonance structures can be used. Se O O 520 CHAPTER 13 BONDING: GENERAL CONCEPTS The other molecules in Exercise 13.76 (PCl3, SCl2, and SiF4) have a tetrahedral arrangement of electron pairs. All have polar bonds; in SiF4 the individual bond dipoles cancel when summed together, and in PCl3 and SCl2 the individual bond dipoles do not cancel. Therefore, SiF4 has no net dipole moment (is nonpolar), and PCl3 and SCl2 have net dipole moments (are polar). For PCl3, the negative end of the dipole moment is between the more electronegative chlorine atoms, and the positive end is around P. For SCl2, the negative end is between the more electronegative Cl atoms, and the positive end of the dipole moment is around S. 82. The molecules in Exercise 13.79 (XeCl2, ICl3, TeF4, and PCl5) all have a trigonal bipyramid arrangement of electron pairs. All of these molecules have polar bonds, but only TeF4 and ICl3 have dipole moments. The bond dipoles from the five P‒Cl bonds in PCl5 cancel each other when summed together, so PCl5 has no dipole moment. The bond dipoles in XeCl2 also cancel: Cl Xe Cl Since the bond dipoles from the two Xe‒Cl bonds are equal in magnitude but point in opposite directions, they cancel each other, and XeCl2 has no dipole moment (is nonpolar). For TeF4 and ICl3, the arrangement of these molecules is such that the individual bond dipoles do not all cancel, so each has an overall net dipole moment. The molecules in Exercise 13.80 (ICl5, XeCl4, and SeCl6) all have an octahedral arrangement of electron pairs. All of these molecules have polar bonds, but only ICl5 has an overall dipole moment. The six bond dipoles in SeCl6 all cancel each other, so SeCl6 has no dipole moment. The same is true for XeCl4: Cl Cl Xe Cl Cl When the four bond dipoles are added together, they all cancel each other and XeCl4 has no overall dipole moment. ICl5 has a structure in which the individual bond dipoles do not all cancel, hence ICl5 has a dipole moment. 83. The two general requirements for a polar molecule are: 1. Polar bonds 2. A structure such that the bond dipoles of the polar bonds do not cancel CHAPTER 13 BONDING: GENERAL CONCEPTS CF4, 4 + 4(7) = 32 valence electrons F 521 XeF4, 8 + 4(7) = 36 e− F F Xe C F F F F F Tetrahedral, 109.5E Square planar, 90E SF4, 6 + 4(7) = 34 e− ≈90E F F ≈120E S See-saw, ≈90E, ≈120E F ≈90E F The arrows indicate the individual bond dipoles in the three molecules (the arrows point to the more electronegative atom in the bond, which will be the partial negative end of the bond dipole). All three of these molecules have polar bonds. To determine the polarity of the overall molecule, we sum the effect of all of the individual bond dipoles. In CF4, the fluorines are symmetrically arranged about the central carbon atom. The net result is for all the individual C−F bond dipoles to cancel each other out, giving a nonpolar molecule. In XeF4, the 4 Xe−F bond dipoles are also symmetrically arranged, and XeF4 is also nonpolar. The individual bond dipoles cancel out when summed together. In SF4, we also have four polar bonds. But in SF4 the bond dipoles are not symmetrically arranged, and they do not cancel each other out. SF4 is polar. It is the positioning of the lone pair that disrupts the symmetry in SF4. CO2, 4 + 2(6) = 16 e− O C O COS, 4 + 6 + 6 = 16 e− S C O CO2 and COS both have linear molecular structures with a 180° bond angle. CO2 is nonpolar because the individual bond dipoles cancel each other out, but COS is polar. By replacing an O with a less electronegative S atom, the molecule is not symmetric any more. The individual bond dipoles do not cancel because the C−S bond dipole is smaller than the C−O bond dipole resulting in a polar molecule. 522 84. CHAPTER 13 a. XeCl4, 8 + 4(7) = 36 e− Cl XeCl2, 8 + 2(7) = 22 e− Cl Xe Cl BONDING: GENERAL CONCEPTS Cl Square planar, 90E, nonpolar Cl Xe Cl Linear, 180E, nonpolar Both compounds have a central Xe atom and terminal Cl atoms, and both compounds do not satisfy the octet rule. In addition, both are nonpolar because the Xe−Cl bond dipoles and lone pairs around Xe are arranged in such a manner that they cancel each other out. The last item in common is that both have 180E bond angles. Although we haven’t emphasized this, the bond angle between the Cl atoms on the diagonal in XeCl4 are 180E apart from each other. b. We didn’t draw the Lewis structures, but all are polar covalent compounds. The bond dipoles do not cancel out each other when summed together. The reason the bond dipoles are not symmetrically arranged in these compounds is that they all have at least one lone pair of electrons on the central atom, which disrupts the symmetry. Note that there are molecules that have lone pairs and are nonpolar, e.g., XeCl4 and XeCl2 in the preceding problem. A lone pair on a central atom does not guarantee a polar molecule. 85. Only statement c is true. The bond dipoles in CF4 and KrF4 are arranged in a manner that they all cancel each other out, making them nonpolar molecules (CF4 has a tetrahedral molecular structure, whereas KrF4 has a square planar molecular structure). In SeF4 the bond dipoles in this see-saw molecule do not cancel each other out, so SeF4 is polar. For statement a, all the molecules have either a trigonal planar geometry or a trigonal bipyramid geometry, both of which have 120E bond angles. However, XeCl2 has three lone pairs and two bonded chlorine atoms around it. XeCl2 has a linear molecular structure with a 180E bond angle. With three lone pairs, we no longer have a 120E bond angle in XeCl2. For statement b, SO2 has a Vshaped molecular structure with a bond angle of about 120E. CS2 is linear with a 180E bond angle and SCl2 is V-shaped but with an approximate 109.5E bond angle. The three compounds do not have the same bond angle. For statement d, central atoms adopt a geometry to minimize electron repulsions, not maximize them. 86. EO3− is the formula of the ion. The Lewis structure has 26 valence electrons. Let x = number of valence electrons of element E. 26 = x + 3(6) + 1, x = 7 valence electrons Element E is a halogen because halogens have seven valence electrons. Some possible identities are F, Cl, Br, and I. The EO3− ion has a trigonal pyramid molecular structure with bond angles of less than 109.5°. 87. The formula is EF2O2−, and the Lewis structure has 28 valence electrons. 28 = x + 2(7) + 6 + 2, x = 6 valence electrons for element E CHAPTER 13 BONDING: GENERAL CONCEPTS 523 Element E must belong to the Group 6A elements since E has six valence electrons. E must also be a Row 3 or heavier element since this ion has more than eight electrons around the central E atom (Row 2 elements never have more than eight electrons around them). Some possible identities for E are S, Se and Te. The ion has a T-shaped molecular structure (see Exercise 13.77) with bond angles of ≈90°. 88. H2O and NH3 have lone pair electrons on the central atoms. These lone pair electrons require more room than the bonding electrons, which tends to compress the angles between the bonding pairs. The bond angle for H2O is the smallest because oxygen has two lone pairs on the central atom, and the bond angle is compressed more than in NH3 where N has only one lone pair. 89. Molecules that have an overall dipole moment are called polar molecules, and molecules that do not have an overall dipole moment are called nonpolar molecules. a. OCl2, 6 + 2(7) = 20 e− KrF2, 8 + 2(7) = 22 e− O Cl O Cl Cl F Kr F Cl V-shaped, polar; OCl2 is polar because the two O−Cl bond dipoles don’t cancel each other. The resulting dipole moment is shown in the drawing. Linear, nonpolar; The molecule is nonpolar because the two Kr‒F bond dipoles cancel each other. BeH2, 2 + 2(1) = 4 e− SO2, 6 + 2(6) = 18 e− S H Be H Linear, nonpolar; Be‒H bond dipoles are equal and point in opposite directions. They cancel each other. BeH2 is nonpolar. O O V-shaped, polar; The S‒O bond dipoles do not cancel, so SO2 is polar (has a net dipole moment). Only one resonance structure is shown. Note: All four species contain three atoms. They have different structures because the number of lone pairs of electrons around the central atom are different in each case. 524 CHAPTER 13 b. SO3, 6 + 3(6) = 24 e− BONDING: GENERAL CONCEPTS NF3, 5 + 3(7) = 26 e- O S O N F F F O Trigonal planar, nonpolar; bond dipoles cancel. Only one resonance structure is shown. Trigonal pyramid, polar; bond dipoles do not cancel. IF3 has 7 + 3(7) = 28 valence electrons. I F F T-shaped, polar; bond dipoles do not cancel. F Note: Each molecule has the same number of atoms but different structures because of differing numbers of lone pairs around each central atom. c. CF4, 4 + 4(7) = 32 e− F F C F SeF4, 6 + 4(7) = 34 e− F F Tetrahedral, nonpolar; bond dipoles cancel. F F Se F See-saw, polar; bond dipoles do not cancel. KrF4, 8 + 4(7) = 36 valence electrons F F Kr F F Square planar, nonpolar; bond dipoles cancel. Note: Again, each molecule has the same number of atoms but different structures because of differing numbers of lone pairs around the central atom. CHAPTER 13 BONDING: GENERAL CONCEPTS d. IF5, 7 + 5(7) = 42 e− F F I F 525 AsF5, 5 + 5(7) = 40 e− F F F F F As Square pyramid, polar; bond dipoles do not cancel. F F Trigonal bipyramid, nonpolar; bond dipoles cancel. Note: Yet again, the molecules have the same number of atoms but different structures because of the presence of differing numbers of lone pairs. 90. a. The C‒H bonds are assumed nonpolar since the electronegativities of C and H are about equal. H C H Cl Cl δ+ δ! C‒Cl is the charge distribution for each C‒Cl bond. In CH2Cl2, the two individual C‒Cl bond dipoles add together to give an overall dipole moment for the molecule. The overall dipole will point from C (positive end) to the midpoint of the two Cl atoms (negative end). In CHCl3 the C‒H bond is essentially nonpolar. The three C‒Cl bond dipoles in CHCl3 add together to give an overall dipole moment for the molecule. The overall dipole will have the negative end at the midpoint of the three chlorines and the positive end around the carbon. H C Cl Cl Cl CCl4 is nonpolar. CCl4 is a tetrahedral molecule where all four C‒Cl bond dipoles cancel when added together. Let’s consider just the C and two of the Cl atoms. There will be a net dipole pointing in the direction of the middle of the two Cl atoms. C Cl Cl 526 CHAPTER 13 BONDING: GENERAL CONCEPTS There will be an equal and opposite dipole arising from the other two Cl atoms. Combining: Cl Cl C Cl Cl The two dipoles cancel, and CCl4 is nonpolar. b. CO2 is nonpolar. CO2 is a linear molecule with two equivalence bond dipoles that cancel. N2O, which is also a linear molecule, is polar because the nonequivalent bond dipoles do not cancel. N δ+ N δ− O c. NH3 is polar. The 3 N‒H bond dipoles add together to give a net dipole in the direction of the lone pair. We would predict PH3 to be nonpolar on the basis of electronegativitity, i.e., P‒H bonds are nonpolar. However, the presence of the lone pair makes the PH3 molecule slightly polar. The net dipole is in the direction of the lone pair and has a magnitude about one third that of the NH3 dipole. δ! N‒H 91. N δ+ H H P H H H H All these molecules have polar bonds that are symmetrically arranged about the central atoms. In each molecule the individual bond dipoles cancel to give no net overall dipole moment. All these molecules are nonpolar even though they all contain polar bonds. Additional Exercises 92. The general structure of the trihalide ions is: X X X Bromine and iodine are large enough and have low-energy, empty d orbitals to accommodate the expanded octet. Fluorine is small, and its valence shell contains only 2s and 2p orbitals (four orbitals) and cannot expand its octet. The lowest-energy d orbitals in F are 3d; they are too high in energy compared with 2s and 2p to be used in bonding. CHAPTER 13 93. BONDING: GENERAL CONCEPTS 527 CO32− has 4 + 3(6) + 2 = 24 valence electrons. 2- O 2- O C C O O 2- O C O O O O HCO3− has 1 + 4 + 3(6) + 1 = 24 valence electrons. H - O H C C O - O O O O H2CO3 has 2(1) + 4 + 3(6) = 24 valence electrons. O C O O H H The Lewis structures for the reactants and products are: O C O H O H + O C O O H H Bonds broken: Bonds formed: 1 C=O (799 kJ/mol) 1 O‒H (467 kJ/mol) 2 C‒O (358 kJ/mol) 1 O‒H (467 kJ/mol) ΔH = 2(358) + 467 ! (799 + 467) = !83 kJ; the carbon-oxygen double bond is stronger than two carbon-oxygen single bonds; hence CO2 and H2O are more stable than H2CO3. 94. TeF5− has 6 + 5(7) + 1 = 42 valence electrons. - F F F Te F F 528 CHAPTER 13 BONDING: GENERAL CONCEPTS The lone pair of electrons around Te exerts a stronger repulsion than the bonding pairs, pushing the four square-planar F's away from the lone pair and thus reducing the bond angles between the axial F atom and the square-planar F atoms. 95. As the halogen atoms get larger, it becomes more difficult to fit three halogen atoms around the small nitrogen atom, and the NX3 molecule becomes less stable. 96. XeF2Cl2, 8 + 2(7) + 2(7) = 36 e− Cl F Cl Xe Cl F Xe F Polar F Cl Nonpolar The two possible structures for XeF2Cl2 are above. In the first structure the F atoms are 90° apart from each other, and the Cl atoms are also 90° apart. The individual bond dipoles would not cancel in this molecule, so this molecule is polar. In the second possible structure the F atoms are 180° apart, as are the Cl atoms. Here, the bond dipoles are symmetrically arranged, so they do cancel out each other, and this molecule is nonpolar. Therefore, measurement of the dipole moment would differentiate between the two compounds. These are different compounds and not resonance structures. 97. The stable species are: a. NaBr: In NaBr2, the sodium ion would have a 2+ charge, assuming that each bromine has a 1− charge. Sodium doesn’t form stable Na2+ compounds. b. ClO4−: ClO4 has 31 valence electrons, so it is impossible to satisfy the octet rule for all atoms in ClO4. The extra electron from the 1− charge in ClO4− allows for complete octets for all atoms. c. XeO4: We can’t draw a Lewis structure that obeys the octet rule for SO4 (30 electrons), unlike with XeO4 (32 electrons). d. SeF4: Both compounds require the central atom to expand its octet. O is too small and doesn’t have low-energy d orbitals to expand its octet (which is true for all Row 2 elements). 98. If we can draw resonance forms for the anion after loss of H+, we can argue that the extra stability of the anion causes the proton to be more readily lost, i.e., makes the compound a better acid. CHAPTER 13 BONDING: GENERAL CONCEPTS 529 a. - O H C - O O H C O b. O CH 3 C - O CH C CH 3 O CH 3 O C CH C CH 3 c. O O O O O In all three cases, extra resonance forms can be drawn for the anion that are not possible when the H+ is present, which leads to enhanced stability. 99. a. Radius: N+ < N < N−; IE: N− < N < N+ N+ has the fewest electrons held by the seven protons in the nucleus whereas N− has the most electrons held by the seven protons. The seven protons in the nucleus will hold the electrons most tightly in N+ and least tightly in N−. Therefore, N+ has the smallest radius with the largest ionization energy (IE), and N− is the largest species with the smallest IE. b. Radius: Cl+ < Cl < Se < Se−; IE: Se− < Se < Cl < Cl+ The general trends tell us that Cl has a smaller radius than Se and a larger IE than Se. Cl+, with fewer electron-electron repulsions than Cl, will be smaller than Cl and have a larger IE. Se−, with more electron-electron repulsions than Se, will be larger than Se and have a smaller IE. c. Radius: Sr2+ < Rb+ < Br−; IE: Br- < Rb+ < Sr2+ 530 CHAPTER 13 BONDING: GENERAL CONCEPTS These ions are isoelectronic. The species with the most protons (Sr2+) will hold the electrons most tightly and will have the smallest radius and largest IE. The ion with the fewest protons (Br−) will hold the electrons least tightly and will have the largest radius and smallest IE. 100. O O CH3CH2O * H + HO Bonds broken (*): O−H C−O * CH3CH2O CCH3 * CCH3 + H * OH Bonds formed (*) O−H C−O We make the same bonds that we have to break in order to convert reactants into products. Therefore, we would predict ΔH = 0 for this reaction using bond energies. This is probably not a great estimate for this reaction because this is not a gas-phase reaction, where bond energies work best. 101. Nonmetals, which form covalent bonds to each other, have valence electrons in the s and p orbitals. Since there are four total s and p orbitals, there is room for only eight valence electrons (the octet rule). The valence shell for hydrogen is just the 1s orbital. This orbital can hold two electrons, so hydrogen follows the duet rule. 102. This molecule has 30 valence electrons. The only C–N bond that can possibly have a double bond character is the N bound to the C with O attached. Double bonds to the other two C–N bonds would require carbon in each case to have 10 valence electrons (which carbon never does). O H C H N C O H H C H N H H C H Assuming 100.00 g of compound: 42.81 g F × H H H H 103. C C H H 1 mol F = 2.253 mol F 19.00 g F The number of moles of X in XF5 is: 2.253 mol F × 1 mol X = 0.4506 mol X 5 mol F This number of moles of X has a mass of 57.19 g (= 100.00 g – 42.81 g). The molar mass of X is: CHAPTER 13 BONDING: GENERAL CONCEPTS 531 57.19 g X = 126.9 g/mol; This is element I. 0.4506 mol X IF5, 7 + 5(7) = 42 e− F F F I F 104. The molecular structure is square pyramid. F For carbon atoms to have a formal charge of zero, each C atom must satisfy the octet rule by forming four bonds (with no lone pairs). For nitrogen atoms to have a formal charge of zero, each N atom must satisfy the octet rule by forming three bonds and have one lone pair of electrons. For oxygen atoms to have a formal charge of zero, each O atom must satisfy the octet rule by forming two bonds and have two lone pairs of electrons. With these bonding requirements in mind, then the Lewis structure of histidine, where all atoms have a formal charge of zero, is: H 2 H C N C C H N H H C H N C C H H O 1 O H We would expect 120° bond angles about the carbon atom labeled 1 and ~109.5° bond angles about the nitrogen atom labeled 2. The nitrogen bond angles should be slightly smaller than 109.5° due to the lone pair of electrons on nitrogen. 105. Yes, each structure has the same number of effective pairs around the central atom, giving the same predicted molecular structure for each compound/ion. (A multiple bond is counted as a single group of electrons.) 106. a. BrFI2, 7 + 7 + 2(7) = 28 e−; two possible structures exist; each has a T-shaped molecular structure. I F Br I I 90° bond angles between I atoms Br F I 180° bond angle between I atoms 532 CHAPTER 13 BONDING: GENERAL CONCEPTS b. XeO2F2, 8 + 2(6) + 2(7) = 34 e−; three possible structures exist; each has a see-saw molecular structure. O O F Xe F O F F F Xe Xe 90° bond angle between O atoms c. O O O 180° bond angle between O atoms F 120° bond angle between O atoms TeF2Cl3−, 6 + 2(7) + 3(7) + 1 = 42 e−; three possible structures exist; each has a square pyramid molecular structure. - F Cl Cl Cl Te Cl F F Te Cl F Cl One F is 180° from the lone pair. Cl Cl Te F Cl Both F atoms are 90° from the lone pair and 90° from each other. F Both F atoms are 90° from the lone pair and 180° from each other. Challenge Problems 107. KrF2, 8 + 2(7) = 22 e-; from the Lewis structure, we have a trigonal bipyramid arrangement of electron pairs with a linear molecular structure. F Kr F Hyperconjugation assumes that the overall bonding in KrF2 is a combination of covalent and ionic contributions (see Section 13.12 of the text for discussion of hyperconjugation). Using hyperconjugation, two resonance structures are possible that keep the linear structure. F F Kr + F - - + Kr F CHAPTER 13 108. BONDING: GENERAL CONCEPTS 533 The skeletal structure of caffeine is: H O H H H C C C N N H H C C C H C O N H N C H H For a formal charge of zero on all atoms, the bonding requirements are: a. b. c. d. Four bonds and no lone pairs for each carbon atom Three bonds and one lone pair for each nitrogen atom Two bonds and two lone pairs for each oxygen atom One bond and no lone pairs for each hydrogen atom Following these guidelines gives a Lewis structure that has a formal charge of zero for all the atoms in the molecule. The Lewis structure is: H O H H C C C N H H H N C C C O C N H C H N H H 534 109. CHAPTER 13 BONDING: GENERAL CONCEPTS 2 Li+(g) + 2 Cl−(g) → 2 LiCl(s) 2 Li(g) → 2 Li+(g) + 2 e− 2 Li(s) → 2 Li(g) 2 HCl(g) → 2 H(g) + 2 Cl(g) 2 Cl(g) + 2 e− → 2 Cl−(g) 2 H(g) → H2(g) ΔH = 2(!829 kJ) ΔH = 2(520. kJ) ΔH = 2(166 kJ) ΔH = 2(427 kJ) ΔH = 2(!349 kJ) ΔH = !(432 kJ) 2 Li(s) + 2 HCl(g) → 2 LiCl(s) + H2(g) ΔH = !562 kJ ____________________________________________________________________________ 110. See Figure 13.11 to see the data supporting MgO as an ionic compound. Note that the lattice energy is large enough to overcome all of the other processes (removing two electrons from Mg, etc.). The bond energy for O2 (247 kJ/mol) and electron affinity (737 kJ/mol) are the same when making CO. However, ionizing carbon to form a C2+ ion must be too large. See Figure 12.35 to see that the first ionization energy for carbon is about 400 kJ/mol greater than the first IE for magnesium. If all other numbers were equal, the overall energy change would be down to ~200 kJ/mol (see Figure 13.11). It is not unreasonable that the second ionization energy for carbon is more than 200 kJ/mol greater than the second ionization energy of magnesium. 111. a. N(NO2)2− contains 5 + 2(5) + 4(6) + 1 = 40 valence electrons. The most likely structures are: N O N N N O O O O N N O O N O O N N N O O O O N N O O There are other possible resonance structures, but these are most likely. b. The NNN and all ONN and ONO bond angles should be about 120°. O CHAPTER 13 BONDING: GENERAL CONCEPTS 535 c. NH4N(NO2)2 → 2 N2 + 2 H2O + O2; break and form all bonds. Bonds broken: 4 N‒H (391 kJ/mol) 1 N‒N (160. kJ/mol) 1 N=N (418 kJ/mol) 3 N‒O (201 kJ/mol) 1 N=O (607 kJ/mol) ___________________________ Bonds formed: 2 N≡N (941 kJ/mol) 4 H‒O (467 kJ/mol) 1 O=O (495 kJ/mol) ___________________________ ΣDformed = 4245 kJ ΣDbroken = 3352 kJ ΔH = ΣDbroken − ΣDformed = 3352 kJ − 4245 kJ = −893 kJ d. To estimate ΔH, we completely ignored the ionic interactions between NH4+ and N(NO2)2−. In addition, we assumed the bond energies in Table 13.6 applied to the N(NO2)- bonds in any one of the resonance structures above. This is a bad assumption since molecules that exhibit resonance generally have stronger overall bonds than predicted. All these assumptions give an estimated ΔH value which is too negative. 112. a. (1) Removing an electron from the metal: IE, positive (ΔH > 0) (2) Adding an electron to the nonmetal: EA, often negative (ΔH < 0) (3) Allowing the metal cation and nonmetal anion to come together: LE, negative (ΔH < 0) b. Often the sign of the sum of the first two processes is positive (or unfavorable). This is especially true due to the fact that we must also vaporize the metal and often break a bond on a diatomic gas. For example, the ionization energy for Na is +495 kJ/mol, and the electron affinity for F is −328 kJ/mol. Overall, the change is +167 kJ/mol (unfavorable). c. For an ionic compound to form, the sum must be negative (exothermic). d. The lattice energy must be favorable enough to overcome the endothermic process of forming the ions; i.e., the lattice energy must be a large, negative quantity. e. While Na2Cl (or NaCl2) would have a greater lattice energy than NaCl, the energy to make a Cl2− ion (or Na2+ ion) must be larger (more unfavorable) than what would be gained by the larger lattice energy. The same argument can be made for MgO compared to MgO2 or Mg2O. The energy to make the ions is too unfavorable or the lattice energy is not favorable enough, and the compounds do not form. 113. a. i. C6H6N12O12 → 6 CO + 6 N2 + 3 H2O + 3/2 O2 The NO2 groups have one N‒O single bond and one N=O double bond, and each carbon atom has one C‒H single bond. We must break and form all bonds. 536 CHAPTER 13 Bonds broken: BONDING: GENERAL CONCEPTS Bonds formed: 3 C‒C (347 kJ/mol) 6 C‒H (413 kJ/mol) 12 C‒N (305 kJ/mol) 6 N‒N (160. kJ/mol) 6 N‒O (201 kJ/mol) 6 N=O (607 kJ/mol) 6 C≡O (1072 kJ/mol) 6 N≡N (941 kJ/mol) 6 H‒O (467 kJ/mol) 3/2 O=O (495 kJ/mol) ______________________________ ΣDformed = 15,623 kJ ____________________________ ΣDbroken = 12,987 kJ ΔH = ΣDbroken − ΣDformed = 12,987 kJ − 15,623 kJ = −2636 kJ ii. C6H6N12O12 → 3 CO + 3 CO2 + 6 N2 + 3 H2O Note: The bonds broken will be the same for all three reactions. Bonds formed: 3 C≡O (1072 kJ/mol) 6 C=O (799 kJ/mol) 6 N≡N (941 kJ/mol) 6___________________________ H‒O (467 kJ/mol) ΣDformed = 16,458 kJ ΔH = 12,987 kJ − 16,458 kJ = −3471 kJ iii. C6H6N12O12 → 6 CO2 + 6 N2 + 3 H2 Bonds formed: 12 C=O (799 kJ/mol) 6 N≡N (941 kJ/mol) 3 H‒H (432 kJ/mol) _________________ ΣDformed = 16,530. kJ ΔH = 12,987 kJ − 16,530. kJ = −3543 kJ b. Reaction iii yields the most energy per mole of CL-20, so it will yield the most energy per kilogram. 1 mol 1000 g − 3543 kJ × × = −8085 kJ/kg mol 438.23 g kg 114. For acids containing the H−O−X grouping, as the electronnegativity of X increases, it becomes more effective at withdrawing electron density from the O−H bond, thereby weakening and polarizing the bond. This increases the tendency for the molecule to produce a proton, and so its acid strength increases. Consider HOBr with Ka = 2 × 10−11. When Br is replaced by the more electronegative Cl, the Ka value increases to 4 × 10−8 for HOCl. CHAPTER 13 BONDING: GENERAL CONCEPTS 537 What determines whether X−O−H type molecules are acids or bases in water lies mainly in the nature of the O−X bond. If X has a relatively high electronegativity, the O−X bond will be covalent and strong. When the compound containing the H−O−X grouping is dissolved in water, the O−X bond will remain intact. It will be the polar and relatively weak H−O bond that will tend to break, releasing a proton. On the other hand, if X has a very low electronegativity, the O−X bond will be ionic and subject to being broken in polar water. Examples of these ionic substances are the strong bases NaOH and KOH. 115. The reaction is: 1/2 I2(s) + 1/2 Cl2(g) → ICl(g) ΔH of = ? Using Hess’s law: 1/2 I2(s) → 1/2 I2(g) 1/2 I2(g) → I (g) 1/2 Cl2(g) → Cl(g) I(g) + Cl(g) → ICl(g) ΔH = 1/2 (62 kJ) ΔH = 1/2 (149 kJ) ΔH = 1/2 (239 kJ) ΔH = −208 kJ (Appendix 4) (Table 13.6) (Table 13.6) (Table 13.6) _______________________________________________________________________________________________ 1/2 I2(s) + 1/2 Cl2(g) → ICl(g) 116. ΔH = 17 kJ so ΔH of = 17 kJ/mol There are four possible ionic compounds we need to consider. They are MX, composed of either M+ and X− ions or M2+ and X2− ions, M2X composed of M+ and X2− ions; or MX2 composed of M2+ and X− ions. For each possible ionic compound, let’s calculate ΔH of , the enthalpy of formation. The compound with the most negative enthalpy of formation will be the ionic compound most likely to form. For MX composed of M2+ and X2−: M(s) → M(g) M(g) → M+(g) + e− M+(g) → M2+(g) + e− 1/2 X2(g) → X(g) X(g) + e− → X−(g) X−(g) + e− → X2−(g) M2+(g) + X2−(g) → MX(s) ΔH = 110. kJ ΔH = 480. kJ ΔH = 4750. kJ ΔH = 1/2 (250. kJ) ΔH = −175 kJ ΔH = 920. kJ ΔH = −4800. kJ M(s) + 1/2 X2(g) → MX(s) ΔH of = 1410 kJ __________________________________________________________________________ For MX composed of M+ and X−, M(s) + 1/2 X2(g) → MX(s): ΔH of = 110. + 480. + 1/2 (250.) – 175 – 1200. = −660. kJ For M2X composed of M+ and X2− , 2 M(s) + 1/2 X2(g) → M2X(s): ΔH of = 2(110.) + 2(480.) + 1/2 (250.) – 175 + 920. – 3600. = –1550. kJ 538 CHAPTER 13 BONDING: GENERAL CONCEPTS For MX2 composed of M2+ and X−, M(s) + X2(g) → MX2(s): ΔH of = 110. + 480. + 4750. + 250 + 2(−175) – 3500. = 1740. kJ Only M+X− and (M+)2X2− have exothermic enthalpies of formation, so these are both theoretically possible. Because M2X has the more negative (more favorable) ΔH of value, we would predict the M2X compound most likely to form. The charges of the ions in M2X are M+ and X2−. Marathon Problem 117. Compound A: This compound is a strong acid (part g). HNO3 is a strong acid and is available in concentrated solutions of 16 M (part c). The highest possible oxidation state of nitrogen is +5, and in HNO3, the oxidation state of nitrogen is +5 (part b). Therefore, compound A is most likely HNO3. The Lewis structures for HNO3 are: O H O N O H N O O O Compound B: This compound is basic (part g) and has one nitrogen (part b). The formal charge of zero (part b) tells us that there are three bonds to the nitrogen and the nitrogen has one lone pair. Assuming compound B is monobasic, then the data in part g tells us that the molar mass of B is 33.0 g/mol (21.98 mL of 1.000 M HCl = 0.02198 mol HCl; thus there are 0.02198 mol of B; 0.726 g/0.02198 mol = 33.0 g/mol). Because this number is rather small, it limits the possibilities. That is, there is one nitrogen, and the remainder of the atoms are O and H. Since the molar mass of B is 33.0 g/mol, then only one O oxygen atom can be present. The N and O atoms have a combined molar mass of 30.0 g/mol; the rest is made up of hydrogens (3 H atoms), giving the formula NH3O. From the list of Kb values for weak bases in Appendix 5 of the text, compound B is most likely NH2OH. The Lewis structure is: H N O H H Compound C: From parts a and f and assuming compound A is HNO3 , then compound C contains the nitrate ion, NO3-. Because part b tells us that there are two nitrogens, the other ion needs to have one N and some H’s. In addition, compound C must be a weak acid (part g), which must be due to the other ion since NO3- has no acidic properties. Also, the nitrogen atom in the other ion must have an oxidation state of -3 (part b) and a formal charge of +1. The ammonium ion fits the data. Thus compound C is most likely NH4NO3. A Lewis structure is: CHAPTER 13 BONDING: GENERAL CONCEPTS + H H N 539 - O N H O H O Note: Two more resonance structures can be drawn for NO3-. Compound D: From part f, this compound has one less oxygen atom than compound C, thus NH4NO2 is a likely formula. Data from part e confirm this. Assuming 100.0 g of compound, we have: 43.7 g N × 1 mol/14.01 g = 3.12 mol N 50.0 g O × 1 mol/16.00 g = 3.12 mol O 6.3 g H × 1 mol/1.008 g = 6.25 mol H There is a 1:1:2 mole ratio of N:O:H. The empirical formula is NOH2, which has an empirical formula mass of 32.0 g/mol. Molar mass = 2.86 g/L(0.08206 L atm K −1 mol −1 )(273 K ) dRT = = 64.1 g/mol P 1.00 atm For a correct molar mass, the molecular formula of compound D is N2O2H4 or NH4NO2. A Lewis structure is: + H N H N H O O H Note: One more resonance structure for NO2− can be drawn. Compound E: A basic solution (part g) that is commercially available at 15 M (part c) is ammonium hydroxide (NH4OH). This is also consistent with the information given in parts b and d. The Lewis structure for NH4OH is: + H H N H H O H CHAPTER 14 COVALENT BONDING: ORBITALS The Localized Electron Model and Hybrid Orbitals 9. The valence orbitals of the nonmetals are the s and p orbitals. The lobes of the p orbitals are 90E and 180E apart from each other. If the p orbitals were used to form bonds, then all bond angles shoud be 90E or 180E. This is not the case. In order to explain the observed geometry (bond angles) that molecules exhibit, we need to make up (hybridize) orbitals that point to where the bonded atoms and lone pairs are located. We know the geometry; we hybridize orbitals to explain the geometry. Sigma bonds have shared electrons in the area centered on a line joining the atoms. The orbitals that overlap to form the sigma bonds must overlap head to head or end to end. The hybrid orbitals about a central atom always are directed at the bonded atoms. Hybrid orbitals will always overlap head to head to form sigma bonds. 10. Geometry linear trigonal planar tetrahedral Hybridization Unhybridized p atomic orbitals sp sp2 sp3 2 1 0 The unhybridized p atomic orbitals are used to form π bonds. Two unhybridized p atomic orbitals each from a different atom overlap side to side, resulting in a shared electron pair occupying the space above and below the line joining the atoms (the internuclear axis). 11. We use d orbitals when we have to, i.e., we use d orbitals when the central atom on a molecule has more than eight electrons around it. The d orbitals are necessary to accommodate the electrons over eight. Row 2 elements never have more than eight electrons around them, so they never hybridize d orbitals. We rationalize this by saying there are no d orbitals close in energy to the valence 2s and 2p orbitals (2d orbitals are forbidden energy levels). However, for Row 3 and heavier elements, there are 3d, 4d, 5d, etc. orbitals that will be close in energy to the valence s and p orbitals. It is Row 3 and heavier nonmetals that hybridize d orbitals when they have to. For phosphorus, the valence electrons are in 3s and 3p orbitals. Therefore, 3d orbitals are closest in energy and are available for hybridization. Arsenic would hybridize 4d orbitals to go with the valence 4s and 4p orbitals, whereas iodine would hybridize 5d orbitals since the valence electrons are in n = 5. 540 CHAPTER 14 COVALENT BONDING: ORBITALS 541 12. Rotation occurs in a bond as long as the orbitals that go to form that bond still overlap when the atoms are rotating. Sigma bonds, with the head-to-head overlap, remain unaffected by rotating the atoms in the bonds. Atoms that are bonded together by only a sigma bond (single bond) exhibit this rotation phenomenon. The π bonds, however, cannot be rotated. The p orbitals must be parallel to each other to form the π bond. If we try to rotate the atoms in a π bond, the p orbitals would no longer have the correct alignment necessary to overlap. Because π bonds are present in double and triple bonds (a double bond is composed of 1 σ and 1 π bond, and a triple bond is always 1 σ and 2 π bonds), the atoms in a double or triple bond cannot rotate (unless the bond is broken). 13. H2O has 2(1) + 6 = 8 valence electrons. O H H H2O has a tetrahedral arrangement of the electron pairs about the O atom that requires sp3 hybridization. Two of the four sp3 hybrid orbitals are used to form bonds to the two hydrogen atoms, and the other two sp3 hybrid orbitals hold the two lone pairs on oxygen. The two O−H bonds are formed from overlap of the sp3 hybrid orbitals from oxygen with the 1s atomic orbitals from the hydrogen atoms. Each O‒H covalent bond is called a sigma (σ) bond since the shared electron pair in each bond is centered in an area on a line running between the two atoms. 14. H2CO has 2(1) + 4 + 6 = 12 valence electrons. O C H H The central carbon atom has a trigonal planar arrangement of the electron pairs that requires sp2 hybridization. The two C−H sigma bonds are formed from overlap of the sp2 hybrid orbitals from carbon with the hydrogen 1s atomic orbitals. The double bond between carbon and oxygen consists of one σ and one π bond. The oxygen atom, like the carbon atom, also has a trigonal planar arrangement of the electrons that requires sp2 hybridization. The σ bond in the double bond is formed from overlap of a carbon sp2 hybrid orbital with an oxygen sp2 hybrid orbital. The π bond in the double bond is formed from overlap of the unhybridized p atomic orbitals. Carbon and oxygen each has one unhybridized p atomic orbital that is parallel with the other. When two parallel p atomic orbitals overlap, a π bond results where the shared electron pair occupies the space above and below a line joining the atoms in the bond. C2H2 has 2(4) + 2(1) = 10 valence electrons. H C C H 542 CHAPTER 14 COVALENT BONDING: ORBITALS Each carbon atom in C2H2 is sp hybridized since each carbon atom is surrounded by two effective pairs of electrons; i.e., each carbon atom has a linear arrangement of the electrons. Since each carbon atom is sp hybridized, then each carbon atom has two unhybridized p atomic orbitals. The two C−H sigma bonds are formed from overlap of carbon sp hybrid orbitals with hydrogen 1s atomic orbitals. The triple bond is composed of one σ bond and two π bonds. The sigma bond between to the carbon atoms is formed from overlap of sp hybrid orbitals from each carbon atom. The two π bonds of the triple bond are formed from parallel overlap of the two unhybridized p atomic orbitals from each carbon. 15. Ethane, C2H6, has 2(4) + 6(1) = 14 valence electrons. The Lewis structure is: H H H C C H H H The carbon atoms are sp3 hybridized. The six C‒H sigma bonds are formed from overlap of the sp3 hybrid orbitals on C with the 1s atomic orbitals from the hydrogen atoms. The carboncarbon sigma bond is formed from overlap of an sp3 hybrid orbital on each C atom. Ethanol, C2H6O has 2(4) + 6(1) + 6 = 20 e− H H H C C O H H H The two C atoms and the O atom are sp3 hybridized. All bonds are formed from overlap with these sp3 hybrid orbitals. The C‒H and O‒H sigma bonds are formed from overlap of sp3 hybrid orbitals with hydrogen 1s atomic orbitals. The C‒C and C‒O sigma bonds are formed from overlap of the sp3 hybrid orbitals on each atom. 16. HCN, 1 + 4 + 5 = 10 valence electrons H C N Assuming N is hybridized, both C and N atoms are sp hybridized. The C‒H σ bond is formed from overlap of a carbon sp hybrid orbital with a hydrogen 1s atomic orbital. The triple bond is composed of one σ bond and two π bonds. The sigma bond is formed from head-to-head overlap of the sp hybrid orbitals from the C and N atoms. The two π bonds in the triple bond are formed from overlap of the two unhybridized p atomic orbitals on each C and N atom. COCl2, 4 + 6 + 2(7) = 24 valence electrons O C Cl Cl CHAPTER 14 COVALENT BONDING: ORBITALS 543 Assuming all atoms are hybridized, the carbon and oxygen atoms are sp2 hybridized, and the two chlorine atoms are sp3 hybridized. The two C‒Cl σ bonds are formed from overlap of sp2 hybrids from C with sp3 hybrid orbitals from Cl. The double bond between the carbon and oxygen atoms consists of one σ and one π bond. The σ bond in the double bond is formed from head-to-head overlap of an sp2 orbital from carbon with an sp2 hybrid orbital from oxygen. The π bond is formed from parallel overlap of the unhybridized p atomic orbitals on each atom of C and O. 17. See Exercises 13.51, 13.52, and 13.54 for the Lewis structures. To predict the hybridization, first determine the arrangement of electron pairs about each central atom using the VSEPR model, then utilize the information in Figure 14.24 of the text to deduce the hybridization required for that arrangement of electron pairs. 13.51 a. HCN; C is sp hybridized. b. PH3; P is sp3 hybridized. c. CHCl3; C is sp3 hybridized. d. NH4+; N is sp3 hybridized. e. H2CO; C is sp2 hybridized. f. g. CO2; C is sp hybridized. h. O2; Each O atom is sp2 hybridized. i. 13.52 SeF2; Se is sp3 hybridized. HBr; Br is sp3 hybridized. a. All the central atoms are sp3 hybridized. b. All the central atoms are sp3 hybridized. c. All the central atoms are sp3 hybridized. 13.54 a. In NO2− and NO3−, N is sp2 hybridized. In N2O4, both central N atoms are also sp2 hybridized. b. In OCN− and SCN−, the central carbon atoms in each ion are sp hybridized, and in N3−, the central N atom is also sp hybridized. 18. For the molecules in Exercise 13.79, all have central atoms with dsp3 hybridization because all are based on the trigonal bipyramid arrangement of electron pairs. See Exercise 13.79 for the Lewis structures. For the molecules in Exercise 13.80, all have central atoms with d2sp3 hybridization because all are based on the octahedral arrangement of electron pairs. See Exercise 13.80 for the Lewis structures. 544 19. CHAPTER 14 COVALENT BONDING: ORBITALS b. a. F N F C F F F F F sp3 nonpolar tetrahedral 109.5E sp3 polar trigonal pyramid < 109.5E The angles in NF3 should be slightly less than 109.5° because the lone pair requires more space than the bonding pairs. c. F d. O F B F F sp3 polar V-shaped < 109°.5 F sp2 nonpolar trigonal planar 120° f. e. H Be F H a b F Te F F linear 180° sp nonpolar g. F b F F As see-saw a) .120°, b) .90° dsp3 polar h. a F F Kr F F trigonal bipyramid a) 90E, b) 120E dsp3 nonpolar linear 180E dsp3 nonpolar CHAPTER 14 COVALENT BONDING: ORBITALS i. F F j. F F F F Se 90o Kr F F F F d2sp3 nonpolar square planar 90° d2sp3 nonpolar octahedral 90° l. k. F F F F I F I F F F d2sp3 polar square pyramid .90E 20. 545 dsp3 polar T-shaped .90E a. S O O V-shaped, sp2, 120E Only one resonance form is shown. Resonance does not change the position of the atoms. We can predict the geometry and hybridization from any one of the resonance structures. b. c. O S O 2- O O S S O O trigonal planar, 120E, sp2 (plus two other resonance structures) tetrahedral, 109.5E, sp3 546 CHAPTER 14 COVALENT BONDING: ORBITALS 2- d. O O e. 2- S O S O O S O O O O O O trigonal pyramid, < 109.5E, sp3 tetrahedral geometry about each S, 109.5E, sp3 hybrids; V-shaped arrangement about peroxide O’s, .109.5E, sp3 hybrids f. 2- O g. S S O F O F O V-shaped, < 109.5E, sp3 tetrahedral, 109.5E, sp3 i. h. ≈ 90o F F S o ≈ 120 F F S F F F j. F b S c a F S F a) .109.5E F F see-saw, .90E and .120E, dsp3 F F b) .90E c) .120E see-saw about S atom with one lone pair (dsp3); bent about S atom with two lone pairs (sp3) octahedral, 90E, d2sp3 CHAPTER 14 COVALENT BONDING: ORBITALS 547 21. H H C H C H For the p orbitals to properly line up to form the π bond, all six atoms are forced into the same plane. If the atoms are not in the same plane, then the π bond could not form since the p orbitals would no longer be parallel to each other. 22. No, the CH2 planes are mutually perpendicular to each other. The center C atom is sp hybridized and is involved in two π bonds. The p orbitals used to form each π bond must be perpendicular to each other. This forces the two CH2 planes to be perpendicular. H H C 23. C C H H To complete the Lewis structures, just add lone pairs of electrons to satisfy the octet rule for the atoms with fewer than eight electrons. Biacetyl (C4H6O2) has 4(4) + 6(1) + 2(6) = 34 valence electrons. O H H C C H sp2 All CCO angles are 120°. The six atoms are not forced to lie in the same plane because of free rotation about the carboncarbon single (sigma) bonds. There are 11 σ and 2 π bonds in biacetyl. O sp3 H C C H H Acetoin (C4H8O2) has 4(4) + 8(1) + 2(6) = 36 valence electrons. sp2 sp3 H O a H H C H C C H O b H C The carbon with the doubly bonded O is sp2 hybridized. The other 3 C atoms are sp3 hybridized. Angle a = 120° and angle b = 109.5°. There are 13 σ and 1 π bonds in acetoin. H H sp3 Note: All single bonds are σ bonds, all double bonds are one σ and one π bond, and all triple bonds are one σ and two π bonds. 548 24. CHAPTER 14 COVALENT BONDING: ORBITALS Acrylonitrile: C3H3N has 3(4) + 3(1) + 5 = 20 valence electrons. a) b) c) 120° 120° 180° b H C a H H C c 6 σ and 3 π bonds C sp2 N sp All atoms of acrylonitrile must lie in the same plane. The π bond in the double bond dictates that the C and H atoms are all in the same plane and the triple bond dictates that N is in the same plane with the other atoms. Methyl methacrylate (C5H8O2) has 5(4) + 8(1) + 2(6) = 40 valence electrons. H H H d) 120° C *H H 14 σ and 2 π bonds e) 120° * * + *d C C sp3 f) ≈109.5° + C* +O C H *H e sp2 O + f H The atoms marked with an asterisk must be coplanar with each other, as well as the atoms marked with a plus. The two planes, however, do not have to coincide with each other due to the rotations about sigma (single) bonds. 25. To complete the Lewis structure, just add lone pairs of electrons to satisfy the octet rule for the atoms that have fewer than eight electrons. H O C C H H H O C H H C C C N O C N H O H H C C C N H N N H H H CHAPTER 14 26. COVALENT BONDING: ORBITALS 549 a. 6 b. 4 c. The center N in ‒N=N=N group d. 33 σ e. 5 π f. g. ≈109.5° h. sp3 180° a. Piperine and capsaicin are molecules classified as organic compounds, i.e., compounds based on carbon. The majority of Lewis structures for organic compounds have all atoms with zero formal charge. Therefore, carbon atoms in organic compounds will usually form four bonds, nitrogen atoms will form three bonds and complete the octet with one lone pair of electrons, and oxygen atoms will form two bonds and complete the octet with two lone pairs of electrons. Using these guidelines, the Lewis structures are: H CH a CH CH b O H CH C c O H e H H piperine g H O H O H f N H H H H3C H d O C H H H H CH2 CH2 C N h i (CH2)3 j CH CH k CH3 CH CH3 H H H O l capsaicin H Note: The ring structures are all shorthand notation for rings of carbon atoms. In piperine the first ring contains six carbon atoms and the second ring contains five carbon atoms (plus nitrogen). Also notice that CH3, CH2, and CH are shorthand for a carbon atoms singly bonded to hydrogen atoms. b. piperine: 0 sp, 11 sp2, and 6 sp3 carbons; capsaicin: 0 sp, 9 sp2, and 9 sp3 carbons c. The nitrogens are sp3 hybridized in each molecule. d. a) b) c) d) 120° 120° 120° 120° e) f) g) h) ≈109.5° 109.5° 120° 109.5° i) j) k) l) 120° 109.5° 120° 109.5° 550 27. CHAPTER 14 CO, 4 + 6 = 10 e−; C O CO2, 4 + 2(6) = 16 e−; O C O COVALENT BONDING: ORBITALS C3O2, 3(4) + 2(6) = 24 e− O C C C O There is no molecular structure for the diatomic CO molecule. The carbon in CO is sp hybridized. CO2 is a linear molecule, and the central carbon atom is sp hybridized. C3O2 is a linear molecule with all the central carbon atoms exhibiting sp hybridization. The Molecular Orbital (MO) Model 28. See Figure 14.34 for the 2s σ bonding and σ antibonding molecular orbitals, and see Figure 14.36 for the 2p σ bonding, σ antibonding, π bonding, and π antibonding molecular orbitals. 29. Bonding and antibonding molecular orbitals are both solutions to the quantum mechanical treatment of the molecule. Bonding orbitals form when in-phase orbitals combine to give constructive interference. This results in enhanced electron probability located between the two nuclei. The end result is that a bonding MO is lower in energy than the atomic orbitals from which it is composed. Antibonding orbitals form when out-of-phase orbitals combine. The mismatched phases produce destructive interference leading to a node of electron probability between the two nuclei. With electron distribution pushed to the outside, the energy of an antibonding orbital is higher than the energy of the atomic orbitals from which it is composed. 30. Bond energy is directly proportional to bond order. Bond length is inversely proportional to bond order. Bond energy and bond length can be measured; bond order is calculated from the molecular orbital energy diagram (bond order is the difference between the number of bonding electrons and the number of antibonding electrons divided by two). Paramagnetic: a kind of induced magnetism, associated with unpaired electrons, that causes a substance to be attracted into an inducing magnetic field. Diamagnetic: a type of induced magnetism, associated with paired electrons, that causes a substance to be repelled from the inducing magnetic field. The key is that paramagnetic substances have unpaired electrons in the molecular orbital diagram, whereas diamagnetic substances have only paired electrons in the MO diagram. To determine the type of magnetism, measure the mass of a substance in the presence and absence of a magnetic field. A substance with unpaired electrons will be attracted by the magnetic field, giving an apparent increase in mass in the presence of the field. A greater number of unpaired electrons will give a greater attraction and a greater observed mass increase. A diamagnetic species will not be attracted by a magnetic field and will not show a mass increase (a slight mass decrease is observed for diamagnetic species). 31. a. H2 has two valence electrons to put in the MO diagram, whereas He2 has four valence electrons. H2: (σ1s)2 He2: (σ1s)2(σ1s*)2 Bond order = B.O. = (2!0)/2 = 1 B.O. = (2!2)/2 = 0 CHAPTER 14 COVALENT BONDING: ORBITALS 551 H2 has a nonzero bond order, so MO theory predicts it will exist. The H2 molecule is stable with respect to the two free H atoms. He2 has a bond order of zero, so it should not form. The He2 molecule is not more stable than the two free He atoms. b. See Figure 14.41 for the MO energy-level diagrams of B2, C2, N2, O2, and F2. B2 and O2 have unpaired electrons in their electron configuration, so they are predicted to be paramagnetic. C2, N2, and F2 have no unpaired electrons in the MO diagrams; they are all diamagnetic. c. From the MO energy diagram in Figure 14.41, N2 maximizes the number of electrons in the lower-energy bonding orbitals and has no electrons in the antibonding 2p molecular orbitals. N2 has the highest possible bond order of three, so it should be a very strong (stable) bond. d. NO+ has 5 + 6 – 1 = 10 valence electrons to place in the MO diagram, and NO− has 5 + 6 + 1 = 12 valence electrons. The MO diagram for these two ions is assumed to be the same as that used for N2. NO+: (σ2s)2(σ2s*)2(π2p)4(σ2p)2 NO−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2 B.O. = (8!2)/2 = 3 B.O. = (8!4)/2 = 2 NO+ has a larger bond order than NO− , so NO+ should be more stable than NO−. 32. The localized electron model does not deal effectively with molecules containing unpaired electrons. We can draw all of the possible structures for NO with its odd number of valence electrons but still not have a good feel for whether the bond in NO is weaker or stronger than the bond in NO−. MO theory can handle odd electron species without any modifications. From the MO electron configurations, the bond order is 2.5 for NO and 2 for NO− (see Exercise 14.31d for the electron configuration of NO−). Therefore, NO should have the stronger bond (and it does). In addition, hybrid orbital theory does not predict that NO− is paramagnetic. The MO theory correctly makes this prediction. 33. + + σ* (out-of-phase; the signs oppose each other) + σ (in-phase; the signs match up) These molecular orbitals are sigma MOs because the electron density is cylindrically symmetric about the internuclear axis. 552 34. CHAPTER 14 COVALENT BONDING: ORBITALS If we calculate a nonzero bond order for a molecule, then we predict that it can exist (is stable). a. H2+: H2: H2−: H22−: (σ1s)1 (σ1s)2 (σ1s)2(σ1s*)1 (σ1s)2(σ1s*)2 B.O. = (1−0)/2 = 1/2, stable B.O. = (2−0)/2 = 1, stable B.O. = (2−1)/2 = 1/2, stable B.O. = (2−2)/2 = 0, not stable B.O. = (8−4)/2 = 2, stable b. N22−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2 2− 2 2 2 4 4 B.O. = (8−6)/2 = 1, stable O2 : (σ2s) (σ2s*) (σ2p) (π2p) (π2p*) 2− 2 2 2 4 4 2 F2 : (σ2s) (σ2s*) (σ2p) (π2p) (π2p*) (σ2p*) B.O. = (8−8)/2 = 0, not stable B.O. = (2−2)/2 = 0, not stable c. Be2: (σ2s)2(σ2s*)2 2 2 2 B.O. = (4−2)/2 = 1, stable B2: (σ2s) (σ2s*) (π2p) 2 2 2 4 4 2 Ne2: (σ2s) (σ2s*) (σ2p) (π2p) (π2p*) (σ2p*) B.O. = (8−8)/2 = 0, not stable 35. N2: (σ2s)2(σ2s*)2(π2p)4(σ2p)2 B.O. = (8 – 2)/2 = 3 We need to decrease the bond order from 3 to 2.5. There are two ways to do this. One is to add an electron to form N2− . This added electron goes into one of the π *2 p orbitals, giving a bond order of (8 – 3)/2 = 2.5. We could also remove a bonding electron to form N2+. The bond order for N2+ is also 2.5 [= (7 – 2)/2]. 36. H2: (σ1s)2 B2: (σ2s)2(σ2s*)2(π2p)2 C22−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2 OF: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3 The bond strength will weaken if the electron removed comes from a bonding orbital. Of the molecules listed, H2, B2, and C22− would be expected to have their bond strength weaken as an electron is removed. OF has the electron removed from an antibonding orbital, so its bond strength increases. 37. CN: (σ2s)2(σ2s*)2(π2p)4(σ2p)1 NO: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)1 O22+: (σ2s)2(σ2s*)2(σ2p)2(π2p)4 N22+: (σ2s)2(σ2s*)2(π2p)4 If the added electron goes into a bonding orbital, the bond order would increase, making the species more stable and more likely to form. Between CN and NO, CN would most likely form CN− since the bond order increases (unlike NO−, where the added electron goes into an antibonding orbital). Between O22+ and N22+, N2+ would most likely form since the bond order increases (unlike O22+ going to O2+). CHAPTER 14 38. COVALENT BONDING: ORBITALS 553 The electron configurations are (assuming the same orbital order as that for N2): CO: (σ2s)2(σ2s*)2(π2p)4(σ2p)2 CO+: (σ2s)2(σ2s*)2(π2p)4(σ2p)1 CO2+: (σ2s)2(σ2s*)2(π2p)4 B.O. = (8-2)/2 = 3, 0 unpaired e− B.O. = (7-2)/2 = 2.5, 1 unpaired e− B.O. = (6-2)/2 = 2, 0 unpaired e− Because bond order is directly proportional to bond energy and, in turn, inversely proportional to bond length, the correct bond length order should be: Shortest → longest bond length: CO < CO+ < CO2+ 39. The π bonds between S atoms and between C and S atoms are not as strong. The atomic orbitals do not overlap with each other as well as the smaller atomic orbitals of C and O overlap. 40. There are 14 valence electrons in the MO electron configuration. Also, the valence shell is n = 3. Some possibilities from Row 3 having 14 valence electrons are Cl2, SCl−, S22−, and Ar22+. 41. Side-to-side overlap of these d orbitals would produce a π molecular orbital. There would be no probability of finding an electron on the axis joining the two nuclei, which is characteristic of π MOs. 42. O2: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)2; O2 should have a lower ionization energy than O. The electron removed from O2 is in a π2p* antibonding molecular orbital, which is higher in energy than the 2p atomic orbitals from which the electron in atomic oxygen is removed. Because the electron removed from O2 is higher in energy than the electron removed from O, it should be easier to remove an electron from O2 than from O. 43. a. The electron density would be closer to F on average. The F atom is more electronegative than the H atom, and the 2p orbital of F is lower in energy than the 1s orbital of H. b. The bonding MO would have more fluorine 2p character since it is closer in energy to the fluorine 2p atomic orbital. c. The antibonding MO would place more electron density closer to H and would have a greater contribution from the higher-energy hydrogen 1s atomic orbital. 44. a. See the illustrations in the solution to Exercise 14.33 for the bonding and antibonding MOs in OH. b. The antibonding MO will have more hydrogen 1s character since the hydrogen 1s atomic orbital is closer in energy to the antibonding MO. c. No, the overall overlap is zero. The px orbital does not have proper symmetry to overlap with a 1s orbital. The 2px and 2py orbitals are called nonbonding orbitals. 554 CHAPTER 14 COVALENT BONDING: ORBITALS + + - d. H HO O σ* ↑ 1s e. Bond order = order. f. 45. ↑↓ ↑ ↑ ↑↓ ↑ 2p x 2py 2p z 2p x 2p y ↑↓ 2s ↑↓ σ ↑↓ 2s 2−0 = 1. Note: The 2s, 2px, and 2py electrons have no effect on the bond 2 To form OH+, a nonbonding electron is removed from OH. Because the number of bonding electrons and antibonding electrons is unchanged, the bond order is still equal to one. Molecules that exhibit resonance have delocalized π bonding. This is a fancy way of saying that the π electrons are not permanently stationed between two specific atoms but instead can roam about over the surface of a molecule. We use the concept of delocalized π electrons to explain why molecules that exhibit resonance have equal bonds in terms of strength. Because the π electrons can roam about over the entire surface of the molecule, the π electrons are shared by all the atoms in the molecule, giving rise to equal bond strengths. The classic example of delocalized π electrons is benzene (C6H6). Figure 14.50 shows the π molecular orbital system for benzene. Each carbon in benzene is sp2 hybridized, leaving one unhybridized p atomic orbital. All six of the carbon atoms in benzene have an unhybridized p orbital pointing above and below the planar surface of the molecule. Instead of just two unhybridized p orbitals overlapping, we say all six of the unhybridized p orbitals overlap, resulting in delocalized π electrons roaming about above and below the entire surface of the benzene molecule. SO2, 6 + 2(6) = 18 e− S O S O O O CHAPTER 14 COVALENT BONDING: ORBITALS 555 In SO2 the central sulfur atom is sp2 hybridized. The unhybridized p atomic orbital on the central sulfur atom will overlap with parallel p orbitals on each adjacent O atom. All three of these p orbitals overlap together, resulting in the π electrons moving about above and below the surface of the SO2 molecule. With the delocalized π electrons, the S−O bond lengths in SO2 are equal (and not different, as each individual Lewis structure indicates). 46. O3 and NO2−are isoelectronic, so we only need consider one of them since the same bonding ideas apply to both. The Lewis structures for O3 are: O O O O O O For each of the two resonance forms, the central O atom is sp2 hybridized with one unhybridized p atomic orbital. The sp2 hybrid orbitals are used to form the two sigma bonds to the central atom. The localized electron view of the π bond utilizes unhybridized p atomic orbitals. The π bond resonates between the two positions in the Lewis structures: In the MO picture of the π bond, all three unhybridized p orbitals overlap at the same time, resulting in π electrons that are delocalized over the entire surface of the molecule. This is represented as: or 556 CHAPTER 14 COVALENT BONDING: ORBITALS The Lewis structures for CO32− are (24 e−): 47. O 2- C O 2- O C O O 2- O C O O O In the localized electron view, the central carbon atom is sp2 hybridized; the sp2 hybrid orbitals are used to form the three sigma bonds in CO32−. The central C atom also has one unhybridized p atomic orbital that overlaps with another p atomic orbital from one of the oxygen atoms to form the π bond in each resonance structure. This localized π bond moves (resonates) from one position to another. In the molecular orbital model for CO32−, all four atoms in CO32− have a p atomic orbital that is perpendicular to the plane of the ion. All four of these p orbitals overlap at the same time to form a delocalized π bonding system where the π electrons can roam above and below the entire surface of the ion. The π molecular orbital system for CO32− is analogous to that for NO3− which is shown in Figure 14.51 of the text. 48. Benzoic acid (C7H6O2) has 7(4) + 6(1) + 2(6) = 46 valence electrons. The Lewis structure for benzoic acid is: O C H H O H C C C C C C H The circle in the ring indicates the delocalized π bonding in the benzene ring. The two benzene resonance Lewis structures have three alternating double bonds in the ring (see Figure 14.48). H H The six carbons in the ring and the carbon bonded to the ring are all sp2 hybridized. The five C−H sigma bonds are formed from overlap of the sp2 hybridized carbon atoms with hydrogen 1s atomic orbitals. The seven C−C σ bonds are formed from head to head overlap of sp2 hybrid orbitals from each carbon. The C−O single bond is formed from overlap of an sp2 hybrid orbital on carbon with an sp3 hybrid orbital from oxygen. The C−O σ bond in the double bond is formed from overlap of carbon sp2 hybrid orbital with an oxygen sp2 orbital. The π bond in the C−O double bond is formed from overlap of parallel p unhybridized atomic orbitals from C and O. The delocalized π bonding system in the ring is formed from overlap of all six unhybridized p atomic orbitals from the six carbon atoms. See Figure 14.50 for delocalized π bonding system in the benzene ring. CHAPTER 14 COVALENT BONDING: ORBITALS 557 Spectroscopy 49. reduced mass = µ = νo = 1 2π 1.66054 × 10 −27 kg m1m 2 (1.0078)(78.918) = amu × amu m1 + m 2 1.0078 + 78.918 = 1.6524 × 10−27 kg c k = = 2.9979 × 1010 cm s−1 × 2650. cm−1 = 7.944 × 1013 s−1 λ μ 1 2π 7.944 × 1013 s−1 = k 1 k = μ 2 π 1.6524 × 10 − 27 kg Solving for k, the force constant: k = 411.7 kg s−2 = 411.7 N m−1 Note: one newton = 1 N = 1 kg m s−2. 50. (14.003)(15.995) 1.66054 × 10 −27 kg k amu × , µ= = 1.2398 × 10−26 kg μ 14.003 + 15.995 amu νo = 1 2π νo = 1 1550. N m −1 1 kg m s −2 × = 5.627 × 1013 s−1 2 π 1.2398 × 10 − 26 kg N λ= 2.9979 × 1010 cm s −1 c = 5.328 × 10−4 cm = ν 5.627 × 1013 s −1 Wave number = 51. 1 1 = = 1877 cm−1 λ 5.328 × 10 − 4 cm a. ΔE = 2hB(Ji + 1) = hν = hc c , = 2B(Ji + 1) λ λ 2.998 × 108 m s −1 c = 1.15 × 1011 s−1 = 2B(0 + 1) = 2B = λ 2.60 × 10 −3 m B= 1.15 × 10 −11 s −1 = 5.75 × 1010 s−1 2 I= 6.626 × 10 −34 J s h = = 1.46 × 10−46 kg m2 8π 2 (5.75 × 1010 s −1 ) 8π 2 B I = µRe2, µ = (12.000)(15.995) 1.66054 × 10 −27 kg m1m 2 = amu × 12.000 + 15.995 amu m1 + m 2 = 1.1385 × 10−26 kg 558 CHAPTER 14 COVALENT BONDING: ORBITALS 1.46 × 10 −46 kg m 2 I = = 1.28 × 10−20 m2, Re = bond length = 1.13 × 10−10 m μ 1.1385 × 10 − 26 kg = 113 pm ΔE = 2B(Ji + 1) = 2B(2 +1) = 6B b. ν = h Re2 = From part a, B = 5.75 × 1010 s−1, so: ν = 6(5.75 × 1010 s−1) = 3.45 × 1011 s−1 52. a. There are three sets of magnetically equivalent hydrogens (marked a, b, and c in the following structure). We are assuming that all the benzene ring hydrogens (marked c) are equivalent and do not exhibit spin-spin coupling. c H c H O H c CH2 b H c O C CH3 a H c Because the hydrogens marked by a and the hydrogens marked by b are separated by more than three sigma bonds, they will also not exhibit spin-spin coupling. Thus we will have three singlet peaks (following our assumptions). Predicting the relative locations of the peaks was not discussed in the text, but they should be in a 3:2:5 relative area ratio (a:b:c). A sketch of the idealized NMR spectrum is: c a b b. b CH3 O a CH3 C C a CH3 CH3 a The different groups of equivalent hydrogen atoms are labeled a and b. Since all H atoms are separated by more than three sigma bonds, we should have no spin-spin coupling. Again, you do not have the information to predict where the two peaks should be in the idealized NMR spectrum, but they should have relative areas of 9:3 (or 3:1). CHAPTER 14 COVALENT BONDING: ORBITALS 559 a b c. c CH2CH3 b a H H c CH2CH3 b a CH3CH2 a b H c The three different groups of equivalent hydrogen atoms are labeled a, b, and c. The H atoms marked c will not exhibit spin-spin coupling, but the H atoms marked a and b will. Since the H atoms marked a in the –CH3 groups neighbor –CH2 groups, a triplet line pattern will result with intensities of 1:2:1. The H atoms marked b in the –CH2 groups neighbor –CH3 groups, so a quartet peak should result with intensities of 1:3:3:1. The relative area ratios of the three different H atoms (a:b:c) should be 9:6:3 (or 3:2:1). c 53. b a a. The –CH2 group neighbors a –CH3 group, so a quartet of peaks should result (iv). b. The –CH3 H atoms are separated by more than three sigma bonds from other H atoms, so no spin-spin coupling should occur. A singlet peak should result (i). c. The –CH2 H atoms neighbor two H atoms, so a triplet peak should result (iii). d. The two H atoms in the –CH2F group neighbor one H atom, so a doublet peak should result (ii). 54. C6H12 NMR spectrum: Because only one peak is present, all the hydrogen atoms are equivalent in their environment. Knowing this, then the process is trial and error to come up with the correct structure for C6H12. Following the “organic rules” in Exercise 14.64 and knowing a double bond is present, the only possibility to explain the NMR pattern is: 560 CHAPTER 14 COVALENT BONDING: ORBITALS CH3 CH3 C C CH3 CH3 Here, all the H atoms have equivalent neighboring atoms and there would be no spin-spin coupling since all H atoms are separated by more than three sigma bonds. C4H10O spectrum: We have a quartet peak and a triplet peak in the spectrum. The quartet peak indicates hydrogen atom(s) that neighbor three H atoms; the triplet peak indicates hydrogen atom(s) that neighbor two H atoms. Again, following the “organic rules” in Exercise 14.64, the only possibility to explain this NMR pattern is: CH3CH2–O–CH2CH3 This compound has two different “types” of hydrogen atoms (–CH3 and –CH2). The –CH3 hydrogen atoms neighbor the –CH2 group, which would produce the triplet peak, and the –CH2 hydrogen atoms neighbor the –CH3 group, that would produce the quartet peak. From inspection, the relative areas of the two different patterns seems to confirm a 6:4 (or 3:2) ratio as it should be. Additional Exercises 55. a. XeO3, 8 + 3(6) = 26 e− b. XeO4, 8 + 4(6) = 32 e− Xe O O O O Xe O O O trigonal pyramid; sp3 tetrahedral; sp3 d. XeOF2, 8 + 6 + 2(7) = 28 e− c. XeOF4, 8 + 6 + 4(7) = 42 e− F F F or Xe F F F O Xe F F 2 square pyramid; d sp 3 F O O Xe O or F F T-shaped; dsp Xe F 3 CHAPTER 14 COVALENT BONDING: ORBITALS 561 e. XeO3F2 has 8 + 3(6) + 2(7) = 40 valence electrons. F F O Xe or O O Xe F or Xe F trigonal bipyramid; dsp3 O O F 56. O F O O O FClO2 + F− → F2ClO2− F3ClO + F− → F4ClO− F2ClO2−, 2(7) + 7 + 2(6) + 1 = 34 e− F4ClO−, 4(7) + 7 + 6 + 1 = 42 e− - F O F Cl O O F Cl F F dsp3 hybridization F d2sp3 hybridization Note: Similar to Exercise 14.55c, d, and e, F2ClO2- has two additional Lewis structures that are possible, and F4ClO- has one additional Lewis structure that is possible, depending on the relative placement of the O and F atoms. The predicted hybridization is unaffected. F3ClO → F− + F2ClO+ F3ClO2 → F− + F2ClO2+ F2ClO+, 2(7) + 7 + 6 − 1 = 26 e− F2ClO2+, 2(7) + 7 + 2(6) − 1 = 32 e− + Cl F F O sp3 hybridization 57. + O Cl F F O sp3 hybridization a. No, some atoms are in different places. Thus these are not resonance structures; they are different compounds. b. For the first Lewis structure, all nitrogens are sp3 hybridized and all carbons are sp2 hybridized. In the second Lewis structure, all nitrogens and carbons are sp2 hybridized. 562 CHAPTER 14 COVALENT BONDING: ORBITALS c. For the reaction: H O H O N O O N C C C C N N N N C H H C H O O H Bonds broken: Bonds formed: 3 C=O (745 kJ/mol) 3 C‒N (305 kJ/mol) 3 N‒H (391 kJ/mol) 3 C=N (615 kJ/mol) 3 C‒O (358 kJ/mol) 3 O‒H (467 kJ/mol) ΔH = 3(745) + 3(305) + 3(391) - [3(615) + 3(358) + 3(467)] ΔH = 4323 kJ − 4320 kJ = 3 kJ The bonds are slightly stronger in the first structure with the carbon-oxygen double bonds since ΔH for the reaction is positive. However, the value of ΔH is so small that the best conclusion is that the bond strengths are comparable in the two structures. 58. Cl H C H H Cl C C Cl H Cl C Cl C Cl H C H In order to rotate about the double bond, the molecule must go through an intermediate stage where the π bond is broken and the sigma bond remains intact. Bond energies are 347 kJ/mol for a C‒C bond and 614 kJ/mol for a C=C bond. If we take the single bond as the strength of the σ bond, then the strength of the π bond is (614 − 347 = ) 267 kJ/mol. Thus 267 kJ/mol must be supplied to rotate about a carbon-carbon double bond. CHAPTER 14 COVALENT BONDING: ORBITALS 563 59. dxz + pz x x z z The two orbitals will overlap side-to-side, so when the orbitals are in-phase, a π bonding molecular orbital would form. 60. N2 (ground state): (σ2s)2(σ2s*)2(π2p)4(σ2p)2, B.O. = 3, diamagnetic (0 unpaired e−) N2 (1st excited state): (σ2s)2(σ2s*)2(π2p)4(σ2p)1(π2p*)1 B.O. = (7 − 3)/2 = 2, paramagnetic (2 unpaired e-) The first excited state of N2 should have a weaker bond and should be paramagnetic. 61. Considering only the 12 valence electrons in O2, the MO models would be: σ2p* ↑ ↑ ↑↓ ↑↓ π 2p * ↑↓ π 2p ↑↓ ↑↓ ↑↓ σ2p ↑↓ ↑↓ σ2s * ↑↓ ↑↓ σ2s ↑↓ O2 ground state Arrangement of electrons consistent with the Lewis structure (double bond and no unpaired electrons). It takes energy to pair electrons in the same orbital. Thus the structure with no unpaired electrons is at a higher energy; it is an excited state. 62. a. Yes, both have four sets of electrons about the P. We would predict a tetrahedral structure for both. See part d for the Lewis structures. 564 CHAPTER 14 COVALENT BONDING: ORBITALS b. The hybridization is sp3 for P in each structure since both structures exhibit a tetrahedral arrangement of electron pairs. c. P has to use one of its d orbitals to form the π bond since the p orbitals are all used to form the hybrid orbitals. d. Formal charge = number of valence electrons of an atom - [(number of lone pair electrons) + 1/2 (number of shared electrons)]. The formal charges calculated for the O and P atoms are next to the atoms in the following Lewis structures. O -1 +1 Cl P O Cl Cl Cl P 0 0 Cl Cl In both structures, the formal charges of the Cl atoms are all zeros. The structure with the P=O bond is favored on the basis of formal charge since it has a zero formal charge for all atoms. 63. a. BH3 has 3 + 3(1) = 6 valence electrons. H trigonal planar, nonpolar, 120E, sp2 B H H b. N2F2 has 2(5) + 2(7) = 24 valence electrons. N V-shaped about both N’s; ≈120E about both N’s; both N atoms: sp2 N F Can also be: F F N F polar nonpolar These are distinctly different molecules. c. C4H6 has 4(4) + 6(1) = 22 valence electrons. H H C H C H C H C H N CHAPTER 14 COVALENT BONDING: ORBITALS 565 All C atoms are trigonal planar with 120E bond angles and sp2 hybridization. Because C and H have about equal electronegativities, the C−H bonds are essentially nonpolar, so the molecule is nonpolar. All neutral compounds composed of only C and H atoms are nonpolar. d. ICl3 has 7 + 3(7) = 28 valence electrons. Cl ≈ 90 o I Cl ≈ 90 o Cl 64. T-shaped, polar .90E, dsp3 For carbon, nitrogen, and oxygen atoms to have formal charge values of zero, each C atom will form four bonds to other atoms and have no lone pairs of electrons, each N atom will form three bonds to other atoms and have one lone pair of electrons, and each O atom will form two bonds to other atoms and have two lone pairs of electrons. Following these bonding requirements gives the following two resonance structures for vitamin B6: O b a H O H g H C H C C f O C H H c C d e C O N C C H H O H H H H C C C H C H H C C C C N O H H H a. 21 σ bonds; 4 π bonds (The electrons in the three π bonds in the ring are delocalized.) b. Angles a), c), and g): ≈109.5°; angles b), d), e), and f): ≈120° c. 6 sp2 carbons; the five carbon atoms in the ring are sp2 hybridized, as is the carbon with the double bond to oxygen. d. 4 sp3 atoms; the two carbons that are not sp2 hybridized are sp3 hybridized, and the oxygens marked with angles a and c are sp3 hybridized. e. Yes, the π electrons in the ring are delocalized. The atoms in the ring are all sp2 hybridized. This leaves a p orbital perpendicular to the plane of the ring from each atom. Overlap of all six of these p orbitals results in a π molecular orbital system where the electrons are delocalized above and below the plane of the ring (similar to benzene in Figure 14.50 of the text). 566 65. CHAPTER 14 COVALENT BONDING: ORBITALS The Lewis structure for caffeine is: H O H H H C C C H H N N C C C C O N H H N C H H The three C atoms each bonded to three H atoms are sp3 hybridized (tetrahedral geometry); the other five C atoms with trigonal planar geometry are sp2 hybridized. The one N atom with the double bond is sp2 hybridized, and the other three N atoms are sp3 hybridized. The answers to the questions are: C C C C 66. 6 C and N atoms are sp2 hybridized 6 C and N atoms are sp3 hybridized 0 C and N atoms are sp hybridized (linear geometry) 25 σ bonds and 4 π bonds We rotated the molecule about some single bonds from the structure given in the question. Following the bonding guidelines outlined in Exercise 14.64, a Lewis structure for aspartame is: H H O H H C H O C C H O C H H N C C N C C H H O H H H O H H C C C C C C H H H Another resonance structure could be drawn having the double bonds in the benzene ring moved over one position. CHAPTER 14 COVALENT BONDING: ORBITALS 567 Atoms that have trigonal planar geometry of electron pairs are assumed to have sp2 hybridization, and atoms with tetrahedral geometry of electron pairs are assumed to have sp3 hybridization. All the N atoms have tetrahedral geometry, so they are all sp3 hybridized (no sp2 hybridization). The oxygens double bonded to carbon atoms are sp2 hybridized; the other two oxygens with two single bonds are sp3 hybridized. For the carbon atoms, the six carbon atoms in the benzene ring are sp2 hybridized, and the three carbons double bonded to oxygen are also sp2 hybridized (tetrahedral geometry). Answering the questions: C C C 67. 9 sp2 hybridized C and N atoms (9 from C’s and 0 from N’s) 7 sp3 hybridized C and O atoms (5 from C’s and 2 from O’s) 39 σ bonds and 6 π bonds (this includes the 3 π bonds in the benzene ring that are delocalized) a. The Lewis structures for NNO and NON are: N N O N N O N N O N O N N O N N O N The NNO structure is correct. From the Lewis structures we would predict both NNO and NON to be linear. However, we would predict NNO to be polar and NON to be nonpolar. Since experiments show N2O to be polar, then NNO is the correct structure. b. Formal charge = number of valence electrons of atoms - [(number of lone pair electrons) + 1/2 (number of shared electrons)]. N N O N N O N N O -1 +1 0 0 +1 -1 -2 +1 +1 The formal charges for the atoms in the various resonance structures are below each atom. The central N is sp hybridized in all the resonance structures. We can probably ignore the third resonance structure on the basis of the relatively large formal charges as compared to the first two resonance structures. c. The sp hybrid orbitals from the center N overlap with atomic orbitals (or appropriate hybrid orbitals) from the other two atoms to form the two sigma bonds. The remaining two unhybridized p orbitals from the center N overlap with two p orbitals from the peripheral N atom to form the two π bonds. 2px sp sp 2py z axis 568 68. CHAPTER 14 COVALENT BONDING: ORBITALS O=N‒Cl: The bond order of the NO bond in NOCl is 2 (a double bond). NO: From molecular orbital theory, the bond order of this NO bond is 2.5 (see Figure 14.43 of the text). Both reactions apparently only involve the breaking of the N‒Cl bond. However, in the reaction ONCl → NO + Cl, some energy is released in forming the stronger NO bond, lowering the value of ΔH. Therefore, the apparent N‒Cl bond energy is artificially low for this reaction. The first reaction only involves the breaking of the N‒Cl bond. Challenge Problems 69. a. F2−(g) → F(g) + F−(g) ΔH = F2− bond energy Using Hess’s law: F2−(g) → F2(g) + e− F2(g) → 2 F(g) F(g) + e− → F−(g) ΔH = 290. kJ (IE for F2−) ΔH = 154 kJ (BE for F2 from Table 13.6) ΔH = !327.8 kJ (EA for F from Table12.8) ____________________________________________________________________________________________ F2−(g) → F(g) + F−(g) ΔH = 116 kJ; BE for F2− = 116 kJ/mol Note that F2− has a smaller bond energy than F2. b. F2: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4 − 2 2 2 4 4 B.O. = (8 – 6)/2 = 1 1 F2 : (σ2s) (σ2s*) (σ2p) (π2p) (π2p*) (σ2p*) B.O. = (8 – 7)/2 = 0.5 MO theory predicts that F2 should have a stronger bond than F2− because F2 has the larger bond order. As determined in part a, F2 indeed has a stronger bond because the F2 bond energy (154 kJ/mol) is greater than the F2− bond energy (116 kJ/mol). 70. The ground state MO electron configuration for He2 is (σ1s)2(σ1s*)2 giving a bond order of 0. Therefore, He2 molecules are not predicted to be stable (and are not stable) in the lowestenergy ground state. However, in a high-energy environment, electron(s) from the antibonding orbitals in He2 can be promoted into higher-energy bonding orbitals, thus giving a nonzero bond order and a “reason” to form. For example, a possible excited-state MO electron configuration for He2 would be (σ1s)2(σ1s*)1(σ2s)1, giving a bond order of (3 – 1)/2 = 1. Thus excited He2 molecules can form, but they spontaneously break apart as the electron(s) fall back to the ground state, where the bond order equals zero. 71. a. E = hc (6.626 × 10 −34 J s)(2.998 × 108 m / s) = = 7.9 × 10−18 J −9 λ 25 × 10 m CHAPTER 14 COVALENT BONDING: ORBITALS 7.9 × 10−18 J × 569 6.022 × 10 23 1 kJ × = 4800 kJ/mol mol 1000 J Using ΔH values from the various reactions, 25-nm light has sufficient energy to ionize N2 and N and to break the triple bond. Thus N2, N2+, N, and N+ will all be present, assuming excess N2. b. To produce atomic nitrogen but no ions, the range of energies of the light must be from 941 kJ/mol to just below 1402 kJ/mol. 941 kJ 1 mol 1000 J × × = 1.56 × 10−18 J/photon 23 mol 1 kJ 6.022 × 10 λ = hc (6.6261 × 10 −34 J s)(2.998 × 108 m / s) = = 1.27 × 10−7 m = 127 nm E 1.56 × 10 −18 J 1402 kJ 1 mol 1000 J × × = 2.328 × 10−18 J/photon 23 mol kJ 6.0221 × 10 λ = hc (6.6261 × 10 −34 J s)(2.9979 × 108 m / s) = = 8.533 × 10−8 m = 85.33 nm E 2.328 × 10 −18 J Light with wavelengths in the range of 85.33 nm < λ < 127 nm will produce N but no ions. c. N2: (σ2s)2(σ2s*)2(π2p)4(σ2p)2; the electron removed from N2 is in the σ2p molecular orbital, which is lower in energy than the 2p atomic orbital from which the electron in atomic nitrogen is removed. Because the electron removed from N2 is lower in energy than the electron removed from N, the ionization energy of N2 is greater than that for N. 72. The electron configurations are: N2: (σ2s)2(σ2s*)2(π2p)4(σ2p)2 O2: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)2 N22−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2 Note: The ordering of the σ2p and π2p orbitals is not important to this question. N2−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)1 O2+: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)1 The species with the smallest ionization energy has the electron that is easiest to remove. From the MO electron configurations, O2, N22−, N2−, and O2+ all contain electrons in the same higher-energy antibonding orbitals ( π *2 p ) , so they should have electrons that are easier to remove as compared to N2, which has no π *2 p electrons. To differentiate which has the easiest π *2 p to remove, concentrate on the number of electrons in the orbitals attracted to the number of protons in the nucleus. 570 CHAPTER 14 COVALENT BONDING: ORBITALS N22− and N2− both have 14 protons in the two nuclei combined. Because N22− has more electrons, one would expect N22− to have more electron repulsions, which translates into having an easier electron to remove. Between O2 and O2+, the electron in O2 should be easier to remove. O2 has one more electron than O2+, and one would expect the fewer electrons in O2+ to be better attracted to the nuclei (and harder to remove). Between N22− and O2, both have 16 electrons; the difference is the number of protons in the nucleus. Because N22− has two fewer protons than O2, one would expect the N22− to have the easiest electron to remove which translates into the smallest ionization energy. 73. The complete Lewis structure follows. All but two of the carbon atoms are sp3 hybridized. The two carbon atoms that contain the double bond are sp2 hybridized (see *). CH 2 CH 3 H H C H 2C H H C H 2C C C *C H H CH 3 C O H H C H H C * C CH CH 3 C CH 2 CH 2 H CH 3 CH CH 3 C CH 2 C CH 2 H CH 2 H No; most of the carbons are not in the same plane since a majority of carbon atoms exhibit a tetrahedral structure. Note: CH, CH2, and CH3 are shorthand for carbon atoms singly bonded to hydrogen atoms. 74. 2p 2s O2 O2− O2+ O The order from lowest IE to highest IE is: O2− < O2 < O2+ < O. CHAPTER 14 COVALENT BONDING: ORBITALS 571 The electrons for O2−, O2, and O2+ that are highest in energy are in the π *2 p MOs. But for O2−, these electrons are paired. O2− should have the lowest ionization energy (its paired π *2 p electron is easiest to remove). The species O2+ has an overall positive charge, making it harder to remove an electron from O2+ than from O2. The highest energy electrons for O (in the 2p atomic orbitals) are lower in energy than the π *2 p electrons for the other species; O will have the highest ionization energy because it requires a larger quantity of energy to remove an electron from O as compared to the other species. 75. a. The CO bond is polar with the negative end around the more electronegative oxygen atom. We would expect metal cations to be attracted to and to bond to the oxygen end of CO on the basis of electronegativity. b. C O FC (carbon) = 4 − 2 − 1/2(6) = −1 FC (oxygen) = 6 − 2 − 1/2(6) = +1 From formal charge, we would expect metal cations to bond to the carbon (with the negative formal charge). c. In molecular orbital theory, only orbitals with proper symmetry overlap to form bonding orbitals. The metals that form bonds to CO are usually transition metals, all of which have outer electrons in the d orbitals. The only molecular orbitals of CO that have proper symmetry to overlap with d orbitals are the π2p* orbitals, whose shape is similar to the d orbitals (see Figure 14.36). Since the antibonding molecular orbitals have more carbon character, one would expect the bond to form through carbon. 76. The molecular orbitals for BeH2 are formed from the two hydrogen 1s orbitals and the 2s and one of the 2p orbitals from beryllium. One of the sigma bonding orbitals forms from overlap of the hydrogen 1s orbitals with a 2s orbital from beryllium. Assuming the z axis is the internuclear axis in the linear BeH2 molecule, then the 2pz orbital from beryllium has proper symmetry to overlap with the 1s orbitals from hydrogen; the 2px and 2py orbitals are nonbonding orbitals since they don’t have proper symmetry necessary to overlap with 1s orbitals. The type of bond formed from the 2pz and 1s orbitals is a sigma bond since the orbitals overlap head to head. The MO diagram for BeH2 is: 572 CHAPTER 14 Be COVALENT BONDING: ORBITALS 2H σ*s σ*p 2px 2p 2py ↑↓ 2s ↑↓ ↑ 1s ↑ 1s σp ↑↓ σs Bond order = (4 - 0)/2 = 2; the MO diagram predicts BeH2 to be a stable species and also predicts that BeH2 is diamagnetic. Note: The σs MO is a mixture of the two hydrogen 1s orbitals with the 2s orbital from beryllium, and the σp MO is a mixture of the two hydrogen 1s orbitals with the 2pz orbital from beryllium. The MOs are not localized between any two atoms; instead, they extend over the entire surface of the three atoms. 77. One of the resonance structures for benzene is: H C H H H C C C C C H H To break C6H6(g) into C(g) and H(g) requires the breaking of 6 C‒H bonds, 3 C=C bonds and 3 C‒C bonds: C6H6(g) → 6 C(g) + 6 H(g) ΔH = 6DC‒H + 3DC=C + 3DC‒C ΔH = 6(413 kJ) + 3(614 kJ) + 3(347 kJ) = 5361 kJ The question asks for ΔH °f for C6H6(g), which is ΔH for the reaction: 6 C(s) + 3 H2(g) → C6H6(g) ΔH = ΔH °f , C6 H 6 ( g ) CHAPTER 14 COVALENT BONDING: ORBITALS 573 To calculate ΔH for this reaction, we will use Hess’s law along with the value ΔH °f for C(g) and the bond energy value for H2 ( D H 2 = 432 kJ/mol). 6 C(g) + 6 H(g) → C6H6(g) ΔH1 = !5361 kJ ΔH2 = 6(717 kJ) 6 C(s) → 6 C(g) 3 H (g) → 6 H(g) ΔH 2 3 = 3(432 kJ) ____________________________________________________________________________________________________________________ ΔH = ΔH1 + ΔH2 + ΔH3 = 237 kJ; ΔH °f , C H ( g ) = 237 kJ/mol 6 6 ° The experimental ΔH f for C6H6(g) is more stable (lower in energy) by 154 kJ than the ΔH °f calculated from bond energies (83 ! 237 = !154 kJ). This extra stability is related to benzene’s ability to exhibit resonance. Two equivalent Lewis structures can be drawn for benzene. The π bonding system implied by each Lewis structure consists of three localized π bonds. This is not correct because all C‒C bonds in benzene are equivalent. We say the π electrons in benzene are delocalized over the entire surface of C6H6 (see Section 14.5 of the text). The large discrepancy between ΔH °f values is due to the delocalized π electrons, whose effect was not accounted for in the calculated ΔH °f value. The extra stability associated with benzene can be called resonance stabilization. In general, molecules that exhibit resonance are usually more stable than predicted using bond energies. 6 C(s) + 3 H2(g) → C6H6(g) 78. a. The NMR indicates that all hydrogens in this compound are equivalent. A possible structure for C2H3Cl2 that explains the NMR data is: H H Cl C C H Cl Cl Note that for organic compounds to have a zero formal charge for all atoms, carbon will always satisfy the octet rule by having four bonds and no lone pairs, oxygen will always satisfy the octet rule by having two bonds and two lone pairs, and Cl will always satisfy the octet rule by having one bond and three lone pairs of electrons. b. We have two “types” of hydrogens. The triplet signal is produced by having two neighboring C−H bonds, whereas the quintet signal is produced by having four neighboring C−H bonds. A possible structure that explains the NMR (number of peaks, types of peaks, and the 2:1 relative intensities of the signals) is: Cl H H H C C C H H H Cl 574 CHAPTER 14 COVALENT BONDING: ORBITALS c. Data: Three “types” of hydrogens; the singlet peak is produced by having no neighboring C−H bonds, the quartet signal indicates three neighboring C−H bonds, and the triplet indicates two neighboring C−H bonds. A possible structure to explain the NMR is: Quartet signal Triplet signal H H H O C C C H H O H Singlet signal d. Data: Three “types” of hydrogens. A possible structure to explain the NMR is: Doublet signal H H H O C C H C H H C C H H H Heptet signal Singlet signal H e. Data: Three “types” of hydrogens. A possible structure to explain the NMR (number of peaks, types of peaks, and relative intensities) is: Triplet H signal H H O C C C H H H Triplet signal Quintet signal Note that the two hydrogens in the –CH2 group have four total neighboring protons (3 + 1), giving the quintet signal. The –CH2 group protons also cause the other “types” of hydrogens to have triplet signals. CHAPTER 14 79. COVALENT BONDING: ORBITALS 575 The two isomers having the formula C2H6O are: H H H C C H H a b H O H H C H O C H c H H The first structure has three types of hydrogens (a, b, and c), so the signals should be seen in three different regions of the NMR spectra. The overall relative areas of the three signals should be in a 3:2:1 ratio due to the number of a, b, and c hydrogens. The signal for the a hydrogens will be split into a triplet signal due to the two neighboring b hydrogens. The signal for the b hydrogens should be split into a quintet signal due to the three neighboring a hydrogens plus the neighboring c hydrogen (four total protons). The c hydrogen signal should be split into a triplet signal due to the two neighboring b hydrogens. In practice, however, the c hydrogen bonded to the oxygen does not behave “normally”. This O−H hydrogen generally behaves as if it was more than three sigma bonds apart from the b hydrogens. Therefore, the spectrum will most likely have a triplet signal for the a hydrogens, a quartet signal for the b hydrogens, and a singlet signal for the c hydrogen. In the second structure, all six hydrogens are equivalent, so only one signal will appear in the NMR spectrum. This signal will be a singlet signal because the hydrogens in the two –CH3 groups are more than three sigma bonds apart from each other (no splitting occurs). Marathon Problem 80. φ1, φ2, and φ3 all must be normalized. ∫φ12 dT = 1 = 4 A2 = 1 − 1 ∫φs2 dT + 4A2∫φpx2 dT + 2 3 1 2 1 = , A2 = , A = 3 3 6 ∫φ22 dT = 1 = 1 6 1 2 1 (−A) ∫φs φpx dT ∫φs dT + A2∫φpx2 dT + B2∫φpy2 dT + 2 3 3 +2 1= 1 1 + 4(A2)(1) + 0 2A ∫φs φpx dT = 3 3 1 1 (B) ∫φs φpy dT − 2AB∫φpx φpy dT = + A2 + B2 + 0 + 0 + 0 3 3 1 1 1 + + B2, = B2, B = 3 6 2 1 2 CHAPTER 15 CHEMICAL KINETICS Reaction Rates 10. The reaction rate is defined as the change in concentration of a reactant or product per unit time. Consider the general reaction: aA → products where rate = − d[A ] dt If we graph [A] vs. t, it would usually look like the solid line in the following plot. a [A] b c 0 t1 t2 time An instantaneous rate is the slope of a tangent line to the graph of [A] vs. t. We can determine the instantaneous rate at any time during the reaction. On the plot, tangent lines at t ≈ 0 and t = t1 are drawn. The slope of these tangent lines would be the instantaneous rates at t ≈ 0 and t = t1. We call the instantaneous rate at t ≈ 0 the initial rate. The average rate is measured over a period of time. For example, the slope of the dashed line connecting points a and c is the average rate of the reaction over the entire length of time 0 to t2 (average rate = Δ[A]/Δt). An average rate is determined over some time period, whereas an instantaneous rate is determined at one specific time. The rate that is largest is generally the initial rate. At t ≈ 0, the slope of the tangent line is greatest, which means the rate is largest at t ≈ 0. The initial rate is used by convention so that the rate of reaction only depends on the forward reaction; at t ≈ 0, the reverse reaction is insignificant because no products are present yet. 576 CHAPTER 15 CHEMICAL KINETICS 577 11. 0.0120/0.0080 = 1.5; reactant B is used up 1.5 times faster than reactant A. This corresponds to a 3 to 2 mole ratio between B and A in the balanced equation. 0.0160/0.0080 = 2; product C is produced twice as fast as reactant A is used up. So the coefficient for C is twice the coefficient for A. A possible balanced equation is 2A + 3B → 4C. 12. The coefficients in the balanced reaction relate the rate of disappearance of reactants to the rate of production of products. From the balanced reaction, the rate of production of P4 will be 1/4 the rate of disappearance of PH3, and the rate of production of H2 will be 6/4 the rate of disappearance of PH3. By convention, all rates are given as positive values. Rate = − Δ[PH 3 ] − (−0.0048 mol / 2.0 L) = = 2.4 × 10−3 mol L−1 s−1 Δt s Δ[P4 ] 1 Δ[ PH 3 ] =− = 2.4 × 10−3/4 = 6.0 × 10−4 mol L−1 s−1 Δt 4 Δt Δ[H 2 ] 6 Δ[PH 3 ] =− = 6(2.4 × 10−3)/4 = 3.6 × 10−3 mol L−1 s−1 Δt 4 Δt 13. Using the coefficients in the balanced equation to relate the rates: d[ NH 3 ] d[H 2 ] d[ N 2 ] d[ N 2 ] and =3 = −2 dt dt dt dt So : d[ NH 3 ] 2 d[H 2 ] 1 d[ H 2 ] 1 d[ NH 3 ] =− =− or dt 3 dt 3 dt 2 dt Ammonia is produced at a rate equal to 2/3 of the rate of consumption of hydrogen. 14. a. The units for rate are always mol L−1 s−1. c. Rate = k[A], mol ⎛ mol ⎞ = k⎜ ⎟ Ls ⎝ L ⎠ b. Rate = k; k has units of mol L−1 s−1. d. Rate = k[A]2, k must have units of s-1. mol ⎛ mol ⎞ = k⎜ ⎟ Ls ⎝ L ⎠ k must have units L mol−1 s−1. e. L2 mol−2 s−1 1/ 2 ⎛ mol ⎞ 1/2 -1/2 1 ⎟ ; k must have units of L mol s− . ⎜ ⎝ L ⎠ 15. mol ⎛ mol ⎞ Rate = k[Cl] [CHCl3], = k⎜ ⎟ Ls ⎝ L ⎠ 16. 1.24 × 10 −12 cm 3 1L 6.022 × 10 23 molecules = 7.47 × 108 L mol−1 s−1 × × molecules s mol 1000 cm 3 1/2 2 578 CHAPTER 15 CHEMICAL KINETICS Rate Laws from Experimental Data: Initial Rates Method 17. a. In the first two experiments, [NO] is held constant and [Cl2] is doubled. The rate also doubled. Thus the reaction is first order with respect to Cl2. Or mathematically: Rate = k[NO]x[Cl2]y 0.36 k (0.10) x (0.20) y (0.20) y = = , 2.0 = 2.0y, y = 1 0.18 k (0.10) x (0.10) y (0.10) y We can get the dependence on NO from the second and third experiments. Here, as the NO concentration doubles (Cl2 concentration is constant), the rate increases by a factor of four. Thus, the reaction is second order with respect to NO. Or mathematically: 1.45 k (0.20) x (0.20) (0.20) x = = , 4.0 = 2.0x, x = 2; so, Rate = k[NO]2[Cl2]. 0.36 k (0.10) x (0.20) (0.10) x Try to examine experiments where only one concentration changes at a time. The more variables that change, the harder it is to determine the orders. Also, these types of problems can usually be solved by inspection. In general, we will solve using a mathematical approach, but keep in mind that you probably can solve for the orders by simple inspection of the data. b. The rate constant k can be determined from the experiments. From experiment 1: 2 0.18 mol ⎛ 0.10 mol ⎞ ⎛ 0.10 mol ⎞ 2 2 1 = k⎜ ⎟ ⎜ ⎟ , k = 180 L mol− min− L L L min ⎝ ⎠ ⎝ ⎠ From the other experiments: k = 180 L2 mol−2 min−1 (second exp.); k = 180 L2 mol−2 min−1 (third exp.) The average rate constant is kmean = 1.8 × 102 L2 mol−2 min−1. 18. a. Rate = k[I−]x[S2O82−]y; 12.5 × 10 −6 k (0.080) x (0.040) y = , 2.00 = 2.0x, x = 1 −6 x y 6.25 × 10 k (0.040) (0.040) k (0.080)(0.040) y 12.5 × 10 −6 = , 2.00 = 2.0y, y = 1; Rate = k[I−][S2O82−] −6 y 6.25 × 10 k (0.080)(0.020) b. For the first experiment: 12.5 × 10 −6 mol ⎛ 0.080 mol ⎞⎛ 0.040 mol ⎞ 3 1 1 = k⎜ ⎟⎜ ⎟ , k = 3.9 × 10− L mol− s− Ls L L ⎝ ⎠⎝ ⎠ CHAPTER 15 CHEMICAL KINETICS The other values are: 579 Initial Rate (mol L−1 s−1) k (L mol−1 s−1) 12.5 × 10−6 6.25 × 10−6 6.25 × 10−6 5.00 × 10−6 7.00 × 10−6 3.9 × 10−3 3.9 × 10−3 3.9 × 10−3 3.9 × 10−3 3.9 × 10−3 kmean = 3.9 × 10−3 L mol−1 s−1 19. a. Rate = k[NOCl]n; using experiments two and three: 2.66 × 10 4 k ( 2.0 × 1016 ) n = , 4.01 = 2.0n, n = 2; Rate = k[NOCl]2 3 16 n 6.64 × 10 k (1.0 × 10 ) 2 b. ⎛ 3.0 × 1016 molecules ⎞ 5.98 × 10 4 molecules ⎜ ⎟ , k = 6.6 × 10−29 cm3 molecules−1 s−1 = k 3 ⎜ ⎟ cm cm 3 s ⎝ ⎠ The other three experiments give (6.7, 6.6, and 6.6) × 10−29 cm3 molecules−1 s−1, respectively. The mean value for k is 6.6 × 10−29 cm3 molecules−1 s−1. c. 20. 6.6 × 10 −29 cm 3 1L 6.022 × 10 23 molecules 4.0 × 10 −8 L = × × molecules s mol mol s 1000 cm 3 a. Rate = k[Hb]x[CO]y Comparing the first two experiments, [CO] is unchanged, [Hb] doubles, and the rate doubles. Therefore, x = 1, and the reaction is first order in Hb. Comparing the second and third experiments, [Hb] is unchanged, [CO] triples, and the rate triples. Therefore, y = 1 and the reaction is first order in CO. b. Rate = k[Hb][CO] c. From the first experiment: 0.619 µmol L−1 s−1 = k(2.21 µmol/L)(1.00 µmol/L), k = 0.280 L µmol−1 s−1 The second and third experiments give similar k values, so kmean = 0.280 L µmol−1 s−1. d. Rate = k[Hb][CO] = 21. 0.280 L 3.36 μmol 2.40 μmol × × = 2.26 µmol L−1 s−1 μmol s L L a. Rate = k[ClO2]x[OH−]y; From the first two experiments: 2.30 × 10−1 = k(0.100)x(0.100)y and 5.75 × 10−2 = k(0.0500)x(0.100)y 580 CHAPTER 15 CHEMICAL KINETICS (0.100) x = 2.00x, x = 2 x (0.0500) Dividing the two rate laws: 4.00 = Comparing the second and third experiments: 2.30 × 10−1 = k(0.100)(0.100)y and 1.15 × 10−1 = k(0.100)(0.0500)y Dividing: 2.00 = (0.100) y = 2.0y, y = 1 (0.050) y The rate law is: Rate = k[ClO2]2[OH−] 2.30 × 10−1 mol L−1 s−1 = k(0.100 mol/L)2(0.100 mol/L), k = 2.30 × 102 L2 mol−2 s−1 = kmean 2 b. Rate = 22. 2.30 × 10 2 L2 ⎛ 0.175 mol ⎞ 0.0844 mol ×⎜ = 0.594 mol L−1 s−1 ⎟ × 2 L L mol s ⎝ ⎠ Rate = k[NO]x[O2]y; Comparing the first two experiments, [O2] is unchanged, [NO] is tripled, and the rate increases by a factor of nine. Therefore, the reaction is second order in NO (32 = 9). The order of O2 is more difficult to determine. Comparing the second and third experiments: 3.13 × 1017 k (2.50 × 1018 ) 2 (2.50 × 1018 ) y = 1.80 × 1017 k (3.00 × 1018 ) 2 (1.00 × 1018 ) y 1.74 = 0.694(2.50)y, 2.51 = 2.50y, y = 1 Rate = k[NO]2[O2]; from experiment 1: 2.00 × 1016 molecules cm−3 s−1 = k(1.00 × 1018 molecules/cm3)2 × (1.00 × 1018 molecules/cm3) k = 2.00 × 10−38 cm6 molecules−2 s−1 = kmean 2 Rate = ⎛ 6.21 × 1018 molecules ⎞ 2.00 × 10 −38 cm 6 7.36 × 1018 molecules ⎜ ⎟ × × ⎜ ⎟ molecules2 s cm 3 cm 3 ⎝ ⎠ Rate = 5.68 × 1018 molecules cm−3 s−1 23. Rate = k[N2O5]x; the rate laws for the first two experiments are: 2.26 × 10−3 = k(0.190)x and 8.90 × 10−4 = k(0.0750)x Dividing: 2.54 = (0.190) x = (2.53)x, x = 1; Rate = k[N2O5] (0.0750) x CHAPTER 15 k = 24. CHEMICAL KINETICS 581 Rate 8.90 × 10 −4 mol L−1 s −1 = = 1.19 × 10−2 s−1; kmean = 1.19 × 10−2 s−1 [ N 2O5 ] 0.0750 mol / L ⎛ [A] ⎞ Rate 2 k[ A]2x = = ⎜⎜ 2 ⎟⎟ x Rate1 k[ A]1 ⎝ [A ]1 ⎠ x The rate doubles as the concentration quadruples: 2 = (4)x, x = 1/2 The order is 1/2 (the square root of the concentration of reactant). For a reactant that has an order of !1 and the reactant concentration is doubled: Rate 2 1 = (2) −1 = Rate1 2 The rate will decrease by a factor of 1/2 when the reactant concentration is doubled for a !1 order reaction. Negative orders are seen for substances that hinder or slow down a reaction. 25. Rate = k[I−]x[OCl−]y[OH−]z; Comparing the first and second experiments: 18.7 × 10 −3 k (0.0026) x (0.012) y (0.10) z = , 2.0 = 2.0x, x = 1 −3 x y z 9.4 × 10 k (0.0013) (0.012) (0.10) Comparing the first and third experiments: 9.4 × 10 −3 k (0.0013)(0.012) y (0.10) z , 2.0 = 2.0y, y = 1 = −3 y z 4.7 × 10 k (0.0013)(0.0060) (0.10) Comparing the first and sixth experiments: 4.8 × 10 −3 k (0.0013)(0.012)(0.20) z , 1/2 = 2.0z, z = !1 = −3 z 9.4 × 10 k (0.0013)(0.012)(0.10) Rate = k[ I − ][OCl − ] ; the presence of OH− decreases the rate of the reaction. − [OH ] For the first experiment: 9.4 × 10 −3 mol (0.0013 mol / L) (0.012 mol / L) =k , k = 60.3 s −1 = 60. s −1 Ls (0.10 mol / L) For all experiments, kmean = 60. s −1 . 582 CHAPTER 15 CHEMICAL KINETICS Integrated Rate Laws 26. Zero order: − d[A] = k, dt [A] [ A ]t [ A ]0 [ A ]t t [ A ]0 0 ∫ d[A] = − ∫ k dt t = − kt , [A]t − [A]0 = − kt , [A]t = !kt + [A]0 0 First order: − d[A] = k[A], dt ln[A] [ A ]t [ A ]0 [ A ]t t d[A] ∫ [A] = − ∫ k dt [ A ]0 0 = − kt , ln[A]t − ln[A]0 = − kt , ln[A]t = !kt + ln[A]0 Second order: − d[A] = k[A]2 , dt − 27. 1 [A] [ A ]t [ A ]t = − kt, − [ A ]0 Zero order: t1/2 = t d[A ] ∫ [A]2 = − ∫ k dt [ A ]0 0 [A]0 ; 2k 1 1 1 1 + = − kt , = kt + [ A ] t [ A ]0 [A]t [ A ]0 first order: t1/2 = ln 2 1 ; second order: t1/2 = k k[ A]0 For a first-order reaction, if the first half-life equals 20. s, the second half-life will also be 20. s because the half-life for a first-order reaction is concentration-independent. The second half-life for a zero-order reaction will be 1/2(20.) = 10. s. This is because the half-life for a zero-order reaction has a direct relationship with concentration (as the concentration decreases by a factor of 2, the half-life decreases by a factor of 2). Because a second-order reaction has an inverse relationship between t1/2 and [A]0, the second half-life will be 40. s (twice the first half-life value). 28. a. Because the 1/[A] versus time plot was linear, the reaction is second order in A. The slope of the 1/[A] versus time plot equals the rate constant k. Therefore, the rate law, the integrated rate law, and the rate constant value are: 1 1 = kt + ; k = 3.60 × 10-2 L mol-1 s-1 [A]0 [A] 1 b. The half-life expression for a second-order reaction is: t1/2 = k[A]0 Rate = k[A]2; CHAPTER 15 CHEMICAL KINETICS For this reaction: t1/2 = 583 1 3.60 × 10 −2 −1 L mol s −1 × 2.80 × 10 −3 mol / L = 9.92 × 103 s Note: We could have used the integrated rate law to solve for t1/2, where [A] = (2.80 × 10−3 /2) mol/L. c. Because the half-life for a second-order reaction depends on concentration, we must use the integrated rate law to solve. 1 3.60 × 10 −2 L 1 1 1 = kt + × t+ , = −4 mol s [A] [A]0 7.00 × 10 M 2.80 × 10 −3 M 1.43 × 103 − 357 = (3.60 × 10−2)t, t = 2.98 × 104 s 29. a. Because the ln[A] versus time plot was linear, the reaction is first order in A. The slope of the ln[A] versus time plot equals −k. Therefore, the rate law, the integrated rate law, and the rate constant value are: Rate = k[A]; ln[A] = −kt + ln[A]0; k = 2.97 × 10−2 min−1 b. The half-life expression for a first order rate law is: t1/2 = 0.6931 ln 2 0.6931 = , t1/2 = = 23.3 min k k 2.97 × 10 − 2 min −1 c. 2.50 × 10−3 M is 1/8 of the original amount of A present initially, so the reaction is 87.5% complete. When a first-order reaction is 87.5% complete (or 12.5% remains), then the reaction has gone through 3 half-lives: 100% → 50.0% → 25.0% → 12.5%; t1/2 t1/2 t1/2 t = 3 × t1/2 = 3 × 23.3 min = 69.9 min Or we can use the integrated rate law: ⎛ 2.50 × 10 −3 M ⎞ ⎛ [A] ⎞ −2 −1 ⎟ ⎟⎟ = − kt , ln⎜ ln⎜⎜ ⎜ 2.00 × 10 − 2 M ⎟ = −(2.97 × 10 min )t [ A ] 0 ⎠ ⎝ ⎝ ⎠ t= 30. ln(0.125) = 70.0 min − 2.97 × 10 − 2 min −1 a. Because the [C2H5OH] versus time plot was linear, the reaction is zero order in C2H5OH. The slope of the [C2H5OH] versus time plot equals -k. Therefore, the rate law, the integrated rate law, and the rate constant value are: Rate = k[C2H5OH]0 = k; [C2H5OH] = −kt + [C2H5OH]0; k = 4.00 × 10−5 mol L−1 s−1 b. The half-life expression for a zero-order reaction is t1/2 = [A]0/2k. 584 CHAPTER 15 t1/2 = CHEMICAL KINETICS [C 2 H 5 OH]0 1.25 × 10 −2 mol / L = = 156 s 2k 2 × 4.00 × 10 −5 mol L−1 s −1 Note: We could have used the integrated rate law to solve for t1/2, where [C2H5OH] = (1.25 × 10−2/2) mol/L. c. [C2H5OH] = −kt + [C2H5OH]0 , 0 mol/L = −(4.00 × 10−5 mol L−1 s−1)t + 1.25 × 10−2 mol/L t= 31. 1.25 × 10 −2 mol / L = 313 s 4.00 × 10 −5 mol L−1 s −1 The first assumption to make is that the reaction is first order. For a first order reaction, a graph of ln[H2O2] versus time will yield a straight line. If this plot is not linear, then the reaction is not first order, and we make another assumption. Time (s) [H2O2] (mol/L) 0 120. 300. 600. 1200. 1800. 2400. 3000. 3600. 1.00 0.91 0.78 0.59 0.37 0.22 0.13 0.082 0.050 ln[H2O2] 0.000 −0.094 −0.25 −0.53 −0.99 −1.51 −2.04 −2.50 −3.00 Note: We carried extra significant figures in some of the natural log values in order to reduce round-off error. For the plots, we will do this most of the time when the natural log function is involved. The plot of ln[H2O2] versus time is linear. Thus the reaction is first order. The differential − d[H 2 O 2 ] = k[H2O2] and ln[H2O2] = −kt + rate law and integrated rate law are Rate = dt ln[H2O2]0. We determine the rate constant k by determining the slope of the ln[H2O2] versus time plot (slope = −k). Using two points on the curve gives: slope = −k = Δy 0 − (3.00) = = −8.3 × 10−4 s−1, k = 8.3 × 10−4 s−1 Δx 0 − 3600. CHAPTER 15 CHEMICAL KINETICS 585 To determine [H2O2] at 4000. s, use the integrated rate law, where [H2O2]0 = 1.00 M. ⎛ [H 2 O 2 ] ⎞ ⎟⎟ = −kt ln[H2O2] = −kt + ln[H2O2]0 or ln ⎜⎜ [ H O ] 2 2 0 ⎝ ⎠ ⎛ [H O ] ⎞ ln ⎜ 2 2 ⎟ = −8.3 × 10−4 s−1 × 4000. s, ln[H2O2] = −3.3, [H2O2] = e−3.3 = 0.037 M ⎝ 1.00 ⎠ 32. The first assumption to make is that the reaction is first order. For a first-order reaction, a graph of ln[C4H6] versus t should yield a straight line. If this isn't linear, then try the second order plot of 1/[C4H6] versus t. The data and the plots follow: Time 195 604 1246 2180 [C4H6] 1.6 × 10−2 1.5 × 10−2 1.3 × 10−2 1.1 × 10−2 ln[C4H6] 1/[C4H6] −4.14 62.5 −4.20 66.7 −4.34 76.9 −4.51 90.9 6210 s 0.68 × 10−2 M −4.99 147 M −1 Note: To reduce round-off error, we carried extra significant figures in the data points. The natural log plot is not linear, so the reaction is not first order. Because the second order plot of 1/[C4H6] versus t is linear, we can conclude that the reaction is second order in butadiene. The differential rate law is: Rate = k[C4H6]2 For a second-order reaction, the integrated rate law is 1 1 . = kt + [C 4 H 6 ] [C 4 H 6 ]0 The slope of the straight line equals the value of the rate constant. Using the points on the line at 1000. and 6000. s: k = slope = 144 L / mol − 73 L / mol = 1.4 × 10−2 L mol−1 s−1 6000. s − 1000. s 586 33. CHAPTER 15 CHEMICAL KINETICS Assume the reaction is first order and see if the plot of ln[NO2] versus time is linear. If this isn’t linear, try the second order plot of 1/[NO2] versus time. The data and plots follow. Time (s) 0 1.20 × 103 3.00 × 103 4.50 × 103 9.00 × 103 1.80 × 104 [NO2] (M) 0.500 0.444 0.381 0.340 0.250 0.174 1/[NO2] (M -1) ln[NO2] −0.693 −0.812 −0.965 −1.079 −1.386 −1.749 2.00 2.25 2.62 2.94 4.00 5.75 The plot of 1/[NO2] versus time is linear. The reaction is second order in NO2. The differential rate law and integrated rate law are Rate = k[NO2]2 and 1 1 = kt + . [ NO 2 ] [ NO 2 ]0 The slope of the plot 1/[NO2] versus. t gives the value of k. Using a couple of points on the plot: slope = k = Δy (5.75 − 2.00) M −1 = = 2.08 × 10−4 L mol−1 s−1 Δx (1.80 × 10 4 − 0) s To determine [NO2] at 2.70 × 104 s, use the integrated rate law, where 1/[NO2]0= 1/0.500 M = 2.00 M −1. 1 1 1 2.08 × 10 −4 L , = kt + = × 2.70 × 104 s + 2.00 M −1 [ NO 2 ] [ NO 2 ]0 [ NO 2 ] mol s 1 = 7.62, [NO2] = 0.131 M [ NO 2 ] 34. a. First, assume the reaction to be first order with respect to O. Hence a graph of ln[O] versus t would be linear if the reaction is first order. CHAPTER 15 CHEMICAL KINETICS t (s) 0 10. × 10−3 20. × 10−3 30. × 10−3 587 [O] (atoms/cm3) ln[O] 5.0 × 109 1.9 × 109 6.8 × 108 2.5 × 108 22.33 21.37 20.34 19.34 Because the graph is linear, we can conclude the reaction is first order with respect to O. b. The overall rate law is: Rate = k[NO2][O] Because NO2 was in excess, its concentration is constant. Thus for this experiment, the rate law is Rate: k′[O], where k′ = k[NO2]. In a typical first-order plot, the slope equals −k. For this experiment, the slope equals −k′ = −k[NO2]. From the graph: slope = 19.34 − 22.23 = −1.0 × 102 s−1, k′ = −slope = 1.0 × 102 s−1 −3 (30. × 10 − 0) s To determine k, the actual rate constant: k′ = k[NO2], 1.0 × 102 s−1 = k(1.0 × 1013 molecules/cm3) k = 1.0 × 10−11 cm3 molecules−1 s−1 35. From the data, the pressure of C2H5OH decreases at a constant rate of 13 torr for every 100. s. Since the rate of disappearance of C2H5OH is not dependent on concentration, the reaction is zero order in C2H5OH. k= 13 torr 1 atm × = 1.7 × 10−4 atm/s 100. s 760 torr The rate law and integrated rate law are: ⎛ 1 atm ⎞ ⎟⎟ = −kt + 0.329 atm Rate = k = 1.7 × 10−4 atm/s; PC 2 H 5OH = −kt + 250. torr ⎜⎜ ⎝ 760 torr ⎠ 588 CHAPTER 15 CHEMICAL KINETICS At 900. s: PC 2 H 5OH = −1.7 × 10−4 atm/s × 900. s + 0.329 atm = 0.176 atm = 0.18 atm = 130 torr 36. a. Because the 1/[A] versus time plot is linear with a positive slope, the reaction is second order with respect to A. The y intercept in the plot will equal 1/[A]0. Extending the plot, the y intercept will be about 10, so 1/10 = 0.1 M = [A]0. b. The slope of the 1/[A] versus time plot will equal k. Slope = k = (60 − 20) L / mol = 10 L mol−1 s−1 (5 − 1) s 1 1 10 L 1 = kt + = = 100, [A] = 0.01 M × 9s+ [A] [ A ]0 mol s 0 .1 M c. For a second-order reaction, the half-life does depend on concentration: t1/2 = First half-life: t1/2 = 1 k[ A]0 1 =1s 10 L 0.1 mol × mol s L Second half-life ([A]0 is now 0.05 M): t1/2 = 1/(10 × 0.05) = 2 s Third half-life ([A]0 is now 0.025 M): t1/2 = 1/(10 × 0.025) = 4 s 37. a. We check for first-order dependence by graphing ln[concentration] versus time for each set of data. The rate dependence on NO is determined from the first set of data since the ozone concentration is relatively large compared to the NO concentration, so it is effectively constant. Time (ms) 0 100. 500. 700. 1000. [NO] (molecules/cm3) 6.0 × 108 5.0 × 108 2.4 × 108 1.7 × 108 9.9 × 107 ln[NO] 20.21 20.03 19.30 18.95 18.41 CHAPTER 15 CHEMICAL KINETICS 589 Because ln[NO] versus t is linear, the reaction is first order with respect to NO. We follow the same procedure for ozone using the second set of data. The data and plot are: [O3] (molecules/cm3) Time (ms) 1.0 × 1010 8.4 × 109 7.0 × 109 4.9 × 109 3.4 × 109 0 50. 100. 200. 300. ln[O3] 23.03 22.85 22.67 22.31 21.95 The plot of ln[O3] versus t is linear. Hence the reaction is first order with respect to ozone. b. Rate = k[NO][O3] is the overall rate law. c. For NO experiment, Rate = k′[NO] and k′ = −(slope from graph of ln[NO] versus t). k′ = −slope = − 18.41 − 20.21 = 1.8 s−1 (1000. − 0) × 10 −3 s For ozone experiment, Rate = k′′[O3] and k′′ = −(slope from ln[O3] versus t). k′′ = −slope = − ( 21.95 − 23.03) = 3.6 s−1 (300. − 0) × 10 −3 s d. From NO experiment, Rate = k[NO][O3] = k′[NO] where k′ = k[O3]. k′ = 1.8 s−1 = k(1.0 × 1014 molecules/cm3), k = 1.8 × 10−14 cm3 molecules−1 s−1 We can check this from the ozone data. Rate = k′′[O3] = k[NO][O3], where k′′ = k[NO]. k′′ = 3.6 s−1 = k(2.0 × 1014 molecules/cm3), k = 1.8 × 10−14 cm3 molecules−1 s−1 Both values of k agree. 590 38. CHAPTER 15 CHEMICAL KINETICS This problem differs in two ways from previous problems: 1. A product is measured instead of a reactant. 2. Only the volume of a gas is given and not the concentration. We can find the initial concentration of C6H5N2Cl from the amount of N2 evolved after infinite time when all the C6H5N2Cl has decomposed (assuming the reaction goes to completion). n= PV 1.00 atm × (58.3 × 10 −3 L) = 2.20 × 10−3 mol N2 = 0.08206 L atm RT × 323 K K mol Because each mole of C6H5N2Cl that decomposes produces one mole of N2, the initial concentration (t = 0) of C6H5N2Cl was: 2.20 × 10 −3 mol = 0.0550 M 40.0 × 10 −3 L We can similarly calculate the moles of N2 evolved at each point of the experiment, subtract that from 2.20 × 10−3 mol to get the moles of C6H5N2Cl remaining, and then calculate [C6H5N2Cl] at each time. We would then use these results to make the appropriate graph to determine the order of the reaction. Because the rate constant is related to the slope of the straight line, we would favor this approach to get a value for the rate constant. There is a simpler way to check for the order of the reaction that saves doing a lot of math. The quantity (V∞ − Vt ), where V∞ = 58.3 mL N2 evolved and Vt = mL of N2 evolved at time t, will be proportional to the moles of C6H5N2Cl remaining; (V∞ − Vt ) will also be proportional to the concentration of C6H5N2Cl. Thus we can get the same information by using (V∞ − Vt ) as our measure of [C6H5N2Cl]. If the reaction is first order, a graph of ln(V∞ − Vt ) versus t would be linear. The data for such a graph are: t (s) Vt (mL) (V∞ − Vt ) ln(V∞ − Vt ) 0 6 9 14 22 30. 0 19.3 26.0 36.0 45.0 50.4 58.3 39.0 32.3 22.3 13.3 7.9 4.066 3.664 3.475 3.105 2.588 2.07 CHAPTER 15 CHEMICAL KINETICS 591 We can see from the graph that this plot is linear, so the reaction is first order. The differential rate law is −d[C6H5N2Cl]/dt = Rate = k[C6H5N2Cl], and the integrated rate law is ln[C6H5N2Cl] = −kt + ln[C6H5N2Cl]0. From separate data, k was determined to be 6.9 × 10−2 s−1. 39. For a first-order reaction, the integrated rate law is ln([A]/[A]0) = −kt. Solving for k: ⎛ 0.250 mol / L ⎞ ⎟⎟ = −k × 120. s, k = 0.0116 s−1 ln⎜⎜ 1 . 00 mol / L ⎝ ⎠ ⎛ 0.350 mol / L ⎞ ⎟⎟ = −0.0116 s−1 × t, t = 150. s ln⎜⎜ ⎝ 2.00 mol / L ⎠ 40. ⎛ [A] ⎞ ln 2 0.6931 ⎟⎟ = −kt; k = = = 4.85 × 10−2 d−1 ln⎜⎜ [ A ] t 14 . 3 d 1/ 2 0 ⎠ ⎝ If [A]0 = 100.0, then after 95.0% completion, [A] = 5.0. ⎛ 5.0 ⎞ 2 1 ln⎜ ⎟ = -4.85 × 10− d− × t, t = 62 days 100 . 0 ⎝ ⎠ 41. Comparing experiments 1 and 2, as the concentration of AB is doubled, the initial rate increases by a factor of 4. The reaction is second order in AB. Rate = k[AB]2, 3.20 × 10 −3 mol L−1 s −1 = k1(0.200 M)2 k = 8.00 × 10 −2 L mol −1 s −1 = kmean For a second-order reaction: 592 CHAPTER 15 t1/2 = 42. CHEMICAL KINETICS 1 1 = = 12.5 s −2 −1 −1 k[AB]0 8.00 × 10 L mol s × 1.00 mol / L a. The integrated rate law for a second order reaction is 1/[A] = kt + 1/[A]0, and the halflife expression is t1/2 = 1/k[A]0. We could use either to solve for t1/2. Using the integrated rate law: 0.555 L 1 1 1.11 L / mol = k H 2.00 s + , k= = (0.900/2) mol/L 0.900 mol / L 2.00 s mol s b. 43. 1 1 8.9 L / mol = 0.555 L mol−1 s−1 × t + , t= = 16 s 0.100 mol / L 0.900 mol / L 0.555 L mol −1 s −1 a. When a reaction is 75.0% complete (25.0% of reactant remains), this represents two halflives (100% → 50%→ 25%). The first-order half-life expression is t1/2 = (ln 2)/k. Because there is no concentration dependence for a first-order half-life, 320. s = two halflives, t1/2 = 320./2 = 160. s. This is both the first half-life, the second half-life, etc. b. t1/2 = ln 2 ln 2 ln 2 = = 4.33 × 10 −3 s −1 , k = k t 1/ 2 160. s At 90.0% complete, 10.0% of the original amount of the reactant remains, so [A] = 0.100[A]0. ⎛ [A] ⎞ 0.100[A]0 ln(0.100) ⎟⎟ = − kt , ln ln ⎜⎜ = − (4.33 × 10 −3 s −1 ) t , t = = 532 s [A]0 − 4.33 × 10 −3 s −1 ⎝ [A ]0 ⎠ 44. Successive half-lives increase in time for a second-order reaction. Therefore, assume reaction is second order in A. t1/2 = a. 1 1 1 = 1.0 L mol−1 min−1 , k= = k[ A]0 t1/ 2 [A]0 10.0 min(0.10 M ) 1 1 1.0 L 1 = kt + = 90. M −1, [A] = 1.1 × 10−2 M = H 80.0 min + [A] [A]0 mol min 0.10 M b. 30.0 min = 2 half-lives, so 25% of original A is remaining. [A] = 0.25(0.10 M) = 0.025 M 45. a. [A] = −kt + [A]0; If k = 5.0 × 10−2 mol L−1 s−1 and [A]0 = 1.00 × 10−3 M, then: [A] = −(5.0 × 10−2 mol L−1 s−1)t + 1.00 × 10−3 mol/L CHAPTER 15 b. CHEMICAL KINETICS 593 [A]0 = −(5.0 × 10−2)t1/2 + [A]0 because at t = t1/2, [A] = [A]o/2. 2 0.50(1.00 × 10 −3 ) −0.50[A]0 = −(5.0 × 10−2)t1/2, t1/2 = = 1.0 × 10−2 s −2 5.0 × 10 [A]0 Note: We could have used the t1/2 expression to solve (t1/2 = ). 2k c. [A] = −kt + [A]0 = −(5.0 × 10−2 mol L−1 s−1)(5.0 × 10−3 s) + 1.00 × 10−3 mol/L [A] = 7.5 × 10−4 mol/L [A]reacted = 1.00 × 10−3 mol/L − 7.5 × 10−4 mol/L = 2.5 × 10−4 mol/L [B]produced = [A]reacted = 2.5 × 10−4 M 46. Because [B]0 >> [A]0, the B concentration is essentially constant during this experiment, so rate = kN[A] where kN = k[B]2. For this experiment, the reaction is a pseudo-first-order reaction in A. a. ⎛ 3.8 × 10 −3 M ⎞ ⎛ [A] ⎞ −1 ⎟ ⎟⎟ = !kNt, ln ⎜ ln ⎜⎜ ⎜ 1.0 × 10 − 2 M ⎟ = !kN × 8.0 s, kN = 0.12 s [ A ] 0 ⎠ ⎝ ⎝ ⎠ For the reaction: kN = k[B]2, k = 0.12 s−1/(3.0 mol/L)2 = 1.3 × 10 −2 L2 mol−2 s−1 b. t1/2 = c. ln 2 0.693 = = 5.8 s k' 0.12 s −1 ⎛ [A] ln ⎜⎜ −2 ⎝ 1.0 × 10 M ⎞ [A] ⎟ = !0.12 s −1 × 13.0 s, = e−0.12(13.0) = 0.21 −2 ⎟ 1.0 × 10 ⎠ [A] = 2.1 × 10 −3 M d. [A] reacted = 0.010 M ! 0.0021 M = 0.008 M [C] reacted = 0.008 M × 2 mol C = 0.016 M ≈ 0.02 M 1 mol A [C]remaining = 2.0 M - 0.02 M = 2.0 M; as expected, the concentration of C basically remains constant during this experiment since [C]0 >> [A]0. 47. a. Because [A]0 << [B]0 or [C]0, the B and C concentrations remain constant at 1.00 M for this experiment. Thus, rate = k[A]2[B][C] = k′[A]2 where k′ = k[B][C]. 594 CHAPTER 15 CHEMICAL KINETICS For this pseudo-second-order reaction: 1 1 1 1 , = k′t + = k′(3.00 min) + −5 [A]0 3.26 × 10 M [A] 1.00 × 10 − 4 M k′ = 6890 L mol–1 min–1 = 115 L mol–1 s–1 k′ = k[B][C], k = k′ 115 L mol −1 s −1 = 115 L3 mol–3 s–1 , k = [B][C] (1.00 M )(1.00 M ) b. For this pseudo-second-order reaction: Rate = k′[A]2, t1/2 = c. 1 1 = = 87.0 s −1 −1 k '[A ]0 115 L mol s (1.00 × 10 − 4 mol / L) 1 1 1 = k′t + = 115 L mol–1 s–1 × 600. s + = 7.90 × 104 L/mol [A] [A]0 1.00 × 10 − 4 mol / L [A] = 1/7.90 × 104 L/mol = 1.27 × 10–5 mol/L From the stoichiometry in the balanced reaction, 1 mol of B reacts with every 3 mol of A. Amount A reacted = 1.00 × 10–4 M – 1.27 × 10–5 M = 8.7 × 10–5 M Amount B reacted = 8.7 × 10–5 mol/L × 1 mol B = 2.9 × 10–5 M 3 mol A [B] = 1.00 M − 2.9 × 10–5 M = 1.00 M As we mentioned in part a, the concentration of B (and C) remain constant because the A concentration is so small compared to the B (or C) concentration. Reaction Mechanisms 48. a. An elementary step (reaction) is one for which the rate law can be written from the molecularity, i.e., from coefficients in the balanced equation. b. The molecularity is the number of species that must collide to produce the reaction represented by an elementary step in a reaction mechanism. c. The mechanism of a reaction is the series of proposed elementary reactions that may occur to give the overall reaction. The sum of all the steps in the mechanism gives the balanced chemical reaction. CHAPTER 15 CHEMICAL KINETICS 595 d. An intermediate is a species that is neither a reactant nor a product but that is formed and consumed in the reaction sequence. e. The rate-determining step is the slowest elementary reaction in any given mechanism. 49. In a unimolecular reaction, a single reactant molecule decomposes to products. In a bimolecular reaction, two molecules collide to give products. The probability of the simultaneous collision of three molecules with enough energy and the proper orientation is very small, making termolecular steps very unlikely. 50. For a mechanism to be acceptable, the sum of the elementary steps must give the overall balanced equation for the reaction, and the mechanism must give a rate law that agrees with the experimentally determined rate law. A mechanism can never be proven absolutely. We can only say it is possibly correct if it follows the two requirements just described. Most reactions occur by a series of steps. If most reactions were one step, then all reactants would appear in the overall rate law, and these orders would be the coefficients in the balanced equation. This is not the case. 51. For elementary reactions, the rate law can be written using the coefficients in the balanced equation to determine the orders. a. Rate = k[CH3NC] b. Rate = k[O3][NO] c. Rate = k[O3] d. Rate = k[O3][O] e. Rate = k 52. [ C] or Rate = kN, where N = the number of 14 6 14 6C atoms (convention) Because the rate of the slowest elementary step equals the rate of a reaction: Rate = Rate of step 1 = k[NO2]2 The sum of all steps in a plausible mechanism must give the overall balanced reaction. Summing all steps gives: NO2 + NO2 → NO3 + NO NO3 + CO → NO2 + CO2 ___________________________________ NO2 + CO → NO + CO2 53. A mechanism consists of a series of elementary reactions in which the rate law for each step can be determined using the coefficients in the balanced equations. For a plausible mechanism, the rate law derived from a mechanism must agree with the rate law determined from experiment. To derive the rate law from the mechanism, the rate of the reaction is assumed to equal the rate of the slowest step in the mechanism. Because step 1 is the rate determining step, the rate law for this mechanism is Rate = k[C4H9Br]. To get the overall reaction, we sum all the individual steps of the mechanism. Summing all steps gives: 596 CHAPTER 15 CHEMICAL KINETICS C4H9Br → C4H9+ + Br− C4H9 + H2O → C4H9OH2+ C4H9OH2+ + H2O → C4H9OH + H3O+ + ___________________________________________________________ C4H9Br + 2 H2O → C4H9OH + Br- + H3O+ Intermediates in a mechanism are species that are neither reactants nor products but that are formed and consumed during the reaction sequence in the mechanism. The intermediates for this mechanism are C4H9+ and C4H9OH2+. 54. The rate law is Rate = k[NO]2[Cl2]. If we assume the first step is rate-determining, we would expect the rate law to be Rate = k1[NO][Cl2]. This isn't correct. However, if we assume the second step to be rate-determining, Rate = k2[NOCl2][NO]. To see if this agrees with experiment, we must substitute for the intermediate NOCl2 concentration. Assuming a fastequilibrium first step (rate reverse = rate forward): k-1[NOCl2] = k1[NO][Cl2], [NOCl2] = Rate = k1 [NO][Cl2]; substituting into the rate equation: k −1 k 2 k1 k k [NO]2[Cl2] = k[NO]2[Cl2] where k = 2 1 k −1 k −1 This is a possible mechanism with the second step the rate-determining step because the derived rate law agrees with the experimentally determined rate law. 55. Let's determine the rate law for each mechanism. If the rate law derived from the mechanism is the same as the experimental rate law, then the mechanism is possible (assuming the sum of all the steps in the mechanism gives the overall balanced equation). When deriving rate laws from a mechanism, we must substitute for all intermediate concentrations. a. Rate = k1[NO][O2]; not possible b. Rate = k2[NO3][NO] and k1[NO][O2] = k−1[NO3] or [NO3] = Rate = k 2 k1 [NO]2[O2]; k −1 c. Rate = k1[NO]2; possible not possible d. Rate = k2[N2O2] and [N2O2] = Rate = k1 [NO][O2] k −1 k1 [NO]2 k −1 k 2 k1 [NO]2 ; not possible k −1 Only the mechanism in b is consistent, so only mechanism b is a possible mechanism for this reaction. CHAPTER 15 56. CHEMICAL KINETICS 597 a. This rate law occurs when the first step is assumed rate determining. Rate = k1[Br2] = k′[Br2] b. This rate law occurs when the second step is assumed rate-determining and the first step is a fast- equilibrium step. Rate = k2[Br][H2]; from the fast-equilibrium first step (rate forward = rate reverse): k1[Br2] = k-1[Br]2, [Br] = (k1/k-1)1/2[Br2]1/2 Substituting into the rate equation: Rate = k2(k1/k-1)1/2[Br2]1/2[H2] = k′′[H2][Br2]1/2 c. From a, k′ = k1; from b, k′′ = k2(k1/k-1)1/2. 57. Rate = k3[Br−][H2BrO3+]; we must substitute for the intermediate concentration. Because steps 1 and 2 are fast-equilibrium steps, rate forward reaction = rate reverse reaction. k2[HBrO3][H+] = k-2[H2BrO3+]; k1[BrO3−][H+] = k-1[HBrO3] [HBrO3] = Rate = 58. k1 k k k [BrO3−][H+]; [H2BrO3+] = 2 [HBrO3][H+] = 2 1 [BrO3−][H+]2 k −1 k −2 k − 2 k −1 k 3 k 2 k1 [Br-][BrO3-][H+]2 = k[Br-][BrO3−][H+]2 k − 2 k −1 Rate = k[BrO3−][SO32−][H+]; first step: SO32− + H+ → HSO3− (fast) The rate law contains BrO3−, SO32− and H+. In order to incorporate all these ions into the rate law, the rate-determining step could contain BrO3− and HSO3− (the intermediate produced in the first step). A possible second step is: BrO3− + HSO3− → products (slow) A likely choice is: BrO3− + HSO3− → SO42− + HBrO2 (slow) Followed by: HBrO2 + SO32− → HBrO + SO42− (fast); HBrO + SO32− → SO42− + H+ + Br− (fast) All species in this mechanism (HSO3−, HBrO2, and HBrO) are known substances. This mechanism also gives the correct experimentally determined rate law. Rate = k2[BrO3−] [HSO3−]; assuming reaction one is a fast equilibrium step: k1[SO32−][H+] = k-1[HSO3−], [HSO3−] = k1 [SO32−][H+] k −1 598 CHAPTER 15 Rate = 59. k 2 k1 [BrO3−][SO32−][H+] = k[BrO3−][SO32−][H+] k −1 Rate = k2[I−][HOCl]; from the fast-equilibrium first step: k1[OCl−] = k-1[HOCl][OH−], [HOCl] = Rate = 60. CHEMICAL KINETICS k1[OCl − ] ; substituting into the rate equation: k −1 [OH − ] k 2 k1[I − ][OCl − ] k[I − ][OCl − ] = k −1[OH − ] [OH − ] a. Rate = k3[COCl][Cl2]; from the fast-equilibrium reactions 1 and 2: [COCl] k k = 2 , [COCl] = 2 [CO][Cl] [Cl][CO] k −2 k −2 1/ 2 ⎛ k ⎞ [Cl]2 k = 1 , [Cl] = ⎜⎜ 1 [Cl 2 ] ⎟⎟ [Cl 2 ] k −1 ⎝ k −1 ⎠ 1/ 2 k ⎛ k ⎞ Thus [COCl] = 2 ⎜⎜ 1 ⎟⎟ k − 2 ⎝ k −1 ⎠ [CO][Cl2]1/2; Substituting into rate law: 1/ 2 k ⎛ k ⎞ Rate = k 3 2 ⎜⎜ 1 ⎟⎟ k − 2 ⎝ k −1 ⎠ [CO][Cl2]3/2 = k[CO][Cl2]3/2 b. Cl and COCl are intermediates. 61. a. MoCl5− − b. Rate = d[ NO 2 ] = k2[NO3−][MoCl5−] dt (Only the last step contains NO2−.) We use the steady-state assumption to substitute for the intermediate concentration in the rate law. The steady-state approximation assumes that the concentration of an intermediate remains constant; i.e., d[intermediate]/dt = 0. To apply the steady-state assumption, we write rate laws for all steps where the intermediate is produced and equate the sum of these rate laws to the sum of the rate laws where the intermediate is consumed. Applying the steady-state approximation to MoCl5−: − d[MoCl 5 ] = 0, so k1[MoCl62−] = k-1[MoCl5−][Cl−] + k2[NO3−][MoCl5−] dt [MoCl5−] = 2− k1[MoCl 6 ] − − k −1[Cl − ] + k 2 [ NO 3 ] ; Rate = − 2− k k [ NO ][MoCl 6 ] d[ NO 2 ] = 1 2 − 3 − dt k −1[Cl ] + k 2 [ NO 3 ] CHAPTER 15 62. Rate = CHEMICAL KINETICS 599 − d[O 3 ] = k1[M][O3] + k2[O][O3] − k-1[M][O2][O]; apply steady-state approx. to O: dt d[O] = 0, so k1[M][O3] = k-1[M][O2][O] + k2[O][O3] dt k1[M][O3] − k-1[M][O2][O] = k2[O][O3] Substitute this expression into the rate law: Rate = − d[O 3 ] = 2k2[O][O3] dt Rearranging the steady-state approximation for [O]: [O] = Substituting into the rate law: Rate = 63. a. Rate = [B*] = k1[O 3 ][M ] k −1[M ][O 2 ] + k 2 [O 3 ] 2 k 2 k 1[O 3 ]2 [ M ] − d[O 3 ] = dt k −1[M ][O 2 ] + k 2 [O 3 ] d[E ] d[B*] = k2[B*]; assume = 0, then k1[B]2 = k-1[B][B*] + k2[B*]. dt dt k1[ B]2 k1k 2 [B]2 d[E ] ; the rate law is: Rate = = k −1[B] + k 2 k −1[B] + k 2 dt b. When k2 << k-1[B], then Rate = k k [ B]2 kk d[E ] = 1 2 [B]. = 1 2 dt k −1[B] k −1 The reaction is first order when the rate of the second step is very slow (when k2 is very small). c. Collisions between B molecules only transfer energy from one B to another. This occurs at a much faster rate than the decomposition of an energetic B molecule (B*). Temperature Dependence of Rate Constants and the Collision Model 64. a. The larger the activation energy, the slower is the rate. b. The higher the temperature, the more molecular collisions there are with sufficient energy to convert to products, and the faster is the rate. c. The greater the frequency of collisions, the greater are the opportunities for molecules to react, and hence the greater is the rate. d. For a reaction to occur, it is the reactive portion of each molecule that must be involved in a collision. Only some of all the possible collisions have the correct orientation to convert reactants to products. 600 65. CHAPTER 15 CHEMICAL KINETICS Two reasons are: 1) The collision must involve enough energy to produce the reaction; i.e., the collision energy must be equal to or exceed the activation energy. 2) The relative orientation of the reactants when they collide must allow formation of any new bonds necessary to produce products. 66. H3O+(aq) + OH−(aq) → 2 H2O(l) should have the faster rate. H3O+ and OH− will be electrostatically attracted to each other; Ce4+ and Hg22+ will repel each other (so Ea is much larger). 67. k = A exp(!Ea/RT) or ln k = − Ea + ln A (the Arrhenius equation) RT ⎛k ⎞ E ⎛1 1 ⎞ ⎟⎟ (Assuming A is temperature independent.) For two conditions: ln ⎜⎜ 2 ⎟⎟ = a ⎜⎜ − k R T T 2 ⎠ ⎝ 1⎠ ⎝ 1 Let k1 = 3.52 × 10 −7 L mol−1s−1, T1 = 555 K; k2 = ?, T2 = 645 K; Ea = 186 × 103 J/mol ⎛ k2 ln ⎜⎜ −7 ⎝ 3.52 × 10 ⎞ 1.86 × 105 J / mol ⎛ 1 1 ⎟= ⎟ 8.3145 J K −1 mol −1 ⎜⎜ 555 K − 645 K ⎝ ⎠ ⎞ ⎟⎟ = 5.6 ⎠ k2 = e5.6 = 270, k2 = 270(3.52 × 10 −7 ) = 9.5 × 10 −5 L mol−1 s−1 3.52 × 10 −7 68. ⎛ 1.7 × 10 −2 s −1 ⎞ ⎛k ⎞ E ⎛ 1 Ea 1 ⎞ ⎟= ln⎜⎜ 2 ⎟⎟ = a ⎜⎜ − ⎟⎟ ; ln⎜⎜ − 4 −1 ⎟ −1 −1 R ⎝ T1 T2 ⎠ ⎝ k1 ⎠ ⎝ 7.2 × 10 s ⎠ 8.3145 J K mol ⎛ 1 1 ⎞ ⎟⎟ ⎜⎜ − ⎝ 660. K 720. K ⎠ Ea = 2.1 × 105 J/mol For k at 325EC (598 K): 5 ⎛ 1.7 × 10 −2 s −1 ⎞ ⎟ = 2.1 × 10 J / mol ln⎜ ⎜ ⎟ 8.3145 J K −1 mol −1 k ⎝ ⎠ ⎛ 1 1 ⎞ ⎜⎜ ⎟⎟ , k = 1.3 × 10 −5 s −1 − ⎝ 598 K 720. K ⎠ For three half-lives, we go from 100% → 50% → 25% → 12.5%. After three half-lives, 12.5% of the original amount of C2H5I remains. Partial pressures are directly related to gas concentrations in mol/L: PC 2 H 5I = 894 torr × 0.125 = 112 torr after 3 half-lives CHAPTER 15 69. CHEMICAL KINETICS ⎛k ⎞ E ⎛ 1 1 ⎞ ln⎜⎜ 2 ⎟⎟ = a ⎜⎜ − ⎟⎟; R ⎝ T1 T2 ⎠ ⎝ k1 ⎠ ln(7.00) = 601 k2 = 7.00, T1 = 295 K, Ea = 54.0 × 103 J/mol k1 54.0 × 103 J / mol ⎛ 1 1 ⎞ ⎜ − ⎟⎟, −1 −1 ⎜ 8.3145 J K mol ⎝ 295 K T2 ⎠ 1 1 − = 3.00 × 10−4 295 K T2 1 = 3.09 × 10−3, T2 = 324 K = 51°C T2 70. ⎛k ⎞ E ln⎜⎜ 2 ⎟⎟ = a R ⎝ k1 ⎠ ln(2.00) = 71. ⎛1 1 ⎞ ⎜⎜ − ⎟⎟; because the rate doubles, assume k2 = 2k1. ⎝ T1 T2 ⎠ ⎛ 1 Ea 1 ⎞ ⎜ ⎟⎟, Ea = 5.3 × 104 J/mol = 53 kJ/mol − −1 −1 ⎜ 298 K 308 K 8.3145 J K mol ⎝ ⎠ From the Arrhenius equation in logarithmic form (ln k = −Ea/RT + ln A), a graph of ln k versus. 1/T should yield a straight line with a slope equal to −Ea/R and a y intercept equal to ln A. a. Slope = −Ea/R, Ea = 1.10 × 104 K × 8.3145 J = 9.15 × 104 J/mol = 91.5 kJ/mol K mol b. The units for A are the same as the units for k (s−1). y intercept = ln A, A = e33.5 = 3.54 × 1014 s−1 c. ln k = −Ea/RT + ln A or k = A exp(−Ea/RT) ⎛ − 9.15 × 10 −4 J / mol k = 3.54 × 1014 s−1 × exp⎜⎜ −1 −1 ⎝ 8.3145 J K mol × 298 K 72. ⎞ ⎟ = 3.24 × 10−2 s−1 ⎟ ⎠ A graph of ln k versus 1/T should be linear with slope = −Ea/R. T (K) 338 318 298 1/T (K−1) k (s−1) ln k 2.96 × 10−3 3.14 × 10−3 3.36 × 10−3 4.9 × 10−3 5.0 × 10−4 3.5 × 10−5 −5.32 −7.60 −10.26 602 CHAPTER 15 Slope = CHEMICAL KINETICS − 10.76 − (−5.85) = −1.2 × 104 K = −Ea/R −3 −3 3.40 × 10 − 3.00 × 10 Ea = −slope × R = 1.2 × 104 K × 8.3145 J K−1 mol−1, Ea = 1.0 × 105 J/mol = 1.0 × 102 kJ/mol 73. The Arrhenius equation is k = Aexp(−Ea/RT) or, in logarithmic form, ln k = −Ea/RT + ln A. Hence a graph of ln k versus 1/T should yield a straight line with a slope equal to -Ea/R since the logarithmic form of the Arrhenius equation is in the form of a straight-line equation, y = mx + b. Note: We carried one extra significant figure in the following ln k values in order to reduce round-off error. T (K) 1/T (K−1) k (L mol−1 s−1) 195 230. 260. 298 369 5.13 × 10−3 4.35 × 10−3 3.85 × 10−3 3.36 × 10−3 2.71 × 10−3 1.08 × 109 2.95 × 109 5.42 × 109 12.0 × 109 35.5 × 109 ln k 20.80 21.81 22.41 23.21 24.29 CHAPTER 15 CHEMICAL KINETICS 603 Using a couple of points on the plot: slope = − Ea 20.95 − 23.65 − 2.70 = = −1.35 × 103 K = −3 −3 −3 R 5.00 × 10 − 3.00 × 10 2.00 × 10 Ea = 1.35 × 103 K × 8.3145 J K−1 mol−1 = 1.12 × 104 J/mol = 11.2 kJ/mol From the best straight line (by calculator): slope = −1.43 × 103 K and Ea = 11.9 kJ/mol 74. When ΔE is positive, the products are at a higher energy relative to reactants, and when ΔE is negative, the products are at a lower energy relative to reactants. The reaction in part a will have the greatest rate because it has the smallest activation energy. 75. In the following reaction profiles R = reactants, P = products, Ea = activation energy, ΔE = overall energy change for the reaction, and RC = reaction coordinate, which is the same as reaction progress. The second reaction profile represents a two-step reaction since an intermediate plateau appears between the reactants and the products. This plateau (see I in plot) represents the energy of the intermediate. The general reaction mechanism for this reaction is: 604 CHAPTER 15 CHEMICAL KINETICS R →I I →P R→ P In a mechanism, the rate of the slowest step determines the rate of the reaction. The activation energy for the slowest step will be the largest energy barrier that the reaction must overcome. Since the second hump in the diagram is at the highest energy, the second step has the largest activation energy and will be the rate determining step (the slow step). 76. 125 kJ/mol R E Ea, reverse 216 kJ/mol P RC The activation energy for the reverse reaction is: Ea, 77. reverse = 216 kJ/mol + 125 kJ/mol = 341 kJ/mol When ΔE is positive (the products have higher energy than the reactants, as represented in the energy profile for Exercise 15.75), then Ea, forward > Ea, reverse. Therefore, this reaction has a positive ΔE value. Catalysis 78. A catalyst increases the rate of a reaction by providing reactants with an alternate pathway (mechanism) to convert to products. This alternate pathway has a lower activation energy, thus increasing the rate of the reaction. A heterogeneous catalyst is in a different phase from the reactants. The catalyst is usually a solid, although a catalyst in a liquid phase can act as a heterogeneous catalyst for some gasphase reactions. Since the catalyzed reaction has a different mechanism than the uncatalyzed reaction, the catalyzed reaction most likely will have a different rate law. 79. A metal-catalyzed reaction can be dependent on the number of adsorption sites on the metal surface. Once the metal surface is saturated with reactant, the rate of reaction becomes independent of concentration. 80. The slope of the ln k versus 1/T (K) plot is equal to !Ea/R. Because Ea for the catalyzed reaction will be smaller than Ea for the uncatalyzed reaction, the slope of the catalyzed plot should be less negative. 81. a. W because it has a lower activation energy than the Os catalyst. CHAPTER 15 CHEMICAL KINETICS 605 b. kw = Aw exp[!Ea(W)/RT]; kuncat = Auncat exp[!Ea(uncat)/RT]; assume Aw = Auncat. kw ⎛ − E a ( W ) E a (uncat ) ⎞ = exp ⎜ + ⎟ k uncat RT ⎝ RT ⎠ ⎛ − 163,000 J / mol + 335,000 J / mol ⎞ kw ⎟ = 1.41 × 1030 = exp ⎜⎜ −1 −1 ⎟ k uncat ⎠ ⎝ 8.3145 J K mol × 298 K The W-catalyzed reaction is approximately 1030 times faster than the uncatalyzed reaction. c. Because [H2] is in the denominator of the rate law, the presence of H2 decreases the rate of the reaction. For the decomposition to occur, NH3 molecules must be adsorbed on the surface of the catalyst. If H2 is also adsorbed on the catalyst surface, then there are fewer sites for NH3 molecules to be adsorbed, and the rate decreases. 82. a. NO is the catalyst. NO is present in the first step of the mechanism on the reactant side, but it is not a reactant since it is regenerated in the second step and does not appear in the overall balanced equation. b. NO2 is an intermediate. Intermediates also never appear in the overall balanced equation. In a mechanism, intermediates always appear first on the product side, whereas catalysts always appear first on the reactant side. c. k = A exp(−Ea/RT); k cat A exp [− E a (cat ) / RT ] ⎡ E (un ) − E a (cat ) ⎤ = = exp ⎢ a ⎥ k un A exp [ − E a (un ) / RT ] RT ⎣ ⎦ ⎞ 0.85 ⎛ k cat 2100 J / mol ⎟ = e = 2.3 = exp ⎜⎜ −1 −1 ⎟ k un ⎝ 8.3145 J K mol × 298 K ⎠ The catalyzed reaction is approximately 2.3 times faster than the uncatalyzed reaction at 25°C. 83. The mechanism for the chlorine catalyzed destruction of ozone is: O3 + Cl → O2 + ClO (slow) ClO + O → O2 + Cl (fast) _________________________ O3 + O → 2 O2 Because the chlorine atom-catalyzed reaction has a lower activation energy, the Cl-catalyzed rate is faster. Hence Cl is a more effective catalyst. Using the activation energy, we can estimate the efficiency that Cl atoms destroy ozone compared to NO molecules (see Exercise 15.82c). At 25°C: ⎡ (−2100 + 11,900) J / mol ⎤ k Cl ⎡ − E (Cl) E a ( NO) ⎤ 3.96 = exp ⎢ a = exp ⎢ + ⎥ = e = 52 ⎥ k NO RT ⎦ ⎣ RT ⎣ (8.3145 × 298) J / mol ⎦ 606 CHAPTER 15 CHEMICAL KINETICS At 25°C, the Cl catalyzed reaction is roughly 52 times faster than the NO-catalyzed reaction, assuming the frequency factor A is the same for each reaction. 84. The reaction at the surface of the catalyst follows the steps: H H D D H C C DCH2 H D D D CH2 D D CH2DCH2D(g) metal surface Thus CH2D‒CH2D should be the product. 85. At high [S], the enzyme is completely saturated with substrate. Once the enzyme is completely saturated, the rate of decomposition of ES can no longer increase, and the overall rate remains constant. 86. Assuming the catalyzed and uncatalyzed reactions have the same form and orders, and because concentrations are assumed equal, the rates will be equal when the k values are equal. k = A exp(!Ea/RT), kcat = kun when Ea,cat/RTcat = Ea,un/RTun 4.20 × 10 4 J / mol 7.00 × 10 4 J / mol = , Tun = 488 K = 215EC 8.3145 J K −1mol −1 × 293 K 8.3145 J K −1 mol −1 × Tun 87. Rate = − d[A] = k[A] x dt Assuming the catalyzed and uncatalyzed reaction have the same form and orders and because concentrations are assumed equal, rate ∝ 1/Δt, where Δt = Δtime. Rate cat Δt rate cat k 2400 yr = un = and = cat Rate un Δt cat Δt cat rate un k un A exp [− E (cat)/RT] Rate cat k ⎡ − E a (cat) + E a (un) ⎤ a = exp ⎢ = cat = ⎥ RT Rate un k un A exp [ − E (un)/RT] ⎣ ⎦ a ⎛ − 5.90 × 10 4 J / mol + 1.84 × 105 J / mol ⎞ k cat ⎟ = 7.62 × 1010 = exp⎜⎜ −1 −1 ⎟ k un 8.3145 J K mol × 600. K ⎝ ⎠ Δt un rate cat k = = cat , Δt cat rate un k un 2400 yr = 7.62 × 1010, Δtcat = 3.15 × 10 −8 yr ≈ 1 s Δt cat CHAPTER 15 CHEMICAL KINETICS 607 Additional Exercises 88. One experimental method to determine rate laws is the method of initial rates. Several experiments are carried out using different initial concentrations of reactants, and the initial rate is determined for each experiment. The results are then compared to see how the initial rate depends on the initial concentrations. This allows the orders in the rate law to be determined. The value of the rate constant is determined from the experiments once the orders are known. The second experimental method utilizes the fact that the integrated rate laws can be put in the form of a straight-line equation. Concentration versus time data are collected for a reactant as a reaction is run. This data are then manipulated and plotted to see which manipulation gives a straight line. From the straight-line plot we get the order of the reactant, and the slope of the line is mathematically related to k, the rate constant. 89. The most common method to experimentally determine the differential rate law is the method of initial rates. Once the differential rate law is determined experimentally, the integrated rate law can be derived. However, sometimes it is more convenient and more accurate to collect concentration versus time data for a reactant. When this is the case, then we do “proof” plots to determine the integrated rate law. Once the integrated rate law is determined, the differential rate law can be determined. Either experimental procedure allows determination of both the integrated and the differential rate law; and which rate law is determined by experiment and which is derived is usually decided by which data are easiest and most accurately collected. 90. For second order kinetics: a. 1 1 − = kt [A] [A]0 1 1 1 1 , = (0.250 L mol−1 s−1)t + = (0.250)(180. s) + [ A ]0 [ A ] [A] 1.00 × 10 − 2 M 1 = 145 M −1, [A] = 6.90 × 10−3 M [A] Amount of A that reacted = 0.0100 − 0.00690 = 0.0031 M [A2] = 1 (3.1 × 10−3 M) = 1.6 × 10−3 M 2 b. After 3 minutes (180. s): [A] = 3.00[B], 6.90 × 10−3 M = 3.00[B], [B] = 2.30 × 10−3 M 1 1 1 1 = k2t + ; , k2 = 2.19 L mol−1 s−1 = k2(180. s) + −2 [B]0 2.30 × 10 −3 M [B] 2.50 × 10 M c. [A]0 = 1.00 × 10−2 M; at t = t1/2, [A] = 5.00 × 10−3 M. 608 CHAPTER 15 CHEMICAL KINETICS 1 1 = (0.250)t + , t1/2 = 4.00 × 102 s −3 5.00 × 10 M 1.00 × 10 − 2 M Or we could have used the equation t1/2 = 1/k[A]0. 91. Heating Time (days) 0.00 1.00 2.00 3.00 6.00 Untreated Deacidifying Antioxidant s ln s s ln s s ln s 100.0 67.9 38.9 16.1 6.8 4.605 4.218 3.661 2.779 1.92 100.1 60.8 26.8 − − 4.606 4.108 3.288 − − 114.6 65.2 28.1 11.3 − 4.741 4.177 3.336 2.425 − a. We used a calculator to fit the data by least squares. The results follow. Untreated: ln s = −(0.465)t + 4.55, k = 0.465 d−1 Deacidifying agent: ln s = −(0.659)t + 4.66, k = 0.659 d−1 Antioxidant: ln s = −(0.779)t + 4.84, k = 0.779 d−1 b. No, the silk degrades more rapidly with the additives since k increases. c. t1/2 = (ln 2)/k; untreated: t1/2 = 1.49 day; deacidifying agent: t1/2 = 1.05 day; antioxidant: t1/2 = 0.890 day. 92. The pressure of a gas is directly proportional to concentration. Therefore, we can use the pressure data to solve the problem because −d[SO2Cl2]/dt is proportional to − dPSO 2Cl 2 /dt. CHAPTER 15 CHEMICAL KINETICS 609 SO2Cl2(g) → SO2(g) + Cl2(g); let P0 = initial partial pressure of SO2Cl2. If x = PSO 2 at some time, then x = PSO 2 = PCl2 and PSO 2Cl 2 = P0 − x. Ptotal = PSO 2Cl 2 + PSO 2 + PCl 2 = P0 – x + x + x, Ptotal = P0 + x, Ptotal − P0 = x At time = 0 hour, Ptotal = P0 = 4.93 atm. The data for other times are: Time (hour) 0.00 1.00 2.00 4.00 8.00 16.00 Ptotal (atm) 4.93 5.60 6.34 7.33 8.56 9.52 PSO 2Cl 2 (atm) 4.93 4.26 3.52 2.53 1.30 0.34 ln PSO 2Cl 2 1.595 1.449 1.258 0.928 0.262 −1.08 Because pressure of a gas is proportional to concentration, and because the ln PSO 2Cl 2 versus time plot is linear, the reaction is first order in SO2Cl2. a. Slope of ln(P) versus t plot is −0.168 h−1 = −k, k = 0.168 h−1 = 4.67 × 10−5 s−1; because concentration units don’t appear in first-order rate constants, this value of k determined from pressure data will be the same as if concentration data in molarity units were used. b. t1/2 = ln 2 0.6931 0.6931 = = = 4.13 h k k 0.168 h −1 c. After 0.500 h: ln PSO 2Cl2 = −kt + ln P0 = −(0.168 h−1)(0.500 h) + ln 4.93 ln PSO 2Cl2 = −0.0840 + 1.595 = 1.511, PSO 2Cl 2 = e1.511 = 4.53 atm 610 CHAPTER 15 CHEMICAL KINETICS PCl 2 = PSO 2 = 4.93 atm − 4.53 atm = 0.40 atm Ptotal = PSO 2Cl2 + PCl2 + PSO 2 = 4.53 + 0.40 + 0.40 = 5.33 atm After 12.0 hours: ln PSO 2Cl 2 = −(0.168 h−1)(12.0 h) + ln(4.93) = −0.42 PSO 2Cl 2 = e−0.42 = 0.66 atm, PSO 2 = 4.93 − 0.66 = 4.27 atm, PCl 2 = 4.27 atm Ptotal = 0.66 + 4.27 + 4.27 = 9.20 atm d. ⎛ PSO 2Cl 2 ln⎜⎜ ⎝ P0 ⎞ ⎟ = −0.168 h−1(20.0 h) = −3.36, ⎟ ⎠ ⎛ PSO 2Cl 2 ⎜ ⎜ P 0 ⎝ ⎞ −3.36 ⎟= e = 3.47 × 10−2 ⎟ ⎠ Fraction left = 0.0347 = 3.47% 93. Carbon cannot form the fifth bond necessary for the transition state because of the small atomic size of carbon and because carbon doesn’t have low-energy d orbitals available to expand the octet. 94. a. Let P0 = initial partial pressure of C2H5OH = 250. torr. If x torr of C2H5OH reacts, then at anytime: PC 2 H 5OH = 250. − x, PC 2 H 4 = PH 2O = x; Ptotal = 250. − x + x + x = 250. + x Therefore, PC 2 H 5OH at any time can be calculated from the data by determining x (= Ptotal − 250.) and then subtracting from 250. torr. Using the PC 2 H 5OH data, a plot of PC 2 H 5OH versus t is linear (plot not included). The reaction is zero order in PC 2 H 5OH . One could also use the Ptotal versus t data since Ptotal increases at the same rate that PC 2 H 5OH decreases. Note: The ln P versus t plot is also linear. The reaction hasn't been followed for enough time for curvature to be seen. However, since PC 2 H 5OH decreases at steady increments of 15 torr for every 10. s of reaction, then we can conclude that the reaction is zero order in C2H5OH. 15 From the data, the integrated rate equation involving Ptotal is Ptotal = t + 250. 10 15 At t = 80. s: Ptotal = (80.) + 250. = 370 torr 10 b. Slope = −k = −15 torr/10. s = 1.5 torr/s (from PC 2 H 5OH versus t plot), k = 1.5 torr/s; the Ptotal versus t plot would give the same rate constant since Ptotal increases at the same rate that PC 2 H 5OH decreases. c. Zero order CHAPTER 15 CHEMICAL KINETICS d. Ptotal = 611 15 (300.) + 250. = 7.0 × 102 torr; this is an impossible answer! 10 Because only 250. torr of C2H5OH are present initially, the maximum pressure can only be 500. torr when all of the C2H5OH is consumed, i.e., Ptotal = PC 2 H 4 + PH 2O = 250. + 250. = 500. torr. Therefore, at 300. s, Ptotal = 500. torr. 95. a. t (s) 0 1000. 2000. 3000. [C4H6] (M) ln[C4H6] 1/[C4H6] (M −1) 0.01000 0.00629 0.00459 0.00361 −4.6052 −5.069 −5.384 −5.624 1.000 × 102 1.59 × 102 2.18 × 102 2.77 × 102 The plot of 1/[C4H6] versus t is linear, thus the reaction is second order in butadiene. From the plot (not included), the integrated rate law is: 1 = (5.90 × 10−2 L mol−1 s−1)t + 100.0 M −1 [C 4 H 6 ] b. When dimerization is 1.0% complete, 99.0% of C4H6 is left. [C4H6] = 0.990(0.01000) = 0.00990 M; 1 = (5.90 × 10−2)t + 100.0 0.00990 t = 17.1 s ≈ 20 s c. 2.0% complete, [C4H6] = 0.00980 M; 1 = (5.90 × 10−2)t + 100.0, 0.00980 t = 34.6 s ≈ 30 s d. 1 1 = kt + ; [C4H6]0 = 0.0200 M; at t = t1/2, [C4H6] = 0.0100 M. [C 4 H 6 ] [ C 4 H 6 ]0 1 1 = (5.90 × 10−2)t1/2 + , t1/2 = 847 s = 850 s 0.0100 0.0200 Or: t1/2 = 1 1 = = 847 s −2 −1 −1 k[ A]0 (5.90 × 10 L mol s )(2.00 × 10 − 2 M ) e. From Exercise 15.32, k = 1.4 × 10−2 L mol−1 s−1 at 500. K. From this problem, k = 5.90 × 10−2 L mol−1 s−1 at 620. K. ⎛ 5.90 × 10 −2 ⎞ ⎛k ⎞ E ⎛ 1 ⎛ 1 Ea 1 ⎞ 1 ⎞ ⎟= ⎟⎟ ⎜ ln⎜⎜ 2 ⎟⎟ = a ⎜⎜ − ⎟⎟, ln⎜⎜ − −2 ⎟ −1 −1 ⎜ R ⎝ T1 T2 ⎠ ⎝ k1 ⎠ ⎝ 1.4 × 10 ⎠ 8.3145 J K mol ⎝ 500. K 620. K ⎠ 12 = Ea(3.9 × 10−4), Ea = 3.1 × 104 J/mol = 31 kJ/mol 612 96. CHAPTER 15 ⎛k ⎞ E We need the value of k at 500. K; ln⎜⎜ 2 ⎟⎟ = a R ⎝ k1 ⎠ ⎛ k2 ln⎜⎜ −12 −1 −1 ⎝ 2.3 × 10 L mol s CHEMICAL KINETICS ⎛1 1 ⎞ ⎜⎜ − ⎟⎟ ⎝ T1 T2 ⎠ ⎞ 1.11 × 105 J / mol ⎛ 1 1 ⎞ ⎟= ⎟ 8.3145 J K −1 mol −1 ⎜⎜ 273 K − 500 K ⎟⎟ = 22.2 ⎠ ⎝ ⎠ k2 = e22.2, k2 = 1.0 × 10−2 L mol−1 s−1 −12 2.3 × 10 Because the decomposition reaction is an elementary reaction, the rate law can be written using the coefficients in the balanced equation. For this reaction, Rate = k[NO2]2. To solve for the time, we must use the integrated rate law for second-order kinetics. The major problem now is converting units so they match. Rearranging the ideal gas law gives n/V = P/RT. Substituting P/RT for concentration units in the second-order integrated rate equation: 1 1 1 1 RT RT RT ⎛ P0 − P ⎜ , , = kt + = kt + = kt , t = − [ NO 2 ] [ NO 2 ]0 P / RT P0 / RT P P0 k ⎜⎝ P × P0 t= 97. ⎞ ⎟⎟ ⎠ (0.08206 L atm K −1 mol −1 )(500. K ) ⎛ 2.5 atm − 1.5 atm ⎞ ⎟⎟ = 1.1 × 103 s × ⎜⎜ 1 . 5 atm 2 . 5 atm × 1.0 × 10 − 2 L mol −1 s −1 ⎠ ⎝ k = A exp(−Ea/RT); + E a , uncat A cat exp (− E a , cat / RT ) ⎛−E k cat = = exp⎜⎜ a , cat k uncat A uncat exp (− E a , uncat / RT ) RT ⎝ ⎞ ⎟ ⎟ ⎠ ⎛ − E a , cat + 5.00 × 10 4 J / mol ⎞ k cat ⎟ 2.50 × 10 = = exp ⎜ ⎜ 8.3145 J K −1 mol −1 × 310. K ⎟ k uncat ⎠ ⎝ 3 ln(2.50 × 103) × 2.58 × 103 J/mol = −Ea, cat + 5.00 × 104 J/mol E a , cat = 5.00 × 104 J/mol − 2.02 × 104 J/mol = 2.98 × 104 J/mol = 29.8 kJ/mol 98. From 338 K data, a plot of ln[N2O5] versus t is linear (plot not included). The integrated rate law is: ln[N2O5] = −(4.86 × 10−3)t − 2.30; k = 4.86 × 10−3 s−1 at 338 K From 318 K data: ln[N2O5] = −(4.98 × 10−4)t − 2.30; k = 4.98 × 10−4 s−1 at 318 K ⎛ 4.86 × 10 −3 ⎞ ⎛k ⎞ E ⎛ 1 ⎛ 1 Ea 1 ⎞ 1 ⎞ ⎟= ⎟⎟ ⎜ ln⎜⎜ 2 ⎟⎟ = a ⎜⎜ − ⎟⎟, ln⎜⎜ − −4 ⎟ −1 −1 ⎜ R ⎝ T1 T2 ⎠ ⎝ k1 ⎠ ⎝ 4.98 × 10 ⎠ 8.3145 J K mol ⎝ 318 K 338 K ⎠ Ea = 1.0 × 105 J/mol = 1.0 × 102 kJ/mol CHAPTER 15 99. a. CHEMICAL KINETICS 1/T (K−1) T (K) 298.2 293.5 290.5 613 k (min−1) −3 3.353 × 10 3.407 × 10−3 3.442 × 10−3 ln k 178 126 100. 5.182 4.836 4.605 A plot of ln k versus 1/T gives a straight line (plot not included). The equation for the straight line is: ln k = -6.48 × 103(1/T) + 26.9 For the ln k versus 1/T plot, slope = −Ea/R = −6.48 × 103 K. −6.48 × 103 K = −Ea/8.3145 J K−1 mol−1, Ea = 5.39 × 104 J/mol = 53.9 kJ/mol b. ln k = −6.48 × 103(1/288.2) + 26.9 = 4.42, k = e4.42 = 83 min−1 About 83 chirps per minute per insect. Note: We carried extra significant figures. c. k gives the number of chirps per minute. The number or chirps in 15 s is k/4. T (°C) T (°F) 25.0 20.3 17.3 15.0 77.0 68.5 63.1 59.0 k (min−1) 42 + 0.80(k/4) 178 126 100. 83 78° F 67°F 62°F 59°F The rule of thumb appears to be fairly accurate, almost ±1°F. 100. a. If the interval between flashes is 16.3 s, then the rate is: 1 flash/16.3 s = 6.13 × 10−2 s−1 = k Interval 16.3 s 13.0 s ⎛k ⎞ E ln⎜⎜ 2 ⎟⎟ = a R ⎝ k1 ⎠ b. k T −2 −1 6.13 × 10 s 7.69 × 10−2 s−1 21.0°C (294.2 K) 27.8°C (301.0 K) ⎛1 1 ⎞ ⎜⎜ − ⎟⎟ ; solving: Ea = 2.5 × 104 J/mol = 25 kJ/mol ⎝ T1 T2 ⎠ ⎞ ⎛ ⎞ k 2.5 × 10 4 J / mol ⎛ 1 1 ⎟ ⎟⎟ = 0.30 ⎜ ln⎜⎜ = − −2 ⎟ −1 −1 ⎜ ⎝ 6.13 × 10 ⎠ 8.3145 J K mol ⎝ 294.2 K 303.2 K ⎠ k = e0.30 × 6.13 × 10−2 = 8.3 × 10−2 s−1; interval = 1/k = 12 seconds. 614 CHAPTER 15 c. T Interval 21.0 °C 27.8 °C 30.0 °C 16.3 s 13.0 s 12 s CHEMICAL KINETICS 54-2(Intervals) 21 °C 28 °C 30. °C This rule of thumb gives excellent agreement to two significant figures. 101. Rate = d[Cl 2 ] d[Cl] = k2[NO2Cl][Cl]; Assume = 0, then: dt dt k1[NO2Cl] = k-1[NO2][Cl] + k2[NO2Cl][Cl], [Cl] = 102. k1[ NO 2 Cl] k −1[ NO 2 ] + k 2 [ NO 2 Cl] Rate = d[Cl 2 ] k1k 2 [ NO 2 Cl]2 = dt k −1[ NO 2 ] + k 2 [ NO 2 Cl] Rate = d[P] d[ ES] = k2[ES]; apply the steady-state approximation to ES, = 0. dt dt k1[E][S] = k-1[ES] + k2[ES]; [E]T = [E] + [ES], so [E] = [E]T − [ES] Substituting: k1[S]([E]T − [ES]) = (k-1 + k2)[ES], k1[E]T[S] = (k-1 + k2 + k1[S]) [ES] [ES] = k1[E ]T [S] ; substituting into the rate equation, Rate = k2[ES]: k −1 + k 2 + k1[S] Rate = k1k 2 [E ]T [S] d[P] = dt k −1 + k 2 + k1[S] Challenge Problems 103. a. Rate = k[CH3X]x[Y]y; for experiment 1, [Y] is in large excess, so its concentration will be constant. Rate = k′[CH3X]x, where k′ = k(3.0 M)y. A plot (not included) of ln[CH3X] versus t is linear (x = 1). The integrated rate law is: ln[CH3X] = −(0.93)t − 3.99; k′ = 0.93 h−1 For experiment 2, [Y] is again constant, with Rate = k′′ [CH3X]x, where k′′ = k(4.5 M)y. The natural log plot is linear again with an integrated rate law: ln[CH3X] = −(0.93)t − 5.40; k′′ = 0.93 h−1 Dividing the rate-constant values: k' 0.93 k (3.0) y = = , 1.0 = (0.67)y, y = 0 y k" 0.93 k (4.5) CHAPTER 15 CHEMICAL KINETICS 615 Reaction is first order in CH3X and zero order in Y. The overall rate law is: Rate = k[CH3X], where k = 0.93 h−1 at 25°C b. t1/2 = (ln 2)/k = 0.6931/(7.88 × 108 h−1) = 8.80 × 10−10 hour c. ⎛ 7.88 × 108 ⎞ ⎛k ⎞ E ⎛ 1 ⎛ 1 Ea 1 ⎞ 1 ⎞ ⎟= ⎟⎟ ⎜ ln⎜⎜ 2 ⎟⎟ = a ⎜⎜ − ⎟⎟ , ln⎜⎜ − −1 −1 ⎜ ⎟ R ⎝ T1 T2 ⎠ ⎝ k1 ⎠ ⎝ 0.93 ⎠ 8.3145 J K mol ⎝ 298 K 358 K ⎠ Ea = 3.0 × 105 J/mol = 3.0 × 102 kJ/mol d. From part a, the reaction is first order in CH3X and zero order in Y. From part c, the activation energy is close to the C-X bond energy. A plausible mechanism that explains the results in parts a and c is: CH3X → CH3 + X (slow) CH3 + Y → CH3Y (fast) Note: This is a possible mechanism because the derived rate law is the same as the experimental rate law (and the sum of the steps gives the overall balanced equation). 104. a. [B] >> [A], so [B] can be considered constant over the experiments. This gives us a pseudo-order rate-law equation. b. Note that in each case the half-life doubles as time increases (in expetiment. 1 the first half-life is 40. s, the second half-life is 80. s; in experiment 2, the first half- life is 20. s, the second half-life is 40. s). This occurs only for a second-order reaction, so the reaction is second order in [A]. Between expt. 1 and expt. 2, we double [B] and the reaction rate doubles, thus it is first order in [B]. The overall rate-law equation is rate = k[A]2[B]. Using t1/2 = 1 1 , we get k = = 0.25 L mol−1 s−1; but this is actually k[A]0 (40.)(10.0 × 10 − 2 ) k′ where Rate = k′[A]2 and k′ = k[B]. k= c. i. k′ 0.25 = 0.050 L2 mol−2 s−1 = [B] 5 .0 This mechanism gives the wrong stoichiometry, so it can’t be correct. ii. Rate = k[E][A] k1[A][B] = k-1[E]; [E] = k 1 [A ][B] k −1 ; Rate = k k1 k −1 [A ]2 [B] 616 CHAPTER 15 CHEMICAL KINETICS This mechanism gives the correct stoichiometry and gives the correct rate law when it is derived from the mechanism. This is a possible mechanism for this reaction. iii. Rate = k[A]2 This mechanism gives the wrong derived rate law, so it can’t be correct. Only mechanism ii is possible. 105. [ A ]t − d[A] = k[A]3, dt n ∫ x dx = ∫ [ A ]0 t d[A] = − ∫ k dt [A ]3 0 xn +1 1 ; so: − n +1 2[A ]2 [ A ]t = − kt , − [ A ]0 1 1 + = − kt 2 2[A]t 2[A ]02 For the half-life equation, [A]t = 1/2[A]0: − − 1 ⎛1 ⎞ 2⎜⎜ [A ]0 ⎟⎟ ⎝2 ⎠ 3 2[A ]02 2 + 1 4 1 = − kt1/ 2 , − + = − kt1/ 2 2 2 2[A ]0 2[ A]02 2[A ]0 = − kt1/ 2 , t1/2 = 3 2[A ]02 k The first half-life is t1/2 = 40. s and corresponds to going from [A]0 to 1/2[A]0. The second half-life corresponds to going from 1/2 [A]0 to 1/4 [A]0 . First half-life = 3 ; second half-life = 2[A ]02 k 3 2 ⎛1 ⎞ 2⎜ [A]0 ⎟ k 2 ⎝ ⎠ = 6 [A]02 k 3 2[A]02 k First half − life = = 3/12 = 1/4 6 Second half − life [A ]02 k Because the first half-life is 40. s, the second half-life will be one-fourth of this, or 10. s. 106. a. Rate = k2[A][M]; assuming a fast equilibrium first step: k1[A][B] = k-1[M], [M] = Rate = k 2 k1 [A]2 [B] k −1 k1[A ][B] k −1 CHAPTER 15 CHEMICAL KINETICS 617 b. Rate = k2[A][M]; applying the steady-state approximation to M: k1[A][B] = k-1[M] + k2[A][M], [M] = Rate = k1[ A][B] k −1 + k 2 [A ] k 2 k1[A]2 [B] k −1 + k 2 [A ] c. If [A] << [B], the rate law from part b reduces to law as in part a. 107. k 2 k1 [A]2[B]. This is the same rate k −1 Rate = k[A]x[B]y[C]z; during the course of experiment 1, [A] and [C] are essentially constant, and Rate = k′[B]y , where k′ = k[A]0x [C]0z . [B] (M) 1.0 × 10−3 2.7 × 10−4 1.6 × 10−4 1.1 × 10−4 8.5 × 10−5 6.9 × 10−5 5.8 × 10−5 time (s) ln[B] 1/[B] (M −1) 0 1.0 × 105 2.0 × 105 3.0 × 105 4.0 × 105 5.0 × 105 6.0 × 105 −6.91 −8.22 −8.74 −9.12 −9.37 −9.58 −9.76 1.0 × 103 3.7 × 103 6.3 × 103 9.1 × 103 12 × 103 14 × 103 17 × 103 A plot of 1/[B] versus t is linear (plot not included). The reaction is second order in B, and the integrated rate equation is: 1/[B] = (2.7 × 10−2 L mol −1 s−1)t + 1.0 × 103 M −1; k′ = 2.7 × 10−2 L mol−1 s−1 For experiment 2, [B] and [C] are essentially constant, and Rate = k′′[A]x, where k′′ = k[B]0y [C]0z = k[B]02 [C]0z . [A] (M) time (s) ln[A] 1/[A] (M −1) 1.0 × 10−2 8.9 × 10−3 7.1 × 10−3 5.5 × 10−3 3.8 × 10−3 2.9 × 10−3 2.0 × 10−3 0 1.0 3.0 5.0 8.0 10.0 13.0 −4.61 −4.72 −4.95 −5.20 −5.57 −5.84 −6.21 1.0 × 102 110 140 180 260 340 5.0 × 102 A plot of ln[A] versus t is linear. The reaction is first order in A, and the integrated rate law is: ln[A] = −(0.123 s−1)t − 4.61; k′′ = 0.123 s−1 618 CHAPTER 15 CHEMICAL KINETICS Note: We will carry an extra significant figure in k′′. Experiment 3: [A] and [B] are constant; Rate = k′′′[C]z The plot of [C] versus t is linear. Thus z = 0. The overall rate law is Rate = k[A][B]2. From experiment 1 (to determine k): k′ = 2.7 × 10−2 L mol−1 s−1 = k[A]0x [C]0z = k[A]o = k(2.0 M), k = 1.4 × 10−2 L2 mol−2 s−1 From experiment 2: k′′ = 0.123 s−1 = k[B]02 , k = 0.123 s −1 = 1.4 × 10−2 L2 mol−2 s−1 2 (3.0 M ) Thus Rate = k[A][B]2 and k = 1.4 × 10−2 L2 mol−2 s−1. 108. First, we need to convert the data from total pressure versus time to pressure of O2 versus time. 3 O2 → Before 1.000 atm Change −3x After 1.000 − 3x 2 O3 0 +2x 2x Ptotal = 1.000 − 3x + 2x = 1.000 − x ; we use this equation to convert the data. For example, at t = 46.89 s, where Ptotal = 0.9500 atm, 1.000 − x = 0.9500 x = 0.0500 (carrying extra sig. fig.) So PO 2 = 1.000 − 3(0.0500) = 0.850 atm. Time 0s 46.89 s 98.82 s 137.9 s 200.0 s 286.9 s 337.9 s 511.3 s The following table contains the other values. PO 2 1.000 atm 0.850 atm 0.710 atm 0.620 atm 0.500 atm 0.370 atm 0.310 atm 0.170 atm A graph of ln PO 2 versus t gives a straight line (plot not included) with a slope of −0.00347. a. Rate = k[O2] or Rate = k(PO 2 ) b. k = 0.00347 s−1 CHAPTER 15 CHEMICAL KINETICS 619 c. When Ptotal = 0.7133 atm: 1.000 – x = 0.7133, x = 0.2867 atm (extra sig. fig.) PO 2 = 1.000 atm – 3(0.2867 atm) = 0.140 atm ln(P/P0) = −kt, ln(0.140/1.000) = –(0.00347 s−1)t, t = 567 s 109. Rate = − d[ N 2 O 5 ] = k1[M][N2O5] − k-1[NO3][NO2][M] dt Assume d[NO3]/dt = 0, so k1[N2O5][M] = k-1[NO3][NO2][M] + k2[NO3][NO2] + k3[NO3][NO]. [NO3] = k1[ N 2 O 5 ][M ] k −1[ NO 2 ][M ] + k 2 [ NO 2 ] + k 3 [ NO] Assume k d[ NO] = 0, so k2[NO3][NO2] = k3[NO3][NO], [NO] = 2 [NO2]. dt k3 k1[ N 2 O 5 ][M ] Substituting: [NO3] = k −1[ NO 2 ][M ] + k 2 [ NO 2 ] + k 2k 3 [ NO 2 ] k3 = k1[ N 2 O 5 ][M ] [ NO 2 ](k −1[M ] + 2k 2 ) Solving for the rate law: Rate = k k [ NO 2 ][ N 2 O 5 ][M ]2 − d[ N 2 O 5 ] = k1[N2O5][M] − −1 1 = k1[N2O5][M] dt [ NO 2 ](k −1[ M ] + 2k 2 ) − 110. Rate = ⎛ − d[ N 2 O 5 ] k −1k1[ M ] ⎞ ⎟ [N2O5][M]; simplifying: = ⎜⎜ k1 − dt k −1[M ] + 2k 2 ⎟⎠ ⎝ Rate = − d[ N 2 O 5 ] 2k1k 2 [M ][ N 2 O 5 ] = dt k −1[M ] + 2k 2 k −1k1[M ]2 [ N 2 O 5 ] k −1[M ] + 2k 2 ⎛k ⎞ E ⎛ 1 rate 2 k 1 ⎞ ln ⎜⎜ 2 ⎟⎟ = a ⎜⎜ − ⎟⎟ ; assuming = 2 = 40.0 : k R T T rate k1 2 ⎠ 1 ⎝ 1⎠ ⎝ 1 ln(40.0) = ⎛ 1 Ea 1 ⎞ ⎜ ⎟ , Ea = 1.55 × 105 J/mol = 155 kJ/mol − −1 −1 ⎜ 8.3145 J K mol ⎝ 308 K 328 K ⎟⎠ (carrying an extra sig. fig.) 620 CHAPTER 15 CHEMICAL KINETICS Note that the activation energy is close to the F2 bond energy. Therefore, the ratedetermining step probably involves breaking the F2 bond. H2(g) + F2(g) → 2 HF(g); for every two moles of HF produced, only one mole of the reactant is used up. Therefore, to convert the data to Preactant versus time, Preactant = 1.00 atm – (1/2)PHF. Preactant Time 1.000 atm 0.850 atm 0.700 atm 0.550 atm 0.400 atm 0.250 atm 0 min 30.0 min 65.8 min 110.4 min 169.1 min 255.9 min The plot of ln Preactant versus time (plot not included) is linear with negative slope, so the reaction is first order with respect to the limiting reagent. For the reactant in excess, because the values of the rate constant are the same for both experiments, one can conclude that the reaction is zero order in the excess reactant. a. For a three-step reaction with the first step limiting, the energy-level diagram could be: E R P Reaction coordinate Note that the heights of the second and third humps must be lower than the first step activation energy. However, the height of the third hump could be higher than the second hump. One cannot determine this absolutely from the information in the problem. b. We know the reaction has a slow first step, and the calculated activation energy indicates that the rate-determining step involves breaking the F2 bond. The reaction is also first order in one of the reactants and zero order in the other reactant. All this points to F2 being the limiting reagent. The reaction is first order in F2, and the rate-determining step in the mechanism is F2 → 2 F. Possible second and third steps to complete the mechanism follow. CHAPTER 15 CHEMICAL KINETICS F2 → 2 F F + H2 → HF + H H + F → HF 621 slow fast fast _____________________________________ F2 + H2 → 2 HF c. F2 was the limiting reactant. 111. a. Rate = (k1 + k2[H+])[I−]m[H2O2]n In all the experiments the concentration of H2O2 is small compared to the concentrations of I− and H+. Therefore, the concentrations of I− and H+ are effectively constant, and the rate law reduces to: Rate = kobs[H2O2]n, where kobs = (k1 + k2[H+])[I−]m Because all plots of ln[H2O2] versus time are linear, the reaction is first order with respect to H2O2 (n = 1). The slopes of the ln[H2O2] versus time plots equal −kobs, which equals −(k1 + k2[H+])[I−]m. To determine the order of I−, compare the slopes of two experiments in which I− changes and H+ is constant. Comparing the first two experiments: − [ k1 + k 2 (0.0400 M )] (0.3000 M ) m slope (exp . 2) − 0.360 = = slope (exp . 1) − 0.120 − [ k1 + k 2 (0.0400 M )] (0.1000 M ) m ⎛ 0.3000 M 3.00 = ⎜⎜ ⎝ 0.1000 M m ⎞ ⎟⎟ = (3.000)m, m = 1 ⎠ The reaction is also first order with respect to I-. b. The slope equation has two unknowns, k1 and k2. To solve for k1 and k2, we must have two equations. We need to take one of the first set of three experiments and one of the second set of three experiments to generate the two equations in k1 and k2. Experiment 1: slope = −(k1 + k2[H+])[I−] −0.120 min−1 = −[k1 + k2(0.0400 M)](0.1000 M) or 1.20 = k1 + k2(0.0400) Experiment 4: −0.0760 min-1 = −[k1 + k2(0.0200 M)](0.0750 M) or 1.01 = k1 + k2(0.0200) Subtracting 4 from 1: 1.20 = k1 + k2(0.0400) −1.01 = �