Transcript
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Complete Solutions Manual
GENERAL CHEMISTRY NINTH EDITION
Ebbing/Gammon
David Bookin Mt. San Jacinto College
Darrell D. Ebbing Wayne State University, Emeritus
Steven D. Gammon Western Washington University
HOUGHTON MIFFLIN COMPANY BOSTON
NEW YORK
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ISBN 13: 9780618952113 ISBN 10: 061895211X
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Contents PREFACE................................................................................................................................................ XI CHAPTER 1 – CHEMISTRY AND MEASUREMENT ................................................................................. 1 Solutions To Exercises ....................................................................................................................... 1 Answers To Concept Checks .............................................................................................................. 3 Answers To SelfAssessment And Review Questions .......................................................................... 4 Answers To Concept Explorations ..................................................................................................... 7 Answers To Conceptual Problems ..................................................................................................... 8 Solutions To Practice Problems ....................................................................................................... 11 Solutions To General Problems ....................................................................................................... 20 Solutions To Strategy Problems ....................................................................................................... 29 Solutions To CumulativeSkills Problems ........................................................................................ 30 CHAPTER 2 – ATOMS, MOLECULES, AND IONS ................................................................................. 36 Solutions To Exercises ..................................................................................................................... 36 Answers To Concept Checks ............................................................................................................ 38 Answers To SelfAssessment And Review Questions ........................................................................ 39 Answers To Concept Explorations ................................................................................................... 42 Answers To Conceptual Problems ................................................................................................... 43 Solutions To Practice Problems ....................................................................................................... 45 Solutions To General Problems ....................................................................................................... 56 Solutions To Strategy Problems ....................................................................................................... 63 Solutions To CumulativeSkills Problems ........................................................................................ 64 CHAPTER 3 – CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS ................................. 66 Solutions To Exercises ..................................................................................................................... 66 Answers To Concept Checks ............................................................................................................ 71 Answers To SelfAssessment And Review Questions ........................................................................ 73 Answers To Concept Explorations ................................................................................................... 75 Answers To Conceptual Problems ................................................................................................... 77 Solutions To Practice Problems ....................................................................................................... 79 Solutions To General Problems ..................................................................................................... 103 Solutions To Strategy Problems ..................................................................................................... 109 Solutions To CumulativeSkills Problems ...................................................................................... 113
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Contents
CHAPTER 4 – CHEMICAL REACTIONS .............................................................................................. 116 Solutions To Exercises .................................................................................................................... 116 Answers To Concept Checks ........................................................................................................... 120 Answers To SelfAssessment And Review Questions ...................................................................... 122 Answers To Concept Explorations .................................................................................................. 124 Answers To Conceptual Problems .................................................................................................. 127 Solutions To Practice Problems ..................................................................................................... 129 Solutions To General Problems ...................................................................................................... 145 Solutions To Strategy Problems ...................................................................................................... 155 Solutions To CumulativeSkills Problems ....................................................................................... 157 CHAPTER 5 – THE GASEOUS STATE ................................................................................................. 164 Solutions To Exercises .................................................................................................................... 164 Answers To Concept Checks ........................................................................................................... 170 Answers To SelfAssessment And Review Questions ...................................................................... 172 Answers To Concept Explorations .................................................................................................. 176 Answers To Conceptual Problems .................................................................................................. 177 Solutions To Practice Problems ..................................................................................................... 179 Solutions To General Problems ...................................................................................................... 192 Solutions To Strategy Problems ...................................................................................................... 200 Solutions To CumulativeSkills Problems ....................................................................................... 202 CHAPTER 6 – THERMOCHEMISTRY .................................................................................................. 207 Solutions To Exercises .................................................................................................................... 207 Answers To Concept Checks ........................................................................................................... 209 Answers To SelfAssessment And Review Questions ...................................................................... 210 Answers To Concept Explorations .................................................................................................. 213 Answers To Conceptual Problems .................................................................................................. 214 Solutions To Practice Problems ..................................................................................................... 216 Solutions To General Problems ...................................................................................................... 225 Solutions To Strategy Problems ...................................................................................................... 233 Solutions To CumulativeSkills Problems ....................................................................................... 235 CHAPTER 7 – QUANTUM THEORY OF THE ATOM............................................................................. 242 Solutions To Exercises .................................................................................................................... 242 Answers To Concept Checks ........................................................................................................... 243 Answers To SelfAssessment And Review Questions ...................................................................... 244 Answers To Concept Explorations .................................................................................................. 246
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Answers To Conceptual Problems ................................................................................................. 248 Solutions To Practice Problems ..................................................................................................... 250 Solutions To General Problems ..................................................................................................... 257 Solutions To Strategy Problems ..................................................................................................... 262 Solutions To CumulativeSkills Problems ...................................................................................... 264 CHAPTER 8 – ELECTRON CONFIGURATIONS AND PERIODICITY ...................................................... 268 Solutions To Exercises ................................................................................................................... 268 Answers To Concept Checks .......................................................................................................... 269 Answers To SelfAssessment And Review Questions ...................................................................... 270 Answers To Concept Explorations ................................................................................................. 273 Answers To Conceptual Problems ................................................................................................. 274 Solutions To Practice Problems ..................................................................................................... 276 Solutions To General Problems ..................................................................................................... 279 Solutions To Strategy Problems ..................................................................................................... 282 Solutions To CumulativeSkills Problems ...................................................................................... 283 CHAPTER 9 – IONIC AND COVALENT BONDING .............................................................................. 286 Solutions To Exercises ................................................................................................................... 286 Answers To Concept Checks .......................................................................................................... 289 Answers To SelfAssessment And Review Questions ...................................................................... 291 Answers To Concept Explorations ................................................................................................. 294 Answers To Conceptual Problems ................................................................................................. 295 Solutions To Practice Problems ..................................................................................................... 298 Solutions To General Problems ..................................................................................................... 317 Solutions To Strategy Problems ..................................................................................................... 328 Solutions To CumulativeSkills Problems ...................................................................................... 330 CHAPTER 10 – MOLECULAR GEOMETRY AND CHEMICAL BONDING THEORY ............................... 336 Solutions To Exercises ................................................................................................................... 336 Answers To Concept Checks .......................................................................................................... 340 Answers To SelfAssessment And Review Questions ...................................................................... 341 Answers To Concept Explorations ................................................................................................. 344 Answers To Conceptual Problems ................................................................................................. 345 Solutions To Practice Problems ..................................................................................................... 348 Solutions To General Problems ..................................................................................................... 361 Solutions To Strategy Problems ..................................................................................................... 368 Solutions To CumulativeSkills Problems ...................................................................................... 370
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CHAPTER 11 – STATES OF MATTER; LIQUIDS AND SOLIDS ............................................................. 375 Solutions To Exercises .................................................................................................................... 375 Answers To Concept Checks ........................................................................................................... 377 Answers To SelfAssessment And Review Questions ...................................................................... 379 Answers To Concept Explorations .................................................................................................. 381 Answers To Conceptual Problems .................................................................................................. 384 Solutions To Practice Problems ..................................................................................................... 387 Solutions To General Problems ...................................................................................................... 399 Solutions To Strategy Problems ...................................................................................................... 407 Solutions To CumulativeSkills Problems ....................................................................................... 410 CHAPTER 12 – SOLUTIONS ............................................................................................................... 413 Solutions To Exercises .................................................................................................................... 413 Answers To Concept Checks ........................................................................................................... 417 Answers To SelfAssessment And Review Questions ...................................................................... 418 Answers To Concept Explorations .................................................................................................. 421 Answers To Conceptual Problems .................................................................................................. 422 Solutions To Practice Problems ..................................................................................................... 424 Solutions To General Problems ...................................................................................................... 435 Solutions To Strategy Problems ...................................................................................................... 445 Solutions To CumulativeSkills Problems ....................................................................................... 448 CHAPTER 13 – RATES OF REACTION ................................................................................................ 453 Solutions To Exercises .................................................................................................................... 453 Answers To Concept Checks ........................................................................................................... 455 Answers To SelfAssessment And Review Questions ...................................................................... 457 Answers To Concept Explorations .................................................................................................. 460 Answers To Conceptual Problems .................................................................................................. 461 Solutions To Practice Problems ..................................................................................................... 464 Solutions To General Problems ...................................................................................................... 478 Solutions To Strategy Problems ...................................................................................................... 490 Solutions To CumulativeSkills Problems ....................................................................................... 492 CHAPTER 14 – CHEMICAL EQUILIBRIUM ......................................................................................... 496 Solutions To Exercises .................................................................................................................... 496 Answers To Concept Checks ........................................................................................................... 500 Answers To SelfAssessment And Review Questions ...................................................................... 501 Answers To Concept Explorations .................................................................................................. 504
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Answers To Conceptual Problems ................................................................................................. 505 Solutions To Practice Problems ..................................................................................................... 507 Solutions To General Problems ..................................................................................................... 521 Solutions To Strategy Problems ..................................................................................................... 536 Solutions To CumulativeSkills Problems ...................................................................................... 541 CHAPTER 15 – ACIDS AND BASES ................................................................................................... 543 Solutions To Exercises ................................................................................................................... 543 Answers To Concept Checks .......................................................................................................... 545 Answers To SelfAssessment And Review Questions ...................................................................... 545 Answers To Concept Explorations ................................................................................................. 548 Answers To Conceptual Problems ................................................................................................. 549 Solutions To Practice Problems ..................................................................................................... 550 Solutions To General Problems ..................................................................................................... 561 Solutions To Strategy Problems ..................................................................................................... 568 Solutions To CumulativeSkills Problems ...................................................................................... 571 CHAPTER 16 – ACIDBASE EQUILIBRIA .......................................................................................... 572 Solutions To Exercises ................................................................................................................... 572 Answers To Concept Checks .......................................................................................................... 582 Answers To SelfAssessment And Review Questions ...................................................................... 583 Answers To Concept Explorations ................................................................................................. 587 Answers To Conceptual Problems ................................................................................................. 589 Solutions To Practice Problems ..................................................................................................... 591 Solutions To General Problems ..................................................................................................... 624 Solutions To Strategy Problems ..................................................................................................... 648 Solutions To CumulativeSkills Problems ...................................................................................... 655 CHAPTER 17 – SOLUBILITY AND COMPLEXION EQUILIBRIA ......................................................... 657 Solutions To Exercises ................................................................................................................... 657 Answers To Concept Checks .......................................................................................................... 661 Answers To SelfAssessment And Review Questions ...................................................................... 662 Answers To Concept Explorations ................................................................................................. 664 Answers To Conceptual Problems ................................................................................................. 666 Solutions To Practice Problems ..................................................................................................... 667 Solutions To General Problems ..................................................................................................... 683 Solutions To Strategy Problems ..................................................................................................... 698 Solutions To CumulativeSkills Problems ...................................................................................... 702
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CHAPTER 18 – THERMODYNAMICS AND EQUILIBRIUM ................................................................... 706 Solutions To Exercises .................................................................................................................... 706 Answers To Concept Checks ........................................................................................................... 710 Answers To SelfAssessment And Review Questions ...................................................................... 710 Answers To Concept Explorations .................................................................................................. 713 Answers To Conceptual Problems .................................................................................................. 714 Solutions To Practice Problems ..................................................................................................... 716 Solutions To General Problems ...................................................................................................... 728 Solutions To Strategy Problems ...................................................................................................... 738 Solutions To CumulativeSkills Problems ....................................................................................... 743 CHAPTER 19 – ELECTROCHEMISTRY ............................................................................................... 750 Solutions To Exercises .................................................................................................................... 750 Answers To Concept Checks ........................................................................................................... 756 Answers To SelfAssessment And Review Questions ...................................................................... 757 Answers To Concept Explorations .................................................................................................. 760 Answers To Conceptual Problems .................................................................................................. 761 Solutions To Practice Problems ..................................................................................................... 764 Solutions To General Problems ...................................................................................................... 795 Solutions To Strategy Problems ...................................................................................................... 808 Solutions To CumulativeSkills Problems ....................................................................................... 811 CHAPTER 20 – NUCLEAR CHEMISTRY ............................................................................................. 816 Solutions To Exercises .................................................................................................................... 816 Answers To Concept Checks ........................................................................................................... 819 Answers To SelfAssessment And Review Questions ...................................................................... 820 Answers To Conceptual Problems .................................................................................................. 822 Solutions To Practice Problems ..................................................................................................... 824 Solutions To General Problems ...................................................................................................... 836 Solutions To Strategy Problems ...................................................................................................... 844 Solutions To CumulativeSkills Problems ....................................................................................... 848 CHAPTER 21 – CHEMISTRY OF THE MAINGROUP ELEMENTS......................................................... 852 Answers To Concept Checks ........................................................................................................... 852 Answers To SelfAssessment And Review Questions ...................................................................... 852 Answers To Conceptual Problems .................................................................................................. 860 Solutions To Practice Problems ..................................................................................................... 861 Solutions To General Problems ...................................................................................................... 880
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Solutions To Strategy Problems ..................................................................................................... 887 CHAPTER 22 – THE TRANSITION ELEMENTS AND COORDINATION COMPOUNDS ........................... 894 Solutions To Exercises ................................................................................................................... 894 Answers To Concept Checks .......................................................................................................... 896 Answers To SelfAssessment And Review Questions ...................................................................... 896 Answers To Conceptual Problems ................................................................................................. 900 Solutions To Practice Problems ..................................................................................................... 901 Solutions To General Problems ..................................................................................................... 912 Solutions To Strategy Problems ..................................................................................................... 915 CHAPTER 23 – ORGANIC CHEMISTRY ............................................................................................. 918 Solutions To Exercises ................................................................................................................... 918 Answers To Concept Checks .......................................................................................................... 922 Answers To SelfAssessment And Review Questions ...................................................................... 922 Answers To Conceptual Problems ................................................................................................. 925 Solutions To Practice Problems ..................................................................................................... 927 Solutions To General Problems ..................................................................................................... 938 Solutions To Strategy Problems ..................................................................................................... 943 CHAPTER 24 – POLYMER MATERIALS: SYNTHETIC AND BIOLOGICAL ........................................... 946 Solutions To Exercises ................................................................................................................... 946 Answers To Concept Checks .......................................................................................................... 946 Answers To SelfAssessment And Review Questions ...................................................................... 947 Answers To Conceptual Problems ................................................................................................. 949 Solutions To Practice Problems ..................................................................................................... 951 Solutions To General Problems ..................................................................................................... 956 Solutions To Strategy Problems ..................................................................................................... 959 APPENDIX A – MATHEMATICAL SKILLS ......................................................................................... 967 Solutions To Exercises ................................................................................................................... 967
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Preface This Complete Solutions Manual provides workedout answers to all of the problems that appear in General Chemistry, 9th Edition, by Darrell D. Ebbing and Steven D. Gammon. The solutions follow the same order as the textbook: Exercises Concept Checks SelfAssessment and Review Questions Concept Explorations Conceptual Problems Practice Problems General Problems Strategy Problems CumulativeSkills Problems For convenience, the short answers for all inchapter Exercises, Concept Checks, and SelfAssessment and Review Questions, as well as for selected oddnumbered problems are provided at the back of the textbook. This Complete Solutions Manual provides detailed solutions for all inchapter Exercises, as well as indepth answers to the Concept Checks, SelfAssessment and Review Questions, Concept Explorations, and Conceptual Problems. Stepbystep workedout solutions are provided for all Practice Problems, General Problems, Strategy Problems, and CumulativeSkills Problems. For each chapter of the text, six additional interactive SelfAssessment and Review Questions are available at the student website (college.hmco.com/pic/ebbing9e). Their indepth answers are available online, and are not included in this manual. In addition, a Strategies and Solutions Guide for the Concept Checks and Conceptual Problems is also available at the website. Please note the following: Significant figures: The answer is first shown with 1 to 2 nonsignificant figures and no units, and the least significant digit is underlined. The answer is then rounded off to the correct number of significant figures, and the units are added. No attempt has been made to round off intermediate answers, but the least significant digit has been underlined wherever possible. Great effort and care have gone into the preparation of this manual. The solutions have been checked and rechecked for accuracy and completeness several times. I would like to express my thanks to TsunMei Chang of the University of WisconsinParkside for her assistance and for her careful work in accuracy reviewing this manuscript. D. B.
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CHAPTER 1
Chemistry and Measurement
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SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 1.1. From the law of conservation of mass, Mass of wood + mass of air = mass of ash + mass of gases Substituting, you obtain 1.85 grams + 9.45 grams = 0.28 grams + mass of gases or, Mass of gases = (1.85 + 9.45 − 0.28) grams = 11.02 grams Thus, the mass of gases in the vessel at the end of the experiment is 11.02 grams. 1.2. Physical properties: soft, silverycolored metal; melts at 64°C. Chemical properties: reacts vigorously with water; reacts with oxygen; reacts with chlorine. 1.3. a.
The factor 9.1 has the fewest significant figures, so the answer should be reported to two significant figures.
5.61 x 7.891 = 4.86 = 4.9 9.1 b.
The number with the least number of decimal places is 8.91. Therefore, round the answer to two decimal places. 8.91 − 6.435 = 2.475 = 2.48
c.
The number with the least number of decimal places is 6.81. Therefore, round the answer to two decimal places. 6.81 − 6.730 = 0.080 = 0.08
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Chapter 1: Chemistry and Measurement
d.
You first do the subtraction within parentheses. In this step, the number with the least number of decimal places is 6.81, so the result of the subtraction has two decimal places. The least significant figure for this step is underlined. 38.91 x (6.81 − 6.730) = 38.91 x 0.080 Next, perform the multiplication. In this step, the factor 0.080 has the fewest significant figures, so round the answer to one significant figure. 38.91 x 0.080 = 3.11 = 3
1.4. a.
1.84 x 10−9 m = 1.84 nm
b.
5.67 x 10−12 s = 5.67 ps
c.
7.85 x 10−3 g = 7.85 mg
d.
9.7 x 103 m = 9.7 km
e.
0.000732 s = 0.732 ms, or 732 µs
f.
0.000000000154 m = 0.154 nm, or 154 pm
a.
Substituting, we find that
1.5.
tC =
5°C 5°C x (tF − 32°F) = x (102.5°F − 32°F) = 39.167°C 9°F 9°F
= 39.2°C b.
Substituting, we find that 1K⎞ 1K⎞ ⎛ ⎛ TK = ⎜ t c x ⎟ + 273.15 K = ⎜ 78°C x ⎟ + 273.15 K = 195.15 K 1°C ⎠ 1°C ⎠ ⎝ ⎝
= 195 K 1.6. Recall that density equals mass divided by volume. You substitute 159 g for the mass and 20.2 g/cm3 for the volume. d =
m 159 g = = 7.871 g/cm3 = 7.87 g/cm3 V 20.2 cm 3
The density of the metal equals that of iron.
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1.7. Rearrange the formula defining the density to obtain the volume. V =
m d
Substitute 30.3 g for the mass and 0.789 g/cm3 for the density. V =
30.3 g = 38.40 cm3 = 38.4 cm3 3 0.789 g/cm
1.8. Since one pm = 10−12 m, and the prefix milli means 10−3, you can write 121 pm x
1 mm 1012 m = 1.21 x 10−7 mm x 103 m 1 pm 3
3
⎛ 1010 m ⎞ ⎛ 1 dm ⎞ −26 3 1.9. 67.6 Å x ⎜ ⎟ x ⎜ 1 ⎟ = 6.76 x 10 dm 1 Å 10 m ⎝ ⎠ ⎝ ⎠ 3
1.10. From the definitions, you obtain the following conversion factors: 1 =
36 in 1 yd
1 =
2.54 cm 1 in
1 =
102 m 1 cm
Then, 3.54 yd x
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36 in 2.54 cm 102 m = 3.236 m = 3.24 m x x 1 cm 1 yd 1 in
ANSWERS TO CONCEPT CHECKS
1.1. Box A contains a collection of identical units; therefore, it must represent an element. Box B contains a compound because a compound is the chemical combination of two or more elements (two elements in this case). Box C contains a mixture because it is made up of two different substances. 1.2. a.
For a person who weighs less than 100 pounds, two significant figures are typically used, although one significant figure is possible (for example, 60 pounds). For a person who weighs 100 pounds or more, three significant figures are typically used to report the weight (given to the whole pound), although people often round to the nearest unit of 10, which may result in reporting the weight with two significant figures (for example, 170 pounds).
b.
165 pounds rounded to two significant figures would be reported as 1.7 x 102 pounds.
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Chapter 1: Chemistry and Measurement
c.
For example, 165 lb weighed on a scale that can measure in 100lb increments would be 200 lb. Using the conversion factor 1 lb = 0.4536 kg, 165 lb is equivalent to 74.8 kg. Thus, on a scale that can measure in 50kg increments, 165 lb would be 50 kg.
a.
If your leg is approximately 32 inches long, this would be equivalent to 0.81 m, 8.1 dm, or 81 cm.
b.
One story is approximately 10 feet, so three stories is 30 feet. This would be equivalent to approximately 9 m.
c.
Normal body temperature is 98.6°F, or 37.0°C. Thus, if your body temperature were 39°C (102°F), you would feel as if you had a moderate fever.
d.
Room temperature is approximately 72°F, or 22°C. Thus, if you were sitting in a room at 23°C (73°F), you would be comfortable in a shortsleeve shirt.
1.3.
1.4. Gold is a very unreactive substance, so comparing physical properties is probably your best option. However, color is a physical property you cannot rely on in this case to get your answer. One experiment you could perform is to determine the densities of the metal and the chunk of gold. You could measure the mass of the nugget on a balance and the volume of the nugget by water displacement. Using this information, you could calculate the density of the nugget. Repeat the experiment and calculations for the sample of gold. If the nugget is gold, the two densities should be equal and be 19.3 g/cm3. Also, you could determine the melting points of the metal and the chunk of pure gold. The two melting points should be the same (1338 K) if the metal is gold.
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ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
1.1. One area of technology that chemistry has changed is the characteristics of materials. The liquidcrystal displays (LCDs) in devices such as watches, cell phones, computer monitors, and televisions are materials made of molecules designed by chemists. Electronics and communications have been transformed by the development of optical fibers to replace copper wires. In biology, chemistry has changed the way scientists view life. Biochemists have found that all forms of life share many of the same molecules and molecular processes. 1.2. An experiment is an observation of natural phenomena carried out in a controlled manner so that the results can be duplicated and rational conclusions obtained. A theory is a tested explanation of basic natural phenomena. They are related in that a theory is based on the results of many experiments and is fruitful in suggesting other, new experiments. Also, an experiment can disprove a theory but can never prove it absolutely. A hypothesis is a tentative explanation of some regularity of nature.
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1.3. Rosenberg conducted controlled experiments and noted a basic relationship that could be stated as a hypothesis—that is, that certain platinum compounds inhibit cell division. This led him to do new experiments on the anticancer activity of these compounds. 1.4. Matter is the general term for the material things around us. It is whatever occupies space and can be perceived by our senses. Mass is the quantity of matter in a material. The difference between mass and weight is that mass remains the same wherever it is measured, but weight is proportional to the mass of the object divided by the square of the distance between the center of mass of the object and that of the earth. 1.5. The law of conservation of mass states that the total mass remains constant during a chemical change (chemical reaction). To demonstrate this law, place a sample of wood in a sealed vessel with air, and weigh it. Heat the vessel to burn the wood, and weigh the vessel after the experiment. The weight before the experiment and that after it should be the same. 1.6. Mercury metal, which is a liquid, reacts with oxygen gas to form solid mercury(II) oxide. The color changes from that of metallic mercury (silvery) to a color that varies from red to yellow depending on the particle size of the oxide. 1.7. Gases are easily compressible and fluid. Liquids are relatively incompressible and fluid. Solids are relatively incompressible and rigid. 1.8. An example of a substance is the element sodium. Among its physical properties: It is a solid, and it melts at 98°C. Among its chemical properties: It reacts vigorously with water, and it burns in chlorine gas to form sodium chloride. 1.9. An example of an element: sodium; of a compound: sodium chloride, or table salt; of a heterogeneous mixture: salt and sugar; of a homogeneous mixture: sodium chloride dissolved in water to form a solution. 1.10. A glass of bubbling carbonated beverage with ice cubes contains three phases: gas, liquid, and solid. 1.11. A compound may be decomposed by chemical reactions into elements. An element cannot be decomposed by any chemical reaction. Thus, a compound cannot also be an element in any case. 1.12. The precision refers to the closeness of the set of values obtained from identical measurements of a quantity. The number of digits reported for the value of a measured or calculated quantity (significant figures) indicates the precision of the value.
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Chapter 1: Chemistry and Measurement
1.13. Multiplication and division rule: In performing the calculation 100.0 x 0.0634 ÷ 25.31, the calculator display shows 0.2504938. We would report the answer as 0.250 because the factor 0.0634 has the least number of significant figures (three). Addition and subtraction rule: In performing the calculation 184.2 + 2.324, the calculator display shows 186.524. Because the quantity 184.2 has the least number of decimal places (one), the answer is reported as 186.5. 1.14. An exact number is a number that arises when you count items or sometimes when you define a unit. For example, a foot is defined to be 12 inches. A measured number is the result of a comparison of a physical quantity with a fixed standard of measurement. For example, a steel rod measures 9.12 centimeters, or 9.12 times the standard centimeter unit of measurement. 1.15. For a given unit, the SI system uses prefixes to obtain units of different sizes. Units for all other possible quantities are obtained by deriving them from any of the seven base units. You do this by using the base units in equations that define other physical quantities. 1.16. An absolute temperature scale is a scale in which the lowest temperature that can be attained theoretically is zero. Degrees Celsius and kelvins have units of equal and are related by the formula tC = (TK − 273.15 K) x
1°C 1K
1.17. The density of an object is its mass per unit volume. Because the density is characteristic of a substance, it can be helpful in identifying it. Density can also be useful in determining whether a substance is pure. It also provides a useful relationship between mass and volume. 1.18. Units should be carried along because (1) the units for the answers will come out in the calculations, and (2), if you make an error in arranging factors in the calculation, this will become apparent because the final units will be nonsense. 1.19. The answer is c, three significant figures. 1.20. The answer is a, 4.43 x 102 mm. 1.21. The answer is e, 75 mL. 1.22. The answer is c, 0.23 mg.
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ANSWERS TO CONCEPT EXPLORATIONS
1.23. a.
First, check the physical appearance of each sample. Check the particles that make up each sample for consistency and hardness. Also, note any odor. Then perform on each sample some experiments to measure physical properties such as melting point, density, and solubility in water. Compare all of these results and see if they match.
b.
It is easier to prove that the compounds were different by finding one physical property that is different, say different melting points. To prove the two compounds were the same would require showing that every physical property was the same.
c.
Of the properties listed in part a, the melting point would be most convincing. It is not difficult to measure, and it is relatively accurate. The density of a powder is not as easy to determine as the melting point, and solubility is not reliable enough on its own.
d.
No. Since neither solution reached a saturation point, there is not enough information to tell if there was a difference in behavior. Many white powders dissolve in water. Their chemical compositions are not the same.
1.24. Part 1 a.
3 g + 1.4 g + 3.3 g = 7.7 g = 8 g
b.
First, 3 g + 1.4 g = 4.4 g = 4 g. Then, 4 g + 3.3 g = 7.3 g = 7 g.
c.
Yes, the answer in part a is more accurate. When you round off intermediate steps, you accumulate small errors and your answer is not as accurate.
d.
The answer 29 g is correct.
e.
This answer is incorrect. It should be 3 x 101 with only one significant figure in the answer. The student probably applied the rule for addition (instead of for multiplication) after the first step.
f.
The answer 28.5 g is correct.
g.
Don’t round off intermediate answers. Indicate the roundoff position after each step by underlining the least significant digit.
Part 2 a.
The calculated answer is incorrect. It should be 11 cm3. The answer given has too many significant figures. There is also a small round off error due to using a roundedoff value for the density.
b.
This is a better answer. It is reported with the correct number of significant figures (three). It can be improved by using all of the digits given for the density.
c.
V =
d.
There was no rounding off of intermediate steps; all the factors are as accurate as possible.
10 ball bearings 1.234 g 1 cm3 x x = 3.90889 = 3.909 cm3 1 3.1569 g 1 ball bearing
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Chapter 1: Chemistry and Measurement
ANSWERS TO CONCEPTUAL PROBLEMS
1.25. a.
Two phases: liquid and solid.
b.
Three phases: liquid water, solid quartz, and solid seashells.
1.26. If the material is a pure compound, all samples should have the same melting point, the same color, and the same elemental composition. If it is a mixture, these properties should differ depending on the composition. 1.27. a.
You need to establish two points on the thermometer with known (defined) temperatures— for example, the freezing point (0°C) and boiling point (100°C) of water. You could first immerse the thermometer in an icewater bath and mark the level at this point as 0°C. Then, immerse the thermometer in boiling water, and mark the level at this point as 100°C. As long as the two points are far enough apart to obtain readings of the desired accuracy, the thermometer can be used in experiments.
b.
You could make 19 evenly spaced marks on the thermometer between the two original points, each representing a difference of 5°C. You may divide the space between the two original points into fewer spaces as long as you can read the thermometer to obtain the desired accuracy.
1.28.
a.
b.
c.
1.29. a.
To answer this question, you need to develop an equation that converts between °F and °YS. To do so, you need to recognize that one degree on the Your Scale does not correspond to one degree on the Fahrenheit scale and that −100°F corresponds to 0° on Your Scale (different “zero” points). As stated in the problem, in the desired range of 100 Your Scale degrees, there are 120 Fahrenheit degrees. Therefore, the relationship can be expressed as 120°F = 100°YS, since it covers the same temperature range. Now you need to “scale” the two systems so that they correctly convert from one scale to the other. You could set up an equation with the known data points and then employ the information from the relationship above.
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9
For example, to construct the conversion between °YS and °F, you could perform the following steps: Step 1:
°F = °YS
Not a true statement, but one you would like to make true. Step 2:
°F = °YS x
120°F 100°YS
This equation takes into account the difference in the size between the temperature unit on the two scales but will not give you the correct answer because it doesn’t take into account the different zero points. Step 3: By subtracting 100°F from your equation from Step 2, you now have the complete equation that converts between °F and °YS. °F = (°YS x
b.
120°F ) − 100°F 100°YS
Using the relationship from part a, 66°YS is equivalent to (66°YS x
120°F ) − 100°F = −20.8°F = −21°F 100°YS
1.30. Some physical properties you could measure are density, hardness, color, and conductivity. Chemical properties of sodium would include reaction with air, reaction with water, reaction with chlorine, reaction with acids, bases, etc. 1.31. The empty boxes are identical, so they do not contribute to any mass or density difference. Since the edge of the cube and the diameter of the sphere are identical, they will occupy the same volume in each of the boxes; therefore, each box will contain the same number of cubes or spheres. If you view the spheres as cubes that have been rounded by removing wood, you can conclude that the box containing the cubes must have a greater mass of wood; hence, it must have a greater density. 1.32. a.
Since the bead is less dense than any of the liquids in the container, the bead will float on top of all the liquids.
b.
First, determine the density of the plastic bead. Since density is mass divided by volume, you get d =
m 3.92 x 102 g = = 0.911 g/mL = 0.91 g/mL V 0.043 mL
Thus, the glass bead will pass through the top three layers and float on the ethylene glycol layer, which is more dense. c.
Since the bead sinks all the way to the bottom, it must be more dense than 1.114 g/mL.
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Chapter 1: Chemistry and Measurement
1.33. a.
A paper clip has a mass of about 1 g.
b.
Answers will vary depending on your particular sample. Keeping in mind that the SI unit for mass is kg, the approximate weights for the items presented in the problem are as follows: a grain of sand, 1 x 10−5 kg; a paper clip, 1 x 10−3 kg; a nickel, 5 x 10−3 kg; a 5.0gallon bucket of water, 2.0 x 101 kg; a brick, 3 kg; a car, 1 x 103 kg.
1.34. When taking measurements, never throw away meaningful information even if there is some uncertainty in the final digit. In this case, you are certain that the nail is between 5 and 6 cm. The uncertain, yet still important, digit is between the 5 and 6 cm measurements. You can estimate with reasonable precision that it is about 0.7 cm from the 5 cm mark, so an acceptable answer would be 5.7 cm. Another person might argue that the length of the nail is closer to 5.8 cm, which is also acceptable given the precision of the ruler. In any case, an answer of 5.7 or 5.8 should provide useful information about the length of the nail. If you were to report the length of the nail as 6 cm, you would be discarding potentially useful length information provided by the measuring instrument. If a higher degree of measurement precision were needed (more significant figures), you would need to switch to a more precise ruler—for example, one that had mm markings. 1.35. a.
The number of significant figures in this answer follows the rules for multiplication and division. Here, the measurement with the fewest significant figures is the reported volume 0.310 m3, which has three. Therefore, the answer will have three significant figures. Since Volume = L x W x H, you can rearrange and solve for one of the measurements, say the length. L =
b.
V 0.310 m 3 = = 0.83496 m = 0.835 m W x H (0.7120 m) (0.52145 m)
The number of significant figures in this answer follows the rules for addition and subtraction. The measurement with the least number of decimal places is the result 1.509 m, which has three. Therefore, the answer will have three decimal places. Since the result is the sum of the three measurements, the third length is obtained by subtracting the other two measurements from the total. Length = 1.509 m − 0.7120 m − 0.52145 m = 0.27555 m = 0.276 m
1.36. The mass of something (how heavy it is) depends on how much of the item, material, substance, or collection of things you have. The density of something is the mass of a specific amount (volume) of an item, material, substance, or collection of things. You could use 1 kg of feathers and 1 kg of water to illustrate that they have the same mass yet have very different volumes; therefore, they have different densities.
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11
SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off.
1.37. By the law of conservation of mass: Mass of sodium carbonate + mass of acetic acid solution = mass of contents of reaction vessel + mass of carbon dioxide Plugging in gives 15.9 g + 20.0 g = 29.3 g + mass of carbon dioxide Mass of carbon dioxide = 15.9 g + 20.0 g − 29.3 g = 6.6 g 1.38. By the law of conservation of mass: Mass of iron + mass of acid = mass of contents of beaker + mass of hydrogen Plugging in gives 5.6 g + 15.0 = 20.4 g + mass of hydrogen Mass of hydrogen = 5.6 g + 15.0 g − 20.4 g = 0.2 g 1.39. By the law of conservation of mass: Mass of zinc + mass of sulfur = mass of zinc sulfide Rearranging and plugging in give Mass of zinc sulfide = 65.4 g + 32.1 g = 97.5 g For the second part, let x = mass of zinc sulfide that could be produced. By the law of conservation of mass: 20.0 g + mass of sulfur = x Write a proportion that relates the mass of zinc reacted to the mass of zinc sulfide formed, which should be the same for both cases. mass zinc 65.4 g 20.0 g = = mass zinc sulfide 97.5 g x
Solving gives x = 29.81 g = 29.8 g
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Chapter 1: Chemistry and Measurement
1.40. By the law of conservation of mass: Mass of aluminum + mass of bromine = mass of aluminum bromide Plugging in and solving give 27.0 g + Mass of bromine = 266.7 g Mass of bromine = 266.7 g − 27.0 g = 239.7 g For the second part, let x = mass of bromine that reacts. By the law of conservation of mass: 15.0 g + x = mass of aluminum bromide Write a proportion that relates the mass of aluminum reacted to the mass of bromine reacted, which should be the same for both cases. mass aluminum 27.0 g 15.0 g = = x mass bromine 239.7 g Solving gives x = 133.1 g = 133 g 1.41. a.
Solid
b.
Liquid
c.
Gas
d.
Solid
a.
Solid
b.
Solid
c.
Solid
d.
Liquid
a.
Physical change
b.
Physical change
c.
Chemical change
d.
Physical change
a.
Physical change
b.
Chemical change
c.
Chemical change
1.42.
1.43.
1.44.
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d.
Physical change
1.45. Physical change: Liquid mercury is cooled to solid mercury. Chemical changes: (1) Solid mercury oxide forms liquid mercury metal and gaseous oxygen; (2) glowing wood and oxygen form burning wood (form ash and gaseous products). 1.46. Physical changes: (1) Solid iodine is heated to gaseous iodine; (2) gaseous iodine is cooled to form solid iodine. Chemical change: Solid iodine and zinc metal are ignited to form a white powder. 1.47. a.
Physical property
b.
Chemical property
c.
Physical property
d.
Physical property
e.
Chemical property
a.
Physical property
b.
Chemical property
c.
Physical property
d.
Chemical property
e.
Physical property
1.48.
1.49. Physical properties: (1) Iodine is solid; (2) the solid has lustrous blueblack crystals; (3) the crystals vaporize readily to a violetcolored gas. Chemical properties: (1) Iodine combines with many metals, such as with aluminum to give aluminum iodide. 1.50. Physical properties: (1) is a solid; (2) has an orangered color; (3) has a density of 11.1 g/cm3; (4) is insoluble in water. Chemical property: Mercury(II) oxide decomposes when heated to give mercury and oxygen. 1.51. a.
Physical process
b.
Chemical reaction
c.
Physical process
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Chapter 1: Chemistry and Measurement
d.
Chemical reaction
e.
Physical process
a.
Chemical reaction
b.
Physical process
c.
Physical process
d.
Physical process
e.
Chemical reaction
a.
Solution
b.
Substance
c.
Substance
d.
Homogeneous or Heterogeneous mixture
a.
Heterogeneous mixture
b.
Substance
c.
Solution
d.
Substance
a.
A pure substance with two phases present, liquid and gas.
b.
A mixture with two phases present, solid and liquid.
c.
A pure substance with two phases present, solid and liquid.
d.
A mixture with two phases present, solid and solid.
a.
A mixture with two phases present, solid and liquid.
b.
A mixture with two phases present, solid and liquid.
c.
A mixture with two phases present, solid and solid.
d.
A pure substance with two phases present, liquid and gas.
1.52.
1.53.
1.54.
1.55.
1.56.
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1.57. a.
six
b.
three
c.
four
d.
five
e.
three
f.
four
a.
three
b.
four
c.
six
d.
four
e.
four
f.
four
1.58.
1.59. 40,000 km = 4.0 x 104 km 1.60. 150,000,000 km = 1.50 x 108 km 1.61. a.
8.71 x 0.0301 = 8.457 = 8.5 0.031
b.
0.71 + 92.2 = 92.91 = 92.9
c.
934 x 0.00435 + 107 = 4.0629 + 107 = 111.06 = 111
d.
(847.89 − 847.73) x 14673 = 0.16 x 14673 = 2347 = 2.3 x 103
a.
0.871 x 0.57 = 0.08456 = 0.085 5.871
b.
8.937 − 8.930 = 0.007
c.
8.937 + 8.930 = 17.867
d.
0.00015 x 54.6 + 1.002 = 0.00819 + 1.002 = 1.0101 = 1.010
1.62.
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Chapter 1: Chemistry and Measurement
1.63. The volume of the first sphere is V1 = (4/3)πr3 = (4/3)π x (5.10 cm)3 = 555.64 cm3
The volume of the second sphere is V2 = (4/3)πr3 = (4/3)π x (5.00 cm)3 = 523.60 cm3
The difference in volume is V1 − V2 = 555.64 cm3 − 523.60 cm3 = 32.04 cm3 = 32 cm3
1.64. The length of the cylinder between the two marks is l = 3.50 cm − 3.10 cm = 0.40 cm
The volume of iron contained between the marks is V = πr2l = π x (1.500 cm)2 x 0.40 cm = 2.82 cm3 = 2.8 cm3
1.65. a.
5.89 x 10−12 s = 5.89 ps
b.
0.2010 m = 20.10 cm
c.
2.560 x 10−9 g = 2.560 ng
d.
6.05 x 103 m = 6.05 km
a.
4.851 x 10−6 g = 4.851 µg
b.
3.16 x 10−2 m = 3.16 cm
c.
2.591 x 10−9 s = 2.591 ns
d.
8.93 x 10−12 g = 8.93 pg
a.
6.15 ps = 6.15 x 10−12 s
b.
3.781 µm = 3.781 x 10−6 m
c.
1.546 Å = 1.546 x 10−10 m
d.
9.7 mg = 9.7 x 10−3 g
a.
6.20 km = 6.20 x 103 m
b.
1.98 ns = 1.98 x 10−9 s
c.
2.54 cm = 2.54 x 10−2 m
d.
5.23 µg = 5.23 x 10−6 g
1.66.
1.67.
1.68.
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1.69. a.
tC =
5°C 5°C x (tF − 32°F) = x (68°F − 32°F) = 20.0°C = 20.°C 9°F 9°F
b.
tC =
5°C 5°C x (tF − 32°F) = x (−23°F − 32°F) = −30.55°C = −31°C 9°F 9°F
c.
tF = (tC x
9°F 9°F ) + 32°F = (26°C x ) + 32°F = 78.8°F = 79°F 5°C 5°C
d.
tF = (tC x
9°F 9°F ) + 32°F = (−70°C x ) + 32°F = −94.0°F = −94°F 5°C 5°C
a.
tC =
5°C 5°C x (tF − 32°F) = x (51°F − 32°F) = 10.555°C = 11°C 9°F 9°F
b.
tC =
5°C 5°C x (tF − 32°F) = x (−7°F − 32°F) = −21.6°C = −22°C 9°F 9°F
c.
tF = (tC x
9°F 9°F ) + 32°F = (−41°C x ) + 32°F = −41.8°F = −42°F 5°C 5°C
d.
tF = (tC x
9°F 9°F ) + 32°F = (22°C x ) + 32°F = 71.6°F = 72°F 5°C 5°C
1.70.
1.71. tF = (tC x
9°F 9°F ) + 32°F = (−21.1°C x ) + 32°F = −5.98°F = −6.0°F 5°C 5°C
1.72. tF = (tC x
9°F 9°F ) + 32°F = (−196°C x ) + 32°F = −320.8°F = −321°F 5°C 5°C
1.73. d =
m 12.4 g = = 7.560 g/cm3 = 7.56 g/cm3 V 1.64 cm3
1.74. d =
m 17.84 g = = 0.7136 g/mL = 0.714 g/mL V 25.0 mL
1.75. First, determine the density of the liquid. d =
m 6.71 g = = 0.7894 = 0.79 g/mL V 8.5 mL
The density is closest to ethanol (0.789 g/cm3).
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Chapter 1: Chemistry and Measurement
1.76. First, determine the density of the mineral sample. d =
m 59.5 g = = 7.531 = 7.5 g/cm3 3 V 7.9 cm
The density is closest to cassiterite (6.99 g/cm3). 1.77. The mass of platinum is obtained as follows. Mass = d x V = 21.4 g/cm3 x 5.9 cm3 = 126 g = 1.3 x 102 g 1.78. The mass of gasoline is obtained as follows. Mass = d x V = 0.70 g/mL x 43.8 mL = 30.66 g = 31 g 1.79. The volume of ethanol is obtained as follows. Recall that 1 mL = 1 cm3. m 19.8 g = = 25.09 cm3 = 25.1 cm3 = 25.1 mL d 0.789 g/cm3
Volume =
1.80. The volume of bromine is obtained as follows. m 88.5 g = 28.54 mL = 28.5 mL = d 3.10 g/mL
Volume =
1.81. Since 1 kg = 103 g, and 1 mg = 10−3 g, you can write 0.480 kg x
103 g 1 mg = 4.80 x 105 mg x 1 kg 103 g
1.82. Since 1 mg = 10−3 g, and 1 µg = 10−6 g, you can write 501 mg x
103 g 1 μg = 5.01 x 105 µg x 1 mg 106 g
1.83. Since 1 nm = 10−9 m, and 1 cm = 10−2 m, you can write 555 nm x
1 cm 109 m = 5.55 x 10−5 cm x 2 10 m 1 nm
1.84. Since 1 Å = 10−10 m, you can write 0.96 Å x
1010 m = 9.6 x 10−11 m 1Å
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1.85. Since 1 km = 103 m, you can write ⎛ 103 m ⎞ 3.73 x 10 km x ⎜ ⎟ ⎝ 1 km ⎠ 8
3
3
= 3.73 x 1017 m3
Now, 1 dm = 10−1 m. Also, note that 1 dm3 = 1 L. Therefore, you can write 3
⎛ 1 dm ⎞ 3.73 x 1017 m3 x ⎜ 1 ⎟ = 3.73 x 1020 dm3 = 3.73 x 1020 L ⎝ 10 m ⎠
1.86. 1 µm = 10−6 m, and 1 dm = 10−1 m. Also, note that 1 dm3 = 1 L. Therefore, you can write ⎛ 106 m ⎞ 1.3 µm x ⎜ ⎟ ⎝ 1 μm ⎠ 3
1.87. 3.58 short ton x
1.88. 3.15 Btu x
3
3
⎛ 1 dm ⎞ x ⎜ 1 ⎟ = 1.3 x 10−15 dm3 = 1.3 x 10−15 L ⎝ 10 m ⎠
2000 lb 16 oz 1g x x = 3.248 x 106 g = 3.25 x 106 g 1 short ton 1 lb 0.03527 oz
252.0 cal 4.184 J x = 3321 J = 3.32 x 103 J 1 Btu 1 cal
1.89. 2425 fathoms x
6 ft 12 in. 2.54 x 102 m = 4434.8 m = 4.435 x 103 m x x 1 fathom 1 ft 1 in.
1.90. 1.3 x 1010 barrels x
42 gal 4 qt 9.46 x 104 m3 = 2.066 x 109 m3 = 2.1 x 109 m3 x x 1 qt 1 barrel 1 gal 3
1L ⎛ 2.54 cm ⎞ 1.91. (20.0 in.)(20.0 in.)(10.0 in.) x ⎜ = 65.54 L = 65.5 L ⎟ x 1000 cm3 ⎝ 1 in. ⎠ 2
⎛ 1000 m ⎞ 25 worms 1.92. (1.00 km)(2.0 km)(1 m) x ⎜ = 5.00 x 107 = 5.0 x 107 worms ⎟ x 3 1m ⎝ 1 km ⎠
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■
Chapter 1: Chemistry and Measurement
SOLUTIONS TO GENERAL PROBLEMS
1.93. From the law of conservation of mass, Mass of sodium + mass of water = mass of hydrogen + mass of solution Substituting, you obtain 19.70 g + 126.22 g = mass of hydrogen + 145.06 g or, Mass of hydrogen = 19.70 g + 126.22 g − 145.06 g = 0.86 g Thus, the mass of hydrogen produced was 0.86 g. 1.94. From the law of conservation of mass, Mass of tablet + mass of acid solution = mass of carbon dioxide + mass of solution Substituting, you obtain 0.853 g + 56.519 g = mass of carbon dioxide + 57.152 g Mass of carbon dioxide = 0.853 g + 56.519 g − 57.152 g = 0.220 g Thus, the mass of carbon dioxide produced was 0.220 g. 1.95. From the law of conservation of mass, Mass of aluminum + mass of iron(III) oxide = mass of iron + mass of aluminum oxide + mass of unreacted iron(III) oxide 5.40 g + 18.50 g = 11.17 g + 10.20 g + mass of iron(III) oxide unreacted Mass of iron(III) oxide unreacted = 5.40 g + 18.50 g − 11.17 g − 10.20 g = 2.53 g Thus, the mass of unreacted iron(III) oxide is 2.53 g. 1.96. From the law of conservation of mass, Mass of sodium bromide + mass of chlorine reacted = mass of bromine + mass of sodium chloride 20.6 g + mass of chlorine reacted = 16.0 g + 11.7 g Mass of chlorine reacted = 16.0 g + 11.7 g − 20.6 g = 7.1 g Thus, the mass of chlorine that reacted is 7.1 g. 1.97. 53.10 g + 5.348 g + 56.1 g = 114.54 g = 114.5 g total 1.98. 68.1 g + 58.2 g + 5.279 g = 131.579 g = 131.6 g total
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21
1.99. a.
Chemical
b.
Physical
c.
Physical
d.
Chemical
1.100. a.
Physical
b.
Chemical
c.
Physical
d.
Chemical
1.101. Compounds always contain the same proportions of the elements by mass. Thus, if we let X be the proportion of iron in a sample, we can calculate the proportion of iron in each sample as follows. Sample A:
X =
mass of iron 1.094 g = = 0.72068 = 0.7207 mass of sample 1.518 g
Sample B:
X =
mass of iron 1.449 g = = 0.70476 = 0.7048 mass of sample 2.056 g
Sample C:
X =
mass of iron 1.335 g = = 0.71276 = 0.7128 mass of sample 1.873 g
Since each sample has a different proportion of iron by mass, the material is not a compound. 1.102. Compounds always contain the same proportions of the elements by mass. Thus, if we let X be the proportion of mercury in a sample, we can calculate the proportion of mercury in each sample as follows. Sample A:
X =
mass of mercury 0.9641 g = = 0.92612 = 0.9261 mass of sample 1.0410 g
Sample B:
X =
mass of mercury 1.4293 g = = 0.92607 = 0.9261 mass of sample 1.5434 g
Sample C:
X =
mass of mercury 1.1283 g = = 0.92612 = 0.9261 mass of sample 1.2183 g
Since each sample has the same proportion of mercury by mass, the data are consistent with the hypothesis that the material is a compound. 1.103. V = (edge)3 = (39.3 cm)3 = 6.069 x 104 cm3 = 6.07 x 104 cm3
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Chapter 1: Chemistry and Measurement
1.104. V = πr2l = π x (2.56 cm)2 x 56.32 cm = 1159 cm3 = 1.16 x 103 cm3
1.105. V = LWH = 47.8 in. x 12.5 in. x 19.5 in. x
1 gal = 50.43 gal = 50.4 gal 231 in 3
1.106. The volume in cubic inches is V = (4/3)πr3 = (4/3)π x (175.0 in.)3 = 2.24492 x 107 in3 = 2.245 x 107 in3
The volume in imperial gallons is V = 2.24492 x 107 in3 x
1 gal = 8.09275 x 104 gal = 8.093 x 104 gal 3 277.4 in
1.107. The volume of the first sphere is given by V1 = (4/3)πr3 = (4/3)π x (5.61 cm)3 = 739.5 cm3
The volume of the second sphere is given by V2 = (4/3)πr3 = (4/3)π x (5.85 cm)3 = 838.6 cm3
The difference in volume between the two spheres is given by V = V2 − V1 = 838.6 cm3 − 739.5 cm3 = 9.91 x 101 = 9.9 x 101 cm3
1.108. The surface area of the first circle is given by S1 = πr2 = π x (7.98 cm)2 = 200.0 cm2
The surface area of the second circle is given by S2 = πr2 = π x (8.50 cm)2 = 226.9 cm2
The difference in surface area between the two circles is Difference = S2 − S1 = 226.9 cm2 − 200.0 cm2 = 26.9 cm2 = 27 cm2 1.109. a.
56.1 51.1 = 7.59 x 10−1 = 7.6 x 10−1 6.58
b.
56.1 + 51.1 = 1.629 x 101 = 1.63 x 101 6.58
c.
(9.1 + 8.6) x 26.91 = 4.763 x 102 = 4.76 x 102
d.
0.0065 x 3.21 + 0.0911 = 1.119 x 10−1 = 1.12 x 10−1
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1.110. a.
9.345  9.005 = 0.03465 = 0.0347 9.811
b.
9.345 + 9.005 = 1.8703 = 1.870 9.811
c.
(7.50 + 7.53) x 3.71 = 55.761 = 55.8
d.
0.71 x 0.36 + 17.36 = 17.6156 = 17.62
1.111. a.
9.12 cg/mL
b.
66 pm
c.
7.1 µm
d.
56 nm
1.112. a.
1.86 cg/mL
b.
77 pm
c.
6.5 nm
d.
0.85 µm
1.113. a.
1.07 x 10−12 s
b.
5.8 x 10−6 m
c.
3.19 x 10−7 m
d.
1.53 x 10−2 s
1.114. a.
6.6 x 10−3 K
b.
2.75 x 10−10 m
c.
2.21 x 10−2 s
d.
4.5 x 10−5 m
1.115. tF = (tC x
9°F 9°F ) + 32°F = (3410°C x ) + 32°F = 6170°F = 6170°F 5°C 5°C
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Chapter 1: Chemistry and Measurement
1.116. tF = (tC x
9°F 9°F ) + 32°F = (1677°C x ) + 32°F = 3050.6°F = 3051°F 5°C 5°C
1.117. tF = (tC x
9°F 9°F ) + 32°F = (825°C x ) + 32°F = 1517°F = 1.52 x 103°F 5°C 5°C
1.118. tF = (tC x
9°F 9°F ) + 32°F = (50°C x ) + 32°F = 122°F = 122°F 5°C 5°C
1.119. The temperature in kelvins is 1K 1K ) + 273.15 K = (29.8°C x ) + 273.15 K = 302.95 K 1°C 1°C = 303.0 K
TK = (tC x
The temperature in degrees Fahrenheit is tF = (tC x
9°F 9°F ) + 32°F = (29.8°C x ) + 32°F = 85.64°F = 85.6°F 5°C 5°C
1.120. The temperature in kelvins is 1K 1K ) + 273.15 K = (−38.9°C x ) + 273.15 K = 234.25 K 1°C 1°C = 234.3 K
TK = (tC x
The temperature in degrees Fahrenheit is tF = (tC x
9°F 9°F ) + 32°F = (−38.9°C x ) + 32°F = −38.02°F = −38.0°F 5°C 5°C
1.121. The temperature in degrees Celsius is tC =
5°C 5°C x (tF − 32°F) = x (1666°F − 32°F) = 907.77°C = 907.8°C 9°F 9°F
The temperature in kelvins is 1K 1K ) + 273.15 K = (907.77°C x ) + 273.15 K = 1180.92 K 1°C 1°C = 1180.9 K
TK = (tC x
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1.122. The temperature in degrees Celsius is tC =
5°C 5°C x (tF − 32°F) = x (236°F − 32°F) = 113.3°C = 113°C 9°F 9°F
The temperature in kelvins is 1K 1K ) + 273.15 K = (113.3°C x ) + 273.15 K = 386.4 K 1°C 1°C
TK = (tC x
= 386 K 3
1.123. Density =
1.74 g 1 kg ⎛ 1 cm ⎞ x x ⎜ 2 ⎟ 3 3 1 cm 10 g ⎝ 10 m ⎠
1.124. Density =
5.96 g 1 kg ⎛ 1 cm ⎞ x x ⎜ 2 ⎟ 3 3 1 cm 10 g ⎝ 10 m ⎠
= 1.74 x 103 kg/m3
3
= 5.96 x 103 kg/m3
1.125. The volume of the quartz is 65.7 mL − 51.2 mL = 14.5 mL. Then, the density is Density =
mass 38.4 g = = 2.648 g/mL = 2.65 g/mL = 2.65 g/cm3 volume 14.5 mL
1.126. First, determine the volume of water in the flask. The mass of water is obtained as follows. 109.3 g − 70.7 g = 38.6 g. Now, using the density (0.997 g/cm3), Volume =
mass 38.6 g = = 38.716 cm3 = 38.716 mL 3 density 0.997 g/ cm
Now, the volume of the ore is 53.2 mL − 38.716 mL = 14.48 mL, or 14.48 cm3. Therefore, the density of the hematite ore is Density =
mass 70.7 g = = 4.881 g/cm3 = 4.88 g/cm3 3 14.48 cm volume
1.127. First, determine the density of the liquid sample. Density =
mass 22.3 g = = 1.486 g/mL = 1.49 g/mL = 1.49 g/cm3 volume 15.0 mL
This density is closest to that of chloroform (1.489 g/cm3), so the unknown liquid is chloroform. 1.128. First, determine the density of the calcite sample. Density =
mass 35.6 g = = 2.759 g/cm3 = 2.76 g/cm3 volume 12.9 cm3
Since a substance will float on the liquids with greater densities, calcite will float on tetrabromoethane (2.96 g/cm3) and methylene iodode (3.33 g/cm3).
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Chapter 1: Chemistry and Measurement
1.129. First, determine the volume of the cube of platinum. V = (edge)3 = (4.40 cm)3 = 85.18 cm3
Now, use the density to determine the mass of the platinum. Mass = d x V = 21.4 g/cm3 x 85.18 cm3 = 1822.9 g = 1.82 x 103 g 1.130. First, determine the volume of the cylinder of silicone. V = πr2l = π x (4.00 cm)2 x 12.40 cm = 623.29 cm3
Now, use the density to determine the mass of the silicon. Mass = d x V = 2.33 g/cm3 x 623.29 cm3 = 1452.2 g = 1.45 x 103 g
1.131. Volume =
mass 35.00 g = = 33.238 mL = 33.24 mL density 1.053 g/ mL
1.132. First, convert kilograms to grams (1 kg = 103 g). Thus, 0.035 kg = 35 g. Then mass 35 g = = 38.80 mL density 0.902 g/ mL
Volume =
Finally, convert the volume to liters (1000 mL = 1 L). Volume = 38.80 mL x
1L = 0.03880 L = 0.039 L 1000 mL
1.133. a.
8.45 kg x
103 g 1 μg = 8.45 x 109 μg x 6 1 kg 10 g
b.
318 µs x
106 s 1 ms x = 3.18 x 10−1 ms 1 μs 103 s
c.
93 km x
103 m 1 nm x = 9.3 x 1013 nm 9 1 km 10 m
d.
37.1 mm x
103 m 1 cm x = 3.71 cm 1 mm 102 m
1.134. a.
125 Å x
1010 m 1 μm = 1.25 x 10−2 µm x 1Å 106 m
b.
32.4 kg x
103 g 1 mg = 3.24 x 107 mg x 1 kg 103 g
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c.
16.8 cm x
d.
2.2 ns x
102 m 1 mm x = 168 mm 1 cm 103 m
109 s 1 μs x = 2.2 x 10−3 µs 1 ns 106 s
1.135. a.
5.91 kg x
103 g 1 mg = 5.91 x 106 mg x 3 1 kg 10 g
b.
753 mg x
103 g 1 μg = 7.53 x 105 µg x 6 1 mg 10 g
c.
90.1 MHz x
d.
498 mJ x
103 J 1 kJ x = 4.98 x 10−4 kJ 1 mJ 103 J
a.
7.19 µg x
106 g 1 mg = 7.19 x 10−3 mg x 3 1 μg 10 g
b.
104 pm x
1012 m 1Å x = 1.04 Å 1 pm 1010 m
c.
0.010 mm x
d.
0.0605 kPa x
106 Hz 1 kHz x = 9.01 x 104 kHz 1 MHz 103 Hz
1.136.
103 m 1 cm x = 1.0 x 10−3 cm 1 mm 102 m 103 Pa 1 cPa x = 6.05 x 103 cPa 1 kPa 102 Pa
⎛ 103 m ⎞ 1.137. Volume = 12,230 km x ⎜ ⎟ ⎝ 1 km ⎠ 3
⎛ 103 m ⎞ 1.138. Volume = 0.501 km3 x ⎜ ⎟ ⎝ 1 km ⎠
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3
3
3
1L ⎛ 1 dm ⎞ x ⎜ 1 ⎟ x = 1.2230 x 1016 L 3 1 dm ⎝ 10 m ⎠ 3
1L ⎛ 1 dm ⎞ x ⎜ 1 ⎟ x = 5.01 x 1011 L 3 10 m 1 dm ⎝ ⎠
27
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Chapter 1: Chemistry and Measurement
1.139. First, calculate the volume of the room in cubic feet. Volume = LWH = 10.0 ft x 11.0 ft x 9.0 ft = 990 ft3 Next, convert the volume to liters. 3
3
1L ⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ V = 990 ft3 x ⎜ = 2.80 x 104 L = 2.8 x 104 L ⎟ x ⎜ ⎟ x 3 3 10 cm 1 ft 1 in ⎝ ⎠ ⎝ ⎠ 1.140. First, calculate the volume of the cylinder in cubic feet. Volume = πr2l = π x (15.0 ft)2 x 5.0 ft = 3534 ft3 Next, convert the volume to liters. 3
3
1L ⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ V = 3534 ft3 x ⎜ = 1.00 x 105 L ⎟ x ⎜ ⎟ x 3 3 1 ft 1 in 10 cm ⎝ ⎠ ⎝ ⎠ = 1.0 x 105 L
1.141. Mass = 275 carats x
200 mg 103 g x = 55.00 g = 55.0 g 1 carat 1 mg
1.142. Mass = 49.6 x 106 troy oz x
31.10 g 1 ton x = 1.542 x 103 ton 1 troy oz 106 g
= 1.54 x 103 ton 1.143. The adhesive is not permanent, is easily removable, and does no harm to the object. 1.144. The scientific question Art Fry was trying to answer was: “Is there an adhesive that will not permanently stick things together?” 1.145. Chromatography depends on how fast a substance moves in a stream of gas or liquid, past a stationary phase to which the substance is slightly attracted. 1.146. The moving stream is a gaseous mixture of vaporized substances plus a gas such as helium, the carrier. The gas is passed through a column containing a stationary phase. As the gas passes through the column, the substances are attracted differently to the stationary column packing and thus are separated. The separated gases pass through a detector, and the results are displayed graphically.
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29
SOLUTIONS TO STRATEGY PROBLEMS
1.147. 5 x 10−2 mg = 0.05 mg. So 4.6 mg − 0.05 mg = 4.55 mg = 4.6 mg
1.148. V =
m 33.0 g = = 41.405 = 41.4 cm3 d 0.797 g/cm3
1.149. V = VA + VB =
mA 175 g + 50.0 mL = 58.33 mL + 50.0 mL + 50.0 mL = 3.00 g/mL dA
= 108.33 = 108.3 mL 1.150. d 832 mi = = 39.619 mi/h = 40. mi/h t 21 h
a.
r =
b.
832 mi 1.609 km x = 63.74 = 64 km/h 21 h 1 mi
c.
832 mi 1.609 km 1 gal 1 qt x x x = 12.19 km/L = 12 km/L 1 mi 0.9464 L 29 gal 4 qt
1.151. a.
Since there is the same number of atoms of the gas in each container, the mass is the same in each container.
b.
Since d = m/V, for the same mass, when the volume is smaller, the density is greater. Since the volume is less in container A, the density is greater.
c.
If the volume of container A was doubled, the density would decrease and become equal to the density in container B.
1.152. m 39.45 g = 0.92011 = 0.920 g/cm3 = V (3.50 cm)3
a.
d =
b.
400.4 mL 1 cm3 39.45 g = 368.41 = 368 g x x 1 1 mL (3.50 cm)3
1.153.
300 million people 1 x 1010 miles 1.609 km 1000 m 1 lightyear x x x x 1 1 person 1 mi 1 km 9.46 x 1015 m = 5.102 x 105 = 5 x 105 lightyears
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Chapter 1: Chemistry and Measurement
3
1.16 g 1 bucket 1L 1 mL ⎛ 2.54 cm ⎞ ⎛ 12 in ⎞ 1.154. x x x x ⎜ ⎟ x ⎜ ⎟ 3 1 bucket 4.08 L 1000 mL 1 cm ⎝ 1 in ⎠ ⎝ 1 ft ⎠ 2.38 x 103 ft 3 x = 1.9161 x 104 g = 1.92 x 104 g 1
3
1.155. a.
Since d = m/V, an increased mass for the same volume means a higher density for the solution.
b.
There would be less water for the same mass, so the density would be higher than in part a.
c.
Since there would be a greater volume of water for the same mass of salt, the density would be lower than in part a.
1.156. The water will have a lower density than the salt solution. It will not conduct electricity. It will have different chemical properties, such as reaction with silver nitrate solution; and different physical properties, such as the boiling point. Also, you could boil away the water from the salt solution and recover the salt.
■
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
1.157. The mass of hydrochloric acid is obtained from the density and the volume. Mass = density x volume = 1.096 g/mL x 50.0 mL = 54.80 g Next, from the law of conservation of mass, Mass of marble + mass of acid = mass of solution + mass of carbon dioxide gas Plugging in gives 10.0 g + 54.80 g = 60.4 g + mass of carbon dioxide gas Mass of carbon dioxide gas = 10.0 g + 54.80 g − 60.4 g = 4.40 g Finally, use the density to convert the mass of carbon dioxide gas to volume. Volume =
mass 4.40 g = = 2.447 L = 2.4 L density 1.798 g/ L
1.158. The mass of sulfuric acid is obtained from the density and the volume. Mass = density x volume = 1.153 g/mL x 50.0 mL = 57.65 g Next, from the law of conservation of mass, Mass of ore + mass of acid = mass of solution + mass of hydrogen sulfide gas
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31
Plugging in gives 10.8 g + 57.65 g = 65.1 g + mass of hydrogen sulfide gas Mass of hydrogen sulfide gas = 10.8 g + 57.65 g − 65.1 g = 3.35 g Finally, use the density to convert the mass of hydrogen sulfide gas to volume. Volume =
mass 3.35 g = = 2.40 L = 2.4 L density 1.393 g/ L
1.159. First, calculate the volume of the steel sphere. 3
⎛ 2.54 cm ⎞ 3 V = (4/3)πr = (4/3)π x (1.58 in) x ⎜ ⎟ = 270.7 cm ⎝ 1 in ⎠ 3
3
Next, determine the mass of the sphere using the density. Mass = density x volume = 7.88 g/cm3 x 270.7 cm3 = 2133 g = 2.13 x 103 g 1.160. First, calculate the volume of the balloon. Note that the radius is onehalf the diameter, or 1.50 ft. 3
3
1L ⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ V = (4/3)πr = (4/3)π x (1.50 ft) x ⎜ ⎟ x ⎜ ⎟ x 103 cm ⎝ 1 ft ⎠ ⎝ 1 in ⎠ = 400.3 L 3
3
Next, determine the mass of the helium using the density. Mass = density x volume = 0.166 g/L x 400.3 L = 66.453 g = 66.5 g 1.161. The area of the ice is 840,000 mi2 − 132,000 mi2 = 708,000 mi2. Now, determine the volume of this ice. Volume = area x thickness 2
3
⎛ 5280 ft ⎞ ⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ = 708,000 mi x 5000 ft x ⎜ ⎟ x ⎜ ⎟ x ⎜ ⎟ ⎝ 1 mi ⎠ ⎝ 1 ft ⎠ ⎝ 1 in ⎠
3
2
= 2.794 x 1021 cm3 Now use the density to determine the mass of the ice. Mass = density x volume = 0.917 g/cm3 x 2.794 x 1021 cm3 = 2.56 x 1021 g = 2.6 x 1021 g
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Chapter 1: Chemistry and Measurement
1.162. The height of the ice is 7500 ft − 1500 ft = 6000 ft. Now, determine the volume of the ice. Volume = area x thickness 2
3
⎛ 5280 ft ⎞ ⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ = 5,500,000 mi x 6000 ft x ⎜ ⎟ x ⎜ ⎟ x ⎜ ⎟ ⎝ 1 mi ⎠ ⎝ 1 ft ⎠ ⎝ 1 in ⎠
3
2
= 2.605 x 1022 cm3 Now use the density to determine the mass of the ice. Mass = density x volume = 0.917 g/cm3 x 2.605 x 1022 cm3 = 2.38 x 1022 g = 2.4 x 1022 g 1.163. Let x = mass of ethanol and y = mass of water. Then, use the total mass to write x + y = 49.6 g, or y = 49.6 g − x. Thus, the mass of water is 49.6 g − x. Next, Total volume = volume of ethanol + volume of water Since the volume is equal to the mass divided by density, you can write Total volume =
mass of ethanol mass of water + density of ethanol density of water
Substitute in the known and unknown values to get an equation for x. 54.2 cm3 =
x 49.6 g  x + 3 0.998 g/cm3 0.789 g/cm
Multiply both sides of this equation by (0.789)(0.998). Also, multiply both sides by g/cm3 to simplify the units. This gives the following equation to solve for x. (0.789)(0.998)(54.2) g = (0.998) x + (0.789)(49.6 g − x) 42.678 g = 0.998 x + 39.134 g − 0.789 x 0.209 x = 3.544 g x = mass of ethanol = 16.95 g
The percentage of ethanol (by mass) in the solution can now be calculated. Percent (mass) =
mass of ethanol 1 6.95 g x 100% = 34.1% = 34% x 100% = mass of solution 49.6 g
To determine the proof, you must first find the percentage by volume of ethanol in the solution. The volume of ethanol is obtained using the mass and the density. Volume =
mass of ethanol 1 6.95 g = = 21.48 cm3 3 density of ethanol 0.789 g/ cm
The percentage of ethanol (by volume) in the solution can now be calculated. Percent (volume) =
volume of ethanol 21.48 cm3 x 100% = x 100% = 39.63% volume of solution 54.2 cm3
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33
The proof can now be calculated. Proof = 2 x Percent (volume) = 2 x 39.63 = 79.27 = 79 proof 1.164. Let x = mass of gold and y = mass of silver. Then, use the total mass to write x + y = 9.35 g, or y = 9.35 g − x. Thus, the mass of silver is 9.35 g − x. Next, Total volume = volume of gold + volume of silver Since the volume is equal to the mass divided by density, you can write Total volume =
mass of gold mass of silver + density of gold density of silver
Substitute in the known and unknown values to get an equation for x. 0.654 cm3 =
x 9.35 g  x + 3 10.5 g/cm3 19.3 g/cm
Multiply both sides of this equation by (19.3)(10.5). Also multiply both sides by g/cm3 to simplify the units. This gives the following equation to solve for x. (19.3)(10.5)(0.654) g = (10.5) x + (19.3)(9.35 g − x) 132.53 g = 10.5 x + 180.45 g − 19.3 x 8.8 x = 47.92 g x = mass of gold = 5.445 g
The percentage of gold (by mass) in the solution can now be calculated. Percent (mass) =
mass of gold 5.445 g x 100% = x 100% = 58.2% = 58% mass of jewlrey 9.35 g
The relative amount of gold in the alloy can now be calculated. The fraction of gold in the alloy is 58.2% / 100% = 0.582. Thus, Proportion of gold = 24 karats x 0.582 = 13.9 karats = 14 karats 1.165. The volume of the mineral can be obtained from the mass difference between the water displaced and the air displaced, and the densities of water and air. Mass difference = 18.49 g − 16.21 g = 2.28 g Volume of mineral = =
mass difference density of water  density of air
2.28 g = 2.286 cm3 0.9982 g/cm3  1.205 x 103 g/cm3
The mass of the mineral is equal to its mass in air plus the weight of the displaced air. The weight of the displaced air is obtained from the volume of the mineral and the density of air.
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Chapter 1: Chemistry and Measurement
Mass of displaced air = density x volume = 1.205 g/L x 2.286 cm3 x
1L = 2.755 x 10−3 g 103 cm3
Mass of mineral = 18.49 g + 2.755 x 10−3 g = 18.4927 g The density of the mineral can now be calculated. Density =
mass 18.4927 g = 8.089 g/cm3 = 8.09 g/cm3 = volume 2.286 cm3
1.166. The volume of the mineral can be obtained from the mass of the water displaced and the density of water. Mass difference = 7.35 g − 5.40 g = 1.95 g Volume of mineral =
mass difference density of water  density of air
Volume of mineral =
1.95 g = 1.955 cm3 0.9982 g/cm3  1.205 x 103 g/cm3
The mass of the mineral is equal to its mass in air plus the weight of the displaced air. The weight of the displaced air is obtained from the volume of the mineral and the density of air. Mass of displaced air = density x volume = 1.205 g/L x 1.955 cm3 x
1L = 2.356 x 10−3 g 103 cm3
Mass of mineral = 7.35 g + 2.356 x 10−3 g = 7.352 g The density of the mineral can now be calculated. Density =
mass 7.352 g = = 3.760 g/cm3 = 3.76 g/cm3 volume 1.955 cm3
1.167. The volume of the object can be obtained from the mass of the ethanol displaced and the density of ethanol. Mass of ethanol displaced = 15.8 g − 10.5 g = 5.3 g Volume of object =
mass of ethanol 5.3 g = = 6.717 cm3 density of ethanol 0.789 g/ cm3
The density of the object can now be calculated. Density =
15.8 g mass = 2.352 g/cm3 = 2.4 g/cm3 = 3 volume 6.717 cm
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1.168. The volume of the metal can be obtained from the mass of the mercury displaced and the density of mercury. Mass of mercury displaced = 255 g − 101 g = 154 g Volume of object =
mass of mercury 154 g = = 11.323 cm3 density of mercury 13.6 g/cm3
The density of the metal can now be calculated. Density =
mass 255 g = = 22.51 g/cm3 = 22.5 g/cm3 volume 11.323 cm3
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CHAPTER 2
Atoms, Molecules, and Ions
■
SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 2.1. The element with atomic number 17 (the number of protons in the nucleus) is chlorine, symbol Cl. The mass number is 17 + 18 = 35. The symbol is 1735 Cl.
2.2. Multiply each isotopic mass by its fractional abundance; then sum: 34.96885 amu x 0.75771
=
26.496247
36.96590 amu x 0.24229
=
8.956467 35.452714 = 35.453 amu
The atomic mass of chlorine is 35.453 amu. 2.3. a.
Se: Group VIA, Period 4; nonmetal
b.
Cs: Group IA, Period 6; metal
c.
Fe: Group VIIIB, Period 4; metal
d.
Cu: Group IB, Period 4; metal
e.
Br: Group VIIA, Period 4; nonmetal
2.4. Take as many cations as there are units of charge on the anion and as many anions as there are units of charge on the cation. Two K+ ions have a total charge of 2+, and one CrO42− ion has a charge of 2−, giving a net charge of zero. The simplest ratio of K+ to CrO42− is 2:1, and the formula is K2CrO4.
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37
2.5. a.
CaO: Calcium, a Group IIA metal, is expected to form only a 2+ ion (Ca2+, the calcium ion). Oxygen (Group VIA) is expected to form an anion of charge equal to the group number minus 8 (O2−, the oxide ion). The name of the compound is calcium oxide.
b.
PbCrO4: Lead has more than one monatomic ion. You can find the charge on the Pb ion if you know the formula of the anion. From Table 2.5, the CrO4 refers to the anion CrO42− (the chromate ion). Therefore, the Pb cation must be Pb2+ to give electrical neutrality. The name of Pb2+ is lead(II) ion, so the name of the compound is lead(II) chromate.
2.6. Thallium(III) nitrate contains the thallium(III) ion, Tl3+, and the nitrate ion, NO3−. The formula is Tl(NO3)3. 2.7. a.
Dichlorine hexoxide
b.
Phosphorus trichloride
c.
Phosphorus pentachloride
a.
CS2
b.
SO3
a.
Boron trifluoride
b.
Hydrogen selenide
2.8.
2.9.
2.10. When you remove one H+ ion from HBrO4, you obtain the BrO4− ion. You name the ion from the acid by replacing ic with ate. The anion is called the perbromate ion. 2.11. Sodium carbonate decahydrate 2.12. Sodium thiosulfate is composed of sodium ions (Na+) and thiosulfate ions (S2O32−), so the formula of the anhydrous compound is Na2S2O3. Since the material is a pentahydrate, the formula of the compound is Na2S2O3•5H2O.
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Chapter 2: Atoms, Molecules, and Ions
2.13. Balance O first in parts (a) and (b) because it occurs in only one product. Balance S first in part (c) because it appears in only one product. Balance H first in part (d) because it appears in just one reactant as well as in the product. a.
Write a 2 in front of POCl3 for O; this requires a 2 in front of PCl3 for final balance: O2 + 2PCl3 → 2POCl3
b.
Write a 6 in front of N2O to balance O; this requires a 6 in front of N2 for final balance: P4 + 6N2O → 6N2 + P4O6
c.
Write 2As2S3 and 6SO2 to achieve an even number of oxygens on the right to balance what will always be an even number of oxygens on the left. The 2As2S3 then requires 2As2O3. Finally, to balance (6 + 12) O's on the right, write 9O2. 2As2S3 + 9O2 → 2As2O3 + 6SO2
d.
Write a 4 in front of H3PO4; this requires a 3 in front of Ca(H2PO4)2 for twelve H's. Ca3(PO4)2 + 4H3PO4 → 3Ca(H2PO4)2
■
ANSWERS TO CONCEPT CHECKS
2.1. CO2 is a compound that is a combination of 1 carbon atom and 2 oxygen atoms. Therefore, the chemical model must contain a chemical combination of 3 atoms stuck together with 2 of the atoms being the same (oxygen). Since each "ball" represents an individual atom, the three models on the left can be eliminated since they don't contain the correct number of atoms. Keeping in mind that balls of the same color represent the same element, only the model on the far right contains two elements with the correct ratio of atoms, 1:2; therefore, it must be CO2. 2.2. If 7999 out of 8000 alpha particles deflected back at the alphaparticle source, this would imply that the atom was a solid, impenetrable mass. Keep in mind that this is in direct contrast to what was observed in the actual experiments, where the majority of the alpha particles passed through without being deflected. 2.3. Elements are listed together in groups because they have similar chemical and/or physical properties. 2.4. a.
This compound is an ether because it has a functional group of an oxygen atom between two carbon atoms (–O–).
b.
This compound is an alcohol because it has an –OH functional group.
c.
This compound is a carboxylic acid because it has the –COOH functional group.
d.
This compound is a hydrocarbon because it contains only carbon and hydrogen atoms.
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39
2.5. A bottle containing a compound with the formula Al2Q3 would have an anion, Q, with a charge of 2−. The total positive charge in the compound due to the Al3+ is 6+ (2 x 3+), so the total negative charge must be 6−; Therefore, each Q ion must have a charge of 2−. Thus, Q would probably be an element from Group VIA on the periodic table.
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ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
2.1. Atomic theory is an explanation of the structure of matter in terms of different combinations of very small particles called atoms. Since compounds are composed of atoms of two or more elements, there is no limit to the number of ways in which the elements can be combined. Each compound has its own unique properties. A chemical reaction consists of the rearrangement of the atoms present in the reacting substances to give new chemical combinations present in the substances formed by the reaction. 2.2. Divide each amount of chlorine, 1.270 g and 1.904 g, by the lower amount, 1.270 g. This gives 1.000 and 1.499, respectively. Convert these to whole numbers by multiplying by 2, giving 2.000 and 2.998. The ratio of these amounts of chlorine is essentially 2:3. This is consistent with the law of multiple proportions because, for a fixed mass of iron (1 gram), the masses of chlorine in the other two compounds are in a ratio of small whole numbers. 2.3. A cathoderay tube consists of a negative electrode, or cathode, and a positive electrode, or anode, in an evacuated tube. Cathode rays travel from the cathode to the anode when a high voltage is turned on. Some of the rays pass through the hole in the anode to form a beam, which is then bent toward positively charged electric plates in the tube. This implies that a cathode ray consists of a beam of negatively charged particles (or electrons) and that electrons are constituents of all matter. 2.4. Millikan performed a series of experiments in which he obtained the charge on the electron by observing how a charged drop of oil falls in the presence and in the absence of an electric field. An atomizer introduces a fine mist of oil drops into the top chamber (Figure 2.6). Several drops happen to fall through a small hole into the lower chamber, where the experimenter follows the motion of one drop with a microscope. Some of these drops have picked up one or more electrons as a result of friction in the atomizer and have become negatively charged. A negatively charged drop will be attracted upward when the experimenter turns on a current to the electric plates. The drop’s upward speed (obtained by timing its rise) is related to its masstocharge ratio, from which you can calculate the charge on the electron. 2.5. The nuclear model of the atom is based on experiments of Geiger, Marsden, and Rutherford. Rutherford stated that most of the mass of an atom is concentrated in a positively charged center called the nucleus around which negatively charged electrons move. The nucleus, although it contains most of the mass, occupies only a very small portion of the space of the atom. Most of the alpha particles passed through the metal atoms of the foil undeflected by the lightweight electrons. When an alpha particle does happen to hit a metalatom nucleus, it is scattered at a wide angle because it is deflected by the massive, positively charged nucleus (Figure 2.8).
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Chapter 2: Atoms, Molecules, and Ions
2.6. The atomic nucleus consists of two kinds of particles, protons and neutrons. The mass of each is about the same, on the order of 1.67 x 10−27 kg, and about 1800 times that of the electron. An electron has a much smaller mass, on the order of 9.11 x 10−31 kg. The neutron is electrically neutral, but the proton is positively charged. An electron is negatively charged. The charges on the proton and the electron are equal in magnitude. 2.7. Protons (hydrogen nuclei) were discovered as products of experiments involving the collision of alpha particles with nitrogen atoms that resulted in a proton being knocked out of the nitrogen nucleus. Neutrons were discovered as the radiation product of collisions of alpha particles with beryllium atoms. The resulting radiation was discovered to consist of particles having a mass approximately equal to that of a proton and having no charge (neutral). 2.8. Oxygen consists of three different isotopes, each having 8 protons but a different number of neutrons. 2.9. The percentages of the different isotopes in most naturally occurring elements have remained essentially constant over time and in most cases are independent of the origin of the element. Thus, what Dalton actually calculated were average atomic masses (relative masses). He could not weigh individual atoms, but he could find the average mass of one atom relative to the average mass of another. 2.10. A mass spectrometer measures the masstocharge ratio of positively charged atoms (and molecules). It produces a mass spectrum, which shows the relative numbers of atoms (fractional abundances) of various masses (isotopic masses). The mass spectrum gives us all the information needed to calculate the atomic weight. 2.11. The atomic mass of an element is the average atomic mass for the naturally occurring element expressed in atomic mass units. The atomic mass would be different elsewhere in the universe if the percentages of isotopes in the element were different from those on earth. 2.12. The element in Group IVA and Period 5 is tin (atomic number 50). 2.13. A metal is a substance or mixture that has characteristic luster, or shine, and is generally a good conductor of heat and electricity. 2.14. The formula for ethane is C2H6. 2.15. A molecular formula gives the exact number of different atoms of an element in a molecule. A structural formula is a chemical formula that shows how the atoms are bonded to one another in a molecule.
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41
2.16. Organic molecules contain carbon combined with other elements such as hydrogen, oxygen, and nitrogen. An inorganic molecule is composed of elements other than carbon. Some inorganic molecules that contain carbon are carbon monoxide (CO), carbon dioxide (CO2), carbonates, and cyanides. 2.17. An ionic binary compound: NaCl; a molecular binary compound: H2O. 2.18. a.
The elements are represented by B, F, and I.
b.
The compounds are represented by A, E, and G.
c.
The mixtures are represented by C, D, and H.
d.
The ionic solid is represented by A.
e.
The gas made up of an element and a compound is represented by C.
f.
The mixtures of elements are represented by D and H.
g.
The solid element is represented by F.
h.
The solids are represented by A and F.
i.
The liquids are represented by E, H, and I.
2.19. In the Stock system, CuCl is called copper(I) chloride, and CuCl2 is called copper(II) chloride. One of the advantages of the Stock system is that more than two different ions of the same metal can be named with this system. In the former (older) system, a new suffix other than ic and ous must be established and/or memorized. 2.20. A balanced chemical equation has the numbers of atoms of each element equal on both sides of the arrow. The coefficients are the smallest possible whole numbers. 2.21. The answer is a: 50 p, 69 n, and 48 e–. 2.22. The answer is d: 65%. 2.23. The answer is c: magnesium hydroxide, Mg(OH)2. 2.24. The answer is b: Li.
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Chapter 2: Atoms, Molecules, and Ions
ANSWERS TO CONCEPT EXPLORATIONS
2.25. Part I a.
Average mass =
2.00 g + 2.00 g + 2.00 g + 2.00 g = 2.00 g 4
Part II
2.00 g + 1.75 g + 3.00 g + 1.25 g = 2.00 g 4
a.
Average mass =
b.
The average mass of a sphere in the two samples is the same. The average does not represent the individual masses. Also, it does not indicate the variability in the individual masses.
Part III
a.
b.
50 blue spheres 2.00 g x = 100.00 g 1 1 blue sphere
If 50 spheres were removed at random, then 50 spheres would remain in the jar. You can use the average mass to calculate the total mass. 50 spheres 2.00 g = 100.00 g x 1 1 sphere
c.
No, the average mass does not represent the mass of an individual sphere.
d.
80.0 g 1 blue sphere = 40.0 blue spheres x 1 2.00 g
e.
60.0 g 1 sphere x = 30.0 spheres 1 2.00 g
The assumption is that the average mass of a sphere (2.00 g) can be used in the calculation. Also, assume the sample is well mixed. Part IV
a.
For green spheres: X = For blue spheres: X =
3 green = 0.750 3 green + 1 blue
1 blue = 0.250 3 green + 1 blue 3.00 g 1.00 g + (0.250) x = 2.50 g 1 green sphere 1 blue sphere
b.
Average mass = (0.750) x
c.
The atomic mass of an element is the weighted average calculated as in part (b) of Part IV above, using fractional abundances and individual masses.
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43
2.26. a.
Atom A has three protons.
b.
The number of protons is the same as the atomic number for that element.
c.
Lithium, Li, has atomic number 3.
d.
The charge on element A is zero. There are three protons, each +1, and three electrons, each −1. This yields a net charge of zero.
e.
The nuclide symbol for A is 73 Li .
f.
Atom B has three protons and thus atomic number 3. It is lithium, with symbol Li.
g.
Atom B has three protons and three neutrons. Its mass number is 6. This is different from the mass number of atom A, which is 7.
h.
Atom B has three protons and three electrons and thus is neutral.
i.
The nuclide symbol for B is 63 Li . The atomic number is 3 and the mass number is 6 for both nuclides.
j.
■
6 + 3 Li
6 3 Li
k.
Two different lithium isotopes are depicted, lithium6 and lithium7.
l.
The mass number of an isotope is the total number of protons and neutrons in its nucleus. Its value is an integer. It is related to the mass of the isotope but not related to the atomic mass, which is a weighted average over the fractional abundances and isotopic masses.
ANSWERS TO CONCEPTUAL PROBLEMS
2.27. If atoms were balls of positive charge with the electrons evenly distributed throughout, there would be no massive, positive nucleus to deflect the beam of alpha particles when it is shot at the gold foil. 2.28. Once the subscripts of the compounds in the original chemical equation are changed (the molecule N2 was changed to the atom N), the substances reacting are no longer the same. Your friend may be able to balance the second equation, but it is no longer the same chemical reaction. 2.29. You could group elements by similar physical properties such as density, mass, color, conductivity, etc., or by chemical properties, such as reaction with air, reaction with water, etc.
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Chapter 2: Atoms, Molecules, and Ions
2.30. You would name the ions with the formulas XO42−, XO32−, and XO2− using the name for XO22− (excite) as the example to determine the root name of the element X (exc). Thus XO42−, with the greatest number of oxygen atoms in the group, would be perexcate; XO32− would be excate; and XO2−, with the fewest oxygen atoms in the group, would be hypoexcite. 2.31. a.
In each case, the total positive charge and the total negative charge in the compounds must cancel. Therefore, the compounds with the cations X+, X2+, and X5+, combined with the SO42− anion, are X2SO4, XSO4, and X2(SO4)5, respectively.
b.
You recognize the fact that whenever a cation can have multiple oxidation states (1+, 2+, and 5+ in this case), the name of the compound must indicate the charge. Therefore, the names of the compounds in part (a) would be exy(I) sulfate, exy(II) sulfate, and exy(V) sulfate, respectively.
a.
This model contains three atoms of two different elements (H and O). Therefore, the model is of H2O.
b.
This model represents a crystal that contains two different elements in a 1:1 ratio (K+ and Cl−). Therefore, the model represents the ionic compound, KCl.
c.
This model contains six atoms, four of which are the same (H), and two others (C and O). Therefore, the model is of CH3OH.
d.
This model contains four atoms of two different elements (N and H). Therefore, the model is of NH3.
2.32.
2.33. A potassium39 atom in this case would contain 19 protons and 20 neutrons. If the charge of the proton were twice that of an electron, it would take twice as many electrons as protons, or 38 electrons, to maintain a charge of zero. 2.34. a.
Since the mass of an atom is not due only to the sum of the masses of the protons, neutrons, and electrons, when you change the element in which you are basing the amu, the mass of the amu must change as well.
b.
Since the amount of material that makes up a hydrogen atom doesn’t change, when the amu gets larger, as in this problem, the hydrogen atom must have a smaller mass in amu.
a.
2Li + Cl2 → 2LiCl
b.
16Na + S8 → 8Na2S
c.
2Al + 3Br2 → 2AlBr3
d.
3Mg + N2 → Mg3N2
2.35.
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e.
6Ca + P4 → 2Ca3P2
2.36.
a.
■
b.
2A + B2 → 2AB
c.
Some possible real elements with formula B2 are F2, Cl2, Br2, and I2.
SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off.
2.37. a.
Argon
b.
Zinc
c.
Silver
d.
Magnesium
a.
Calcium
b.
Copper
c.
Mercury
d.
Tin
a.
K
b.
S
c.
Fe
d.
Mn
2.38.
2.39.
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45
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Chapter 2: Atoms, Molecules, and Ions
2.40. a.
C
b.
Na
c.
Ni
d.
Pb
2.41. The mass of the electron is found by multiplying the two values: 1.602 x 10−19 C x
5.64 x 1012 kg = 9.035 x 10−31 kg = 9.04 x 10−31 kg 1C
2.42. The mass of the fluorine atom is found by multiplying the two values: 1.602 x 10−19 C x
1.97 x 107 kg = 3.155 x 10−26 kg = 3.16 x 10−26 kg 1C
2.43. The isotope of atom A is the atom with 18 protons, atom C; the atom that has the same mass number as atom A (37) is atom D. 2.44. The isotope of atom A is the atom with 32 protons, atom B; the atom that has the same mass number as atom A (71) is atom D. 2.45. Each isotope of chlorine (atomic number 17) has 17 protons. Each neutral atom will also have 17 electrons. The number of neutrons for Cl35 is 35 – 17 = 18 neutrons. The number of neutrons for Cl37 is 37 – 17 = 20 neutrons. 2.46. Each isotope of lithium (atomic number 3) has three protons. Each neutral atom will also have three electrons. The number of neutrons for Li6 is 6 – 3 = 3 neutrons. The number of neutrons for Li7 is 7 – 3 = 4 neutrons. 2.47. The element with 14 protons in its nucleus is silicon (Si). The mass number = 14 + 14 = 28. The notation for the nucleus is 1428 Si . 2.48. The element with 11 protons in its nucleus is sodium (Na). The mass number = 11 + 11 = 22. The notation for the nucleus is
22 11
Na .
2.49. Since the atomic ratio of nitrogen to hydrogen is 1:3, divide the mass of N by onethird of the mass of hydrogen to find the relative mass of N. Atomic mass of N 7.933 g 13.901 g N 13.90 = = = Atomic mass of H 1/3 x 1.712 g 1gH 1
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2.50. Since the atomic ratio of hydrogen to sulfur is 2:1, divide the mass of S by onehalf of the mass of hydrogen to find the relative mass of S. Atomic mass of S 9.330 g 31.78 g S 31.8 = = = Atomic mass of H 1 1/2 x 0.587 g 1gH 2.51. Multiply each isotopic mass by its fractional abundance, and then sum: X63: 62.930 x 0.6909
=
43.4783
X65: 64.928 x 0.3091
=
20.0692 63.5475 = 63.55 amu
The element is copper, atomic mass 63.546 amu. 2.52. Multiply each isotopic mass by its fractional abundance, and then sum: 49.9472 x 0.002500
=
50.9440 x 0.9975
=
0.124868 50.81664
= 50.94150 = 50.94 amu The atomic mass of this element is 50.94 amu. The element is vanadium (V). 2.53. Multiply each isotopic mass by its fractional abundance, and then sum: 38.964 x 0.9326
=
36.3378
39.964 x 1.00 x 10−4
=
0.0039964
40.962 x 0.0673
=
2.75674
= 39.09853 = 39.10 amu The atomic mass of this element is 39.10 amu. The element is potassium (K). 2.54. Multiply each isotopic mass by its fractional abundance, and then sum: 27.977 x 0.9221
=
25.798
28.976 x 0.0470
=
1.362
29.974 x 0.0309
=
0.9262
= 28.086 = 28.09 amu The atomic mass of this element is 28.09 amu. The element is silicon (Si). 2.55. According to the picture, there are 20 atoms, 5 of which are brown and 15 of which are green. Using the isotopic masses in the problem, the atomic mass of element X is 5 15 (23.02 amu) + (25.147 amu) = 5.755 + 18.8602 = 24.6152 = 24.615 amu 20 20
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Chapter 2: Atoms, Molecules, and Ions
2.56. According to the picture, there are 24 atoms, 8 of which are brown and 16 of which are green. Using the isotopic masses in the problem, the atomic mass of element X is 8 16 (47.510 amu) + (51.126 amu) = 15.8366 + 34.084 = 49.9206 = 49.921 amu 24 24 2.57. a.
C: Group IVA, Period 2; nonmetal
b.
Po: Group VIA, Period 6; metal
c.
Cr: Group VIB, Period 4; metal
d.
Mg: Group IIA, Period 3; metal
e.
B: Group IIIA, Period 2; metalloid
a.
S: Group VIA, Period 3; nonmetal
b.
Fe: Group VIIIB, Period 4; metal
c.
Ba: Group IIA, Period 6; metal
d.
Cu: Group IB, Period 4; metal
e.
Ne: Group VIIIA, Period 2; nonmetal
a.
Tellurium
b.
Aluminum
a.
Bismuth
b.
Magnesium
2.58.
2.59.
2.60.
2.61. Examples are: a.
O (oxygen)
b.
Na (sodium)
c.
Fe (iron)
d.
Ce (cerium)
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49
2.62. Examples are: a.
Zr (zirconium)
b.
Cl (chlorine)
c.
B (boron)
d.
U (uranium)
2.63. They are different in that the solid sulfur consists of S8 molecules, whereas the hot vapor consists of S2 molecules. The S8 molecules are four times as heavy as the S2 molecules. Hot sulfur is a mixture of S8 and S2 molecules, but at high enough temperatures only S2 molecules are formed. Both hot sulfur and solid sulfur consist of molecules with only sulfur atoms. 2.64. They are different in that the solid phosphorus consists of P4 molecules, whereas the hot vapor consists of P2 molecules. The P4 molecules are twice as heavy as the P2 molecules. Hot phosphorus is a mixture of P4 and P2 molecules above the boiling point, but at high temperatures only P2 molecules are formed. Both solid phosphorus and phosphorus vapor consist of molecules with only phosphorus atoms. 2.65. The number of nitrogen atoms in the 1.50g sample of N2O is 2.05 x 1022 N2O molecules x
2 N atoms = 4.10 x 1022 N atoms 1 N 2 O molecule
The number of nitrogen atoms in 1.00 g of N2O is 1.00 g x
4.10 x 1022 N atoms = 2.733 x 1022 N atoms = 2.73 x 1022 N atoms 1.50 g N 2 O
2.66. Since each HNO3 molecule contains one N atom, in 4.30 x 1022 HNO3 molecules there are 4.30 x 1022 N atoms. The number of oxygen atoms in 2.81 g of HNO3 is obtained as follows. 2.81 g x
4.30 x 1022 HNO3 molecules 3 O atoms x 4.50 g HNO3 1 HNO3 molecule = 8.0553 x 1022 O atoms = 8.06 x 1022 O atoms
2.67. 3.3 x 1021 H atoms x
1 NH 3 molecule = 1.1 x 1021 NH3 molecules 3 H atoms
2.68. 4.2 x 1023 H atoms x
1 C2 H 5OH molecule = 7.0 x 1022 C2H5OH molecules 6 H atoms
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Chapter 2: Atoms, Molecules, and Ions
2.69. a.
N2H4
b.
H2O2
c.
C3H8O
d.
PCl3
a.
C3H8O3
b.
Si2H6
c.
NH3O
d.
SF4
a.
PCl5
b.
NO2
c.
C3H6O2
a.
H2SO4
b.
C6H6
c.
C3H6O
2.70.
2.71.
2.72.
2.73.
1 NO3 ion 1 Fe(NO3 ) 2 unit 1 Fe atom 1 Fe atom 1 x = = x 6 O atoms 6 3 O atoms 1 Fe(NO3 ) 2 unit 2 NO3 ions Thus, the ratio of iron atoms to oxygen atoms is one Fe atom to six O atoms.
2.74.
1 (NH 4 )3 PO 4 unit 3 NH 4 + ions 4 H atoms 12 H atoms 3 x x = = + 4 O atoms 1 NH 4 ion 4 O atoms 1 1 (NH 4 )3 PO 4 unit
Thus, the ratio of hydrogen atoms to oxygen atoms is three H atoms to one O atom. 2.75. a.
Fe(CN)3
b.
K2SO4
c.
Li3N
d.
Ca3P2
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51
2.76. a.
Co3N2
b.
(NH4)3PO4
c.
Na2CO3
d.
Fe(OH)3
a.
Na2SO4: sodium sulfate (Group IA forms only 1+ cations.)
b.
CaS: calcium sulfide (Group IIA forms only 2+ cations.)
c.
CuCl: copper(I) chloride (Group IB forms 1+ and 2+ cations.)
d.
Cr2O3: chromium(III) oxide (Group VIB forms numerous oxidation states.)
a.
Na2O: sodium oxide (Group IA forms only 1+ cations.)
b.
Mn2O3: manganese(III) oxide (Group VIIB forms numerous oxidation states.)
c.
NH4HCO3: ammonium bicarbonate or ammonium hydrogen carbonate.
d.
Cu(NO3)2: copper(II) nitrate (Group IB forms 1+ and 2+ cations.)
a.
Lead(II) permanganate: Pb(MnO4)2 (Permanganate is in Table 2.5.)
b.
Barium hydrogen carbonate: Ba(HCO3)2 (The HCO3− ion is in Table 2.5.)
c.
Cesium sulfide: Cs2S (Group 1A ions form 1+ cations.)
d.
Iron(II) acetate: Fe(C2H3O2)2 (The acetate ion = 1− [from Table 2.5]; for the sum of charges to be zero, two must be used.)
a.
Sodium thiosulfate: Na2S2O3 (The S2O32− is in Table 2.5.)
b.
Copper(I) hydroxide: CuOH (Cu must be 1+ to balance one OH−.)
c.
Calcium hydrogen carbonate: Ca(HCO3)2 (The HCO3− ion is in Table 2.5.)
d.
Nickel(II) phosphide: Ni3P2 (Two P3− must be used to balance three Ni2+.)
2.77.
2.78.
2.79.
2.80.
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Chapter 2: Atoms, Molecules, and Ions
2.81. a.
Molecular
b.
Ionic
c.
Molecular
d.
Ionic
a.
Ionic
b.
Molecular
c.
Molecular
d.
Ionic
a.
Dinitrogen monoxide
b.
Tetraphosphorus dec(a)oxide
c.
Arsenic trichloride
d.
Dichlorine hept(a)oxide
a.
Dinitrogen difluoride
b.
Carbon tetrachloride
c.
Dinitrogen pent(a)oxide
d.
Tetr(a)arsenic hex(a)oxide
a.
NBr3
b.
XeO4
c.
OF2
d.
Cl2O5
a.
ClF3
b.
NO2
c.
N2F4
d.
PF5
2.82.
2.83.
2.84.
2.85.
2.86.
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2.87. a.
Selenium trioxide
b.
Disulfur dichloride
c.
Carbon monoxide
a.
Nitrogen trifluoride
b.
Diphosphorus tetrahydride
c.
Oxygen difluoride
a.
Bromic acid: HBrO3
b.
Hyponitrous acid: H2N2O2
c.
Disulfurous acid: H2S2O5
d.
Arsenic acid: H3AsO4
a.
Selenous acid: H2SeO3
b.
Sulfurous acid: H2SO3
c.
Hypoiodous acid: HIO
d.
Nitrous acid: HNO2
2.88.
2.89.
2.90.
2.91. Na2SO4•10H2O is sodium sulfate decahydrate. 2.92. NiSO4•6H2O is nickel(II) sulfate hexahydrate. 2.93. Iron(II) sulfate heptahydrate is FeSO4•7H2O. 2.94. Cobalt(II) chloride hexahydrate is CoCl2•6H2O.
2.95. Pb(NO3)2 x
6 O atoms 3 O atoms + K2CO3 x = 9 O atoms 1 Pb(NO3 ) 2 unit 1 K 2 CO3 unit
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Chapter 2: Atoms, Molecules, and Ions
2.96. 2 PbO x
1 O atom 2 O atoms + 2 SO2 x = 6 O atoms 1 PbO unit 1 SO 2 unit
The equation is not balanced as written. 2.97. a.
Balance: Sn + NaOH → Na2SnO2 + H2 If Na is balanced first by writing a 2 in front of NaOH, the entire equation is balanced. Sn + 2NaOH → Na2SnO2 + H2
b.
Balance: Al + Fe3O4 → Al2O3 + Fe First balance O (it appears once on each side) by writing a 3 in front of Fe3O4 and a 4 in front of Al2O3: Al + 3Fe3O4 → 4Al2O3 + Fe Now balance Al against the 8 Al's on the right and Fe against the 9 Fe's on the left: 8Al + 3Fe3O4 → 4Al2O3 + 9Fe
c.
Balance: CH3OH + O2 → CO2 + H2O First balance H (it appears once on each side) by writing a 2 in front of H2O: CH3OH + O2 → CO2 + 2H2O To avoid fractional coefficients for O, multiply the equation by 2: 2CH3OH + 2O2 → 2CO2 + 4H2O Finally, balance O by changing 2O2 to "3O2"; this balances the entire equation: 2CH3OH + 3O2 → 2CO2 + 4H2O
d.
Balance: P4O10 + H2O → H3PO4 First balance P (it appears once on each side) by writing a 4 in front of H3PO4: P4O10 + H2O → 4H3PO4 Finally, balance H by writing a 6 in front of H2O; this balances the entire equation: P4O10 + 6H2O → 4H3PO4
e.
Balance: PCl5 + H2O → H3PO4 + HCl First balance Cl (it appears once on each side) by writing a 5 in front of HCl: PCl5 + H2O → H3PO4 + 5HCl Finally, balance H by writing a 4 in front of H2O; this balances the entire equation: PCl5 + 4H2O → H3PO4 + 5HCl
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55
2.98. a.
Balance: Cl2O7 + H2O → HClO4 Balance Cl (appears only once on each side) by writing a 2 in front of HClO4; this balances the entire equation: Cl2O7 + H2O → 2HClO4
b.
Balance: MnO2 + HCl → MnCl2 + Cl2 + H2O First balance O (appears only once on each side) by writing a 2 in front of H2O: MnO2 + HCl → MnCl2 + Cl2 + 2H2O Finally, balance H and Cl by writing a 4 in front of HCl to balance the entire equation: MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
c.
Balance: Na2S2O3 + I2 → NaI + Na2S4O6 First balance S by writing a 2 in front of Na2S2O3: 2Na2S2O3 + I2 → NaI + Na2S4O6 Finally, balance Na by writing a 2 in front of NaI; this balances the entire equation: 2Na2S2O3 + I2 → 2NaI + Na2S4O6
d.
Balance: Al4C3 + H2O → Al(OH)3 + CH4 First balance Al with a 4 in front of Al(OH)3, and balance C with a 3 in front of CH4: Al4C3 + H2O → 4Al(OH)3 + 3CH4 Finally, balance H and O with a 12 in front of H2O; this balances the entire equation: Al4C3 + 12H2O → 4Al(OH)3 + 3CH4
e.
Balance: NO2 + H2O → HNO3 + NO First balance H with a 2 in front of HNO3: NO2 + H2O → 2HNO3 + NO Finally, balance N with a 3 in front of NO2; this balances the entire equation: 3NO2 + H2O → 2HNO3 + NO
2.99. Balance: Ca3(PO4)2(s) + H2SO4(aq) → CaSO4(s) + H3PO4(aq) Balance Ca first with a 3 in front of CaSO4: Ca3(PO4)2(s) + H2SO4(aq) → 3CaSO4(s) + H3PO4(aq) Next, balance the P with a 2 in front of H3PO4: Ca3(PO4)2(s) + H2SO4(aq) → 3CaSO4(s) + 2H3PO4(aq) Finally, balance the S with a 3 in front of H2SO4; this balances the equation: Ca3(PO4)2(s) + 3H2SO4(aq) → 3CaSO4(s) + 2H3PO4(aq)
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Chapter 2: Atoms, Molecules, and Ions
2.100. Balance: Na(s) + H2O(l) → NaOH(aq) + H2(g) Balance H first with a 2 in front of H2O and NaOH: Na + 2H2O → 2NaOH + H2 Then, balance Na with a 2 in front of Na; this balances the equation: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) 2.101. Balance: NH4Cl(aq) + Ba(OH)2(aq) → NH3(g) + BaCl2(aq) + H2O(l) Balance O first with a 2 in front of H2O: NH4Cl + Ba(OH)2 → NH3 + BaCl2 + 2H2O Balance H with a 2 in front of NH4Cl and a 2 in front of NH3; this balances the equation: Δ 2NH4Cl(aq) + Ba(OH)2(aq) ⎯⎯ → 2NH3(g) + BaCl2(aq) + 2H2O(l)
2.102. Balance: PbS(s) + PbSO4(s) → Pb(l) + SO2(g) Balance S first with a 2 in front of SO2: PbS + PbSO4 → Pb + 2SO2 Balance Pb with a 2 in front of Pb; this balances the equation: Δ PbS(s) + PbSO4(s) ⎯⎯ → 2Pb(l) + 2SO2(g)
■
SOLUTIONS TO GENERAL PROBLEMS
2.103. Calculate the ratio of oxygen for 1 g (fixed amount) of nitrogen in both compounds: A:
2.755 g O 2.2844 g O = 1.206 g N 1gN
B:
4.714 g O 2.8552 g O = 1.651 g N 1gN
Next, find the ratio of oxygen per gram of nitrogen for the two compounds. g O in B/1 g N 2.8552 g O 1.2498 g O = = g O in A/1 g N 2.2844 g O 1g O B contains 1.25 times as many O atoms as A does (there are five O's in B for every four O's in A). 2.104. Calculate the ratio of oxygen for 1 g (fixed amount) of sulfur in both compounds: A:
1.811 g O 1.4966 g O = 1.210 g S 1gS
B:
1.779 g O 0.99775 g O = 1.783 g S 1gS
Next, find the ratio of oxygen per gram of sulfur for the two compounds. g O in A/1 g S 1.4966 g O 1.4999 g O = = g O in B/1 g S 0.99775 g O 1g O
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57
A contains 1.50 times as many O atoms as B does (there are three O's in A for every two O's in B). 2.105. The smallest difference is between −1.12 x 10−18 C and 9.60 x 10−19 C and is equal to −1.6 x 10−19 C. If this charge is equivalent to one electron, the number of excess electrons on a drop may be found by dividing the negative charge by the charge of one electron. Drop 1:
3.20 x 1019 C = 2.0 ≅ 2 electrons 1.6 x 1019 C
Drop 2:
6.40 x 1019 C = 4.0 ≅ 4 electrons 1.6 x 1019 C
Drop 3:
9.60 x 1019 C = 6.0 ≅ 6 electrons 1.6 x 1019 C
Drop 4:
1.12 x 1018 C = 7.0 ≅ 7 electrons 1.6 x 1019 C
2.106. The smallest difference in charge for the oil drop is −1.85 x 10−19; assume this is the fundamental unit of negative charge. Use this to divide into each drop's charge: Drop 1:
5.55 x 1019 C = 3.0 ≅ 3 electrons 1.85 x 1019 C
Drop 2:
9.25 x 1019 C = 5.0 ≅ 5 electrons 1.85 x 1019 C
Drop 3:
1.11 x 1018 C = 6.0 ≅ 6 electrons 1.85 x 1019 C
Drop 4:
1.48 x 1018 C = 8.0 ≅ 8 electrons 1.85 x 1019 C
2.107. For the Eu atom to be neutral, the number of electrons must equal the number of protons, so a neutral europium atom has 63 electrons. The +3 charge on the Eu3+ indicates there are three more protons than electrons, so the number of electrons is 63 − 3 = 60. 2.108. For the Cs atom to be neutral, the number of electrons must equal the number of protons, so a neutral cesium atom has 55 electrons. The +1 charge on the Cs+ indicates there is one more proton than electrons, so the number of electrons is 55 − 1 = 54. 2.109. The number of protons = mass number − number of neutrons = 81 − 46 = 35. The element with Z = 35 is bromine (Br). The ionic charge = number of protons − number of electrons = 35 − 36 = −1. Symbol:
81 35
Br  .
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Chapter 2: Atoms, Molecules, and Ions
2.110. The number of protons = mass number − number of neutrons = 80 − 55 = 25. The element with Z = 25 is manganese (Mn). The ionic charge = number of protons − number of electrons = 25 − 23 = +2. Symbol:
80 25
Mn 2 + .
2.111. The sum of the fractional abundances must equal 1. Let y equal the fractional abundance of 63 Cu. Then the fractional abundance of 65Cu equals (1 − y). We write one equation in one unknown: Atomic mass = 63.546 = 62.9298y + 64.9278(1 − y) 63.546 = 64.9278 − 1.9980y 64.9278  63.546 y = = 0.69159 1.9980 The fractional abundance of 63Cu = 0.69159 = 0.6916. The fractional abundance of 65Cu = 1 − 0.69159 = 0.30841 = 0.3084. 2.112. As in the previous problem, the sum of the fractional abundances must equal 1. Thus, the abundance of one isotope can be expressed in terms of the other. Let y equal the fractional abundance of Ag107. Then the fractional abundance of Ag109 equals (1 − y). We can write one equation in one unknown: Atomic mass = 107.87 = 106.91y + 108.90(1 − y) 107.87 = 108.90 − 1.99y y =
108.90  107.87 = 0.51758 1.99
The fractional abundance of Ag107 = 0.51758 = 0.518. The fractional abundance of Ag109 = 1 − 0.51758 = 0.48241 = 0.482. 2.113. a.
Bromine, Br
b.
Hydrogen, H
c.
Niobium, Nb
d.
Fluorine, F
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2.114. a.
Bromine, Br
b.
Mercury, Hg
c.
Gallium, Ga
d.
Rubidium, Rb
2.115. a.
Chromium(III) ion
b.
Lead(IV) ion
c.
Copper(I) ion
d.
Copper(II) ion
2.116. a.
Manganese(II) ion
b.
Nickel(II) ion
c.
Cobalt(II) ion
d.
Cobalt(III) ion
2.117. All possible ionic compounds: Na2SO4, NaCl, NiSO4, and NiCl2. 2.118. All possible ionic compounds: CaO, Ca(NO3)2, Cr2O3, and Cr(NO3)3. 2.119. a.
Tin(II) phosphate
b.
Ammonium nitrite
c.
Magnesium hydroxide
d.
Chromium(II) sulfate
2.120. a.
Copper(II) nitrite
b.
Ammonium phosphide
c.
Sodium sulfate
d.
Mercury(II) chloride
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Chapter 2: Atoms, Molecules, and Ions
2.121. a.
Hg2S [Mercury(I) exists as the polyatomic Hg22+ ion (Table 2.5).]
b.
Co2(SO3)3
c.
(NH4)2Cr2O7
d.
AlN
2.122. a.
H2O2
b.
AlPO4
c.
Pb3P4
d.
BF3
2.123. a.
Arsenic tribromide
b.
Hydrogen selenide (dihydrogen selenide)
c.
Diphosphorus pent(a)oxide
d.
Silicon dioxide
2.124. a.
Chlorine tetrafluoride
b.
Carbon disulfide
c.
Nitrogen trifluoride
d.
Sulfur hexafluoride
2.125. a.
Balance the C and H first: C2H6 + O2 → 2CO2 + 3H2O Avoid a fractional coefficient for O on the left by doubling all coefficients except O2's, and then balance the O's: 2C2H6 + 7O2 → 4CO2 + 6H2O
b.
Balance the P first: P4O6 + H2O → 4H3PO3 Then balance the O (or H), which also gives the H (or O) balance: P4O6 + 6H2O → 4H3PO3
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c.
61
Balancing the O first is the simplest approach. (Starting with K and Cl and then proceeding to O will cause the initial coefficient for KClO3 to be changed in balancing O last.) 4KClO3 → KCl + 3KClO4
d.
Balance the N first: (NH4)2SO4 + NaOH → 2NH3 + H2O + Na2SO4 Then balance the Na, followed by O; this also balances the H: (NH4)2SO4 + 2NaOH → 2NH3 + 2H2O + Na2SO4
e.
Balance the N first: 2NBr3 + NaOH → N2 + NaBr + HOBr Note that NaOH and HOBr each have one O and that NaOH and NaBr each have one Na; thus the coefficients of all three must be equal; from 2NBr3, this coefficient must be 6Br/2 = 3: 2NBr3 + 3NaOH → N2 + 3NaBr + 3HOBr
2.126. a.
Balance the Na first: 3NaOH + H3PO4 → Na3PO4 + H2O Then balance the O; this also balances the H: 3NaOH + H3PO4 → Na3PO4 + 3H2O
b.
Balance the Cl with a 4 in front of the HCl; then balance the O's with a 2 in front of H2O: SiCl4 + 2H2O → SiO2 + 4HCl
c.
Balance the O first with an 8 in front of CO; then balance the C with an 8 in front of C: Ca3(PO4)2 + 8C → Ca3P2 + 8CO
d.
Balance the O by multiplying O2 by 3 and doubling both products to give a total of six O's on both sides of the equation: H2S + 3O2 → 2SO2 + 2H2O Then balance H and S with a 2 in front of H2S: 2H2S + 3O2 → 2SO2 + 2H2O
e.
Since the reaction has two N's on the left and one N on the right, try a tentative Nbalancing by writing a 2 in front of NO2: N2O5 → 2NO2 + O2 Now there are five O's on the left and six O's on the right. Begin to balance the O's with a 6 in front of N2O5; this gives 6N2O5 → 2NO2 + O2 Because changing NO2 will change the oxygen balance, first balance the N's using 12NO2: 6N2O5 → 12NO2 + O2
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Chapter 2: Atoms, Molecules, and Ions
Now there are only (24 + 2) O's on the right, so balance the O's by writing 3O2: 6N2O5 → 12NO2 + 3O2 This can be reduced to 2N2O5 → 4NO2 + O2 2.127. Let: x = number of protons. Then 1.21x is the number of neutrons. Since the mass number is 62, you get 62 = x + 1.21x = 2.21x Thus, x = 28.054, or 28. The element is nickel (Ni). Since the ion has a +2 charge, there are 26 electrons. 2.128. Let: x = number of protons. Then 1.30x is the number of neutrons. Since the mass number is 85, you get 85 = x + 1.30x = 2.30x Thus, x = 36.95, or 37. The element is rubidium (Rb). Since the ion has a +1 charge, there are 36 electrons. 2.129. The average atomic mass would be Natural carbon: 12.011 x 1/2 = Carbon13:
13.00335 x 1/2 = Average
=
6.005500 6.501675 12.507175
The average atomic mass of the sample is 12.507 amu. 2.130. The average atomic mass would be Natural chlorine: 35.4527
x 1/2 =
17.7263500
Chlorine35: 34.96885
x 1/2 =
17.4844250
Average
=
35.2107750
The average atomic mass of the sample is 35.2108 amu. 2.131. The island of stability is a region of the periodic table where a relatively stable superheavy nuclide is at the peak of stability and is surrounded by foothills consisting of less stable nuclides. It is centered around the most stable nuclide, which is predicted to have an atomic number of 114 and a mass number of 298. 2.132.
70 30
Zn +
208 82
Pb →
277 112
Uub +
1 0
n
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SOLUTIONS TO STRATEGY PROBLEMS
2.133. SO3, sulfur trioxide NO2, nitrogen dioxide PO43−, phosphate ion N2, nitrogen Mg(OH)2, magnesium hydroxide 2.134. The unknown metal, M, is a cation with a +3 charge. An example is aluminum. 2.135. The name of the product is aluminum sulfide. The reaction is 16Al(s) + 3S8(s) → 8Al2S3(s) 2.136. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l) 2.137. (0.7721)(37.24 amu) + (1 − 0.7721)(x) = 37.45 amu x =
37.45  (0.7721)(37.24) = 38.161 amu = 38.2 amu (1  0.7721)
2.138. Sulfur has atomic number 16, so S2+ has 16 minus 2 or 14 electrons, which is the number of neutrons in the unknown ion. The number of protons is 27 minus 14 = 13, which is the atomic number, so the element is aluminum. The number of electrons is 13 − 3 = 10.
2.139. 6.5 x 1020 formula units CaCl2 x
2.140. MgCO3, magnesium carbonate Mg3P2, magnesium phosphide Pb(CO3)2, lead(IV) carbonate Pb3P4, lead(IV) phosphide
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3 ions = 1.95 x 1021 ions 1 formula unit
63
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Chapter 2: Atoms, Molecules, and Ions
2.141. SO3, sulfur trioxide HNO2, nitrous acid Mg3N2, magnesium nitride HBr, hydrogen bromide Cu3(PO4)2, copper(II) phosphate CuSO4•5H2O, copper(II) sulfate pentahydrate 2.142. Pb(IO3)2, lead(II) iodate KIO4, potassium periodate Zn(IO)2, zinc hypoiodite Al(IO2)3, aluminum iodite
■
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
2.143. The spheres occupy a diameter of 2 x 1.86 Å = 3.72 Å. The line of sodium atoms would stretch a length of Length =
3.72 A° x 2.619 x 1022 Na atoms = 9.742 x 1022 Å 1 Na atom
Now, convert this to miles. 1010 m 1 mile x = 6.055 x 109 miles 1 A° 1.609 x 103 m = 6.06 x 109 miles
9.742 x 1022 Å x
2.144. The spheres occupy a diameter of 2 x 0.99 Å = 1.98 Å. The line of chlorine atoms would stretch a length of Length =
1.98 A° x 1.699 x 1022 Cl atoms = 3.364 x 1022 Å 1 Cl atom
Now, convert this to miles. 1 mile 1010 m x = 2.090 x 109 miles ° 1.609 x 103 m 1A = 2.09 x 109 miles
3.364 x 1022 Å x
2.145. NiSO4•7H2O(s) → NiSO4•6H2O(s) + H2O(g) [8.753]
=
[8.192 g
+ (8.753 − 8.192 = 0.561 g) ]
The 8.192 g of NiSO4•6H2O must contain 6 x 0.561 = 3.366 g H2O. Mass of anhydrous NiSO4 = 8.192 g NiSO4•6H2O − 3.366 g 6H2O = 4.826 g
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2.146. CoSO4•7H2O(s) → CoSO4•H2O(s) [3.548 g]
=
[2.184 g
+ 6H2O(g) + (3.548 − 2.184 = 1.364 g)]
Mass of one H2O per 3.548 g of CoSO4•7H2O = 1.364 g ÷ 6 = 0.22733 g Mass of anhydrous CoSO4 = 2.184 g CoSO4•H2O − 0.22733 g H2O = 1.9566 g = 1.957 g
2.147. Mass of O = 0.6015 L x 15.9994 amu O x
1.330 g O = 0.799995 g 1L
3.177 g X = 63.538 amu X = 63.54 amu 0.799995 g O
The atomic mass of X is 63.54 amu; X is copper.
2.148. Mass of Cl = 0.4810 L x 35.453 amu Cl x
2.948 g Cl = 1.41799 g Cl 1L
4.315 g X = 107.88 amu X = 107.9 amu 1.41799 g Cl
The atomic mass of X is 107.9 amu; X is silver.
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CHAPTER 3
Calculations with Chemical Formulas and Equations
■
SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 3.1. a.
b.
c.
d.
NO2
C6H12O6
NaOH
1 x AM of N
=
14.0067 amu
2 x AM of O
=
2 x 15.9994
=
31.9988 amu
MM of NO2
=
46.0055
=
46.0 amu (3 s.f.)
6 x AM of C
=
6 x 12.011
=
72.066 amu
12 x AM of H
=
12 x 1.0079
=
12.0948 amu
6 x AM of O
=
6 x 15.9994
=
95.9964 amu
MM of C6H12O6 = = 180. amu (3 s.f.) 1 x AM of Na =
180.1572 amu
1 x AM of O
=
15.9994 amu
1 x AM of H
=
22.98977 amu 1.0079 amu
MM of NaOH = = 40.0 amu (3 s.f.) Mg(OH)2 1 x AM of Mg =
39.9971 amu 24.305 amu
2 x AM of O
=
2 x 15.9994
=
31.9988 amu
2 x AM of H
=
2 x 1.0079
=
2.0158 amu
MM of Mg(OH)2 = = 58.3 amu (3 s.f.)
58.3196 amu
3.2. a.
The molecular model represents a molecule made up of one S and three O. The chemical formula is SO3. Calculating the formula mass by using the same approach as in Example 3.1 in the text yields 80.07 amu.
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b.
The molecular model represents one S, four O, and two H. The chemical formula is then H2SO4. The formula mass is 98.09 amu.
a.
The atomic mass of Ca = 40.08 amu; thus, the molar mass = 40.08 g/mol, and 1 mol Ca = 6.022 x 1023 Ca atoms.
67
3.3.
Mass of one Ca =
40.08 g 1 mol = 6.6556 x 10−23 = 1 mol Ca 6.022 x 1023 atoms
= 6.656 x 10−23 g/atom b.
The molecular mass of C2H5OH, or C2H6O, = (2 x 12.01) + (6 x 1.008) + 16.00 = 46.068. Its molar mass = 46.07 g/mol, and 1 mol = 6.022 x 1023 molecules of C2H6O. Mass of one C2H6O =
46.07 g 1 mol = 6.022 x 1023 atoms 1 mol C 2 H 6 O
= 7.6503 x 10−23 = 7.650 x 10−23 g/molecule 3.4. The molar mass of H2O2 is 34.02 g/mol. Therefore,
0.909 mol H 2 O 2 x
34.02 g H 2 O 2 = 30.92 = 30.9 g H 2 O 2 1 mol H 2 O 2
3.5. The molar mass of HNO3 is 63.01 g/mol. Therefore, 28.5 g HNO3 x
1 mol HNO3 = 0.4523 = 0.452 mol HNO3 63.01 g HNO3
3.6. Convert the mass of HCN from milligrams to grams. Then convert grams of HCN to moles of HCN. Finally, convert moles of HCN to the number of HCN molecules. 56 mg HCN x
1g 1 mol HCN 6.022 x 1023 HCN molecules x x 1 mol HCN 1000 mg 27.02 g HCN
= 1.248 x 1021 = 1.2 x 1021 HCN molecules 3.7. The molecular mass of NH4NO3 = 80.05; thus, its molar mass = 80.05 g/mol. Hence Percent N =
28.02 g x 100% = 35.00 = 35.0% 80.05 g
Percent H =
4.032 g x 100% = 5.036 = 5.04% 80.05 g
Percent O =
48.00 g x 100% = 59.96 = 60.0% 80.05 g
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Chapter 3: Calculations with Chemical Formulas and Equations
3.8. From the previous exercise, NH4NO3 is 35.0% N (fraction N = 0.350), so the mass of N in 48.5 g of NH4NO3 is 48.5 g NH4NO3 x (0.350 g N/1 g NH4NO3) = 16.975 = 17.0 g N 3.9. First, convert the mass of CO2 to moles of CO2. Next, convert this to moles of C (1 mol CO2 is equivalent to 1 mol C). Finally, convert to mass of carbon, changing milligrams to grams first: 5.80 x 10−3 g CO2 x
1 mol CO 2 1 mol C 12.01 g C x = 1 583 x 10−3g C x 1 mol C 1 mol CO 2 44.01 g
Do the same series of calculations for water, noting that 1 mol H2O contains 2 mol H. 1.58 x 10−3 g H2O x
1 mol H 2 O 2 mol H 1.008 g H x = 1.767 x 10−4g H x 1 mol H 1 mol H 2 O 18.02 g
The mass percentages of C and H can be calculated using the masses from the previous calculations: Percent C =
1.583 mg x 100% = 40.90 = 40.9% C 3.87 mg
Percent H =
0.1767 mg x 100% =4.5658 = 4.57% H 3.87 mg
The mass percentage of O can be determined by subtracting the sum of the above percentages from 100%: Percent O = 100.000% − (40.90 + 4.5658) = 54.5342 = 54.5% O 3.10. Convert the masses to moles that are proportional to the subscripts in the empirical formula: 33.4 g S x
1 mol S = 1.0414 mol S 32.07 g S
(83.5 − 33.4) g O x
1 mol O = 3.1312 mol O 16.00 g O
Next, obtain the smallest integers from the moles by dividing each by the smallest number of moles: For O:
3.1312 = 3.01 1.0414
For S:
1.0414 = 1.00 1.0414
The empirical formula is SO3.
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69
3.11. For a 100.0g sample of benzoic acid, 68.8 g are C, 5.0 g are H, and 26.2 g are O. Using the molar masses, convert these masses to moles: 68.8 g C x 5.0 g H x 26.2 g O x
1 mol C = 5.729 mol C 12.01 g C 1 mol H = 4.96 mol H 1.008 g H
1 mol O = 1.638 mol O 16.00 g O
These numbers are in the same ratio as the subscripts in the empirical formula. They must be changed to integers. First, divide each one by the smallest number of moles: For C:
5.729 = 3.497 1.638
For H:
4.96 1.638 = 3.03 For O: = 1.000 1.638 1.638
Rounding off, we obtain C3.5H3.0O1.0. Multiplying the numbers by 2 gives whole numbers, for an empirical formula of C7H6O2. 3.12. For a 100.0g sample of acetaldehyde, 54.5 g are C, 9.2 g are H, and 36.3 g are O. Using the molar masses, convert these masses to moles: 54.5 g C x 9.2 g H x 36.3 g O x
1 mol C = 4.537 mol C 12.01 g C 1 mol H = 9.12 mol H 1.008 g H 1 mol O = 2.268 mol O 16.00 g O
These numbers are in the same ratio as the subscripts in the empirical formula. They must be changed to integers. First, divide each one by the smallest number of moles: For C:
4.537 = 2.000 2.268
For H:
9.12 2.268 = 4.02 For O: = 1.000 2.268 2.268
Rounding off, we obtain C2H4O, the empirical formula, which is also the molecular formula. →
2HCl
1 molec. (mol) H2 + 1 molec. (mol) Cl2
→
2 molec. (mol) HCl (molec., mole interp.)
2.016 g H2
→
2 x 36.5 g HCl (mass interp.)
3.13. H 2
+ Cl 2 + 70.9 g Cl2
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Chapter 3: Calculations with Chemical Formulas and Equations
3.14. Equation: Na + H2O → 1/2H2 + NaOH, or 2Na + 2H2O → H2 + 2NaOH.From this equation, one mole of Na corresponds to onehalf mole of H2, or two moles of Na corresponds to one mole of H2. Therefore, 7.81 g H2 x
1 mol H 2 2 mol Na 22.99 g Na x = 178.1 = 178 g Na x 1 mol Na 1 mol H 2 2.016 g H 2
3.15. Balanced equation: 2ZnS + 3O2 → 2ZnO + 2SO2 Convert grams of ZnS to moles of ZnS. Then determine the relationship between ZnS and O2 (2ZnS is equivalent to 3O2). Finally, convert to mass of O2. 5.00 x 103g ZnS x
3 mol O 2 32.00 g O 2 1 mol ZnS 1 kg x x x 97.46 g ZnS 2 mol ZnS 1 mol O 2 1000 g
= 2.463 = 2.46 kg O2 3.16. Balanced equation: 2HgO → 2Hg + O2 Convert the mass of O2 to moles of O2. Using the fact that one mole of O2 is equivalent to two moles of Hg, determine the number of moles of Hg, and convert to mass of Hg. 6.47 g O2 x
1 mol O 2 2 mol Hg 200.59 g Hg x = 81.11 = 81.1 g Hg x 1 mol O 2 1 mol Hg 32.00 g O 2
3.17. First, determine the limiting reactant by calculating the moles of AlCl3 that would be obtained if Al and HCl were totally consumed: 0.15 mol Al x 0.35 mol HCl x
2 mol AlCl3 = 0.150 mol AlCl3 2 mol Al 2 mol AlCl3 = 0.1166 mol AlCl3 6 mol HCl
Because the HCl produces the smaller amount of AlCl3, the reaction will stop when HCl is totally consumed but before all the Al is consumed. The limiting reactant is therefore HCl. The amount of AlCl3 produced must be 0.1166, or 0.12 mol. 3.18. First, determine the limiting reactant by calculating the moles of ZnS produced by totally consuming Zn and S8: 7.36 g Zn x
1 mol Zn 8 mol ZnS x = 0.11256 mol ZnS 65.39 g Zn 8 mol Zn
6.45 g S8 x
1 mol S8 8 mol ZnS = 0.2011 mol ZnS x 1 mol S8 256.56 g S8
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The reaction will stop when Zn is totally consumed; S8 is in excess, and not all of it is converted to ZnS. The limiting reactant is therefore Zn. Now convert the moles of ZnS obtained from the Zn to grams of ZnS: 0.11256 mol ZnS x
97.46 g ZnS = 10.97 = 11.0 g ZnS 1 mol ZnS
3.19. First, write the balanced equation: CH3OH + CO → HC2H3O2 Convert grams of each reactant to moles of acetic acid: 15.0 g CH 3OH x
10.0 g CO x
1 mol CH 3 OH 1 mol HC2 H 3O 2 x = 0.4681 mol HC 2 H 3O 2 32.04 g CH 3OH 1 mol CH 3OH
1 mol HC 2 H 3O 2 1 mol CO x = 0.3570 mol HC 2 H 3O 2 28.01 g CO 1 mol CO
Thus, CO is the limiting reactant, and 0.03570 mol HC2H3O2 is obtained. The mass of product is 0.3570 mol HC 2 H 3O 2 x
60.05 g HC 2 H 3O 2 = 21.44 g HC 2 H 3O 2 1 mol HC2 H3O 2
The percentage yield is 19.1 g actual yield x 100% = 89.08 = 89.1% 21.44 g theoretical yield
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ANSWERS TO CONCEPT CHECKS
3.1. a.
Each tricycle has one seat, so you have a total of 1.5 mol of seats.
b.
Each tricycle has three tires, so you have 1.5 mol x 3 = 4.5 mol of tires.
c.
Each Mg(OH)2 has two OH− ions, so there are 1.5 mol x 2 = 3.0 mol OH− ions.
a.
When conducting this type of experiment, you are assuming that all of the carbon and hydrogen show up in the CO2 and H2O, respectively. In this experiment, where all of the carbon and hydrogen do not show up, when you analyze the CO2 for carbon and H2O for hydrogen, you find that the masses in the products are less than those in the carbon and hydrogen you started with.
3.2.
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Chapter 3: Calculations with Chemical Formulas and Equations
b.
Since you collected less carbon and hydrogen than were present in the original sample, the calculated mass percentage will be less than the expected (real) value. For example, say you have a 10.0g sample that contains 7.5 g of carbon. You run the experiment on the 10.0g sample and collect only 5.0 g of carbon. The calculated percent carbon based on your experimental results would be 50% instead of the correct amount of 75%.
a.
C2H8O2 is not an empirical formula because each of the subscripts can be divided by 2 to obtain a possible empirical formula of CH4O. (The empirical formula is not the smallest integer ratio of subscripts.)
b.
C1.5H4 is not a correct empirical formula because one of the subscripts is not an integer. Multiply each of the subscripts by 2 to obtain the possible empirical formula C3H8. (Since the subscript of carbon is the decimal number 1.5, the empirical formula is not the smallest integer ratio of subscripts.)
c.
Yes, the empirical formula and the molecular formula can be the same, as is the case in this problem, where the formula is written with the smallest integer subscripts.
a.
Correct. Coefficients in balanced equations can represent amounts in atoms and molecules.
b.
Incorrect. The coefficients in a balanced chemical equation do not represent amounts in grams. One gram of carbon and one gram of oxygen represent different molar amounts.
c.
Incorrect. The coefficients in a balanced chemical equation do not represent amounts in grams.
d.
Correct. You might initially think this is an incorrect representation; however, 12 g of C, 32 g of O2, and 44 g of CO2 all represent one mole of the substance, so the relationship of the chemical equation is obeyed.
e.
Correct. The coefficients in balanced equations can represent amounts in moles.
f.
Incorrect. The amount of O2 present is not enough to react completely with one mole of carbon. Only onehalf of the carbon would react, and onehalf mole of CO2 would form.
g.
Incorrect. In this representation, oxygen is being shown as individual atoms of O, not as molecules of O2, so the drawings are not correctly depicting the chemical reaction.
h.
Correct. The molecular models correctly depict a balanced chemical reaction since the same number of atoms of each element appears on both sides of the equation.
a.
X2(g) + 2Y(g) → 2XY(g)
b.
Since the product consists of a combination of X and Y in a 1:1 ratio, it must consist of two atoms hooked together. If you count the total number of X atoms (split apart the X2 molecules) and Y atoms present prior to the reaction, you find that there are four X atoms and three Y atoms. From these starting quantities, you are limited to three XY molecules and left with an unreacted X. Option #1 represents this situation and is therefore the correct answer.
3.3.
3.4.
3.5.
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c.
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73
Since Y(g) was completely used up during the course of the reaction, it is the limiting reactant.
ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
3.1. The molecular mass is the sum of the atomic masses of all the atoms in a molecule of the substance whereas the formula mass is the sum of the atomic masses of all the atoms in one formula unit of the compound, whether the compound is molecular or not. A given substance could have both a molecular mass and a formula mass if it existed as discrete molecules. 3.2. To obtain the formula mass of a substance, sum up the atomic masses of all atoms in the formula of the compound. 3.3. A mole of N2 contains Avogadro's number (6.02 x 1023) of N2 molecules and 2 x 6.02 x 1023 N atoms. One mole of Fe2(SO4)3 contains three moles of SO42− ions, and it contains twelve moles of O atoms. 3.4. A sample of the compound of known mass is burned, and CO2 and H2O are obtained as products. Next, you relate the masses of CO2 and H2O to the masses of carbon and hydrogen. Then you calculate the mass percentages of C and H. You find the mass percentage of O by subtracting the mass percentages of C and H from 100. 3.5. The empirical formula is obtained from the percentage composition by assuming for the purposes of the calculation a sample of 100 g of the substance. Then the mass of each element in the sample equals the numerical value of the percentage. Convert the masses of the elements to moles of the elements using the atomic mass of each element. Divide the moles of each by the smallest number to obtain the smallest ratio of each atom. If necessary, find a wholenumber factor to multiply these results by to obtain integers for the subscripts in the empirical formula. 3.6. The empirical formula is the formula of a substance written with the smallest integer (wholenumber) subscripts. Each of the subscripts in the formula C6H12O2 can be divided by 2, so the empirical formula of the compound is C3H6O. 3.7. The number of empirical formula units in a compound, n, equals the molecular mass divided by the empirical formula mass. n =
34.0 amu = 2.00 17.0 amu
The molecular formula of hydrogen peroxide is therefore (HO)2, or H2O2.
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Chapter 3: Calculations with Chemical Formulas and Equations
3.8. The coefficients in a chemical equation can be interpreted directly in terms of molecules or moles. For the mass interpretation, you will need the molar masses of CH4, O2, CO2, and H2O, which are 16.0, 32.0, 44.0, and 18.0 g/mol, respectively. A summary of the three interpretations is given below the balanced equation: CH4 1 molecule 1 mole 16.0 g
+ + + +
2O2 2 molecules 2 moles 2 x 32.0 g
→ → → →
CO2 1 molecule 1 mole 44.0 g
+ + + +
2H2O 2 molecules 2 moles 2 x 18.0 g
3.9. A chemical equation yields the mole ratio of a reactant to a second reactant or product. Once the mass of a reactant is converted to moles, this can be multiplied by the appropriate mole ratio to give the moles of a second reactant or product. Multiplying this number of moles by the appropriate molar mass gives mass. Thus, the masses of two different substances are related by a chemical equation. 3.10. The limiting reactant is the reactant that is entirely consumed when the reaction is complete. Because the reaction stops when the limiting reactant is used up, the moles of product are always determined by the starting number of moles of the limiting reactant. 3.11. Two examples are given in the book. The first involves making cheese sandwiches. Each sandwich requires two slices of bread and one slice of cheese. The limiting reactant is the cheese because some bread is left unused. The second example is assembling automobiles. Each auto requires one steering wheel, four tires, and other components. The limiting reactant is the tires, since they will run out first. 3.12. Since the theoretical yield represents the maximum amount of product that can be obtained by a reaction from given amounts of reactants under any conditions, in an actual experiment you can never obtain more than this amount. 3.13. The answer is a, 3.27 g of NH3. 3.14. The answer is b, 1 g of formaldehyde. 3.15. The answer is d, 4.85 x 1024 atoms. 3.16. The answer is a, C3H4O3.
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75
ANSWERS TO CONCEPT EXPLORATIONS
3.17. Part 1 500 nails = 100 nails = 1 x 102 nails 1 kg
a.
0.2 kg x
b.
10 dozen x
c.
1 nail x
d.
2.0 mol x
12 nails 1 kg x = 0.240 kg 1 dozen 500 nails
1 kg = 0.00200 kg = 2.00 g 500 nails 6.02 x 1023 nails 1 kg x = 2.408 x 1021 = 2.4 x 1021 kg 1 mol 500 nails
Part 2 1000 g 1 mol 6.02 x 1023 atoms x x = 3.00 x 1025 = 3 x 1025 atoms 1 mol 1 kg 4.003 g
a.
0.2 kg x
b.
10 dozen x
c.
1 atom x
d.
2.0 mol x
12 atoms 1 mol 4.003 g x x = 7.979 x 10−22 = 7.98 x 10−22 g 23 1 dozen 6.02 x 10 atoms 1 mol
1 mol 4.003 g x = 6.646 x 10−24 = 6.65 x 10−24 g 23 6.02 x 10 atoms 1 mol
4.003 g = 8.006 = 8.0 g 1 mol
Part 3
a.
A nail and a helium atom have very different masses. You would not expect an equal mass of each, 1.0 kg, to have an equal number of binkles.
b.
A binkle is a fixed number of objects, 3 x 1012. You would expect 3.5 binkles of nails and 3.5 binkles of helium atoms to have equal counts. A nail has a larger mass than a helium atom, so a binkle of nails would have more mass.
c.
A helium atom has a smaller mass than a lithium atom, so 3.5 g of helium contains more atoms than 3.5 g of lithium atoms.
3.18. Part 1 a.
There are two hydrogen atoms and one oxygen atom in each molecule of H2O.
b.
There are two moles of hydrogen atoms and one mole of oxygen atoms in each mole of H2O.
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c.
d.
Mass H = 1.0 mol H2O x
2 mol H 1.008 g x = 2.016 = 2.0 g H 1 mol H 1 mol H 2 O
Mass O = 1.0 mol H2O x
1 mol O 16.00 g x = 16.00 = 16 g O 1 mol H 2 O 1 mol O
The mass of 1.0 mol H2O is 2.016 + 16.00 = 18.016 = 18 g/mol.
Part 2
a.
Molar mass XCl2 =
100.0 g = 400 = 4.0 x 102 g/mol 0.25 mol
Molar mass YCl2 =
125.0 g = 250 = 2.5 x 102 g/mol 0.50 mol
b.
Since there are two chloride ions in each compound, 1.0 mol XCl2 and 1.0 mol YCl2 contain the same number of chloride ions.
c.
Element X is heavier than element Y. Therefore, for 1.0 mol XCl2 and 1.0 mol YCl2, the mass of element X is more than the mass of Y.
d.
The mass of chloride ion is the same for both. In 1.0 mole, 1.0 mol x
e.
2 mol Cl ions 35.45 g x = 70.90 = 71 g 1 mol XCl2 1 mol Cl
Molar mass X = 400.0 − 70.90 = 329.1 = 3.3 x 102 g/mol Molar mass Y = 250.0 − 70.90 = 179.1 = 1.8 x 102 g/mol 200.0 g x
1 mol XCl2 1 mol X ions = 0.500 = 0.50 mol X ions x 1 mol XCl2 400 g
200.0 g x
1 mol XCl2 2 mol Cl ions = 1.00 = 1.0 mol Cl− ions x 1 mol XCl2 400 g
g.
250.0 g x
1 mol YCl2 1 mol Y ions 179.10 g x x = 179.10 = 1.8 x 102g Y ions 1 mol YCl2 1 mol Y 250 g
h.
Molar mass YBr3 = 179.10 + 3 x 79.90 = 418.8 = 4.2 x 102 g/mol
f.
Part 3
If the sample of AlCl3 contains 12 chloride ions, it must also contain 4 aluminum ions. This makes a total of 4 AlCl3 units. a.
4 AlCl3 units x
1 mol AlCl3 133.33 g x = 8.859 x 10−22 = 8.86 x 10−22 g 23 6.02 x 10 units 1 mol AlCl3
b.
4 AlCl3 units x
1 mol AlCl3 = 6.644 x 10−24 = 6.64 x 10−24 mol 6.02 x 1023 units
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77
ANSWERS TO CONCEPTUAL PROBLEMS
3.19. a.
3H2(g) + N2(g) → 2NH3(g)
b.
Since there is no H2 present in the container, it was entirely consumed during the reaction, which makes it the limiting reactant.
c.
According to the chemical reaction, three molecules of H2 are required for every molecule of N2. Since there are two molecules of unreacted N2, you would need six additional molecules of H2 to complete the reaction.
a.
The limiting reactant in cooking with a gas grill would be the propane. This makes sense because propane is the material you must purchase in order to cook your food.
b.
Since the chemical reaction requires only propane and oxygen, if the grill will not light with ample propane present, the limiting reactant must be the oxygen.
c.
Once again, here is a case where you have adequate propane, so you can conclude that a yellow flame indicates that not enough oxygen is present to combust all of the propane. If there is not enough O2 available for complete combustion, a reasonable assumption is that some of the products will have fewer oxygen atoms than CO2. Therefore, a mixture of products would be obtained in this case, including carbon monoxide (CO) and soot (carbon particles).
a.
This answer is unreasonable because 1.0 x 10−3 g is too small a mass for 0.33 mol of an element. For example, 0.33 mol of hydrogen, the lightest element, would have a mass of 0.33 g.
b.
This answer is unreasonable because 1.80 x 10−10 g is too large for one water molecule. (The mass of one water molecule is 2.99 x 10−23 g.)
c.
This answer is reasonable because 3.01 x 1023 is onehalf of Avogadro’s number.
d.
This answer is unreasonable because the units for molar mass should be g/mol, so this quantity is 1000 times too large.
a.
In order to have a complete reaction, a ratio of 2 moles of hydrogen to every mole of oxygen is required. In this case, there is not enough oxygen in the air outside of the bubble for the complete reaction of hydrogen.
b.
In this case, you have a ratio of 1 mole of H2 to 1 mole of O2. According to the balanced chemical reaction, every mole of O2 can react with 2 moles of H2. In this case, when 0.5 mole of O2 has reacted, all of the H2 (1 mole) will be consumed, leaving behind 0.5 mole of unreacted O2.
3.20.
3.21.
3.22.
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c.
In this case, you have a ratio of 2 moles of H2 to 1 mole of O2, which is the correct stoichiometric amount, so all of the hydrogen and all of the oxygen react completely.
d.
In order for reaction to occur, both oxygen and hydrogen must be present. Oxygen does not combust, and there is no hydrogen present to burn, so no reaction occurs.
a.
The limiting reactant would be the charcoal because the air would supply as much oxygen as is needed.
b.
The limiting reactant would be the magnesium because the beaker would contain much more water than is needed for the reaction (approximately 18 mL of water is 1 mole).
c.
The limiting reactant would be the H2 because the air could supply as much nitrogen as is needed.
a.
Since the balanced chemical equation for the reaction is 2H2 + O2 → 2H2O in order to form the water, you need 2 molecules of hydrogen for every 1 molecule of oxygen. Given the quantities of reactants present in the container and applying the 2:1 ratio, you can produce a maximum of twelve molecules of water.
b.
The drawing of the container after the reaction should contain 12 H2O molecules and 2 O2 molecules.
a.
The problem is that Avogadro’s number was inadvertently used for the molar mass of calcium, which should be 40.08 g/mol. The correct calculation is
3.23.
3.24.
3.25.
27.0 g Ca x b.
1 mol Ca = 0.6736 = 0.674 mol Ca 40.08 g Ca
The problem here is an incorrect mole ratio. There are 2 mol K+ ions per 1 mol K2SO4. The correct calculation is 2.5 mol K2SO4 x
2 mol K + ions 6.022 x 1023 K + ions x 1 mol K 2SO 4 1 mol K + ions
= 3.01 x 1024 = 3.0 x 1024 K+ ions c.
The problem here is an incorrect mole ratio. The result should be 0.50 mol Na x
2 mol H 2 O = 0.50 mol H2O 2 mol Na
3.26. a.
The missing concept is the mole ratio. His reasoning would be correct only if the reactants reacted in a onetoone mole ratio. Here, the ratio is 5 mol O2/2 mol C2H2. This means that, for every 2 moles of C2H2, 5 moles of O2 are required. Since there are only 4 moles here, there is insufficient O2, and it is the limiting reactant.
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b.
■
The missing concept again is the mole ratio. Since in this problem the reactants react in a onetoone mole ratio, the reactant with the fewest moles is the limiting reactant, and the lucky guess works.
SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off.
3.27. a.
FM of CH3OH
= AM of C + 4(AM of H) + AM of O = 12.01 amu + (4 x 1.008 amu) + 16.00 amu = 32.042 = 32.0 amu (3 s.f.)
b.
FM of NO3
= AM of N + 3(AM of O) = 14.01 amu + (3 x 16.00 amu) = 62.01 = 62.0 amu (3 s.f.)
c.
FM of K2CO3
= 2(AM of K) + AM of C + 3(AM of O) = (2 x 39.10 amu) + 12.01 amu + (3 x 16.00 amu) = 138.210 = 138 amu (3 s.f.)
d.
FM of Ni3(PO4)2
= 3(AM of Ni) + 2(AM of P) + 8(AM of O) = (3 x 58.70 amu) + (2 x 30.97 amu) + (8 x 16.00 amu) = 366.040 = 366 amu (3 s.f.)
3.28. a.
FM of H2SO4
= 2(AM of H) + AM of S + 4(AM of O) = (2 x 1.008 amu) + 32.07 amu + (4 x 16.00 amu) = 98.086 = 98.1 amu (3 s.f.)
b.
FM of PCl5
= AM of P + 5(AM of Cl) = 30.97 amu + (5 x 35.45 amu) = 208.220 = 208 amu (3 s.f.)
c.
d.
FM of K2SO3
= 2(AM of K) + AM of S + 3(AM of O)
FM of Ca(OH)2
= (2 x 39.10 amu) + 32.07 amu + (3 x 16.00 amu) = 158.270 = 158 amu (3 s.f.) = AM of Ca + 2(AM of H) + 2(AM of O) = 40.08 amu + (2 x 1.008 amu) + (2 x 16.00 amu) = 74.096 = 74.1 amu (3 s.f.)
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3.29. a.
SO2 1 x AM of S
=
2 x AM of O = b.
2 x 16.00
=
MM of NO2
=
64.07 amu = 64.1 amu (3 s.f.)
=
30.97 amu
3 x 35.45
=
106.35 amu
MM of PCl3
=
137.32 amu = 137 amu (3 s.f.)
=
1.008 amu
=
14.01 amu
PCl3 1 x AM of P 3 x AM of Cl
=
32.07 amu 32.00 amu
3.30. a.
HNO2
1 x AM of H 1 x AM of N 2 x AM of O
b.
=2 x 16.00
=
MM of HNO2
=
47.018 amu = 47.0 amu (3 s.f.)
CO 1 x AM of C
=
12.01 amu
1 x AM of O
= MM of CO
=
32.00 amu
16.00 amu 28.01 amu = 28.0 amu (3 s.f.)
3.31. First, find the formula mass of NH4NO3 by adding the respective atomic masses. Then convert it to the molar mass: FM of NH4NO3
= 2(AM of N) + 4(AM of H) + 3(AM of O) = (2 x 14.01 amu) + (4 x 1.008 amu) + (3 x 16.00 amu) = 80.052 amu
The molar mass of NH4NO3 is 80.05 g/mol. 3.32. First, find the formula mass of H3PO4 by adding the respective atomic masses. Then convert it to the molar mass: FM of H3PO4
= 3(AM of H) + AM of P + 4(AM of O) = (3 x 1.008) + 30.97 + (4 x 16.00) = 97.994 amu
The molar mass of H3PO4 is 97.99 g/mol. 3.33. a.
The atomic mass of Na equals 22.99 amu; thus, the molar mass equals 22.99 g/mol. Because 1 mol of Na atoms equals 6.022 x 1023 Na atoms, we calculate Mass of one Na atom =
22.99 g/mol = 3.8177 x 10−23 6.022 x 1023 atom/mol
= 3.818 x 10−23 g/atom
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b.
81
The atomic mass of N equals 14.01 amu; thus, the molar mass equals 14.01 g/mol. Because 1 mol of N atoms equals 6.022 x 1023 N atoms, we calculate Mass of one N atom =
14.01 g/mol = 2.3264 x 10−23 6.022 x 1023 atom/mol
= 2.326 x 10−23 g/atom c.
The formula mass of CH3Cl = [12.01 + (3 x 1.008) + 35.45] = 50.48 amu; thus, the molar mass equals 50.48 g/mol. Because 1 mol of CH3Cl molecules equals 6.022 x 1023 CH3Cl molecules, we calculate Mass of one CH3Cl molecule =
50.48 g/mol = 8.3826 x 10−23 23 6.022 x 10 molecules/mol
= 8.383 x 10−23 g/molecule d.
The formula mass of Hg(NO3)2 = 200.59 + (2 x 14.01) + (6 x 16.00)] = 324.61 amu; thus, the molar mass equals 324.61 g/mol. Because 1 formula mass of Hg(NO3)2 equals 6.022 x 1023 Hg(NO3)2 formula units, we calculate Mass of one Hg(NO3)2 =
324.61 g/mol = 5.3904 x 10−22 6.022 x 1023 units/mol
= 5.390 x 10−22 g/unit 3.34. a.
Mass of one Ar atom =
39.95 g/mol = 6.634 x 10−23 = 6.63 x 10−23 g/atom 23 6.022 x 10 atom/mol
b.
Mass of one Te atom =
127.60 g/mol = 2.1188 x 10−22 = 2.119 x 10−22 g/atom 6.022 x 1023 atom/mol
c.
Mass of one PBr3 molecule =
d.
270.67 g/mol = 4.4947 x 10−22 6.022 x 1023 molecules/mol
= 4.495 x 10−22 g/molecule 106.87 g/mol Mass of one Fe(OH)3 unit = = 1.7747 x 10−22 6.022 x 1023 unit/mol = 1.775 x 10−22 g/unit
3.35. First, find the formula mass (in amu) using the periodic table (inside front cover): FM of (CH3CH2)2O = (4 x 12.01 amu) + (10 x 1.008 amu) + 16.00 amu = 74.12 amu Mass of one (CH3CH2)2O molecule =
74.12 g/mol 6.022 x 1023 molecules/mol
= 1.2308 x 10−22 = 1.231 x 10−22 g/molecule
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3.36. First, find the formula mass (in amu) using the periodic table (inside front cover): FM of glycerol = (3 x 12.01 amu) + (8 x 1.008 amu) + (3 x 16.00 amu) = 92.09 amu Mass of one glycerol molecule =
92.09 g/mol 6.022 x 1023 molecules/mol
= 1.529 x 10−22 = 1.53 x 10−22 g/molecule 3.37. From the table of atomic masses, we obtain the following molar masses for parts a through d: Na = 22.99 g/mol; S = 32.07 g/mol; C = 12.01 g/mol; H = 1.008 g/mol; Cl = 35.45 g/mol; and N = 14.01 g/mol. a.
0.15 mol Na x
22.99 g = 3.448 = 3.4 g Na 1 mol Na
b.
0.594 mol S x
32.07 g S = 19.04 = 19.0 g S 1 mol S
c.
Using molar mass = 84.93 g/mol for CH2Cl2, we obtain 84.93 g CH 2 Cl2 = 236.1 = 236 g CH2Cl2 1 mol CH 2 Cl2
2.78 mol CH2Cl2 x d.
Using molar mass = 68.14 g/mol for (NH4)2S, we obtain 38 mol (NH4)2S x
68.14 g (NH 4 ) 2S = 2.58 x 103 = 2.6 x 103 g (NH4)2S 1 mol (NH 4 ) 2S
3.38. From the table of atomic masses, we obtain the following molar masses for parts a through d: C = 12.01 g/mol; O = 16.00 g/mol; K = 39.10 g/mol; Cr = 52.00 g/mol; Fe = 55.85 g/mol; and F = 19.00 g/mol. 55.85 g Fe = 11.44 = 11.4 g Fe 1 mol Fe
a.
0.205 mol Fe x
b.
0.79 mol F x
c.
Using molar mass = 44.01 g/mol for CO2, we obtain
19.00 g F = 15.01 = 15 g F 1 mol F
5.8 mol CO2 x d.
44.01 g CO 2 = 255.2 = 2.6 x 102 g CO2 1 mol CO 2
Using molar mass 194.20 g/mol for K2CrO4, we obtain 48.1 mol K 2 CrO 4 x
194.20 g K 2 CrO 4 = 9341.02 = 9.34 x 103 g K2CrO4 1 mol K 2 CrO 4
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3.39. First, find the molar mass of H3BO3: (3 x 1.008 amu) + 10.81 amu + (3 x 16.00 amu) = 61.83. Therefore, the molar mass of H3BO3 = 61.83 g/mol. The mass of H3BO3 is calculated as follows: 0.543 mol H 3 BO3 x
61.83 g H3 BO3 = 33.57 = 33.6 g H 3 BO3 1 mol H 3 BO3
3.40. First, find the molar mass of CS2: 12.01 amu + (2 x 32.07 amu) = 76.15 amu. Therefore, the molar mass of CS2 = 76.15 g/mol. The mass of CS2 is calculated as follows: 0.0205 mol CS2 x
76.15 g CS2 = 1.5610 = 1.56 g CS2 1 mol CS2
3.41. From the table of atomic masses, we obtain the following rounded molar masses for parts a through d: C = 12.01 g/mol; Cl = 35.45 g/mol; H = 1.008 g/mol; Al = 26.98 g/mol; and O = 16.00 g/mol. 1 mol C = 0.2381 = 0.238 mol C 12.01 g C
a.
2.86 g C x
b.
7.05 g Cl2 x
c.
The molar mass of C4H10 = (4 x 12.01) + (10 x 1.008) = 58.12 g C4H10/mol C4H10. The mass of C4H10 is calculated as follows:
1 mol Cl2 = 0.09943 = 0.0994 mol Cl2 70.90 g Cl2
76 g C 4 H10 x
d.
1 mol C4 H10 = 1.307 = 1.3 mol C 4 H10 58.12 g C 4 H10
The molar mass of Al2(CO3)3 = (2 x 26.98) + (3 x 12.01) + (9 x 16.00 g) =233.99 g/mol Al2(CO3)3. The mass of Al2(CO3)3 is calculated as follows: 26.2 g Al2(CO3)3 x
1 mol Al2 (CO3 )3 = 0.1119 = 0.112 mol Al2(CO3)3 233.99 g Al2 (CO3 )3
3.42. From the table of atomic masses, we obtain the following rounded molar masses for parts a through d: As = 74.92 g/mol; S = 32.07 g/mol; N = 14.01 g/mol; H = 1.008 g/mol; Al = 26.98 g/mol; and O = 16.00 g/mol. a.
2.57 g As x
1 mol As = 0.03430 = 0.0343 mol As 74.92 g As
b.
7.83 g S8 x
1 mol S8 = 0.03051 = 0.0305 mol S8 256.56 g S8
c.
The molar mass of N2H4 = (2 x 14.01) + (4 x 1.008) = 32.052 g N2H4/mol N2H4. The moles of N2H4 is calculated as follows: 36.5 g x
1 mol N 2 H 4 = 1.1387 = 1.14 mol N2H4 32.052 g N 2 H 4
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d.
The molar mass of Al2(SO4)3 = (2 x 26.98) + (3 x 32.07) + (12 x 16.00) = 342.17 g Al2(SO4)3/mol Al2(SO4)3. The mass of Al2(SO4)3 is calculated as follows: 227 g Al2(SO4)3 x
1 mol Al2 (SO 4 )3 = 0.6634 = 0.663 mol Al2(SO4)3 342.17 g Al2 (SO 4 )3
3.43. Calculate the formula mass of calcium sulfate: 40.08 amu + 32.07 amu + (4 x 16.00 amu) = 136.15 amu. Therefore, the molar mass of CaSO4 is 136.15 g/mol. Use this to convert the mass of CaSO4 to moles: 0.791 g CaSO 4 x
1 mol CaSO 4 = 5.809 x 10−3 = 5.81 x 10−3 mol CaSO4 136.15 g CaSO 4
Calculate the molecular mass of water: (2 x 1.008 amu) + 16.00 amu = 18.02 amu. Therefore, the molar mass of H2O equals 18.02 g/mol. Use this to convert the rest of the sample to moles of water: 0.209 g H 2 O x
1 mol H 2 O = 1.159 x 10−2 = 1.16 x 10−2 mol H2O 18.02 g H 2 O
Because 0.01159 mol is about twice 0.005811 mol, both numbers of moles are consistent with the formula, CaSO4•2H2O. 3.44. Calculate the formula mass of copper(II) sulfate: 63.55 amu + 32.07 amu + (4 x 16.00 amu) = 159.62 amu. Thus, the molar mass of CuSO4 is 159.62 g/mol. From the previous problem, the molar mass of H2O is 18.02 g/mol. Use this to convert the rest of the sample to moles of water: 0.558 g H 2 O x
1 mol H 2 O = 0.03096 mol H 2 O 18.02 g H 2 O
Calculate the moles of CuSO4 in order to be able to compare relative molar amounts of CuSO4 and H2O. Then, divide the moles of H2O by the moles of CuSO4: 0.989 g CuSO 4 x
1 mol CuSO 4 = 0.006196 mol CuSO 4 159.62 g CuSO 4
0.03096 mol H2O ÷ 0.006196 mol CuSO4 = 4.99/1, or about 5:1 (consistent with CuSO4•5H2O) 3.45. The following rounded atomic masses are used: Li = 6.94 g/mol; Br = 79.90 g/mol; N = 14.01 g/mol; H = 1.008 g/mol; Pb = 207.2 g/mol; Cr = 52.00 g/mol; O = 16.00 g/mol; and S = 32.07 g/mol. Also, Avogadro’s number is 6.022 x 1023 atoms, so a.
No. Li atoms = 8.21 g Li x
b.
No. Br atoms = 32.0 g Br2 x
6.022 x 1023 atoms = 7.122 x 1023 = 7.12 x 1023 atoms 6.941 g Li 2 x 6.022 x 1023 atoms = 2.412 x 1023 (2 x 79.90) g Br2
= 2.41 x 1023 atoms
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c.
No. NH3 molecules = 45 g NH3 x
6.022 x 1023 molecules = 1.59 x 1024 17.03 g NH 3
= 1.6 x 1024 molecules d.
6.022 x 1023 units = 3.745 x 1023 = 3.75 x 1023 323.2 g PbCrO 4
No. PbCrO 4 units = 201 g PbCrO 4 x units
e.
3 x 6.022 x 1023 ions = 6.587 x 1022 392.21 g Cr2 (SO 4 )3
No. SO42− ions = 14.3 g Cr2(SO4)3 x = 6.59 x 1022 ions
3.46. These rounded atomic masses are used: Al = 26.98 g/mol; I = 126.90 g/mol; N = 14.01 g/mol; O = 16.00 g/mol; Na = 22.99 g/mol; Cl = 35.45 g/mol; Ca = 40.08 g/mol; and P = 30.97 g/mol. Also, Avogadro’s number is 6.022 x 1023 atoms, so 6.022 x 1023 atoms = 5.736 x 1023 = 5.74 x 1023 atoms 26.98 g Al
a.
No. Al atoms = 25.7 g Al x
b.
No. I atoms = 8.71 g I2 x
c.
No. N2O5 molecules = 14.9 g N2O5 x
2 x 6.022 x 1023 atoms = 4.133 x 1022 = 4.13 x 1022 atoms 2 x 126.90 g I 2 6.022 x 1023 molecules = 8.306 x 1022 108.02 g N 2 O5
= 8.31 x 1022 molecules d.
No. NaClO 4 units = 3.31 g NaClO 4 x
6.022 x 1023 units = 1.628 x 1022 122.44 g NaClO 4
= 1.63 x 1022 units e.
No. Ca2+ ions = 4.71 g Ca3(PO4)2 x
3 x 6.022 x 1023 ions = 2.743 x 1022 310.18 g Ca 3 (PO 4 ) 2
= 2.74 x 1022 ions 3.47. Calculate the molecular mass of CCl4: 12.01 amu + (4 x 35.45 amu) = 153.81 amu. Use this and Avogadro's number to express it as 153.81 g/NA in order to calculate the number of molecules: 7.58 mg CCl4 x
1g 6.022 x 1023 molecules x = 2.968 x 1019 153.81 g CCl4 1000 mg
= 2.97 x 1019 molecules
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3.48. Calculate the molecular mass of ClF3: 35.45 amu + (3 x 19.00 amu) = 92.45 amu. Use this and Avogadro's number to express it as 92.45 g/NA in order to calculate the number of molecules: 5.88 mg ClF3 x
1g x 1000 mg
6.022 x 1023 molecules = 3.830 x 1019 92.45 g ClF3
= 3.83 x 1019 molecules 3.49. Mass percentage carbon = Percent carbon =
mass of C in sample x 100% mass of sample 1.584 g x 100% = 86.274 = 86.27% 1.836 g
3.50. Mass percentage alcohol =
mass of alcohol in solution x 100% mass of sample
Percent alcohol =
4.01 g x 100% = 66.72 = 66.7% 6.01 g mass of P in sample x 100% mass of sample
3.51. Mass percentage phosphorus = Percent P =
3.52. Mass percentage sulfur = Percent sulfur =
1.72 mg x 100% = 20.16 = 20.2% 8.53 mg
mass of S in sample x 100% mass of sample 1.64 mg x 100% = 51.73 = 51.7% 3.17 mg
3.53. Start with the definition for percentage nitrogen, and rearrange this equation to find the mass of N in the fertilizer. Mass percentage nitrogen = Mass N =
mass of N in fertilizer x 100% mass of fertilizer
mass % N 14.0% x mass of fertilizer = x 4.15 kg = 0.5810 = 0.581 kg N 100% 100%
3.54. Start by finding the mass of 2.50 L of seawater using the density of 1.025 g/cm3. Mass seawater = 2.50 L x
103 cm3 1.025 g = 2.5625 x 103 g x 1L 1 cm3
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Continue with the definition for percentage of bromine in the seawater, and rearrange this equation to find the mass of Br in the seawater. Mass percentage Br = Mass Br =
mass of Br in seawater x 100% mass of seawater
mass % Br 0.0065% x mass seawater = x 2.5625 x 103 g 100% 100% = 0.166 = 0.17 g Br
3.55. Convert moles to mass using the molar masses from the respective atomic masses. Then calculate the mass percentages from the respective masses. 0.0898 mol Al x 0.0381 mol Mg x Percent Al = Percent Mg =
26.98 g Al = 2.422 g Al 1 mol Al
24.31 g Mg = 0.9262 g Mg 1 mol Mg
mass of Al 2.422 g Al = x 100% = 72.34 = 72.3% Al mass of alloy 3.349 g alloy
mass of Mg 0.9262 g Mg = x 100% = 27.655 = 27.7% Mg mass of alloy 3.349 g alloy
3.56. Convert moles to mass using the molar masses from the respective atomic masses of 20.18 g/mol for Ne and 83.80 g/mol for Kr. Then calculate the mass percentages from the respective masses. 0.0856 mol Ne x
20.18 g Ne = 1.727 g Ne 1 mol Ne
0.0254 mol Kr x
83.80 g Kr = 2.129 g Kr 1 mol Kr
Percent Ne =
mass of Ne 1.727 g Ne = x 100% = 44.79 = 44.8% Ne mass of mix 3.856 g mix
Percent Kr =
mass of Kr 2.129 g Kr = x 100% = 55.21 = 55.2% Kr mass of mix 3.856 g mix
3.57. In each part, the numerator consists of the mass of the element in one mole of the compound; the denominator is the mass of one mole of the compound. Use the atomic weights of C = 12.01 g/mol; O = 16.00 g/mol; Na = 22.99 g/mol; H = 1.008 g/mol; P = 30.97 g/mol; Co = 58.93 g/mol; and N = 14.01 g/mol. a.
Percent C =
mass of C 12.01 g C = x 100% = 42.878 = 42.9% mass of CO 28.01 g CO
Percent O = 100.000% − 42.878%C = 57.122 = 57.1%
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b.
mass of C 12.01 g C = x 100% = 27.289 = 27.3% mass of CO 2 44.01 g CO 2
Percent C =
Percent O = 100.000% − 27.289% C = 72.711 = 72.7% c.
mass of Na 22.99 g Na = x 100% = 19.161 = 19.2% mass of NaH 2 PO 4 119.98 g NaH 2 PO 4
Percent Na = Percent H =
mass of H 2.016 g H = x 100% = 1.6802 = 1.68% mass of NaH 2 PO 4 119.98 g NaH 2 PO 4
Percent P =
mass of P 30.97 g P = x 100% = 25.812 = 25.8% mass of NaH 2 PO 4 119.98 g NaH 2 PO 4
Percent O = 100.000% − (19.161 + 1.6802 + 25.812)% = 53.346 = 53.3% d.
Percent Co = Percent N =
mass of Co 58.93 g Co = x 100% = 32.211 = 32.2% mass of Co(NO3 ) 2 182.95 g Co(NO3 ) 2 mass of N 2 x 14.01 g N = x 100% = 15.316 = 15.3% mass of Co(NO3 ) 2 182.95 g Co(NO3 ) 2
Percent O = 100.000% − (32.211 + 15.316)% = 52.473 = 52.5% 3.58. In each part, the numerator consists of the mass of the element in one mole of the compound; the denominator is the mass of one mole of the compound. Answers are rounded to three significant figures. a.
Percent N =
mass of N 14.01 g N = x 100% = 30.449 = 30.4% mass of NO 2 46.01 g NO 2
Percent O = 100.000% − 30.449% N = 69.551 = 69.6% b.
Percent H =
mass of H 2.016 g H = x 100% = 5.9259 = 5.93% mass of H 2 O 2 34.02 g H 2 O 2
Percent O = 100.000% − 5.2959% H = 94.07 = 94.1% c.
Percent K =
mass of K 39.10 g K = x 100% = 28.221 = 28.2% mass of KClO 4 138.55 g KClO 4
Percent Cl =
mass of Cl 35.45 g Cl = x 100% = 25.586 = 25.6% mass of KClO 4 138.55 g KClO 4
Percent O = 100.000% − (28.221 + 25.586)% = 46.193 = 46.2% d.
Percent Mg = Percent N =
mass of M g 24.31 g Mg = = 20.897 = 20.9% mass of M g (NO 2 ) 2 116.33 g M g (NO 2 ) 2 mass of N 2 x 14.01 g N = = 24.086 = 24.1% mass of Mg(NO 2 )2 116.33 g Mg(NO 2 ) 2
Percent O = 100.000% − (20.897 + 24.086)% = 55.017 = 55.0%
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3.59. The molecular model of toluene contains seven carbon atoms and eight hydrogen atoms, so the molecular formula of toluene is C7H8. The molar mass of toluene is 92.134 g/mol. The mass percentages are Percent C =
mass of C 7 x 12.01 g = x 100% = 91.247 = 91.2% mass of C7 H8 92.134 g
Percent H = 100% − 91.247 = 8.753 = 8.75% 3.60. The molecular model of 2propanol contains three carbon atoms, eight hydrogen atoms, and one oxygen atom, so the molecular formula of 2propanol is C3H8O. The molar mass of 2propanol is 60.094 g/mol. The mass percentages are Percent C =
mass of C 3 x 12.01 g = x 100% = 59.956 = 60.0% mass of C3 H8 O 60.094 g
Percent H =
mass of H 8 x 1.008 g = x 100% = 13.418 = 13.4% mass of C3 H8 O 60.094 g
Percent O = 100% − (59.956 + 13.418) = 26.626 = 26.6% 3.61. Find the moles of C in each amount in onestep operations. Calculate the moles of each compound using the molar mass; then multiply by the number of moles of C per mole of compound: Mol C (glucose) = 6.01 g x
1 mol 6 mol C x = 0.200 mol 180.2 g 1 mol glucose
Mol C (ethanol) = 5.85 g x
1 mol 2 mol C x = 0.254 mol (more C) 46.07 g 1 mol ethanol
3.62. Find the moles of S in each amount in onestep operations. Calculate the moles of each compound using the molar mass; then multiply by the number of moles of S per mole of compound. Mol S (CaSO 4 ) = 40.8 g x
1 mol CaSO 4 1 mol S x = 0.2997 mol (more S) 136.15 g 1 mol CaSO 4
Mol S (Na 2SO3 ) = 35.2 g x
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1 mol Na 2SO3 1 mol S x = 0.2793 mol 126.05 g 1 mol Na 2SO3
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3.63. First, calculate the mass of C in the glycol by multiplying the mass of CO2 by the molar mass of C and the reciprocal of the molar mass of CO2. Next, calculate the mass of H in the glycol by multiplying the mass of H2O by the molar mass of 2H and the reciprocal of the molar mass of H2O. Then use the masses to calculate the mass percentages. Calculate O by difference. 9.06 mg CO 2 x
1 mol CO 2 12.01 g C x = 2.472 mg C 44.01 g CO 2 1 mol C
5.58 mg H 2 O x
1 mol H 2 O 2H 1.008 g H x x = 0.6243 mg H 18.02 g H 2 O 1 H 2O 1 mol H
Mass O = 6.38 mg − (2.472 + 0.6243) = 3.284 mg O Percent C = (2.472 mg C/6.38 mg glycol) x 100% = 38.74 = 38.7% Percent H = (0.6243 mg H/6.38 mg glycol) x 100% = 9.785 = 9.79% Percent O = (3.284 mg O/6.38 mg glycol) x 100% = 51.47 = 51.5% 3.64. First, calculate the mass of C in the phenol by multiplying the mass of CO2 by the molar mass of C and the reciprocal of the molar mass of CO2. Next, calculate the mass of H in the phenol by multiplying the mass of H2O by the molar mass of 2H and the reciprocal of the molar mass of H2O. Then use the masses to calculate the mass percentages. Calculate O by difference. 14.67 mg CO 2 x 3.01 mg H 2 O x
1 mol CO 2 12.01 g C x = 4.0033 mg C 44.01 g CO 2 1 mol C 1 mol H 2 O 2H 1.008 g H x x = 0.3368 mg H 18.02 g CO 2 1 H 2O 1 mol H
Mass O = 5.23 mg − (4.0033 + 0.3368) = 0.8899 mg O Percent C = (4.0033 mg/5.23 mg) x 100% = 76.54 = 76.5% Percent H = (0.3368 mg/5.23 mg) x 100% = 6.439 = 6.44% Percent O = (0.8899 mg/5.23 mg) x 100% = 17.0 = 17% 3.65. Start by calculating the moles of Os and O; then divide each by the smaller number of moles to obtain integers for the empirical formula. Mol Os = 2.16 g Os x
1 mol Os = 0.01136 mol (smaller number) 190.2 g Os
Mol O = (2.89 − 2.16) g O x
1 mol O = 0.0456 mol 16.00 g O
Integer for Os = 0.01136 ÷ 0.01136 = 1.000 Integer for O = 0.0456 ÷ 0.01136 = 4.01 Within experimental error, the empirical formula is OsO4.
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3.66. Start by calculating the moles of W and O; then divide each by the smaller number of moles to obtain integers for the empirical formula. Mol W = 4.23 g W x
1 mol W = 0.02301 mol (smaller number) 183.85 g W
Mol O = (5.34 − 4.23) g O x
1 mol O = 0.06938 mol 16.00 g O
Integer for W = 0.02301 ÷ 0.02301 = 1.000 Integer for O = 0.06938 ÷ 0.02301 = 3.015 Because 3.015 = 3.0 within experimental error, the empirical formula is WO3 3.67. Assume a sample of 100.0 g of potassium manganate. By multiplying this by the percentage composition, we obtain 39.6 g of K, 27.9 g of Mn, and 32.5 g of O. Convert each of these masses to moles by dividing by molar mass. Mol K = 39.6 g K x
1 mol K = 1.013 mol 39.10 g K
Mol Mn = 29.7 g Mn x Mol O = 32.5 g O x
1 mol Mn = 0.5078 mol (smallest number) 54.94 g Mn
1 mol O = 2.031 mol 16.00 g O
Now, divide each number of moles by the smallest number to obtain the smallest set of integers for the empirical formula. Integer for K = 1.013 ÷ 0.5078 = 1.998, or 2 Integer for Mn = 0.5078 ÷ 0.5078 = 1.000, or 1 Integer for O = 2.031 ÷ 0.5078 = 3.999, or 4 The empirical formula is thus K2MnO4. 3.68. Assume a sample of 100.0 g of hydroquinone. By multiplying this by the percentage composition, we obtain 65.4 g of C, 5.5 g of H, and 29.1 g of O. Convert each of these masses to moles by dividing by the molar mass. Mol C = 65.4 g C x Mol H = 5.5 g H x Mol O = 29.1 g O x
1 mol C = 5.445 mol 12.01 g C 1 mol H = 5.46 mol 1.008 g H
1 mol O = 1.819 mol (smallest number) 16.00 g O
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Now, divide each number of moles by the smallest number to obtain the smallest set of integers for the empirical formula. Integer for C = 5.445 ÷ 1.819 = 2.99, or 3 Integer for H = 5.46 ÷ 1.819 = 3.0, or 3 Integer for O = 1.819 ÷ 1.819 = 1.00, or 1 The empirical formula is thus C3H3O. 3.69. Assume a sample of 100.0 g of acrylic acid. By multiplying this by the percentage composition, we obtain 50.0 g C, 5.6 g H, and 44.4 g O. Convert each of these masses to moles by dividing by the molar mass. Mol C = 50.0 g C x Mol H = 5.6 g H x Mol O = 44.0 g O x
1 mol C = 4.163 mol 12.01 g C 1 mol H = 5.56 mol 1.008 g H
1 mol O = 2.775 mol (smallest number) 16.00 g O
Now, divide each number of moles by the smallest number to obtain the smallest number of moles and the tentative integers for the empirical formula. Tentative integer for C = 4.163 ÷ 2.775 = 1.50, or 1.5 Tentative integer for H = 5.56 ÷ 2.775 = 2.00, or 2 Tentative integer for O = 2.775 ÷ 2.775 = 1.00, or 1 Because 1.5 is not a whole number, multiply each tentative integer by 2 to obtain the final integer for the empirical formula: C: 2 x 1.5 = 3 H: 2 x 2 = 4 O: 2 x 1 = 2 The empirical formula is thus C3H4O2.
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3.70. Assume a sample of 100.0 g of malonic acid. By multiplying this by the percentage composition, we obtain 34.6 g C, 3.9 g H, and 61.5 g O. Convert each of these masses to moles by dividing by the molar mass. Mol C = 34.6 g C x Mol H = 3.9 g H x Mol O = 61.5 g O x
1 mol C = 2.881 mol (smallest number) 12.01 g C 1 mol H = 3.87 mol 1.008 g H
1 mol O = 3.844 mol 16.00 g O
Now, divide each number of moles by the smallest number to obtain the smallest number of moles and the tentative integers for the empirical formula. Tentative integer for C = 2.881 ÷ 2.881 = 1.00, or 1 Tentative integer for H = 3.87 ÷ 2.881 = 1.34, or 4/3 Tentative integer for O = 3.884 ÷ 2.881 = 1.334, or 4/3 Because 4/3 is not a whole number, multiply each tentative integer by 3 to give the final integer for the empirical formula: C: 3 x 1 = 3 H: 3 x 4/3 = 4 O: 3 x 4/3 = 4 The empirical formula is thus C3H4O4. 3.71. a.
Assume for the calculation that you have 100.0 g; of this quantity, 92.25 g is C and 7.75 g is H. Now, convert these masses to moles: 92.25 g C x 7.75 g H x
1 mol C = 7.68109 mol C 12.01 g C 1 mol H = 7.688 mol H 1.008 g H
Usually, you divide all the mole numbers by the smaller one, but in this case the mole numbers are equal, so the ratio of the number of C atoms to the number of H atoms is 1:1. Thus, the empirical formula for both compounds is CH.
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b.
Obtain n, the number of empirical formula units in the molecule, by dividing the molecular mass of 52.03 amu and 78.05 amu by the empirical formula mass of 13.018 amu: For 52.03: n =
52.03 amu = 3.9968, or 4 13.018 amu
For 78.05: n =
78.05 amu = 5.9955, or 6 13.018 amu
The molecular formulas are as follows: for 52.03, (CH)4 or C4H4; for 78.05, (CH)6 or C6H6. 3.72. a.
Assume for the calculation that you have 100.0 g; of this quantity, 85.62 g is C and 14.38 g is H. Now convert these masses to moles: 85.62 g C x
1 mol C = 7.12905 mol C 12.01 g C
14.38 g H x
1 mol H = 14.26 mol H 1.008 g H
Divide both mole numbers by the smaller one: For C:
7.129 mol = 1.00 7.129 mol
For H:
14.26 mol = 2.0002 7.129 mol
The empirical formula is obviously CH2. b.
Obtain n, the number of empirical formula units in the molecule, by dividing the molecular masses of 28.03 amu and 56.06 amu by the empirical formula weight of 14.026 amu: For 28.03: n =
28.03 amu = 1.9984, or 2 14.026 amu
For 56.06: n =
56.06 amu = 3.9968, or 4 14.026 amu
The molecular formulas are as follows: for 28.03, (CH2)2 or C2H4; for 56.06, (CH2)4 or C4H8. 3.73. The formula mass corresponding to the empirical formula C2H6N may be found by adding the respective atomic masses. Formula mass = (2 x 12.01 amu) + (6 x 1.008 amu) + 14.01 amu = 44.08 amu
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Dividing the molecular mass by the formula mass gives the number of times the C2H6N unit occurs in the molecule. Because the molecular mass is an average of 88.5 ([90 + 87] ÷ 2), this quotient is 88.5 amu ÷ 44.1 amu = 2.006, or 2 Therefore, the molecular formula is (C2H6N)2, or C4H12N2. 3.74. The formula mass corresponding to the empirical formula BH3 may be found by adding the respective atomic masses. Formula mass = 10.81 amu + (3 x 1.008 amu) = 13.83 amu Dividing the molecular mass by the formula mass gives the number of times the BH3 unit occurs in the molecule. Because the molecular mass is 28 amu, this quotient is 28 amu ÷ 13.83 amu = 2.02 Therefore, the molecular formula is (BH3)2, or B2H6. 3.75. Assume a sample of 100.0 g of oxalic acid. By multiplying this by the percentage composition, we obtain 26.7 g C, 2.2 g H, and 71.1 g O. Convert each of these masses to moles by dividing by the molar mass. Mol C = 26.7 g C x Mol H = 2.2 g H x Mol O = 71.1 g O x
1 mol C = 2.223 mol 12.01 g C
1 mol H = 2.18 mol (smallest number) 1.008 g H 1 mol O = 4.443 mol 16.00 g O
Now, divide each number of moles by the smallest number to obtain the smallest set of integers for the empirical formula. Integer for C = 2.223 ÷ 2.18 = 1.02, or 1 Integer for H = 2.18 ÷ 2.18 = 1.00, or 1 Integer for O = 4.443 ÷ 2.18 = 2.038, or 2 The empirical formula is thus CHO2. The formula mass corresponding to this formula may be found by adding the respective atomic masses: Formula mass = 12.01 amu + 1.008 amu + (2 x 16.00 amu) = 45.02 amu Dividing the molecular mass by the formula mass gives the number of times the CHO2 unit occurs in the molecule. Because the molecular mass is 90 amu, this quotient is 90 amu ÷ 45.02 amu = 2.00, or 2 The molecular formula is thus (CHO2)2, or C2H2O4.
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3.76. Assume a sample of 100.0 g of adipic acid. By multiplying this by the percentage composition, we obtain 49.3 g C, 6.9 g H, and 43.8 g O. Convert each of these masses to moles by dividing by the molar mass. Mol C = 49.3 g C x Mol H = 6.9 g H x Mol O = 43.8 g O x
1 mol C = 4.105 mol 12.01 g C 1 mol H = 6.85 mol 1.008 g H
1 mol O = 2.738 mol (smallest number) 16.00 g O
Now, divide each number of moles by the smallest number to obtain the smallest set of integers for the empirical formula. Tentative integer for C = 4.105 ÷ 2.738 = 1.499, or 1.5 Tentative integer for H = 6.85 ÷ 2.738 = 2.50, or 2.5 Tentative integer for O = 2.738 ÷ 2.738 = 1.000, or 1 Because 1.5 and 2.5 are not whole numbers, multiply each tentative integer by 2 to give the final integers for the empirical formula: C: 2 x 1.5 = 3 H: 2 x 2.5 = 5 O: 2 x 1 = 2 The empirical formula is thus C3H5O2. The formula mass corresponding to this formula may be found by adding the respective atomic masses: Formula mass = (3 x 12.01 amu) + (5 x 1.008 amu) + (2 x 16.00 amu) = 73.1 amu Dividing the molecular mass by the formula mass gives the number of times the CHO2 unit occurs in the molecule. Because the molecular mass is 146 amu, this quotient is 146 amu ÷ 73.1 amu = 2.00, or 2 The molecular formula is thus (C3H5O2)2, or C6H10O4. 3.77. C2H4 1 molecule C2H4 1 mole C2H4 28.052 g C2H4
+ + + +
3O2 3 molecules O2 3 moles O2 3 x 32.00 g O2
→ → → →
2CO2 2 molecules CO2 2 moles CO2 2 x 44.01 g CO2
+ + + +
2H2O 2 molecules H2O 2 moles H2O 2 x 18.016 g H2O
2H2S 2 molecules H2S 2 moles H2S 2 x 34.09 g H2S
+ + + +
3O2 3 molecules O2 3 moles O2 3 x 32.00 g O2
→ → → →
2SO2 2 molecules SO2 2 moles SO2 2 x 64.07 g CO2
+ + + +
2H2O 2 molecules H2O 2 moles H2O 2 x 18.016 g H2O
3.78.
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3.79. By inspecting the balanced equation, obtain a conversion factor of eight mol CO2 to two mol C4H10. Multiply the given amount of 0.30 mole of C4H10 by the conversion factor to obtain the moles of CO2. 0.30 mol C4H10 x
8 mol CO 2 = 1.20 = 1.2 mol CO2 2 mol C4 H10
3.80. By inspecting the balanced equation, obtain a conversion factor of three mol H2O to one mol C2H5OH. Multiply the given amount of 0.69 mol of C2H5OH by the conversion factor to obtain the moles of H2O. 0.69 mol C2H5OH x
3 mol H 2 O = 2.07 = 2.1 mol H2O 1 mol C2 H 5 OH
3.81. By inspecting the balanced equation, obtain a conversion factor of three mol O2 to two mol Fe2O3. Multiply the given amount of 3.91 mol Fe2O3 by the conversion factor to obtain moles of O2. 3.91 mol Fe2O3 x
3 mol O 2 = 5.865 = 5.87 mol O2 2 mol Fe 2 O3
3.82. By inspecting the balanced equation, obtain a conversion factor of three mol NiCl2 to one mol Ni3(PO4)2. Multiply the given amount of 0.517 mol Ni3(PO4)2 by the conversion factor to obtain moles of NiCl2. 0.517 mol Ni3(PO4)2 x
3 mol NiCl2 = 1.551 = 1.55 mol NiCl2 1 mol Ni3 (PO 4 ) 2
3.83. 3NO2 + H2O → 2HNO3 + NO Three moles of NO2 are equivalent to two moles of HNO3 (from equation). One mole of NO2 is equivalent to 46.01 g NO2 (from molecular mass of NO2). One mole of HNO3 is equivalent to 63.02 g HNO3 (from molecular mass of HNO3). 7.50 g HNO3 x
1 mol HNO3 3 mol NO 2 46.01 g NO 2 = 8.213 x x 63.02 g HNO3 2 mol HNO3 1 mol NO 2
= 8.21 g NO2 3.84. 2Ca3(PO4)2 + 6SiO2 + 10C → P4 + 6CaSiO3 + 10CO Two moles of Ca3(PO4)2 are equivalent to one mole of P4 (from equation). One mol of P4 is equivalent to 123.9 g P4 (from molecular mass of P4). One mol of Ca3(PO4)2 is equivalent to 310.2 g Ca3(PO4)2 [from molecular mass of Ca3(PO4)2].
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15.0 g P4 x
2 mol Ca 3 (PO 4 ) 2 310.18 g Ca 3 (PO 4 ) 2 1 mol P4 x x 1 mol P4 1 mol Ca 3 (PO 4 ) 2 123.88 g P4
= 75.11 = 75.1 g Ca3(PO4)2 3.85. WO3 + 3H2 → W + 3H2O One mole of W is equivalent to three moles of H2 (from equation). One mole of H2 is equivalent to 2.016 g H2 (from molecular mass of H2). One mole of W is equivalent to 183.8 g W (from atomic mass of W). 4.81 kg of H2 is equivalent to 4.81 x 103 g of H2. 4.81 x 103 g H2 x
1 mol H 2 1 mol W 183.85 g W x x 3 mol H 2 1 mol W 2.016 g H 2
= 1.462 x 105 = 1.46 x 105 g W 3.86. 4C3H6 + 6NO → 4C3H3N + 6H2O + N2 Four moles of C3H6 are equivalent to four moles of C3H3N (from equation). One mole of C3H6 is equivalent to 42.08 g C3H6 (from molecular mass of C3H6). One mole of C3H3N is equivalent to 53.06 g C3H3N (from molecular mass of C3H3N). 452 kg of C3H6 are equivalent to 4.52 x 105 g C3H6. 4.52 x 105 g C3H6 x
1 mol C3 H 6 4 mol C3 H 3 N 53.06 g C3 H 3 N x x 42.08 g C3 H 6 4 mol C3 H 6 1 mol C3 H 3 N
= 5.699 x 105 = 5.70 x 105 g C3H3N 3.87. Write the equation, and set up the calculation below the equation (after calculating the two molecular masses): CS2 + 3Cl2 → CCl4 + S2Cl2 62.7 g Cl2 x
1 mol Cl2 1 mol CS2 76.15 g CS2 x x 22.448 = 22.4 g CS2 70.90 g Cl2 3 mol Cl2 1 mol CS2
3.88. From the molecular models, the balanced chemical equation is Pt 4NH3 + 5O2 ⎯⎯ → 4NO + 6H2O
The molar mass of NH3 is 17.03 g/mol, and for O2, it is 32.00 g/mol. This gives 6.1 g NH3 x
1 mol NH 3 5 mol O 2 32.00 g O 2 x x = 14.32 = 14 g O2 17.03 g NH 3 4 mol NH 3 1 mol O 2
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3.89. Write the equation, and set up the calculation below the equation (after calculating the two molecular masses): 2N2O5 → 4NO2 + O2 1.315 g O2 x
1 mol O 2 4 mol NO 2 46.01 g NO 2 x x = 7.5628 = 7.563 g NO2 32.00 g O 2 1 mol O 2 1 mol NO 2
3.90. Write the equation, and set up the calculation below the equation (after calculating the two formula masses): 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O 6.01 g Cu(NO3)2 x
1 mol Cu(NO3 ) 2 2 mol NO 30.01 g NO x x 187.56 g Cu(NO3 ) 2 3 mol Cu(NO3 ) 2 1 mol NO
= 0.6410 = 0.641 g NO 3.91. First determine whether KO2 or H2O is the limiting reactant by calculating the moles of O2 that each would form if it were the limiting reactant. Identify the limiting reactant by the smaller number of moles of O2 formed. 0.15 mol H 2 O x
3 mol O 2 = 0.225 mol O2 2 mol H 2 O
0.25 mol KO 2 x
3 mol O 2 = 0.187 mol O2 (KO2 is the limiting reactant.) 4 mol KO 2
Moles of O2 produced = 0.19 mol 3.92. First, determine whether NaOH or Cl2 is the limiting reactant by calculating the moles of NaClO that each would form if it were the limiting reactant. Identify the limiting reactant by the smaller number of moles of NaClO formed. 1.23 mol NaOH x 1.47 mol Cl2 x
1 mol NaClO = 0.615 mol NaClO (smaller number) 2 mol NaOH
1 mol NaClO = 1.47 mol NaClO 1 mol Cl2
NaOH is the limiting reactant, and 0.615 mol of NaClO will form.
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Chapter 3: Calculations with Chemical Formulas and Equations
3.93. First determine whether CO or H2 is the limiting reactant by calculating the moles of CH3OH that each would form if it were the limiting reactant. Identify the limiting reactant by the smaller number of moles of CH3OH formed. Use the molar mass of CH3OH to calculate the mass of CH3OH formed. Then calculate the mass of the unconsumed reactant. CO + 2H2 → CH3OH 10.2 g H2 x 35.4 g CO x
1 mol CH 3OH 1 mol H 2 x = 2.529 mol CH3OH 2.016 g H 2 2 mol H 2 1 mol CH 3OH 1 mol CO = 1.263 mol CH3OH x 1 mol CO 28.01 g CO
CO is the limiting reactant. Mass CH3OH formed = 1.263 mol CH3OH x
32.042 g CH 3OH 1 mol CH 3OH
= 40.47 = 40.5g CH3OH Hydrogen is left unconsumed at the end of the reaction. The mass of H2 that reacts can be calculated from the moles of product obtained: 1.263 mol CH 3OH x
2 mol H 2 2.016 g H 2 x = 5.092 g H 2 1 mol CH 3OH 1 mol H 2
Unreacted H2 = 10.2 g total H2 − 5.092 g reacted H2 = 5.108 = 5.1 g H2 3.94. First, determine whether CS2 or O2 is the limiting reactant by calculating the moles of SO2 that each would form if it were the limiting reactant. Identify the limiting reactant by the smaller number of moles of SO2 formed. Use the molar mass of SO2 to calculate the mass of SO2 formed. Then calculate the mass of the unconsumed reactant. CS2 + 3O2 → CO2 + 2SO2 30.0 g O2 x 35.0 g CS2 x
1 mol O 2 2 mol SO 2 x = 0.6250 mol SO2 32.00 g O 2 3 mol O 2
1 mol CS2 2 mol SO 2 x = 0.9192 mol SO2 76.15 g CS2 1 mol CS2
O2 is the limiting reactant. Mass SO2 formed = 0.6250 mol SO2 x
64.07 g SO 2 = 40.04 = 40.0 g 1 mol SO 2
CS2 is left unconsumed at the end of the reaction. The mass of CS2 that reacts can be calculated from the moles of product obtained: 0.6250 mol SO2 x
1 mol CS2 76.15 g CS2 x = 23.79 g CS2 2 mol SO 2 1 mol CS2
Unreacted CS2 = 35.0 g total CS2 − 23.79 g reacted CS2 = 11.20 = 11.2 g CS2
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3.95. First, determine which of the three reactants is the limiting reactant by calculating the moles of TiCl4 that each would form if it were the limiting reactant. Identify the limiting reactant by the smallest number of moles of TiCl4 formed. Use the molar mass of TiCl4 to calculate the mass of TiCl4 formed. 3TiO2 + 4C + 6Cl2 → 3TiCl4 + 2CO2 + 2CO 4.15 g TiO2 x 5.67 g C x
1 mol TiO 2 3 mol TiCl4 x = 0.05195 mol TiCl4 79.88 g TiO 2 3 mol TiO 2
3 mol TiCl4 1 mol C x = 0.35407 mol TiCl4 4 mol C 12.01 g C 1 mol Cl2 3 mol TiCl4 x = 0.04781 mol TiCl4 70.90 g Cl2 6 mol Cl2
6.78 g Cl2 x
Cl2 is the limiting reactant. 189.68 g TiCl4 = 9.068 = 9.07 g TiCl4 1 mol TiCl4
Mass TiCl4 formed = 0.04781 mol TiCl4 x
3.96. First, determine which of the three reactants is the limiting reactant by calculating the moles of HCN that each would form if it were the limiting reactant. Identify the limiting reactant by the smallest number of moles of HCN formed. Use the molar mass of HCN to calculate the mass of HCN formed. 2NH3 + 3O2 + 2CH4 → 2HCN + 6H2O 11.5 g NH3 x
1 mol NH3 2 mol HCN = 0.675 mol HCN x 2 mol NH 3 17.03 g NH 3
10.5 g CH4 x
1 mol CH 4 2 mol HCN = 0.654 mol HCN x 2 mol CH 4 16.04 g CH 4
12.0 g O2 x
1 mol O 2 2 mol HCN = 0.2500 mol HCN x 3 mol O 2 32.00 g O 2
O2 is the limiting reactant. Mass HCN formed = 0.2500 mol HCN x
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27.03 g HCN = 6.757 = 6.76 g HCN 1 mol HCN
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3.97. First, determine which of the two reactants is the limiting reactant by calculating the moles of aspirin that each would form if it were the limiting reactant. Identify the limiting reactant by the smallest number of moles of aspirin formed. Use the molar mass of aspirin to calculate the theoretical yield in grams of aspirin. Then calculate the percentage yield. C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2 4.00 g C 4 H 6 O3 x
1 mol C4 H 6 O3 1 mol C9 H8 O 4 x = 0.03918 mol C9H8O4 102.09 g C4 H 6 O3 1 mol C4 H 6 O3
2.00 g C7 H 6 O3 x
1 mol C7 H 6 O3 1 mol C9 H8 O 4 x = 0.01448 mol C9H8O4 138.12 g C7 H 6 O3 1 mol C7 H 6 O3
Thus, C7H6O3 is the limiting reactant. The theoretical yield of C9H8O4 is 0.01448 mol C9H8O4 x
180.15 g C9 H8 O 4 = 2.609 g C9H8O4 1 mol C9 H8 O 4
The percentage yield is Percentage yield =
actual yield 1.86 g x 100% = x 100% = 71.29 = 71.3% theoretical yield 2.609 g
3.98. First, determine which of the two reactants is the limiting reactant by calculating the moles of methyl salicylate that each would form if it were the limiting reactant. Identify the limiting reactant by the smallest number of moles of methyl salicylate formed. Use the molar mass of methyl salicylate to calculate the theoretical yield in grams of methyl salicylate. Then calculate the percentage yield. C7H6O3 + CH3OH → C8H8O3 + H2O 11.20 g CH3OH x
1 mol CH3OH 1 mol C8 H8 O3 x = 0.3496 mol C8H8O3 32.04 g CH 3OH 1 mol CH 3OH
1.50 g C7H6O3 x
1 mol C7 H 6 O3 1 mol C8 H8 O3 x = 0.01086 mol C8H8O3 138.12 g C7 H 6 O3 1 mol C7 H 6 O3
Thus, C7H6O3 is the limiting reactant. The theoretical yield of C8H8O3 is 0.01086 mol C8H8O3 x
152.14 g C8 H8 O3 = 1.652 g C8H8O3 1 mol C8 H8 O3
The percentage yield is Percentage yield =
actual yield 1.27 g x 100% = x 100% = 76.87 = 76.9% theoretical yield 1.652 g
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103
SOLUTIONS TO GENERAL PROBLEMS
3.99. For 1 mol of caffeine, there are eight mol of C, ten mol of H, four mol of N, and two mol of O. Convert these amounts to masses by multiplying them by their respective molar masses: 8 mol C x 12.01 g C/1 mol C
=
96.08 g C
10 mol H x 1.008 g H/1 mol H
=
10.08 g H
4 mol N x 14.01 g N/1 mol N
=
56.04 g N
2 mol O x 16.00 g O/1 mol O
=
1 mol of caffeine (total)
=
32.00 g O 194.20 g (molar mass)
Each mass percentage is calculated by dividing the mass of the element by the molar mass of caffeine and multiplying by 100 percent: Mass percentage = (mass element ÷ mass caffeine) x 100%. Mass percentage C = (96.08 g ÷ 194.20 g) x 100% = 49.5% (3 s.f.) Mass percentage H = (10.08 g ÷ 194.20 g) x 100% = 5.19% (3 s.f.) Mass percentage N = (56.04 g ÷ 194.20 g) x 100% = 28.9% (3 s.f.) Mass percentage O = (32.00 g ÷ 194.20 g) x 100% = 16.5% (3 s.f.) 3.100. For each mole of morphine, there are 17 mol C, 19 mol H, 1 mol N, and 3 mol O. Convert these amounts to masses by multiplying by the respective molar masses: 17 mol C x 12.01 g C/1 mol C
=
204.17 g C
19 mol H x 1.008 g H/1 mol H
=
19.15 g H
1 mol N x 14.01 g N/1 mol N
=
14.01 g N
3 mol O x 16.00 g O/1 mol O
=
1 mol of morphine (total)
=
48.00 g O 285.33 g (molar mass)
Each mass percentage is calculated by dividing the mass of the element by the molar mass of morphine and multiplying by 100%: Mass percentage = (mass element ÷ mass morphine) x 100%. Mass percentage C = (204.17 g ÷ 285.33 g) x 100% = 71.6% (3 s.f.) Mass percentage H = (19.15 g ÷ 285.33 g) x 100% = 6.71% (3 s.f.) Mass percentage N = (14.01 g ÷ 285.33 g) x 100% = 4.91% (3 s.f.) Mass percentage O = (48.00 g ÷ 285.33 g) x 100% = 16.8% (3 s.f.)
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Chapter 3: Calculations with Chemical Formulas and Equations
3.101. Assume a sample of 100.0 g of dichlorobenzene. By multiplying this by the percentage composition, we obtain 49.1 g C, 2.7 g of H, and 48.2 g of Cl. Convert each mass to moles by dividing by the molar mass: 1 mol C = 4.088 mol C 12.01 g C
49.1 g C x 2.7 g H x
1 mol H = 2.68 mol H 1.008 g H
48.2 g Cl x
1 mol Cl = 1.360 mol Cl 35.45 g Cl
Divide each number of moles by the smallest number to obtain the smallest set of integers for the empirical formula. Integer for C = 4.088 mol ÷ 1.360 mol = 3.00, or 3 Integer for H = 2.68 mol ÷ 1.360 mol = 1.97, or 2 Integer for Cl = 1.360 mol ÷ 1.360 mol = 1.00, or 1 The empirical formula is thus C3H2Cl. Find the formula mass by adding the atomic masses: Formula mass = (3 x 12.01 amu) + (2 x 1.008 amu) + 35.45 amu = 73.496 = 73.50 amu Divide the molecular mass by the formula mass to find the number of times the C3H2Cl unit occurs in the molecule. Because the molecular mass is 147 amu, this quotient is 147 amu ÷ 73.50 amu = 2.00, or 2 The molecular formula is (C3H2Cl)2, or C6H4Cl2. 3.102. Assume a sample of 100.0 g of sorbic acid. By multiplying this by the percentage composition, we obtain 64.3 g C, 7.2 g H, and 28.5 g O. Convert each mass to moles by dividing by the molar mass. 64.3 g C x 7.2 g H x 28.5 g O x
1 mol C = 5.353 mol C 12.01 g C 1 mol H = 7.14 mol H 1.008 g H 1 mol O = 1.781 mol O 16.00 g O
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Divide each number of moles by the smallest number to obtain the smallest set of integers for the empirical formula. Integer for C = 5.353 mol ÷ 1.781 mol = 3.01, or 3 Integer for H = 7.14 mol ÷ 1.781 mol = 4.0, or 4 Integer for O = 1.781 mol ÷ 1.781 mol = 1.00, or 1 The empirical formula is thus C3H4O. Find the formula mass by adding the atomic masses: Formula mass = (3 x 12.01 amu) + (4 x 1.008 amu) + 16.00 amu = 56.062 = 56.06 amu Divide the molecular mass by the formula mass to find the number of times the C3H4O unit occurs in the molecule. Because the molecular mass was given as 112 amu, this quotient is 112 amu ÷ 56.06 amu = 2.00, or 2 The molecular formula is (C3H4O)2, or C6H8O2. 3.103. Find the percentage composition of C and S from the analysis: 0.01665 g CO 2 x
1 mol CO 2 1 mol C 12.01 g C x x = 0.004544 g C 44.01 g CO 2 1 mol CO 2 1 mol C
Percent C = (0.004544 g C ÷ 0.00796 g comp.) x 100% = 57.09% 0.01196 g BaSO4 x
1 mol BaSO 4 1 mol S 32.07 g S x = 0.001643 g S x 1 mol S 1 mol BaSO 4 233.39 g BaSO 4
Percent S = (0.001643 g S ÷ 0.00431 g comp.) x 100% = 38.12% Percent H = 100.00% − (57.09 + 38.12)% = 4.79% We now obtain the empirical formula by calculating moles from the grams corresponding to each mass percentage of element: 57.09 g C x
1 mol C = 4.754 mol C 12.01 g C
38.12 g S x
1 mol S = 1.189 mol S 32.07 g S
4.79 g H x
1 mol H = 4.752 mol H 1.008 g H
Dividing the moles of the elements by the smallest number (1.189), we obtain for C: 3.997, or 4; for S: 1.000, or 1; and for H: 3.996, or 4. Thus, the empirical formula is C4H4S (formula mass = 84). Because the formula mass was given as 84 amu, the molecular formula is also C4H4S.
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3.104. Find the percentage composition of H and N from the analysis: 0.00663 g H2O x
1 mol H 2 O 2 mol H 1.008 g H x = 0.0007417 g H x 1 mol H 1 mol H 2 O 18.02 g H 2 O
1.46 mg N2 from the analysis is equivalent to 0.00146 g N. Percent H = (0.0007417 g H ÷ 0.00971 g comp.) x 100% = 7.64% Percent N = (0.00146 g N ÷ 0.00971 g comp.) x 100% = 15.0% Percent C = 100.00% − (7.64 + 15.0)% = 77.4% Calculate the moles from grams to obtain the empirical formula: 77.4 g C x
1 mol C = 6.44 mol C 12.01 g C
7.64 g H x
1 mol H = 7.58 mol H 1.008 g H
15.0 g N x
1 mol N = 1.07 mol N 14.01 g N
Dividing the moles of elements by the smallest number (1.07) gives for C: 6.02, or 6; for H: 7.08, or 7; and for N: 1.00, or 1. The empirical formula is thus C6H7N (formula mass = 93 amu). Because the molecular mass was given as 93 amu, the molecular formula is also C6H7N. 3.105. For g CaCO3, use this equation: CaCO3 + H2C2O4 → CaC2O4 + H2O + CO2. 0.472 g CaC 2 O 4 x
1 mol CaCO3 100.09 g CaCO3 1 mol CaC2 O 4 x x 128.10 g CaC2 O 4 1 mol CaC 2 O 4 1 mol CaCO3
= 0.3688 g CaCO3 Mass percentage CaCO3 =
mass CaCO3 0.3688 g x 100% = x 100% mass limestone 0.438 g
= 84.19 = 84.2% 3.106. For the mass of TiO2, use this equation: TiO2 + C + 2Cl2 → TiCl4 + CO2. 35.4 g TiCl4 x
1 mol TiCl4 1 mol TiO 2 79.88 g TiO 2 x x = 14.91 g TiO2 189.68 g TiCl4 1 mol TiCl4 1 mol TiO 2
Mass percentage TiO2 =
mass TiO 2 14.91 g x 100% = 85.68 = 85.7% x 100% = mass rutile 17.4 g
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3.107. Calculate the theoretical yield using this equation: 2C2H4 + O2 → 2C2H4O. 10.6 g C2 H 4 x
1 mol C2 H 4 2 mol C2 H 4 O 44.05 g C2 H 4 O x x = 16.65 g C2H4O 28.05 g C 2 H 4 2 mol C 2 H 4 1 mol C2 H 4 O
Percentage yield =
actual yield 9.91 g x 100% = x 100% = 59.53 = 59.5% theoretical yield 16.65 g
3.108. Calculate the theoretical yield using this equation: C6H6 + HNO3 → C6H5NO2 + H2O 22.4 g C6 H 6 x
1 mol C6 H 6 1 mol C6 H 5 NO 2 123.11 g C6 H 5 NO 2 x x 78.11 g C6 H 6 1 mol C6 H 6 1 mol C6 H 5 NO 2 = 35.30 g C6H5NO2
Percentage yield =
actual yield 31.6 g x 100% = x 100% = 89.51 = 89.5% theoretical yield 35.30 g
3.109. To find Zn, use these equations: 2C + O2 → 2CO and ZnO + CO → Zn + CO2 Two moles of C produces 2 mol CO; because 1 mol ZnO reacts with 1 mol CO, 2 mol ZnO will react with 2 mol CO. Thus, 2 mol C is equivalent to 2 mol ZnO, or 1 mol C is equivalent to 1 mol ZnO. Using this to calculate mass of C from mass of ZnO, we have 75.0 g ZnO x
1 mol ZnO 1 mol C 12.01 g C x x = 11.07 g C 81.39 g ZnO 1 mol ZnO 1 mol C
Thus, all of the ZnO is used up in reacting with just 11.07 g of C, making ZnO the limiting reactant. Use the mass of ZnO to calculate the mass of Zn formed: 75.0 g ZnO x
1 mol ZnO 1 mol Zn 65.39 g Zn x x = 60.256 = 60.3 g Zn 81.39 g ZnO 1 mol ZnO 1 mol Zn
3.110. To find CH4, use these equations: 4NH3 + 5O2 → 4NO + 6H2O 2NO + 2CH4 → 2HCN + 2H2O + H2 Four moles of NH3 produces 4 mol NO; because 2 mol CH4 reacts with 2 mol NO, 4 mol CH4 will react with 4 mol NO. Thus, 4 mol NH3 is equivalent to 4 mol CH4. Using this to calculate the mass of CH4 from the mass of NH3, we have 24.2 g NH 3 x
1 mol NH 3 4 mol CH 4 16.04 g CH 4 x x = 22.8 g CH 4 17.03 g NH 3 4 mol NH 3 1 mol CH 4
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Chapter 3: Calculations with Chemical Formulas and Equations
Thus, all of the NH3 is used up in reacting with just 22.8 g of CH4, making NH3 the limiting reactant. Use the mass of NH3 to calculate the mass of HCN formed: 24.2 g NH3 x
1 mol NH 3 4 mol HCN 27.03 g HCN x x = 38.41 = 38.4 g HCN 17.03 g NH 3 4 mol NH 3 1 mol HCN
3.111. For CaO + 3C → CaC2 + CO, find the limiting reactant in terms of moles of CaC2 obtainable: Mol CaC2 = 2.60 x 103 g C x
1 mol CaC2 1 mol C x = 72.16 mol 12.01 g C 3 mol C
Mol CaC2 = 2.60 x 103 g CaO x
1 mol CaC2 1 mol CaO x = 46.362 mol 1 mol CaO 56.08 g CaO
Because CaO is the limiting reactant, calculate the mass of CaC2 from it: Mass CaC2 = 46.362 mol CaC2 x
64.10 g CaC2 = 2.971 x 103 = 2.97 x 103 g CaC2 1 mol CaC 2
3.112. For CaF2 + H2SO4 → 2HF + CaSO4, find the limiting reactant in terms of moles of HF obtainable: Mol HF = 11.9 g CaF2 x Mol HF = 10.3 g H2SO4 x
1 mol CaF2 2 mol HF = 0.3048 mol x 1 mol CaF2 78.08 g CaF2 1 mol H 2SO 4 2 mol HF x = 0.2100 mol 1 mol H 2SO 4 98.09 g H 2SO 4
Because H2SO4 is the limiting reactant, calculate the mass of HF from it: Mass HF = 0.2100 mol HF x
20.01 g HF = 4.202 = 4.20 g HF 1 mol HF
3.113. From the equation 2Na + H2O → 2NaOH + H2, convert mass of H2 to mass of Na, and then use the mass to calculate the percentage: 0.108 g H2 x
1 mol H 2 2 mol Na 22.99 g Na x x = 2.463 g Na 1 mol H 2 1 mol Na 2.016 g H 2
Percent Na =
mass Na 2.463 g x 100% = x 100% = 16.17 = 16.2% mass amalgam 15.23 g
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3.114. From the equation CaCO3 → CaO + CO2, convert the mass of CO2 to mass of CaCO3, and then use the mass to calculate the percentage: 0.00395 g CO2 x
1 mol CaCO3 100.09 g CaCO3 1 mol CO 2 x x 1 mol CO 2 1 mol CaCO3 44.01 g CO 2
= 0.008983 g CaCO3 Percent CaCO3 =
mass CaCO3 0.008983 g x 100% = x 100% mass sandstone 0.0187 g
= 48.039 = 48.0% Because the sandstone contains only SiO2 and CaCO3, the difference between 100 percent and the percentage of CaCO3 is the percentage of SiO2: Percent SiO2(silica) = 100.00% − 48.039% = 51.96 = 52.0% 3.115. The mass spectrometer measures the masses of positive ions produced from a very small sample and displays the data as a mass spectrum. This mass spectrum can be used to identify a substance or to obtain the molecular formula of a compound. The mass spectrum contains a wealth of information about molecular structure. 3.116. The mass spectrum of a molecule is much more complicated for several reasons. First, the molecular ions produced often break into fragments, giving several different kinds of positive ions. Second, many of the atoms in any ion can occur with different isotopic masses, so each ion often has many peaks.
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SOLUTIONS TO STRATEGY PROBLEMS
3.117. 2 H 2 molec = 96 H2 molecules 1 O 2 molec
a.
48 O2 molecules x
b.
5 mol H2O x
1 mol O 2 = 2.5 mol O2 2 mol H 2 O
5 mol H2O x
2 mol H 2 = 5.0 mol H2 2 mol H 2 O
37.5 g O2 x
4.0 g H 2 = 4.68 = 4.7 g H2 32 g O 2
c.
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Chapter 3: Calculations with Chemical Formulas and Equations
d.
The mass of water formed is 4.0 g + 32 g = 36.0 g 30.0 g H2O x
32 g O 2 = 26.6 = 27 g O2 36 g H 2 O
30.0 g H2O x
4.0 g H 2 = 3.33 = 3.3 g H2 36 g H 2 O
3.118. The balanced equation is 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s). 1 mol Fe 2 O3 1 mol Al x = 0.7598 = 0.76 mol Fe2O3 2 mol Al 26.98 g
a.
41 g Al x
b.
First, determine the limiting reactant. 99.1 g Al x
1 mol Al 2 mol Fe x = 3.673 = 3.67 mol Fe 26.98 g 2 mol Al
3.14 mol Fe2O3 x
2 mol Fe = 6.28 mol Fe 1 mol Fe 2 O3
Since the limiting reactant is Al, 3.67 mol Fe will form. c.
7.0 g Fe x
1 mol Fe 2 mol Al 6.02 x 1023 atoms x x = 7.54 x 1022 2 mol Fe 1 mol 55.85 g = 7.5 x 1022 atoms
3.119. a.
First, determine the limiting reactant. 25 g H2SO4 x
1 mol H 2SO 4 1 mol K 2SO 4 174.27 g x x = 44.4 = 44 g 1 mol K 2SO 4 98.086 g 1 mol H 2SO 4
7.7 g KOH x
1 mol K 2SO 4 1 mol KOH 174.27 g x x = 11.95 = 12 g 56.108 g 1 mol K 2SO 4 2 mol KOH
KOH is the limiting reactant, and 12 g K2SO4 is produced. b.
KOH is the limiting reactant. The mass of H2SO4 that reacts is 7.7 g KOH x
1 mol H 2SO 4 1 mol KOH 98.086 g x x = 6.73 g reacted 56.108 g 1 mol H 2SO 4 2 mol KOH
The mass remaining after reaction is 25 g − 6.73 g = 18.27 = 18 g. c.
The theoretical yield was determined in part a to be 11.9 g. For an actual yield of 47.2%, the amount of K2SO4 formed is 11.95 g x 0.472 = 5.644 = 5.6 g
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3.120. a.
First, determine the mass of C, H, and O in the compound. Mass C = 0.561 g CO2 x
1 mol CO 2 1 mol C 12.01 g C x = 0.1531 g x 1 mol C 1 mol CO 2 44.01 g
Mass H = 0.306 g H2O x
1 mol H 2 O 2 mol H 1.008 g H x = 0.03423 g x 1 mol H 1 mol H 2 O 18.018 g
Mass O = 0.255 g − 0.1531 g − 0.03423 g = 0.0676 g Next, determine the empirical formula. mol C = 0.1531 g x
1 mol C = 0.01275 mol C 12.01 g C
1 mol H = 0.03396 mol H 1.008 g H
mol H = 0.03423 g x mol O = 0.0676 g x
1 mol O = 0.00423 mol O 16.00 g O
Because the smallest number of moles is 0.00423 mol O, divide each molar quantity by this factor. 0.01275 mol C = 3.0; 0.00423 mol O
0.03396 mol H = 8.0; 0.00423 mol O
0.00423 mol O = 1 0.00423 mol O
Therefore, the empirical formula is C3H8O. b.
The empirical formula mass is 60.094 g/formula unit. Thus, the multiplication factor is n =
180 g/mol = 2.99 = 3 60.094 g/unit
The molecular formula of the compound is C9H24O3. 3.121. The balance chemical equation is 4NH3 + 5O2 → 4NO + 6H2O. The theoretical yield of NO is 8.5 g NH3 x
1 mol NH 3 4 mol NO 30.01 g NO x = 14.97 g NO x 1 mol NO 4 mol NH 3 17.034 g NH 3
The percentage yield is Percentage Yield =
12.0 g x 100% = 80.1% = 80.% 14.97 g
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Chapter 3: Calculations with Chemical Formulas and Equations
3.122. The fraction of Pb in Pb3N2 is X =
621.6 g Pb = 0.95687 649.62 g Pb3 N 2
The mass of lead in the paint is Mass Pb = 27.5 g Pb3N2 x 0.95687 = 26.31 = 26.3 g Pb 3.123. The fraction of O in Na2SO3 is X1 =
48.00 g O = 0.38086 126.03 g Na 2SO3
The fraction of O in MgSO4 is X2 =
64.00 g O = 0.53165 120.38 g MgSO 4
The total mass of oxygen present is Mass O = (0.38086 x 12.1 g) + (0.53165 x 15.5 g) = 4.608 g + 8.240 g = 12.848 = 12.85 g 3.124. First, calculate the mass of KO2 actually used in the reaction. 195 g O2 x
1 mol O 2 4 mol KO 2 71.10 g KO 2 x x = 577.68 g KO2 used 32.00 g O 2 3 mol O 2 1 mol KO 2
Therefore, the KO2 was not completely consumed in the reaction. The mass of KO2 left over is 750 g − 577.68 g = 172.3 = 172 g. The mass of additional oxygen that could be produced is 172.3 g KO2 x
3.125. 175 tablets x
1 mol KO 2 3 mol O 2 32.00 g O 2 x x = 58.16 = 58.2 g O2 71.10 g KO 2 4 mol KO 2 1 mol O 2
1 mol CaCO3 68.4 x 103 g = 0.1196 = 0.120 mol CaCO3 x 100.09 g CaCO3 1 tablet
3.126. First, calculate the molar mass of the compound. Note that 6.61 x 105 mg = 661 g. Molar mass =
661 g = 299.09 g/mol 2.21 mol
Lead has two main ionic forms, Pb2+ and Pb4+. Either way, there is only one lead atom per formula unit. Thus the mass of nitrite per formula unit is 299.09 g/mol − 207.2 g Pb/mol = 91.89 g nitrite per formula unit.
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113
The percent composition of nitrite is %NO2– =
■
91.89 g NO 2  /mol x 100% = 30.73 = 31% 299 g/mol
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
3.127. Let y equal the mass of CuO in the mixture. Then 0.500 g − y equals the mass of Cu2O in the mixture. Multiplying the appropriate conversion factors for Cu times the mass of each oxide will give one equation in one unknown for the mass of 0.425 g Cu: ⎡ 127.10 g Cu ⎤ ⎡ 63.55 g Cu ⎤ 0.425 = y ⎢ ⎥ ⎥ + (0.500 − y) ⎢ ⎣ 79.55 g CuO ⎦ ⎣143.10 g Cu 2 O ⎦
Simplifying the equation by dividing the conversion factors and combining terms gives 0.425 = 0.79887 y + 0.888190 (0.500 − y) 0.08932 y = 0.019095 y = 0.2139 = 0.21 g = mass of CuO
3.128. Let y equal the mass of Fe2O3 in the mixture. Then 0.500 g − y equals the mass of FeO in the mixture. The mass of Fe in the mixture is 0.720 x 0.500 g = 0.360 g. Multiplying the appropriate conversion factors for Fe times the mass of each oxide will give one equation in one unknown for the mass of 0.360 g Fe in the mixture: ⎡ 111.70 g Fe ⎤ ⎡ 55.85 g Fe ⎤ 0.360 = y ⎢ ⎥ + (0.500 − y) ⎢ ⎥ ⎣ 71.85 g FeO ⎦ ⎣159.70 g Fe2 O3 ⎦ Simplifying the equation by dividing the conversion factors and combining terms gives 0.360 = 0.6994 y + 0.7773 (0.500 − y)
0.07790 y = 0.02865 y = 0.3677 = 0.368 g = mass of Fe2O3
3.129. If one heme molecule contains one iron atom, then the number of moles of heme in 35.2 mg of heme must be the same as the number of moles of iron in 3.19 mg of iron. Start by calculating the moles of Fe (equals moles of heme): 3.19 x 10−3 g Fe x
1 mol Fe = 5.712 x 10−5 mol Fe or heme 55.85 g Fe
Molar mass of heme =
35.2 x 103 g = 616.2 = 616 g/mol 5.712 x 105 mol
The molecular mass of heme is 616 amu.
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Chapter 3: Calculations with Chemical Formulas and Equations
3.130. Convert the mass of BaSO4 to mass of S to find the percentage of sulfur: 0.00546 g BaSO4 x
1 mol BaSO 4 1 mol S 32.07 g S x x 1 mol S 1 mol BaSO 4 233.40 g BaSO 4
= 0.7502 x 10−3 g S 0.7502 x 103 g S x 100% = 9.160 = 9.16% Mass percentage S = 8.19 x 103 g pen. V Convert the mass of S to moles; then, recognizing that moles of S equals moles of pen. V, use that number of moles to calculate the molar mass: 0.7502 x 10−3 g S x Molar mass pen. V =
1 mol S = 2.339 x 10−5 mol S 32.07 g S 8.19 x 103 g = 350.1 = 350. g/mol 2.339 x 105 mol
The molecular mass of penicillin V is 350 amu. 3.131. Use the data to find the molar mass of the metal and anion. Start with X2. Mass X2 in MX = 4.52 g − 3.41 g = 1.11 g Molar mass X2 = 1.11 g ÷ 0.0158 mol = 70.25 g/mol Molar mass X = 70.25 ÷ 2 = 35.14 = 35.1 g/mol Thus X is Cl, chlorine. Moles of M in 4.52 g MX = 0.0158 x 2 = 0.0316 mol Molar mass of M = 3.41 g ÷ 0.0316 mol = 107.9 = 108 g/mol Thus M is Ag, silver. 3.132. Use the data to find the molar mass of the metal and the anion. Start with M2+. Mol M2+ in MX2 = 0.158 mol ÷ 2 = 0.0790 mol Molar mass of M2+ = 1.92 g ÷ 0.0790 mol = 24.30 = 24.3 g/mol Thus M2+ is Mg2+. Now, find the mass of MX2, and then find the molar mass of X: (1 − 0.868) mass MX2 = 1.92 g, or mass MX2 = 14.545 g Mass of X in MX2 = 14.545 g − 1.92 g = 12.625 g Moles of X in MX2 = 0.158 mol Molar mass of X = 12.625 g ÷ 0.158 mol = 79.905 = 79.9 g/mol Thus X is Br (molar mass 79.90).
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3.133. After finding the volume of the alloy, convert it to mass Fe using density and percent Fe. Then use Avogadro's number and the atomic mass for the number of atoms. Volume = 10.0 cm x 20.0 cm x 15.0 cm = 3.00 x 103 cm3 Mass Fe = 3.00 x 103 cm3 x
8.17 g alloy 54.7 g Fe x = 1.3407 x 104 g 3 1 cm 100.0 g alloy
No. of Fe atoms = 1.3407 x 10−4 g Fe x
1 mol Fe 6.022 x 1023 Fe atoms x 1 mol Fe 55.85 g Fe
= 1.4456 x 1026 = 1.45 x 1026 Fe atoms 3.134. After finding the volume of the cylinder, convert it to mass Co using density and percent Co. Then use Avogadro's number and the atomic mass for the number of atoms. Volume = 3.1416 x (2.50 cm)2 x 10 cm = 196.35 cm3 Mass Co = 196.35 cm3 x
8.20 g alloy 12.0 g Co x = 1.932 x 102 g 3 1 cm 100.0 g alloy
No. of Co atoms = 1.932 x 102 g Co x
1 mol Co 6.022 x 1023 Co atoms x 1 mol Co 58.93 g Co
= 1.974 x 1024 = 1.97 x 1024 Co atoms
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CHAPTER 4
Chemical Reactions
■
SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 4.1. a.
According to Table 4.1, all compounds that contain sodium, Na+, are soluble. Thus, NaBr is soluble in water.
b.
According to Table 4.1, most compounds that contain hydroxides, OH−, are insoluble in water. However, Ba(OH)2 is listed as one of the exceptions to this rule, so it is soluble in water.
c.
Calcium carbonate is CaCO3. According to Table 4.1, most compounds that contain carbonate, CO32−, are insoluble. CaCO3 is not one of the exceptions, so it is insoluble in water.
a.
The problem states that HNO3 is a strong electrolyte, but Mg(OH)2 is a solid, so retain its formula. On the product side, Mg(NO3)2 is a soluble ionic compound, but water is a nonelectrolyte, so retain its formula. The resulting complete ionic equation is
4.2.
2H+(aq) + 2NO3−(aq) + Mg(OH)2(s) → 2H2O(l) + Mg2+(aq) + 2NO3−(aq) The corresponding net ionic equation is 2H+(aq) + Mg(OH)2(s) → 2H2O(l) + Mg2+(aq) b.
Both reactants are soluble ionic compounds, and on the product side, NaNO3 is also a soluble ionic compound. PbSO4 is a solid, so retain its formula. The resulting complete ionic equation is Pb2+(aq) + 2NO3−(aq) + 2Na+(aq) + SO42−(aq) → PbSO4(s) + 2Na+(aq) + 2NO3−(aq) The corresponding net ionic equation is Pb2+(aq) + SO42−(aq) → PbSO4(s)
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4.3. The formulas of the compounds are NaI and Pb(C2H3O2)2. Exchanging anions, you get sodium acetate, NaC2H3O2, and lead(II) iodide, PbI2. The equation for the exchange reaction is NaI + Pb(C2H3O2)2 → NaC2H3O2 + PbI2 From Table 4.1, you see that NaI is soluble, Pb(C2H3O2)2 is soluble, NaC2H3O2 is soluble, and PbI2 is insoluble. Thus, lead(II) iodide precipitates. The balanced molecular equation with phase labels is Pb(C2H3O2)2(aq) + 2NaI(aq) → PbI2(s) + 2NaC2H3O2(aq) To get the net ionic equation, you write the soluble ionic compounds as ions and cancel the spectator ions, (C2H3O2− and Na+). The final result is Pb2+(aq) + 2I−(aq) → PbI2(s) 4.4. a.
H3PO4 is not listed as a strong acid in Table 4.3, so it is a weak acid.
b.
Hypochlorous acid, HClO, is not one of the strong acids listed in Table 4.3, so we assume that HClO is a weak acid.
c.
As noted in Table 4.3, HClO4 is a strong acid.
d.
As noted in Table 4.3, Sr(OH)2 is a strong base.
4.5. The salt consists of the cation from the base (Li+) and the anion from the acid (CN−); its formula is LiCN. You will need to add H2O as a product to complete and balance the molecular equation: HCN(aq) + LiOH(aq) → LiCN(aq) + H2O(l) Note that LiOH (a strong base) and LiCN (a soluble ionic substance) are strong electrolytes; HCN is a weak electrolyte (it is not one of the strong acids in Table 4.3). After eliminating the spectator ions (Li+ and CN−), the net ionic equation is HCN(aq) + OH−(aq) → H2O(l) + CN−(aq) 4.6. The first step in the neutralization is described by the following molecular equation: H2SO4(aq) + KOH(aq) → KHSO4(aq) + H2O(l) The corresponding net ionic equation is H+(aq) + OH−(aq) → H2O(l) The reaction of the acid salt KHSO4 is given by the following molecular equation: KHSO4(aq) + KOH(aq) → K2SO4(aq) + H2O(l) The corresponding net ionic equation is HSO4−(aq) + OH−(aq) → H2O(l) + SO42−(aq)
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Chapter 4: Chemical Reactions
4.7. First, write the molecular equation for the exchange reaction, noting that the products of the reaction would be soluble Ca(NO3)2 and H2CO3. The carbonic acid decomposes to water and carbon dioxide gas. The molecular equation for the process is CaCO3(s) + 2HNO3(aq) → Ca(NO3)2(aq) + H2O(l) + CO2(g) The corresponding net ionic equation is CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g) 4.8. a.
For potassium dichromate, K2Cr2O7, 2 x (oxidation number of K) + 2 x (oxidation number of Cr) + 7 x (oxidation number of O) = 0 For oxygen, the oxidation number is −2 (rule 3), and for potassium ion, the oxidation number is +1 (rule 2) [2 x (+1)] + 2 x (oxidation number of Cr) + [7 x (−2)] = 0 Therefore, 2 x oxidation number of Cr = − [2 x (+1)] − [7 x (−2)] = +12 or, oxidation number of Cr = +6.
b.
For the permanganate ion, MnO4−, (Oxidation number of Mn) + 4 x (oxidation number of O) = −1 For oxygen, the oxidation number is −2 (rule 3). (oxidation number of Mn) + [4 x (−2)] = −1 Therefore, Oxidation number of Mn = −1 − [4 x (−2)] = +7
4.9. Identify the oxidation states of the elements.
0 0 Ca + Cl2
+2 1 → CaCl2
Break the reaction into two halfreactions, making sure that both mass and charge are balanced. Ca → Ca2+ + 2e− Cl2 + 2e− → 2Cl− Since each halfreaction has two electrons, it is not necessary to multiply the reactions by any factors to cancel them out. Adding the two halfreactions together and canceling out the electrons, you get Ca(s) + Cl2(g) → CaCl2(s)
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4.10. Convert mass of NaCl (molar mass, 58.44 g) to moles of NaCl. Then divide moles of solute by liters of solution. Note that 25.0 mL = 0.0250 L. 0.0678 g NaCl x
Molarity =
1 mol NaCl = 1.160 x 10−3 mol NaCl 58.44 g NaCl
1.160 x 103 mol NaCl = 0.04641 = 0.0464 M 0.0250 L soln
4.11. Convert grams of NaCl (molar mass, 58.44 g) to moles NaCl and then to volume of NaCl solution. 0.0958 g NaCl x
1 mol NaCl 1 L solution 1000 mL x x 0.163 mol NaCl 1L 58.44 g NaCl
= 10.06 = 10.1 mL NaCl 4.12. One (1) liter of solution is equivalent to 0.15 mol NaCl. The amount of NaCl in 50.0 mL of solution is 50.0 mL x
1L 0.15 mol NaCl x = 0.00750 mol NaCl 1000 mL 1 L soln
Convert to grams using the molar mass of NaCl (58.44 g/mol). 0.00750 mol NaCl x
58.4 g NaCl = 0.438 = 0.44 g NaCl 1 mol NaCl
4.13. Use the rearranged version of the dilution formula from the text to calculate the initial volume of 1.5 M sulfuric acid required: Vi =
M f Vf Mi
=
0.18 M x 100.0 mL = 12.0 = 12 mL 1.5 M
4.14. There are two different reactions taking place in forming the CaC2O4 (molar mass 128.10 g/mol) precipitate. These are CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) CaCl2(aq) + Na2C2O4(aq) → CaC2O4(s) + 2NaCl(aq) The overall stoichiometry of the reactions is one mol CaCO3/one mol CaC2O4. Also note that each CaCO3 contains one Ca atom, so this gives an overall conversion factor of one mol Ca/one mol CaC2O4. The mass of Ca can now be calculated. 0.1402 g CaC2O4 x
1 mol CaC 2 O 4 1 mol Ca 40.08 g Ca x x 1 mol CaC2 O 4 1 mol Ca 128.10 g CaC 2 O 4
= 0.043866 g Ca
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Chapter 4: Chemical Reactions
Now, calculate the percentage of calcium in the 128.3 mg (0.1283 g) limestone: 0.043866 g Ca x 100% = 34.190 = 34.19% 0.1283 g limestone 4.15. Convert the volume of Na3PO4 to moles using the molarity of Na3PO4. Note that 45.7 mL = 0.0457 L. 0.265 mol Na 3 PO 4 = 0.01211 mol Na3PO4 1L
0.0457 L Na3PO4 x
Finally, calculate the amount of NiSO4 required to react with this amount of Na3PO4: 0.1211 mol Na3PO4 x
3 mol NiSO 4 1 L NiSO 4 x = 0.04844 L (48.4 mL) 2 mol Na 3 PO 4 0.375 mol NiSO 4
4.16. Convert the volume of NaOH solution (0.0391 L) to moles NaOH (from the molarity of NaOH). Then convert moles NaOH to moles HC2H3O2 (from the chemical equation). Finally, convert moles of HC2H3O2 (molar mass 60.05 g/mol) to grams HC2H3O2. 0.0391 L NaOH x
1 mol HC2 H 3O 2 60.05 g HC2 H 3O 2 0.108 mol NaOH x x 1 mol NaOH 1L 1 mol HC 2 H 3O 2 = 0.25359 g
The mass percentage of acetic acid in the vinegar can now be calculated. Percentage mass =
■
0.25359 g HC2 H3O 2 x 100% = 5.071 = 5.07% 5.00 g vinegar
ANSWERS TO CONCEPT CHECKS
4.1. The left beaker contains two types of individual atoms (ions) and no solid; therefore, it must represent the soluble, LiI. Because LiI is a soluble ionic compound, it is an electrolyte. The beaker on the right represents a molecular compound that is soluble but not dissociated in solution. Therefore, it must be the CH3OH. Because the CH3OH is not dissociated in solution, and no ions are present, it is a nonelectrolyte. 4.2. a.
In order to solve this part of the problem, keep in mind that this is an exchange (metathesis) reaction. Since you are given the products in the picture, you need to work backward to determine the reactants. Starting with the solid SrSO4(s), you know that the SO42− anion started the reaction with a different cation (not Sr2+). Since Na+ is the only option, you can conclude that one of the reactants must be Na2SO4. Based on solubility rules, you know that Na2SO4 is soluble, so you represent it as Na2SO4(aq). The remaining cation and anion indicate that the other reactant is the soluble Sr(C2H3O2)2. Observing the soluble and insoluble species in the picture, you can conclude that the molecular equation is Na2SO4(aq) + Sr(C2H3O2)2(aq) → SrSO4(s) + 2NaC2H3O2(aq)
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b.
121
Writing the strong electrolytes in the form of ions and the solid with its molecular formula, the complete ionic equation for the reaction is 2Na+(aq) + SO42−(aq) + Sr2+(aq) + 2C2H3O2−(aq) → SrSO4(s) + 2Na+(aq) + 2C2H3O2−(aq)
c.
After canceling the spectator ions, the net ionic equation for the reaction is Sr2+(aq) + SO42−(aq) → SrSO4(s)
4.3. a.
MOH must be a base since OH− is being produced in solution. It must be a strong base because the reaction indicates that MOH is completely soluble (strong electrolyte). In order to maintain charge balance in the formula, the element M must be a 1+ cation, probably a metal from Group IA of the periodic table. Examples of bases that fall into this category include NaOH and KOH.
b.
This must be an acid since H+ is being produced in solution. It is a weak acid because the double arrow is used, indicating only a partial ionization in solution. From the chemical reaction, A− represents an anion with a 1− charge. Acetic acid, HC2H3O2, is a weak acid of this type.
c.
This must be an acid since H+ is being produced in solution. H2A(aq) is a weak acid because the equation indicates only partial ionization in solution. A2− represents an anion with a 2− charge. Carbonic acid, H2CO3, is a weak acid of this type.
d.
Examples of M include Na+, K+, and Li+. Examples of A for reaction b include F−, C2H3O2−, and CN−. Examples of A for reaction c. include S2−, CO32−, and C4H4O62−.
a.
In order to answer this question, you need to compare the number of atoms of X per unit of volume. In order to compare volumes, use the lines on the sides of the beakers. Beaker A has concentration of five atoms per two volume units, 5/2 or 2.5/1. Beaker B has a concentration of ten atoms per one volume unit, 10/1. Beaker C has a concentration of ten atoms per two volume units, 10/2 or 5/1. Beaker D has a concentration of five atoms per volume unit, 5/1. Comparing the concentrations reveals that the ranking from lowest to highest concentration is Beaker A < Beaker C = Beaker D < Beaker B.
b.
To make the concentrations of X equal in each beaker, they all have to be made to match the beaker with the lowest concentration. This is Beaker A, which has five atoms of X in onehalf a beaker of solution. To make the concentrations equal, do the following: double the volume of Beakers C and D, and quadruple the volume of Beaker B by adding the solvent. Overall, Beakers A and B will contain a full beaker of solution, and Beakers C and D will contain a halfbeaker of solution.
4.4.
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Chapter 4: Chemical Reactions
4.5.
■
a.
Since flask C required three times the amount of titrant (NaOH) as of acid A, you have learned that acid C has three times as many acidic protons as acid A. Since flask B required two times the amount of titrant as of acid A, you have also learned that acid B has two times as many acidic protons as acid A.
b.
If you assume that acid A contains a monoprotic acid, then you know the number of moles of A in the flask. After performing the titration, you know that the moles of NaOH must equal the moles of acid in flask A. You take the number of moles of NaOH and divide it by the volume of NaOH added during the titration to determine the concentration of the NaOH solution.
ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
4.1. Some electrolyte solutions are strongly conducting because they are almost completely ionized, and others are weakly conducting because they are weakly ionized. The former solutions will have many more ions to conduct electricity than will the latter solutions if both are present at the same concentrations. 4.2. A strong electrolyte is an electrolyte that exists in solution almost entirely as ions. An example is NaCl. When NaCl dissolves in water, it dissolves almost completely to give Na+ and Cl− ions. A weak electrolyte is an electrolyte that dissolves in water to give a relatively small percentage of ions. An example is NH3. When NH3 dissolves in water, it reacts very little with the water, so the level of NH3 is relatively high, and the level of the NH4+ and OH− ions is relatively low. 4.3. Soluble means the ability of a substance to dissolve in water. A compound is insoluble if it does not dissolve appreciably in water. An example of a soluble ionic compound is sodium chloride, NaCl, and an example of an insoluble ionic compound is calcium carbonate, CaCO3. 4.4. The advantage of using a molecular equation to represent an ionic equation is that it states explicitly what chemical species have been added and what chemical species are obtained as products. It also makes stoichiometric calculations easy to perform. The disadvantages are (1) the molecular equation does not represent the fact that the reaction actually involves ions, and (2) the molecular equation does not indicate which species exist as ions and which exist as molecular solids or molecular gases. 4.5. A spectator ion is an ion that does not take part in the reaction. In the following ionic reaction, the Na+ and Cl− are spectator ions: Na+(aq) + OH−(aq) + H+(aq) + Cl−(aq) → Na+(aq) + Cl−(aq) + H2O(l)
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123
4.6. A net ionic equation is an ionic equation from which spectator ions have been canceled. The value of such an equation is that it shows the reaction that actually occurs at the ionic level. An example is the ionic equation representing the reaction of calcium chloride (CaCl2) with potassium carbonate(K2CO3). CaCl2(aq) + K2CO3(aq) → CaCO3(s) + 2 KCl(aq): Ca2+(aq) + CO32−(aq) → CaCO3(s)
(net ionic equation)
4.7. The three major types of chemical reactions are precipitation reactions, acid–base reactions, and oxidation–reduction reactions. Oxidation–reduction reactions can be further classified as combination reactions, decomposition reactions, displacement reactions, and combustion reactions. Brief descriptions and examples of each are given below. A precipitation reaction is a reaction that involves the formation of an insoluble solid compound. An example is 2KCl (aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbI2(s). An acid–base reaction, or neutralization reaction, results in an ionic compound and possibly water. An example is HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l). A combination reaction is a reaction in which two substances combine to form a third substance. An example is 2Na(s) + Cl2(g) → 2NaCl(s). A decomposition reaction is a reaction in which a single compound reacts to give two or more Δ substances. An example is 2HgO(s) ⎯⎯ → 2Hg(l) + O2(g). A displacement reaction, or single replacement reaction, is a reaction in which an element reacts with a compound displacing an element from it. An example is Cu(s) + 2AgNO3(aq) → 2Ag(s) + Cu(NO3)2(aq). A combustion reaction is a reaction of a substance with oxygen, usually with rapid release of heat to produce a flame. The products include one or more oxides. An example is CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). 4.8. To prepare crystalline AgCl and NaNO3, first make solutions of AgNO3 and NaCl by weighing equivalent molar amounts of both solid compounds. Then mix the two solutions together, forming a precipitate of silver chloride and a solution of soluble sodium nitrate. Filter off the silver chloride, and wash it with water to remove the sodium nitrate solution. Then allow it to dry to obtain pure crystalline silver chloride. Finally, take the filtrate containing the sodium nitrate and evaporate it, leaving pure crystalline sodium nitrate. 4.9. An example of a neutralization reaction is HBr
acid
+
KOH
→
base
KBr
salt
+
H2O(l)
4.10. An example of a polyprotic acid is carbonic acid, H2CO3. The successive neutralization is given by the following molecular equations: H2CO3(aq) + NaOH(aq) → NaHCO3(aq) + H2O(l) NaHCO3(aq) + NaOH(aq) → Na2CO3(aq) + H2O(l)
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Chapter 4: Chemical Reactions
4.11. Since an oxidation–reduction reaction is an electron transfer reaction, one substance must lose the electrons and be oxidized while another substance must gain electrons and be reduced. 4.12. A displacement reaction is an oxidation–reduction reaction in which a free element reacts with a compound, displacing an element from it. Cu(s) + 2AgNO3(aq) → 2Ag(s) + Cu(NO3)2(aq) Ag+ is the oxidizing agent, and Cu is the reducing agent. 4.13. The number of moles present does not change when the solution is diluted. 4.14. The reaction is HCl + NaOH → NaCl + H2O After titration, the volume of hydrochloric acid is converted to moles of HCl using the molarity. Since the stoichiometry of the reaction is 1 mol HCl to 1 mol NaOH, these quantities are equal. Moles HCl = moles NaOH = molarity x volume You could then multiply by the molar mass of NaOH to obtain the amount in the mixture. 4.15. The answer is e, HF(aq) + OH−(aq) → F−(aq) + H2O(l). 4.16. The answer is c, magnesium hydroxide(s). 4.17. The answer is d, MgCl2. 4.18. The answer is d, HNO2.
■
ANSWERS TO CONCEPT EXPLORATIONS
4.19. Part 1 a.
NH3(aq) + H2O(l) → NH4+(aq) + OH−(aq)
b.
H2O → C12H22O11(aq) C12H22O11(s) ⎯⎯⎯
c.
NH3 forms ions by reacting with water. But it reacts very little with water, and only a small percentage of ions are formed. Thus, it is a weak electrolyte. Sucrose dissolves in water but does not form any ions. Thus, it is a nonelectrolyte.
d.
HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq)
e.
NH3 and HCl both react with water to form ions. Therefore, both compounds are electrolytes.
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f.
HCl reacts completely with water to form ions, so it is a strong electrolyte. NH3 reacts very little with water, so it is a weak electrolyte.
g.
The ionic compounds are KCl, MgBr2, Ca(OH)2, and PbS. The molecular compounds are NH3, CO2, HCl, and HC2H3O2.
h.
KCl and MgBr2 are both soluble by Rule 3, that most chlorides, bromides, and iodides are soluble. Neither compound is listed as an exception. Ca(OH)2 is soluble by Rule 8, that most hydroxides are insoluble, but Ca(OH)2 is listed as an exception. PbS is insoluble by Rule 7. It is not listed as an exception.
i.
If you placed a soluble ionic compound in water it would mix with the water and form ions. After a while, the solid would no longer be evident and only ions would be in the solution. If you placed an insoluble ionic compound in water, it would mix with the water and dissolve only partially to form ions. After a while, the undissolved solid would settle to the bottom of the beaker. No, these substances do not react with water when they are added to water.
j.
H2O KCl(s) ⎯⎯⎯ → K+(aq) + Cl−(aq) H2O → Mg2+(aq) + 2Br−(aq) MgBr2(s) ⎯⎯⎯ H2O → Ca2+(aq) + 2OH−(aq) Ca(OH)2(s) ⎯⎯⎯
k.
All three soluble ionic compounds are strong electrolytes because they dissolve in water completely to form ions.
l.
NaCl(aq) → Na+(aq) + Cl−(aq) HI(aq) + H2O(l) → H3O+(aq) + I−(aq)
m.
NaCl and HI have different behavior in aqueous solution. NaCl dissolves in water completely to form ions. HI reacts with water completely to form ions. The reactions are NaCl(aq) → Na+(aq) + Cl−(aq) HI(aq) + H2O(l) → H3O+(aq) + I−(aq)
Part 2
a.
Since AX is a strong electrolyte, it would ionize completely. The solution would be made up primarily of ions, with no AX molecules present. Since AY is a weak electrolyte, it would ionize only partially. The solution would consist primarily of AY molecules with relatively few ions present. Therefore, AY(aq) is much larger than AX(aq).
b.
Since AX is a strong electrolyte, it would ionize completely. There would be a large amount of X−(aq) present. Since AY is a weak electrolyte, it would ionize only partially. There would be a very small amount of Y−(aq) present. Therefore, X−(aq) is much larger than Y−(aq).
4.20. Part 1 a.
NaCl(aq) → Na+(aq) + Cl−(aq) MgCl2(aq) → Mg2+(aq) + 2Cl−(aq)
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Chapter 4: Chemical Reactions
b.
The picture on the left has eight larger spheres and four smaller spheres. The picture on the right has four larger spheres and four smaller spheres. Since Cl− ions are the largest of the ions present, they are represented by the larger spheres. Therefore, there is a 2to1 ratio of Cl− ions to cations in the picture on the left and a 1–to–1 ratio of Cl− ions to cations in the picture on the right. This means that MgCl2 is shown in the picture on the left, and NaCl is shone in the picture on the right.
c.
1.0 L of 1.0 M of each compound contains 1.0 mol.
d.
For NaCl there is a 1–to–1 mole ratio of chloride ions to NaCl. Therefore 1.0 mol of NaCl, contains 1.0 mol of Cl− ions. For MgCl2, there is a 2–to–1 mole ratio of chloride ions to MgCl2. Therefore 1.0 mol of MgCl2 contains 2.0 mol of Cl− ions.
e.
Since the volumes of the solutions are 1.0 L, the molarity and the moles are numerically the same. Thus, for the NaCl solution the concentration of Cl− is 1.0 M, and in the MgCl2 beaker the concentration of Cl− is 2.0 M.
f.
Since the mole ratio of chloride ion to original compound is different for each compound, different concentrations of chloride ions result from equal amounts of the original compounds.
Part 2
a.
Since the solution is homogeneous, the concentrations are the same throughout the solution. Both the part remaining and the part removed still have the same concentration as before, so MgCl2(aq) is 1.0 M and Cl−(aq) is 2.0 M.
b.
Since half of the solution was removed, half of the moles were also removed. Thus, there is 0.50 mol MgCl2 and 1.0 mol Cl−.
c.
The solution is homogeneous and the concentration is uniform throughout the solution. Onehalf the volume contains onehalf the amount of substance. Thus, onehalf the amount of substance in onehalf the volume is equivalent to the same concentration as before.
Part 3
a.
Since M1V1 = M2V2, doubling the volume reduces the concentration by an equal factor. Thus the concentration of NaCl(aq) would be 0.50 M.
b.
Since no solution was removed, there is still the same number of moles of NaCl as before, 1.0 mol.
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c.
■
127
The number of moles does not change because nothing was removed from the solution. The concentration does change because the volume of the solution changed upon addition of more water. Having the same number of moles in a larger volume is equivalent to a smaller concentration.
ANSWERS TO CONCEPTUAL PROBLEMS
4.21. a.
Any soluble salt that will form a precipitate when reacted with Ag+ ions in solution will work, for example: CaCl2, Na2S, (NH4)2CO3.
b.
No, no precipitate will form.
c.
You would underestimate the amount of silver present in the solution.
4.22. a.
NH4 +(aq)
NH4 +(aq)
Pb2+(aq)
NO3 (aq)
Cl(aq) NO3 (aq) t=0
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PbCl2(s) t>0
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Chapter 4: Chemical Reactions
b.
Na+(aq)
Na+(aq)
NO3 (aq)
NO3 (aq)
FeS(s)
FeS(s)
t=0
t>0
c.
Li+(aq)
Li+(aq)
Na+(aq)
Na+(aq)
I(aq)
I(aq)
CO3 2(aq)
CO3 2(aq)
t=0
t>0
4.23. a.
3Ca(C2H3O2)2(aq) + 2(NH4)3PO4(aq) → Ca3(PO4)2(s) + 6NH4C2H3O2(aq)
b.
3Ca2+(aq) + 6C2H3O2−(aq) + 6NH4+(aq) + 2PO43−(aq) → Ca3(PO4)2(s) + 6NH4+(aq) + 6C2H3O2−(aq)
c.
3Ca2+(aq) + 2PO43−(aq) → Ca3(PO4)2(s)
a.
Sample 1, Sample 2, Sample 3
b.
They all will require the same volume of 0.1 M NaOH.
4.24.
4.25. Probably not, since the ionic compound that is a nonelectrolyte is not soluble.
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4.26. A good starting point is to identify the solution that contains the base. Since bases produce OH− in aqueous solution, we would expect to see OH− present in the BZ solution. The center beaker depicts OH− in the solution, so it must be the base. By default, the remaining two beakers must contain acid. This is confirmed by the presence of H3O+ in both the left and right beakers. Keeping in mind that weak acids only partially dissociate, for the weak acid HA, we would expect to observe HA, H3O+, and A− in the solution. In the case of the strong acid, HX, that completely dissociates, we would expect to observe only H3O+ and X− in the solution. The beaker on the right has only H3O+ and one other species in solution, so it must be the strong acid HX. Examining the beaker on the left reveals that there are three species present, which indicates that it must be the weak acid. 4.27. a.
Since both solutions are made with compounds that contain chloride ions, the total chloride ion concentration is highest.
b.
First, determine the concentrations of the compounds after mixing them together. Use the dilution relationship, M1V1 = M2V2. Since equal volumes of equal molar solutions are mixed, the resulting concentrations are 0.50 M KBr and 0.50 M K3PO4.There is one Br− ion per KBr, so the concentration of Br− is 0.50 M. There is also one PO43− ion per K3PO4, so its concentration is also 0.50 M. Potassium ion can be determined as follows: 0.50 M KBr x
1 mol K + 3 mol K + + 0.50 M K3PO4 x = 2.0 M K+ 1 mol KBr 1 mol K 3 PO 4
4.28. Since equal moles of the compounds are used, the highest Cl− concentration is for the compound(s) with the largest subscript on Cl in the formula. Thus the order, from highest to lowest, is AlCl3 > KCl = NaCl = HCl > PbCl2 = NH3 = KOH = HCN.
■
SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 4.29. a.
Insoluble
b.
Soluble
c.
Soluble
d.
Soluble
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Chapter 4: Chemical Reactions
4.30. a.
Insoluble
b.
Soluble
c.
Soluble
d.
Soluble
a.
Insoluble
b.
Soluble; The ions present would be NH4+ and SO42−.
c.
Insoluble
d.
Soluble; The ions present would be Na+ and CO32−.
a.
Soluble; The ions present would be Li+ and SO42−.
b.
Insoluble
c.
Insoluble
d.
Soluble; The ions present would be Ca2+ and NO3−.
a.
H+(aq) + OH−(aq) → H2O(l)
b.
Ag+(aq) + Br −(aq) → AgBr(s)
c.
S2−(aq) + 2H+(aq) → H2S(g)
d.
OH−(aq) + NH4+(aq) → NH3(g) + H2O(l)
a.
H+(aq) + NH3(aq) → NH4+(aq)
b.
H+(aq) + OH−(aq) → H2O(l)
c.
Pb2+(aq) + 2Br−(aq) → PbBr2(s)
d.
MgCO3(s) + 2H+(aq) → Mg2+(aq) + H2O(l) + CO2(g)
4.31.
4.32.
4.33.
4.34.
4.35. Molecular equation: Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq) Net ionic equation: Pb2+(aq) + SO42−(aq) → PbSO4(s) 4.36. Molecular equation: K2CO3(aq) + 2HBr(aq) → CO2(g) + H2O(l) + 2KBr(aq) Net ionic equation: CO32−(aq) + 2H+(aq) → CO2(g) + H2O(l)
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4.37. a.
FeSO4(aq) + NaCl(aq) → NR
b.
Na2CO3(aq) + MgBr2(aq) → MgCO3(s) + 2NaBr(aq) CO32−(aq) + Mg2+(aq) → MgCO3(s)
c.
MgSO4(aq) + 2NaOH(aq) → Mg(OH)2(s) + Na2SO4(aq) Mg2+(aq) + 2OH−(aq) → Mg(OH)2(s)
d.
NiCl2(aq) + NaBr(aq) → NR
a.
AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)
4.38. Ag+(aq) + I−(aq) → AgI(s) b.
Ba(NO3)2(aq) + K2SO4(aq) → BaSO4(s) + 2KNO3(aq) Ba2+(aq) + SO42−(aq) → BaSO4(s)
c.
Mg(NO3)2(aq) + K2SO4(aq) → NR
d.
CaCl2(aq) + Al(NO3)3(aq) → NR
a.
Ba(NO3)2(aq) + Li2SO4(aq) → BaSO4(s) + 2LiNO3(aq)
4.39. Ba2+(aq) + SO42−(aq) → BaSO4(s) b.
Ca(NO3)2(aq) + NaBr(aq) → NR
c.
Al2(SO4)3(aq) + 6NaOH(aq) → 2Al(OH)3(s) + 3Na2SO4(aq) Al3+(aq) + 3OH−(aq) → Al(OH)3(s)
d.
3CaBr2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaBr(aq) 3Ca2+(aq) + 2PO43−(aq) → Ca3(PO4)2(s)
4.40. a.
ZnCl2(aq) + Na2S(aq) → ZnS(s) + 2NaCl(aq) Zn2+(aq) + S2−(aq) → ZnS(s)
b.
CaCl2(aq) + Na2S(aq) → CaS(s) + 2NaCl(aq) Ca2+(aq) + S2−(aq) → CaS(s)
c.
MgSO4(aq) + KBr(aq) → NR
d.
MgSO4(aq) + K2CO3(aq) → MgCO3(s) + K2SO4(aq) Mg2+(aq) + CO32−(aq) → MgCO3(s)
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Chapter 4: Chemical Reactions
4.41. a.
Weak acid
b.
Strong base
c.
Strong acid
d.
Weak acid
a.
Weak base
b.
Weak acid
c.
Strong base
d.
Strong acid
a.
NaOH(aq) + HNO3(aq) → H2O(l) + NaNO3(aq)
4.42.
4.43. H+(aq) + OH−(aq) → H2O(l) b.
2HCl(aq) + Ba(OH)2(aq) → 2H2O(l) + BaCl2(aq) H+(aq) + OH−(aq) → H2O(l)
c.
2HC2H3O2(aq) + Ca(OH)2(aq) → 2H2O(l) + Ca(C2H3O2)2(aq) HC2H3O2(aq) + OH−(aq) → H2O(l) + C2H3O2−(aq)
d.
NH3(aq) + HNO3(aq) → NH4NO3(aq) NH3(aq) + H+(aq) → NH4+(aq)
4.44. a.
Al(OH)3(s) + 3HCl(aq) → AlCl3(aq) + 3H2O(l) Al(OH)3(s) + 3H+(aq) → Al3+(aq) + 3H2O(l)
b.
2HBr(aq) + Sr(OH)2(aq) → 2H2O(l) + SrBr2(aq) H+(aq) + OH−(aq) → H2O(l)
c.
Ba(OH)2(aq) + 2HC2H3O2(aq) → Ba(C2H3O2)2(aq) + 2H2O(l) OH−(aq) + HC2H3O2(aq) → C2H3O2−(aq) + H2O(l)
d.
HNO3(aq) + KOH(aq) → H2O(l) + KNO3(aq) H+(aq) + OH−(aq) → H2O(l)
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4.45. a.
2HBr(aq) + Ca(OH)2(aq) → 2H2O(l) + CaBr2(aq) H+(aq) + OH−(aq) → H2O(l)
b.
3HNO3(aq) + Al(OH)3(s) → 3H2O(l) + Al(NO3)3(aq) 3H+(aq) + Al(OH)3(s) → 3H2O(l) + Al3+(aq)
c.
2HCN(aq) + Ca(OH)2(aq) → 2H2O(l) + Ca(CN)2(aq) HCN(aq) + OH−(aq) → H2O(l) + CN−(aq)
d.
HCN(aq) + LiOH(aq) → H2O(l) + LiCN(aq) HCN(aq) + OH−(aq) → H2O(l) + CN−(aq)
4.46. a.
HClO4(aq) + LiOH(aq) → H2O(l) + LiClO4(aq) H+(aq) + OH−(aq) → H2O(l)
b.
2HNO2(aq) + Ba(OH)2(aq) → 2H2O(l) + Ba(NO2)2(aq) HNO2(aq) + OH−(aq) → H2O(l) + NO2−(aq)
c.
HNO2(aq) + NaOH(aq) → H2O(l) + NaNO2(aq) HNO2(aq) + OH−(aq) → H2O(l) + NO2−(aq)
d.
2HCN(aq) + Sr(OH)2(aq) → 2H2O(l) + Sr(CN)2(aq) HCN(aq) + OH−(aq) → H2O(l) + CN−(aq)
4.47. a.
2KOH(aq) + H3PO4(aq) → K2HPO4(aq) + 2H2O(l) 2OH−(aq) + H3PO4(aq) → HPO42−(aq) + 2H2O(l)
b.
3H2SO4(aq) + 2Al(OH)3(s) → 6H2O(l) + Al2(SO4)3(aq) 3H+(aq) + Al(OH)3(s) → 3H2O(l) + Al3+(aq)
c.
2HC2H3O2(aq) + Ca(OH)2(aq) → 2H2O(l) + Ca(C2H3O2)2(aq) HC2H3O2(aq) + OH−(aq) → H2O(l) + C2H3O2−(aq)
d.
H2SO3(aq) + NaOH(aq) → H2O(l) + NaHSO3(aq) H2SO3(aq) + OH−(aq) → HSO3−(aq) + H2O(l)
4.48. a.
Ca(OH)2(aq) + 2H2SO4(aq) → 2H2O(l) + Ca(HSO4)2(aq) H+(aq) + OH−(aq) → H2O(l)
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Chapter 4: Chemical Reactions
b.
2H3PO4(aq) + Ca(OH)2(aq) → 2H2O(l) + Ca(H2PO4)2(aq) H3PO4(aq) + OH−(aq) → H2O(l) + H2PO4−(aq)
c.
NaOH(aq) + H2SO4(aq) → NaHSO4(aq) + H2O(l) OH−(aq) + H+(aq) → H2O(l)
d.
Sr(OH)2(aq) + 2H2CO3(aq) → Sr(HCO3)2(aq) + 2H2O(l) OH−(aq) + H2CO3(aq) → HCO3−(aq) + H2O(l)
4.49. Molecular equations: Ionic equations:
2H2SO3(aq) + Ca(OH)2(aq) → 2H2O(l) + Ca(HSO3)2(aq)
Ca(HSO3)2(aq) + Ca(OH)2(aq) → 2H2O(l) + 2CaSO3(s) H2SO3(aq) + OH−(aq) → H2O(l) + HSO3−(aq) Ca2+(aq) + HSO3−(aq) + OH−(aq) → CaSO3(s) + H2O(l)
4.50. Molecular equations:
2H3PO4(aq) + Ca(OH)2(aq) → 2H2O(l) + Ca(H2PO4)2(aq) Ca(H2PO4)2(aq) + Ca(OH)2(aq) → 2H2O(l) + 2CaHPO4(aq) 2CaHPO4(aq) + Ca(OH)2(aq) → 2H2O(l) + Ca3(PO4)2(s)
Ionic equations:
H3PO4(aq) + OH−(aq) → H2O(l) + H2PO4−(aq) H2PO4−(aq) + OH−(aq) → H2O(l) + HPO42−(aq) 2HPO42−(aq) + 2OH−(aq) + 3Ca2+(aq) → 2H2O(l) + Ca3(PO4)2(s)
4.51. a.
Molecular equation: CaS(s) + 2HBr(aq) → CaBr2(aq) + H2S(g) Ionic equation: CaS(s) + 2H+(aq) → Ca2+(aq) + H2S(g)
b.
Molecular equation: MgCO3(s) + 2HNO3(aq) → CO2(g) + H2O(l) + Mg(NO3)2(aq) Ionic equation: MgCO3(s) + 2H+(aq) → CO2(g) + H2O(l) + Mg2+(aq)
c.
Molecular equation: K2SO3(aq) + H2SO4(aq) → K2SO4(aq) + SO2(g) + H2O(l) Ionic equation: SO32−(aq) + 2H+(aq) → SO2(g) + H2O(l)
4.52. a.
Molecular equation: BaCO3(s) + 2HNO3(aq) → CO2(g) + H2O(l) + Ba(NO3)2(aq) Ionic equation: BaCO3(s) + 2H+(aq) → CO2(g) + H2O(l) + Ba2+(aq)
b.
Molecular equation: K2S(aq) + 2HCl(aq) → H2S(g) + 2KCl(aq) Ionic equation: S2−(aq) + 2H+(aq) → H2S(g)
c.
Molecular equation: CaSO3(s) + 2HI(aq) → SO2(g) + H2O(l) + CaI2(aq) Ionic equation: CaSO3(s) + 2H+(aq) → SO2(g) + H2O(l) + Ca2+(aq)
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4.53. Molecular equation: FeS(s) + 2HCl(aq) → H2S(g) + FeCl2(aq) Ionic equation: FeS(s) + 2H+(aq) → H2S(g) + Fe2+(aq) 4.54. Molecular equation: BaCO3(s) + 2HBr(aq) → BaBr2(aq) + CO2(g) + H2O(l) Ionic equation: BaCO3(s) + 2H+(aq) → Ba2+(aq) + CO2(g) + H2O(l) 4.55. a.
Because all three O's = a total of −6, both Ga's = +6; thus, the oxidation number of Ga = +3.
b.
Because both O's = a total of −4, the oxidation number of Nb = +4.
c.
Because the four O's = a total of −8 and K = +1, the oxidation number of Br = +7.
d.
Because the four O's = a total of −8 and the 2 K's = +2, the oxidation number of Mn = +6.
a.
Because the three O's = a total of −6, the oxidation number of Cr = +6.
b.
Because the two Cl's = a total of −2, both Hg's = +2; thus, the oxidation number of Hg = +1.
c.
Because the three O's = a total of −6 and the 3 H's = a total of +3, the oxidation number of Ga = +3.
d.
Because the four O's = a total of −8 and the 3 Na's = a total of +3, the oxidation number of P = +5.
a.
Because the charge of −1 = [xN + 2 (from 2 H's)], xN must equal −3.
b.
Because the charge of −1 = [xI − 6 (from 3 O's)], xI must equal +5.
c.
Because the charge of −1 = [xAl − 8 (4 O's) + 4 (4 H's)], xAl must equal +3.
d.
Because the charge of 0 = [xCl − 8 (4 O's) + 1 (1 H's)], xCl must equal +7.
a.
Because the charge of −1 = [xN − 4 (from 2 O's)], xN must equal +3.
b.
Because the charge of −2 = [xCr − 8 (from 4 O's)], xCr must equal +6.
c.
Because the charge of −2 = [xZn − 8 (4 O's) + 4 (4 H's)], xZn must equal +2.
d.
Because the charge of −1 = [xAs − 6 (3 O's) + 2 (2 H's)], xAs must equal +3.
a.
From the list of common polyatomic anions in Table 2.5, the formula of the ClO3 anion must be ClO3−. Thus, the formula of Mn is Mn2+ (see also Table 4.5). Since the oxidation state of O is −2 and the net ionic charge is −1, the oxidation state of chlorine is determined by xCl − 6 = −1, so xCl must equal +5.
4.56.
4.57.
4.58.
4.59.
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Chapter 4: Chemical Reactions
b.
From the list of common polyatomic anions in Table 2.5, the formula of the CrO4 anion must be CrO42−. Thus, the formula of Fe is Fe3+. Since the oxidation state of O is −2 and the net ionic charge is −2, the oxidation state of Cr is determined by xCr − 8 = −2, so xCr must equal +6.
c.
From the list of common polyatomic anions in Table 2.5, the formula of the Cr2O7 anion must be Cr2O72−. Thus, the formula of Hg is Hg2+. Since the oxidation state of O is −2 and the net ionic charge is −2, the oxidation state of Cr determined by 2xCr − 14 = −2, so xCr must equal +6.
d.
From the list of common polyatomic anions in Table 2.5, the formula of the PO4 anion must be PO43−. Thus, the formula of Co is Co2+. Since the oxidation state of O is −2 and the net ionic charge is −3, the oxidation state of P is determined by xP − 8 = −3, so xP must equal +5.
a.
From the formula of ClO3− in the list of common polyatomic anions in Table 2.5, the formula of the BrO3 anion must be BrO3−. Thus, the oxidation state of Hg is +1, and its formula must be Hg22+ (from Table 4.5). Since the oxidation state of O is −2 and the net ionic charge is −1, the oxidation state of Br is determined by xBr − 6 = −1, so xBr must equal +5.
b.
From the list of common polyatomic anions in Table 2.5, the formula of the SO4 anion must be SO42−. Thus, the oxidation state of Cr is +3. Since the oxidation state of O is −2 and the net ionic charge is −2, the oxidation state of S is determined by xS − 8 = −2, so xS must equal +6.
c.
From the formula of SO42− in the list of common polyatomic anions in Table 2.5, the formula of the SeO4 anion must be SeO42−. Thus, the oxidation state of Co is +2. Since the oxidation state of O is −2 and the net ionic charge is −2, the oxidation state of Se is determined by xSe − 8 = −2, so xSe must be +6.
d.
The formula of the hydroxide anion is OH−. Thus, the oxidation state of Pb is +2. As usual, the oxidation state of H is +1 and the oxidation state of O is −2.
a.
Phosphorus changes from an oxidation number of zero in P4 to +5 in P4O10, losing electrons and acting as a reducing agent. Oxygen changes from an oxidation number of zero in O2 to −2 in P4O10, gaining electrons and acting as an oxidizing agent.
b.
Cobalt changes from an oxidation number of zero in Co(s) to +2 in CoCl2, losing electrons and acting as a reducing agent. Chlorine changes from an oxidation number of zero in Cl2 to −1 in CoCl2, gaining electrons and acting as an oxidizing agent.
a.
Carbon changes from an oxidation number of zero in C to +2 in CO, losing electrons and acting as a reducing agent. Zinc changes from an oxidation number of +2 in ZnO to zero in Zn, gaining electrons and acting as an oxidizing agent.
b.
Iron changes from an oxidation number of zero in Fe(s) to +2 in FeS, losing electrons and acting as a reducing agent. Sulfur changes from an oxidation number of 0 in S8 to −2 in FeS, gaining electrons and acting as an oxidizing agent.
4.60.
4.61.
4.62.
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137
4.63. a.
Al changes from oxidation number zero to +3; Al is the reducing agent. F changes from oxidation number zero to −1; F2 is the oxidizing agent.
b.
Hg changes from oxidation state +2 to 0; Hg2+ is the oxidizing agent. N changes from oxidation state +3 to +5; NO2− is the reducing agent.
4.64. a.
C changes from oxidation number +2 to +4; the CO is the reducing agent. Fe changes from oxidation number +3 to 0; the Fe2O3 is the oxidizing agent.
b.
S changes from oxidation number −2 to +6; the PbS is the reducing agent. O changes from oxidation number −1 to −2; the H2O2 is the oxidizing agent.
4.65. a.
First, identify the species being oxidized and reduced, and assign the appropriate oxidation states. Since CuCl2 and AlCl3 are both soluble ionic compounds, Cl− is a spectator ion and can be removed from the equation. The resulting net ionic equation is
0 +2 2+ Cu (aq) + Al(s)
→
+3 Al3+(aq)
0 + Cu(s)
Next, write the halfreactions in an unbalanced form. Al → Al3+
(oxidation)
Cu2+ → Cu
(reduction)
Next, balance the charge in each equation by adding electrons to the more positive side to create balanced halfreactions. Al → Al3+ + 3e− Cu
2+
−
+ 2e
→ Cu
(oxidation halfreaction) (reduction halfreaction)
Multiply each halfreaction by a factor that will cancel out the electrons. 2 x (Al → Al3+ + 3e−) 3 x (Cu2+ + 2e− → Cu) ___________________________________ 3Cu2+ + 2Al + 6e− → 2Al3+ + 3Cu + 6e− Therefore, the balanced oxidationreduction reaction is 3Cu2+ + 2Al → 2Al3+ + 3Cu Finally, add six Cl− ions to each side, and add phase labels. The resulting balanced equation is 3CuCl2(aq) + 2Al(s) → 2AlCl3(aq) + 3Cu(s)
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Chapter 4: Chemical Reactions
b.
First, identify the species being oxidized and reduced, and assign the appropriate oxidation states.
+3 Cr3+(aq)
+
0 Zn(s)
→
0 Cr(s)
+2 Zn2+(aq)
+
Next, write the halfreactions in an unbalanced form. Zn → Zn2+
(oxidation)
Cr3+ → Cr
(reduction)
Next, balance the charge in each equation by adding electrons to the more positive side to create balanced halfreactions. Zn → Zn2+ + 2e−
(oxidation halfreaction)
Cr3+ + 3e− → Cr
(reduction halfreaction)
Multiply each halfreaction by a factor that will cancel out the electrons. 3 x (Zn → Zn2+ + 2e−) 2 x (Cr3+ + 3e− → Cr) ___________________________________ 2Cr3+ + 3Zn + 6e− → 2Cr + 3Zn2+ + 6e− Therefore, the balanced oxidationreduction reaction, including phase labels, is 2Cr3+(aq) + 3Zn(s) → 2Cr(s) + 3Zn2+(aq) 4.66. a.
First, identify the species being oxidized and reduced, and assign the appropriate oxidation states. Since FeI3 and MgI2 are both soluble ionic compounds, I− is a spectator ion and can be removed from the equation. The resulting net ionic equation is
+3 Fe3+(aq)
+
0 Mg(s)
→
0 Fe(s)
+
+2 Mg 2+(aq)
Next, write the halfreactions in an unbalanced form. Mg → Mg2+
(oxidation)
Fe3+ → Fe
(reduction)
Next, balance the charge in each equation by adding electrons to the more positive side to create balanced halfreactions. Mg → Mg2+ + 2e− (oxidation halfreaction) Fe3+ + 3e− → Fe
(reduction halfreaction)
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139
Multiply each halfreaction by a factor that will cancel out the electrons. 3 x (Mg → Mg2+ + 2e−) 2 x (Fe3+ + 3e− → Fe) ___________________________________ 2Fe3+ + 3Mg + 6e− → 2Fe + 3Mg2+ + 6e− Therefore, the balanced oxidationreduction reaction is 2Fe3+ + 3Mg → 2Fe + 3Mg2+ Finally, add six I− ions to each side, and add phase labels. The resulting balanced equation is 2FeI3(aq) + 3Mg(s) → 2Fe(s) + 3MgI2(aq) b.
First, identify the species being oxidized and reduced, and assign the appropriate oxidation states.
0 H2(g)
+
+1 Ag+(aq) →
0 Ag(s)
+
+1 H+(aq)
Next, write the halfreactions in an unbalanced form, making sure that mass is balanced. H2 → 2H+
(oxidation)
Ag+ → Ag
(reduction)
Next, balance the charge in each equation by adding electrons to the more positive side to create balanced halfreactions. H2 → 2H+ + 2e− +
−
Ag + e
→ Ag
(oxidation halfreaction) (reduction halfreaction)
Multiply each halfreaction by a factor that will cancel out the electrons. 1 x (H2 → 2H+ + 2e−) 2 x (Ag+ + e− → Ag) __________________________________ H2 + 2Ag+ + 2e− → 2Ag + 2H+ + 2e− Therefore, the balanced oxidationreduction reaction, including phase labels, is H2(g) + 2Ag+(aq) → 2Ag(s) + 2H+(aq)
4.67. Molarity =
moles solute 0.0512 mol = = 2.048 = 2.05 M liters solution 0.0250 L
4.68. Molarity =
moles solute 0.0285 mol = = 0.5700 = 0.570 M liters solution 0.0500 L
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Chapter 4: Chemical Reactions
4.69. Find the number of moles of solute (KMnO4) using the molar mass of 158.03 g KMnO4 per 1 mol KMnO4: 0.798 g KMnO4 x Molarity =
1 mol KMnO 4 = 5.0497 x 10−3 mol KMnO4 158.03 g KMnO 4
moles solute 5.0497 x 103 mol = = 0.10099 = 0.101 M liters of solution 0.0500 L
4.70. Find the number of moles of solute (H2C2O4) using the molar mass of 90.04 g H2C2O4 per 1 mol H2C2O4: 1.192 g H2C2O4 x Molarity =
1 mol H 2 C 2 O 4 = 0.013238 mol H2C2O4 90.04 g H 2 C2 O 4
moles solute 0.013238 mol = = 0.13238 = 0.1324 M liters solution 0.1000 L
4.71. 0.150 mol CuSO4 x
1 L solution = 1.250 = 1.25 L solution 0.120 mol CuSO 4
4.72. 0.102 mol HClO4 x
1 L solution = 0.80952 = 0.810 L = 8.10 x 102 mL 0.126 mol HClO 4
4.73. 0.0353 g KOH x
1 mol KOH 1 L solution x = 0.035751 L = 35.8 mL 0.0176 mol KOH 56.10 g KOH
4.74. 0.949 g H2SO4 x
1 mol H 2SO 4 1 L soln = 0.044999 L (45.0 mL soln) x 0.215 mol H 2SO 4 98.09 g H 2SO 4
4.75. From the molarity, one L of heme solution is equivalent to 0.0019 mol of heme solute. Before starting the calculation, note that 150 mL of solution is equivalent to 150 x 10−3 L of solution: 150 x 10−3 L soln x
0.0019 mol heme = 2.850 x 10−4 = 2.9 x 10−4 mol heme 1 L solution
4.76. From the molarity, one L of insulin solution is equivalent to 0.0048 mol of insulin solute. Before starting the calculation, note that 28 mL of solution is equivalent to 28 x 10−3 L of solution: 28 x 10−3 L soln x
0.0048 mol insulin = 1.344 x 10−4 = 1.3 x 10−4 mol insulin 1 L soln
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141
4.77. Multiply the volume of solution by molarity to convert it to moles; then convert to mass of solute by multiplying by the molar mass: 100.0 x 10−3 L soln x
0.025 mol Na 2 Cr2 O 7 262.0 g Na 2 Cr2 O 7 x 1 L solution 1 mol Na 2 Cr2 O7 = 0.6550 = 0.66 g Na2Cr2O7
4.78. Multiply the desired volume of solution by the molarity to convert it to moles; then convert to mass of solute by multiplying by the molar mass: 0.20 mol Na 2SO 4 142.05 g Na 2SO 4 x = 7.10 = 7.1 g Na2SO4 1 L solution 1 mol Na 2SO 4
0.250 L soln x
Weigh out 7.1 g of pure Na2SO4, and place it in a 250mL volumetric flask. Add enough water to fill the flask to the mark on the neck. 4.79. Use the rearranged version of the dilution formula to calculate the initial volume of 15.8 M HNO3 required: Vi =
M f Vf Mi
=
0.12 M x 1000 mL = 7.59 = 7.6 mL 15.8 M
4.80. Use the rearranged version of the dilution formula to calculate the initial volume of 12.4 M HCl required: Vi =
M f Vf Mi
=
0.50 M x 1500 mL = 60.4 = 60. mL 12.4 M
4.81. The initial concentration of KCl (molar mass, 74.55 g/mol) is 3.50 g KCl x
1 mol KCl 1 x = 4.694 M 0.0100 L 74.55 g KCl
Using the dilution factor, M1V1 = M2V2, with V2 = 10.0 mL + 60.0 mL = 70.0 mL, after the solutions are mixed, the concentration of KCl is M2 =
M 1V1 4.694 M x 10.0 mL = = 0.67069 M KCl 70.0 mL V2
For CaCl2, the concentration is M2 =
M 1V1 = V2
0.500 M x 10.0 mL = 0.42857 M CaCl2 70.0 mL
Therefore, the concentrations of the ions are 0.671 M K+ and 0.429 M Ca2+. For Cl−, it is 0.67069 M + 2 x 0.42857 M = 1.5278 M = 1.528 M
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Chapter 4: Chemical Reactions
4.82. Using the dilution factor, M1V1 = M2V2, with V2 = 50.0 mL + 25.0 mL = 75.0 mL, after the solutions are mixed, the concentration of NaClO3 is M2 =
M 1V1 0.20 M x 50.0 mL = = 0.1333 M NaClO3 75.0 mL V2
For Na2SO4, the concentration is M2 =
M 1V1 = V2
0.20 M x 25.0 mL = 0.06666 M Na2SO4 75.0 mL
Therefore, the concentrations of the ions are 0.13 M ClO3− and 0.067 M SO42−. For Na+, it is 0.1333 M + 2 x 0.06666 M = 0.2666 = 0.27 M. 4.83. Use the appropriate conversion factors to convert the mass of BaSO4 to the mass of Ba2+ ions: 0.513 g BaSO 4 x
1 mol BaSO 4 1 mol Ba 2+ 137.33 g Ba 2+ x x 233.40 g BaSO 4 1 mol BaSO 4 1 mol Ba 2+ = 0.30184 g Ba2+
Then calculate the percentage of barium in the 458 mg (0.458 g) compound: 0.30184 g Ba 2+ x 100% = 65.9039 = 65.9% Ba 2+ 0.458 g 4.84. Use the appropriate conversion factors to convert the mass of AgI to the mass of I− ions: 2.185 g AgI x
1 mol AgI 1 mol I126.90 g Ix x = 1.18105 g I– 1 mol AgI 1 mol I234.77 g AgI
Then calculate the percentage of iodine in the 1.545 g compound: 1.18105 g Ix 100% = 76.443 = 76.44% I– 1.545 g 4.85. a.
The mass of chloride ion in the AgCl from the copper chloride compound is 86.00 mg AgCl x
35.45 mg Cl= 21.271 mg Cl143.32 mg AgCl
The percentage of chlorine in the 59.40 mg sample is 21.271 mg Clx 100% = 35.809 = 35.81% Cl59.40 mg sample
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b.
143
Of the various approaches, it is as easy to calculate the theoretical percentage of Cl− in both CuCl and CuCl2 as it is to use another approach: CuCl: CuCl 2 :
35.45 mg Clx 100% = 35.808% 99.00 mg CuCl 70.90 mg Clx 100% = 52.733% 134.45 mg CuCl2
The compound is obviously CuCl. 4.86. a.
The mass of chloride ion in the AgCl from the gold chloride compound is 100.3 mg AgC l x
35.45 mg Cl= 24.809 mg Cl143.32 mg AgCl
The percentage of chloride in the 162.7 mg sample is 24.809 mg Clx 100% = 15.248 = 15.25% Cl162.7 mg sample b.
Of the various approaches, it is as easy to calculate the theoretical percentage of Cl− in both AuCl and AuCl3 as it is to use another approach: AuCl: AuCl3 :
35.45 mg Clx 100% = 15.253% 232.4 mg AuCl 106.35 mg Clx 100% = 35.0620% 303.32 mg AuCl3
The compound is obviously AuCl. 4.87. First, calculate the moles of chlorine in the compound: 0.3048 g AgCl x
1 mol AgCl 1 mol Clx = 0.0021267 mol Cl– 1 mol AgCl 143.32 g AgCl
Then, calculate the g Fex+ from the g Cl−: ⎛ 35.45 g Cl  ⎞ x+ g Fex+ = 0.01348 g comp − ⎜ 0.0021267 mol Cl  x ⎟ = 0.059408 g Fe 1 mol Cl ⎝ ⎠ Now, calculate the moles of Fex+ using the molar mass: 0.059408 g Fex+ x
1 mol Fe x + = 0.0010637 mol Fex+ 55.85 g Fe x +
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Chapter 4: Chemical Reactions
Finally, divide the mole numbers by the smallest mole number: For Cl:
0.002127 mol Cl 0.0010637 mol Fe x + = 2.00; for Fex+: = 1.00 0.0010637 mol 0.0010637 mol
Thus, the formula is FeCl2. 4.88. First, calculate the moles of Ba2+ in the compound: 2.012 g BaCrO4 x
1 mol BaCrO 4 1 mol Ba 2+ = 0.0079422 mol Ba2+ x 1 mol BaCrO 4 253.33 g BaCrO 4
Next, calculate the g O from the g Ba2+: ⎛ 137.33 g Ba 2+ ⎞ g O = 1.345 g comp − ⎜ 0.0079422 mol Ba 2+ x ⎟ = 0.25430 mol O 1 mol Ba 2+ ⎠ ⎝ Now, calculate the moles of O using the molar mass: 0.25430 g O x
1 mol O = 0.015894 mol O 16.00 g O
Finally, divide the mole numbers by the smallest mole number: For O:
0.015894 mol O 0.0079422 mol Ba 2+ = 2.00; for Ba 2+ : = 1.00 0.0079422 mol 0.0079422 mol
Thus, the formula is BaO2 (barium peroxide). 4.89. Using molarity, convert the volume of Na2CO3 to moles of Na2CO3; then use the equation to convert to moles of HNO3 and finally to volume: 2HNO3 + Na2CO3 → 2NaNO3 + H2O + CO2 44.8 x 10−3 L Na2CO3 x
0.150 mol Na 2 CO3 1 L soln
x
2 mol HNO3 1 mol Na 2 CO3
x
1 L HNO3 0.250 mol HNO3
= 0.05376 L = 0.0538 L = 53.8 mL 4.90. As in the previous problem, use molarity to convert to volume; then use the equation to convert to moles of the other reactant and finally to volume: Na2CO3 + Ca(OH)2 → CaCO3 + 2NaOH 49.8 x 10−3 L Ca(OH)2 x
1 mol Na 2 CO3 1 L Na 2 CO3 0.150 mol Ca(OH) 2 x x 1 mol Ca(OH) 2 0.350 mol Na 2 CO3 1 L soln
= 0.02134 L (21.3 mL) Na2CO3
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4.91. The reaction is H2SO4 + 2NaHCO3 → Na2SO4 + 2H2O + CO2. 8.20 g NaHCO3 x
1 mol NaHCO3 1 mol H 2SO 4 1 L soln x x 84.01 g NaHCO3 0.150 mol H 2SO 4 2 mol NaHCO3
= 0.3253 L (325 mL) soln 4.92. The reaction is 10FeSO4 + 2KMnO4 + 8H2SO4 → 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 4H2O 3.36 g FeSO4 x
1 mol FeSO 4 2 mol KMnO 4 1 L soln x x 0.250 mol KMnO 4 151.92 g FeSO 4 10 mol FeSO 4 = 0.01769 L (17.7 mL) soln
4.93. First, find the mass of H2O2 required to react with KMnO4. 5H2O2 + 2KMnO4 + 3H2SO4 → 5O2 + 2MnSO4 + K2SO4 + 8H2O 51.7 x 10−3 L KMnO4 x
0.145 mol KMnO 4 5 mol H 2 O 2 34.02 g H 2 O 2 x x 1 L soln 2 mol KMnO 4 1 mol H 2 O 2
= 0.6375 g H2O2 Percent H2O2 = (mass H2O2 ÷ mass sample) x 100% = (0.6375 g ÷ 20.0 g) x 100% = 3.187 = 3.19% 4.94. First, find the mass of Fe2+ required to react with the K2Cr2O7. 6FeSO4 + K2Cr2O7 + 7H2SO4 → 3Fe2(SO4)3 + Cr2(SO4)3 + 7H2O + K2SO4 43.7 x 10−3 L K2Cr2O7 x x
0.150 mol K 2 Cr2 O 7 6 mol FeSO 4 x 1 L soln 1 mol K 2 Cr2 O7
1 mol Fe 2+ 1 mol FeSO 4
x
55.85 g Fe 2+ = 2.196 g Fe2+ 1 mol Fe 2+
2.196 g Fe2+ in reaction = 2.196 g Fe2+ in ore, so Percent Fe = (mass Fe2+ ÷ mass ore) x 100% = (2.196 g ÷ 3.33 g) x 100% = 65.96 = 66.0%
■
SOLUTIONS TO GENERAL PROBLEMS
4.95. For the reaction of magnesium metal and hydrobromic acid, the equations are as follows. Molecular equation: Mg(s) + 2HBr(aq) → H2(g) + MgBr2(aq) Ionic equation: Mg(s) + 2H+(aq) → H2(g) + Mg2+(aq)
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Chapter 4: Chemical Reactions
4.96. For the reaction of aluminum metal and perchloric acid, the equations are as follows. Molecular equation: 2Al(s) + 6HClO4(aq) → 3H2(g) + 2Al(ClO4)3(aq) Ionic equation: 2Al(s) + 6H+(aq) → 3H2(g) + 2Al3+(aq) 4.97. For the reaction of nickel(II) sulfate and lithium hydroxide, the equations are as follows. Molecular equation: NiSO4(aq) + 2LiOH(aq) → Ni(OH)2(s) + Li2SO4(aq) Ionic equation: Ni2+(aq) + 2OH−(aq) → Ni(OH)2(s) 4.98. For the reaction of potassium sulfate and barium bromide, the equations are as follows. Molecular equation: BaBr2(aq) + K2SO4(aq) → BaSO4(s) + 2KBr(aq) Ionic equation: Ba2+(aq) + SO42−(aq) → BaSO4(s) 4.99. a.
Molecular equation: LiOH(aq) + HCN(aq) → LiCN(aq) + H2O(l) Ionic equation: OH−(aq) + HCN(aq) → CN−(aq) + H2O(l)
b.
Molecular equation: Li2CO3(aq) + 2HNO3(aq) → 2LiNO3(aq) + CO2(g) + H2O(l) Ionic equation: CO32−(aq) + 2H+(aq) → CO2(g) + H2O(l)
c.
Molecular equation: LiCl(aq) + AgNO3(aq) → LiNO3(aq) + AgCl(s) Ionic equation: Cl−(aq) + Ag+(aq) → AgCl(s)
d.
Molecular equation: MgSO4(aq) + LiCl(aq) → NR (Li2SO4 and MgCl2 are soluble.)
4.100. a.
Molecular equation: Al(OH)3(s) + 3HNO3(aq) → Al(NO3)3(aq) + 3H2O(l) Ionic equation: Al(OH)3(s) + 3H+(aq) → Al3+(aq) + 3H2O(l)
b.
Molecular equation: NaBr(aq) + HClO4(aq) → NR (HBr and NaClO4 are soluble.)
c.
Molecular equation: CaCl2(aq) + NaNO3(aq) → NR (Ca(NO3)2 and NaCl are soluble.)
d.
Molecular equation: MgSO4(aq) + Ba(NO3)2(aq) → Mg(NO3)2(aq) + BaSO4(s) Ionic equation: SO42−(aq) + Ba2+(aq) → BaSO4(s)
4.101. a.
Molecular equation: Sr(OH)2(aq) + 2HC2H3O2(aq) → Sr(C2H3O2)2(aq) + 2H2O(l) Ionic equation: HC2H3O2(aq) + OH−(aq) → C2H3O2−(aq) + H2O(l)
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b.
147
Molecular equation: NH4I(aq) + CsCl(aq) → NR (NH4Cl and CsI are soluble.)
c.
Molecular equation: NaNO3(aq) + CsCl(aq) → NR (NaCl and CsNO3 are soluble.)
d.
Molecular equation: NH4I(aq) + AgNO3(aq) → NH4NO3(aq) + AgI(s) Ionic equation: I−(aq) + Ag+(aq) → AgI(s)
4.102. a.
Molecular equation: 2HClO4(aq) + BaCO3(s) → Ba(ClO4)2(aq) + H2CO3(aq) [CO2(g) + H2O(l)] Ionic equation: 2H+(aq) + BaCO3(s) → Ba2+(aq) + CO2(g) + H2O(l)
b.
Molecular equation: H2CO3(aq) + Sr(OH)2(aq) → 2H2O(l) + SrCO3(s) Ionic equation: H2CO3(aq) + Sr2+(aq) + 2OH−(aq) → 2H2O(l) + SrCO3(s)
c.
Molecular equation: 2K3PO4(aq) + 3MgCl2(aq) → 6KCl(aq) + Mg3(PO4)2(s) Ionic equation: 2PO43−(aq) + 3Mg2+(aq) → Mg3(PO4)2(s)
d.
Molecular equation: FeSO4(aq) + MgCl2(aq) → NR (FeCl2 and MgSO4 are soluble.)
4.103. For each preparation, the compound to be prepared is given first, followed by the compound from which it is to be prepared. Then the method of preparation is given, followed by the molecular equation for the preparation reaction. Steps such as evaporation are not given in the molecular equation. a.
To prepare CuCl2 from CuSO4, add a solution of BaCl2 to a solution of the CuSO4, precipitating BaSO4. The BaSO4 can be filtered off, leaving aqueous CuCl2, which can be obtained in solid form by evaporation. Molecular equation: CuSO4(aq) + BaCl2(aq) → BaSO4(s) + CuCl2(aq)
b.
To prepare Ca(C2H3O2)2 from CaCO3, add a solution of acetic acid, HC2H3O2, to the solid CaCO3, forming CO2, H2O, and aqueous Ca(C2H3O2)2. The aqueous Ca(C2H3O2)2 can be converted to the solid form by evaporation, which also removes the CO2 and H2O products. Molecular equation: CaCO3(s) + 2HC2H3O2(aq) → Ca(C2H3O2)2(aq) + CO2(g) + H2O(l)
c.
To prepare NaNO3 from Na2SO3, add a solution of nitric acid, HNO3, to the solid Na2SO3, forming SO2, H2O, and aqueous NaNO3. The aqueous NaNO3 can be converted to the solid by evaporation, which also removes the SO2 and H2O products. Molecular equation: Na2SO3(s) + 2HNO3(aq) → 2NaNO3(aq) + SO2(g) + H2O(l)
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Chapter 4: Chemical Reactions
d.
To prepare MgCl2 from Mg(OH)2, add a solution of hydrochloric acid (HCl) to the solid Mg(OH)2, forming H2O and aqueous MgCl2. The aqueous MgCl2 can be converted to the solid form by evaporation. Molecular equation: Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l)
4.104. For each preparation, the compound to be prepared is given first, followed by the compound from which it is to be prepared. Then, the method of preparation is given, followed by the molecular equation for the preparation reaction. Steps such as evaporation are not given in the molecular equation. a.
To prepare MgCl2 from MgCO3, add a solution of hydrochloric acid (HCl) to the solid MgCO3, forming CO2, H2O, and aqueous MgCl2. The aqueous MgCl2 can be converted to the solid form by evaporation. Molecular equation: MgCO3(s) + 2HCl(aq) → MgCl2(aq) + CO2(g) + H2O(l)
b.
To prepare NaNO3 from NaCl, add a solution of AgNO3 to a solution of the NaCl, precipitating AgCl. The AgCl can be filtered off, leaving aqueous NaNO3, which can be obtained in solid form by evaporation. Molecular equation: NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)
c.
To prepare Al(OH)3 from Al(NO3)3, add a solution of NaOH to a solution of Al(NO3)3, precipitating Al(OH)3. The Al(OH)3 can be filtered off and dried to remove any water adhering to it. Molecular equation: Al(NO3)3(aq) + 3NaOH(aq) → Al(OH)3(s) + 3NaNO3(aq)
d.
To prepare HCl from H2SO4, add a solution of BaCl2 to the solution of H2SO4, precipitating BaSO4. The BaSO4 can be filtered off, leaving the desired solution of aqueous HCl. Molecular equation: H2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2HCl(aq)
4.105. a.
Decomposition
b.
Decomposition
c.
Combination
d.
Displacement
4.106. a.
Combination
b.
Displacement
c.
Decomposition
d.
Combination
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4.107. Pb(NO3)2 + H2SO4
[ → PbSO4(s) + HNO3(aq) ]
Pb(NO3)2 + MgSO4
[ → PbSO4(s) + Mg(NO3)2(aq) ]
Pb(NO3)2 + Ba(OH)2
[ → Pb(OH)2(s) + Ba(NO3)2(aq) ]
b.
Ba(OH)2 + MgSO4
[ → BaSO4(s) + Mg(OH)2(s) ]
c.
Ba(OH)2 + H2SO4
[ → BaSO4(s) + H2O(l) ]
AgNO3 + H3PO4
[ → Ag3PO4(s) + HNO3(aq) ]
AgNO3 + Sr(OH)2
[ → AgOH(s) + Sr(NO3)2(aq) ]
CuSO4 + H3PO4
[ → Cu3(PO4)2(s) + H2SO4(aq) ]
b.
Sr(OH)2 + CuSO4
[ → SrSO4(s) + Cu(OH)2(s) ]
c.
Sr(OH)2 + H3PO4
[ → Sr3(PO4)2(s) + H2O(l) ]
a.
4.108. a.
4.109. Divide the mass of CaCl2 by its molar mass and volume to find molarity: 4.50 g CaCl2 x
1 mol CaCl2 1 = 0.04054 = 0.0405 M CaCl2 x 1.000 L soln 110.98 g CaCl2
The CaCl2 dissolves to form Ca2+ and 2Cl− ions. Therefore, the molarities of the ions are 0.0405 M Ca2+ and 2 x 0.04054, or 0.0811, M Cl− ions. 4.110. Divide the mass of Fe2(SO4)3 by its molar mass and volume to find molarity: 3.45 g Fe 2 (SO 4 )3 x
1 mol Fe 2 (SO 4 )3 1 x = 0.008627 399.91 g Fe 2 (SO 4 )3 1.000 L soln
= 0.00863 M Fe2(SO4)3 The Fe2(SO4)3 dissolves to form 2Fe3+ and 3SO42− ions. Therefore, the molarities of the ions are 2 x 0.00863, or 0.0173, M Fe3+ and 3 x 0.00863, or 0.0259, M SO42−. 4.111. Divide the mass of K2Cr2O7 by its molar mass and volume to find molarity. Then calculate the volume needed to prepare 1.00 L of a 0.100 M solution. 89.3 g K2Cr2O7 x Molarity = Vi =
0.3035 mol K 2 Cr2 O 7 = 0.3035 M 1.00 L
M f x Vf Mi
1 mol K 2 Cr2 O 7 = 0.3035 mol K2Cr2O7 294.20 g K 2 Cr2 O 7
=
0.100 M x 1.00 L = 0.3294 L (329 mL) 0.3035 M
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Chapter 4: Chemical Reactions
4.112. Divide the mass by the molar mass and then by volume to find molarity. Then, calculate the volume needed to prepare 1.00 L of 0.1150 M. 71.2 g H2C2O4 x Molarity = Vi =
1 mol H 2 C2 O 4 = 0.7907 mol H2C2O4 90.04 g H 2 C 2 O 4
0.7907 mol H 2 C 2 O 4 = 0.7907 = 0.791 M 1.00 L
M f x Vf Mi
=
0.150 M x 1.00 L = 0.1896 L (190. mL) 0.7907 M
Place 190. mL of the 0.791 M solution in a 1 L volumetric flask, and dilute to 1.00 L. 4.113. Assume a volume of 1.000 L (1000 cm3) for the 6.00% NaBr solution, and convert to moles and then to molarity. 1000 cm3 x
1.046 g soln 6.00 g NaBr 1 mol NaBr x x = 0.6099 mol 3 1 cm 100 g soln 102.89 g NaBr
Molarity NaBr =
0.6099 mol NaBr = 0.6099 = 0.610 M 1.000 L
4.114. Assume a volume of 1.000 L (1000 mL) for the 4.00% NH3 solution, and convert to moles and then to molarity. 1000 mL x
4.00 g NH 3 1 mol NH 3 0.979 g soln x = 2.299 mol NH3 x 1 mL 100 g soln 17.03 g NH 3
Molarity NH3 =
2.299 mol NH 3 = 2.299 = 2.30 M 1.000 L
4.115. First, calculate the moles of BaCl2: 1.128 g BaSO4 x
1 mol BaSO 4 1 mol BaCl2 x = 0.0048329 mol BaCl2 233.40 g BaSO 4 1 mol BaSO 4
Then calculate the molarity from the moles and volume (0.0500 L): Molarity =
0.0048329 mol BaCl2 = 0.096658 = 0.0967 M 0.0500 L
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4.116. First, calculate the moles of CaCl2: 1 mol CaC2 O 4 1 mol CaCl2 x = 0.0112177 mol CaCl2 128.10 g CaC2 O 4 1 mol CaC2 O 4
1.437 g CaC 2 O 4 x
Then calculate the molarity from the moles and volume (0.0500 L): Molarity =
0.0112177 mol CaCl2 = 0.22435 = 0.224 M 0.0500 L
4.117. First, calculate the grams of thallium(I) sulfate: 0.2122 g TlI x
1 mol Tl2SO 4 504.83 g Tl2SO 4 1 mol TlI x = 0.16168 g Tl2SO4 x 2 mol TlI 331.28 g TlI 1 mol Tl2SO 4
Then calculate the percent Tl2SO4 in the rat poison: Percent Ti2SO4 =
0.16168 g x 100% = 21.301 = 21.30% 0.7590 g
4.118. First, calculate the grams of CaCO3: 1 mol CaCO3 100.1 g CaCO3 1 mol CaC 2 O 4 x x 1 mol CaC2 O 4 1 mol CaCO3 128.10 g CaC2 O 4
0.629 g CaC2O4 x
= 0.49151 g CaCO3 Then calculate the percent CaCO3 in the antacid: Percent CaCO3 =
0.49151 g x 100% = 72.28 = 72.3% 0.680 g
4.119. The mass of copper(II) ion and the mass of sulfate ion in the 98.77mg sample are 0.09877 g x 0.3250 = 0.032100 g Cu2+ ion 0.11666 g BaSO4 x
96.07 g SO 24 = 0.048019 = 0.04802 g SO42− 233.40 g BaSO 4
Mass of water left = 0.09877 g − (0.03210 g Cu2+ + 0.04802 g SO42− ) = 0.01865 g H2O Moles of water left = 0.01865 g ÷ 18.02 g/mol = 1.035 x 10−3 mol Moles of Cu2+ = 0.03210 g ÷ 63.55 g/mol = 5.0511 x 10−4 mol Ratio of water to Cu2+, or CuSO4 = 1.035 x 10−3 ÷ 5.0511 x 10−4 = 2.05, or 2 The formula is thus CuSO4•2H2O.
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Chapter 4: Chemical Reactions
4.120. The mass of copper(II) ion and the mass of sulfate ion in the 85.42mg sample are 0.08542 g x 0.2976 = 0.025421 g Cu2+ ion 0.09333 g BaSO4 x
96.07 g SO 24 = 0.038416 = 0.03842 g SO42− 233.40 g BaSO 4
Mass of water left = 0.08542 g − (0.025421 g Cu2+ + 0.038416 g SO42−) = 0.021583 g H2O Moles of water left = 0.021583 ÷ 18.02 g/mol = 1.1977 x 10−3 mol Moles of Cu2+ = 0.025421 g ÷ 63.55 g/mol = 4.0002 x 10−4 mol Ratio of water to Cu2+, or CuSO4 = 1.1977 x 10−3 ÷ 4.000 x 10−4 = 2.99, or 3 The formula is CuSO4•3H2O. 4.121. For these calculations, the relative numbers of moles of gold and chlorine must be determined. These can be found from the masses of the two elements in the sample: Total mass = mass of Au + mass of Cl = 328 mg The mass of chlorine in the precipitated AgCl is equal to the mass of chlorine in the compound of gold and chlorine. The mass of Cl in the 0.464 g of AgCl is 0.464 g AgCl x
1 mol AgCl 1 mol Cl 35.45 g Cl x x 143.32 g AgCl 1 mol AgCl 1 mol Cl
= 0.11476 g Cl (114.76 mg Cl) Mass percentage Cl =
mass Cl 114.76 mg x 100% = x 100% = 35.0% Cl mass comp 328 mg
To find the empirical formula, convert each mass to moles: Mass percentage Au = 328 mg − 114.76 mg Cl = 213.23 mg Au 0.11476 g Cl x
1 mol Cl = 0.003238 mol Cl 35.45 g Cl
0.21323 g Au x
1 mol Au = 0.001082 mol Au 196.97 g Au
Divide both numbers of moles by the smaller number (0.001082) to find the integers. Integer for Cl: 0.003238 mol ÷ 0.001082 mol = 2.99, or 3 Integer for Au: 0.001082 mol ÷ 0.001082 mol = 1.00, or 1 The empirical formula is AuCl3.
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4.122. To find the mass percentages of Cl and Sc, find the mass of the chlorine in the precipitated AgCl, which is equal to the mass of the chlorine in the compound of scandium and chlorine. The mass of Cl in the 167.4 mg (0.1674 g) of AgCl is 0.1674 g AgCl x
1 mol AgCl 1 mol Cl 35.45 g Cl x x = 0.04141 g Cl 1 mol Cl 143.32 g AgCl 1 mol AgCl
The mass of Sc equals the difference between the sample mass and the mass of Cl: Mass Sc = 0.0589 g − 0.04141 g = 0.01749 g Sc The mass percentage of each element is found by dividing the mass of the element by the mass of the compound and multiplying by 100%: Mass percentage = (mass elem. ÷ mass comp.) x 100%. Mass percentage Cl = (0.04141 g ÷ 0.0589 g) x 100% = 70.305 = 70.3% Mass percentage Sc = (0.01749 g ÷ 0.0589 g) x 100% = 29.69 = 29.7% To find the empirical formula, convert each mass to moles: 0.04141 g Cl x
1 mol Cl = 0.0011680 mol Cl 35.45 g Cl
0.01749 g Sc x
1 mol Sc = 0.0003890 mol Sc 44.96 g Sc
Divide both numbers of moles by the smaller number (0.0003890) to find the integers: Integer for Cl = 0.0011680 mol ÷ 0.0003890 mol = 3.00, or 3 Integer for Sc = 0.0003890 mol ÷ 0.0003890 mol = 1.00, or 1 The empirical formula is ScCl3. 4.123. From the equations NH3 + HCl → NH4Cl and NaOH + HCl → NaCl + H2O, we write Mol NH3 = mol HCl(NH3) Mol NaOH = mol HCl(NaOH) We can calculate the mol NaOH and the sum [mol HCl(NH3) + mol HCl(NaOH)] from the titration data. Because the sum equals mol NH3 plus mol NaOH, we can calculate the unknown mol NH3 from the difference: Mol NH3 = sum − mol NaOH. Mol HCl (NaOH) + mol HCl (NH 3 ) = 0.0463 L x
0.213 mol HCl 1.000 L
= 0.009862 mol HCl Mol NaOH = 0.0443 L x
0.128 mol NaOH = 0.005670 mol NaOH 1.000 L
Mol HCl(NH3) = 0.009862 mol − 0.005670 mol = 0.004192 mol Mol NH3 = mol HCl(NH3) = 0.004192 mol NH3
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Chapter 4: Chemical Reactions
Because all the N in the (NH4)2SO4 was liberated as and titrated as NH3, the amount of N in the fertilizer is equal to the amount of N in the NH3. Thus, the moles of NH3 can be used to calculate the mass percentage of N in the fertilizer: 0.004192 mol NH 3 x
1 mol N 14.01 g N x = 0.05873 g N 1 mol NH 3 1 mol N
Mass percentage N =
mass N 0.05873 g N x 100% = x 100% = 9.659 = 9.66% mass fert. 0.608 g
4.124. From NaHCO3 + HCl → NaCl + H2O + CO2 and HCl + NaOH → NaCl + H2O, we write Mol NaHCO3 = mol HCl(NaHCO3) Mol NaOH = mol HCl(NaOH) We can calculate the mol NaOH and the sum [mol HCl(NaHCO3) + mol HCl(NaOH)] from the titration data. Because the sum equals mol NaHCO3 plus mol NaOH, we can calculate the unknown mol NaHCO3 from the difference: Mol NaHCO3 = sum − mol NaOH. Mol HCl (NaOH) + mol HCl (NaHCO3 ) = 0.0500 L x
0.190 mol HCl 1.000 L
= 0.009500 mol HCl Mol NaOH = 0.0471 L x
0.128 mol NaOH = 0.006029 mol NaOH 1.000 L
Mol HCl(NaHCO3) = 0.009500 mol − 0.006029 mol = 0.003471 mol Mol NaHCO3 = mol HCl(NaHCO3) = 0.003471 mol Because all the NaHCO3 in the tablet was titrated as NaHCO3, the moles of NaHCO3 can be used to calculate the mass percentage of NaHCO3 in the tablet: 0.003471 mol NaHCO3 x
84.01 g NaHCO3 = 0.2916 g NaHCO3 1 mol NaHCO3
Mass percentage NaHCO3 =
mass NaHCO3 0.2916 g N x 100% = x 100% mass tab. 0.500 g
= 58.32 = 58.3%
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155
SOLUTIONS TO STRATEGY PROBLEMS
4.125. The final volume to which the solution must be diluted is V2 =
M 1V1 (3.47 M) (100 mL) = = 194.94 mL 1.78 M M2
The amount of water that must be added is 194.94 mL − 100 mL = 94.94 mL. Since the density of water is 1.00 g/mL, this is equivalent to 94.94 g of water, so the amount of ice is 94.94 g water x
3 g ice = 284.83 = 285 g ice 1 g water
4.126. The balanced molecular equation is 2AgNO3(aq) + Na2CO3(aq) → Ag2CO3(s) + 2NaNO3(aq) Calculate the mass of silver nitrate required to react with the sodium carbonate solution. 30.6 x 10−3 L x 0.511 M x
2 mol AgNO3 x 1 mol Na 2 CO3
169.88 g = 5.3126 g 1 mol AgNO3
The mass of AgNO3 remaining is 8.30 g – 5.3126 g = 2.987 g. The volume of the solution is 45.1 mL + 30.6 mL = 75.7 mL (75.7 x 10−3 L). The molarity of the silver ion can now be determined. 2.987 g AgNO3 x 75.7 x 103 L
1 mol AgNO3 = 0.2322 = 0.232 M 169.88 g
4.127. 0.248 M x 38.2 x 10−3 L x
2 mol Al3+ 26.98 g x = 0.51119 = 0.511 g 1 mol Al2 (SO 4 )3 1 mol Al3+
4.128. The ratio of nitrate ion to aluminum nitrate (Al(NO3)3) is 3 to 1. The molarity of the nitrate ion after dilution is M2 =
3 mol NO3M 1V1 (0.256 M) (31.6 mL) x = = 0.3809 = 0.381 M 63.7 mL 1 mol Al2 (NO3 )3 V2
4.129. 3.33 x 1023 ions x
1 mol Zn(C 2 H 3O 2 ) 2 1 mol 183.478 g x x 23 2 mol C 2 H 3O 2 6.022 x 10 ions 1 mol
= 50.72 = 50.7 g Zn(C2H3O2)2
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Chapter 4: Chemical Reactions
4.130. The balanced molecular equation is H3AsO4(aq) + 3NaOH(aq) → Na3AsO4(aq) + 3H2O(l) The molarity of the arsenic acid is M2 =
1 mol H 3 AsO 4 (0.6441 M ) (53.07 mL) x = 0.43588 = 0.4359 M 3 mol NaOH 26.14 mL
4.131. The reactions are 6 x (K → K+ + e−) 1 x (N2 + 6e− → 2N3− ) 6K(s) + N2(g) + 6e− → 2K3N(s) + 6e− Thus six electrons are canceled. 4.132. a.
oxidation–reduction reaction
b.
precipitation reaction
c.
neutralization reaction
d.
precipitation reaction
4.133. Since the solution is homogeneous, the concentration of the solution in the beaker is the same as the original solution, 0.196 M. The concentration of the solution in the large flask is also the same. 4.134. Calculate the mol of chloride ion for each compound in solution. PbCl2 is insoluble, as such it will produce no chloride ions in solution. 1.0 g KCl x
1 mol KCl 1 mol Cl= 0.01341 mol Cl− x 1 mol KCl 74.55 g
1.0 g CaCl2 x
1 mol CaCl2 2 mol Cl= 0.01802 mol Cl− x 1 mol CaCl2 110.98 g
Therefore, the CaCl2 mixture would have the highest concentration of Cl ion.
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157
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
4.135. For this reaction, the formulas are listed first, followed by the molecular and net ionic equations, the names of the products, and the molecular equation for another reaction giving the same precipitate. Lead(II) nitrate is Pb(NO3)2, and cesium sulfate is Cs2SO4. Molecular equation: Pb(NO3)2(aq) + Cs2SO4(aq) → PbSO4(s) + 2CsNO3(aq) Net ionic equation: Pb2+(aq) + SO42−(aq) → PbSO4(s) PbSO4 is lead(II) sulfate, and CsNO3 is cesium nitrate. Molecular equation: Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq) 4.136. For this reaction, the formulas are listed first, followed by the molecular and net ionic equations, the names of the products, and the molecular equation for another reaction giving the same precipitate. Silver nitrate is AgNO3, and strontium chloride is SrCl2. Molecular equation: 2AgNO3(aq) + SrCl2(aq) → 2AgCl(s) + Sr(NO3)2(aq) Net ionic equation: Ag+(aq) + Cl−(aq) → AgCl(s) AgCl is silver(I) chloride, and Sr(NO3)2 is strontium nitrate. Molecular equation: AgClO4(aq) + NaCl(aq) → AgCl(s) + NaClO4(aq) 4.137. Net ionic equation: 2Br−(aq) + Cl2(g) → 2Cl−(aq) + Br2(l) Molecular equation:
CaBr2(aq) + Cl2(g) → CaCl2(aq) + Br2(l)
Mass Br2 = 40.0 g + 14.2 g − 22.2 g = 32.0 g Combining the three known masses gives the unknown mass of Br2. Now, use a ratio of the known masses of CaBr2 to Br2 to convert pounds of Br2 to grams of CaBr2: 10.0 lb Br2 x
40.0 g CaBr2 453.6 g x = 5670 = 5.67 x 103 g CaBr2 1 lb 32.0 g Br2
4.138. Net ionic equation: Ba2+(aq) + CO32−(aq) → BaCO3(s) Molecular equation: BaS(aq) + Na2CO3(aq) → Na2S(aq) + BaCO3(s) MassBaCO3 = 33.9 g + 21.2 g − 15.6 g = 39.5 g Combining the three known masses gives the unknown mass of BaCO3. Now, use a ratio of the known masses of BaS to BaCO3 to ultimately obtain grams of BaS:
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Chapter 4: Chemical Reactions
5.00 ton BaCO3 x 4.291 ton BaS x
33.9 g BaS = 4.291 ton BaS 39.5 g BaCO3
2000 lb 453.6 g x = 3.892 x 106 = 3.89 x 106 g BaS 1 ton 1 lb
4.139. Molecular equation: Hg(NO3)2 + H2S(g) → HgS(s) + 2HNO3(aq) Net ionic equation: Hg2+ + H2S(g) → HgS(s) + 2H+(aq) The acid formed is nitric acid, a strong acid. The other product is mercury(II) sulfide. Mass HNO3 = (81.15 g + 8.52 g) − 58.16 g = 31.51 g Mass of solution = 550.0 g H2O + 31.51 g HNO3 = 581.51 = 581.5 g 4.140. Molecular equation: Hg(NO3)2 + H2S → HgS(s) + 2HNO3(aq) Net ionic equation: Hg2+ + H2S → HgS(s) + 2H+(aq) The acid formed is nitric acid, a strong acid. The other product is mercury(II) sulfide. Mass HNO3 = (65.65 g + 4.26 g) − 54.16 g = 15.75 g Mass of solution = 395.0 g H2O + 15.75 g HNO3 = 410.75 = 410.8 g 4.141. Let the number of Fe3+ ions equal y; then the number of Fe2+ ions equals (7 − y). Since Fe7S8 is neutral, the number of positive charges must equal the number of negative charges. If the signs are omitted, then Total charge on both Fe2+ and Fe3+ = total charge on all eight sulfide ions 3y + 2(7 − y) = 8 x 2 y + 14 = 16 y = 2
Thus, the ratio of Fe2+ to Fe3+ is 5/2. 4.142. To define the problem in terms of percentages, use 100 X2O3 oxides. Then the number of X3+ ions equals 100, and the sum of the X2+ and X5+ ions also equals 100. Let the number of X2+ ions equal y; then the number of X5+ ions equals (100 − y). Since X2O3 is neutral, the number of positive charges must be equal to the number of negative charges. If the signs are omitted, then; Total charge on X2+, X3+, and X5+ = total charge of all 300 oxide ions 2y + (100 x 3) + 5 (100 − y) = 300 x 2 −3y + 800 = 600 y = (200 ÷ 3) = 66.67
The percentage of X2+ in the 100 X2O3 ‘s is (66.67 ÷ 200) x 100 = 33.3%.
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4.143. Use the density, formula mass,, and percentage to convert to molarity. Then combine the 0.200 mol with mol/L to obtain the volume in liters. 0.807 g soln 0.940 g ethanol 1 mol ethanol 1000 mL x x x 1 mL 1L 1.00 g soln 46.07 g ethanol =
16.465 mol ethanol 1 L ethanol
L ethanol = 0.200 mol ethanol x
1 L ethanol = 0.012146 = 0.0121 L 16.465 mol ethanol
4.144. Use the density, formula mass, and percentage to convert to molarity. Then combine the 0.350 mol with mol/L to obtain the volume in liters. 1.072 g soln 0.560 g glycol 1 mol glycol 1000 mL 9.672 mol glycol x x x = 1 mL 1L 1.00 g soln 62.07 g glycol 1 L glycol L glycol = 0.350 mol glycol x
1 L glycol = 0.03618 = 0.0362 L 9.672 mol glycol
4.145. Convert the 2.183 g of AgI to mol AgI, which is chemically equivalent to mol KI. Use that to calculate the molarity of the KI. 2.183 g AgI x Molarity =
1 mol AgI = 9.2984 x 10−3 mol AgI (equivalent to mol KI) 234.77 g AgI
9.2984 x 103 mol KI = 0.9298 = 0.930 M 0.0100 L
4.146. Convert the 5.719 g of BaSO4 to moles of BaSO4, which is chemically equivalent to moles of Na2SO4. Use that to calculate the molarity of the Na2SO4. 5.719 g BaSO4 x Molarity =
1 mol BaSO 4 = 0.024502 mol BaSO4 (eq. to mol Na2SO4) 233.40 g BaSO 4
0.024502 mol Na 2SO 4 = 0.9801 = 0.980 M 0.0250 L
4.147. Convert the 6.026 g of BaSO4 to moles of BaSO4; then, from the equation, deduce that 3 mol of BaSO4 is equivalent to 1 mol of M2(SO4)3 and is equivalent to 2 mol of M. Use that with 1.200 g of the metal M to calculate the atomic mass of M. 6.026 g BaSO 4 x
1 mol BaSO 4 2 mol M x = 0.017212 mol M 233.40 g BaSO 4 3 mol BaSO 4
Atomic mass of M in g/mol =
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1.200 g M = 69.719 g/mol (gallium) 0.017212 mol M
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4.148. Convert the 7.964 g of AgCl to moles of AgCl; then, from the equation, deduce that 2 mol of AgCl is equivalent to 1 mol of MCl2 and is equivalent to 1 mol of M. Use that with 2.434 g of the metal M to calculate the atomic mass of M. 1 mol AgCl 1 mol M x = 0.027784 mol M 143.32 g AgCl 2 mol AgCl
7.964 g AgCl x
Atomic mass M =
2.434 g M = 87.604 = 87.60 g/mol (strontium) 0.027784 mol M
4.149. Use the density, formula mass, percentage, and volume to convert to moles of H3PO4. Then, from the equation P4O10 + 6H2O → 4H3PO4, deduce that 4 mol H3PO4 is equivalent to 1 mol of P4O10, and use that to convert to moles of P4O10. 1500 mL x
0.0500 g H 3 PO 4 1 mol H 3 PO 4 1.025 g soln x x = 0.7844 mol H3PO4 1 g soln 98.00 g H 3 PO 4 1 mL
0.7844 mol H3PO4 x
1 mol P4 O10 = 0.1961 mol P4O10 4 mol H 3 PO 4
Mass P4O10 = 0.1961 mol P4O10 x
283.92 g P4 O10 = 55.677 = 55.7 g P4O10 1 mol P4 O10
4.150. Use the density, formula mass, percentage, and volume to convert to moles of FeCl3, which is chemically equivalent to moles of Fe, from the equation 2Fe + 3Cl2(g) → 2FeCl3. 0.0900 g FeCl3 1 mol FeCl3 1.067 g soln x x 3000 mL = 1.7761 mol FeCl3 x 1 mL 1 g soln 162.20 g FeCl3 1.7761 mol FeCl3 = 1.7761 mol Fe Mass Fe = 1.7761 mol Fe x
55.85 g Fe = 99.19 = 99.2 g Fe 1 mol Fe
4.151. Convert the 0.1068 g of hydrogen to moles of H2; then deduce from the equation that 3 moles of H2 is equivalent to 2 mol of Al. Use the moles of Al to calculate mass of Al and the percentage Al. 0.1068 g H2 x
1 mol H 2 2 mol Al = 0.0353174 mol Al x 3 mol H 2 2.016 g H 2 26.98 g Al 1 mol Al x 100% = 85.229 = 85.23% 1.118 g alloy
0.0353174 mol Al x Percentage Al =
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4.152. Convert the 0.1152 g of H2 to moles of H2; then deduce from the equation that 3 mol of H2 is equivalent to 2 mol of Fe. Use the moles of Fe to calculate mass of Fe and the percentage Fe. 0.1152 g H2 x
1 mol H 2 2 mol Fe = 0.038095 mol Fe x 3 mol H 2 2.016 g H 2 55.85 g Fe 1 mol Fe x 100% = 90.229 = 90.23% 2.358 g alloy
0.038095 mol Fe x Percentage Fe =
4.153. Use the formula mass of Al2(SO4)3 to convert to moles of Al2(SO4)3. Then deduce from the equation that 1 mol of Al2(SO4)3 is equivalent to 3 mol of H2SO4, and calculate the moles of H2SO4 needed. Combine density, percentage, and formula mass to obtain molarity of H2SO4. Then combine molarity and moles to obtain volume. 37.4 g Al2(SO4)3 x
1 mol Al2 (SO 4 )3 3 mol H 2SO 4 x = 0.3279 mol H2SO4 342.17 g Al2 (SO 4 )3 1 mol Al2 (SO 4 )3
15.0 g H 2SO 4 1 mol H 2SO 4 1.104 g soln 1000 mL x x x = 1.688 mol H2SO4 /L 1 mL 1L 100 g soln 98.09 g H 2SO 4
0.3279 mol H2SO4 x
1 L H 2SO 4 = 0.1942 = 0.194 L (194 mL) 1.688 mol H 2SO 4
4.154. Use the formula mass of Na3PO4 to convert to moles of Na3PO4. Then deduce from the equation that 1 mol of Na3PO4 is equivalent to 3 mol of NaOH, and calculate the moles of NaOH needed. Combine density, percentage, and formula mass to obtain molarity of NaOH. Then combine molarity and moles to obtain the volume. 26.2 g Na3PO4 x
1 mol Na 3 PO 4 3 mol NaOH x = 0.4794 mol NaOH 163.94 g Na 3 PO 4 1 mol Na 3 PO 4
1.133 g soln 0.120 g NaOH 1 mol NaOH 1000 mL x x x = 3.399 mol NaOH 1 mL 1 g soln 40.00 g NaOH 1L
0.4794 mol NaOH x
1 L NaOH = 0.14104 L (141 mL) 3.399 mol NaOH
4.155. The equations for the neutralization are 2HCl + Mg(OH)2 → MgCl2 + 2H2O and 3HCl + Al(OH)3 → AlCl3 + 3H2O. Calculate the moles of HCl, and set it equal to the total moles of hydroxide ion, OH−. 0.0485 L HCl x
0.187 mol HCl = 0.0090695 mol HCl 1 L HCl
0.0090695 mol HCl = 2 [mol Mg(OH)2] + 3 [mol Al(OH)3] Rearrange the last equation, and solve for the moles of Al(OH)3. (Eq 1) mol Al(OH)3 = 0.0030231 mol HCl − 2/3 [mol Mg(OH)2]
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Chapter 4: Chemical Reactions
The total mass of chloride salts is equal to the sum of the masses of MgCl2 (molar mass = 95.21 g/mol) and AlCl3 (molar mass = 133.33 g/mol). [95.21 g/mol x mol MgCl2] + [133.33 g/mol x mol AlCl3] = 0.4200 g Since the moles of Mg(OH)2 equals the moles of MgCl2, and the moles of Al(OH)3 equals the moles of AlCl3, you can substitute these quantities into the last equation and get [95.21 g/mol x mol Mg(OH)2] + [133.33 g/mol x mol Al(OH)3] = 0.4200 g Substitute Equation 1 into this equation for the moles of Al(OH)3: [95.21 g/mol x mol Mg(OH)2] + [133.33 g/mol x (0.0030231 mol − 2/3 mol Mg(OH)2)] = 0.4200 g Rearrange the equation, and solve for the moles of Mg(OH)2. 6.323 mol Mg(OH)2 + 0.403070 = 0.4200 mol Mg(OH)2 = 0.01693 ÷ 6.323 = 0.0026728 mol Calculate the mass of Mg(OH)2 in the antacid tablet (molar mass = 58.33 g/mol). 0.0026728 mol Mg(OH)2 x
58.33 g Mg(OH) 2 = 0.15590 g Mg(OH)2 1 mol Mg(OH) 2
Use Eq 1 to find the moles and mass of Al(OH)3 (molar mass 78.00 g/mol) in a similar fashion. mol Al(OH)3 = 0.0030231 − 2/3[0.0026728 mol Mg(OH)2] = 0.0012412 mol 0.0012412 mol Al(OH)3 x
78.00 g Al(OH)3 = 0.096681 g Al(OH)3 1 mol Al(OH)3
The mass percentage Mg(OH)2 in the antacid is Mass percentage of Mg(OH)2 = [0.15590 g ÷ (0.15590 + 0.096681) g] x 100% = 61.72 = 61.7% 4.156. The equations for the neutralization are 2HCl + MgCO3 → MgCl2 + H2O + CO2 and 2HCl + CaCO3 → CaCl2 + H2O + CO2. Calculate the moles of HCl, and set it equal to the total moles of carbonate ion, CO32−. 0.04133 L HCl x
0.08750 mol HCl = 0.0036164 mol HCl 1 L HCl
0.0036164 mol HCl = 2 [mol CaCO3] + 2 [mol MgCO3] Rearrange this equation, and solve for the moles of MgCO3. (Eq 1) mol MgCO3 = 0.0018082 mol HCl − mol CaCO3 The total mass of chloride salts equals the sum of the masses of MgCl2 (molar mass = 95.21 g/mol) and CaCl2 (molar mass = 110.98 g/mol). [110.98 g/mol x mol CaCl2] + [95.21 g/mol x mol MgCl2] = 0.1900 g Since the moles of CaCO3 equals the moles of CaCl2, and moles of MgCO3 equal the moles of MgCl2, you can substitute these quantities into the last equation and get [110.98 g/mol x mol CaCO3] + [95.21 g/mol x mol MgCO3] = 0.1900 g
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Substitute Eq 1 into this equation for the moles of MgCO3: [110.98 g/mol x mol CaCO3] + [95.21 g/mol x (0.0018082 mol − mol CaCO3)] = 0.1900 g Rearrange the equation and solve for the moles of MgCO3. 15.77 mol CaCO3 + 0.172159 = 0.1900 mol CaCO3 = 0.017841 ÷ 15.77 = 0.0011313 mol Calculate the mass of CaCO3 (molar mass = 100.09 g/mol) in the antacid tablet. 0.0011313 mol CaCO3 x
100.09 g CaCO3 = 0.113234 g CaCO3 1 mol CaCO3
Use Eq 1, and find the moles and mass of MgCO3 (molar mass = 84.32 g/mol) in a similar fashion. mol MgCO3 = 0.0018082 mol − 0.0011313 mol CaCO3 = 0.0006769 mol 0.0006769 mol MgCO3 x
84.32 g MgCO3 = 0.057076 g MgCO3 1 mol MgCO3
The mass percentage CaCO3 in the antacid is Mass percentage of CaCO3 = [0.113234 g ÷ (0.113234 + 0.057076) g] x 100% = 66.487 = 66.5%
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CHAPTER 5
The Gaseous State
■
SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 5.1. First, convert to atm (57 kPa = 57 x 103 Pa). 57 x 103 Pa x
1 atm = 0.562 = 0.56 atm 1.01325 x 105 Pa
Next, convert to mmHg. 57 x 103 Pa x
760 mmHg = 427.5 = 4.3 x 102 mmHg 1.01325 x 105 Pa
5.2. Application of Boyle’s law gives Vf = Vi x
Pi Pf
= 20.0 L x
1.00 atm = 24.096 = 24.1 L 0.830 atm
5.3. First, convert the temperatures to the Kelvin scale.
Ti = (19 + 273) = 292 K Tf = (25 + 273) = 298 K Following is the data table.
Vi = 4.38 dm3
Pi = 101 kPa
Ti = 292 K
Vf = ?
Pf = 101 kPa
Tf = 298 K
Apply Charles’s law to obtain
Vf = Vi x
Tf Ti
= 4.38 dm3 x
298 K = 4.470 = 4.47 dm3 292 K
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165
5.4. First, convert the temperatures to kelvins.
Ti = (24 + 273) = 297 K Tf = (35 + 273) = 308 K Following is the data table.
Vi = 5.41 dm3
Pi = 101.5 kPa
Ti = 297 K
Vf = ?
Pf = 102.8 kPa
Tf = 308 K
Apply both Boyle’s law and Charles’s law combined to get
Vf = Vi x
Pi Pf
x
Tf Ti
= 5.41 dm3 x
101.5 kPa 308 K x = 5.539 = 5.54 dm3 102.8 kPa 297 K
5.5. Use the ideal gas law, PV = nRT, and solve for n:
n =
PV ⎛ V ⎞ = ⎜ ⎟P RT ⎝ RT ⎠
Note that everything in parentheses is constant. Therefore, you can write
n = constant x P Or, expressing this as a proportion,
n ∝ P 5.6. First, convert the mass of O2 to moles of O2 (molar mass 32.00 g/mol) and convert temperature to kelvins.
T = 23 + 273 = 296 K 3.03 kg O2 x
1 mol O 2 1000 g x = 94.688 mol O2 1 kg 32.00 g O 2
Summarize the data in a table. Variable
Value
P
?
V
50.0 L
T
296 K
n
94.688 mol
Solve the ideal gas equation for P, and substitute the data to get P =
nRT (94.688)(0.08206 L • atm/K • mol)(296 K) = = 46.00 = 46.0 atm V 50.0 L
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Chapter 5: The Gaseous State
5.7. The data given are Variable
Value
1 atm = 0.98947 atm 760 mmHg
P
752 mmHg x
V
1 L (exact number)
T
(21 + 273) = 294 K
n
?
Using the ideal gas law, solve for n, the moles of helium. n =
PV (0.98947 atm)(1 L) = 0.04101 mol = RT (0.08206 L • atm/K • mol)(294 K)
Now convert mol He to grams. 0.04101 mol He x
4.00 g He = 0.16404 g He 1 mol He
Therefore, the density of He at 21°C and 752 mmHg is 0.164 g/L. The difference in mass between one liter of air and one liter of helium is Mass air − mass He = 1.188 g − 0.16404 g = 1.02396 = 1.024 g difference 5.8. Tabulate the values of the variables. Variable
Value
P
0.862 atm
V
1 L (exact number)
T
(25 + 273) = 298 K
n
?
From the ideal gas law, PV = nRT, you obtain n =
PV (0.862 atm)(1 L) = = 0.03525 mol RT (0.08206 L • atm/K • mol)(298 K)
Dividing the mass of the vapor by moles gives you the mass per mole (the molar mass). Molar mass =
grams vapor 2.26 g = = 64.114 g/mol moles vapor 0.03525 mol
Therefore, the molecular mass is 64.1 amu.
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167
5.9. First, determine the number of moles of Cl2 from the mass of HCl (molar mass36.46 g/mol) and from the stoichiometry of the chemical equation: 9.41 g HCl x
5 mol Cl2 1 mol HCl x = 0.080653 mol Cl2 16 mol HCl 36.46 g HCl
Tabulate the values of the variables: Variable
Value 1 atm = 1.0355 atm 760 mmHg
P
787 mmHg x
T
(40 + 273) K = 313 K
n
0.080653 mol
? V Rearrange the ideal gas law to obtain V: V =
nRT (0.080653 mol)(0.08206 L • atm/K • mol)(313 K) = 2.001 = 2.00 L = P 1.0355 atm
5.10. Each gas obeys the ideal gas law. In each case, convert grams to moles and substitute into the ideal gas law to determine the partial pressure of each. 1.031 g O2 x P =
nRT (0.0322188)(0.08206 L • atm/K • mol)(291 K) = 0.076936 atm = V 10.0 L
0.572 g CO2 x P =
1 mol O 2 = 0.0322188 mol O2 32.00 g O 2
1 mol CO 2 = 0.012997 mol CO2 44.01 g CO 2
nRT (0.012997)(0.08206 L • atm/K • mol)(291 K) = 0.031036 atm = V 10.0 L
The total pressure is equal to the sum of the partial pressures: P = PO2 + PCO2 = 0.076936 + 0.031036 = 0.10797 = 0.1080 atm The mole fraction of oxygen in the mixture is Mole fraction O2 =
PO2 P
=
0.07694 atm = 0.7122 = 0.712 0.10802 atm
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Chapter 5: The Gaseous State
5.11. Determine the number of moles of O2 from the mass of KClO3 and from the stoichiometry of the chemical reaction. 1.300 g KClO3 x
1 mol KClO3 3 mol O 2 x = 0.0159184 mol O 2 122.5 g KClO3 2 mol KClO3
The vapor pressure of water at 23°C is 21.1 mmHg (Table 5.6). Find the partial pressure of O2 using Dalton's law:
P = PO2 + PH2 O PO2 = P − PH2 O = (745 − 21.1) mmHg = 723.9 mmHg Solve for the volume using the ideal gas law. Variable
Value
1 atm = 0.9525 atm 760 mmHg
P
723.9 mmHg x
V
?
T
(23 + 273) = 296 K
n
0.0159184
From the ideal gas law, PV = nRT, you have
V =
nRT (0.0159184 mol)(0.08206 L • atm/K • mol) (296 K) = = 0.4059 = 0.406 L P 0.9525 atm
5.12. The absolute temperature is (22 + 273) = 295 K. In SI units, the molar mass of carbon tetrachloride, CCl4, is 153.8 x 10−3 kg/mol. Therefore, 1/ 2
⎛ 3RT ⎞ u = ⎜ ⎟ ⎝ M ⎠
1/ 2
⎛ 3 x 8.31 kg • m 2 /(s 2 • K • mol) x 295 K ⎞ = ⎜ ⎟ 153.8 x 103 kg/mol ⎝ ⎠
= 218.7 = 219 m/s
5.13. Determine the rms molecular speed for N2 at 455°C (728 K): 1/ 2
⎛ 3RT ⎞ u = ⎜ ⎟ ⎝ M ⎠
1/ 2
⎛ 3 x 8.31 kg • m 2 /(s 2 • K • mol) x 728 K ⎞ = ⎜ ⎟ 28.02 x 103 kg/mol ⎝ ⎠
= 804.81 m/s
After writing this equation with the same speed for H2, square both sides and solve for T. The molar mass of H2 in SI units is 2.016 x 10−3 kg/mol. Therefore,
T =
u2M (804.81) 2 (2.016 x 103 kg/mol) = = 52.379 = 52.4 K 3R (3)(8.31 kg • m 2 /s 2 • K • mol)
Because the average kinetic energy of a molecule is proportional to only T, the temperature at which an H2 molecule has the same average kinetic energy as an N2 molecule at 455°C is exactly the same temperature, 455°C.
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5.14. The two rates of effusion are inversely proportional to the square roots of their molar masses, so you can write Rate of effusion of O 2 = Rate of effusion of He
M m (He) M m (O 2 )
where Mm(He) is the molar mass of He (4.00 g/mol) and Mm(O2) is the molar mass of O2 (32.00 g/mol). Substituting these values into the formula gives Rate of effusion of O 2 = Rate of effusion of He
4.00 g/mol = 0.35355 32.00 g/mol
Rearranging gives Rate of effusion of O2 = 0.35355 x rate of effusion of He.
Now, the problem states that the rate of effusion can be given in terms of volume of gas effused per second, so Volume of O 2 Volume of He = 0.35355 x Time for He Time for O 2 Substituting in the values gives 10.0 mL 10.0 mL = 0.35355 x Time for O 2 3.52 s
Rearranging gives Time for O2 =
3.52 s = 9.956 = 9.96 s 0.35355
5.15. The problem states that the rate of effusion is inversely proportional to the time it takes for a gas to effuse, so you can write Rate of effusion of H 2 time for gas = = Rate of effusion of gas time for H 2
M m (gas) = 4.67 M m (H 2 )
Rearranging and solving for Mm(H2) gives Mm(gas) = (4.67)2 x Mm(H2) = (4.67)2 x 2.016 g/mol = 43.96 = 44.0 g/mol
Thus, the molecular mass of the gas is 44.0 amu.
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Chapter 5: The Gaseous State
5.16. From Table 5.7, a = 5.570 L2•atm/mol2 and b = 0.06499 L/mol. Substitute these values into the van der Waals equation, along with R = 0.08206 L•atm/K•mol, T = 273.2 K, n = 1.000 mol, and V = 22.41 L. P =
nRT n2 a − V2 (V  nb)
P =
(1.00 mol) (0.08206 L • atm/K • mol)(273.2 K) (1.00 mol) 2 (5.570 L2 • atm/mol2 ) − (22.41 L) 2 22.41 L  (1.00 mol)(0.06499 L/mol)
= 1.0033 − 0.011091 = 0.99221 = 0.992 atm Using the ideal gas law, P = 1.0004 atm (larger).
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ANSWERS TO CONCEPT CHECKS
5.1. The general relationship between pressure (P) and the height (h) of a liquid column in a barometer is P = gdh, where g is the constant acceleration of gravity and d is the density. Examination of the relationship indicates that, for a given pressure, as the density of the liquid in the barometer decreases, the height of the liquid must increase. In order to make this relationship more apparent, you can rearrange the equation to gh =
P d
Because you are conducting the experiment at constant pressure and gravity is a constant, this mathematical relationship demonstrates that the height of the liquid in the barometer is inversely proportional to the density of the liquid in the barometer. (h ∝
1 ) d
This inverse relationship means that as the height of the liquid decreases, the density of the liquid must increase. Since the density of mercury is greater than the density of water, the barometer with the water will have the higher column. 5.2. a.
In the first step, when the temperature decreases, the pressure will also decrease. This is because, according to the combined gas law, the pressure is directly proportional to the temperature (P ∝ T). In the second step, when the volume increases, the pressure will decrease, since, according to Boyle’s law, pressure and volume are inversely related (P ∝ 1/V). Both changes result in the pressure decreasing, so the final pressure will be less than the starting pressure.
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171
b.
In the first step, when the temperature increases, the pressure will also increase. This is because, according to the combined gas law, pressure is directly proportional to the temperature (P ∝ T). In the second step, when the volume decreases, the pressure will increase, since, according to the ideal gas law, pressure and volume are inversely related (P ∝ 1/V). Both changes result in the pressure increasing, so the final pressure will be greater than the starting pressure.
a.
According to Avogadro’s law, equal volumes of any two (or more) gases at the same temperature and pressure contain the same number of molecules (or atoms in this case). Therefore, all three flasks contain the same number of atoms.
b.
Since density is mass divided by volume, and all three flasks have the same volume (3.0 L), the gas with the largest molar mass, xenon (Xe), will have the greatest density.
c.
According to the ideal gas law, PV = nRT, pressure is directly proportional to the temperature. Since the helium flask is being heated, it will have the highest pressure.
d.
Since the three flasks started with the same number of atoms, and hence the same number of moles, they would all still have the same number of moles no matter how the temperatures of the flasks were changed.
a.
In a mixture of gases, each gas exerts the pressure it would exert if it were the only gas in the flask. The pressure of H2 is the same whether it is in the flask by itself or with the Ar. Therefore, the pressure of H2 does not change.
b.
According to the ideal gas law, PV = nRT, pressure (P) is directly proportional to the number of moles (n). Since the number of moles of H2 and the number of moles of Ar are equal, their pressures are also equal.
c.
The total pressure is equal to the sum of the pressures of the H2 gas and the Ar gas in the container. The total pressure will also be equal to twice the pressure of the H2 gas when it was in the flask by itself. It is also equal to twice the pressure that the Ar gas would exert if it were in the flask by itself.
a.
The rate of effusion is inversely proportional to the square root of the molecular mass of the gas at constant temperature and pressure. Thus, He (molecular mass 4.00 amu) will diffuse faster than Ar (molecular weight 39.95 amu) and reach the end of the tube first.
b.
The speed of an atom is directly proportional to the absolute temperature. If you raise the temperature of the Ar, you can make it reach the end of the tube at the same time as the He.
a.
If the real gas molecules occupy a relatively small volume, then the volume of the gas is essentially equal to the volume of the container, the same as for an ideal gas. However, if there were large intermolecular attractions, the pressure would be less than for an ideal gas. Therefore, the pressure would be greater for the ideal gas.
5.3.
5.4.
5.5.
5.6.
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Chapter 5: The Gaseous State
b.
If the real gas molecules occupy a relatively large volume, then the volume available for the gas is less than for an ideal gas, and the pressure will be greater. If there are negligible intermolecular attractions, then the pressure is essentially the same as for an ideal gas. Overall, the pressure would be less for the ideal gas.
c.
Since the effect of molecular volume and intermolecular attractions on the pressure of a real gas are opposite, you cannot determine how the pressure of the two gases compare.
ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
5.1. Pressure is the force exerted per unit area of surface. Force is further defined as mass multiplied by acceleration. The SI unit of mass is kg, of acceleration is m/s2, and of area is m2. Therefore, the SI unit of pressure (Pascal) is given by force mass x acceleration kg x m/s 2 kg = = = = Pa 2 m area area m • s2
Pressure =
5.2. A manometer is a device that measures the pressure of a gas in a vessel. The gas pressure in the flask is proportional to the difference in height between the liquid levels in the manometer (Figure 5.4). 5.3. The general relationship between the pressure (P) and the height (h) of the liquid in a manometer is P = gdh. Therefore, the variables that determine the height of the liquid in a manometer are the density (d) of the liquid and the pressure of the gas being measured. The acceleration of gravity (g) is a constant, 9.81 m/s2. 5.4. From Boyle's law, PV = constant. Because this is true for conditions Pi and Vi as well as conditions Pf and Vf, we can write PfVf = PiVi = constant
Dividing both sides of this equation by Pf gives Vf = Vi x
Pi Pf
5.5. A linear relationship between variables such as x and y is given by the mathematical relation y = a + bx
The variable y is directly proportional to x only if a = 0. 5.6. First, find the equivalent of absolute zero on the Fahrenheit scale. Converting −273.15°C to degrees Fahrenheit, you obtain −459.67°F. Since the volume of a gas varies linearly with the temperature, you get the following linear relationship: V = a + btF
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173
where tF is the temperature on the Fahrenheit scale. Since the volume of a gas is zero at absolute zero, you get 0 = a + b(−459.67), or a = 459.67b The equation can now be rewritten as V = 459.67b + btF = b(459.67 + tF) = bTF
where TF is the temperature in the new absolute scale based on the Fahrenheit scale. The relationship is TF = tF + 459.67
5.7. From Charles's law, V = constant x T. Because this is true for conditions Ti and Vi as well as conditions Tf and Vf, we can write Vf Tf
=
Vi = constant Ti
Multiplying both sides of the equation by Tf gives Vf = Vi x
Tf Ti
5.8. Avogadro's law states that equal volumes of any two gases at the same temperature and pressure contain the same number of molecules. The law of combining volumes states that the volumes of reactant gases at a given pressure and temperature are in ratios of small whole numbers. The combiningvolume law may be explained from Avogadro's law using the reaction N2 + 3H2 → 2NH3 as follows. In this equation, the molecules are in a ratio of small whole numbers, 1 to 3 to 2. In Avogadro’s law, these numbers represent volumes. But since equal volumes contain the same number of molecules, the ratios for volumes are the same as the ratios for molecules. 5.9. The standard conditions are 0°C and 1 atm pressure (STP). 5.10. The molar gas volume, Vm, is the volume of one mole of gas at any given temperature and pressure. At standard conditions (STP), the molar gas volume equals 22.4 L. 5.11. Boyle's law (V ∝1/P) and Charles's law (V ∝ T) can be combined and expressed in a single statement: the volume occupied by a given amount of gas is proportional to the absolute temperature divided by the pressure. In equation form, this is V = constant x
T P
The constant is independent of temperature and pressure but does depend on the amount of gas. For one mole, the constant will have a specific value, denoted as R. The molar volume, Vm, is Vm = R x
T P
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Chapter 5: The Gaseous State
Because Vm has the same value for all gases, we can write this equation for n moles of gas if we multiply both sides by n. This yields the equation nVm =
nRT P
Because Vm is the volume per mole, nVm is the total volume V. Substituting gives V =
nRT P
or
PV = nRT
5.12. The variables in the ideal gas law are P, V, n, and T. The SI units of these variables are pascals (P), cubic meters (V), moles (n), and kelvins (T). 5.13. Use the value of R from Table 5.5 and the conversion factor 1 atm = 760 mmHg. This gives 0.082058
L • atm 760 mmHg L • mmHg x = 62.3640 = 62.364 K • mol 1 atm K • mol
5.14. Six empirical gas laws can be obtained. They can be stated as follows: P x V = constant
(T and n constant)
P = constant T
(V and n constant)
P = constant n
(T and V constant)
V = constant T
(P and n constant)
V = constant n
(P and T constant)
n x T = constant
(P and V constant)
5.15. The postulates and supporting evidence are the following: (1)
Gases are composed of molecules whose sizes are negligible compared with the distance between them.
(2)
Molecules move randomly in straight lines in all directions and at various speeds.
(3)
The forces of attraction or repulsion between two molecules (intermolecular forces) in a gas are very weak or negligible, except when they collide.
(4)
When molecules collide with one another, the collisions are elastic.
(5)
The average kinetic energy of a molecule is proportional to the absolute temperature.
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One example of evidence that supports the kinetic theory of gases is Boyle’s law. Constant temperature means that the average molecular force from collision remains constant. If you increase the volume, you decrease the number of collisions per unit wall area, thus lowering the pressure in accordance with Boyle’s law. Another example is Charles’s law. If you raise the temperature, you increase the average molecular force from a collision with the wall, thus increasing the pressure. For the pressure to remain constant, it is necessary for the volume to increase so that the frequency of collisions with the wall decreases. Thus, when you raise the temperature of a gas while keeping the pressure constant, the volume increases in accordance with Charles’s law. 5.16. Boyle's law requires the temperature to be constant. Postulate 5 of the kinetic theory holds that the average kinetic energy of a molecule is constant at constant temperature. Therefore, the average molecular force from collisions is constant. If we increase the volume of a gas, this decreases the number of molecules per unit volume and so decreases the frequency of collisions per unit wall area, causing the pressure to decrease in accordance with Boyle’s law. 5.17. According to kinetic theory, the pressure of a gas results from the bombardment of container walls by molecules. 5.18. The rms speed of a molecule equals (3RT/Mm)1/2. Therefore, it varies directly with the square root of the absolute temperature. The rms speed does not depend on the molar volume. 5.19. A gas appears to diffuse more slowly because it never travels very far in one direction before it collides with another molecule and moves in another direction. Thus, it must travel a very long, crooked path as the result of collisions. 5.20. Effusion is the process in which a gas flows through a small hole in a container. It results from the gas molecules encountering the hole by chance, rather than by colliding with the walls of the container. The faster the molecules move, the more likely they are to encounter the hole. Thus, the rate of effusion depends on the average molecular speed, which depends inversely on molecular mass. 5.21. The behavior of a gas begins to deviate significantly from that predicted by the ideal gas law at high pressures and at relatively low temperatures. 5.22. The constant a is the proportionality constant in the van der Waals equation related to intermolecular forces The term nb represents the volume occupied by n moles of molecules. 5.23. The answer is a; “the greater the volume occupied by a given amount of gas, the higher the intermolecular force” is not part of the kineticmolecular theory. 5.24. The answer is c, the container volume didn’t change.
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Chapter 5: The Gaseous State
5.25. The answer is b, to occupy the entire 22.4 L volume of the box. 5.26. The answer is d; it is not true that the Ar atoms hit the walls of the flask with the greatest force of the three gases.
■
ANSWERS TO CONCEPT EXPLORATIONS
5.27. a.
For the same volume and temperature, pressure is proportional to the number of molecules in the sample. Since container B contains twice as many particles, the pressure in container B is double the pressure in container A.
b.
Since all the containers have the same volume and temperature, the pressure in containers C, D, and E are all the same. Since container F contains twice as many particles, the pressure is double the pressure in the other containers.
c.
Since pressure is inversely proportional to volume, and since the number of particles is the same for both containers, the pressure in container H is onehalf the pressure in container G.
d.
There would be twice the number of particles in double the volume, so overall, the pressures in containers G and H are now equal.
e.
Pressure is directly proportional to temperature and inversely proportional to volume. Here, the temperature is doubled, effectively doubling the pressure. But the volume is also doubled, effectively reducing the pressure by onehalf. Overall, the pressures in containers J and L are the same.
a.
The gas atom is creating pressure by colliding with the walls of the container. The collisions exert a force on the wall, which is observed as pressure.
b.
For constant volume and temperature, pressure is proportional to the number of particles present. Since there are four times as many particles in container B, the pressure in container B is also four times the pressure in container A.
c.
Since there are four times as many particles in container B, the pressure in container B is also four times the pressure in container A. The pressure is not related to the mass of the particles, only to the number of particles.
d.
The rootmeansquare (rms) speed is inversely proportional to the square root of the particle’s mass. Since the atom is lighter, it will have the larger rms speed. The difference in rms speeds neither supports nor contradicts the answer concerning the pressure in the container, because the force on the wall does not only depend on the speed of the particle alone. The force depends on the momentum, which is the product of mass times speed. This in turn depends only on the temperature. At constant temperature there is a constant force.
e.
For an equal volume and an equal number of particles, the pressure varies directly with the absolute temperature. Since the temperature in container F is double the temperature in container E, the pressure in container F is double the pressure in container E.
5.28.
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177
f.
Since the temperature in container H is double the pressure in container G, the pressure in container H is double the pressure in container G. The pressure is not related to the mass of the particles, only to the number of particles.
g.
Pressure is directly related to the number of particles and directly related to the absolute temperature. Here, the number of particles in container J is double the number of particles in container I, effectively doubling the pressure. But the temperature in container J is onehalf the temperature in container I, effectively reducing the pressure by onehalf. Overall, the pressures in the two containers are the same.
h.
Here, the number of particles in container K is three times the number of particles in container L, and the temperature in container K is double the temperature in container L, so overall the pressure in container K is six times the pressure in container L.
ANSWERS TO CONCEPTUAL PROBLEMS
5.29. a.
The volume of the tire and the amount of air in the tire remain constant. From the ideal gas law, PV = nRT, under these conditions the pressure will vary directly with the temperature (P ∝ T). Thus, on a cold day, you would expect the pressure in the tires to decrease, and they would appear flatter.
b.
Aerosol cans are filled with a fixed amount of gas in a constant volume. From the ideal gas law, under these conditions the pressure will vary directly with the temperature (P ∝ T). If you put an aerosol can in a fire, you will increase the temperature and thus the pressure. If the pressure gets high enough, the can will explode.
c.
As the water bottle sits in the sun, the liquid water warms up. As the temperature of the water increases, so does its vapor pressure (Table 5.6). If the pressure gets high enough, it will pop the lid off the bottle.
d.
The amount of air in the balloon and the temperature remain constant. From the ideal gas law, under these conditions the pressure is inversely proportional to the volume (P ∝ 1/V). Thus, as you squeeze the balloon, you decrease the volume, resulting in an increase in pressure. If you squeeze hard enough and make the volume small enough, the balloon will pop.
5.30. Since the flasks are identical and contain the same amount of gas, the initial pressures in the N2 flask, the O2 flask, and the He flask will be the same. After the changes, the pressure in the He flask would be highest, with a pressure equal to three times the original. Next would be the O2 flask, with a pressure equal to two times the original. Last would be the N2 flask, with a pressure equal to onehalf the original. 5.31. a.
The pressure and volume of a gas are inversely proportional; therefore, an increase by a factor of 2 in pressure would decrease the volume by ½ (C to D).
b.
The pressure and volume of a gas are inversely proportional; therefore, a decrease by a factor of 2 in pressure would double the volume (C to A).
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Chapter 5: The Gaseous State
c.
The volume and temperature of a gas are directly proportional; therefore an increase in kelvin temperature by a factor of 1.5 would result in an increase in volume by a factor of 1.5 (C to B).
d.
Since the piston can move, the pressure would not change (it would be equal to the starting pressure). The volume of gas is directly proportional to the number of moles; therefore, an increase in the number of moles by a factor of 2 would cause the volume increase by a factor of 2 (C to A).
a.
Since 1.0 out of the 3.0 moles of gas in the container is N2, the fraction of the pressure due to N2 is 1/3.
b.
Mole fractions are not a function of temperature, so nothing would happen.
c.
You would expect the pressure to be higher for two reasons. First, the water would occupy some volume, reducing the volume available for the gas to occupy. Thus, according to Boyle’s law, as volume decreases, pressure increases. Second, after a time, the water would evaporate, and the vapor pressure due to the water would contribute to the total pressure, thereby increasing it.
d.
Yes, there is enough information in the problem to calculate the pressure in the flask, but you would also need to know the vapor pressure of water at 22.0°C (Table 5.6).
a.
The container with the O2 has the greater density, since the molar mass of O2 (32.00 g/mol) is greater than that of H2 (2.016 g/mol).
b.
Since the H2 molecules are lighter, they will be moving faster.
c.
Both containers have the same number of molecules (Avogadro’s law).
d.
The pressure in each of the containers will not change when the valve is opened. Each container starts with the same pressure. Since the total volume remains constant, the pressure will not change.
e.
The fraction of the total pressure due to the H2 would now be 1/4.
a.
Pressure decreases as you increase in altitude. Thus, the pressure at 6000 m is less than the pressure at 3000 m. For two identical balloons, the balloon at 6000 m will have the greater volume (Boyle’s law).
b.
In order to calculate the volume of each balloon, you would need the temperature and pressure on the ground and the temperature and pressure at their respective heights.
5.32.
5.33.
5.34.
5.35. In order to double the volume, you could reduce the pressure by 1/2. You could also double the temperature in Kelvin scale, but you cannot determine the final temperature without knowing the initial temperature.
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179
5.36.
■
a.
If you assume that the flasks are at ordinary room temperature, say 25°C, then there would be approximately 1 mol of He (at 1.0 atm) and 1.5 mol of Xe (at 1.5 atm). Thus, the F2 flask (with 2.0 mol) would contain the greater number of moles of gas.
b.
You could decrease the temperature or remove some gases from the other two containers.
SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off.
5.37. Use the conversion factor 1 atm = 760 mmHg. 0.047 atm x
760 mmHg = 35.7 = 36 mmHg 1 atm
5.38. Use the conversion factors 1 atm = 760 mmHg and 1 atm = 101325 Pa, or 101.325 kPa. 259 mmHg x
1 atm 101.325 kPa = 34.53 = 34.5 kPa x 1 atm 760 mmHg
5.39. Using Boyle's law, solve for Vf of the neon gas at 1.292 atm pressure. Vf = Vi x
Pi Pf
= 3.15 L x
0.951 atm = 2.318 = 2.32 L 1.292 atm
5.40. Using Boyle's law, solve for Vf of the helium gas at 632 mmHg. Vf = Vi x
Pi Pf
= 2.68 L x
789 mmHg = 3.3457 = 3.35 L 632 mmHg
5.41. Using Boyle's law, let Vf = volume at 0.974 atm (Pf), Vi = 50.0 L, and Pi = 19.8 atm. Vf = Vi x
Pi Pf
= 50.0 L x
19.8 atm = 1016.4 = 1.02 x 103 L 0.974 atm
5.42. Using Boyle's law, let Vf = volume at 1.212 atm (Pf), Vi = 8.58 m3, and Pi = 1.020 atm. Vf = Vi x
Pi Pf
= 8.58 m3 x
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1.020 atm = 7.220 = 7.22 m3 1.212 atm
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Chapter 5: The Gaseous State
5.43. Using Boyle's law, let Pi = pressure of 315 cm3 of gas, and solve for it. Pi = Pf x
Vf Vi
= 2.51 kPa x
0.0457 cm3 = 3.641 x 10−4 = 3.64 x 10−4 kPa 3 315 cm
5.44. Using Boyle's law, let Pf = final pressure of 27.0 dm3 of gas, and solve for it. Pf = Pi x
Vi Vf
456 dm3 = 1705.7 = 1.71 x 103 kPa 27.0 dm3
= 101 kPa x
5.45. Use Charles's law: Ti = 18°C + 273 = 291 K, and Tf = 0°C + 273 = 273 K. Vf = Vi x
Tf Ti
= 3.92 mL x
273 K = 3.677 = 3.68 mL 291 K
5.46. Use Charles's law: Ti = 0°C + 273 = 273 K, and Tf = 20°C + 273 = 293 K. Vf = Vi x
Tf Ti
= 22.41 L x
293 K = 24.051 = 24.1 L 273 K
5.47. Use Charles's law: Ti = 22°C + 273 = 295 K, and Tf = −197°C + 273 = 76 K. Vf = Vi x
Tf Ti
= 2.54 L x
76 K = 0.654 = 0.65 L 295 K
5.48. Use Charles's law: Ti = 0°C + 273 = 273 K, and Tf = 25°C + 273 = 298 K. Vf = Vi x
Tf Ti
= 4.83 L x
298 K = 5.272 = 5.27 L 273 K
5.49. Use Charles's law: Ti = 25°C + 273 = 298 K, and Vf is the difference between the vessel's volume of 39.5 cm3 and the 7.7 cm3 of ethanol that is forced into the vessel. Tf = Ti x
Vf Vi
= 298 K x
(39.5  7.7) cm3 = 239.9 K (−33.2 or −33°C) 39.5 cm3
5.50. Use Charles's law: Ti = 18°C + 273 = 291 K, Vi = 62.3 cm3, and Vf = 61.2 cm3. Tf = Ti x
Vf Vi
= 291 K x
61.2 cm3 = 285.8 = 286 K (12.7 or 13°C) 62.3 cm3
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181
5.51. Use the combined law: Ti = 31°C + 273 = 304 K, and Tf = 0°C + 273 = 273 K. Vf = Vi x
Pi Pf
Tf
x
Ti
= 35.5 mL x
753 mmHg 273 K x = 31.58 = 31.6 mL 760 mmHg 304 K
5.52. Use the combined law: Ti = 23°C + 273 = 296 K, and Tf = 0°C + 273 = 273 K. Vf = Vi x
Pi Pf
Tf
x
Ti
= 3.84 mL x
785 mmHg 273 K x = 3.658 = 3.66 mL 760 mmHg 296 K
5.53. The balanced equation is 4NH3 + 5O2 → 4NO + 6H2O The ratio of moles of NH3 to moles of NO is 4 to 4, or 1 to 1, so one volume of NH3 will produce one volume of NO at the same temperature and pressure. 5.54. The balanced equation is CO2 + 3H2 → CH3OH + H2O The ratio of moles of H2 to moles of CO2 is 3 to 1, so three volumes of H2 are required to react with one volume of CO2 at the same temperature and pressure. 5.55. Solve the ideal gas law for V: V =
nRT ⎛ 1⎞ = nRT ⎜ ⎟ P ⎝ P⎠
If the temperature and number of moles are held constant, then the product nRT is constant, and volume is inversely proportional to pressure: V = constant x
1 P
5.56. Solve the ideal gas law for V: V =
nR x T P
If n and P are held constant, the nR/P quotient is a constant. The equation may now be written V = constant x T
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Chapter 5: The Gaseous State
5.57. Calculate the moles of oxygen, and then solve the ideal gas law for P: n = 91.3 g x P =
1 mol O 2 = 2.853 mol O2 32.00 g O 2
nRT (2.853 mol) (0.08206 L • atm / K • mol) (294 K) = = 8.022 = 8.02 atm V 8.58 L
5.58. Calculate the moles of methane, and then solve the ideal gas law for P: n = 7.13 g x P =
1 mol CH 4 = 0.4456 mol 16.0 g CH 4
nRT (0.4456 mol)(0.08206 L • atm/K • mol)(292 K) = V 5.00 L
= 2.1355 = 2.14 atm 5.59. Using the moles of chlorine, solve the ideal gas law for V: V =
nRT (3.50 mol)(0.08206 L • atm/K • mol) (307 K) = = 22.04 = 22.0 L P 4.00 atm
5.60. Calculate the moles of oxygen, and then solve the ideal gas law for V: n = 5.67 g x V =
1 mol O 2 = 0.17718 mol 32.00 g O 2
nRT (0.17718 mol)(0.08206 L • atm/K • mol)(296 K) = P 0.985 atm
= 4.369 = 4.37 L 5.61. Solve the ideal gas law for temperature in K, and convert to °C: T =
PV (3.50 atm) (4.00 L) = = 416.1 = 416 K nR (0.410 mol)(0.08206 L • atm/K • mol)
°C = 416 − 273 = 143°C 5.62. Find the moles of C3H8, solve the ideal gas law for temperature in K, and convert to °C: n = 5.65 g x T =
1 mol C3 H8 = 0.12814 mol 44.09 g C3 H8
PV (741/760 atm) (2.50 L) = = 231.79 K nR (0.12814 mol)(0.08206 L • atm/K • mol)
°C = 231.79 − 273 = −41.20 = −41°C
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5.63. Because density equals mass per unit volume, calculating the mass of 1 L (exact number) of gas will give the density of the gas. Start from the ideal gas law, and calculate n; then convert the moles of gas to grams using the molar mass. n =
PV (751/760 atm) (1 L) = = 0.03961 mol RT (0.08206 L • atm/K • mol) (304 K) 17.03 g = 0.67458 = 0.675 g 1 mol
0.03961 mol x
Therefore, the density of NH3 at 31°C is 0.675 g/L. 5.64. As in the previous problem, calculate the mass of 1 L (exact number) of gas to obtain the density of the gas. Start from the ideal gas law, and calculate n; then convert to grams. n =
PV (967/760 atm) (1 L) = = 0.04712 mol RT (0.08206 L • atm/K • mol) (329 K)
34.08 g = 1.6058 g 1 mol
0.04712 mol x
Therefore, the density of H2S at 56°C is 1.61 g/L. 5.65. Calculate the mass of 1 L (exact number) using the ideal gas law; convert to mass. n =
PV (0.897 atm) (1 L) = = 0.036804 mol RT (0.08206 L • atm/K • mol) (297 K)
0.036804 mol x
58.12 g = 2.139 = 2.14 g 1 mol
Therefore, the density of C4H10 is 2.14 g/L. 5.66. Calculate the mass of 1 L (exact number) using the ideal gas law; convert to mass. n =
PV (797/760 atm) (1 L) = = 0.034446 mol RT (0.08206 L • atm/K • mol) (371 K)
0.034446 mol x
119.5 g = 4.116 = 4.12 g 1 mol
Therefore, the density of CHCl3 at 98°C is 4.12 g/L. 5.67. The ideal gas law gives n moles, which then are divided into the mass of 1.585 g for molar mass. n =
PV (753/760 atm) (1 L) = = 0.033262 mol RT (0.08206 L • atm/K • mol) (363 K)
Molar mass =
1.585 g = 47.651 g/mol 0.033262 mol
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Chapter 5: The Gaseous State
The molecular weight is 47.7 amu. 5.68. The moles in 237 mL (0.237 L) of the compound are obtained from the ideal gas law. Dividing the mass (0.548 g) of the gas by the moles gives the molar mass and molecular mass. n =
PV (755/760 atm) (0.237 L) = = 0.007692 mol RT (0.08206 L • atm/K • mol) (373 K)
Molar mass =
0.548 g = 71.24 g/mol 0.007692 mol
The molecular mass is 71.2 amu. 5.69. The moles in 250 mL (0.250 L) of the compound are obtained from the ideal gas law. The 2.56g mass of the gas then is divided by the moles to obtain the molar mass and molecular mass. n =
PV (786/760 atm) (0.250 L) = = 0.007996 mol RT (0.08206 L • atm/K • mol) (394 K)
Molar mass =
2.56 g = 3.201 x 102 = 3.20 x 102 g/mol 0.007996 mol
The molecular mass is 3.20 x 102 amu. 5.70. The moles in 345 mL (0.345 L) of the compound are obtained from the ideal gas law. The 2.30g mass of the gas then is divided by the moles to obtain the molar mass and molecular mass. n =
PV (985/760 atm) (0.345 L) = = 0.01294 mol RT (0.08206 L • atm/K • mol) (421 K)
Molar mass =
2.30 g = 177.7 g/mol 0.01294 mol
The molecular mass is 178 amu. 5.71. For a gas at a given temperature and pressure, the density depends on molecular mass (or, for a mixture, on average molecular mass). Thus, at the same temperature and pressure, the density of NH4Cl gas would be greater than that of a mixture of NH3 and HCl, because the average molecular mass of NH3 and HCl would be lower than that of NH4Cl. 5.72. For a gas at a given temperature and pressure, the density depends on molecular mass (or, for a mixture, on average molecular mass). Thus, at the same temperature and pressure, the density of PCl5 gas would be greater than that of a mixture of PCl3 and Cl2 formed by the decomposition of PCl5, because the average molecular mass of PCl3 and Cl2 would be lower than that of PCl5. 5.73. The 0.050 mol CaC2 will form 0.050 mol C2H2. The volume is found from the ideal gas law: V =
(0.050 mol)(0.08206 L • atm/K • mol)(299 K) = 1.36 = 1.4 L 684/760 atm
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5.74. The 0.0840 mol Mg will form 0.0840 mol of H2. The volume is found from the ideal gas law: V =
(0.0840 mol)(0.08206 L • atm/K • mol)(306 K) = 2.410 = 2.41 L 665/760 atm
5.75. Use the equation to obtain the moles of CO2 and then the ideal gas law to obtain the volume. 2LiOH(s) + CO2(g) → Li2CO3(aq) + H2O(l) 1 mol CO 2 1 mol LiOH = 6.8267 mol CO2 x 2 mol LiOH 23.95 g LiOH
327 g LiOH x V =
(6.8267 mol)(0.08206 L • atm/K • mol)(294 K) = 160.2 = 160. L 781/760 atm
5.76. Use the equation to obtain the moles of ammonia and then the ideal gas law to obtain the volume. Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g) 4.56 g Mg3N2 x V =
1 mol Mg 3 N 2 2 mol NH3 x = 0.090386 mol NH3 100.9 g Mg 3 N 2 1 mol Mg 3 N 2
(0.090386 mol)(0.08206 L • atm/K • mol)(297 K) = 2.223 = 2.22 L (753/760) atm
5.77. Use the equation to obtain the moles of ammonia and then the ideal gas law to obtain the volume. 2NH3(g) + CO2(g) → NH2CONH2(aq) + H2O(l) 908 g urea x V =
2 mol NH 3 1 mol urea = 30.23 mol NH3 x 1 mol urea 60.06 g urea
(30.23 mol)(0.08206 L • atm/K • mol)(298 K) = 246.4 = 246 L 3.00 atm
5.78. Use the equation to obtain the moles of ammonia and then the ideal gas law to obtain the volume. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) 50.0 g NO x V =
5 mol O 2 1 mol NO x = 2.083 mol O2 4 mol NO 30.0 g NO
(2.083 mol)(0.08206 L • atm/K • mol)(308 K) = 24.487 = 24.5 L 2.15 atm
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Chapter 5: The Gaseous State
5.79. Use the equation to obtain the moles of ammonia and then the ideal gas law to obtain the volume. 2NH3(g) + H2SO4(aq) → (NH4)2SO4(aq) 150.0 g (NH4)2SO4 x
2 mol NH 3 1 mol ( NH 4 ) 2SO 4 x 132.1 g ( NH 4 )2SO 4 1 mol ( NH 4 ) 2SO 4
= 2.2710 mol NH3 nRT (2.2710 mol) (0.08206 L • atm/K • mol) (288 K) V = = = 46.67 = 46.7 L P 1.15 atm 5.80. Use the equation to obtain the moles of ammonia, and then the ideal gas law to obtain the volume. 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) 26.8 g NaHCO3 x V =
1 mol NaHCO3 1 mol CO 2 = 0.15952 mol CO2 x 2 mol NaHCO3 84.0 g NaHCO3
nRT (0.15952 mol)(0.08206 L • atm/K • mol)(350 K) = = 4.6059 = 4.61 L P (756/760) atm
5.81. Calculate the partial pressure of each gas; then add the pressures since the total pressure is equal to the sum of the partial pressures: P(He) =
(0.0200 mol)(0.08206 L • atm/K • mol)(283 K) = 0.18578 atm 2.50 L
P(H2) =
(0.0100 mol)(0.08206 L • atm/K • mol)(283 K) 2.50 L
= 0.09289 atm
Total pressure = 0.18587 = 0.09289 = 0.27867 = 0.279 5.82. Calculate the partial pressure of each gas; then add the pressures since the total pressure is equal to the sum of the partial pressures: P(He) =
(0.0300 mol)(0.08206 L • atm/K • mol)(293 K) = 0.1803 atm 4.00 L
P(O2) =
(0.0200 mol)(0.08206 L • atm/K • mol)(293 K) = 0.1202 atm 4.00 L
Total pressure = 0.1803 + 0.1202 = 0.3005 = 0.301 atm
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187
5.83. Convert mass of O2 and mass of He to moles. Use the ideal gas law to calculate the partial pressures, and then add to obtain the total pressures. 0.00103 g O2 x
1 mol O 2 = 3.219 x 10−5 mol O2 32.00 g O 2
0.00056 g He x
1 mol He = 1.40 x 10−4 mol 4.00 g He
P =
nRT (3.219 x 105 mol)(0.08206 L • atm/K • mol)(288 K) = 0.2000 L V
= 0.003804 atm O2 nRT (1.40 x 104 mol)(0.08206 L • atm/K • mol)(288 K) P = = 0.2000 L V P = PO2 + PHe
= 0.01654 atm He = 0.003804 atm + 0.01654 atm = 0.0203 = 0.020 atm
5.84. Obtain the partial pressure of helium using Dalton's law; then use the ideal gas law to obtain the mass of helium. PHe = P − PO2 = 3.00 atm − 0.200 atm = 2.800 atm nHe =
PHeV (2.80 atm) (10.0 L) = = 1.1645 mol He RT (0.08206 L • atm/K • mol)(293 K)
MassHe = 1.1645 mol He x
4.003 g He = 4.661 = 4.66 g He 1 mol He
5.85. For each gas, P(gas) = P x (mole fraction of gas). P(H2)
= 760 mmHg x 0.250 = 190.0
= 190 mmHg
P(CO2)
= 760 mmHg x 0.650 = 494.0
= 494 mmHg
P(HCl)
= 760 mmHg x 0.054 = 41.04
= 41 mmHg
P(HF)
= 760 mmHg x 0.028 = 21.28
= 21 mmHg
P(SO2)
= 760 mmHg x 0.017 = 12.92
= 13 mmHg
P(H2S)
= 760 mmHg x 0.001 = 0.76
= 0.8 mmHg
5.86. For each gas, P(gas) = P x (mole fraction of gas). P(He)
= 6.91 atm x 0.790
= 5.4589
= 5.46 atm
P(N2)
= 6.91 atm x 0.170
= 1.1747
= 1.17 atm
P(O2)
= 6.91 atm x 0.040
= 0.2764
= 0.28 atm
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Chapter 5: The Gaseous State
5.87. The total pressure is the sum of the partial pressures of CO and H2O, so PCO = P − Pwater = 689 mmHg − 23.8 mmHg = 665.2 mmHg PCOV (665/760 atm)(3.85 L) = = 0.1378 mol CO RT (0.08206 L • atm/K • mol)(298 K)
nCO =
1 mol HCOOH 46.03 g HCOOH x = 6.342 = 6.34 g HCOOH 1 mol CO 1 mol HCOOH
0.1378 mol CO x
5.88. The total pressure is the sum of the partial pressures of N2 and H2O, so ⎛ 1 atm ⎞ ⎛ 1 atm ⎞ PN2 = P − PH2 O = 97.8 kPa x ⎜ ⎟ − 25.2 mmHg x ⎜ ⎟ ⎝ 101.3 kPa ⎠ ⎝ 760 mmHg ⎠ = 0.9322 atm nN2 =
PN2 V
=
RT
(0.9322 atm) (3.75 L) = 0.1424 mol N2 0.08206 L • atm/ K • mol) (299 K) 1 mol NH 4 NO2 64.04 g NH 4 NO2 x = 9.124 = 9.12 g NH4NO3 1 mol N 2 1 mol NH 4 NO2
0.1424 mol N2 x
5.89. Substitute 298 K (25°C) and 398 K (125°C) into Maxwell's distribution: 1/ 2
1/ 2
u25
⎛ 3 x 8.31 kg • m 2 /(s 2 • K • mol) x 298 K ⎞ = ⎜ ⎟ 28.02 x 103 kg/mol ⎝ ⎠
⎛ 3RT ⎞ = ⎜ ⎟ ⎝ M ⎠
= 5.149 x 102 = 5.15 x 102 m/s 1/ 2
1/ 2
⎛ 3 x 8.31 kg • m 2 /(s 2 • K • mol) x 398 K ⎞ = ⎜ ⎟ 28.02 x 103 kg/mol ⎝ ⎠
⎛ 3RT ⎞ u125 = ⎜ ⎟ ⎝ M ⎠
= 5.9507 x 102 = 5.95 x 102 m/s Graph as in Figure 5.25. 5.90. Substitute 296 K (23°C) into Maxwell's distribution: 1/ 2
1/ 2
u23
⎛ 3 x 8.31 kg • m 2 /(s 2 • K • mol) x 296 K ⎞ = ⎜ ⎟ 159.8 x 103 kg/mol ⎝ ⎠
⎛ 3RT ⎞ = ⎜ ⎟ ⎝ M ⎠
= 214.8 = 215 m/s
The rms speed is the same at 1.50 atm as at 1.00 atm because the temperature is the same. 5.91. Substitute 330 K (57°C) into Maxwell's distribution: 1/ 2
u330 K
⎛ 3RT ⎞ = ⎜ ⎟ ⎝ M ⎠
1/ 2
⎛ 3 x 8.31 kg • m 2 /(s 2 • K • mol) x 330 K ⎞ = ⎜ ⎟ 352 x 103 kg/mol ⎝ ⎠ = 1.528 x 102 = 1.53 x 102 m/s
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189
5.92. Substitute 365 K into Maxwell's distribution: 1/ 2
⎛ 3RT ⎞ u365 K = ⎜ ⎟ ⎝ M ⎠
1/ 2
⎛ 3 x 8.31 kg • m 2 /(s 2 • K • mol) x 365 K ⎞ = ⎜ ⎟ 2.016 x 103 kg/mol ⎝ ⎠ = 2.124 x 103 = 2.12 x 103 m/s
This is equal to 2.12 km/s, which is less than the escape velocity. 5.93. Because u(CO2) = u(H2), we can equate the two righthand sides of the Maxwell distributions: 1/ 2
⎛ 3RT (CO 2 ) ⎞ ⎜ ⎟ ⎝ M m (CO 2 ) ⎠
1/2
⎛ 3RT (H 2 ) ⎞ = ⎜ ⎟ ⎝ M m (H 2 ) ⎠
Squaring both sides, rearranging to solve for T(CO2), and substituting numerical values, we have T(CO2) = T(H2) x
M m (CO 2 ) 44.01 g/mol = 298 K x = 6505.4 K M m (H 2 ) 2.016 g/mol
Thus the temperature is T = 6505.4 − 273 = 6232 = 6.23 x 103 °C 5.94. Substitute 375 m/s into the Maxwell distribution, square both sides, and solve for T: 1/ 2
⎛ 3 x 8.31 kg • m 2 /(s 2 • K • mol) x T ⎞ u = 400 m/s = ⎜ ⎟ 32.00 x 103 kg/ mol ⎝ ⎠ (400)2 = 160000 =
3 x 8.31 x T 32.00 x 103 K
T = 205.37 K = −67.77 = −67.8 °C 5.95. The rates of effusion are directly related to the rms speed. Assume the temperature is the same for both, and write a ratio of two Maxwell distributions. 1/ 2
uN uO
2
2
⎛ 3RT ⎞ 1/ 2 1/ 2 ⎜ ⎟ M m (N 2 ) ⎠ ⎛ M m (O 2 ) ⎞ ⎛ 32.00 g/mol ⎞ ⎝ = = ⎜ ⎟ =⎜ ⎟ 1/ 2 ⎝ 28.02 g/mol ⎠ ⎛ 3RT ⎞ ⎝ M m (N 2 ) ⎠ ⎜ ⎟ ⎝ M m (O 2 ) ⎠ = 1.0686 = 1.069
Thus the ratio of the rates of effusion of N2 and O2 is 1.069 to 1.
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Chapter 5: The Gaseous State
5.96. The rates of effusion are directly related to the rms speed. Assume the temperature is the same for both, and write a ratio of two Maxwell distributions. 1/ 2
uH uH
2 Se
2
=
⎛ 3RT ⎞ ⎜ ⎟ ⎝ M m (H 2 ) ⎠
1/ 2
1/ 2
⎛ 3RT ⎞ ⎜ ⎟ ⎝ M m (H 2Se) ⎠
⎛ M (H Se) ⎞ = ⎜ m 2 ⎟ ⎝ M m (H 2 ) ⎠
1/ 2
⎛ 80.976 g/mol ⎞ =⎜ ⎟ ⎝ 2.016 g/mol ⎠
= 6.3377 = 6.338
Thus the ratio of the rates of effusion of H2 and H2Se is 6.338 to 1. 5.97. The rates of effusion are directly related to the rms speed. Assume the temperature is the same for both, and write a ratio of two Maxwell distributions.
uH uI
1/ 2
2
2
⎛ M (I ) ⎞ = ⎜ m 2 ⎟ ⎝ M m (H 2 ) ⎠
1/ 2
⎛ 253.8 g/mol ⎞ =⎜ ⎟ ⎝ 2.016 g/mol ⎠
= 11.220 = 11.22
Because hydrogen diffuses 11.22 times as fast as iodine, the time it would take would be 1/11.22 of the time required for iodine:
t(H2) = 39 s x (1/11.22) = 3.47 = 3.5 s 5.98. The rates of effusion are directly related to the rms speed. Assume the temperature is the same for both, and write a ratio of two Maxwell distributions.
uHe uN
1/ 2
2
2
⎛ M (N ) ⎞ = ⎜ m 2 ⎟ ⎝ M m (He) ⎠
1/ 2
⎛ 28.02 g/mol ⎞ =⎜ ⎟ ⎝ 4.003 g/mol ⎠
= 2.6457 = 2.646
Because helium diffuses 2.646 times as fast as nitrogen, the time it would take would be 1/2.646 of the time required for nitrogen:
t(He) = 10.6 hr x (1/2.646) = 4.006 = 4.01 hr 5.99. Since the volume of gas produced is directly related to the rate of effusion, the ratio of the volumes of gas can be written in terms of the ratio of the rms speeds. This in turn can be written in terms of the masses of the particles. 1/ 2 1/ 2 ugas ⎛ M m (Ar) ⎞ ⎛ 39.95 g/mol ⎞ 4.83 mL gas = ⎜ = = ⎜ ⎟ ⎟ uAr 9.23 m L Ar ⎝ M m (gas) ⎠ ⎝ M m (gas) ⎠
Squaring both sides gives 0.27383 =
39.95 g/mol M m (gas)
Solving for the molar mass gives
Mm(gas) =
39.95 g/mol = 145.8 g/mol; molecular mass = 146 amu 0.27383
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191
5.100. The time it takes for a fixed volume of gas to effuse through a hole is inversely related to the rate of effusion. The ratio of the times can thus be written in terms of the rms speeds and, in turn, the molar masses. 1/ 2 1/ 2 ugas ⎛ M m (N 2 ) ⎞ ⎛ 28.02 g/mol ⎞ 68.3 s (N 2 ) = ⎜ = ⎜ = ⎟ ⎟ uN 85.6 s (gas) M m (gas) ⎠ ⎝ ⎝ M m (gas) ⎠ 2
Squaring both sides gives 0.636639 =
28.02 g/mol M m (gas)
Solving for the molar mass gives
Mm(gas) =
28.02 g/mol = 44.01 g/mol; molecular mass = 44.0 amu 0.636639
5.101. Solving the van der Waals equation for n = 1 and T = 355.2 K for P gives
P =
RT a (0.08206 L • atm/K • mol)(355.2 K) 12.56 L2 • atm − 2 = − (30.00 L) 2 V (30.00 L  0.08710 L) (V  b)
P = 0.974419 − 0.0139555 = 0.96046 = 0.9605 atm P(ideal gas law) = 0.97159 atm 5.102. Solving the van der Waals equation for n = 1 and T = 393.2 K for P gives
P =
RT a (0.08206 L • atm/K • mol)(393.2 K) 5.537 L2 • atm − 2 = − (32.50 L) 2 V (32.50 L  0.03049 L) (V  b)
P = 0.99373 − 0.0052421 = 0.988490 = 0.9885 atm P(ideal gas law) = 0.99279 atm 5.103. To calculate a/V2 in the van der Waals equation, we obtain V from the ideal gas law at 1.00 atm:
V =
RT (0.08206 L • atm/K • mol)(273 K) = = 22.40 L P 1.00 atm
a 5.570 = 1.110 x 10−2 = V2 (22.4 L) 2 At 1.00 atm, V = 22.4 L, and a/V2 = 1.110 x 10−2. Substituting into the van der Waals equation:
V =
RT a P + 2 V
+b =
(0.08206 L • atm/ K • mol) (273 K) + 0.06499 = 22.22 = 22.2 L (1.00 + 1.110 x 102 ) atm
At 10.0 atm, the van der Waals equation gives 2.08 L. The ideal gas law gives 22.4 L for 1.00 atm and 2.24 L for 10.0 atm.
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Chapter 5: The Gaseous State
5.104. To substitute for a/V2 in the van der Waals equation, we obtain V from the ideal gas law at 1.00 atm: V =
RT (0.08206 L • atm/K • mol)(273 K) = = 22.40 L P 1.00 atm
a 1.382 = = 2.754 x 10−3 2 V (22.4 L) 2
At 1.00 atm, V = 22.4 L, and a/V2 = 2.75 x 10−3. Substituting into the van der Waals equation: V =
RT a P + 2 V
+b =
(0.08206 L • atm/ K • mol) (273 K) + 0.03186 = 22.37 = 22.4 L (1.00 + 2.75 x 103 ) atm
At 10.0 atm, the van der Waals equation gives 2.21 L. The ideal gas law gives 22.4 L for 1.00 atm and 2.24 L for 10.0 atm.
■
SOLUTIONS TO GENERAL PROBLEMS
5.105. Calculate the mass of 1 cm2 of the 20.5 m of water above the air in the glass. The volume is the product of the area of 1 cm2 and the height of 20.5 x 102 cm (20.5 m) of water. The density of 1.00 g/cm3 must be used to convert volume to mass: m = dxV m = 1.00 g/cm3 x (1.00 cm2 x 20.5 x 102 cm) = 2.05 x 103 g, or 2.05 kg The pressure exerted on an object at the bottom of the column of water is P =
force (m)(g) (2.05 kg)(9.807 m/s 2 ) = 2.01 x 105 kg/ms2 = = 2 2 area area ⎛ 10 m ⎞ (1.00 cm 2 ) ⎜ ⎟ ⎝ 1 cm ⎠
= 2.01 x 105 Pa The total pressure on the air in the tumbler equals the sum of the barometric pressure and the water pressure: P = 1.00 x 102 kPa + 2.01 x 102 kPa = 3.01 x 102 kPa
Multiply the initial volume by a factor accounting for the change in pressure to find Vf : Vf = Vi x
Pi Pf
⎛ 1.00 x 102 kPa ⎞ 3 = 243 cm3 x ⎜ ⎟ = 80.73 = 80.7 cm 2 3.01 x 10 kPa ⎝ ⎠
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5.106. The volume of 1 m2 of the 40.0 m of water above the air is 40.0 m3. The mass of water above this area is m = d x V =
1.025 g ⎛ 1 cm ⎞ x ⎜ 2 ⎟ 3 1 cm ⎝ 10 m ⎠
3
x 40.0 m3 = 4.1000 x 107 g (4.1000 x 104 kg)
The pressure on this area is P =
force (m)(g) (4.100 x 104 kg)(9.807 m/s 2 ) = = = 4.0208 x 105 kg/ ms2 or Pa area area 1 m2 1 atm = 3.9692 atm 1.013 x 105 Pa
P = 4.0208 x 105 Pa x
The total pressure at 40.0 m is P = Pair + Pwater = 1.00 atm + 3.9692 atm = 4.9692 atm
The density of a gas is directly proportional to the pressure, so d2 = d1 x
P2 ⎛ 4.9692 atm ⎞ = (1.205 g/L) x ⎜ ⎟ = 5.987 = 5.99 g/L P1 ⎝ 1.00 atm ⎠
5.107. Use the combined gas law and solve for Vf : Vf = Vi x
Pi Pf
x
Tf Ti
= 201 mL x
738 mmHg 273 K x = 181.24 = 181 mL 760 mmHg 294 K
5.108. Use the combined gas law and solve for Vf : Vf = Vi x
Pi Pf
x
Tf Ti
11.0 atm 273 K x = 122.98 = 123 L 1.00 atm 293 K
= 12.0 L x
5.109. Use the combined gas law and solve for Vf : Vf = Vi x
Pi Pf
x
Tf Ti
= 5.0 dm3 x
100.0 kPa 293 K x = 6.46 = 6.5 dm3 79.0 kPa 287 K
5.110. Let Vi = 1 volume (vol) in the combined gas law; solving for Vf will give the relative final volume, or factor, for the increase: Vf = Vi x
Pi Pf
x
Tf Ti
= 1 vol x
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1.00 atm 252 K x = 875.0 vol, or 875 times 3 1.00 x 10 atm 288 K
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Chapter 5: The Gaseous State
5.111. Use the ideal gas law to calculate the moles of helium, and combine this with Avogadro's number to obtain the number of helium atoms: n =
PV (765/760 atm) (0.01205 L) = = 4.993 x 10−4 mol RT (0.08206 L • atm/K • mol)(296 K)
4.993 x 10−4 mol He x
6.022 x 1023 He 2+ ions 1 atom x = 3.0067 x 1020 1 mol He 1 He 2+ ion
= 3.01 x 1020 atoms 5.112. Use the ideal gas law to calculate the moles of nitrogen, and combine this with the molar mass to obtain the mass of nitrogen. Then calculate the mass percentage: n =
PV (749/760 atm)(0.00159 L) = = 6.407 x 10−5 mol • • RT (0.08206 L atm/K mol)(298 K)
6.407 x 10−5 mol N2 x Percent N =
28.02 g N 2 = 1.7955 x 10−3 g N (1.7955 mg N) 1 mol N 2
mass N 1.7955 mg N x 100% = x 100% = 20.52 = 20.5% mass comp. 8.75 mg
5.113. Calculate the molar mass, Mm, by dividing the mass of 1 liter of air by the moles of the gas from the ideal gas equation: Mm =
mass (0.082058 L • atm / K • mol) (273.15 K) = 1.2929 g air x (1 atm)(1 L) n
= 28.9792 g/mol = 28.979 g/ mol (amu) 5.114. First, calculate the molar mass, Mm, by dividing the mass of 1 liter of air by the moles of the gas from the ideal gas equation: Mm =
mass (0.08206 L • atm / K • mol) (293 K) = 1.22 g gas x = 29.33 g/mol (1 atm)(1 L) n
Find the empirical formula from the 80.0% C and 20% H by assuming 1 g of compound and calculating the moles: 0.800 g C x
1 mol C = 0.06661 mol C 12.01 g C
0.200 g H x
1 mol H = 0.1984 mol H 1.008 g H
0.06661 mol C = 1.00, or 1 mol C; 0.06661
0.1984 mol H = 2.98, or 3 mol H 0.06661
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195
The simplest formula is CH3, whose empirical formula mass is 15.03 g/mol. The number of CH3 units contained in the molecular mass of 29.33 is Molecular mass 29.33 = = 1.95, or 2 Formula mass 15.03 The molecular formula is (CH3)2, or C2H6. 5.115. Use the ideal gas law to calculate the moles of CO2. Then convert to mass of LiOH. n =
PV (1.00 atm) (5.8 x 102 L) = = 25.89 mol CO2 RT (0.08206 L • atm/K • mol)(273 K)
25.89 mol CO2 x
2 mol LiOH 23.95 g LiOH x = 1240 = 1.2 x 103 g LiOH 1 mol LiOH 1 mol CO 2
5.116. Use the ideal gas law to calculate moles of CO2. Then convert to mass of pyruvic acid. n =
PV (349/760 atm) (0.0212 L) = = 3.915 x 10−4 mol CO2 RT (0.08206 L • atm/K • mol)(303 K)
3.915 x 10−4 mol CO2 x
1 mol C3 H 4 O3 88.06 g C3 H 4 O3 x 1 mol CO 2 1 mol C3 H 4 O3
= 0.03447 = 0.0345 g C3H4O3 5.117. Convert mass to moles of KClO3, and then use the equation below to convert to moles of O2. Use the ideal gas law to convert moles of O2 to pressure at 25°C (298 K). 2KClO3(s) → 2KCl(s) + 3O2(g) 170.0 g KClO3 x P =
1 mol KClO3 3 mol O 2 x = 2.080 mol O2 2 mol KClO3 122.55 g KClO3
nRT (2.080 mol) ( 0.08206 L • atm/ K • mol) (298 K) = = 20.353 = 20.4 atm O2 V 2.50 L
5.118. Convert mass to moles of KCHO2, and then use the equation below to convert to moles of H2. Use the ideal gas law to convert moles of H2 to pressure at 25°C (298 K). KCHO2(s) + KOH(s) → K2CO3(s) + H2(g) 50.0 g KCHO2 x
1 mol KCHO 2 = 0.59438 mol KCHO2 84.12 g KCHO 2
From the equation, 0.59438 mol of KCHO2 produces 0.59438 mol of H2. P =
(0.59438 mol O 2 )(0.08206 L • atm/K • mol)(298 K) nRT = = 5.814 = 5.81 atm H2 2.50 L V
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Chapter 5: The Gaseous State
5.119. Find the number of moles of CO2 first. Then convert this to moles of HCl and molarity of HCl. n =
( 727 / 760 atm )
x 0.141 L
0.08206 L • atm/K • mol x 300 K
= 0.0054788 mol
mol HCl = 0.0054788 mol CO2 x (2 mol HCl /1 mol CO2 ) = 0.010957 mol M HCl = 0.010957 mol HCl ÷ 0.0249 L HCl = 0.4400 = 0.440 mol/L HCl
5.120. Find the number of moles of CO2 first. Then convert this to moles of HCl and molarity of HCl. n =
731/760 atm x 0.159 L = 0.006296 mol 0.08206 L • at m/ K • mol x 296 K
mol HCl = 0.006296 mol CO2 x Molarity =
2 mol HCl = 0.01259 mol 1 mol CO 2
0.01259 mol = 0.6770 = 0.677 M HCl 0.0186 L
5.121. The number of moles of carbon dioxide is n =
646 / 760 atm x 0.1500 L = 0.005179 mol 0.08206 L • at m/ K • mol x 300 K
The number of moles of molecular acid used is 0.1250 mol/L x 0.04141 L = 0.005176 mol acid Thus, the acid is H2SO4 since 1 mole of H2SO4 reacts to form 1 mole of CO2 5.122. The number of moles of carbon dioxide is n =
722 / 760 atm x 0.1250 L = 0.004990 mol 0.08206 L • at m/ K • mol x 290 K
The number of moles of molecular acid used is 0.2040 mol/L x 0.04990 L = 0.0099756 mol acid Thus, the acid is HCl since the ratio of acid to CO2 is 0.0099756 ÷ 0.004990 = 2.00 to 1. 5.123. Use Maxwell's distribution to calculate the temperature in kelvins; then convert to °C. T =
u2M m (0.605 x 103 m/s)2 (17.03 x 103 kg/mol) = = 250.0 = 250. K (−23°C) 3R 3 (8.31 kg • m 2 /s 2 • K • mol)
5.124. Use Maxwell's distribution to calculate the temperature in kelvins. T =
u2M m (3.53 x 103 m/s) 2 (4.003 x 103 kg/mol) = = 2.0008 x 103 = 2.00 x 103 K 2 2 3R 3 (8.31 kg • m / s • K • mol )
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5.125. Calculate the ratio of the rootmeansquare (rms) molecular speeds, which is the same as the ratio of the rates of effusion: 1/ 2
⎛ M [U(238)F6 ] ⎞ u [U(235)F6 ] = ⎜ m ⎟ u [U(238)F6 ] ⎝ M m [U(235)F6 ] ⎠
1/ 2
⎛ 352.04 g/ mol ⎞ = ⎜ ⎟ ⎝ 349.03 g/ mol ⎠
= 1.004302 = 1.0043
5.126. Calculate the ratio of the rootmeansquare (rms) molecular speeds, which is the same as the ratio of the rates of effusion through the barrier: 1/ 2
⎛ M [H (1,2)] ⎞ u [H 2 (1,1)] = ⎜ m 2 ⎟ u [H 2 (1,2)] ⎝ M m [H 2 (1,1)] ⎠
1/ 2
⎛ 3.0219 g/ mol ⎞ = ⎜ ⎟ ⎝ 2.0156 g/ mol ⎠
= 1.22444 = 1.1224
5.127. First, calculate the apparent molar masses at each pressure using the ideal gas law. Only the calculation of the apparent molar mass for 0.2500 atm will be shown; the other values will be summarized in a table. n =
PV (0.2500 atm) (3.1908 L) = = 3.55896 x 10−2 mol RT (0.082057 L • atm/K • mol) (273.15 K)
Apparent molar mass =
1.000 g = 28.098 = 28.10 g/mol 3.55896 x 102 mol
The following table summarizes the apparent molar masses calculated as above for all P's; these data are plotted in the graph to the right of the table. App. Molar Mass (g/mol)
0.2500
28.10
0.5000
28.14
0.7500
28.19
1.0000
28.26
Apparent Molar Mass
P (atm)
28.3 28.2 28.1 28.0 0.00 0.25 0.50 0.75 1.00 1.25 P (atm)
Extrapolation back to P = 0 gives 28.07 g/mol for the molar mass of the unknown gas (CO).
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Chapter 5: The Gaseous State
5.128. First, calculate the apparent molar mass at each pressure using the ideal gas law. Only the calculation of the molar mass at 0.2500 atm will be shown; the other values will be summarized in a table. n =
PV (0.2500 atm) (2.801 L) = = 3.1242 x 10−2 mol RT (0.082057 L • atm/K • mol) (273.15 K)
Apparent molar mass =
1.000 g = 32.008 = 32.01 g/mol 3.1242 x 102 mol
The following table summarizes the apparent molar masses calculated as above for all P's; these data are plotted in the graph to the right of the table. P (atm)
App. Molar Mass (g/mol)
0.2500
32.008
0.5000
32.020
0.7500
32.021
1.0000
32.029
Extrapolation back to P = 0 gives close to 32.00 g/mol for the molar mass of O2. Apparent Molar Mass
32.03 32.02 32.01 32.00 31.99 0.00
0.25
0.50
0.75
1.00
P (atm)
5.129. Use CO + 1/2O2 → CO2, instead of 2CO. First, find the moles of CO and O2 by using the ideal gas law. nCO =
PV (0.500 atm) (2.00 L) = = 0.04062 mol RT (0.08206 L • atm/K • mol) (300 K)
nO2 =
PV (1.00 atm) (1.00 L) = = 0.04062 mol (0.08206 L • atm/K • mol) (300 K) RT
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There are equal amounts of CO and O2, but (from the equation) only half as many moles of O2 as CO are required for the reaction. Therefore, when 0.04062 mol CO has been consumed, only 0.04062/2 mol O2 will have been used up. Then 0.04062/2 mol O2 will remain, and 0.04062 mol CO2 will have been produced. At the end, nCO = 0 mol; nO2 = 0.0202 mol; and nCO2 = 0.04062 mol
However, the total volume with the valve open is 3.00 L, so the partial pressures of O2 and CO2 must be calculated from the ideal gas law for each: (0.0203 mol O 2 ) (0.08206 L • atm/K • mol)(300 K) nRT = = 0.16658 = 0.167 atm O2 3.00 L V (0.04062 mol CO 2 ) (0.08206 L • atm/K • mol) (300 K) nRT = V 3.00 L
= 0.33332 = 0.333 atm CO2 5.130. Use H2 + 1/2O2 → H2O. First, find the moles of H2 and O2 by using the ideal gas law. nH2 =
PV (0.500 atm)(2.00 L) = = 0.04062 mol RT (0.08206 L • atm/K • mol)(300 K)
nO2 =
PV (1.00 atm)(1.00 L) = = 0.04062 mol (0.08206 L • atm/K • mol)(300 K) RT
When all the H2 has been consumed, half of the O2 will remain. At the end, there will be 0.04062 mol H2O formed. However, the final volume is 3.00 L, so the partial pressures of H2O and O2 must each be calculated from the ideal gas law: (0.04062 mol H 2 O)(0.08206 L • atm/K • mol)(300 K) nRT = V 3.00 L
= 0.3333 = 0.333 atm H2O (0.02031 mol O 2 )(0.08206 L • atm/K • mol)(300 K) nRT = V 3.00 L
= 0.16666 = 0.167 atm O2 5.131. Cells in the lining of arteries detect increased blood pressure and respond by producing nitric oxide. NO rapidly diffuses through the artery wall to cells in the surrounding muscle tissue. In response, the muscle tissue relaxes, the blood vessel expands, and the blood pressure drops. 5.132. Angina results from reduced blood flow to the heart muscle. Nitroglycerin breaks down in the body to form nitric oxide, which relaxes the arteries, allowing greater blood flow to the heart. 5.133. Svante Arrhenius was the first to show how sensitive the temperature of the earth might be to the percentage of carbon dioxide in the atmosphere.
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Chapter 5: The Gaseous State
5.134. The principal gases in the atmosphere are oxygen, O2, and nitrogen, N2. These gases are transparent to visible light from the sun, and when the light reaches the surface of the earth, it is absorbed and converted to heat. The heat causes atoms in the earth’s surface to vibrate, which then radiate the heat energy as infrared radiation, or heat rays. Neither oxygen nor nitrogen absorbs infrared radiation, but other gases, especially carbon dioxide, do absorb infrared radiation, and it is this absorption that warms the atmosphere, giving a greenhouse effect.
■
SOLUTIONS TO STRATEGY PROBLEMS
5.135. First calculate the moles of carbon dioxide that formed. n =
PV (738/760 atm)(183 x 103 L) = = 0.007287 molCO2 RT (0.08206 L • atm/K • mol)(297.15 K)
Now, find the molarity. 0.007287 mol CO 2 2 mol HCl x = 0.5924 = 0.592 M 3 24.6 x 10 L 1 mol CO 2 5.136. a.
The taller, narrower curve, with a maximum value near 500 m/s, is for the heaver particles, neon. The flatter, wider curve, with a maximum value near 1500 m/s, is for the lighter particles, hydrogen. 1/ 2
b.
uNe
⎛ 3 x 8.31 kg • m 2 /(s 2 • K • mol) x 200 K ⎞ = ⎜ ⎟ 20.18 x 103 kg/mol ⎝ ⎠
uH 2
⎛ 3 x 8.31 kg • m 2 /(s 2 • K • mol) x 200 K ⎞ = ⎜ ⎟ 2.016 x 103 kg/mol ⎝ ⎠
= 497.06 = 497 m/s 1/ 2
= 1573 = 1.57 x 103 m/s
c.
Since hydrogen has the higher rms speed, it will have a greater effusion rate. Molecules moving faster collide with the hole in the container more often and have a higher probability of effusing.
d.
T =
5.137. Pf =
5.138. Vf =
u2M m (497.06 m/s) 2 (2.016 x 103 kg/mol) = = 19.98 = 20.0 K 3R 3 (8.31 kg • m 2 / s 2 • K • mol )
PV i iT f V f Ti PV i iT f Pf Ti
=
(1 atm) (162 L) (279.07 K) = 8.1752 = 8.18 atm (18.8 L) (294.15 K)
=
(740 mmHg) (435 mL) (273.15 K) = 388.03 = 388 mL (760 mmHg) (298.15 K)
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201
5.139. First note that 16.0 g O2 represents 0.500 mol O2 and that 14.0 g N2 represents 0.500 mol N2. Altogether, then, there is 1.00 mol of gas present in the container. a.
At STP 1.00 mol of gas occupies 22.4 L.
b.
PO2 = X O2 x Ptot =
c.
X N2 =
0.500 mol O 2 x 1 atm = 0.500 atm 0.500 mol O 2 + 0.500 mol N 2
0.500 mol N 2 = 0.500 (50.0% mole percent) 0.500 mol O 2 + 0.500 mol N 2
5.140. a.
The molar mass of C4H7S is 87.166 g/mol. 1/ 2
⎛ 3 x 8.31 kg • m 2 /(s 2 • K • mol) x 298.15 K ⎞ u = ⎜ ⎟ 87.166 x 103 kg/mol ⎝ ⎠
= 292.08 = 292 m/s
150 m = 0.5135 = 0.514 s 292 m/s
b.
t =
c.
This is probably not a good estimate because molecules do not travel in straight lines over such distances without undergoing numerous collisions. The actual path is random and irregular, and the time it would take cannot be estimated accurately.
5.141. The density of a gas can be calculated by using d = PMm/RT. For two different gases under the same conditions of temperature and pressure, this reduces to d2 M2 146.07 g/mol = = = 5.036 = 5.04 29.0 g/mol M1 d1
Thus, the density of SF6 is 5.04 times the density of air. 5.142. a.
P =
(1.0 mol)(0.08206 L • atm/K • mol)(348.15 K) 760 mmHg x 1.0 L 1 atm = 2.171 x 104 = 2.2 x 104 mmHg
b.
The molecules are undergoing constant random motion and are colliding with each other and with the walls of the container. The pressure in the container is caused by collisions of the molecules with the walls.
c.
If the temperature of the container is increased to 150°C, the average speed of the molecules will increase, and the number of collisions with other molecules and with the walls of the container will also increase. This will result in an increase in the pressure.
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Chapter 5: The Gaseous State
5.143. The moles of H2S gas formed is n =
(12.0 atm)(0.600 L) = 0.22045 mol H2S (0.08206 L • atm/K • mol)(398 K)
The mass can now be calculated. 0.22045 mol H2S x
1 mol S8 256.56 g x =7.069 = 7.07 g S8 1 mol S8 8 mol H 2S
5.144.
■
a.
Since the piston is movable, the pressure inside each flask is the same as the pressure outside, 1.0 atm.
b.
Since containers A and B have the same volume, and since container B has twice as many particles, the absolute temperature in container B is onehalf the temperature in container A. Container C has twice the volume of container A and also twice as many particles. Thus, the temperature in container C is the same as the temperature in container A.
c.
Since the piston is movable, the pressure inside each flask will remain the same as before, 1.0 atm. Now the volume is directly proportional to the amount of gas present, so the volumes of containers B and C will be equal, and the volume of container A will be onehalf as much.
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
5.145. Assume a 100.0g sample, giving 85.2 g CH4 and 14.8 g C2H6. Convert each to moles: 85.2 g CH4 x 14.8 g C2H6 x
1 mol CH 4 = 5.311 mol CH4 16.04 g CH 4 1 mol CH 4 = 0.4921 mol C2H6 16.04 g CH 4
VCH4 =
(5.311 mol)(0.08206 L • atm/K • mol) (291 K) = 125.03 L (771/760) atm
VC2 H6 =
(0.4921 mol) (0.08206 L • atm/K • mol) (291 K) = 11.58 L (771/760) atm
The density is calculated as follows: d =
85.2 g CH 4 + 14.8 g C2 H 6 = 0.7319 = 0.732 g/L (125.03 + 11.58) L
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203
5.146. Assume a 100.0g sample, giving 34.3 g He, 51.7 g N2, and 14.0 g O2. Convert to moles: 34.3 g He x
1 mol He = 8.5750 mol He 4.00 g
51.7 g N2 x
1 mol N 2 = 1.8451 mol N2 28.02 g
14.0 g O2 x
1 mol O 2 = 0.43750 mol O2 32.00 g O 2
VHe =
(8.575 mol ) (0.08206 L • atm/K • mol) (295 K) = 208.9 L (755/760) atm
VN2 =
(1.8451 mol ) (0.08206 L • atm/K • mol)(295 K) = 44.96 L (755/760) atm
VO2 =
(0.4375 mol) (0.08206 L • atm/K • mol) (295 K) = 10.66 L (755/760) atm
The density is calculated as follows: d =
34.3 g He + 51.7 g N 2 + 14.0 g O2 = 0.37802 = 0.378 g/L (208.9 + 44.96 + 10.66) L
5.147. First, subtract the height of mercury equivalent to the 25.00 cm (250 mm) of water inside the tube from 771 mmHg to get Pgas. Then subtract the vapor pressure of water, 18.7 mmHg, from Pgas to get PO 2
hHg =
(hW ) (d W ) 250 mm x 0.99987 g / cm3 = = 18.38 mmHg d Hg 13.596 g / cm3
Pgas = P − P25 cm water = 771 mmHg − 18.38 mmHg = 752.62 mmHg PO = 752.62 mmHg − 18.7 mmHg = 733.92 mmHg 2
n =
PV (733.92/760 atm) (0.0310 L) = = 0.0012408 mol O2 (0.08206 L • atm/K • mol) (294 K) RT
Mass = (2 x 0.0012408) mol Na2O2 x
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77.98 g Na 2 O2 = 0.1935 = 0.194 g Na2O2 1 mol Na 2 O 2
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Chapter 5: The Gaseous State
5.148. Proceed as in the previous problem, subtracting the height of the water and then the vapor pressure to find PH . 2
hHg =
(hW ) (d W ) 310 mm x 0.99987 g / cm3 = 22.80 mmHg 13.596 g / cm3 d Hg
Pgas = P − P31 cm water = 751 mmHg − 22.80 mmHg = 728.2 mmHg PH
2
n =
= (728.2 − 14.5) mmHg = 713.7 mmHg PV (713.7 / 760 atm) (0.0221 L) = = 0.00087209 mol H2 RT (0.08206 L • atm/K • mol)(290 K)
Mass = 0.00087209 mol H2 x
1 mol Zn 65.39 g Zn x = 0.057026 = 0.0570 g Zn 1 mol H 2 1 mol Zn
5.149. First find the moles of CO2: n =
PV (785/760 atm) (1.94 L) = = 0.08194 mol CO RT (0.08206 L • atm/K • mol) (298 K)
Set up one equation in one unknown: x = mol CaCO3; (0.08194 − x) = mol MgCO3. 7.85 g = (100.1 g/mol)x + (84.32 g/mol)(0.08194 − x) x =
(7.85  6.9092) = 0.05962 mol CaCO3 (100.1  84.32)
(0.08194 − x) = 0.02232 mol MgCO3 CaCO3 =
0.05962 mol CaCO3 x 100.1 g/mol x 100% = 76.02 = 76% 7.85 g
Percent MgCO3 = 100.00% − 76.02% = 23.98 = 24% 5.150. First find the moles of H2S: n =
PV (745/760 atm) (1.049 L) = = 0.04233 mol H2S RT (0.08206 L • atm/K • mol) (296 K)
Set up one equation in one unknown: x = mol ZnS; (0.04233 − x) = mol PbS. 6.12 g = (97.46 g/mol)x + (239.25 g/mol)(0.04233 − x) x =
(6.12  10.127) = 0.02826 mol ZnS (97.46  239.25)
(0.04233 − x) = 0.014070 mol PbS Percent ZnS =
0.02826 mol ZnS x 97.46 g/mol x 100% = 45.00 = 45.0% 6.12 g
Percent PbS = 55.0%
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205
5.151. Write a massbalance equation to solve for the moles of each gas, using 28.01 g/mol for the molar mass of N2 and 20.18 g/mol for the molar mass of Ne. Let y equal moles of each gas: 28.01 y + 20.18 y = 10.0 g y = (10.0 ÷ 48.19) = 0.20751 mol
Total moles (n) = 0.41502 mol (use below) Use the ideal gas law to calculate the volume, which can then be used to calculate density. (0.41502 mol)(0.08206 L • atm/K • mol)(500 K) = 1.1352 L 15.00 atm d = 10.0 g ÷ 1.1352 L = 8.808 = 8.81 g/L
5.152. Write a massbalance equation to solve for moles of each, using 39.95 g/mol for the molar mass of Ar and 20.18 g/mol for the molar mass of Ne. Let y equal moles of Ar and 2y equal moles of Ne: 39.95 y + 20.18 (2y) = 50.0 g y = (50.0 ÷ 80.31) = 0.6225 mol
Total moles (n) = mol Ar + mol Ne = 0.6225 + (2)(0.6225) = 1.8677 mol (use below) Volume of the mixture = 50.0 g ÷ 4.00 g/L = 12.5 L Use the ideal gas law to calculate the total pressure. P(total) =
(1.8677 mol)(0.08206 L • atm/K • mol)(350 K) = 4.292 atm 12.5 L
Partial pressure of Ne = 2/3 P(total) = 2/3(4.292 atm) = 2.861 = 2.86 atm 5.153. Rearrange the equation PMm = dRT to find the quantity RT/Mm. Then plug into the equation for the rootmeansquare speed. RT P = d Mm 1/ 2
1/ 2
⎛ 3RT ⎞ u = ⎜ ⎟ ⎝ Mm ⎠
1/ 2
⎛ 3P ⎞ = ⎜ ⎟ ⎝ d ⎠
⎛ ⎛ 675 ⎞ ⎞ 2 2 ⎜ 3 ⎜ 760 ⎟ atm x (8.31 kg • m / s • K • mol) ⎟ ⎠ ⎟ = ⎜ ⎝ 3 ⎜ (3.00 x 10 kg/L)(0.08206 L • atm / K • mol) ⎟ ⎜ ⎟ ⎝ ⎠ = 2.9990 x 102 = 3.00 x 102 m/s
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Chapter 5: The Gaseous State
5.154. Rearrange the equation for the rootmeansquare speed to get Mm = 3RT/u2.
Mm =
(3) (8.31 kg • m 2 / (s 2 • mol • K) (298.2 K) = 0.029737 kg/mol = 29.74 g/mol (5.00 x 102 m/s) 2
Now, rearrange the equation PMm = dRT to find the density, d = PMm/RT.
d =
PM m (2.50 atm) (29.74 g/ mol) = = 3.038 = 3.04 g/L RT (0.08206 L • atm / K • mol) (298.2 K)
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CHAPTER 6
Thermochemistry
■
SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 6.1. Substitute into the formula Ek = 1/2 mv2 using SI units: Ek = 1/2 x 9.11 x 10−31 kg x (5.0 x 106 m/s)2 = 1.13 x 10−17 = 1.1 x 10−17 J 1.13 x 10−17 J x
1 cal = 2.72 x 10−18 = 2.7 x 10−18 cal 4.184 J
6.2. Heat is evolved; therefore, the reaction is exothermic. The value of q is −1170 kJ. 6.3. The thermochemical equation is 2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g); ΔH = −1049 kJ ΔH equals −1049 kJ for 1 mol N2H4 and for 2 mol N2H4.
6.4. a.
N2H4(l) + 1/2N2O4(l) → 3/2N2(g) + 2H2O(g); ΔH = −524.5 kJ
b.
4H2O(g) + 3N2(g) → 2N2H4(l) + N2O4(l); ΔH = 1049 kJ
6.5. The reaction is 2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g); ΔH = −1049 kJ 10.0 g N2H4 x
1 mol N 2 H 4 1 mol N 2 O 4 1049 kJ x x 32.02 g 2 mol N 2 H 4 1 mol N 2 O 4
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= −163.80 = −164 kJ
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Chapter 6: Thermochemistry
6.6. Substitute into the equation q = s x m x Δt to obtain the heat transferred. The temperature change is Δt = tf − ti = 100.0°C − 20.0°C = 80.0°C
Therefore, q = s x m Δt = 0.449J/(g•°C) x 5.00 g x 80.0°C = 1.796 x 102 = 1.80 x 102 J
6.7. The total mass of the solution is obtained by adding the volumes together and by using the density of water (1.000 g/mL). This gives 33 + 42 = 75 mL, or 75 g. The heat absorbed by the solution is q = s x m x Δt = 4.184 J/(g•°C) x 75 g x (31.8°C − 25.0°C) = 2133.8 J
The heat released by the reaction, qrxn, is equal to the negative of this value, or −2133.8 J. To obtain the enthalpy change for the reaction, you need to calculate the moles of HCl that reacted. This is Mol HCl = 1.20 mol/L x 0.033 L = 0.0396 mol The enthalpy change for the reaction can now be calculated. ΔH =
2133.8 J = −53884 J/mol = −54 kJ/mol 0.0396 mol
Expressing this result as a thermochemical equation, you have HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l); ΔH = −54 kJ
6.8. Use Hess's law to find ΔH for 4Al(s) + 3MnO2(s) → 2Al2O3(s) + 3Mn(s) from the following data for equations 1 and 2: 2Al(s) + 3/2O2(g) → Al2O3(s); Mn(s) + O2(g) → MnO2(s);
ΔH = −1676 kJ (1)
ΔH = −520 kJ
(2)
If you take equation 1 and multiply it by 2, you obtain 4Al(s) + 3O2(g) → 2Al2O3(s);
ΔH = 2 x (−1676 kJ) = −3352 kJ
Since the desired reaction has three MnO2 on the left side, reverse equation 2 and multiply it by 3. The result is 3MnO2(s) → 3Mn(s) + 3O2(g)
ΔH = −3 x (−520 kJ) = 1560 kJ
If you add the two equations and corresponding enthalpy changes, you obtain the enthalpy change of the desired equation. 4Al(s) + 3O2(g) 3MnO2(s)
→ →
2Al2O3(s) 3Mn(s) + 3O2(g)
ΔH = −3352 kJ ΔH = 1560 kJ
4Al(s) +3MnO2(s)
→
2Al2O3(s) + 3Mn(s)
ΔH = −1792 kJ
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209
6.9. The vaporization process, with the ΔH°f values given below the substances, is
H2O(l) →
H2O(g)
285.8
241.8
(kJ)
The calculation is o ΔH vap = Σn ΔH of (products) − Σm ΔH of (reactants) = ΔH of [H2O(g)] − ΔH of [H2O(l)]
= (−241.8 kJ) − (−285.8 kJ) = 44.0 kJ 6.10. The reaction, with the ΔH°f values given below the substances, is
3NO2(g)
+
H2O(l)
33.10
→
285.8
2HNO3(aq)
+ NO(g)
207.4
90.29
(kJ)
The calculation is o ΔH rxn = Σn ΔH of (products) − Σm ΔH of (reactants)
= [2 ΔH of (HNO3) + ΔH of (NO)] − [3 ΔH of (NO2) + ΔH of (H2O)] = [2(−207.4) + (90.29)] kJ − [3(33.10) + (−285.8)] kJ = −138.01 = −138.0 kJ 6.11. The net chemical reaction, with the ΔH°f values given below the substances, is
2 NH4+(aq) 2(132.5)
+
2OH(aq) 2(230.0)
→
2NH3(g) 2(45.90)
+
2H2O(l) 2(285.8)
(kJ)
The calculation is o ΔH rxn = [2 ΔH of (NH3) + 2 ΔH of (H2O)] − [2 ΔH of (NH4+) + 2 ΔH of (OH−)]
= [2(−45.90) + 2(−285.8)] − [2(−132.5) + 2(−230.0)] = 61.60 = 61.6 kJ
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ANSWERS TO CONCEPT CHECKS
6.1. The photovoltaic cells collect the sun’s energy, converting it to electrical energy. This electrical energy is stored in the battery as chemical energy, which is later changed back to electrical energy that runs a motor. As the motor rotates, it changes the electrical energy to kinetic energy (energy of motion) of the motor, then of water, which in turn is changed to potential energy (energy of position) of water as the water moves upward in the gravitational field of earth.
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6.2. a.
This reaction is the one shown in the problem, and it has a positive ΔH, so the reaction is endothermic.
b.
This reaction is simply twice reaction a, so it is also endothermic.
c.
This reaction is the reverse of reaction a, so it is exothermic.
d.
This reaction is simply twice reaction c, so it is more exothermic than reaction c. Thus, reaction d is the most exothermic reaction.
6.3. You can think of the sublimation of ice as taking place in two stages. First, the solid melts to liquid; then the liquid vaporizes. The first process has an enthalpy ΔHfus. The second process has an enthalpy ΔHvap. Therefore, the total enthalpy, which is the enthalpy of sublimation, is the sum of these two enthalpies:
ΔHsub = ΔHfus + ΔHvap
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ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
6.1. Energy is the potential or capacity to move matter. Kinetic energy is the energy associated with an object by virtue of its motion. Potential energy is the energy an object has by virtue of its position in a field of force. Internal energy is the sum of the kinetic and potential energies of the particles making up a substance. 6.2. In terms of SI base units, a joule is kg•m2/s2. 6.3. Originally, a calorie was defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius. At present, the calorie is defined as 4.184 J. 6.4. At either of the two highest points above the earth in a pendulum's cycle, the energy of the pendulum is all potential energy and is equal to the product mgh (m = mass of pendulum, g = constant acceleration of gravity, and h = height of pendulum). As the pendulum moves downward, its potential energy decreases from mgh to near zero, depending on how close it comes to the earth's surface. During the downward motion, its potential energy is converted to kinetic energy. When it reaches the lowest point (middle) of its cycle, the pendulum has its maximum kinetic energy and minimum potential energy. As it rises above the lowest point, its kinetic energy begins to be converted to potential energy. When it reaches the other high point in its cycle, the energy of the pendulum is again all potential energy. By the law of conservation of energy, this energy cannot be lost, only converted to other forms. At rest, the energy of the pendulum has been transferred to the surroundings in the form of heat. 6.5. As the heat flows into the gas, the gas molecules gain energy and move at a greater average speed. The internal energy of the gas increases.
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6.6. An exothermic reaction is a chemical reaction or a physical change in which heat is evolved (q is negative). For example, burning one mole of methane, CH4(g), yields carbon dioxide, water, and 890 kJ of heat. An endothermic reaction is a chemical reaction or physical change in which heat is absorbed (q is positive). For example, the reaction of one mole of barium hydroxide with ammonium nitrate absorbs 170.8 kJ of heat in order to form ammonia, water, and barium nitrate. 6.7. Changes in internal energy depend only on the initial and final states of the system, which are determined by variables such as temperature and pressure. Such changes do not depend on any previous history of the system. 6.8. The enthalpy change equals the heat of reaction at constant pressure. 6.9. At constant pressure, the enthalpy change is positive (the enthalpy increases) for an endothermic reaction. 6.10. It is important to give the states when writing an equation for ΔH because ΔH depends on the states of all reactants and products. If any state changes, ΔH changes. 6.11. When the equation for the reaction is doubled, the enthalpy is also doubled. When the equation is reversed, the sign of ΔH is also reversed. 6.12. First, convert the 10.0 g of water to moles of water, using its molar mass (18.02 g/mol). Next, using the equation, multiply the moles of water by the appropriate mole ratio (1 mol CH4 / 2 mol H2O). Finally, multiply the moles of CH4 by the heat of the reaction (−890.3 kJ/mol CH4). 6.13. The heat capacity (C) of a substance is the quantity of heat needed to raise the temperature of the sample of substance one degree Celsius (or one kelvin). The specific heat of a substance is the quantity of heat required to raise the temperature of one gram of a substance by one degree Celsius (or one kelvin) at constant pressure. 6.14. A simple calorimeter consists of an insulated container (for example, a pair of styrene coffee cups as in Figure 6.12) with a thermometer. The heat of the reaction is obtained by conducting the reaction in the calorimeter. The temperature of the mixture is measured before and after the reaction, and since this is a constantpressure process, the heat is directly related to the enthalpy change, ΔH. 6.15. Hess’s law states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps. In other words, no matter how you go from reactants to products, the enthalpy change for the overall chemical change is the same. This is because enthalpy is a state function.
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6.16. No, you can still obtain the enthalpy for the desired reaction. You will need to come up with a system of reactions that can be combined to give the desired reaction and will need to know the enthalpy changes for each of the steps. Then, using Hess’s law, you can obtain the enthalpy change for the reaction under study. 6.17. The thermodynamic standard state consists of the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: one atm pressure and the specified temperature (usually 25°C). 6.18. The reference form of an element is the most stable form (physical state and allotrope) of the element under standard thermodynamic conditions. The standard enthalpy of formation of an element in its reference form is zero. 6.19. The standard enthalpy of formation of a substance, ΔH°f , is the enthalpy change for the formation of one mole of the substance in its standard state from its elements in their reference form and in their standard states. 6.20. The equation for the formation of H2S(g) is H2(g) + 1/8S8(rhombic) → H2S(g) 6.21. The reaction of C(g) + 4H(g) → CH4(g) is not an appropriate equation for calculating the ΔH°f of methane because the most stable form of each element is not used. Both H2(g) and C(graphite) should be used instead of H(g) and C(g), respectively. 6.22. A fuel is any substance that is burned or similarly reacted to provide heat and other forms of energy. The fossil fuels are petroleum (oil), gas, and coal. They were formed millions of years ago when aquatic plants and animals were buried and compressed by layers of sediment at the bottoms of swamps and seas. Over time, this organic matter was converted by bacterial decay and pressure to fossil fuels. 6.23. One of the ways of converting coal to methane involves the watergas reaction. C(s) + H2O(g) → CO(g) + H2(g) In this reaction, steam is passed over hot coal. This mixture is then reacted over a catalyst to give methane. CO(g) + 3H2(g) → CH4(g) + H2O(g) 6.24. Some possible rocket fuel/oxidizer combinations are H2/O2 and hydrazine/dinitrogen tetroxide. The chemical equations for their reactions are H2(g) + 1/2O2(g) → H2O(g); ΔH° = −242 kJ 2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g); ΔH° = −1049 kJ
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6.25. The answer is d, burning enough butane to produce one mole of water. 6.26. The answer is b, 40.°C. 6.27. The answer is c, endothermic, and the temperature of the resulting solution falls. 6.28. The answer is c, the water with the mercury will be the hottest.
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ANSWERS TO CONCEPT EXPLORATIONS
6.29. Part 1 a.
The temperature change for the hot water is Δt = 60°C − 80°C = −20°C. The temperature change for the cold water is Δt = 60°C − 20°C = 40°C. The magnitude of the temperature change for the cold water is twice the magnitude of the temperature change for the hot water.
b.
The heat transferred from the hot water to the cold water.
c.
q = s x m x Δt = (4.18 J/g•°C)(100. g)(40°C) = 1.672 x 104 = 1.67 x 104 J = 16.7 kJ
d.
The quantity of heat transferred from the hotwater sample is equal to the quantity of heat transferred to the coldwater sample.
e.
Since both samples are water, the specific heat capacities of both samples are the same. Also, the quantity of heat is the same for both. Thus mhotΔthot = mcoldΔtcold. Since the cold water has onehalf the mass of the hot water, it will have twice the temperature change.
f.
Since the hot water has onehalf the mass of the cold water, it will have twice the temperature change of the cold water.
g.
Since the first sample (cold water) increases by onethird the temperature change of the second sample (hot water), the first sample (cold water) has three times the mass of the second sample (hot water).
h.
Δtcold mhot 7g = = = 2.33 Δthot mcold 3g
Therefore, the cold water will have a temperature change 2.3 times the temperature change for the hot water. Part 2 The amount of heat added cannot be calculated. In order to perform this calculation, it is necessary to know the mass of the water. Part 3 Since q = s x m x Δt, for two samples of water with an identical temperature change, the heat required is directly proportional to the mass of the solution. If the second solution requires double the heat of the first solution, then it has double the mass of the first solution.
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6.30. a.
q = (4.18 J/g•°C) x (100. g) x (41°C − 21°C) = 8360 J = 8.4 x 103 J = 8.4 kJ
b.
qrxn = − qwater = −(4.18 J/g•°C) x (100. g) x (31°C − 21°C) = −4180 J = −4.2 x 103 J = −4.2 kJ This is onehalf of the value of q in part a.
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c.
In parts a and b, you see that for each 10°C temperature change, 4.2 kJ of heat is required. Thus, for a temperature change of 30°C (from 21°C to 51°C), three times as much heat is required. This is 4.2 kJ x 3, or 12.6 kJ.
d.
You should need three times as much as for part b, or 0.06 mol of X and 0.03 mol of Y.
e.
There is 4.2 kJ for each 0.01 mol of Y. This gives 420 kJ for one mole of Y, which is the enthalpy of the reaction.
ANSWERS TO CONCEPTUAL PROBLEMS
6.31. Kinetic energy is proportional to mass and to speed squared. Compare the kinetic energy of the smaller car with that of the larger car (twice the mass), assuming both are traveling at the same speed. The larger car would have twice the kinetic energy of the smaller car. Or we could say that the smaller car has only half the kinetic energy of the larger car. Now suppose the speed of the smaller is increased by a factor of 2 (so it is now moving at twice its original speed). Its kinetic energy is increased by a factor of 4. Therefore, the smaller car now has onehalf times four times, or twice, the kinetic energy of the larger car. The smaller car has the greater kinetic energy. 6.32. The equation says that 1 mol of butane reacts with 13/2 mol of oxygen to yield 4 mol of carbon dioxide and 5 mol of water. The reaction yields a certain amount of heat, which you can symbolize as q. So choice a yields heat q. On the other hand, choice b is only 1 mol of oxygen, not 13/2 mol. So, choice b yields heat equal to 2/13 q. This result might be easier to see by first looking at choice c. Note that the 1 mol of carbon dioxide stated in choice c is only onefourth that given in the equation. This means that choice c yields onefourth q (just the inverse of the coefficient of the equation). Similarly, choice d yields onefifth q. Therefore, choice b yields the least heat. 6.33. a.
After the water is placed into the freezer, it will lose heat to the freezer, so qsys is negative.
b.
The water will have turned to ice.
c.
The initial enthalpy (of the water) is higher than the final enthalpy (of the ice).
d.
After several hours, the temperature of the water will be −20°C.
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6.34. a.
Heat will flow from the iron block to the aluminum block.
Fe Al b.
The aluminum block will be absorbing heat, so qsys will be positive.
c.
The temperatures of the aluminum block and the iron block will both be the same.
d.
The heat gained by the aluminum will equal the heat lost by the iron. Each heat term will be equal to s x m x Δt. This gives (0.901 J/g•°C)(20.0 g)(tf − 45.0°C) = (0.449 J/g•°C)(20.0 g)(50.0°C − tf) You can cancel the mass and the units from both sides to give a simplified expression. (0.901)(tf − 45.0°C) = (0.449)(50.0°C − tf) This can be simplified to give 1.350 tf = 62.995, or tf = 46.66 = 46.7°C.
6.35. You can imagine this process taking place in two steps: first, the preparation of water vapor from the elements, and second, the change of the vapor to liquid. Here are the equations: H2(g) + ½O2(g) → H2O(g); ΔHf H2O(g) → H2O(l); −ΔHvap The last equation is the reverse of the vaporization of water, so the enthalpy of the step is the negative of the enthalpy of vaporization. The enthalpy change for the preparation of one mole of liquid water, ΔH, is the sum of the enthalpy changes for these two steps: ΔH = ΔHf + (−ΔHvap) = ΔHf − ΔHvap 6.36. The expression for the heat is q = s x m x Δt. For the same amount of heat and mass, the product s x Δt must be constant. The metal with the smaller specific heat will have the larger Δt. Since the specific heat for aluminum (0.901 J/g•°C) is larger than that for iron (0.449 J/g•°C), the block of iron will have the larger Δt and be warmer. 6.37. a.
The heat lost by the metal is equal to the heat gained by the water. Since q = s x m x Δt, the heat gained by the water is directly proportional to Δt. Since Δt is larger for water with metal A, metal A lost more heat. Now, each metal has the same mass and Δt, so the specific heat is directly proportional to q. Since q is larger for A, the specific heat is larger for A.
b.
The metal with the higher specific heat will have absorbed more heat to reach the starting temperature of 95oC; therefore, it will release more heat to the water, causing the water to reach a higher temperature. The beaker with metal A will rise to the higher temperature.
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6.38. Silver metal reacts with a halogen to produce the corresponding silver halide. For example, silver reacts with fluorine to produce silver fluoride. Each reaction corresponds to the reaction forming the silver halide, so you look in the table of enthalpies of formation of compounds (Table 6.2). The most exothermic reaction would be the one with the most negative enthalpy of formation. That would be the reaction for the formation of silver fluoride. 6.39. Let us write ΔHrxn for the enthalpy change when one mole of P4S3 burns in O2 to give P4O10 and SO2. In principle, you could calculate ΔHrxn from enthalpies of formation for the reactants and products of this reaction. You would require the values for P4S3, O2 (which equals zero), P4O10, and SO2. Enthalpies of formation for the products, P4O10 and SO2, are given in Table 6.2 in the text. This means that if you have measured ΔHrxn, you can use the enthalpies of formation of P4O10 and SO2 to calculate the enthalpy of formation for P4S3. What you have done is this: you have used enthalpies of combustion to calculate enthalpies of formation. This is the idea most often used to obtain enthalpies of formation. 6.40.
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a.
Since ΔH is positive, the reaction is endothermic, and the solution will be colder.
b.
While the salt dissolves, heat will flow into the beaker to raise the temperature of the water back to the initial temperature.
c.
After the water returns to room temperature, q for the system will be zero.
SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 6.41. The heat released, in kilocalories, is −445.1 kJ x
1000 J 1 cal 1 kcal x x = −106.38 = −106.4 kcal 1 kJ 4.184 J 1000 cal
6.42. The heat released, in kilocalories, is −309.1 kJ x
1000 J 1 cal 1 kcal x x = −73.876 = −73.88 kcal 1 kJ 4.184 J 1000 cal
6.43. The kinetic energy, in joules, is Ek = 1/2 x 4.85 x 103 lb x
0.4536 kg ⎡ 57 mi ⎤ x ⎢ ⎥ 1 lb ⎣ 1h ⎦
2
⎡1609 m ⎤ x ⎢ ⎥ ⎣ 1 mi ⎦
2
⎡ 1h ⎤ x ⎢ ⎥ ⎣ 3600 s ⎦
2
= 7.13 x 105 = 7.1 x 105 J
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The kinetic energy, in calories, is Ek = 7.13 x 105 J x
1 cal = 1.70 x 105 = 1.7 x 105 cal 4.184 J
6.44. The kinetic energy, in joules, is 0.0000648 kg ⎡ 2520 ft ⎤ x ⎢ Ek = 1/2 x 245 gr x ⎥ 1 gr ⎣ 1s ⎦
2
⎡1 yd ⎤ x ⎢ ⎥ ⎣ 3 ft ⎦
2
⎡ 0.9144 m ⎤ x ⎢ ⎥ ⎣ 1 yd ⎦
2
= 4683 = 4.68 x 103 J The kinetic energy, in calories, is Ek = 4683 J x
1 cal = 1119 = 1.12 x 103 cal 4.184 J
6.45. To insert the mass of one molecule of ClO2 in the formula, multiply the molar mass by the reciprocal of Avogadro's number. The kinetic energy in joules is 67.45 g 1 mol 1 kg ⎡ 306 m ⎤ Ek = 1/2 x x x x ⎢ 23 ⎥ 1 mol 6.022 x 10 molec. 1000 g ⎣ 1s ⎦
2
= 5.243 x 10−21 = 5.24 x 10−21 J/molec. 6.46. To insert the mass of one molecule of N2O in the formula, multiply the molar mass by the reciprocal of Avogadro's number. The kinetic energy in joules is Ek = 1/2 x
44.02 g 1 mol 1 kg ⎡ 379 m ⎤ x x x ⎢ 23 ⎥ 1 mol 6.022 x 10 molec. 1000 g ⎣ 1s ⎦
2
= 5.249 x 10−21 = 5.25 x 10−21 J/molec. 6.47. Endothermic reactions absorb heat, so the sign of q will be positive because energy must be gained by the system from the surroundings. The flask will feel cold to the touch. 6.48. Exothermic reactions evolve heat, so the sign of q will be negative because energy is lost from the system to the surroundings. The flask will feel hot to the touch. 6.49. The gain of 66.2 kJ of heat per 2 mol NO2 means the reaction is endothermic. Because energy is gained by the system from the surroundings, q is positive and is +66.2 kJ for 2 mol NO2 reacting. 6.50. The release of 939 kJ of heat per 2 mol HCN means the reaction is exothermic. Because energy is lost from the system to the surroundings, q is negative, and because the reaction involves 2 mol HCN, q for the reaction is −939 kJ.
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6.51. The reaction of Fe(s) with HCl must yield H2 and FeCl2. To balance the hydrogen, 2HCl must be written as a reactant: Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g) To write a thermochemical equation, the sign of ΔH must be negative because heat is evolved: Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g); ΔH = −89.1 kJ 6.52. The decomposition of 2 mol KClO3 to KCl and O2 must yield 2 mol KCl to balance the potassium and chlorine. To balance the oxygen, 3 mol O2 must be written as a product: 2KClO3(s) → 2KCl(s) + 3O2(g) To write a thermochemical equation, the sign of ΔH must be negative because heat is evolved: 2KClO3(s) → 2KCl(s) + 3O2(g); ΔH = −78.0 kJ 6.53. The first equation is P4(s) + 5O2(g) → P4O10(s); ΔH = −3010 kJ The second equation is P4O10(s) → P4(s) + 5O2(g); ΔH = ? The second equation has been obtained by reversing the first equation. Therefore, to obtain ΔH for the second equation, ΔH for the first equation must be reversed in sign: −(−3010) = +3010 kJ. 6.54. The first equation is CS2(l) + 3O2(g) → CO2(g) + 2SO2(g); ΔH = −1077 kJ The second equation is 1/2CS2(l) + 3/2O2(g) → 1/2CO2(g) + SO2(g); ΔH = ? The second equation has been obtained by dividing each coefficient in the first equation by 2. Therefore, to obtain ΔH for the second equation, ΔH for the first equation must be divided by 2; that is, (−1077) ÷ 2 = −538.5 kJ. 6.55. The first equation is 1/4P4O10(s) + 3/2H2O(l) → H3PO4(aq); ΔH = −96.2 kJ The second equation is P4O10(s) + 6H2O(l) → 4H3PO4(aq); ΔH = ? The second equation has been obtained from the first by multiplying each coefficient by 4. Therefore, to obtain the ΔH for the second equation, the ΔH for the first equation must be multiplied by 4; that is, −96.2 x 4 = −384.8 = 385 kJ
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6.56. The first equation is 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g); ΔH = −906 kJ The second equation is NO(g) + 3/2H2O(g) → NH3(g) + 5/4O2(g); ΔH = ? The second equation has been obtained by reversing the first equation and dividing each coefficient by 4. Therefore, to obtain ΔH for the second equation, the ΔH for the first equation must be reversed in sign and divided by 4; that is, +906 ÷ 4 = 226.50 = 227 kJ. 6.57. Because nitric oxide is written as NO in the equation, the molar mass of NO equals 30.01 g per mol NO. From the equation, 2 mol NO evolve 114 kJ heat. Divide the 114 kJ by the 2 mol NO and by the 30.01 g/mol NO to obtain the amount of heat evolved per gram of NO: 114 kJ 1 mol NO 1.90 kJ x = 1.899 = 2 mol NO 30.01 g NO g NO 6.58. Because hydrogen is written as H2 in the equation, the molar mass of H2 equals 2.016 g per mol H2. From the equation, 2 mol H2 evolve 484 kJ heat. Divide this by the 2 mol H2 and by the 2.016 g/mol H2 to obtain the amount of heat evolved per gram of hydrogen: 1 mol H 2 484 kJ 120 kJ x = −120.03 = 2 mol H 2 g H2 2.016 g H 2 6.59. The molar mass of ammonia is 17.03 g/mol. From the equation, 4 mol NH3 evolve 1267 kJ of heat. Divide 35.8 g NH3 by its molar mass and the 4 mol NH3 in the equation to obtain the amount of heat evolved: 35.8 g NH3 x
1 mol NH 3 1267 kJ x = −6.658 x 102 = −6.66 x 102 kJ 17.03 g NH 3 4 mol NH 3
6.60. The molar mass of H2S is 34.08 g/mol. From the equation, 2 mol H2S evolve 1036 kJ of heat. This gives 28.5 g H2S x
1 mol H 2S 1036 kJ = −433.18 = −433 kJ x 2 mol H 2S 34.08 g H 2S
6.61. The molar mass of C3H8 is 44.06 g/mol. From the equation, 1 mol C3H8 evolves 2043 kJ heat. This gives −369 kJ x
1 mol C3 H8 44.06 g C3 H8 x 2043 kJ 1 mol C3 H8
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= 7.957 = 7.96 g C3H8
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6.62. The molar mass of C2H5OH is 46.05 g/mol. From the equation, 1 mol C2H5OH evolves 1235 kJ heat. This gives −358 kJ x
1 mol C 2 H 5 OH 46.05 g C 2 H 5OH x = 13.348 = 13.3 g C2H5OH 1235 kJ 1 mol C2 H 5 OH
6.63. Multiply the 180 g (0.180 kg) of water by the specific heat of 4.18 J/(g•°C) and by Δt to obtain heat in joules: 180 g x (96°C − 19°C) x
4.18 J = 57934 = 5.8 x 104 J 1 g • °C
6.64. Multiply the 1.28 x 103 g (1.28 kg) iron by the specific heat of 0.449 J/(g•°C) and by Δt to obtain heat in joules: 1.28 x 103 g x (178°C − 21°C) x
0.449 J = 90231 = 9.02 x 104 J 1 g • °C
6.65. Use the 2.26 x 103 J/g (2.26 kJ/g) heat of vaporization to calculate the heat of condensation. Then use it to calculate Δt, the temperature change. Heat of condensation =
2.26 x 103 J x 168 g = 3.7968 x 105 J 1g
Temperature change = Δt =
3.7968 x 105 J 1 g • °C = 5.8085 = 5.81°C x 4 6.44 x 10 g 1.015 J
6.66. Use the 334 J/g (0.334 kJ/g) heat of fusion to calculate the heat of melting. Then use the heat of melting to calculate Δt, the temperature change. Heat of melting = x 31.5 g x
334 J = 10521 J 1g
Heat melting + heat warming = heat lost by 210 g H2O Let t = the final temperature, and substitute into the above equation: ⎡ 4.18 J ⎤ ⎡ 4.18 J ⎤ 10521 J + ⎢ x 31.5 g x (t − 0°C) = ⎢ ⎥ x 210 g x (21.0°C − t) ⎥ ⎣1 g • °C⎦ ⎣1 g • °C⎦ Solve the above equation for t. 10521 J + 131.67t = 18433.8 J − 877.8t t =
18433.8 J  10521 J = 7.838 = 7.84°C (131.67 + 877.8) J/ °C
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6.67. The enthalpy change for the reaction is equal in magnitude and opposite in sign to the heatenergy change occurring from the cooling of the solution and calorimeter. qcalorimeter = (1071 J/°C)(21.56°C − 25.00°C) = −3684.2 J
Thus, 15.3 g NaNO3 is equivalent to −3684.2 J heat energy. The amount of heat absorbed by 1.000 mol NaNO3 is calculated from +3684.2 J (opposite sign): 1.000 mol NaNO3 x
85.00 g NaNO3 3684.2 J x = 2.0468 x 104 J 1 mol NaNO3 15.3 g NaNO3
Thus, the enthalpy change, ΔH, for the reaction is 2.05 x 104 J, or 20.5 kJ, per mol NaNO3. 6.68. The enthalpy change for the reaction is equal in magnitude and opposite in sign to the heat energy produced from the warming of the solution and the calorimeter. qcalorimeter = (1258 J/°C)(38.7°C − 25.0°C) = +1.7234 x 104 J
Thus, 23.6 g CaCl2 is equivalent to 1.7234 x 104 J heat energy. The amount of heat released by 1.20 mol CaCl2 is calculated from −1.7234 x 104 J (opposite sign): 1.20 mol CaCl2 x
110.98 g CaCl2 1.7234 x 104 J = −9.7255 x 104 J x 23.6 g CaCl2 1 mol CaCl2
Thus, the enthalpy change, ΔH, for the reaction is −9.73 x 104 J, or −97.3 kJ, per mol CaCl2. 6.69. The energy change for the reaction is equal in magnitude and opposite in sign to the heat energy produced from the warming of the solution and the calorimeter. qcalorimeter = (9.63 kJ/°C)(33.73 − 25.00°C) = +84.069 kJ
Thus, 2.84 g C2H5OH is equivalent to 84.069 kJ heat energy. The amount of heat released by 1.000 mol C2H5OH is calculated from −84.069 kJ (opposite sign): 1.00 mol C2H5OH x
46.07 g C 2 H 5 OH 84.069 kJ = −1363.75 kJ x 2.84 g C 2 H 5 OH 1 mol C2 H 5 OH
Thus, the enthalpy change, ΔH, for the reaction is −1.36 x 103 kJ/mol ethanol. 6.70. The energy change for the reaction is equal in magnitude and opposite in sign to the heat energy produced from the warming of the solution and the calorimeter. qcalorimeter = (12.05 kJ/°C)(37.18°C − 25.00°C) = +146.769 kJ
Thus, 3.51 g C6H6 is equivalent to 146.769 kJ heat energy. The amount of heat released by 1.25 mol C6H6 is calculated from −146.769 kJ (opposite sign): 1.25 mol C6H6 x
78.11 g C6 H 6 1 mol C6 H 6
x
146.769 kJ = −4082.6 3.51 g C6 H 6
Thus, the enthalpy change, ΔH, for the reaction is −4.08 x 103 kJ/mol benzene.
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Chapter 6: Thermochemistry
6.71. Using the equations in the data, reverse the direction of the first reaction, and reverse the sign of its ΔH. Then multiply the second equation by 2, multiply its ΔH by 2, and add. Setup: N2(g)
+
2H2O(l)
→
N2H4(l) + O2(g);
ΔH = (−622.2 kJ) x (−1)
2H2(g)
+
O2(g)
→
2H2O(l);
ΔH = (−285.8 kJ) x (2)
N2(g)
+
2H2(g)
→
N2H4(l);
ΔH = 50.6 kJ
6.72. Using the equations in the data, reverse the direction of the first reaction, and reverse the sign of its ΔH. Then divide the reversed first equation by 2, divide its ΔH by 2, and add to the second equation. Setup: H2O(l) + /2O2(g)
→
H2O2(l);
ΔH = (−196.0 kJ) x (−1/2)
H2(g) + 1/2O2(g)
→
H2O(l);
ΔH = (−285.8 kJ)
H2(g) + O2(g)
→
H2O2(l);
ΔH = −187.8 kJ
6.73. Using the equations in the data, multiply the second equation by 2, and reverse its direction; do the same to its ΔH. Then multiply the first equation by 2 and its ΔH by 2. Finally, multiply the third equation by 3 and its ΔH by 3. Then add. Setup: 4NH3(g)
→ 2N2(g) + 6H2(g);
ΔH = (−91.8 kJ) x (−2)
2N2(g) + 2O2(g)
→ 4NO(g);
ΔH = (180.6 kJ) x (2)
6H2(g) + 3O2(g)
→ 6H2O(g);
ΔH = (−483.7 kJ) x (3)
4NH3(g) + 5O2(g)
→ 4NO(g) + 6H2O(g);
ΔH = −906.3 kJ
6.74. Using the equations in the data, reverse the direction of the second equation and reverse the sign of its ΔH. Then, reverse the direction of the first equation and multiply by 1/2; do the same to its ΔH. Finally, multiply the third equation and its ΔH by 1/2. Add all three equations. Setup: CH4(g)
→ C(graph) + 2H2(g);
ΔH = (−74.9 kJ) x (−1)
NH3(g)
→ 1/2N2(g) + 3/2H2(g);
ΔH = (−91.8 kJ) x (−1/2)
1/2H2(g) + C(graph) + 1/2N2(g)
→ HCN(g);
ΔH = (270.3 kJ) x (1/2)
CH4(g) + NH3(g)
→ HCN(g) + 3H2(g);
ΔH = 255.95 = 256.0 kJ
6.75. After reversing the second equation in the data, add all the equations. Setup: C2H4(g) + 3O2(g)
→
2CO2(g) + 2H2O(l);
ΔH = (−1411 kJ)
2CO2(g) + 3H2O(l) H2(g) + 1/2O2(g)
→ →
C2H6(g) + 7/2O2(g); H2O(l);
ΔH = (−1560) x (−1) ΔH = (−286 kJ)
C2H4(g) + H2(g)
→
C2H6(g);
ΔH = −137 kJ
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223
6.76. After reversing the first equation, double the second and third equations, and then add. Setup: 2CO2(g) + 2H2O(l)
→ CH3COOH(l) + 2O2(g);
ΔH = (−874 kJ) x (−1)
2C(graph) + 2O2(g) 2H2(g) + O2(g)
→ 2CO2(g); → 2H2O(l);
ΔH = (−394 kJ) x (2) ΔH = (−286 kJ) x (2)
2C(graph) + 2H2(g) + O2(g)
→ CH3COOH(l);
ΔH = −486 kJ
6.77. Write the ΔH° values (Appendix C) underneath each compound in the balanced equation:
C2H5OH(l) → C2H5OH(g) (kJ) 277.7 235.4 ΔH°vap = [ ΔH°f(C2H5OH(g))] − [ ΔH°f(C2H5OH(l))] = [−235.4] − [−277.7] = +42.3 kJ 6.78. Write the ΔH° values (Table 6.2) underneath each compound in the balanced equation:
CCl4(l)
→
135.4
CCl 4(g) 95.98
(kJ)
ΔH°vap = [ ΔH°f(CCl4(g))] − [ ΔH°f(CCl4(l))] = [−95.98] − [−135.4] kJ = +39.4 kJ 6.79. Write the ΔH° values (Table 6.2) underneath each compound in the balanced equation:
2H2S(g) + 2(20.50)
3O2(g)
→
3(0)
2H2O(l)
+
2(285.8)
2SO2(g) 2(296.8)
(kJ)
ΔH° = ΣnΔH°f(products) − ΣmΔH°f(reactants) = [2(−285.8) + 2(−296.8)] − [2(−20.50) + 3(0)] kJ = −1124.20 = −1124.2 kJ 6.80. Write the ΔH° values (Table 6.2) underneath each compound in the balanced equation:
CS2(l)
89.70
+
3O2(g)
→
3(0)
CO2 (g)
+
2SO2(g)
2(296.8) (kJ)
393.5
ΔH° = ΣnΔH°f(products) − ΣmΔH°f(reactants) = [(−393.5) + 2(−296.8)] − [(89.70) + 3(0)] = −1076.8 kJ 6.81. Write the ΔH° values (Appendix C) underneath each compound in the balanced equation:
Fe2O3(s) 825.5
+ 3CO(g)
→
3(110.5)
2Fe(s)
+
2(0)
3CO2(g) 3(393.5)
(kJ)
ΔH° = ΣnΔH°f(products) − ΣmΔH°f(reactants) = [2(0) + 3(−393.5)] − [(−825.5) + 3(−110.5)] = −23.5 kJ
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Chapter 6: Thermochemistry
6.82. Write the ΔH° values (Appendix C) underneath each compound in the balanced equation:
2PbS(s)
3O2(g)
2(98.32)
3(0)
→
2SO2(g)
2PbO(s)
2(296.8)
2(219.4)
(kJ)
ΔH° = ΣnΔH°f(products) − ΣmΔH°f(reactants) = [2(−296.8) + 2(−219.4)] − [2(−98.32) + 3(0)] = −835.76 = −835.8 kJ 6.83. Write the ΔH° values (Table 6.2) underneath each compound in the balanced equation:
HCl(g)
→
92.31
H+(aq)
Cl(aq)
0
167.2 (kJ)
ΔH° = ΣnΔH°f(products) − ΣmΔH°f(reactants) = [(0) + (−167.2)] − [−92.31] = −74.89 = −74.9 kJ 6.84. Write the ΔH° values (Table 6.2) underneath each compound in the balanced equation:
CaCO3(s) 1206.9
+ CO2(g) + H2O(l) 285.8 393.5
→
Ca2+(aq) 542.8
+
2HCO3(aq) 2(692.0) (kJ)
ΔH° = ΣnΔH°f(products) − ΣmΔH°f(reactants) = [(−542.8) + 2(−692.0)] − [(−1206.9) + (−393.5) + (−285.8)] = −40.6 kJ 6.85. Calculate the molar heat of formation from the equation with the ΔH° values below each substance; then convert to the heat for 10.0 g of MgCO3 using the molar mass of 84.3.
MgCO3(s)
→
1111.7
MgO(s) 601.2
+
CO2(g) 393.5
(kJ)
ΔH° = ΣnΔH°f(products) − ΣmΔH°f(reactants) ΔH° = −601.2 kJ + (−393.5 kJ) − (−1111.7 kJ) = 117.0 kJ Heat = 10.0 g x
1 mol 117.0 kJ x = 13.87 = 13.9 kJ mol 84.3 g
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225
6.86. Calculate the molar heat of formation from the equation with the ΔH° values below each substance; then convert to the heat for 10.0 g of BaCO3 using the molar mass of 197.3.
BaCO3 (s)
→
BaO(s)
+
548.1
1216.3
CO2(g) 393.5
(kJ)
ΔH° = ΣnΔH°f(products) − ΣmΔH°f(reactants) ΔH° = −548.1 kJ + (−393.5 kJ) − (−1216.3 kJ) = 274.7 kJ 1 mol 274.7 kJ x = 13.92 = 13.9 kJ 197.3 g mol
Heat = 10.0 g x
■
SOLUTIONS TO GENERAL PROBLEMS
6.87. The SI units of force must be kg•m/s2 (= newton, N) to be consistent with the joule, the SI unit of energy: kg • m kg • m 2 x m = = joule, J s2 s2
6.88. Solving Ep = mgh, you obtain g = Ep/mh. Thus, the SI unit of g is (kg•m2/s2)/(kg•m) = m/s2. 6.89. Using Table 1.5 and 4.184 J/cal, convert the 686 Btu/lb to J/g: 686 Btu 252 cal 4.184 J 1 lb 1 kg x x x x 1 lb 1 Btu 1 cal 0.4536 kg 103 g
= 1.594 x 103 = 1.59 x 103 J/g 6.90. Using Table 1.5 and 4.184 J/cal, convert the 51600 Btu/lb to J/g: 51600 Btu 252 cal 4.184 J 1 lb 1 kg x x x x 1 lb 1 Btu 1 cal 0.4536 kg 103 g
= 1.199 x 105 = 1.20 x 105 J/g 6.91. Substitute into the equation Ep = mgh, and convert to SI units. Ep = 1.00 lb x
9.807 m 0.4536 kg 0.9144 m x 167 ft x x s2 1 lb 3 ft =
226.433 kg • m 2 = 226 J s2
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Chapter 6: Thermochemistry
At the bottom, all the potential energy is converted to kinetic energy, so Ek = 226.433 kg•m2/s2. Because Ek = 1/2mv2, solve for v, the speed (velocity): v =
Ek = 1/2 x m
226.433 kg • m 2/ s 2 = 31.60 = 31.6 m/s 1/2 x 1.00 lb x 0.4536 kg/lb
6.92. Substitute into the equation Ek = 1/2mv2, and convert to SI units. Ek = 1/2 x 2354 lb x
⎛ 11.2 km 0.4536 kg 103 m ⎞ x x ⎜ ⎟ 1 lb s 1 km ⎠ ⎝
2
= 6.697 x 1010 kg•m2/s2 = 6.70 x 1010 J 6.93. The equation is CaCO3(s) → CaO(s) + CO2(g); ΔH = 177.9 kJ Use the molar mass of 100.08 g/mol to convert the heat per mole to heat per 27.3 g. 21.3 g CaCO3 x
1 mol CaCO3 177.9 kJ = 37.86 = 37.9 kJ x 1 mol CaCO3 100.08 g CaCO3
6.94. The equation is CaO(s) + H2O(l) → Ca(OH)2(s); ΔH = −65.2 kJ Use the molar mass in the conversion. 24.5 g CaO x
1 mol CaO 65.2 kJ x = −28.48 = −28.5 kJ 1 mol CaO 56.08 g CaO
The heat released is 28.5 kJ. 6.95. The equation is 2HCHO2(l) + O2(g) → 2CO2(g) + 2H2O(l) Use the molar mass of 46.03 g/mol to convert −30.3 kJ/5.48 g to ΔH per mole of acid. 46.03 g HCHO 2 30.3 kJ x 5.48 g HCHO 2 1 mol HCHO 2
= −254.508 = −255 kJ/mol
6.96. The equation is HC2H3O2(l) + 2O2(g) → 2CO2(g) + 2H2O(l) Use the molar mass of 60.05 g/mol to convert −52.0 kJ/3.58 g to ΔH per mole of acid. 60.05 g HC2 H 3O 2 52.0 kJ x = −872.2 = −872 kJ/mol 3.58 g HC 2 H 3O 2 1 mol HC 2 H 3O 2
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6.97. The heat gained by the water at the lower temperature equals the heat lost by the water at the higher temperature. Each heat term is s x m x Δt. This gives (4.184 J/g•°C)(54.9 g)(tf − 31.5°C) = (4.184 J/g•°C)(21.0 g)(52.7°C − tf) The specific heat and the units can be canceled from both sides to give (54.9)(tf − 31.5°C) = (21.0)(52.7°C − tf) After rearranging, you get 75.9 tf = 2836.05. This gives tf = 37.36 = 37.4°C. 6.98. The heat gained by the water at the lower temperature equals the heat lost by the water at the higher temperature. Each heat term is s x m x Δt. This gives (4.184 J/g•°C)(45.4 g)( tf − 35.7°C) = (4.184 J/g•°C)(20.5 g)(66.2°C − tf) The specific heat and the units can be canceled from both sides to give (45.4)( tf − 35.7°C) = (20.5)(66.2°C − tf) After rearranging, you get 65.9 tf = 2977.88. This gives tf = 45.18 = 45.2°C. 6.99. Divide the 235 J heat by the mass of lead and the Δt to obtain the specific heat. Specific heat =
235 J 0.128 J = 0.12798 = 121.6 g (35.5°C  20.4°C) g • °C
6.100. Divide the 47.0 J heat by the mass of copper and the Δt to obtain the specific heat. Specific heat =
47.0 J 0.385 J = 0.3848 = 35.4 g (3.45°C) g • °C
6.101. The energy used to heat the Zn comes from cooling the water. Calculate q for water: qwat = specific heat x mass x Δt qwat =
4.18 J x 50.0 g x (96.68°C − 100.00°C) = −693.88 J g • °C
The sign of q for the Zn is the reverse of the sign of q for water because the Zn is absorbing heat: qmet = −(qwat) = −(−693.88) = 693.88 J
Specific heat =
693.88 J 0.383 J = 0.3826 = 25.3 g (96.68°C  25.00°C) g • °C
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Chapter 6: Thermochemistry
6.102. The energy given up by the metal is used to heat the water. Calculate q for water: qwat = specific heat x mass x Δt qwat =
4.18 J x 26.7 g x (30.00°C − 25.00°C) = 558.03 J g • °C
The sign of q for the metal is the reverse of the sign of q for water because the metal is giving up heat: qmet = −(qwat) = −558.03 J
Specific heat =
558.03 J 0.899 J = 0.8989 = 19.6 g (30.00°C  61.67°C) g • °C
6.103. First, multiply the molarities by the volumes, in liters, to get the moles of NaOH and the moles of HCl, and thus determine the limiting reactant. Mol NaOH = M x V = 0.996 M x 0.0141 L = 0.01404 mol Mol HCl = M x V = 0.905 M x 0.0323 L = 0.02923 mol Therefore, NaOH is the limiting reactant. The total volume of the system is 14.1 mL + 32.3 mL = 46.4 mL. Since the density of water is 1.00 g/mL, the total mass of the system is 46.4 g. Next, set the heat released by the reaction (mol x enthalpy of reaction) equal to the heat absorbed by the water (s x m x Δt). This gives (0.01404 mol)(55.8 x 103 J/mol) = (4.184 J/g•°C)(46.4 g)(tf − 21.6°C) After dividing, this gives tf − 21.6°C = 4.035°C, or tf = 25.63 = 25.6°C. 6.104. First, multiply the molarities by the volumes, in liters, to get the moles of KOH and the moles of HBr and thus determine the limiting reactant. Mol KOH = M x V = 1.05 M x 0.0291 L = 0.030555 mol Mol HBr = M x V = 1.07 M x 0.0209 L = 0.022363 mol Therefore, HBr is the limiting reactant. The total volume of the system is 29.1 mL + 20.9 mL = 50.0 mL. Since the density of water is 1.00 g/mL, the total mass of the system is 50.0 g. Next, set the heat released by the reaction (mol x enthalpy of reaction) equal to the heat absorbed by the water (s x m x Δt). This gives (0.022363 mol)(55.8 x 103 J/mol) = (4.184 J/g•°C)(50.0 g)( tf − 21.8°C) After dividing, this gives tf − 21.8°C = 5.964°C, or tf = 27.76 = 27.8°C. 6.105. Use Δt and the heat capacity of 547 J/°C to calculate q: q = CΔt = (547 J/°C)(36.66 − 25.00)°C = 6.378 x 103 J (6.378 kJ)
Energy is released in the solution process in raising the temperature, so ΔH is negative: ΔH =
6.378 kJ 23.95 g LiOH x = −23.57 = −23.6 kJ/mol 1 mol LiOH 6.48 g LiOH
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229
6.106. Use Δt and the heat capacity of 682 J/°C to calculate q: q = CΔt = (682 J/°C)(14.14 − 25.00)°C = −7.4065 x 103 J (−7.4065 kJ)
Energy is absorbed in the solution process in lowering the temperature, so the sign of ΔH must be reversed, making the heat positive: ΔH =
101.1 g KNO3 7.4065 kJ x = 34.91 = 34.9 kJ/mol 21.45 g KNO3 1 mol KNO3
6.107. Use Δt and the heat capacity of 13.43 kJ/°C to calculate q: q = CΔt = (13.43 kJ/°C)(35.84 − 25.00)°C = 145.581 kJ
As in the previous two problems, the sign of ΔH must be reversed, making the heat negative: ΔH =
60.05 g HC 2 H 3O 2 145.581 kJ x 10.00 g HC2 H 3O 2 1 mol HC2 H 3O 2
= −874.21 = −874.2 kJ/mol
6.108. Use Δt and the heat capacity of 15.8 kJ/°C to calculate q: q = CΔt = (15.8 kJ/°C)(20.54 − 20.00)°C = 8.53 kJ
As in the previous three problems, the sign of ΔH must be reversed, making the heat negative: ΔH =
8.53 kJ 150.1 g sugar x = −2.336 x 103 = −2.3 x 103 kJ/mol 0.548 g sugar 1 mol sugar
6.109. Using the equations in the data, reverse the direction of the first reaction, and reverse the sign of its ΔH. Then add the second and third equations and their ΔH's. H2O(g) + SO2(g)
→
H2S(g) + 3/2O2(g);
ΔH = (−518 kJ) x (−1)
H2(g) + 1/8S8(rh.) +
1/2O2(g) O2(g)
→ →
H2O(g); SO2(g);
ΔH = (−242 kJ) ΔH = (−297 kJ)
H2(g)
1/8S8(rh.)
→
H2S(g);
ΔH = −21 kJ
+
6.110. Using the equations in the data, reverse the direction of the reaction involving oxidation of the glycol, and reverse the sign of its ΔH. Then add half of the reaction involving oxidation of C2H4O, and half of its ΔH. 2CO2(g) + 3H2O(l)
→ HOC2H4OH(l) + 5/2O2(g);
ΔH = (−1189.8 kJ) x (−1)
C2H4O(g) + 5/2O2(g)
→ 2CO2(g) + 2H2O(l);
ΔH = (−2612.2 kJ) x (1/2)
C2H4O(g) + H2O(l)
→ HOC2H4OH(l);
ΔH = −116.3 kJ
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Chapter 6: Thermochemistry
6.111. Write the ΔH° values (Table 6.2) underneath each compound in the balanced equation.
CH4 (g)
+
→
H2 O(g)
CO(g) + 3H2(g) 110.5
241.8
74.9
3(0) (kJ)
ΔH° = [−110.5 + 3(0)] − [(−74.87) + (−241.8)] = 206.17 = 206.2 kJ 6.112. Write the ΔH° values (Table 6.2) underneath each compound in the balanced equation.
2CH4 (g)
O2(g)
2(74.87)
2CO(g)
4H2 (g)
2(110.5)
4(0)
→
0
(kJ)
ΔH° = [2(−110.5) + 4(0)] − [2(−74.87) + 0] = −71.26 = −71.3 kJ 6.113. Write the ΔH° values (Table 6.2) underneath each compound in the balanced equation. The ΔHf° of −635 kJ/mol CaO is given in the problem.
→
CaCO3 (s)
CaO(s) + CO2(g) 635
1206.9
(kJ)
393.5
ΔH° = [(−635) + (−393.5)] − [−1206.9] = 178.4 = 178 kJ 6.114. Write the ΔH° values (Table 6.2) underneath each compound in the balanced equation.
→
2NaHCO3(s)
Na2CO3(s) 1130.8
2(950.8)
+
H2O(g) + CO2(g) 241.8
393.5
(kJ)
ΔH° = [(−1130.8) + (−241.8) + (−393.5)] − [2(−950.8)] = 135.5 kJ 6.115. Calculate the molar heat of reaction from the equation with the ΔH° values below each substance; then convert to the heat for the reaction at 25°C.
2H2(g) + 0.0 kJ
O2(g) 0.0 kJ
→
2H2O(l) 285.8 kJ
ΔH° = 2 x (−285.8 kJ) − 0 − 0 = −571.6 kJ The moles of oxygen in 2.000 L with a density of 1.11 g/L is 2.000 L O2 x
1.11 g O 2 1 mol O 2 x = 0.06937 mol O2 1 L O2 32.0 g O 2
The heat of reaction from 0.06937 mol O2 is 0.6937 mol O2 x
571.6 kJ = −39.654 = −39.7 kJ mol O 2
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231
6.116. Calculate the molar heat of reaction from the equation with the ΔH° values below each substance; then convert to the heat for the reaction at 25°C.
Cl2(g) + 2Na(s) → 2NaCl(s) 0.0 kJ
0.0 kJ
 411.1 kJ
ΔH° = 2 x (− 411.1 kJ) − 0 − 0 = −822.2 kJ The moles of chlorine in 4.000 L Cl2 with a density of 2.46 g/L is 4.000 L Cl2 x
2.46 g Cl2 1 mol Cl2 × = 0.1387 mol Cl2 L Cl2 70.9 g Cl2
The heat of reaction from 0.1387 mol Cl2 is 822.2 kJ = 114.03 = 114 kJ mol Cl2
0.1387 mol Cl2 x
6.117. Let ΔHf° = the unknown standard enthalpy of formation for sucrose. Then write this symbol and the other ΔHf° values underneath each compound in the balanced equation, and solve for ΔHf° using the ΔH° of −5641 kJ for the reaction. C12H22O11(s)
+
ΔH f °
12O2(g) 12(0)
→ 12CO2(g)
+ 11H2O(l)
12(393.5)
11(285.8)
(kJ)
ΔH° = −5641 = [12(−393.5) + 11(−285.8)] − [ΔHf° + 12(0)] ΔHf° = −2225 kJ/mol sucrose 6.118. Let ΔHf° = the unknown standard enthalpy of formation for acetone. Then write this symbol and the other ΔHf° values underneath each compound in the balanced equation, and solve for ΔHf° using the ΔH° of −1791 kJ for the reaction.
CH3COCH3(l) ΔH f °
+
4O2(g)
→
4(0)
3CO2(g) 3(393.5)
+
3H2O(l) 3(285.8)
ΔH° = −1791 = [3(−393.5) + 3(−285.8)] − [ΔHf° + 4(0)] ΔHf° = −246.9 = −247 kJ/mol acetone
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(kJ)
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Chapter 6: Thermochemistry
6.119. First, calculate the heat evolved from the molar amounts represented by the balanced equation. From Appendix C, obtain the individual heats of formation, and write those below the reactants and products in the balanced equation. Then multiply the molar heats of formation by the number of moles in the balanced equation, and write those products below the molar heats of formation. 2Al(s)
+
3NH4 NO3(s)
→
3N2(g)
+
Al2O3(s)
6H2 O(g) +
0.0 kJ
365.6 kJ
0.0 kJ
241.8
1675.7 kJ (kJ/mol)
0.0 kJ
1096.8 kJ
0.0 kJ
1450.8
1675.7 kJ (kJ/eqn.)
The total heat of reaction of 2 mol Al and 3 mol NH4NO3 is ΔH = −1450.8 − 1675.7 − (−1096.8 kJ) = −2029.7 kJ Now, 245 kJ represents the following fraction of the total heat of reaction: 245 kJ = 0.1207 2029.7 kJ Thus, 245 kJ requires the fraction 0.1207 of the moles of each reactant: 0.1207 x 2, or 0.2414, mol of Al and 0.1207 x 3, or 0.3621, mol of NH4NO3. The mass of each reactant and the mass of the mixture are as follows: 0.2414 mol Al x 26.98 g/mol Al = 6.513 g of Al 0.3621 mol NH4NO3 x 80.04 g/mol NH4NO3 = 28.98 g NH4NO3 6.513 g Al + 28.98 g NH4NO3 = 35.49 = 35.5 g of the mixture 6.120. First, calculate the heat evolved from the molar amounts represented by the balanced equation. From Appendix C, obtain the individual heats of formation, and write those below the reactants and products in the balanced equation. Then multiply the molar heats of formation by the number of moles in the balanced equation, and write those products below the molar heats of formation.
2Al(s) + Fe2O3(s) 0.0 kJ 0.0 kJ
825.5 kJ 825.5 kJ
→
2Fe(l)
+
+12.40 kJ +24.80 kJ
Al2O3(s) 1675.7 (kJ) (kJ/mol) 1675.7 (kJ) (kJ/eqn)
The total heat of reaction of 2 mol Al and 1 mol Fe2O3 is ΔH = +24.80 − 1675.7 − (−825.5 kJ) = −825.40 kJ Now, 348 kJ represents the following fraction of the total heat of reaction: 348 kJ = 0.4216 825.40 kJ
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233
Thus, 348 kJ requires the fraction 0.4216 of the moles of each reactant: 0.4216 x 2, or 0.8432, mol of Al and 0.4216 x 1, or 0.4216, mol of Fe2O3. The mass of each reactant and the mass of the mixture are as follows: 0.8432 mol Al x 26.98 g/mol Al = 22.750 g Al 0.4216 mol Fe2O3 x 159.7 g/mol Fe2O3 = 67.331 g Fe2O3 22.750 g Al + 67.331 g Fe2O3 = 90.081 = 90.1 g of the mixture 6.121. White phosphorus is a yellowishwhite, waxy substance often sold in the form of sticks looking something like fat crayons. It is not found in modern matches. It was outlawed in the early 1900s because workers began to show symptoms of white phosphorus poisoning, in which the jawbone disintegrated. 6.122. The phosphorus compound in “strike anywhere” matches is tetraphosphorus trisulfide, P4S3. It burns in air in the very exothermic reaction P4S3(s) + 8O2(g) → P4O10(s) + 3SO2(s)
■
SOLUTIONS TO STRATEGY PROBLEMS 1/ 2
⎛ 2E ⎞ 6.123. v = ⎜ ⎟ ⎝ m ⎠
1/ 2
⎛ 2 x 15.75 J ⎞ = ⎜ ⎟ 3 ⎝ 56.6 x 10 kg ⎠
= 23.591 = 23.6 m/s
6.124. No other information is needed to answer either question. 2SO2(g) + O2(g) → 2SO3(g);
ΔH = −198 kJ
3SO3(g) → 3SO2(g) + 3/2O2(g);
ΔH = 297 kJ
6.125. The reaction is 4Fe(s) + 3O2(g) → 2Fe2O3(s);
ΔH = −1651 kJ
The heat released is ⎛ 1651 kJ ⎞ ⎛ 1 mol Fe 2 O3 ⎞ 10.3 g Fe2O3 x ⎜ ⎟ = 53.24 = 53.2 kJ ⎟ x ⎜ ⎝ 159.69 g ⎠ ⎝ 2 mol Fe 2 O3 ⎠ If ozone gas was burned instead, you would need to know the heat of the following reaction: 2O3(g) → 3O2(g) ⎛ 32.00 g ⎞ ⎛ 1 mol O 2 ⎞ 6.126. 7.60 kJ x ⎜ ⎟ x ⎜ 1 mol O ⎟ = 0.5024 = 0.502 g ⎝ 484 kJ ⎠ ⎝ 2 ⎠
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Chapter 6: Thermochemistry
6.127. For 2 mol water produced, 2 mol H2 reacted. This is 2 x 2.016 g/mol = 4.032 g. 6.128. The heat lost by metal is equal to the heat gained by calorimeter. (0.449 J/g•°C)(m)(95.4°C − 51.9°C) = (4.18 J/g•°C)(284 g)(51.9°C − 32.2°C) Solve for the mass of iron.
m =
(4.18 J/ g • °C)( 284 g)(51.9°C  32.2°C) = 1197.3 = 1.20 x 103 g (0.449 J/ g • °C)(95.4°C  51.9°C)
6.129. The heat released in the reaction is equal to the heat gained by the calorimeter. ⎛ 1 mol C6 H 5 COOH ⎞ m x ⎜ ⎟ x (3226 kJ/mol) = (12.41 kJ/°C)(27.65°C − 23.44°C) 122.12 g ⎝ ⎠ Solve for the mass of benzoic acid.
m =
(12.41 kJ/ °C)(27.65°C  23.44°C)(122.12 g/mol) = 1.977 = 1.98 g 3226 kJ/mol
6.130. Multiply the first reaction by 2. Then reverse the second reaction, and multiply it by 2. Finally, multiply the third reaction by 3. Add the resulting equations together to get the reaction of interest. 2A + 2B
→
4C
ΔH = (−447 kJ) x (2)
4E 6D + 3B
→ →
2A + 6D 6F
ΔH = (−484 kJ) x (−2) ΔH = (−429 kJ) x (3)
4E + 5B
→
4C + 6F
ΔH = −1213 kJ
6.131. ΔH°comb = Σn ΔH of (products) − Σm ΔH of (reactants) = [ ΔH of (P4O10) + 3 ΔH of (SO2)] − [ ΔH of (P4S3) + 8 ΔH of (O2)] −3651 kJ = [(−3009.9 kJ) + 3(−296.8 kJ)] − [(1 mol) ΔH of (P4S3) + 8(0 kJ)] Solve for the enthalpy of formation of P4S3.
ΔH of (P4S3) =
(3009.9 kJ) + (3)(296.8 kJ)  (3651 kJ) = −249.3 = −249 kJ/mol 1 mol
The value from Appendix C is −224.6 kJ/mol, which is a difference of 10.9%. 6.132. ΔH°comb = Σn ΔH of (products) − Σm ΔH of (reactants) = [7 ΔH of (CO2) + 4 ΔH of (H2O)] − [ ΔH of (C6H5CH3) + 9 ΔH of (O2)] −3908 kJ = [7(−393.5 kJ) + 4(−285.8 kJ)] − [(1 mol) ΔH of (C6H5CH3) + 9(0 kJ)]
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235
Solve for the enthalpy of formation of C6H5CH3.
ΔH of ( C6H5CH3) =
■
(7)(393.5 kJ) + (4)(285.8 kJ)  (3908 kJ) = 10.3 = 10. kJ/mol 1 mol
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
6.133. The heat lost, q, by the water(s) with a temperature higher than the final temperature must be equal to the heat gained, q, by the water(s) with a temperature lower than the final temperature.
q(lost by water at higher temp) = q(gained by water at lower temp) Σ (s x m x Δt) = Σ (s x m x Δt) Divide both sides of the equation by the specific heat, s, to eliminate this term, and substitute the other values. Since the mass of the water at 50.0°C is greater than the sum of the masses of the other two waters, assume the final temperature will be greater than 37°C and greater than 15°C. Use this to set up the three Δt expressions, and write one equation in one unknown, letting t equal the final temperature. Simplify by omitting the “grams” from 45.0 g, 25.0 g, and 15.0 g. Σ (m x Δt) = Σ (m x Δt) 45.0 x (50.0°C − t) = 25.0 x (t − 15.0°C) + 15.0 x (t − 37°C) 2250°C − 45.0 t = 25.0 t − 375°C + 15.0 t − 555.0°C −85.0 t = −3180°C
t = 37.41 = 37.4°C (the final temperature) 6.134. The heat lost, q, by the substances(s) with a temperature higher than the final temperature must be equal to the heat gained, q, by the substances(s) with a temperature lower than the final temperature.
q(lost) = q(gained) Σ (s x m x Δt) = Σ (s x m x Δt) Since the temperature of the iron, 95.0°C, is far larger than the temperatures of the other two substances, assume the final temperature will be greater than 35.5°C and greater than 25°C. Then, write the heat term for the iron on the left and the sum of the heat terms for water and ethanol on the right. Since the masses, m, are equal, divide both sides by m to eliminate this term, and solve one equation in one unknown, letting t equal the final temperature. (To simplify the setup, only the temperature unit is retained.) Σ (s x Δt) = Σ (s x Δt) 0.449 x (95.0°C − t) = 2.43 x (t − 35.5°C) + 4.18 x (t − 25.0°C) 42.655°C − 0.449 t = 2.43 t − 86.265°C + 4.18 t − 104.5°C −7.059 t = −233.42°C
t = 33.06 = 33.1°C (the final temperature)
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Chapter 6: Thermochemistry
The final temperature is actually less than that of the ethanol, but rewriting the equation to account for this still gives the same final temperature. 6.135. First, calculate the mole fraction of each gas in the product, assuming 100 g product: Mol CO = 33 g CO x 1 mol CO/28.01 g CO = 1.178 mol CO Mol CO2 = 67 g CO2 x 1 mol CO2/44.01 g CO2 = 1.522 mol CO2 1.178 mol CO = 0.4363 (1.178 + 1.522) mol
Mol frac. CO = Mol frac. CO2 =
1.522 mol CO 2 = 0.5637 (1.178 + 1.522) mol
Now, calculate the starting moles of C, which equal the total moles of CO and CO2: Starting mol C(s) = 1.00 g C x 1 mol C/12.01 g C = 0.08326 mol C = mol CO + CO2 Use the mole fractions to convert mol CO + CO2 to mol CO and mol CO2: 0.4363 mol CO x 0.08326 mol total = 0.03633 mol CO 1 mol total
Mol CO = Mol CO2 =
0.5637 mol CO 2 x 0.08326 mol total = 0.04693 mol CO2 1 mol total
Now, use enthalpies of formation (Table 6.2) to calculate the heat of combustion for both:
C(s)
+
0.03633 0 0 C(s) 0.04693 0 0
1/2O2 (g)
→
0.03633 (mol) 110.5 (kJ/mol) 110.5 (kJ/mol)
excess 0 0 +
O2(g) excess 0 0
CO(g)
→
CO2(g) 0.04693 (mol) 393.5 (kJ/mol) 18.47 (kJ/0.04693 mol)
Total ΔH = −4.014 + (−18.47) = −22.48 = −22 kJ Heat released = 22 kJ
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237
6.136. The balanced equations are CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) C2H6(g) + 3.5O2(g) → 2CO2(g) + 3H2O(l) First, calculate the moles of CH4 and C2H6 (let 80.0% = 0.800 g and 20.0% = 0.200 g). Mol CH4 = 0.800 g CH4 x
1 mol CH 4 = 0.04987 mol CH4 16.04 g CH 4 1 mol C2 H 6 = 0.006651 mol C2H6 30.07 g C 2 H 6
Mol C2H6 = 0.200 g C2H6 x
Now, use the enthalpies of formation (Table 6.2) to calculate the total heat: CH4(g)
+
→
2O2(g)
CO 2(g)
+
2H2O(l)
0.04987
excess
0.04987
2(0.04987) (mol)
74.87 3.734
0 0
393.5 19.62
285.8 (kJ/mol)
C2H6(g) 0.006651 84.68 0.5632
+
3.5O 2
→
2CO2(g)
28.51 (kJ/eqn.)
+
3H 2O(l)
excess
2(0.006651)
3(0.006651) (mol)
0 0
393.5 5.234
285.8 (kJ/mol) 5.703 (kJ/eqn.)
For the combustion of CH4: ΔH = [(−28.51) + (−19.62)] − [(−3.734) + 0] = −44.395 kJ For the combustion of C2H6: ΔH = [(−5.234) + (−5.703)] − [(−0.5632) + 0] = −10.373 kJ For the combustion of the total (1.00 g) of CH4 and C2H6: ΔH = −44.395 + (−10.373) = −54.768 = −54.8 kJ Heat evolved = 54.8 kJ 6.137. The ΔH° for the reactions to produce CO and CO2 in each case is equal to the ΔHf° of the respective gases. Thus ΔH° to produce CO is −110.5 kJ/mol, and ΔH° to produce CO2 is −393.5 kJ/mol. Since 1 mol graphite is needed to produce 1 mol either CO or CO2, one equation in one unknown can be written for the heat produced, using m for the moles of CO and (2.00 − m) for moles of CO2. − 481 kJ = m x (−110.5 kJ) + (2.00 − m) x (−393.5 kJ) − 481 = −110.5 m − 787 + 393.5 m mol CO = m = 1.081; mass CO = 1.081 x 28.0 g/mol = 30.26 = 30.3 g mol CO2 = (2.00 − m) = 0.919; mass CO2 = 0.919 mol x 44.0 g/mol = 40.4 = 40. g
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Chapter 6: Thermochemistry
6.138. The ΔH° values for each reaction are as follows: CH4(g) + 2O2(g)
→
CO2(g) + 2H2O(l)
ΔH° = −393.5 kJ + 2(−285.8 kJ) − (−74.87 kJ) − 0 = −890.23 kJ C2H4(g) + 3O2(g)
→
2CO2(g) + 2H2O(l)
ΔH° = 2(−393.5 kJ) + 2(−285.8 kJ) − (52.47 kJ) − 0 = −1411.07 kJ Since 10.0 g of CH4 and C2H4 produce 520 kJ of heat, one equation in one unknown can be written for the heat produced, using m for the mass of CH4 and (10.00 − m) for the mass of C2H4. Use m /16.0 for the moles of CH4 and (10.00 − m) /28.0 for the moles of C2H4. −520 kJ = (m /16.0) x (−890.23 kJ) + [(10.00 − m) /28.0] x (−1411.07 kJ) −520 = −55.63 m + (−503.95) + 50.39 m Mass CH4 = m = 16.05 ÷ 5.24 = 3.062 = 3.1 g Mass percentage CH4 = (3.06 g ÷ 10.0 g) x 100% = 3.06 = 3.1% 6.139. The equation is 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g); ΔH° = −906 kJ First, determine the limiting reactant by calculating the moles of NH3 and of O2; then, assuming one of the reactants is totally consumed, calculate the moles of the other reactant needed for the reaction. 10.0 g NH3 x 20.0 g O2 x
1 mol NH 3 = 0.5872 mol NH3 17.03 g NH 3
1 mol O 2 = 0.6250 mol O2 32.00 g O 2
0.6250 mol O2 x
4 mol NH 3 = 0.500 mol NH3 needed 5 mol O 2
Because NH3 is present in excess of what is needed, O2 must be the limiting reactant. Now calculate the heat released on the basis of the complete reaction of 0.622250 mol O2: ΔH =
906 kJ x 0.6250 mol O2 = −113.25 = −113 kJ 5 mol O 2
The heat released by the complete reaction of the 20.0 g (0.6250 mol) of O2(g) is 113 kJ.
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239
6.140. The equation is CS2(g) + 3Cl2(g) → S2Cl2(g) + CCl4(g); ΔH° = −230 kJ First, determine the limiting reactant by calculating the moles of CS2 and of Cl2; then, assuming one of the reactants is totally consumed, calculate the moles of the other reactant needed for the reaction. 10.0 g CS2 x
1 mol CS2 = 0.1314 mol CS2 76.13 g CS2
10.0 g Cl2 x
1 mol Cl2 = 0.1410 mol Cl2 70.91 g Cl2
0.1410 mol Cl2 x
1 mol CS2 = 0.04700 mol CS2 3 mol Cl2
Because CS2 is present in excess of what is needed, Cl2 must be the limiting reactant. Now calculate the heat released on the basis of the complete reaction of 0.1410 mol Cl2. ΔH =
230 kJ x 0.1410 mol Cl2 = −10.810 = −10.8 kJ 3 mol Cl2
The heat released by the complete reaction of 10.0 g of Cl2(g) is 10.8 kJ. 6.141. The equation is N2(g) + 3H2(g) → 2NH3(g); ΔH° = −91.8 kJ a.
To find the heat evolved from the production of 1.00 L of NH3, convert the 1.00 L to mol NH3 using the density and molar mass (17.03 g/mol). Then convert the moles to heat (ΔH) using ΔH°. Mol NH3 = ΔH =
b.
d xV 0.696 g/L x 1 L = = 0.04087 mol NH3 17.03 g/mol MM
91.8 kJ x 0.04087 mol NH3 = −1.876 (1.88 kJ heat evolved) 2 mol NH 3
First, find the moles of N2 using the density and molar mass (28.02 g/mol). Then convert to the heat needed to raise the N2 from 25°C to 400°C. Mol N2 =
d xV 1.145 g/L x 0.500 L = 0.02043 mol N2 = MM 28.02 g/mol
0.02043 mol N2 x
29.12 J x (400 − 25)°C = 223.2 J (0.2232 kJ) mol • °C
Percent heat for N2 =
0.2232 kJ x 100% = 11.89 = 11.9% 1.876 kJ
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Chapter 6: Thermochemistry
6.142. The equation is 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g); ΔH° = −1124 kJ a.
First, calculate the moles of SO2 using the density and molar mass (64.07 g/mol). Then convert the moles to heat (ΔH) using ΔH°. Mol SO2 = ΔH =
b.
d xV 2.62 g/L x 1 L = 0.04089 mol SO2 = MM 64.07 g/mol
1124 kJ x 0.04089 mol SO2 = −22.98 (23.0 kJ heat evolved) 2 mol SO 2
Recall that 1.000 L of SO2 = 0.04089 mol SO2. Use this with the molar heat capacity of 30.2 J/(mol•°C) to calculate the heat needed to raise the temperature of SO2 from 25°C to 500°C. 0.04089 mol SO2 x
30.2 J x (500 − 25)°C = 586.6 J (0.5866 kJ) (mol • °C )
Percent heat for SO2 =
0.5866 kJ x 100% = 2.552 = 2.55% 22.98 kJ
6.143. The glucose equation is C6H12O6 + 6O2 → 6CO2 + 6H2O; ΔH° = −2802.8 kJ Convert the 2.50 x 103 kcal to mol of glucose using the ΔH° of −2802.8 kJ for the reaction and the conversion factor of 4.184 kJ/kcal: 2.50 x 103 kcal x
4.184 kJ 1 mol glucose x = 3.7319 mol glucose 1.000 kcal 2802.8 kJ
Next, convert mol glucose to mol LiOH using the above equation for glucose and the equation for LiOH: 2LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l) 3.7319 mol glucose x
6 mol CO 2 2 mol LiOH x = 44.783 mol LiOH 1 mol CO 2 1 mol glucose
Finally, use the molar mass of LiOH to convert moles to mass: 44.783 mol LiOH x
23.95 g LiOH = 1.0725 x 103 g (1.07 kg) LiOH 1 mol LiOH
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241
6.144. Using ΔHf° = −1273 kJ for glucose, the conversion of glucose to heat and ΔH° in the body is C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l); ΔH° = −2802.8 kJ Convert the 1.00 x 102 kcal to mol of glucose, using the ΔH° of −2802.8 kJ: 100 kcal x
4.184 kJ 1 mol glucose x = 0.14927 mol glucose 1.000 kcal 2802.8 kJ
Next, convert mol glucose to mol KO2 using the above equation for glucose and the equation for KO2: 4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) 0.14927 mol glucose x 1.1942 mol KO2 x
6 mol O 2 4 mol KO 2 x 1 mol glucose 3 mol O 2
71.1 g KO 2 = 84.91 = 84.9 g KO2 1 mol KO 2
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= 1.1942 mol KO2
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CHAPTER 7
Quantum Theory of the Atom
■
SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 7.1. Rearrange the equation c = νλ, which relates wavelength to frequency and the speed of light (3.00 x 108 m/s):
λ =
c
ν
=
3.00 x 108 m/s = 7.672 x 10−7 = 7.67 x 10−7 m, or 767 nm 3.91 x 1014 /s
7.2. Rearrange the equation c = νλ, which relates frequency to wavelength and the speed of light (3.00 x 108 m/s). Recognize that 456 nm = 4.56 x 10−7 m.
ν =
c
λ
=
3.00 x 108 m/s = 6.578 x 1014 = 6.58 x 1014 /s 4.56 x 107 m
7.3. First, use the wavelengths to calculate the frequencies from c = νλ. Then calculate the energies using E = hν.
ν = ν = ν =
c
λ c
λ c
λ
=
3.00 x 108 m/s = 3.00 x 1014 /s 1.0 x 106 m
=
3.00 x 108 m/s = 3.00 x 1016 /s 1.0 x 108 m
=
3.00 x 108 m/s = 3.00 x 1018 /s 1.0 x 1010 m
E = hν = 6.63 x 10−34 J•s x 3.00 x 1014 /s = 1.989 x 10−19 = 2.0 x 10−19 J (IR) E = hν = 6.63 x 10−34 J•s x 3.00 x 1016 /s = 1.989 x 10−17 = 2.0 x 10−17 J (UV) E = hν = 6.63 x 10−34 J•s x 3.00 x 1018 /s = 1.989 x 10−15 = 2.0 x 10−15 J (x ray)
The xray photon (shortest wavelength) has the greatest amount of energy; the infrared photon (longest wavelength) has the least amount of energy.
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243
7.4. From the formula for the energy levels, E = −RH/n2, obtain the expressions for both Ei and Ef. Then calculate the energy change for the transition from n = 3 to n = 1 by subtracting the lower value from the upper value. Set this equal to hν. The result is 8RH ⎡  RH ⎤ ⎡  RH ⎤ ⎢ 9 ⎥ − ⎢ 1 ⎥ = 9 = hν ⎣ ⎦ ⎣ ⎦
The frequency of the emitted radiation is
ν =
8RH 8 2.179 x 1018 J = x = 2.921 x 1015 = 2.92 x 1015 /s 9 9 6.63 x 1034 J • s
Since λ = c/ν,
λ =
3.00 x 108 m/s = 1.027 x 10−7 = 1.03 x 10−7 m, or 103 nm 15 2.92 x 10 /s
7.5. Calculate the frequency from c = νλ, recognizing that 589 nm is 5.89 x 10−7 m.
ν =
c
λ
=
3.00 x 108 m/s = 5.093 x 1014 = 5.09 x 1014 /s 7 5.89 x 10 m
Finally, calculate the energy difference. E = hν = 6.63 x 10−34 J•s x 5.093 x 1014 /s = 3.3766 x 10−19 = 3.38 x 10−19 J 7.6. To calculate wavelength, use the mass of an electron (m = 9.11 x 10−31 kg), and Planck's constant (h = 6.63 x 10−34 J•s, or 6.63 x 10−34 kg•m2/s).
λ =
h 6.63 x 1034 kg • m 2 /s = 9.11 x 1031 kg x 2.19 x 106 m/s mv
= 3.323 x 10−10 m (332 pm)
7.7.
■
a.
The value of n must be a positive whole number greater than zero. Here, it is zero. Also, if n is zero, there are no allowed values for l and ml.
b.
The values for l can range only from zero to (n − 1). Here, l has a value greater than n.
c.
The values for ml range from −l to +l. Here, ml has a value greater than l.
d.
The value for ms is either + 1/2 or −1/2. Here, it is zero.
ANSWERS TO CONCEPT CHECKS
7.1. The frequency and wavelength are inversely related. Therefore, if the frequency is doubled, the wavelength is halved. Red light has a wavelength around 700 nm, so doubling its frequency halves its wavelength to about 350 nm, which is in the ultraviolet range just beyond the visible spectrum.
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Chapter 7: Quantum Theory of the Atom
7.2. Since the transitions are between adjacent levels, the energylevel diagram must look something like the following diagram, with the red transition between two close levels and the blue transition between two more widely spaced levels. (The three levels could be spaced so the red and blue transitions are interchanged, with the blue transition above the red one.)
Level 3 750 nm (red line) Level 2 400 nm (blue line) Level 1 The transition from the top level to the lowest level would correspond to a transition that is greater in energy change than either of the other two transitions. Thus, the three transitions, from lowest to highest energy change, are in this order: red, blue, and the transition from the highest to lowest level. The last transition would have the highest frequency and therefore the shortest wavelength. It would lie just beyond the blue portion of the visible spectrum in the ultraviolet region. 7.3. The de Broglie relation says the wavelength of a particle is inversely proportional to both mass and speed. So, to maintain the wavelength constant while the mass increases would mean the speed would have to decrease. In going from a particle with the mass of an electron to one with that of a proton, the speed would have to decrease by a factor of about 2000 in order to maintain the same wavelength. The proton would have to have a speed approximately 2000 times slower than an electron of the same wavelength.
■
ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
7.1. Light is a wave, which is a form of electromagnetic radiation. In terms of waves, light can be described as a continuously repeating change, or oscillation, in electric and magnetic fields that can travel through space. Two characteristics of light are wavelength (often given in nanometers, nm) and frequency. 7.2. The relationship among the different characteristics of light waves is c = νλ, where ν is the frequency, λ is the wavelength, and c is the speed of light. 7.3. Starting with the shortest wavelengths, the electromagnetic spectrum consists of gamma rays, x rays, far ultraviolet (UV), near UV, visible light, near infrared (IR), far IR, microwaves, radar, and TV/FM radio waves (longest wavelengths). 7.4. The term quantized means the possible values of the energies of an atom are limited to only certain values. Planck was trying to explain the intensity of light of various frequencies emitted by a hot solid at different temperatures. The formula he arrived at was E = nhν, where E is energy, n is a whole number (n = 1, 2, 3, …), h is Planck's constant, and ν is frequency.
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7.5. Photoelectric effect is the term applied to the ejection of electrons from the surface of a metal or from other materials when light shines on it. Electrons are ejected only when the frequency (or energy) of light is larger than a certain minimum, or threshold, value that is constant for each metal. If a photon has a frequency equal to or greater than this minimum value, then it will eject one electron from the metal surface. 7.6. The wave–particle picture of light regards the wave and particle depictions of light as complementary views of the same physical entity. In the equation E = hν, E is the energy of a light particle (photon) and ν is the frequency of the associated wave. 7.7. The equation that relates the particle properties of light is E = hν. The symbol E is energy, h is Planck's constant, and ν is the frequency of the light. 7.8. According to physical theory at Rutherford's time, an electrically charged particle revolving around a center would continuously lose energy as electromagnetic radiation. As an electron in an atom lost energy, it would spiral into the nucleus (in about 10−10 s). Thus, the stability of the atom could not be explained. 7.9. According to Bohr, an electron in an atom can have only specific energy values. An electron in an atom can change energy only by going from one energy level (of allowed energy) to another energy level (of allowed energy). An electron in a higher energy level can go to a lower energy level by emitting a photon of an energy equal to the difference in energy. However, when an electron is in its lowest energy level, no further changes in energy can occur. Thus, the electron does not continuously radiate energy as was thought at Rutherford's time. These features solve the difficulty alluded to in Question 7.8. 7.10. Emission of a photon occurs when an electron in a higher energy level undergoes a transition to a lower energy level. The energy lost is emitted as a photon. 7.11. Absorption, the reverse of emission, occurs when a photon of a certain required energy is absorbed by a certain electron in an atom. The energy of the photon must be equal to the energy necessary to excite the electron of the atom from a lower energy level, usually the lowest, to a higher energy level. 7.12. The quantum corral, shown in Figure 7.21, shows evidence of electron waves. A practical example of diffraction is the operation of the scanning tunneling microscope. 7.13. The square of a wave function equals the probability of finding an electron within a region of space. 7.14. The uncertainty principle says we can no longer think of the electron as having a precise orbit in an atom similar to the orbit of the planets around the sun. This principle says it is impossible to know with absolute precision both the position and the speed of a particle such as an electron.
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Chapter 7: Quantum Theory of the Atom
7.15. Quantum mechanics vastly changes Bohr's original picture of the hydrogen atom in that we can no longer think of the electron as having a precise orbit around the nucleus in this atom. Recall that Bohr's theory depended on the hydrogen electron having specific energy values and thus specific positions and speeds around the nucleus. But quantum mechanics and the uncertainty principle say it is impossible to know with absolute precision both the speed and the position of an electron. So Bohr's energy levels are only the most probable paths of the electrons. 7.16. a.
The principal quantum number can have an integer value between one and infinity.
b.
The angular momentum quantum number can have any integer value between zero and (n − 1).
c.
The magnetic quantum number can have any integer value between −l and +l.
d.
The spin quantum number can be either +1/2 or −1/2.
7.17. The notation is 4f. This subshell contains seven orbitals. 7.18. An s orbital has a spherical shape. A p orbital has two lobes positioned along a straight line through the nucleus at the center of the line (a dumbbell shape). 7.19. The answer is d, an electron in the n = 1 level is higher in energy than an electron in the n = 4 level. 7.20. The answer is a, I only. 7.21. The answer is c, transition from the n = 3 to the n = 1 level. 7.22. The answer is a, 8.66 x 10−6 m.
■
ANSWERS TO CONCEPT EXPLORATIONS
7.23. a.
Red light has a greater wavelength than blue light.
b.
The frequency of blue light is greater than that of red light.
c.
The energy of blue light is greater than that of red light.
d.
No, the speed of red light and the speed of blue light are the same, the constant speed of light.
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e.
247
The energy of three photons of blue light is three times the energy of one photon of blue light. To compare the two energies, first note that the energy of n photons of light is E = nhν. Also, since ν = c/λ, the energy can be written as E = nhc/λ. Now find the ratio as follows. nblue hc
Eblue λblue n λ (2)(704 nm) = = blue red = = 1.040 = 1.04 n hc nred λblue (3)(451 nm) Ered red
λred
Thus, the energy of two photons corresponding to a wavelength of 451 nm (blue light) is 1.04 times the energy of three photons corresponding to a wavelength of 704 nm (red light). f.
The frequency and energy of the light are
ν =
c
λ
=
3.00 x 108 m/s = 2.459 x 1014/s 1.22 x 106 m
E = hν = (6.63 x 10−34 J•s)(2.459 x 1014/s) = 1.630 x 10−19 J
Now use the appropriate formula to determine the energy for a transition from n = 1 to n = 2, which is the minimum energy required for a transition to occur.
⎛ 1 1 ⎞ 1⎞ ⎛ 1 E = −RH ⎜ 2  2 ⎟ = −(2.179 x 10−18J) ⎜ 2  2 ⎟ = 1.6345 x 10−18 J. ⎜n ⎟ 1 ⎠ ni ⎠ ⎝2 ⎝ f Since the available energy is less than the minimum energy required, no transition occurs. g.
One mole of hydrogen atoms would require a minimum of one mole of photons. First, use the appropriate formula to calculate the energy required for the transition. ⎛ 1 1 ⎞ E = −RH ⎜ 2  2 ⎟ = −(2.179 x 10−18J) ⎜n ni ⎟⎠ ⎝ f
1⎞ ⎛1 −18 ⎜ 2  2 ⎟ = 1.9368 x 10 J. 3 1 ⎝ ⎠
The wavelength of the light is
ν = λ =
E 1.9368 x 1018 J = = 2.921 x 1015/s 6.63 x 1034 J • s h c
ν
=
3.00 x 108 m/s = 1.026 x 10−7 m = 1.03 x 10−7 m (103 nm) 2.921 x 1015 /s
7.24. a.
The groundstate energy levels of X and H are the same.
b.
The energy of an electron in H would be lower (more negative) because 1/n3 is smaller than 1/n2.
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Chapter 7: Quantum Theory of the Atom
c.
The spacing of the energy levels of hydrogen atoms is larger, because 1 1 is greater than − 2 n (n + 1) 2
1 1 − 3 n (n + 1)3
Find the ratio of the energies as follows.
EH  RH / n 2 = = n  RH /n3 EX Thus, EH = nEX.
■
d.
Since its energylevel spacings are larger, hydrogen would involve the emission of a higher frequency of light.
e.
It will take the same energy to remove the electron from the ground state of H and X atoms. However, since its energylevel spacings are larger for an H atom, hydrogen would involve more energy to completely remove its electron from states other than the ground state.
f.
No, since the energy spacings are different for these atoms, the amounts of energy required for a transition between the n = 2 and n = 5 levels are also different. The atoms will require photons of different energy.
ANSWERS TO CONCEPTUAL PROBLEMS
7.25. Wavelength and frequency are inversely related. Moreover, ultraviolet light is at higher frequency than yellow light. Doubling the frequency of a beam of light would give that beam a higher frequency than yellow light, whereas doubling the wavelength would give that beam a lower frequency than yellow light. Consequently, the beam with frequency doubled must be the one in the ultraviolet region. Here is another way to look at the problem. Energy is directly related to the frequency and inversely related to the wavelength. Thus, the beam whose frequency is doubled will increase in energy, whereas the beam whose wavelength is doubled will decrease in energy. Since yellow light is in the visible region of the spectrum, which is lower in energy than the ultraviolet region (Figure 7.5), the beam whose frequency is doubled will be higher in energy and thus in the UV region of the spectrum. 7.26. Frequency is inversely related to wavelength. Thus, the infrared radiation with a wavelength that is one thousand times larger than the visible light would have a frequency one thousand times smaller than the visible light. But, the visible light has a frequency that is one thousand times smaller than that of the X radiation. This makes the frequency of the X radiation one million times larger that the frequency of the infrared radiation. Therefore, since energy is directly related to the frequency, the energy of the X radiation would be one million times as large as the energy of the infrared radiation.
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7.27. That one color of light does not result in an ejection of electrons implies that that color has too little energy per photon. Of the two colors, red and green, red light has less energy per photon. Thus, you expect the experiment with red light to result in no ejection of photons, whereas the experiment with green light must be the one that ejects electrons. (Two red photons have more than enough energy to eject an electron, but this energy needs to be concentrated in only one photon to be effective.) In the photoelectric effect, one photon of light ejects at most one electron. Therefore, in the experiment with green light, one electron is ejected. 7.28. Ultraviolet radiation is higher in energy than red light (Figure 7.5). Since an atom that started in the ground state cannot emit more energy than it absorbed, the absorbed photon must be higher in energy than the emitted photon. This makes the emitted photon (photon 2) the red light. 7.29. Energy is inversely proportional to the wavelength of the radiation. The transition from the highest energy level to the lowest energy level would involve the greatest energy change and thus the shortest wavelength, x nm. 7.30. In a transition from the n = 1 to the n = 5 energy level, an atom will absorb a photon with the same energy as the photon that was emitted in the transition from the n = 5 to the n = 1 energy level. Since yellow light was emitted, the experiment using yellow light will promote the electron to the n = 5 level. 7.31. A proton is approximately 2000 times the weight of an electron. Also, from the de Broglie relation, l = h/mv, you see that the wavelength is inversely proportional to both the mass and the speed of the particle. Considering the protons in parts b and c, since the mass is the same in both parts, the proton with the smaller speed, part b, will have a longer wavelength. Now, comparing the electron in part a with the proton in part b, since both have the same speed, the electron in part a with the smaller mass will have the longer wavelength. Therefore, the electron in part a will have the longest wavelength. 7.32. For the first shell, the quantum numbers would have the following allowable values:
n = 1
l = 1 ml = 0, +1, −1
For the second shell, the quantum numbers would have the following allowable values:
n = 2
l = 1 ml = 0, +1, −1 l = 2
ml = 0, +1, −1, +2, −2
7.33. a.
The frequency is directly proportional to the energy difference between the two transition levels (ΔE = hν). The lowest frequency corresponds with the smallest energy difference. Thus the transition from n = 3 to n = 2 will emit the lowestfrequency light.
b.
The highest frequency corresponds with the largest energy difference. Thus the n = 3 to n = 5 transition will require (absorb) the highestfrequency light.
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Chapter 7: Quantum Theory of the Atom
c.
Since the frequency is proportional to the energy difference between two transition levels, the energy difference is the same for absorption and transmission, and the color light is the same for both. Thus, green light is absorbed.
a.
The higher energy would correspond with the larger orbital, which is on the right.
b.
Since the orbital on the right is higher in energy than the orbital on the left, the transition of an electron from the orbital on the right to the one on the left would be accompanied by the release of energy.
c.
An orbital of the same type that is higher in energy would have the same width but would be taller.
7.34.
■
SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. Starting with Problem 7.35, the value 2.998 x 108 m/s will be used for the speed of light. 7.35. Solve c = λν for λ:
λ =
c
ν
=
2.998 x 108 m/s = 219.63 = 219.6 m 1.365 x 106 /s
7.36. Solve c = λν for λ:
λ =
c
ν
=
2.998 x 108 m/s = 0.023831 = 0.02383 m (2.383 cm) 1.258 x 1010 /s
7.37. Solve c = λν for ν. Recognize that 478 nm = 478 x 10−9 m, or 4.78 x 10−7 m.
ν =
c
λ
=
2.998 x 108 m/s = 6.271 x 1014 = 6.27 x 1014 /s 7 4.78 x 10 m
7.38. Solve c = λν for ν. Recognize that 656 nm = 656 x 10−9 m = 6.56 x 10−7 m.
ν =
c
λ
=
2.998 x 108 m/s = 4.570 x 1014 = 4.57 x 1014 /s 6.56 x 107 m
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7.39. Radio waves travel at the speed of light, so divide the distance by c: 56 x 109 m x
1s = 186.7 = 1.9 x 102 s 2.998 x 108 m
7.40. Electromagnetic signals travel at the speed of light, so divide the distance by c: 998 x 109 m x
1s = 3328.8 = 3.33 x 103 s (0.925 hr) 8 2.998 x 10 m
7.41. To do the calculation, divide 1 meter by the number of wavelengths in 1 meter to find the wavelength of this transition. Then use the speed of light (with nine digits for significant figures) to calculate the frequency:
λ =
1m = 6.057802106 x 10−7 m 1,650,763.73
ν =
c
λ
=
2.99792458 x 108 m/s = 4.948865162 x 1014 = 4.94886516 x 1014 /s 7 6.057802106 x 10 m
7.42. To find the wavelength, divide the speed of light (with nine digits for significant figures) by the frequency:
λ =
c
ν
=
2.99792458 x 108 m/s = 0.03261225572 = 0.0326122557 m 9,192,631,770 /s
7.43. Solve for E, using E = hν, and use four significant figures for h: E = hν = (6.626 x 10−34 J•s) x (1.365 x 106 /s) = 9.0444 x 10−28
= 9.044 x 10−28 J 7.44. Solve for E, using E = hν, and use four significant figures for h: E = hν = (6.626 x 10−34 J•s) x (1.258 x 1010 /s) = 8.33550 x 10−24
= 8.336 x 10−24 J 7.45. Recognize that 535 nm = 535 x 10−9 m = 5.35 x 10−7 m. Then calculate ν and E.
ν =
c
λ
=
2.998 x 108 m/s = 5.6037 x 1014 /s 7 5.35 x 10 m
E = hν = (6.626 x 10−34 J•s) x (5.6037 x 1014/s) = 3.713 x 10−19 = 3.71 x 10−19 J
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Chapter 7: Quantum Theory of the Atom
7.46. Recognize that 451 nm = 451 x 10−9 m = 4.51 x 10−7 m. Then calculate ν and E.
ν =
c
λ
=
2.998 x 108 m/s = 6.6474 x 1014 /s 7 4.51 x 10 m
E = hν = (6.626 x 10−34 J•s) x (6.6474 x 1014/s) = 4.404 x 10−19 = 4.40 x 10−19 J
7.47. First, calculate the wavelength of this transition from the frequency using the speed of light:
λ =
c
ν
=
2.998 x 108 m/s = 7.8072 x 10−7 = 7.81 x 10−7 m (781 nm) 3.84 x 1014 /s
Using Figure 7.5, note that 781 nm is just on the edge of the red end of the spectrum and is barely visible to the eye. 7.48. First, calculate the wavelength of this transition from the frequency using the speed of light:
λ =
c
ν
=
2.998 x 108 m/s = 5.541 x 10−7 = 5.54 x 10−7 m (554 nm) 5.41 x 1014 /s
Using Figure 7.5, note that 554 nm is in the yellowgreen region of the spectrum and is visible to the eye. 7.49. Solve the equation E = −RH/n2 for both E5 and E3; equate to hν, and solve for ν. E = hν = = E5 − E3 =
16RH  RH  RH  RH  RH − = − = 2 2 5 3 25 9 225
The frequency of the emitted radiation is
ν =
16RH 225h
=
16 2.179 x 1018 J x = 2.338 x 1014 = 2.34 x 1014 /s 34 • 225 6.626 x 10 J s
7.50. Solve the equation E = −RH/n2 for both E4 and E3; equate to hν, and solve for ν. E = hν = = E4 − E3 =
7RH  RH  RH  RH  RH − = − = 2 2 4 3 16 9 144
The frequency of the emitted radiation is
ν =
7RH 144h
=
7 2.179 x 1018 J x = 1.5986 x 1014 = 1.60 x 1014 /s 34 144 6.626 x 10 J • s
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7.51. Solve the equation E = −RH/n2 for both E2 and E1; solve for ν, and convert to λ. E = hν = = E2 − E1 =
3RH  RH  RH  RH  RH − = − = 2 2 2 1 4 1 4
The frequency of the emitted radiation is
ν =
3RH 4h
=
3 2.179 x 1018 J x = 2.466 x 1015 /s 4 6.626 x 1034 J • s
The wavelength can now be calculated.
λ =
c
ν
=
2.998 x 108 m/s = 1.2155 x 10−7 = 1.22 x 10−7 m (near UV) 2.466 x 1015 /s
7.52. Solve the equation E = −RH/n2 for both E5 and E4; solve for ν, and convert to λ. E = hν = = E5 − E4 =
9 RH  RH  RH  RH  RH − = − = 52 42 25 16 400
The frequency of the emitted radiation is
ν =
9 RH 400h
=
9 2.179 x 1018 J x = 7.399 x 1013 /s 400 6.626 x 1034 J • s
The wavelength can now be calculated.
λ =
c
ν
=
2.998 x 108 m/s = 4.0517 x 10−6 = 4.05 x 10−6 m (near IR) 7.399 x 1013 /s
7.53. This is the highest energy transition from the n = 6 level, so the electron must undergo a transition to the n = 1 level. Solve the Balmer equation using Bohr's approach: E = hν = = E6 − E1 =
35 RH  RH  RH  RH  RH − = − = 2 2 36 6 1 36 1
The frequency of the emitted radiation is
ν =
35 RH 36h
=
35 2.179 x 1018 J x = 3.197 x 1015 /s 6.626 x 1034 J • s 36
The wavelength can now be calculated.
λ =
c
ν
=
2.998 x 108 m/s = 9.3769 x 10−8 = 9.38 x 10−8 m (93.8 nm) 3.197 x 1015 /s
7.54. This is the lowest energy transition from the n = 7 level, so the electron must undergo a transition to the n = 6 level. Solve the Balmer equation using Bohr's approach: E = hν = = E7 − E6 =
13RH  RH  RH  RH  RH − = − = 2 2 7 6 49 36 1764
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Chapter 7: Quantum Theory of the Atom
The frequency of the emitted radiation is
ν =
13RH 1764h
=
13 2.179 x 1018 J x = 2.4235 x 1013 /s 34 • 6.626 x 10 J s 1764
The wavelength can now be calculated.
λ =
c
ν
=
2.998 x 108 m/s = 1.2370 x 10−5 = 1.237 x 10−5 m (12.37 μm) 2.4235 x 1013 /s
7.55. Noting that 422.7 nm = 4.227 x 10−7 m, convert the 422.7 nm to frequency. Then convert the frequency to energy using E = hν.
ν =
c
λ
=
2.998 x 108 m/s = 7.0925 x 1014 /s 4.227 x 107 m
E = hν = (6.626 x 10−34 J•s) x 7.0925 x 1014 /s) = 4.6994 x 10−19
= 4.699 x 10−19 J 7.56. Noting that 285.2 nm = 2.852 x 10−7 m, convert the 285.2nm wavelength to frequency. Then convert the frequency to energy using E = hν.
ν =
c
λ
=
2.998 x 108 m/s = 1.0511 x 1015 /s 2.852 x 107 m
E = hν = (6.626 x 10−34 J•s) x 1.0511 x 1015 /s) = 6.96458 x 10−19
= 6.965 x 10−19 J 7.57. The mass of a neutron = 1.67493 x 10−27 kg. Its speed or velocity, v, of 4.15 km/s equals 4.15 x 103 m/s. Substitute these parameters into the de Broglie relation, and solve for λ:
λ =
h 6.626 x 1034 kg • m 2 /s = = 9.532 x 10−11 = 9.53 x 10−11 m 27 3 mv 1.67493 x 10 kg x 4.15 x 10 m/s
A wavelength of 9.53 x 10−11 m (95.3 pm) would be in the xray region of the spectrum. 7.58. The mass of a proton = 1.67262 x 10−27. Its speed or velocity, v, of 6.58 km/s equals 6.58 x 103 m/s. Substitute these parameters into the de Broglie relation, and solve for λ:
λ =
h 6.626 x 1034 kg • m 2 /s = = 6.020 x 10−11 = 6.02 x 10−11 m mv 1.67262 x 1027 kg x 6.58 x 103 m/s
A wavelength of 6.02 x 10−11 m (60.2 pm) would be in the xray region of the spectrum.
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7.59. The mass of an electron equals 9.109 x 10−31 kg. The wavelength, λ, given as 10.0 pm, is equivalent to 1.00 x 10−11 m. Substitute these parameters into the de Broglie relation, and solve for the frequency, ν: v =
h 6.626 x 1034 kg • m 2 /s = = 7.274 x 107 = 7.27 x 107 m/s mλ 9.109 x 1031 kg x 1.00 x 1011 m
7.60. The mass of a neutron equals 1.67493 x 10−27 kg. The wavelength, λ, given as 12.0 pm, is equivalent to 1.20 x 10−11 m. Substitute these parameters into the de Broglie relation, and solve for the frequency, ν: v =
h 6.626 x 1034 kg • m 2 /s = = 3.296 x 104 = 3.30 x 104 m/s mλ 1.67493 x 1027 kg x 1.20 x 1011 m
7.61. Substitute the 1.45 x 10−1 kg mass of the baseball and the 30.0 m/s velocity, v, into the de Broglie relation, and solve for wavelength (recall that 1 pm = 10−12 m).
λ =
h 6.626 x 1034 kg • m 2 /s = = 1.523 x 10−34 =1.52 x 10−34 m 1 mv 1.45 x 10 kg x 30.0 m/s
= 1.52 x 10−22 pm Because this is much smaller than 100 pm, the wavelength is much smaller than the diameter of one atom. 7.62. The mass of O2 to three significant figures is 32.00 ÷ 6.022 x 1023 = 5.313 x 10−23 g = 5.313 x 10−26 kg. Substitute this mass and the 521 m/s velocity, v, into the de Broglie relation, and solve for wavelength (recall that 1 pm = 10−12 m).
λ =
h 6.626 x 1034 kg • m 2/s = = 2.3937 x 10−11 = 2.39 x 10−11 m 26 mv 5.313 x 10 kg x 521 m/s
= 23.9 pm Because this is on the order of 100 pm, the wavelength is on the order of an atomic diameter. 7.63. The possible values of l range from zero to (n − 1), so l may be 0, 1, 2, or 3. The possible values of ml range from −l to +l, so ml may be −3, −2, −1, 0, +1, +2, or +3. 7.64. The possible l values are 0, 1, 2, 3, 4, and 5. The possible ml values are −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, or 5. 7.65. For the M shell, n = 3; there are three subshells in this shell (l = 0, 1, and 2). An f subshell has l = 3; the number of orbitals in this subshell is 2(3) + 1 = 7 (ml = −3, −2, −1, 0, 1, 2, and 3).
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Chapter 7: Quantum Theory of the Atom
7.66. For the N shell, n = 4; there are four subshells in this shell (l = 0, 1, 2, and 3). For a g subshell, the value of l = 4; the number of orbitals in this subshell is 2(4) + 1 = 9 (ml = −4, −3, −2, −1, 0, 1, 2, 3, and 4). 7.67. a.
6d
b.
5g
c.
4f
d.
6p
a.
3d
b.
4s
c.
4p
d.
5f
a.
Not permissible; ms can be only +1/2 or −1/2.
b.
Not permissible; l can be only as large as (n − 1).
c.
Not permissible; ml cannot exceed +2 in magnitude.
d.
Not permissible; n cannot be zero.
e.
Not permissible; ms can be only + 1/2 or −1/2.
a.
Permissible.
b.
Not permissible; l can be only as large as (n − 1).
c.
Not permissible; n starts at 1, not at zero.
d.
Permissible.
e.
Not permissible; ml cannot exceed l in magnitude.
7.68.
7.69.
7.70.
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SOLUTIONS TO GENERAL PROBLEMS
7.71. Use c = νλ to calculate frequency; then use E = hν to calculate energy.
ν =
c
2.998 x 108 m/s = 6.503 x 1014 = 6.50 x 1014 /s 7 4.61 x 10 m
=
λ
E = hν = (6.626 x 10−34 J•s) x (6.503 x 1014/s) = 4.309 x 10−19 = 4.31 x 10−19 J
7.72. Use c = νλ to calculate frequency; then use E = hν to calculate energy.
ν =
c
2.998 x 108 m/s = 5.4115 x 1014 = 5.41 x 1014 /s 5.54 x 107 m
=
λ
E = hν = (6.626 x 10−34 J•s) x (5.4115 x 1014/s) = 3.5856 x 10−19 = 3.59 x 10−19 J
7.73. Calculate the frequency corresponding to 4.10 x 10−19 J. Then convert that to wavelength.
ν = λ =
E 4.10 x 1019 J = = 6.187 x 1014 /s 34 • 6.626 x 10 J s h c
ν
=
2.998 x 108 m/s = 4.845 x 10−7 = 4.85 x 10−7 m = 485 nm (blue) 14 6.187 x 10 /s
7.74. Calculate the frequency corresponding to 3.34 x 10−19 J. Then convert that to wavelength.
ν = λ =
E 3.34 x 1019 J = = 5.0407 x 1014 /s 6.626 x 1034 J • s h c
ν
=
2.998 x 108 m/s = 5.947 x 10−7 = 5.95 x 10−7 m = 595 nm (yellow) 5.0407 x 1014 /s
7.75. Solve for frequency using E = hν.
ν =
E 4.34 x 1019 J = = 6.549 x 1014 = 6.55 x 1014 /s 34 • 6.626 x 10 J s h
7.76. Solve for frequency using E = hν.
ν =
E 5.90 x 1019 J = = 8.904 x 1014 = 8.90 x 1014 /s 6.626 x 1034 J • s h
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Chapter 7: Quantum Theory of the Atom
7.77. First calculate Ep, the energy of the 345nm photon, noting that it is equivalent to 3.45 x 10−7 m. hc
Ep =
λ
(6.626 x 1034 J • s)(2.998 x 108 m/s) = 5.7578 x 10−19 J 7 3.45 x 10 m
=
Now, subtract the work function of Ca = 4.34 x 10−19 J (Problem 7.75) from Ep: 5.7578 x 10−19 J − 4.34 x 10−19 J = 1.4178 x 10−19 J Note that for this situation, E = 1/2mv2. Recall that the mass of the electron is 9.1095 x 10−31 kg. Now calculate speed, v: 2 x 1.4178 x 1019 J = 5.579 x 105 = 5.58 x 105 m/s 9.1095 x 1031 kg
2E = m
v =
7.78. First calculate Ep, the energy of the 276nm photon, noting that it is equivalent to 2.76 x 10−7 m. hc
Ep =
λ
(6.626 x 1034 J • s)(2.998 x 108 m/s) = 7.197 x 10−19 J 7 2.76 x 10 m
=
Now, subtract the work function of Mg = 5.90 x 10−19 J (Problem 7.76) from Ep: 7.197 x 10−19 J − 5.90 x 10−19 J = 1.297 x 10−19 J Note that for this situation, E = 1/2mv2. Recall that the mass of the electron is 9.1095 x 10−31 kg. Now calculate speed, v: 2E = m
v =
2 x 1.297 x 1019 J = 5.337 x 105 = 5.34 x 105 m/s 31 9.1095 x 10 kg
7.79. This is a transition from the n = 5 level to the n = 2 level. Solve the Balmer equation using Bohr's approach. E = hν = = E5 − E2 =
21RH  RH  RH  RH  RH − = − = 2 2 100 5 2 25 4
The wavelength can now be calculated.
ν = λ =
21RH 100h c
ν
=
=
21 2.179 x 1018 J x = 6.9059 x 1014 /s 34 • 6.626 x 10 J s 100
2.998 x 108 m/s = 4.341 x 10−7 = 4.34 x 10−7 m (434 nm) 14 6.9059 x 10 /s
7.80. This is a transition from the n = 6 level to the n = 2 level. Solve the Balmer equation using Bohr's approach. E = hν = = E6 − E2 =
32 RH  RH  RH  RH  RH − = − = 2 2 6 2 36 4 144
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The wavelength can now be calculated.
ν = λ =
32 RH 144h c
ν
=
=
32 2.179 x 1018 J x = 7.3079 x 1014 /s 34 • 6.626 x 10 J s 144
2.998 x 108 m/s = 4.1024 x 10−7 = 4.102 x 10−7 m (410.2 nm) 14 7.3079 x 10 /s
7.81. Use 397 nm = 3.97 x 10−7 m, and convert to frequency and then to energy.
ν =
c
λ
=
2.998 x 108 m/s = 7.551 x 1014 /s 3.97 x 107 m
E = hν = (6.626 x 10−34 J•s) x (7.551 x 1014) = 5.0037 x 10−19 J
Substitute this energy into the Balmer formula, recalling that the Balmer series is an emission spectrum, so ΔE is negative:  RH  RH  RH  RH − = − = 5.0037 x 10−19 J 2 2 2 2 4 ni ni
E = E2 − Ei =
1 1 ΔE 5.0037 x 1019 J − 2 = = 0.22963 = 4 2.179 x 1018 J ni  RH 1 1 = − 0.22963 = 0.02036 2 ni 4
⎡ 1 ⎤ ni = ⎢ ⎥ ⎣ 0.02036 ⎦
1/2
= 7.007 = 7.0 (= n)
7.82. Convert 9.50 x 10−8 m to frequency and then to energy.
ν =
c
λ
=
2.998 x 108 m/s = 3.1557 x 1015 /s 9.50 x 108 m
E = hν = (6.626 x 10−34 J•s) x (3.1557 x 1015) = 2.0910 x 10−18 J
Substitute this energy into the Balmer formula, recalling that the Lyman series is an emission spectrum, so ΔE is negative, and nf = 1:  RH  RH  RH  RH − = − = 2.0910 x 10−18 J 2 2 2 1 1 ni ni
E = E1 − Ei =
1 −
1 ΔE 2.0910 x 1018 J = 0.95962 = = 2.179 x 1018 J ni 2  RH
1 = 1 − 0.95962 = 0.04037 ni 2 12
⎡ 1 ⎤ ni = ⎢ ⎥ 0.04037 ⎣ ⎦
= 4.976 = 5 (= n)
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Chapter 7: Quantum Theory of the Atom
7.83. Employ the Balmer formula using Z = 2 for the He+ ion. E = hν = E3 − E2 =
(22 )RH (22 )RH 4RH 4RH 20 RH − = − = 2 2 3 2 9 4 36
The frequency of the radiation is
ν = λ =
20 RH 20 2.179 x 1018 J = x = 1.8269 x 1015 /s 34 36h 6.626 x 10 J • s 36 c
ν
=
2.998 x 108 m/s = 1.6409 x 10−7 = 1.641 x 10−7 m (164.1 nm; near UV) 15 1.8269 x 10 /s
7.84. Employ the Balmer formula using Z = 3 for the Li2+ ion. E = hν = E5 − E3 =
ν = λ =
(32 )RH (32 )RH 9RH 9RH 144 RH − = − = 2 2 25 9 225 5 3
144 RH 144 2.179 x 1018 J = x = 2.1046 x 1015 /s 225h 6.626 x 1034 J • s 225 c
ν
=
2.998 x 108 m/s = 1.4244 x 10−7 = 1.424 x 10−7 m (142.4 nm; UV) 2.1046 x 1015 /s
7.85. First, use the wavelength of 10.0 pm (1.00 x 10−11 m) and the mass of 9.1095 x 10−31 kg to calculate the velocity, v. Then use the kinetic energy equation to calculate kinetic energy from velocity. v =
h 6.626 x 1034 kg • m 2 / s = = 7.2737 x 107 m/s mλ 9.1095 x 1031 kg x 1.00 x 1011 m
E = 1/2mv2 = 1/2 x (9.1095 x 10−31 kg) x (7.273 x 107 m/s)2 = 2.409 x 10−15 J EeV = 2.409 x 10−15 J x
1 eV = 1.503 x 104 = 1.50 x 104 eV 1.602 x 1019 J
7.86. First, use the wavelength of 10.0 pm (1.00 x 10−11 m) and the mass of 1.675 x 10−27 kg to calculate the velocity, v. Then use the kinetic energy equation to calculate kinetic energy from velocity. v =
h 6.626 x 1034 kg • m 2 / s = = 3.9558 x 104 m/s 1.675 x 1027 kg x 1.00 x 1011 m mλ
E = 1/2mv2 = 1/2 x (1.675 x 10−27 kg) x (3.9558 x 104 m/s)2 = 1.3105 x 10−18 J EeV = 1.3105 x 10−18 J x
1 eV = 8.1803 = 8.18 eV 1.602 x 1019 J
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261
7.87. a.
Five
b.
Seven
c.
Three
d.
One
a.
Nine
b.
Seven
c.
One
d.
Three
7.88
7.89. The possible subshells for the n = 6 shell are 6s, 6p, 6d, 6f, 6g, and 6h. 7.90. The possible subshells for the n = 7 shell are 7s, 7p, 7d, 7f, 7g, 7h, and 7i. 7.91. Gamma rays are a form of electromagnetic radiation similar to x rays, but the photons of gamma rays have higher energy. Highenergy radiation kills bacteria and molds in foods by breaking up the DNA molecules within their cells. 7.92. The gamma rays come from the radioactive decay of cobalt60. Food is not made radioactive by irradiating it with gamma rays. 7.93. When a photon of appropriate wavelength is used to flash the laser crystal, it stimulates a transition, and a photon of exactly the same wavelength as the original photon is emitted. In place of just one photon, there are now two photons. The net effect is to increase the intensity of the light at this wavelength. 7.94. Music is encoded on a CD in the form of pits, or indentations, on a spiral track. When the disc is played, a small laser beam scans the track and is reflected back to a detector. Light reflected from an indentation is out of phase with light from the laser and interferes with it, reducing the intensity of the wave. As the disc moves around, light is reflected either from the main surface at full intensity (on) or from a pit at reduced intensity (off). A lightsensitive detector converts this reflected light to a digital signal, and a converter changes the signal to one that speakers can accept. 7.95. Tunneling depends on the probability interpretation of quantum mechanics. The probability of an electron in an atom being at a location far from atom A and near atom B is very small but not zero. This means that an electron that normally belongs to atom A can find itself near atom B without any extra energy having been supplied. The electron is said to have tunneled from one atom to another.
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Chapter 7: Quantum Theory of the Atom
7.96. If the probe is close enough to the sample, electrons can tunnel from the probe to the sample, which gives rise to a measurable electric current. The magnitude of the current depends on the distance between the probe and the sample. By adjusting this distance, the current can be maintained at a fixed value. As the probe scans the sample, it moves toward or away from the sample to maintain a fixed current, following the contours of the sample.
■
SOLUTIONS TO STRATEGY PROBLEMS
7.97. λ =
c
ν
=
3.00 x 108 m/s = 0.03865 m = 0.0387 m 7.76 x 109 /s
7.98. E = hν = (6.63 x 10−34 J•s)(1.69 x 106 /s) = 1.120 x 10−27 = 1.12 x 10−27 J
7.99. ν =
E 4.60 x 1019 J = = 6.938 x 1014 /s h 6.63 x 1034 Js
λ =
c
ν
=
3.00 x 108 m/s = 4.323 x 10−7 = 4.32 x 10−7 m = 432 nm 6.938 x 1014 /s
7.100. The shortest wavelength of visible light is for the n = 6 to n = 2 transition. E = hν = = E6 − E2 =
32 RH  RH  RH  RH  RH − = − = 2 2 144 6 2 36 4
The wavelength can now be calculated.
ν = λ =
32 RH 144h c
ν
=
=
32 2.179 x 1018 J x = 7.303 x 1014 /s 6.63 x 1034 J • s 144
3.00 x 108 m/s = 4.107 x 10−7 = 4.11 x 10−7 m (411 nm) 7.303 x 1014 /s
Figure 7.11 includes all the visible lines in the hydrogen spectrum. 7.101. Light of wavelength 1.03 x 10−7 m (103 nm) is in the ultraviolet region of the spectrum. The frequency and energy of this radiation are
ν =
c
λ
=
3.00 x 108 m/s = 2.912 x 1015 /s 7 1.03 x 10 m
E = hν = (6.63 x 10−34 J•s)(2.912 x 1015 /s) = 1.931 x 10−18 J
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263
Now use the appropriate formula to determine the quantum number of the excited (ni) level. Remember, ΔE is negative for an emission spectrum.
⎛ 1 ⎛1 ⎛ 1 ⎞ 1 ⎞ 1 ⎞ ΔE = −RH ⎜ 2  2 ⎟ = −RH ⎜ 2  2 ⎟ = −RH ⎜ 1  2 ⎟ ⎜n ⎟ ni ⎠ ni ⎠ ni ⎠ ⎝1 ⎝ ⎝ f 1 −
1 ΔE 1.931 x 1018 J = = = 0.8862 2.179 x 1018 J  RH ni 2 1 = 2.964 = 2.96 1  0.8862
ni =
Thus, the quantum number of the excited level is 3. 7.102. First, determine the energy of one photon. E =
799 kJ 1000 J 1 mol = 1.326 x 10−18 J x x 1 mol 1 kJ 6.022 x 1023 photons
The wavelength can now be determined.
ν = λ =
E 1.326 x 1018 J = 2.001 x 1015 /s = h 6.63 x 1034 J • s c
ν
=
3.00 x 108 m/s = 1.499 x 10−7 = 1.50 x 10−7 m (150. nm) 2.001 x 1015 /s
This corresponds to a photon in the near ultra violet region of the spectrum, which is represented by the Lyman series in hydrogen. The minimum energy for a transition in this region corresponds to the n = 2 to n = 1 transition. ⎛ 1 3RH (3)(2.179 x 1018 J) 1 ⎞ 1⎞ ⎛1 ΔE = −RH ⎜ 2  2 ⎟ = −RH ⎜ 2  2 ⎟ = = ⎜ nf 4 4 ni ⎟⎠ 2 ⎠ ⎝1 ⎝
= −1.6342 x 10−18 J This is more than the energy required per photon to break the bond. 7.103. First, calculate the mass of one oxygen molecule. m =
32.00 g O 2 1 mol 1 kg = 5.313 x 10−26 kg x x 6.02 x 1023 molec 1 mol 1000 g
Now the de Broglie wavelength can be calculated.
λ =
h 6.63 x 1034 J • s = = 2.603 x 10−11 = 2.60 x 10−11 m (26.0 pm) (5.313 x 1026 kg)(479 m/s) mv
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Chapter 7: Quantum Theory of the Atom
This wavelength can be compared to the length of the molecule (242 pm) as follows. 26.03 pm x 100% = 10.75% = 10.8% 242 pm Thus, the de Broglie wavelength is 10.8% of the length of the molecule. 7.104. The energy required to heat the water is E = s x m x Δt = (4.18 J/mol•°C)(1.00 g)(1.00°C) = 4.18 J The microwave radiation has a wavelength of 12.2 cm (0.122 m), so the frequency and energy are
ν =
c
λ
=
3.00 x 108 m/s = 2.459 x 109 /s 0.122 m
E = hν = (6.63 x 10−34 J•s)(2.459 x 109 /s) = 1.630 x 10−24 J/photon Therefore, the number of photons needed to heat the water is n =
4.18 J = 2.563 x 1024 = 2.56 x 1024 photons 1.630 x 1024 J/photon
7.105. a.
Change ml to 0 (n = 3, l = 0, ml = 0) for a count of 1.
b.
Change l to 4 (n = 5, l = 4, ml = 4) for a count of 2.
c.
Change n to 4 (n = 4, l = 3, ml = −3) for a count of 3.
d.
Change l to 4 (n = 5, l = 4, ml = 3) for a count of 2. The total count is 1 + 2 + 3 + 2 = 8.
7.106. For the n = 5 level, the sublevels are 5s, 5p, 5d, 5f, and 5g. The numbers of orbitals for the sublevels are 1, 3, 5, 7, and 9, respectively. The total number of orbitals for the n = 5 level is therefore 1 + 3 + 5 + 7 + 9 = 25.
■
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
7.107. First, use Avogadro's number to calculate the energy for one Cl2 molecule. 239 kJ 1000 J 1 mol x x = 3.9687 x 10−19 J/molecule 23 1 mol 1 kJ 6.022 x 10 molecules
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Then convert energy to frequency and finally to wavelength.
ν = λ =
E 3.9687 x 1019 J = = 5.9897 x 1014 /s 34 • 6.626 x 10 J s h c
ν
=
2.998 x 108 m/s = 5.0052 x 10−7 m (501 nm; visible region) 14 5.9897 x 10 /s
7.108. First, use Avogadro's number to calculate the energy per H2 molecule. 432 kJ 1000 J 1 mol x x = 7.1736 x 10−19 J/molecule 23 1 mol 1 kJ 6.022 x 10 molecules Then convert energy to frequency and finally to wavelength.
ν = λ =
E 7.1736 x 1019 J = 1.0826 x 1015 /s = h 6.626 x 1034 J • s c
ν
=
2.998 x 108 m/s = 2.7691 x 10−7 m (277 nm; UV) 1.0826 x 1015 /s
7.109. First, calculate the energy needed to heat the 0.250 L of water from 20.0°C to 100.0°C. 0.250 L x
1000 g 4.184 J x x (100.0 °C − 20.0°C) = 8.368 x 104 J • 1L (g °C)
Then calculate the frequency, the energy of one photon, and the number of photons.
ν =
c
λ
=
2.998 x 108 m/s = 2.398 x 109 /s 0.125 m
E of one photon = hν = (6.626 x 10−34 J•s) x (2.398 x 109/s) = 1.589 x 10−24 J n = 8.368 x 104 J x
1 photon = 5.2656 x 1028 = 5.27 x 1028 photons 24 1.589 x 10 J
7.110. First, calculate the energy needed to heat the 1.00 L of water from 20.0°C to 30.0°C. 1000 g 4.184 J x x (30.0 °C − 20.0°C) = 4.184 x 104 J 1L (g • °C)
1.00 L x
Then calculate the frequency, the energy of one photon, and the number of photons.
ν =
c
λ
=
2.998 x 108 m/s = 1.0707 x 1014 /s 2.80 x 106 m
E of one photon = hν = (6.626 x 10−34 J•s) x (1.0707 x 1014/s) = 7.0945 x 10−20 J n = 4.184 x 104 J x
1 photon = 5.8974 x 1023 = 5.90 x 1023 photons 7.0945 x 1020 J
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Chapter 7: Quantum Theory of the Atom
7.111. First, write the following equality for the energy to remove one electron, Eremoval: Eremoval = E425 nm − Ek of ejected photon Use E = hν to calculate the energy of the photon. Then recall that Ek, the kinetic energy, is 1/2mv2. Use this to calculate Ek. E425 nm =
hc
λ
=
(6.626 x 1034 J • s)(2.998 x 108 m/s) = 4.6740 x 10−19 J 4.25 x 107 m
Ek = 1/2mv2 = 1/2 x (9.1095 x 10−31 kg) x (4.88 x 105 m/s)2 = 1.0846 x 10−19 J Subtract to find Eremoval, and convert it to kJ/mol: Eremoval = 4.6740 x 10−19 J − (1.0847 x 10−19 J) = 3.5893 x 10−19 = 3.59 x 10−19 J/electron Eremoval =
3.5893 x 1019 J 6.022 x 1023 e1 kJ x x = 2.161 x 102 1e 1 mol 1000 J
= 2.16 x 102 kJ/mol 7.112. First, write the following equality for the energy to remove one electron, Eremoval: Eremoval = E405 nm − Ek of ejected electron Use E = hν to calculate the energy of the photon. Then recall that Ek, the kinetic energy, is 1/2mv2. Use this to calculate Ek. E405 nm =
hc
λ
=
(6.626 x 1034 J • s)(2.998 x 108 m/s) = 4.9048 x 10−19 J 4.05 x 107 m
Ek = 1/2mv2 = 1/2 x (9.1095 x 10−31 kg) x (3.36 x 105 m/s)2 = 5.142 x 10−20 J Subtract to find Eremoval, and convert it to kJ/mol: Eremoval = 4.9048 x 10−19 − 5.142 x 10−20 = 4.3906 x 10−19 = 4.39 x 10−19 J/electron Eremoval =
4.3906 x 1019 J 6.022 x 1023 e1 kJ x x = 264.4 1e 1 mol 1000 J
= 264 kJ/mol 7.113. First, calculate the energy, E, in joules, using the product of voltage and charge: E = (4.00 x 103 V) x (1.602 x 10−19 C) = 6.408 x 10−16 J Now, use the kinetic energy equation, Ek = 1/2mv2, and solve for velocity: v =
λ =
2 Ek m
=
2 x 6.408 x 1016 J = 3.7508 x 107 m/s 9.1095 x 1031 kg
h 6.626 x 1034 J • s = = 1.939 x 10−11 31 7 (9.1095 x 10 kg) x (3.7508 x 10 m/s) mv
= 1.94 x 10−11 m (19.4 pm)
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7.114. First, calculate the energy, E, in joules, using the product of voltage and charge: E = (1.00 x 104 V) x (1.602 x 10−19 C) = 1.602 x 10−15 J Now, use the kinetic energy equation Ek = 1/2mv2, and solve for velocity: v =
λ =
2 Ek m
=
2 x 1.602 x 1015 J = 5.9306 x 107 m/s 31 9.1095 x 10 kg
h 6.626 x 1034 J • s = = 1.226 x 10−11 mv (9.1095 x 1031 kg) x (5.9306 x 107 m/s)
= 1.23 x 10−11 m (12.3 pm)
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CHAPTER 8
Electron Configurations and Periodicity
■
SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 8.1. a.
Possible orbital diagram.
b.
Possible orbital diagram.
c.
Impossible orbital diagram; there are two electrons in a 2p orbital with the same spin.
d.
Possible electron configuration.
e.
Impossible electron configuration; only two electrons are allowed in an s subshell.
f.
Impossible electron configuration; only six electrons are allowed in a p subshell.
8.2. Look at the periodic table. Start with hydrogen and go through the periods, writing down the subshells being filled, stopping with manganese (Z = 25). You obtain the following order: Order:
1s
2s2p
3s3p
Period:
first
second third
4s3d4p fourth
Now fill the subshells with electrons, remembering that you have a total of twentyfive electrons to distribute. You obtain 1s22s22p63s23p64s23d5, or
1s22s22p63s23p63d54s2
8.3. Arsenic is a maingroup element in Period 4, Group VA, of the periodic table. The five outer electrons should occupy the 4s and 4p subshells; the five valence electrons have the configuration 4s24p3. 8.4. Because the sum of the 6s2 and 6p2 electrons gives four outer (valence) electrons, lead should be in Group IVA, which it is. Looking at the table, you find lead in Period 6. From its position, it would be classified as a maingroup element.
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269
8.5. The electron configuration of phosphorus is 1s22s22p63s23p3. The orbital diagram is
1s
2s
2p
3s
3p
8.6. The radius tends to decrease across a row of the periodic table from left to right, and it tends to increase from the top of a column to the bottom. Therefore, in order of increasing radius, Be < Mg < Na 8.7. It is more likely that answer a, 1000 kJ/mol, is the ionization energy for iodine because ionization energies tend to decrease with atomic number in a group (I is below Cl in Group VIIA). 8.8. Fluorine should have a more negative electron affinity because (1) carbon has only two electrons in the p subshell, (2) the −1 fluoride ion has a stable noblegas configuration, and (3) the electron can approach the fluorine nucleus more closely than the carbon nucleus. This follows the general trend, which is toward more negative electron affinities from left to right in any period.
■
ANSWERS TO CONCEPT CHECKS
8.1. The secondperiod elements are those in which the 2s and 2p orbitals fill. Each orbital can hold only one electron, so all four orbitals will be filled after four electrons. Therefore, the second period will have four elements. 8.2. The s orbital fills in the first two elements of the period (Groups IA and IIA); then the p orbital starts to fill (Group IIIA). Thus, the first element is in Group IIA (Mg), and the next element is in Group IIIA (Al). 8.3. From the information given, the element must be in Group IIA. These elements have positive electron affinities and also have large third ionization energies. 8.4. A metalloid is an element near the staircase line in the periodic table (the green elements in the periodic table on the inside front cover of the text). The formula R2O5 suggests a Group VA element. There are two metalloids in Group VA, arsenic and antimony. That this is an acidic oxide indicates this metalloid has considerable nonmetal character. So, of the two metalloids, the one nearer the top of the column, arsenic, seems more likely. This agrees with the text, which notes that arsenic(V) oxide is acidic, whereas antimony(V) oxide is amphoteric.
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Chapter 8: Electron Configurations and Periodicity
ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
8.1. In the original SternGerlach experiment, a beam of silver atoms is directed into the field of a specially designed magnet. (The same can be done with hydrogen atoms.) The beam of atoms is split into two by the magnetic field; half are bent toward one magnetic pole face and the other half toward the other magnetic pole face. This effect shows that the atoms themselves act as magnets with a positive or a negative component, as indicated by the positive or negative spin quantum numbers. 8.2. In effect, the electron acts as though it were a sphere of spinning charge (Figure 8.3). Like any circulating electric charge, it creates a magnetic field with a spin axis that has more than one possible direction relative to a magnetic field. Electron spin is subject to a quantum restriction to one of two directions corresponding to the ms quantum numbers +1/2 and −1/2. 8.3. The Pauli exclusion principle limits the configurations of an atom by excluding configurations in which two or more electrons have the same four quantum numbers. For example, each electron in the same orbital must have different ms values. This also implies that only two electrons occupy one orbital. 8.4. According to the principles discussed in Section 7.5, the number of orbitals in the g subshell (l = 4) is given by 2l + 1 and is thus equal to 9 Because each orbital can hold a maximum of two electrons, the g subshell can hold a maximum of eighteen electrons. 8.5. The orbitals, in order of increasing energy up to and including the 3p orbitals (but not including the 3d orbitals), are as follows: 1s, 2s, 2p, 3s, and 3p (Figure 8.7). 8.6. The noblegas core is an innershell configuration corresponding to one of the noble gases. The pseudonoblegas core is an innershell configuration corresponding to one of the noble gases together with (n − 1)d10 electrons. Like the noblegas core electrons, the d10 electrons are not involved in chemical reactions. The valence electron is an electron (of an atom) located outside the noblegas core or pseudonoblegas core. It is an electron primarily involved in chemical reactions. 8.7. The orbital diagram for the 1s22s22p4 ground state of oxygen is
1s
2s
2p
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271
Another possible oxygen orbital diagram, but not a ground state, is
1s
2s
2p
8.8. A diamagnetic substance is a substance that is not attracted by a magnetic field or is very slightly repelled by such a field. This property generally indicates the substance has only paired electrons. A paramagnetic substance is a substance that is weakly attracted by a magnetic field. This property generally indicates the substance has one or more unpaired electrons. Groundstate oxygen has two unpaired 2p electrons and is therefore paramagnetic. 8.9. In Groups IA and IIA, the outer s subshell is being filled: s1 for Group IA and s2 for Group IIA. In Groups IIIA to VIIIA, the outer p subshell is being filled: p1 for IIIA, p2 for IVA, p3 for VA, p4 for VIA, p5 for VIIA, and p6 for VIIIA. In the transition elements, the (n − 1)d subshell is being filled from d1 to d10 electrons. In the lanthanides and actinides, the f subshell is being filled from f1 to f14 electrons. 8.10. Mendeleev arranged the elements in order of increasing atomic weight, an arrangement that was later changed to atomic numbers. His periodic table was divided into rows (periods) and columns (groups). In his first attempt, he left spaces for what he believed to be undiscovered elements. In his row 5, under aluminum and above indium in Group III, he left a blank space. This Group III element he called ekaaluminum, and he predicted its properties from those of aluminum and indium. Later, the French chemist de Boisbaudran discovered this element and named it gallium. 8.11. In a plot of atomic radii versus atomic number (Figure 8.16), the major trends that emerge are the following: (1) Within each period (horizontal row), the atomic radius tends to decrease with increasing atomic number or nuclear charge. The largest atom in a period is thus the Group 1A atom, and the smallest atom in a period is thus the noblegas atom. (2) Within each group (vertical column), the atomic radius tends to increase with the period number. In a plot of ionization energy versus atomic number (Figure 8.18), the major trends are that (1) the ionization energy within a period increases with atomic number, and (2) the ionization energy within a group tends to decrease going down the group. 8.12. The alkaline earth element with the smallest radius is beryllium (Be). 8.13. Group VIIA (halogens) is the main group with the most negative electron affinities. Configurations with filled subshells (ground states of the noblegas elements) would form unstable negative ions when adding one electron per atom. 8.14. The Na+ and Mg2+ ions are stable because they have the same electronic configuration as the noble gas neon. If Na2+ and Mg3+ ions were to exist, they would be very unstable because they would not have the same electronic configuration as a noblegas structure and because of the energy needed to remove an electron from an inner shell.
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Chapter 8: Electron Configurations and Periodicity
8.15. The elements tend to increase in metallic character from right to left in any period. They also tend to increase in metallic character down any column (group) of elements. 8.16. A basic oxide is an oxide that reacts with acids. An example is calcium oxide, CaO. An acidic oxide is an oxide that reacts with bases. An example is carbon dioxide, CO2. 8.17. Rubidium is the alkali metal atom with a 5s1 configuration. 8.18. Atomic number equals 117 (protons in last known element plus those needed to reach Group VIIA). 8.19. The following elements are in Groups IIIA to VIA: Group IIIA
Group IVA
Group VA
Group VIA
B: metalloid
C: nonmetal
N: nonmetal
O: nonmetal
Al: metal
Si: metalloid
P: nonmetal
S: nonmetal
Ga: metal
Ge: metalloid
As: metalloid
Se: nonmetal
In: metal
Sn: metal
Sb: metalloid
Te: metalloid
Tl: metal
Pb: metal
Bi: metal
Po: metal
Yes, each column displays the expected increasing metallic character. 8.20. The oxides of the following elements are listed as acidic, basic, or amphoteric: Group IIIA
Group IVA
Group VA
Group VIA
B: acidic
C: acidic
N: acidic
O: amphoteric (H2O)
Al: amphoteric
Si: acidic
P: acidic
S: acidic
Ga: amphoteric
Ge: acidic
As: acidic
Se: acidic
In: basic
Sn: amphoteric
Sb: amphoteric
Te: amphoteric
Tl: basic
Pb: amphoteric
Bi: basic
Po: amphoteric
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273
8.21. 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g) 8.22. Barium should be a soft, reactive metal. Barium should form the basic oxide, BaO. Barium metal, for example, would be expected to react with water according to the equation Ba(s) + 2H2O(l) → Ba(OH)2(aq) + H2(g) 8.23. The two oxides of carbon are carbon monoxide, CO, and carbon dioxide, CO2. 8.24. a.
White phosphorus
b.
Sulfur
c.
Bromine
d.
Sodium
8.25. The answer is c, 3. 8.26. The answer is b, [Ne]3s23p3 8.27. The answer is a, First IE 900 kJ/mol, second IE 1750 kJ/mol, third IE 15,000 kJ/mol. 8.28. The answer is b, II only.
■
ANSWERS TO CONCEPT EXPLORATIONS
8.29. a.
Since there is a big jump between the second and the third ionization energies, X be most likely to belong to Group IIA.
b.
An ion of element X would be most likely to have a +2 charge.
c.
Element Y would not be in the same period as element X. The elements belong in the same group, with Y above X in the group.
d.
An atom of Y is smaller than an atom of X. Both elements are in the same group, and Y is above X, so it is smaller.
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Chapter 8: Electron Configurations and Periodicity
8.30.
■
a.
Removing an electron from a −1 ion is the reverse of electron affinity. Since Z has a less negative electron affinity, it will have the smaller energy for the reverse process. Thus, it is easier to remove an electron from Z−. It would require more energy to remove an electron from a W− ion.
b.
Electron affinity is inversely related to the size of the atom. Since Z has a smaller negative electron affinity, it has the larger atomic radius.
c.
The trend for ionization energy parallels the trend for electron affinity. A smaller negative electron affinity (harder to do) means a smaller positive ionization energy (easier to do). Thus, Z has the smaller first ionization energy.
d.
The valence electrons in element Z feel a weaker effective nuclear charge than those in element W. The lower negative electron affinity and lower positive ionization energy mean electrons are not attracted as much to the nucleus in an atom of Z compared to an atom of W.
ANSWERS TO CONCEPTUAL PROBLEMS
8.31. This statement of the Pauli principle implies there can be two electrons with the same spin in a given orbital. Because an electron can have either one of two spins, any orbital can hold a maximum of four electrons. The first six elements of the periodic table would have the following electron configurations: (1)
1s1
(2)
1s2
(3)
1s3
(4)
1s4
(5)
1s42s1
(6)
1s42s2
8.32. The first period of the periodic table would have the following allowed quantum numbers: n = 1; l = 1; ml = 0, +1, −1; ms = +1/2, −1/2. There are six different possible combinations. Therefore, there would be six elements in the first period. 8.33. The elements are in Group IIA (only s electrons) and IIIB (d electrons). They are also in Period 5. Therefore, the elements are strontium (Sr) and yttrium (Y). 8.34. The elements are in Group IIA. They must also be in Periods 4 (no d electrons) and 5 (d electrons). Therefore, the elements are calcium (Ca) and strontium (Sr).
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8.35. Keeping in mind that a filled orbital is usually a stable configuration for an atom, an element in this universe with five electrons would probably lose the two electrons in the second orbital and form a cation with a charge of positive two. The other possible option is for the atom to gain seven additional electrons to fill the second orbital. However, this is unlikely given that the nuclear charge would be relatively small, and electronelectron repulsions in such an atom would be large. 8.36. Keep in mind that the ionization energy of an atom provides a measure of how strongly an electron is attracted to that atom. The electron affinity of an atom provides a measure of how strongly attracted an additional electron is to the atom. Both electron affinity and ionization energy provide information about the strength of the attraction between electrons and a particular nucleus. An element that forms an anion easily has an electron affinity much less than zero and a very large first ionization energy. Examples are the elements on the upper right of the periodic table, such as fluorine, with an electron affinity of −328 kJ/mol and a first ionization energy of 1681 kJ/mol. 8.37. The elements that form oxides of the form RO2 are in Groups IVA and VIA. However, the metalloid oxide in Group VIA is amphoteric. Therefore, the elements are in Group IVA. The metalloid is germanium (Ge), and the metal is tin (Sn). GeO2 is the acidic oxide, and SnO2 is the amphoteric oxide. 8.38. Oxides of Groups IIIA and VA have oxides of the form R2O3. However, Group VA oxides can also be of the form R2O5, so the element is in Group IIIA. It is also acidic. Therefore, it must be boron oxide, B2O3. 8.39. a.
Elements of Group IIA and Group VIIIA do not form negative ions; thus they have electron affinities that are greater than zero.
b.
The large difference between the second and third ionization energies means this would be a Group IIA element.
c.
Luster and conductivity are properties of metals. Group IIA is the only main group that contains just metallic elements.
a.
These elements would form +2 ions and have oxides with formula AO.
b.
2A + O2 → 2AO
c.
The oxides with formula AO would be basic. The reaction with water would be
8.40.
AO + H2O → A(OH)2
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Chapter 8: Electron Configurations and Periodicity
SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 8.41. a.
Not allowed; the paired electrons in the 2p orbital should have opposite spins.
b.
Allowed; electron configuration is 1s22s22p4.
c.
Not allowed; the electrons in the 1s orbital must have opposite spins.
d.
Not allowed; the 2s orbital can hold at most two electrons, with opposite spins.
a.
Not allowed; the two electrons in the 1s orbital must have opposite spins.
b.
Allowed; 1s22s22p4.
c.
Not allowed; the 2s orbital can hold at most two electrons, with opposite spins.
d.
Not allowed; the unpaired electrons in the 2p orbitals should have parallel spins.
a.
Impossible state; the 2p orbitals can hold no more than six electrons.
b.
Impossible state; the 3s orbital can hold no more than two electrons.
c.
Possible state.
d.
Possible state; however, the 3p and 4s orbitals should be filled before the 3d orbital.
a.
Possible state; however, the 2s orbital should be filled with two electrons before the 2p orbitals are filled completely.
b.
Impossible; the 2p orbitals can hold no more than six electrons.
c.
Impossible; the 2s orbital can hold no more than two electrons.
d.
Possible state; however, the 3s and 3p orbitals should be filled before the 3d orbital.
8.42.
8.43.
8.44.
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8.45. The six possible orbital diagrams for 1s22p1 are
1s2
2p 1
8.46. The twelve possible orbital diagrams for 1s12p1 are 1s1
2p 1
1s1
2p 1
8.47. Iodine (Z = 53): 1s22s22p63s23p63d104s24p64d105s25p5 8.48. Phosphorus (Z = 15): 1s22s22p63s23p3 8.49. Manganese (Z = 25): 1s22s22p63s23p63d54s2 8.50. Cobalt (Z = 27): 1s22s22p63s23p63d74s2 8.51. Bromine (Z = 35): 4s24p5 8.52. Antimony (Z = 51): 5s25p3
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1s1
2p 1
277
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Chapter 8: Electron Configurations and Periodicity
8.53. Zirconium (Z = 40): 4d25s2 8.54. Titanium (Z = 22): 3d24s2 8.55. The highest value of n is six, so thallium (Tl) is in the sixth period. The 5d subshell is filled, and there is a 6p electron, so Tl belongs in an A group. There are three valence electrons, so Tl is in Group IIIA. It is a maingroup element. 8.56. The highest value of n is six, so iridium (Ir) is in Period 6. The 5d subshell is not completely filled, so Ir belongs to a B group. There are nine valence electrons, so Ir is in Group VIIIB. Iridium is a dtransition element. 8.57. Cobalt (Z = 27): [Ar]
3d
4s
8.58. Terbium (Z = 65): [Xe]
4f
6s
8.59. Potassium (Z = 19): [Ar]
4s All the subshells are filled in the argon core; however, the 4s electron is unpaired, causing the ground state of the potassium atom to be a paramagnetic substance. 8.60. Calcium (Z = 20): [Ar]
4s All the subshells through the 4s subshell are filled, so the ground state of a calcium atom is a diamagnetic substance.
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279
8.61. Atomic radius increases going down a column (group), from S to Se, and increases going from right to left in a row, from Se to As. Thus, the order by increasing atomic radius is S, Se, As. 8.62. Atomic radius increases going down a column (group), from O to S, and increases going from right to left in a row, from S to P. Thus, the order by increasing atomic radius is O, S, P. 8.63. Ionization energy increases going left to right in a row. Thus, the order by increasing ionization energy is Na, Al, Cl, Ar. 8.64. Ionization energy increases going up a column (group), from Ca to Mg, and increases going left to right in a row, from Mg to S. Thus, the order by increasing ionization energy is Ca, Mg, S. 8.65. a.
In general, the electron affinity becomes more negative going from left to right within a period. Thus, Br has a more negative electron affinity than As.
b.
In general, a nonmetal has a more negative electron affinity than a metal. Thus, F has a more negative electron affinity than Li.
a.
In general, the electron affinity becomes more negative going from left to right within a period. Thus, Cl has a more negative electron affinity than S.
b.
In general, the electron affinity of a nonmetal is more negative than that of a metal. Thus, Se has a more negative electron affinity than K.
8.66.
8.67. Chlorine forms the ClO3− ion, so bromine should form the BrO3− ion, and potassium forms the K+ ion, so lithium should be Li+. Thus, the expected formula of lithium bromate is LiBrO3. 8.68. The expected positive oxidation states of selenium are +4 and +6. The corresponding oxides have the simplest formulas of SeO2 and SeO3.
■
SOLUTIONS TO GENERAL PROBLEMS
8.69. Strontium: 1s22s22p63s23p63d104s24p65s2 8.70. Tin: 1s22s22p63s23p63d104s24p64d105s25p2 8.71. Polonium: 6s26p4 8.72. Thallium: 6s26p1
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Chapter 8: Electron Configurations and Periodicity
8.73. The orbital diagram for arsenic is
1s
2s
2p
3d
3s
4s
3p
4p
8.74. The orbital diagram for germanium is
1s
2s
2p
3d
3s
4s
3p
4p
8.75. For ekalead: [Rn] 5f146d107s27p2. It is a metal; the oxide is ekaPbO or ekaPbO2. 8.76. For ekabismuth: [Rn] 5f146d107s27p3. It is a metal; the oxide is ekaBi2O3 or ekaBi2O5. 8.77. The ionization energy of Fr is ~370 kJ/mol (slightly less than that of Cs). 8.78. The ionization energy of At is ~900 kJ/mol (between the ionization energy of Po and Rn). 8.79. Niobium: [Kr]
4d
5s
4d
5s
8.80. Ruthenium: [Kr]
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281
8.81. a.
Cl2
b.
Na
c.
Sb
d.
Ar
a.
F2
b.
Ba
c.
Ga
d.
O2
8.82.
8.83. Element with Z = 23: 1s22s22p63s23p63d34s2. The element is in Group VB (three of the five valence electrons are d electrons) and in Period 4 (largest n is 4). It is a dblock transition element. 8.84. Element with Z = 33: 1s22s22p63s23p63d104s24p3. The element is in Group VA (five valence electrons, none of them d electrons) and in Period 4 (largest n is 4). It is a maingroup element. 8.85. Nuclear magnetic resonance depends upon the property of nuclear spin. NMR uses the frequency range 300 MHz to 900 MHz. 8.86. Chemical shift depends on the chemical environment of the proton, so the external magnetic field needed to bring a proton into resonance with the radiation varies with the chemical (bonding) environment. 8.87. When an electron in the cathode ray hits a metal atom in the target, it can (if it has sufficient energy) knock an electron from an inner shell of the atom. This produces a metal ion with an electron missing from an inner orbital. This electron configuration is unstable, so an electron from an orbital of higher energy drops into the halffilled orbital, and a photon is emitted. The photon corresponds to electromagnetic radiation in the xray region. 8.88. Instead of irradiating a sample with an electron beam and analyzing the frequencies of emitted x rays, you irradiate a sample with x rays and analyze the kinetic energies of ejected electrons. When you scan the various kinetic energies of ejected electrons, you see a spectrum with peaks corresponding to the different occupied orbitals. These ionization energies are approximately equal to the positive values of the orbital energies, so this spectrum provides direct experimental verification of the discrete energy levels associated with the electrons of the atom.
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Chapter 8: Electron Configurations and Periodicity
8.89. When you place a diamagnetic material in an external magnetic field, its electrons move so as to induce, or generate, a smaller magnetic field that is opposite in direction to the external field. This results in a repulsive force between the diamagnetic material and the external field. This repulsive force can be made upward to balance the downward force of gravity so the diamagnetic material can be levitated by a magnetic field. 8.90. A frog was used.
■
SOLUTIONS TO STRATEGY PROBLEMS
8.91. The element is in Group IA, so it is rubidium. The groundstate electron configuration is [Kr]5s1. 8.92. The element is in Group VA, so it is phosphorus. The groundstate valenceshell configuration is 3s23p3. 8.93. The element is in Group VIA, so it is tellurium. The valenceshell configuration is 5s25p4. 8.94. The element is in the sixth period and is the tenth element from the left, which is platinum. Its symbol is Pt. 8.95. The element is in Period 5, so it is tellurium. Its symbol is Te. The value of b must be 4. 8.96. The maingroup element in Period 5 with five valence electrons is antimony. Its symbol is Sb. The oxide with the most oxygens is Sb2O5. 8.97. The maingroup atom in Period 5 with the smallest radius is xenon. The trend in atomic radius is to decrease as you go from left to right across a period. This is because electrons are being added to the same valence shell, but the charge on the nucleus is increasing. The net effect is to pull the valence electrons closer to the nucleus. The orbital diagram for xenon is
5s
5p
8.98. Hydrogen is the smallest atom. All of the other atoms are in the fourth period. The trend is the atomic radius increases from right to left in a period. Thus, K (potassium), which is in Group IA, has the largest radius.
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283
8.99. The element in Period 6 that forms the ion R+2 and forms the oxides RO and RO2 is lead (Group IVA). The element that immediately follows lead is bismuth (Z = 83) and is in Group VA. The orbital diagram for the groundstate valenceshell electrons is
6s
6p
8.100. In order of increasing ionization energy, Se < Br < Cl.
■
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
8.101. The equation is Ba(s) + 2H2O(l) → Ba(OH)2(aq) + H2(g) Using the equation, calculate the moles of H2; then use the ideal gas law to convert to volume. mol H2 = 2.50 g Ba x V =
1 mol H 2 1 mol Ba x = 0.018204 mol 1 mol Ba 137.33 g Ba
nRT (0.018204 mol)(0.082057 L • atm/K • mol)(294.2 K) = P (748/760) atm
= 0.44651 L (447 mL) 8.102. The equation is 2Cs(s) + 2H2O(l) → 2CsOH(aq) + H2(g) Use the ideal gas law to calculate moles of H2 from 48.1 mL (0.0481 L). Then use the equation to convert moles of H2 to moles and mass of Cs. n =
[(768/760) atm] (0.0481 L) PV = = 0.002027 mol H2 RT (0.082057 L • atm/K • mol)(292.2 K)
mol Cs = 0.002027 mol H2 x mass Cs = 0.004054 mol Cs x
2 mol Cs = 0.004054 mol Cs 1 mol H 2 132.9 g Cs = 0.5387 = 0.539 g Cs 1 mol Cs
8.103. Radium is in Group IIA; hence, the radium cation is Ra2+, and its oxide is RaO. Use the atomic weights to calculate the percentage of Ra in RaO. Percent Ra =
226 amu Ra x 100% = 93.38 = 93.4% Ra 226 amu Ra + 16.00 amu O
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Chapter 8: Electron Configurations and Periodicity
8.104. Tellurium is in Group VIA; hence, its anion is Te2−, and its hydrogen compound is H2Te. Use the atomic weights to calculate the percentage of Te in H2Te. Percent Te =
127.60 amu Te x 100% = 98.4447 127.60 amu Te + (2)(1.00794 amu H) = 98.445% Te
8.105. Convert 5.00 mg (0.00500 g) Na to moles of Na; then convert to energy using the first ionization energy of 496 kJ/mol Na. 1 mol Na = 2.174 x 10−4 mol Na 22.99 g Na
mol Na = 0.00500 g Na x 2.174 x 10−4 mol Na x
496 kJ = 0.1078 = 0.108 kJ = 108 J 1 mol Na
8.106. Convert 2.65 mg (0.00265 g) Cl atoms to moles of Cl(g); then convert to energy using the electron affinity of 349 kJ/mol Cl(g). mol Cl = 0.00265 g Cl x 7.4753 x 10−5 mol Cl x
1 mol Cl = 7.4753 x 10−5 mol Cl 35.45 g Cl
349 kJ = −2.608 x 10−2 = −2.61 x 10−2 kJ 1 mol Cl
8.107. Use the Bohr formula, where nf = ∞ and ni = 1. 1⎤ ⎡ 1 ΔE = −RH ⎢ 2  2 ⎥ 1 ⎦ ⎣∞
I.E. =
= −RH[−1] = RH =
2.179 x 1018 J 1 H atom
2.179 x 1018 J 6.022 x 1023 H atoms 1.31219 x 106 J x = 1 H atom 1 mol H 1 mol H
= 1.312 x 103 kJ/mol H 8.108. Use the Bohr formula, where nf = ∞, ni = 1, and Z = 2. ⎡ 22 22 ⎤ ΔE = −RH ⎢ 2  2 ⎥ 1 ⎦ ⎣∞ I.E. =
= −RH[−4] = 4RH =
4 x 2.179 x 1018 J 1 He + ion
4 x 2.179 x 1018 J 6.022 x 1023 He + ions 5.2487 x 106 J x = 1 He + ion 1 mol He + 1 mol He +
= 5.25 x 103 kJ/mol He+
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8.109. Add the three equations after reversing the equation for the lattice energy and its ΔH: e−
→ →
Na+(g) Cl−(g)
Na+(g) +
Cl−(g)
→
NaCl(s)
−1(ΔH = 786 kJ/mol)
Na(g)
Cl(g)
→
NaCl(s)
ΔH = −639 kJ/mol
Na(g) Cl(g)
+
+
+
e−
ΔH = +496 kJ/mol ΔH = −349 kJ/mol
8.110. Add the three equations after reversing the equation for the lattice energy and its ΔH: K+(g) Br−(g)
e−
ΔH = +419 kJ/mol ΔH = −325 kJ/mol
+
e
→ →
+
K (g)
+
Br−(g)
→
KBr(s)
−1(ΔH = 689 kJ/mol)
K(g)
+
Br(g)
→
KBr(s)
ΔH = −595 kJ/mol
K(g) Br(g)
−
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+
285
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CHAPTER 9
Ionic and Covalent Bonding
■
SOLUTIONS TO EXERCISES .
. . . 9.1. The Lewis symbol for oxygen is : O . . and the Lewis symbol for magnesium is Mg . The magnesium atom loses two electrons, and the oxygen atom accepts two electrons. You can represent this electron transfer as follows: . Mg .
+
.. . O : .
Mg
2+
+
.. : O: ..
2
9.2. The electron configuration of the Ca atom is [Ar]4s2. By losing two electrons, the atom assumes a 2+ charge and the argon configuration, [Ar]. The Lewis symbol is Ca2+. The S atom has the configuration [Ne]3s23p4. By gaining two electrons, the atom assumes a 2− charge and the argon configuration [Ne]3s23p6 and is the same as [Ar]. The Lewis symbol is .. : .. S:
2
9.3. The electron configuration of lead (Pb) is [Xe]4f145d106s26p2. The electron configuration of Pb2+ is [Xe]4f145d106s2. 9.4. The electron configuration of manganese (Z = 25) is [Ar]3d54s2. To find the ion configuration, remove first the 4s electrons, then the 3d electrons. In this case, only two electrons need to be removed. The electron configuration of Mn2+ is [Ar]3d5. 9.5. S2− has a larger radius than S. The anion has more electrons than the atom. The electronelectron repulsion is greater; hence, the valence orbitals expand. The anion radius is larger than the atomic radius. 9.6. The ionic radii increase down any column because of the addition of electron shells. All of these ions are from the Group IIA family; therefore, Mg2+ < Ca2+ < Sr2+. 9.7. Cl−, Ca2+, and P3− are isoelectronic with an electron configuration equivalent to [Ar]. In an isoelectronic sequence, the ionic radius decreases with increasing nuclear charge. Therefore, in order of increasing ionic radius, we have Ca2+, Cl−, and P3−.
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287
9.8. The absolute values of the electronegativity differences are C–O, 1.0; C–S, 0.0; and H–Br, 0.7. Therefore, C–O is the most polar bond. 9.9. First, calculate the total number of valence electrons. C has four, Cl has seven, and F has seven. The total number is 4 + (2 x 7) + (2 x 7) = 32. The expected skeleton consists of a carbon atom surrounded by Cl and F atoms. Distribute the electron pairs to the surrounding atoms to satisfy the octet rule. All 32 electrons (16 pairs) are accounted for.
Cl F
C
F
Cl 9.10. The total number of electrons in CO2 is 4 + (2 x 6) = 16. Because carbon is more electropositive than oxygen, it is expected to be the central atom. Distribute the electrons to the surrounding atoms to satisfy the octet rule.
O
C
O
All sixteen electrons have been used, but note that there are only four electrons on carbon. This is four electrons short of a complete octet, which suggests the existence of double bonds. Move a pair of electrons from each oxygen to the carbonoxygen bonds.
O
C
O
or
O
C
O
9.11. a.
There are (3 x 1) + 6 = 9 valence electrons in H3O. The H3O+ ion has one less electron than is provided by the neutral atoms because the charge on the ion is 1+. Hence, there are eight valence electrons in H3O+. The electrondot formula is H .. H :O:H ..
+
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b.
Cl has seven valence electrons, and O has six valence electrons. The total number of valence electrons from the neutral atoms is 7 + (2 x 6) = 19. The charge on the ClO2− is −1, which provides one more electron than the neutral atoms. This makes a total of 20 valence electrons. The electrondot formula for ClO2− is
O
Cl O
9.12. The resonance formulas for NO3− are

O O


O
N O
O
N O
O O
N
O
9.13. The number of valence electrons in SF4 is 6 + (4 x 7) = 34. The skeleton structure is a sulfur atom surrounded by fluorine atoms. After the electron pairs are placed on the F atoms to satisfy the octet rule, two electrons remain.
F
F S
F
F
These additional two electrons are put on the sulfur atom because it had d orbitals and, therefore, can expand its octet.
F
F S
F
F
9.14. Be has two valence electrons, and Cl has seven valence electrons. The total number of valence electrons is 2 + (2 x 7) = 16 in the BeCl2 molecule. Be, a Group IIA element, can have fewer than eight electrons around it. The electrondot formula of BeCl2 is
Cl Be Cl
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9.15. The total number of electrons in H3PO4 is 3 + 5 + 24 = 32. Assume a skeleton structure in which the phosphorus atom is surrounded by the more electronegative four oxygen atoms. The hydrogen atoms are then attached to the oxygen atoms. Distribute the electron pairs to the surrounding atoms to satisfy the octet rule. If you assume all single bonds (structure on the left), the formal charge on the phosphorus is 1+ and the formal charge on the top oxygen is 1−. Using the principle of forming a double bond with a pair of electrons on the atom with the negative formal charge, you obtain the structure on the right. The formal charge on all oxygens in this structure is zero; the formal charge on phosphorus is 5 − 5 = 0. This is the better structure.
O
O P
O
H
O
H
H
O
P
O
O
H
H
O
H
9.16. The bond length can be predicted by adding the covalent radii of the two atoms. For O–H, we have 66 pm + 37 pm = 103 pm. 9.17. As the bond order increases, the bond length decreases. Since the C=O is a double bond, we would expect it to be the shorter one, 123 pm.
H
H C
9.18. H
C
O + 3 O2
2 O
C
O
+ 2 H
H
H
One C=C bond, four C–H bonds, and three O2 bonds are broken. There are four C=O bonds and four O–H bonds formed. ΔH = {[602 + (4 x 411) + (3 x 494)] − [(4 x 799) + (4 x 459)]} kJ = −1304 kJ
■
ANSWERS TO CONCEPT CHECKS
9.1. a.
The +2 ions are common in transition elements, but it is the outer s electrons that are lost to form these ions in compounds. Iron, whose configuration is [Ar]3d64s2, would be expected to lose two 4s electrons to give the configuration [Ar]3d6 for the Fe2+ ion in compounds. The configuration given in the problem is for an excited state; you would not expect to see it in compounds.
b.
Nitrogen, whose groundstate atomic configuration is [He]2s22p3, would be expected to form an anion with a noblegas configuration by gaining three electrons. This would give the anion N3− with the configuration [He]2s22p6. You would not expect to see the anion N2− in compounds.
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c.
The zinc atom has the groundstate configuration [Ar]3d104s2. The element is often considered to be a transition element. In any case, you would expect the atom to form compounds by losing its 4s electrons to give Zn2+ with the pseudonoblegas configuration [Ar]3d10. This is the ion configuration given in the problem.
d.
The configuration of the groundstate sodium atom is [He]2s22p63s1. You would expect the atom to lose one electron to give the Na+ ion with the noblegas configuration [He]2s22p6. You would not expect to see compounds with the Na2+ ion.
e.
The ground state of the calcium atom is [Ne]3s23p64s2. You would expect the atom to lose its two outer electrons to give Ca2+ with the noblegas configuration [Ne]3s23p6, which is the configuration given in the problem.
a.
There are two basic points to consider in assessing the validity of each of the formulas given in the problem. One is whether the formula has the correct skeleton structure. You expect the F atoms to be bonded to the central N atoms because the F atoms are more electronegative. The second point is the number of dots in the formula. This should equal the total number of electrons in the valence shell of the atoms (five for each nitrogen atom and seven for each fluorine atom), which is (2 x 5) + (2 x 7) = 24, or twelve pairs. The number showing in the formula here is thirteen, which is incorrect.
b.
This formula has the correct skeleton structure and the correct number of dots. All of the atoms have octets, so the formula would appear to be correct. As a final check, however, you might try drawing the formula beginning with the skeleton structure. In drawing an electrondot formula, after connecting atoms by single bonds (a single electron pair), you would place electron pairs around the outer atoms (the F atoms in this formula) to give octets. After doing that, you would have used up nine electron pairs (three for the single bonds and three for each F atom to fill out its octet). This leaves three pairs, which you might distribute as follows:
9.2.
F
N
N
F
One of the nitrogen atoms (the one on the right) does not have an octet. The lack of an octet on this atom suggests trying for a double bond. This suggests that you move one of the lone pairs on the left N atom into one of the adjacent bonding regions. Moving this lone pair into the N–N region would give a symmetrical result, whereas moving the lone pair into the F–N region would not. The text notes that the atoms often showing multiple bonds are C, N, O, and S. The formula given here with a nitrogennitrogen double bond appears quite reasonable. c.
This formula is similar to the previous one, b, but the double bond is between the F and N atoms. Multiple bonds to F are less likely than those between two N atoms, so this is not the preferred formula. You could also apply the rules of formal charge in this case, and you would come to the same conclusion. The formula here gives a 1− charge to the left F atom and a 1+ charge to the right N atom, whereas each of the atoms in the b formula has zero formal charge. Rule A says that whenever you can write several Lewis formulas for a molecule, you should choose the formula having the lowest magnitudes of formal charges. In this case, this is the formula b. Strictly speaking, both b and c could be regarded as resonance formulas, but b would have much more importance than c in describing the electronic structure of the molecule.
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d.
This formula is similar to the one you drew earlier in describing how you would get to the formula in b. The left N atom does not have an octet, which suggests you move a lone pair on the other N atom into the N–N bond region to give a double bond.
e.
This formula does not have the correct skeleton structure.
f.
This formula has the correct skeleton structure and the correct total number of electron dots, but neither N atom has an octet. In fact, there is no bond between the two N atoms.
a.
This model and corresponding Lewis structure, H:C:::N:, has the expected skeleton structure. (H must be an exterior atom, but either C or N might be in the center; however, you expect the more electropositive, or less electronegative, atom, C, to be in this position). This formula also has the correct number of electron dots (1 + 4 + 5 = 10, or 5 pairs). Finally, the formal charge of each atom is zero. Therefore, this model should be an accurate representation of the HCN molecule.
b.
This structure has the more electronegative atom, N, in the central position; you don't expect this to be the correct structure. You can also look at this from the point of view of formal charges. This formula has a 1+ charge on the N atom and a 1− charge on the C atom. You would not expect the more electronegative atom to have the positive formal charge. Moreover, the previous formula, a, has zero charges for each atom, which would be preferred.
c.
If you draw the Lewis structure, you will see that each atom has an octet, but the formula has too many electron pairs (seven instead of five). You can remove two pairs and still retain octets if you move two lone pairs into the bonding region, to give a triple bond.
d.
If you draw the Lewis structure, you will see that each atom has an octet, but the formula has too many electron pairs (six instead of five). You would arrive at the same conclusion using formal charges. The formal charges are 2− for H, 1+ for C, and 1+ for N. Since you already know that formula a has zero formal charge for each atom, such high formal charges for the atoms in formula d mean it is not a very good representation of the molecule.
9.3.
■
ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
9.1. As an Na atom approaches a Cl atom, the outer electron of the Na atom is transferred to the Cl atom. The result is an Na+ and a Cl− ion. Positively charged ions attract negatively charged ions, so, finally, the NaCl crystal consists of Na+ ions surrounded by six Cl− ions surrounded by six Na+ ions. 9.2. Ions tend to attract as many ions of opposite charge about them as possible. The result is that ions tend to form crystalline solids rather than molecular substances. 9.3. The energy terms involved in the formation of an ionic solid from atoms are the ionization energy of the metal atom, the electron affinity of the nonmetal atom, and the energy of the attraction of the ions forming the ionic solid. The energy of the solid will be low if the ionization energy of the metal is low, the electron affinity of the nonmetal is high, and the energy of the attraction of the ions is large.
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9.4. The lattice energy for potassium bromide is the change in energy that occurs when KBr(s) is separated into isolated K+(g) and Br −(g) ions in the gas phase. KBr(s) → K+(g) + Br −(g) 9.5. A monatomic cation with a charge equal to the group number corresponds to the loss of all valence electrons. This loss of electrons would give a noblegas configuration, which is especially stable. A monatomic anion with a charge equal to the group number minus eight would have a noblegas configuration. 9.6. Most of the transition elements have configurations in which the outer s subshell is doubly occupied. These electrons will be lost first, and we might expect each to be lost with almost equal ease, resulting in +2 ions. 9.7. If we assume the ions are spheres that are just touching, the distances between centers of the spheres will be related to the radii of the spheres. For example, in LiI, we assume that the −1 ions are large spheres that are touching. The distance between centers of the I− ions equals two times the radius of the I− ion. 9.8. In going across a period, the cations decrease in radius. When we reach the anions, there is an abrupt increase in radius, and then the radii again decrease. Ionic radii increase going down any column of the periodic table. 9.9. As the H atoms approach one another, their 1s orbitals begin to overlap. Each electron can then occupy the space around both atoms; that is, the two electrons are shared by the atoms. 9.10.
Potential energy
Cl + Cl
240 kJ/mol Bond dissociation energy Cl2 Bond length 194 pm Distance between the nuclei
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9.11. An example is thionyl chloride, SOCl2:
Cl Cl
S
O coordinate covalent bond
Note that the O atom has eight electrons around it; that is, it has two more electrons than the neutral atom. These two electrons must have come from the S atom. Thus, this bond is a coordinate covalent bond. 9.12. In many atoms of the maingroup elements, bonding uses an s orbital and the three p orbitals of the valence shell. These four orbitals are filled with eight electrons, thus accounting for the octet rule. 9.13. Electronegativity increases from left to right (with the exception of the noble gases) and decreases from top to bottom in the periodic table. 9.14. The absolute difference in the electronegativities of the two atoms in a bond gives a rough measure of the polarity of the bond. 9.15. Resonance is used to describe the electron structure of a molecule in which bonding electrons are delocalized. In a resonance description, the molecule is described in terms of two or more Lewis formulas. If we want to retain Lewis formulas, resonance is required because each Lewis formula assumes that a bonding pair of electrons occupies the region between two atoms. We must imagine that the actual electron structure of the molecule is a composite of all resonance formulas. 9.16. Molecules having an odd number of electrons do not obey the octet rule. An example is nitrogen monoxide, NO. The other exceptions fall into two groups. In one group are molecules with an atom having fewer than eight valence electrons around it. An example is borane, BH3. In the other group are molecules with an atom having more than eight valence electrons around it. An example is sulfur hexafluoride, SF6. 9.17. As the bond order increases, the bond length decreases. For example, the average carboncarbon singlebond length is 154 pm, whereas the carboncarbon doublebond length is 134 pm, and the carboncarbon triplebond length is 120 pm. 9.18. Bond energy is the average enthalpy change for the breaking of a bond in a molecule. The enthalpy of a reaction for gaseous reactions can be determined by summing the bond energies of all the bonds that are broken and subtracting the sum of the bond energies of all the bonds that are formed. 9.19. The answer is a, BaCO3.
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9.20. The answer is e, S− > S > S+. 9.21. The answer is d, 1s22s22p5. 9.22. The answer is a, sulfur.
■
ANSWERS TO CONCEPT EXPLORATIONS
9.23. a.
Since a cation is smaller than the neutral atom, the smallest sphere (far right) is the correct representation of M+.
b.
Since an anion is larger than the neutral atom, the largest sphere is the correct representation of X−.
c.
The chemical equation for the reaction is M(s) + X(g) → MX(s)
d.
The product of the reaction is an ionic compound. The M atom readily forms the M+ ion, and the X atom readily forms the X− ion. This is characteristic of an electrontransfer reaction, in which ionic compounds are formed.
e.
Since MX is an ionic compound, the best representation of its structure is the crystal.
a.
The formation of a bond is an energyreleasing process and is exothermic.
b.
Bond energy is the energy required to break a bond. It is the reverse process from part a, so it is an endothermic process and a positive quantity.
c.
ΔHrxn ≅ BE(reactants) − BE(products) = 0 − BE(X–X)
9.24.
BE(X–X) = −ΔHrxn = −(−500 kJ/mol) = 500 kJ/mol d.
ΔHrxn ≅ BE(reactants) − BE(products) = BE(X–X) + BE(Y–Y) − 2BE(X–Y) ≅ 500 kJ/mol + 750 kJ/mol − 2(1500 kJ/mol) = −1750 kJ/mol
e.
The approach here is to use the BornHaber cycle to calculate ΔH for the reaction. The reactions and energies are as follows. A(g) → A+(g) + e− 1/2X2(g) → X(g) X(g) + e− → X−(g) A+(g) + X−(g) → AX(s)
ΔH1 ΔH2 ΔH3 ΔH4
= = = =
I1 = 400 kJ 1/2BE = 1/2(500 kJ) = 250 kJ EA = −525 kJ −U = −(100 kJ) = −100 kJ
A(g) + 1/2X2(g) → AX(s) ΔHrxn = ΔH1 + ΔH2 + ΔH3 + ΔH4 ΔHrxn = 400 kJ + 250 kJ + −525 kJ + −100 kJ = 25 kJ Since the reaction is endothermic, it is not likely that AX will be formed in the reaction.
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f.
■
295
You would predict that the formation of AX(s) would be likely if the reaction were exothermic, so ΔHrxn must be negative. Therefore, the minimum amount of AX(s) lattice energy required would be 100 kJ + 25 kJ = 125 kJ.
ANSWERS TO CONCEPTUAL PROBLEMS
9.25. Because the compound that forms is a combination of a metal and a nonmetal, we would expect it to be ionic. If we assume that metal atoms tend to lose electrons to obtain filled shells, then the metal atom X would lose three electrons from the n = 2 level, forming the X3+ cation. We can expect the nonmetal atom Y to gain electrons to obtain a filled shell, so it requires an additional electron to fill the n = 1 level, forming the Y− anion. To produce an ionic compound with an overall charge of zero, the compound formed from these two elements would be XY3. 9.26. Elements in higher periods in Group IIIA tend to form ions either by losing the outer p electron to form +1 ions or by losing the outer s and p electrons to form +3 ions. In the case of thallium, these ions have the configurations in d and b, respectively. The +2 ion given in configuration a represents a loss of the two outer s electrons, leaving the outer p electrons of higher energy. This configuration would represent an excited state of the +2 ion. You would not expect a +2 ion in thallium compounds, and an excited configuration is even less likely. The +4 ion given in c is formed by loss of the outer s and p electrons plus a d electron from the closed d subshell. This would be an unlikely ion for thallium compounds. 9.27. The smaller atom on the left (yellow) becomes a larger ion on the right, while the larger atom on the left (blue) becomes the smaller ion on the right. Since cations are smaller than their parent atom, and anions are larger than their parent atom, the cation on the right is the smaller ion (blue), and the anion is the larger ion (yellow). Finally, since metals tend to form cations, and nonmetals form anions, the metal on the left is the larger atom (blue), and the nonmetal is the smaller atom (yellow). 9.28. Element 117 would fall in Group VII A, the halogens. The monatomic ion for this element would probably be like the other halogens, an anion with a 1− charge. Since anions are larger than the parent atom, the larger sphere (on the right) would be the ion, and the smaller sphere (on the left) would be the atom. 9.29. a.
Incorrect. The atoms in this formula do not obey the octet rule. The formula has the correct number of valence electrons, so this suggests a multiple bond between the N atoms.
b.
Correct. The central atom is surrounded by more electronegative atoms, as you would expect, and each atom obeys the octet rule.
c.
Incorrect. The skeleton structure is acceptable (the central atom is surrounded by more electronegative atoms), but you would expect the double bond to be between C and O rather than between C and F (C, N, O, and S form multiple bonds). You would come to this same conclusion using rules of formal charge. (The formula has a formal charge of 1+ on F and 1− on O, whereas you would expect these formal charges to be interchanged, with the negative charge on the F atom, which is more electronegative.)
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d.
Incorrect. The skeleton structure is OK, but the carbon atom has ten valence electrons about it. This suggests that you replace the two carbonoxygen double bonds by one carbonoxygen double bond (because there is one extra pair of electrons on the C atom). The most symmetrical location of the double bond uses the oxygen atom not bonded to an H atom. Also, only this formula has zero formal charges on all atoms.
9.30. a.
H
S
H
b.
O O
H c.
O
N
O H
C
H
d.
H
C
C
a.
To arrive at a skeleton structure, you decide which is the central atom. (It cannot be H). The C atom is less electronegative than the Cl atom, so you place it as the central atom and surround it by the other atoms.
H
9.31.
Cl H
Cl
C H
b.
HNO2 is an oxyacid in which O atoms bond to the central atom with the H bonded to O. The central atom must be N (the only other atom), so the skeletal structure is
H c.
N
O
You place the least electronegative atom (N) as the central atom and bond it to the other atoms.
F d.
O
N
O
N is less electronegative than O. The most symmetrical structure would be the two N atoms in the center with two O atoms bonded to each N atom.
O
O N N
O
O
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9.32. The ranking is a, c, b (best) for the following reasons: a.
This formula has a 2− formal charge on the outer N and 1+ charges on the other two atoms. This formula has a larger magnitude of formal charges than either of the other two formulas, so it ranks last as a representation of the electron structure of the N2O molecule.
b.
This formula has a 1− formal charge on O and a 1+ formal charge on the central N. This places the negative formal charge on the more electronegative atom, O, making this the best representation of the electron structure of the molecule.
c.
This formula has a 1− formal charge on the outer N atom and a 1+ formal charge on the central N atom. This is better than resonance formula a, but not as good as b.
a.
In order to be a neutral ionic compound, element X must have a charge of 3−. In an ionic compound, calcium forms the Ca2+ cation. The combination of Ca2+ and X3− would form the ionic compound with the formula Ca3X2.
b.
Since element X formed an ionic compound with sodium metal, it is probably a nonmetal with a high electron affinity. When a nonmetal with a high electron affinity combines with a metal such as calcium, an ionic compound is formed.
a.
In general, the enthalpy of a reaction is (approximately) equal to the sum of the bond energies for the bonds broken minus the sum of the bond energies for the bonds formed. Therefore, the bond energy (BE) for each of the reactions can be calculated by the following relationships:
9.33.
9.34.
ΔH = BE(X–X) + BE(O=O) – 2 x BE(X–O) – BE(O–O) ΔH = BE(Y–Y) + BE(O=O) – 2 x BE(Y–O) – BE(O–O) ΔH = BE(Z–Z) + BE(O=O) – 2 x BE(Z–O) – BE(O–O) Because the bond energies of X–O, Y–O, and Z–O are all equal, the difference in the values of ΔH for each reaction is solely a function of the X–X, Y–Y, or Z–Z bonds. Therefore, the compound with the most positive value of ΔH has the strongest bonds. The ranking of strongest to weakest bond is Y–Y > Z–Z > X–X. b.
Since the bond strengths of all the products are equal, the same amount of energy would be required to dissociate each of the products into atoms.
c.
In this case, less energy would be required to break the bonds, so the ΔH for each reaction would decrease.
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Chapter 9: Ionic and Covalent Bonding
SOLUTIONS TO PRACTICE PROBLEMS
9.35. a.
P has the electron configuration [Ne]3s23p3. It has five electrons in its valence shell. The Lewis formula is
P b.
P3− has three more valence electrons than P. It now has eight electrons in its valence shell. The Lewis formula is
P c.
3
Ga has the electron configuration [Ar]3d104s24p1. It has three electrons in its valence shell. The Lewis formula is
Ga d.
3+
Ga has three fewer valence electrons than Ga. It now has zero electrons in its valence shell. The Lewis formula is Ga3+
9.36. a.
Br has the valence configuration [Ar]3d104s24p5. It has seven electrons in its valence shell. The Lewis formula is
Br b.
Br− has one more electron than Br. It now has eight electrons in its valence shell. The Lewis formula is
Br c.

Strontium has the electron configuration [Kr]5s2. It has two valence electrons in its valence shell. The Lewis formula is
Sr d.
2+
Sr has two fewer electrons than Sr. It now has zero electrons in its valence shell. The Lewis formula is Sr2+
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9.37. a.
If the calcium atom loses two electrons, and the bromine atoms gain one electron each, all three atoms will assume noblegas configurations. This can be represented as follows. Br
b.
+
Ca
+

Br
Br
+ Ca2+ +
Br
If the potassium atom loses one electron, and the iodine atom gains one electron, both atoms will assume a noblegas configuration. This can be represented as follows. K
K+
I
+
+

I
9.38. a.
If the magnesium atom loses two electrons, and the sulfur atom gains two electrons, both atoms will assume a noblegas configuration. This can be represented as follows:
Mg
+
S
Mg2+ +
S
2
b.
If the barium atom loses two electrons, and both iodine atoms gain one electron, all three atoms will have noblegas configurations. This can be represented as I I I + Ba + I + Ba2+ + follows.
a.
As:
b.
As3+:
c.
Se:
9.39.
d.
1s22s 22p 63s23p 63d 104s24p 3 1s 22s22p 63s23p 63d 104s2
1s 22s22p 63s23p 63d 104s24p 4
As 3+
As
Se 2
Se2:
1s22s 22p 63s23p 63d 104s24p 6
Se
9.40. a.
In:
1s22s22p 63s 23p 63d 104s24p 64d 105s25p 1
b.
In:
1s22s22p 63s 23p 63d 104s24p 64d 105s25p 2
c.
K+:
1s22s22p 63s23p 6
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In
In
K+
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d.
I:
1s22s 22p 63s23p 63d 104s24p 64d 105s 25p 6
a.
Bi:
[Xe]4f
Bi3+:
The three 6p electrons are lost f rom the valence shell. [Xe]4f 145d 106s2
I
9.41. 14
5d 106s26p 3
b.
9.42. a.
Sn:
[Kr]4d 105s25p 2
Sn2+:
The two 5p electrons are lost f rom the valence shell. [Kr]4d 105s2
b.
9.43. The +2 ion is formed by the loss of electrons from the 4s subshell. Ni2+:
[Ar]3d8
The +3 ion is formed by the loss of electrons from the 4s and 3d subshells. Ni3+:
[Ar]3d7
9.44. The +1 ion is formed by the loss of the one electron in the 4s subshell. Cu+:
[Ar]3d10
The +2 ion is formed by the loss of a 4s and a 3d electron. Cu2+:
[Ar]3d9
9.45. a.
Sr2+ < Sr The cation is smaller than the neutral atom because it has lost all its valence electrons; hence, it has one less shell of electrons. The electronelectron repulsion is reduced, so the orbitals shrink because of the increased attraction of the electrons to the nucleus.
b.
Br < Br− The anion is larger than the neutral atom because it has more electrons. The electronelectron repulsion is greater, so the valence orbitals expand to give a larger radius.
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9.46. a.
Te < Te2− The anion is larger than the neutral atom because it has more electrons. The electronelectron repulsion is greater, so the valence orbitals expand to give a larger radius.
b.
Al3+ < Al The cation is smaller than the neutral atom because it has lost all its valence electrons; hence, it has one less shell of electrons. The electronelectron repulsion is reduced, so the orbitals shrink because of the increased attraction of the electrons to the nucleus.
9.47. S2− < Se2− < Te2− All have the same number of electrons in the valence shell. The radius increases with the increasing number of filled shells. 9.48. P3− is larger. It has the same number of valence electrons as N3−, but the valence shell has n = 3 for P3− and n = 2 for N3−. The radius increases with the increasing number of filled shells. 9.49. Smallest Na+ (Z = 11), F− (Z = 9), N3− (Z = 7) Largest These ions are isoelectronic. The atomic radius increases with the decreasing nuclear charge (Z). 9.50. Smallest Na+ (Z = 11), Cl− (Z = 17), S2− (Z = 16) Largest These ions are all from elements in the third row of the periodic table. The atomic radius increases with increasing negative charge.
lone pairs 2H
+
H Se H
Se
Bonding electron pairs
9.51.
H 3H
+
Bonding electron pairs
As H
As
H
9.52.
lone pair
9.53. Arsenic is in Group VA on the periodic table and has five valence electrons. Bromine is in Group VIIA and has seven valence electrons. Arsenic forms three covalent bonds, and bromine forms one covalent bond. Therefore, the simplest compound would be AsBr3.
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9.54. Silicon is in Group IVA on the periodic table and has four valence electrons. Chlorine is in Group VIIA and has seven valence electrons. Silicon forms four covalent bonds, and chlorine forms one covalent bond. Therefore, the simplest compound would be SiCl4. 9.55. a.
P, N, O Electronegativity increases from left to right and from bottom to top in the periodic table.
b.
Na, Mg, Al Electronegativity increases from left to right within a period.
c.
Al, Si, C Electronegativity increases from left to right and from bottom to top in the periodic table.
9.56. a.
Rb, Sr, Ca Electronegativity increases from left to right and from bottom to top in the periodic table.
b.
Ca, Ga, Ge Electronegativity increases from left to right within a period.
c.
Sb, As, Se Electronegativity increases from left to right and from bottom to top in the periodic table.
9.57. XO − XP = 3.5 − 2.1 = 1.4 XCl − XC = 3.0 − 2.5 = 0.5 XBr − XAs = 2.8 − 2.0 = 0.8 The bonds arranged by increasing difference in electronegativity are C–Cl, As–Br, P–O. 9.58. XN − XH = 3.0 − 2.1 = 0.9 XBr − XSi = 2.8 − 1.8 = 1.0 XCl − XN = 3.0 − 3.0 = 0.0 The difference in electronegativity is smallest for the N–Cl bond; hence, it is the least polar. 9.59. a.
P–O δ+
b.
δ−
C–Cl δ+
δ−
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c.
303
As–Br δ+
δ−
9.60. The atom with the greater electronegativity has the partial negative charge. a.
H–N δ+
b.
δ−
Si–Br δ+
δ−
c.
Nonpolar
a.
Total number of valence electrons = 7 + 7 = 14. Br–Br is the skeleton. Distribute the remaining 12 electrons.
9.61.
Br b.
Total valence electrons = (2 x 1) + 6 = 8. The skeleton is H–Se–H. Distribute the remaining 4 electrons.
H c.
Br
S
H
Total valence electrons = (3 x 7) + 5 = 26. The skeleton is
F F N Distribute the remaining 20 electrons. F
F F
F
N
9.62. a.
Total valence electrons = 7 + 5 + 6 = 18. The skeleton is
O
N
Cl
Distribute the remaining 14 electrons so that the O and Cl atoms have an octet.
O
N
Cl
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Chapter 9: Ionic and Covalent Bonding
Draw a nitrogenoxygen double bond to achieve an N octet.
O b.
N
Cl
Total valence electrons = 5 + (3 x 7) = 26. The skeleton is
Br Br
P
Br Distribute the remaining 20 electrons.
Br Br c.
P
Br
Total valence electrons = 7 + 7 = 14. F–I is the skeleton. Distribute the remaining 12 electrons.
F
I
9.63. a.
Total valence electrons = 2 x 5 = 10. The skeleton is P–P. Distribute the remaining electrons symmetrically:
P
P
Neither P atom has an octet. There are 4 fewer electrons than needed. This suggests the presence of a triple bond. Make one lone pair from each P a bonding pair.
P b.
P
Total valence electrons = 4 + 6 + (2 x 7) = 24. The skeleton is
Br O
C
Br
Distribute the remaining 18 electrons.
Br O
C
Br
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305
Note that carbon is two electrons short of an octet. This suggests the presence of a double bond. The most likely double bond is between C and O.
Br O c.
Br
C
Total valence electrons = 1 + 5 + (2 x 6) = 18. The skeleton is most likely H–O–N–O. Distribute the remaining 12 electrons:
H
O
N
O
Note that the N atom does not have an octet. It is two electrons short. The most likely double bond is between N and O.
H
O
N
O
9.64. a.
Total valence electrons = 4 + 6 = 10. The skeleton is C–O. Distribute the remaining 8 electrons:
C
O
Neither atom has an octet. There are 4 fewer electrons than needed. This suggests a triple bond.
C b.
O
Total valence electrons = 7 + 4 + 5 = 16. The skeleton is Br–C–N. Distribute the remaining 12 electrons:
Br
C
N
Note that the carbon atom is four electrons short of an octet. Make a triple bond between C and N from the four nonbonding electrons on nitrogen.
Br c.
C
N
Total valence electrons = (2 x 5) + (2 x 7) = 24. The skeleton is F–N–N–F. Distribute the remaining 18 electrons.
F
N
N
F
Note that one of the nitrogens is two electrons short of an octet. This suggests the presence of a double bond. The most likely double bond is between the nitrogens.
F
N
N
F
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Chapter 9: Ionic and Covalent Bonding
9.65. a.
Total valence electrons = 7 + 6 + 1 = 14. The skeleton is Cl–O. Distribute the remaining 12 electrons. −
O
Cl b.
Total valence electrons = 4 + (3 x 7) + 1 = 26. The skeleton is
Cl Cl
Sn
Cl
Distribute the remaining 20 electrons so that each atom has an octet. −
Cl Cl c.
Sn
Cl
Total valence electrons = (2 x 6) + 2 = 14. The skeleton is S–S. Distribute the remaining 12 electrons. .. :S ..
2.. S: ..
9.66. a.
Valence electrons = 7 + (2 x 7) − 1 = 20. The skeleton is Br–I–Br. Distribute the remaining 16 electrons.
+ Br b.
I
Br
Total valence electrons = 7 + (2 x 7) − 1 = 20. The skeleton is F–Cl–F. Distribute the remaining 16 electrons.
+ F c.
Cl
F
Total valence electrons = 4 + 5 + 1 = 10. The skeleton is C–N. Distribute the remaining 8 electrons.
C
N
Note that neither atom has an octet. There are four fewer electrons than needed, suggesting a triple bond:
C
N

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9.67. a.
There are two possible resonance structures for HNO3.
O
O
N
N
H O
O
H O
O
One electron pair is delocalized over the nitrogen atom and the two oxygen atoms. b.
There are three possible resonance structures for SO3.
O
O
O
S
S
S
O
O
O
O
O
O
One pair of electrons is delocalized over the region of the three sulfuroxygen bonds. 9.68. a.
There are two resonance forms for NO2−.

N O
O

N O
O
One pair of electrons is delocalized over the region of both nitrogenoxygen bonds. b.
There are two possible resonance structures for FNO2.
O
O
N
N O
F
O
F
One electron pair is delocalized over the nitrogen atom and the two oxygen atoms. 9.69.
H H
C H
H
O H
N O
C H
One pair of electrons is delocalized over the O–N–O bonds.
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O N O
307
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Chapter 9: Ionic and Covalent Bonding
9.70.

O
C
C H

O
O
H
O
The two carbonoxygen bonds are expected to be the same. This means one pair of electrons is delocalized over the region of the O–C–O bonds. 9.71. a.
Total valence electrons = 8 + (2 x 7) = 22. The skeleton is F–Xe–F. Place six electrons around each fluorine atom to satisfy its octet.
Xe
F
F
There are three electron pairs remaining. Place them on the xenon atom.
Xe
F b.
F
Total valence electrons = 6 + (4 x 7) = 34. The skeleton is
F F
Se
F
F Distribute 24 of the remaining 26 electrons on the fluorine atoms. The remaining pair of electrons is placed on the selenium atom. F F
Se
F
F c.
Total valence electrons = 6 + (6 x 7) = 48. The skeleton is
F
F Te
F F
F F
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309
Distribute the remaining 36 electrons on the fluorine atoms.
F
F Te
F
F
F d.
F
Total valence electrons = 8 + (5 x 7) − 1 = 42. The skeleton is
F
F Xe
F
F F Use 30 of the remaining 32 electrons on the fluorine atoms to complete their octets. The remaining 2 electrons form a lone pair on the xenon atom. +
F
F Xe F
F F
9.72. a.
Total valence electrons = (3 x 7) + 1 = 22. The skeleton is I–I–I. Distribute the remaining 18 electrons. The central iodine atom may expand its octet.
I b.
I
I
Total valence electrons = 7 + (3 x 7) = 28. The skeleton is
F
Cl
F
F Distribute 18 of the remaining 22 electrons to complete the octets of the fluorine atoms. The four remaining electrons form two sets of lone pairs on the chlorine atom.
F
Cl
F
F
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Chapter 9: Ionic and Covalent Bonding
c.
Total valence electrons = 7 + (4 x 7) + 1 = 36. The skeleton is
F I
F
F
F Distribute 24 of the 28 remaining electrons to complete the octets of the fluorine atoms. The 4 electrons remaining form two sets of lone pairs on the iodine atom.

F I
F
F
F d.
Total valence electrons = 7 + (5 x 7) = 42. The skeleton is
F
F Br
F
F F Use 30 of the remaining 32 electrons to complete the octets of the fluorine atoms. The 2 electrons remaining form a lone pair on the bromine atom. F
F Br F
F F
9.73. a.
Total valence electrons = 3 + (3 x 7) = 24. The skeleton is
Cl Cl
B
Cl
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311
Distribute the remaining 18 electrons.
Cl Cl
Cl
B
Although boron has only 6 electrons, it has the normal number of covalent bonds. b.
Total valence electrons = 3 + (2 x 7) − 1 = 16. The skeleton is Cl–Tl–Cl. Distribute the remaining 12 electrons.
+ Cl
Cl
TI
Tl has only 4 electrons around it. c.
Total valence electrons = 2 + (2 x 7) = 16. The skeleton is Br–Be–Br. Distribute the remaining 12 electrons.
Br
Be
Br
In covalent compounds, beryllium frequently has two bonds, even though it does not have an octet. 9.74. a.
Total valence electrons = 3 + (3 x 7) = 24. The skeleton is
I Al I I Distribute the remaining 18 electrons to the iodine atoms. I I b.
Al
I
Total valence electrons = 2 + (2 x 7) = 16. The skeleton is Cl–Be–Cl. Distribute the remaining 12 electrons.
Cl
Be
Cl
In covalent compounds, beryllium commonly has two bonds, even though it does not have an octet.
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Chapter 9: Ionic and Covalent Bonding
c.
Total valence electrons = 2 + (3 x 7) + 1 = 24. The skeleton is
F Be F F Distribute the remaining 18 electrons on the fluorine atoms. F Be
F
F
This ion is isoelectronic with BF3, and the beryllium, like boron, has only 6 electrons around it. 9.75. a.
The total number of electrons in O3 is 3 x 6 = 18. Assume a skeleton structure in which one oxygen atom is singly bonded to the other two oxygen atoms. This requires four electrons for the two single bonds, leaving fourteen electrons to be used. Distribute three electron pairs to each of the outer oxygen atoms to complete their octets. This requires twelve more electrons, leaving two electrons to distribute. Since this is two electrons short of an octet, move a pair of electrons from one of the outer oxygen atoms to give an oxygenoxygen double bond. One of the possible resonance structures is shown below; the other structure would have the double bond written between the left and central oxygen atoms.
O
O
O
Starting with the left oxygen, the formal charge of this oxygen is 6 − 1 − 6 = −1. The formal charge of just the central oxygen is 6 − 3 − 2 = +1. The formal charge of the right oxygen is 6 − 2 − 4 = 0. The sum of all three is 0. b.
The total number of electrons in CO is 4 + 6 = 10. Assume a skeleton structure in which the oxygen atom is singly bonded to the carbon atom. This requires two electrons for the single bond and leaves eight electrons. Distribute these remaining electrons around the atoms. Since four more electrons are needed to complete octets on both atoms, move two electron pairs to give a carbonoxygen triple bond. The structure is
C
O
The formal charge of the carbon is 4 − 3 − 2 = −1. The formal charge of the oxygen is 6 − 3 − 2 = +1. The sum of both is 0.
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c.
313
The total number of electrons in HNO3 is 1 + 5 + 18 = 24. Assume a skeleton structure in which the nitrogen atom is singly bonded to two oxygen atoms and doubly bonded to one oxygen. This requires two electrons for the O–H single bond and leaves eight electrons to be used for the N bonds.
H
O
N
O
O The formal charge of the nitrogen is 5 − 4 − 0 = +1. The formal charge of the hydrogen is 1 − 1 − 0 = 0. The formal charge of the oxygen bonded to the hydrogen is 6 − 2 − 4 = 0. The formal charge of the other singly bonded oxygen is 6 − 1 − 6 = −1. The formal charge of the doubly bonded oxygen is 6 − 2 − 4 = 0. 9.76. a.
The total number of electrons in ClNO is 7 + 5 + 6 = 18. Assume a skeleton structure in which one chlorine atom is singly bonded to the nitrogen atom. If a N–O single bond is assumed, twenty electrons are needed to fill the outer octets. Hence, a N=O double bond must be used, the structure being
Cl
N
O
The formal charge of the chlorine is 7 – 1 – 6 = 0. The formal charge of the nitrogen is 5 – 3 – 2 = 0. The formal charge of the oxygen is 6 – 2 – 4 = 0. b.
The total number of electrons in POCl3 is 5 + 6 + 21 = 32. Assume a skeleton structure in which the phosphorus atom is singly bonded to each of the three chlorines and doubly bonded to the oxygen. This structure has a lower formal charge than one with a single P–O bond.
Cl Cl
P
O
Cl The formal charge of phosphorus is 5 – 5 – 0 = 0. The formal charge on each chlorine is 7 – 1 – 6 = 0. The formal charge of oxygen is 6 – 2 – 4 = 0. c.
The total number of electrons in N2O is 2 x 5 + 6 = 16. Assume a skeleton structure in which the two nitrogen atoms are double bonded to each other. The oxygen atom can be singly bonded or doubly bonded to the central nitrogen. Using a N–O single bond gives the lowest formal charge.
N
N
O
The formal charge of the end nitrogen is 5 − 2 − 3 = 0. The formal charge of the nitrogen between the end nitrogen and oxygen is 5 − 4 = +1. The formal charge of the oxygen is 6 − 6 − 1 = −1.
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Chapter 9: Ionic and Covalent Bonding
9.77. a.
The total number of electrons in SOF2 is 6 + 6 + 14 = 26. Assume a skeleton structure in which the sulfur atom is singly bonded to the two fluorine atoms. If a S–O single bond is assumed, 26 electrons are needed. However, there would be formal charges of +1 on the sulfur and −1 on the oxygen. Using a S=O double bond requires only 26 electrons and results in zero formal charge.
F
S
F
O The formal charge on each of the two fluorine atoms is 7 − 1 − 6 = 0. The formal charge on the oxygen is 6 − 2 − 4 = 0. The formal charge on the sulfur is 6 − 4 − 2 = 0. b.
The total number of electrons in H2SO3 is 2 + 6 + 18 = 26. Assume a skeleton structure in which the sulfur atom is singly bonded to the three oxygen atoms. Then form single bonds from the two hydrogen atoms to each of two oxygen atoms. If a S–O single bond is assumed, 26 electrons are needed. However, there would be formal charges of +1 on the sulfur and −1 on the oxygen. A S=O double bond also requires 26 electrons but results in zero formal charge on all atoms.
H
O
S
O
H
O The formal charge of each hydrogen is 1 − 1 = 0. The formal charge of each oxygen bonded to a hydrogen is 6 − 2 − 4 = 0. The formal charge of the oxygen doubly bonded to the sulfur is 6 − 2 − 4 = 0. The formal charge of the sulfur is 6 − 4 − 2 = 0. c.
The total number of electrons in HClO2 is 1 + 7 + 12 = 20. Assume a skeleton structure in which the chlorine atom is singly bonded to the two oxygen atoms, and the hydrogen is singly bonded to one of the oxygen atoms. If all Cl–O single bonds are assumed, 20 electrons are needed, but the chlorine exhibits a formal charge of +1 and one oxygen exhibits a formal charge of −1. Hence, a pair of electrons on the oxygen without the hydrogen is used to form a Cl=O double bond.
H
O
Cl O
The formal charge of hydrogen is 1 − 1 = 0. The formal charge of the oxygen bonded to hydrogen is 6 − 2 − 4 = 0. The formal charge of the oxygen that is not bonded to the hydrogen is 6 − 2 − 4 = 0. The formal charge of the chlorine is 7 − 3 − 4 = 0.
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315
9.78. a.
The total number of electrons in ClO2F is 7 + 12 + 7 = 26. Assume a skeleton structure in which the chlorine atom is singly bonded to the two oxygen atoms and the fluorine atom. If all single bonds are assumed, the chlorine has a formal charge of +2, and each oxygen has a formal charge of −1. The general principle to be followed whenever two atoms on a bond have opposite charges is to move an electron pair on the atom with the negative charge into the bond and form a double bond. Doing this once gives the structure on the left below. In this structure, the chlorine now has a formal charge of +1, and the oxygen on the left has a formal charge of −1. Following the general principle again results in the formation of a second double bond, as shown in the structure on the right.
F Cl
O
F O
O
Cl
O
In the structure on the right, the formal charge of each oxygen is 6 − 2 − 4 = 0. The formal charge of the fluorine is 7 − 1 − 6 = 0. The formal charge of the chlorine is 7 − 5 − 2 = 0. b.
The total number of electrons in SO3 is 6 + 18 = 24. Assume a skeleton structure in which the sulfur atom is singly bonded to the three oxygen atoms. If all single bonds are assumed, the sulfur atom has a formal charge of +3, and each of the oxygen atoms has a formal charge of −1. As in part a, the general principle to be followed whenever two atoms on a bond have opposite charges is to move an electron pair on the atom with the negative charge into the bond and form a double bond. Doing this once gives the structure on the left below. In this structure, the sulfur now has a formal charge of +2, and the other oxygen atoms have a formal charge of −1. Following the general principle again two more times results in the formation of two more double bonds, as shown in the structure on the right.
O
O
O
S
S O
O
O
The formal charge of the oxygen atoms in the structure on the right is 6 − 2 − 4 = 0. The formal charge of the sulfur is 6 − 6 = 0. c.
The total number of electrons in BrO3−, including the charge, is 26. Assume a skeleton structure in which the bromine atom is singly bonded to the three oxygen atoms. If all single bonds are assumed, the bromine atom has a formal charge of +2, and each of the oxygen atoms has a formal charge of −1. As in part a, the general principle to be followed whenever two atoms on a bond have opposite charges is to move an electron pair on the atom with the negative charge into the bond and form a double bond. Doing this once gives the structure on the left below. In this structure, the bromine now has a formal charge of +1, and the two oxygen atoms have a formal charge of −1.
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Chapter 9: Ionic and Covalent Bonding
Following the general principle again results in the formation of a second double bond, as shown in the structure on the right.
O O
Br
O

O O
O
Br
In the righthand structure, the formal charge of the two doubly bonded oxygen atoms is 6 − 2 − 4 = 0. The formal charge of the singly bonded oxygen is 6 − 1 − 6 = −1. The formal charge of the bromine is 7 − 5 − 2 = 0. 9.79. rF = 64 pm rP = 110 pm dP–F = rF + rP = 64 pm + 110 pm = 174 pm 9.80. rB = 88 pm rF = 64 pm dB–F = rB + rF = 88 pm + 64 pm = 152 pm 9.81. a.
dC–H = rC + rH = 77 pm + 37 pm = 114 pm
b.
dS–Cl = rS + rCl = 104 pm + 99 pm = 203 pm
c.
dBr–Cl = rBr + rCl = 114 pm + 99 pm = 213 pm
d.
dSi–O = rSi + rO = 117 pm + 66 pm = 183 pm
9.82. dC–H = rC + rH = 77 pm + 37 pm = 114 pm
dexp = 107 pm
dC–Cl = rC + rCl = 77 pm + 99 pm = 176 pm
dexp = 177 pm
The calculated bond distances agree very well with the experimental values. 9.83. Methylamine Acetonitrile
147pm
Single C–N bond is longer.
116pm
Triple C–N bond is shorter.
9.84. Formaldehyde has a shorter carbonoxygen bond than methanol. In methanol, the carbonoxygen bond is a single bond, and in formaldehyde it is a double bond.
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H
H C
+
C
9.85. H
H
Br
H
H
H
H
C
C
H
H
317
Br
In the reaction, a C=C double bond is converted to a C–C single bond. An H–Br bond is broken, and one C–H bond and one C–Br bond are formed. ΔH ≅ BE(C=C) + BE(H–Br) − BE(C–C) − BE(C–H) − BE(C–Br)
= (602 + 362 − 346 − 411 − 285) kJ = −78 kJ
H
H C 9.86. H
+
C
H
OH
H
H
H
H
C
C
H
H
OH
In the reaction, a C=C bond and an O–H bond are converted to a C–C single bond, a C–O bond, and a C–H bond. ΔH ≅ [BE(C=C) + BE(O–H)] − [BE(C–C) + BE(C–O) + BE(C–H)]
= [(602 + 459) − (346 + 358 + 411)] kJ = −54 kJ
■
SOLUTIONS TO GENERAL PROBLEMS
9.87. a.
Strontium is a metal, and oxygen is a nonmetal. The binary compound is likely to be ionic. Strontium, in Group IIA, forms Sr2+ ions; oxygen, from Group VIA, forms O2− ions. The binary compound has the formula SrO and is named strontium oxide.
b.
Carbon and bromine are both nonmetals; hence, the binary compound is likely to be covalent. Carbon usually forms four bonds, and bromine usually forms one bond. The formula for the binary compound is CBr4. It is called carbon tetrabromide.
c.
Gallium is a metal, and fluorine is a nonmetal. The binary compound is likely to be ionic. Gallium is in Group IIIA and forms Ga3+ ions. Fluorine is in Group VIIA and forms F− ions. The binary compound is GaF3 and is named gallium(III) fluoride.
d.
Nitrogen and bromine are both nonmetals; hence, the binary compound is likely to be covalent. Nitrogen usually forms three bonds, and bromine usually forms one bond. The formula for the binary compound is NBr3. It is called nitrogen tribromide.
a.
Sodium is a metal, and sulfur is a nonmetal. The binary compound is likely to be ionic. Sodium, in Group IA, forms Na+ ions. Sulfur, in Group VIA, forms S2− ions. The binary compound is Na2S. It is named sodium sulfide.
9.88.
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Chapter 9: Ionic and Covalent Bonding
b.
Aluminum is a metal, and fluorine is a nonmetal. The binary compound is likely to be ionic. Aluminum, in Group IIIA, forms Al3+ ions. Fluorine, in Group VIIA, forms F− ions. The binary compound is AlF3. It is named aluminum fluoride.
c.
Calcium is a metal, and chlorine is a nonmetal. The binary compound is likely to be ionic. Calcium, in Group IIA, forms Ca2+ ions. Chlorine, in Group VIIA, forms Cl− ions. The binary compound is CaCl2. It is named calcium chloride.
d.
Silicon and bromine are both nonmetals; hence, the binary compound is likely to be covalent. Silicon usually forms four bonds, and bromine usually forms one bond. The formula for the binary compound is SiBr4. It is called silicon tetrabromide.
9.89. Total valence electrons = 5 + (4 x 6) + 3 = 32. The skeleton is
O O
As
O
O Distribute the remaining 24 electrons to complete the octets around the oxygen atoms. .. :O: .. :O ..
As
3.. O: ..
3
: O: or
.. : O ..
:O: ..
As
.. O: ..
: O: ..
The formula for lead(II) arsenate is Pb3(AsO4)2. 9.90. Total valence electrons = 6 + (3 x 6) + 2 = 26. The skeleton is
O O
Se
O
Distribute the remaining 20 electrons to complete the octets of oxygen and selenium atoms. .. :O: .. :O ..
Se ..
2
2:O: .. O: ..
or
.. :O ..
Se ..
.. O: ..
The formula for aluminum selenite is Al2(SeO3)3.
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319
9.91. Total valence electrons = 1 + 7 + (3 x 6) = 26. The skeleton is
H
O
I
O
O Distribute the remaining 18 electrons to satisfy the octet rule. The structure on the right has no formal charge.
H
O
I
O
H
O
I
or O
O
O
9.92. Total valence electrons = (2 x 1) + 6 + (4 x 6) = 32. The skeleton is
O H
O
Se
H
O
O Distribute the remaining 20 electrons to satisfy the octet rule. The structure on the right has no formal charge. O H
O
O
Se
H
O
O
or
H
O
Se
O
H
O
9.93. Total valence electrons = 5 + (2 x 1) + 1 = 8. The skeleton is H–N–H. Distribute the remaining 4 electrons to complete the octet of the nitrogen atom.
H
N

H
9.94. Total valence electrons = 3 + (4 x 1) + 1 = 8. The skeleton structure will use all the electrons. Because there are no electrons left, the Lewis structure is

H H
Al
H
H
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Chapter 9: Ionic and Covalent Bonding
9.95. Total valence electrons = 5 + (2 x 6) − 1 = 16. The skeleton is O–N–O. Distribute the remaining electrons on the oxygen atoms.
O
N
O
The nitrogen atom is short four electrons. Use two double bonds, one with each oxygen.
O
N
O
+
9.96. Total valence electrons = 5 + (4 x 7) − 1 = 32. The skeleton is
Br P
Br
Br
Br Distribute the 24 remaining electrons to complete the octets on the bromine atoms.
+
Br P
Br
Br
Br
9.97.
O
O a.
b.
Cl
Se
Se
C
or
Cl
Se
Cl
Se 
Cl Cl
Cl
Ga
Cl
Cl c.
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C
[
d.
[C
321
2
9.98.
O Br
a.
O Br
P
or
Br
H b.
Br
P
Br
Br
H
H
Si
Si
H
H
H
c.
[
F
I
d.
[
N
O
a.
Total valence electrons = 5 + (3 x 7) = 26. The skeleton is
[
F
+
+
[
9.99.
Cl
Sb
Cl
Cl Distribute the remaining 20 electrons to the chlorine atoms and the antimony atom to complete their octets.
Sb
Cl
Cl
Cl b.
Total valence electrons = 7 + 4 + 5 = 16. The skeleton is I–C–N. Distribute the remaining 12 electrons.
I
C
N
Notice the carbon atom is four electrons short of an octet. Make a triple bond between C and N from the four nonbonding electrons on the nitrogen.
I
C
N
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Chapter 9: Ionic and Covalent Bonding
c.
Total valence electrons = 7 + (3 x 7) = 28. The skeleton is
Cl
Cl
I Cl
Distribute 18 of the remaining 22 electrons to complete the octets of the chlorine atoms. The 4 remaining electrons form two sets of lone pairs on the iodine atom.
Cl
Cl
I Cl
d.
Total valence electrons = 7 + (5 x 7) = 42. The skeleton is
F
F I
F F
F
Use 30 of the remaining 32 electrons to complete the octets of the fluorine atoms. The 2 electrons remaining form a lone pair on the iodine atom:
F
F I F
F F
9.100. a.
Total valence electrons = 3 + (4 x 7) + 1 = 32. The skeleton is
Cl Cl
Al
Cl
Cl
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323
Distribute the remaining 24 electrons to complete the octets of the Cl atoms.

Cl Cl
Al
Cl
Cl b.
Total valence electrons = 3 + (6 x 7) + 3 = 48. The skeleton is
F
F Al
F
F F
F
Distribute the remaining 36 electrons to complete the octets of the F atoms. .. :F: .. :F ..
.. :F:
c.
.. F: ..
Al :F .. :
3
:F: ..
Total valence electrons = 7 + (3 x 7) = 28. The skeleton is
F
Br
F
F Distribute 18 of the remaining 22 electrons to complete the octets of the fluorine atoms. The 4 remaining electrons form two sets of lone pairs on the bromine atom.
F
Br
F
F d.
Total valence electrons = 7 + (6 x 7) − 1 = 48. The skeleton is
F F
F F
I F
F
Distribute the remaining 36 electrons to complete the octets of the fluorine atoms.
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Chapter 9: Ionic and Covalent Bonding
F F
+
F I
F F
F
9.101. a.
One possible electrondot structure is
Se
O
O
Because the seleniumoxygen bonds are expected to be equivalent, the structure must be described in resonance terms. Se
O
O
Se
O
O
Se
O
O
The electrons are delocalized over the selenium atom and the two oxygen atoms. b.
The possible electrondot structures are O
O N
N
N O
O
O
O
O
O
O N
N O
O
O N
N O
O
N O
O
At each end of the molecule, a pair of electrons is delocalized over the region of the nitrogen atom and the two oxygen atoms. 9.102.
H H a.
C H
H
O H
N O
C H
O N O
One pair of electrons is delocalized over the O–N–O bonds.
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b.
325
The possible electrondot structures are .. :O
.. :O: C
C
C
:O: ..
O: ..
.. :O:
.. O: C
: .. O
.. 2:O:
.. :O:
2
C O .. :
:O ..
.. :O
2
.. O: C
C : .. O:
:O: ..
2
C :O: ..
At each end of the molecule, a pair of electrons is delocalized over the region of the two carbonoxygen bonds. 9.103. The compound S2N2 will have a fourmembered ring structure. Calculate the total number of valence electrons of sulfur and nitrogen, which will be 6 + 6 + 5 + 5 = 22. Writing the skeleton of the fourmembered ring with single S–N bonds will use up eight electrons, leaving fourteen electrons for electron pairs. After writing an electron pair on each atom, there will be six electrons left for three electron pairs. No matter where these electrons are written, either a nitrogen or a sulfur will be left with less than eight electrons. This suggests one or more double bonds are needed. If only one S=N double bond is written, writing the remaining twelve electrons as six electron pairs will give the sulfur in the S=N double bond a formal charge of 1+ (see second line of formulas on next page). Writing two S=N double bonds using the same sulfur for both double bonds will give a formal charge of zero for all four atoms (see first line of formulas). The first two resonance formulas below have a zero formal charge on all atoms. However, one of the sulfur atoms does not obey the octet rule. S
N
S
N
N
S
N
S
S
N
S
N
+ S
N
S
N
N
S +
N
S
N
S
N
S +
+
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Chapter 9: Ionic and Covalent Bonding
In the four resonance formulas at the bottom of the previous page, all of the atoms obey the octet rule, but there is a positive formal charge on one sulfur atom and a negative formal charge on one nitrogen atom. 9.104. If you try to write an electrondot formula for the acetate ion, you will find you can write two formulas.
O
O
H
C
C
H
O
O
H
C
C
H
H
H
According to theory, the C=O double bond is delocalized; that is, a bonding pair of electrons is spread over the carbon atom and the two oxygen atoms, rather than localized between a carbon and one oxygen. This results in both bonds being of identical length. 9.105. The possible electrondot structures are
O
O N
O
N
O
O N
O
N O
O O
O
O
O
N O
O
O N
N O
O
O
O
N O
Because double bonds are shorter, the terminal N–O bonds are 118 pm, and the central N–O bonds are 136 pm.
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327
9.106. The best possible electrondot structure is
H H
C
O
N
O
H Because double bonds are shorter, the terminal N=O bond is 122 pm, and the inner N–O bond is 137 pm. 9.107. ΔH = BE(H–H) + BE(O=O) − 2BE(H–O) − BE(O–O) = (432 + 494 − 2 x 459 − 142) kJ = −134 kJ 9.108. ΔH = 2BE(H–H) + BE( N
N ) − BE(N–N) − 4BE(N–H)
= (2 x 432 + 942 − 167 − 4 x 386) kJ = +95 kJ 9.109. ΔH = BE(N=N) + BE(F–F) − BE(N–N) − 2BE(N–F) = (418 + 155 − 167 − 2 x 283) kJ = −1.60 x 102 kJ 9.110. ΔH = BE( C
N ) + 2BE(H–H) − BE(C–N) − 2BE(C–H) − 2BE(N–H)
= (887 + 2 x 432 − 305 − 2 x 411 − 2 x 386) kJ = −148 kJ 9.111. Ionic materials like NaCl are solids at room temperature, with high melting points. The molten liquid, which is clear, is very corrosive. Ionic materials consist of small, spherical ions that pack closely together. Thus, the ions interact strongly, giving a solid with a high melting point. Roomtemperature ionic liquids, which are also clear, consist of large, nonspherical cations with various anions. The large, bulky cation keeps the ions from packing closely, yielding a substance with weak interactions and a low melting point. 9.112. Organic solvents are volatile liquids that evaporate easily into the surrounding air, where they can contribute to air pollution. Organic solvents are often flammable. Ionic liquids are neither volatile nor flammable. Also, the proper choice of ionic liquid may improve the yield and lower the costs of a chemical process. 9.113. The decomposition of nitroglycerin is given by 4C3H5(ONO2)3(l) → 6N2(g) + 12CO2(g) + 10H2O(g) + O2(g) The stability of the products results from their strong bonds, which are much stronger than those in nitroglycerin. Nitrogen has a strong nitrogennitrogen triple bond, and carbon dioxide has two strong carbonoxygen double bonds.
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Chapter 9: Ionic and Covalent Bonding
9.114. Nobel discovered that nitroglycerin behaved better when absorbed on diatomaceous earth, a crumbly rock, giving an explosive mixture called dynamite. 9.115. A chemical bond acts like a stiff spring connecting nuclei. As a result, the nuclei in a molecule vibrate relative to each other. The vibration of molecules is revealed in their absorption of infrared radiation. The frequency of radiation absorbed equals the frequencies of nuclear vibrations. 9.116. The IR spectrum can act as a compound’s fingerprint. It can also yield structural information. Certain structural features of molecules appear as absorption peaks in definite regions of the infrared spectrum. Knowledge of where various bands absorb provide clues to the structure of the molecule.
■
SOLUTIONS TO STRATEGY PROBLEMS
9.117. Structure a is incorrect. Al3+ ion has zero valence electrons. The rest are valid Lewis structures. 9.118. Both structure a (Ar) and structure d (Cl−) are isoelectronic with the potassium ion, K+. The electron configuration is [Ne]3s23p6. 9.119. The ion is Ti2+ (Z = 22) and the atom is Ca (Z = 20). 9.120. a.
Cl−
b.
Ca+
c.
Ni2+
d.
Ge2+
e.
Br−
9.121. Both structure c (COCl2) and structure d (C2H4) possess a double bond. CN− has an atom with a nonzero formal charge. 9.122. Only structure b (CO2) contains only double bonds.
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329
9.123. The resonance structures for the azide ion are
N
N
N
N
N
N
N
N
N
In the structure on the left, the nitrogen on the right has a formal charge of 2− and the central nitrogen has a formal charge of 1+. In the center structure, the nitrogen atoms on each end have formal charges of 1−, and the central nitrogen has a formal charge of 1+. In the structure on the right, the nitrogen on the left has a formal charge of 2− and the central nitrogen has a formal charge of 1+. The center structure is the most likely because it has smaller formal charges. The resonance structures for the nitronium ion are
O
N
O
+
O
N
O
+
O
N
O
+
In the structure on the left, the oxygen on the right has a formal charge of 1−, the nitrogen has a formal charge of 1+, and the oxygen on the left has a formal charge of 1+. In the center structure, the oxygen atoms on each end have formal charges of zero, and the nitrogen has a formal charge of 1+. In the structure on the right, the oxygen on the left has a formal charge of 1−, the nitrogen has a formal charge of 1+, and the oxygen on the right has a formal charge of 1+. The center structure is the most likely because it has smaller formal charges. 9.124. If M is Fe, then the electron configuration of X is [Ar]3d104s24p4. If M is Co, the electron configuration of X is exactly the same as for iron. 9.125. The reaction is X2 + Y2 → 2XY. The enthalpy of the reaction is ΔH°rxn = 2ΔH°f[XY] − ΔH°f[X2] − ΔH°f[Y2] = 2ΔH°f[XY] − 0 − 0 = 2ΔH°f[XY] ≅ BE[X–X] + BE[Y–Y] − 2BE[X–Y]≅ BE[X–X] + BE[Y–Y] − 2BE[X–Y]
Combining the two expressions gives 2ΔH°f[XY] = BE[X–X] + BE[Y–Y] − 2BE[X–Y] BE[X–Y] =
BE[XX] + BE[YY]  2ΔH°f [XY] 414 kJ + 159 kJ  2(336 kJ) = 2 mol 2 mol
= 622.5 = 623 kJ/mol= 622.5 = 623 kJ/mol
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Chapter 9: Ionic and Covalent Bonding
9.126. The resonance formulas for nitrous oxide are
O
N
N
O
N
O
N
N
N
In the structure on the left, the oxygen has a formal charge of 1+, the nitrogen in the center has a formal charge of 1+, and the nitrogen on the left has a formal charge of 2−. In the center structure, the oxygen has a formal charge of zero, the nitrogen in the center has a formal charge of 1+, and the nitrogen on the right has a formal charge of 1−. In the structure on the right, the oxygen has a formal charge of 1−, the nitrogen in the center has a formal charge of 1+, and the nitrogen on the right has a formal charge of zero. The structure on the right is the most likely because it has smaller formal charges, and the negative formal charge is on the oxygen atom. The structure in the middle also has smaller formal charges, but here, nitrogen has the negative formal charge, which is less favorable than to have it on the oxygen atom. The structure on the left is the least likely structure. Thus, the NN bond length should be between the triplebond length, which is 109 pm, and the doublebond length, which is 122 pm. Since the most likely structure places a triplebond between the nitrogen atoms, the value should be closer to the triple bond value of 109 pm.
■
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
Note on significant figures: The final answer to each cumulativeskills problem is given first with one nonsignificant figure (the rightmost significant figure is underlined) and then is rounded to the correct number of figures. Intermediate answers usually also have at least one nonsignificant figure. Atomic weights, except for that of hydrogen, are rounded to two decimal places. 9.127. The electronegativity differences and bond polarities are P–H
0.0
nonpolar
O–H
1.4
polar (acidic)
H3PO3(aq) + 2NaOH(aq) → Na2HPO3(aq) + 2H2O(l) 1 mol H 3 PO3 0.1250 mol NaOH 1 x 0.02250 L x x 2 mol NaOH 1L 0.2000 L H 3 PO3
= 0.0070312 = 0.007031 M H3PO3 9.128. The electronegativity differences and bond polarities are P–H
0.0
nonpolar
O–H
1.4
polar (acidic)
H3PO2(aq) + NaOH(aq) → NaH2PO2(aq) + H2O(l) 1 mol H 3 PO 2 0.1250 mol NaOH 1 x 0.02250 L x x 1 mol NaOH 1L 0.2000 L H 3 PO 2
= 0.014062 = 0.01406 M H3PO2
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9.129. After assuming a 100.0g sample, convert to moles: 10.9 g Mg x
1 mol Mg = 0.4485 mol Mg 24.3 g Mg
31.8 g Cl x
1 mol Cl = 0.89696 mol Cl 35.453 g Cl 1 mol O = 3.581 mol O 16.00 g O
57.3 g O x Divide by 0.4485: Mg:
0.4485 0.89696 3.581 = 1; Cl: = 2.00; O: = 7.98 0.4485 0.4485 0.4485
The simplest formula is MgCl2O8. However, since Cl2O82− is not a wellknown ion, write the simplest formula as Mg(ClO4)2, magnesium perchlorate. The Lewis formulas are Mg2+ and
O
Cl


O
O O
or
O
O
Cl
O
O
9.130. After assuming a 100.0g sample, convert to moles: 1 mol Ca = 0.7559 mol Ca 40.08 g Ca
30.3 g Ca x 21.2 g N x
1 mol N = 1.513 mol N 14.01 g N
48.5 g O x
1 mol O = 3.031 mol O 16.00 g O
Divide by 0.7559: Ca:
0.7559 1.513 3.031 = 1; N: = 2.00; O: = 4.01 0.7559 0.7559 0.7559
The simplest formula is CaN2O4. However, since N2O42− is not a wellknown ion, write the simplest formula as Ca(NO2)2, calcium nitrite. The Lewis formulas are Ca2+ and
N
O
[
[
O

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Chapter 9: Ionic and Covalent Bonding
9.131. After assuming a 100.0g sample, convert to moles: 1 mol C = 2.081 mol C 12.01 g C
25.0 g C x 2.1 g H x
1 mol H = 2.08 mol H 1.008 g H
39.6 g F x
1 mol F = 2.085 mol F 18.99 g F
33.3 g O x
1 mol O = 2.081 mol O 16.00 g O
The simplest formula is CHOF. Because the molecular mass of 48.0 divided by the formula mass of 48.0 is one, the molecular formula is also CHOF. The Lewis formula is
H C
O
F 9.132. After assuming a 100.0g sample, convert to moles: 14.5 g C x 85.5 g Cl x
1 mol C = 1.207 mol C 12.01 g C 1 mol Cl = 2.4116 mol Cl 35.453 g Cl
Divide by 1.207:Divide by 1.207: C:
1.207 2.4116 = 1; Cl: = 1.998 1.207 1.207
The simplest formula is CCl2. Dividing the molecular mass by the formula mass gives 166 amu ÷ 82.92 amu = 2.0. The molecular formula is C2Cl4. The Lewis formula is
Cl
Cl C
Cl
C Cl
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333
9.133. First, calculate the number of moles in one liter: n =
PV RT
MW =
=
1.00 atm x 1.00 L = 0.028727 mol 0.0821 L • atm/ (K • mol) x 424 K
g 7.49 g = = 260.7 g/mol mol 0.028727 mol
MW = 260.7 amu = 118.69 amu Sn + n x (35.453 amu Cl) n =
260.7  118.69 = 4.0055 35.453
The formula is SnCl4. It is molecular because the electronegativity difference is 1.2, and because it is a liquid, and is volatile at 151°C. The Lewis formula is
Cl Cl
Sn
Cl
Cl 9.134. First, calculate the number of moles present: n =
PV RT
MW =
=
(765/ 760) atm x 0.0142 L = 5.8816 x 10−4 mol 0.0821 L • atm/ (K • mol) x 296 K
g 0.100 g = = 170.02 g/mol mol 5.8816 x 104 mol
MW = 170.02 amu = 74.92 amu As + n x (18.998 amu F) n =
170.02  74.92 = 5.0057 18.998
The formula is AsF5. The Lewis formula is
F
F
F
As F
F
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Chapter 9: Ionic and Covalent Bonding
9.135.
HCN(g)
→
(135.1
H(g) +
C(g) + N(g)
218.0
716.7
472.7)
ΔHf(kJ/mol)
ΔHrxn = 1272.3 kJ/mol N ) = ΔHrxn − BE(C–H) = [1272.3 − 411] kJ/mol
BE( C
= 861.3 kJ/mol (Table 9.5 has 887 kJ/mol.) 9.136.
H C
O
4 H (g) + 2 C (g)
+
O (g)
H3C (166.1
4 × 218
249.2)
2 × 716.7
ΔHf(kJ/mol)
ΔHrxn = 2720.7 kJ/mol BE(C=O) = ΔHrxn − [4 x BE(C–H)] − BE(C–C) = [2720.7 − (4 x 411) − 346] kJ/mol = 730.7 kJ/mol (Table 9.5 has 745 kJ/mol.) 9.137. Use the O–H bond and its bond energy of 459 kJ/mol to calculate XO. BE(O–H) = 1/2[BE(H−) + BE(O−)] + k(XO − XH)2 459 kJ/mol = 1/2(432 kJ/mol + 142 kJ/mol) + 98.6 kJ (XO − XH)2 Collecting the terms gives 459  287 = (XO − XH)2 98.6
Taking the square root of both sides gives 1.3207 = 1.32 = (XO − XH) Because XH = 2.1, XO = 2.1 + 1.32 = 3.42 = 3.4. 9.138. BE(N–I) = 1/2[BE(N–N) + BE(I–I)] + k x (XN − XI)2 BE(N–I) = 1/2(167 + 149) + (98.6)(3.0 − 2.5)2 BE(N–I) = 158 + 24.65 = 182.65 = 183 kJ/mol
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9.139. X =
I.E.  E.A. [1250  (349)] kJ/mol = = 799.5 kJ/mol 2 2 799.5 kJ/mol = 3.47 (Pauling’s X = 3.0) 230 kJ/mol
9.140. X =
I.E.  E.A. [1314  (141)] kJ/mol = = 727.5 kJ/mol 2 2 727.5 kJ/mol = 3.16 (Pauling’s X = 3.5) 230 kJ/mol
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CHAPTER 10
Molecular Geometry and Chemical Bonding Theory
■
SOLUTIONS TO EXERCISES
10.1. a.
A Lewis structure of ClO3− is
O O
Cl
O
There are four electron pairs in a tetrahedral arrangement about the central atom. Three pairs are bonding, and one pair is nonbonding. The expected geometry is trigonal pyramidal. b.
The Lewis structure of OF2 is
F
O
F
There are four electron pairs in a tetrahedral arrangement about the central atom. Two pairs are bonding, and two pairs are nonbonding. The expected geometry is bent. c.
The Lewis structure of SiF4 is
F
F Si F
F
There are four bonding electron pairs in a tetrahedral arrangement around the central atom. The expected geometry is tetrahedral.
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10.2. First, distribute the valence electrons to the bonds and the chlorine atoms. Then distribute the remaining electrons to iodine.
Cl
I
Cl
Cl The five electron pairs around iodine should have a trigonal bipyramidal arrangement with two lone pairs occupying equatorial positions. The molecule is Tshaped. 10.3. Both trigonal pyramidal (b) and Tshaped (c) geometries are consistent with a nonzero dipole moment. In trigonal planar geometry, the Br–F contributions to the dipole moment would cancel. 10.4. On the basis of symmetry, SiF4 (b) would be expected to have a dipole moment of zero. The bonds are all symmetrical about the central atom. 10.5. The Lewis structure for ammonia, NH3, is
H
N
H
H There are four pairs of electrons around the nitrogen atom. According to the VSEPR model, these are arranged tetrahedrally around the nitrogen atom, and you should use sp3 hybrid orbitals. Each N–H bond is formed by the overlap of a 1s orbital of a hydrogen atom with one of the singly occupied sp3 hybrid orbitals of the nitrogen atom. This gives the following bonding description for NH3: 2p 3
sp
3
sp
Energy
2s
1s
1s
N atom
N atom
(ground state)
(hybridized)
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N  H bonds lone pair
1s N atom (in NH3)
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Chapter 10: Molecular Geometry and Chemical Bonding Theory
10.6. The Lewis structure for PCl5 is
Cl
Cl P Cl
Cl Cl
The phosphorus atom has five single bonds and no lone pairs around it. This suggests that you use sp3d hybrid orbitals on phosphorus. Each chlorine atom (valenceshell configuration 3s23p5) has one singly occupied 3p orbital. The P–Cl bonds are formed by the overlap of a phosphorus sp3d hybrid orbital with a singly occupied chlorine 3p orbital. The hybridization of phosphorus can be represented as follows:
3d
3d
3p
Energy
sp3d
3s P atom
P atom
(ground state)
(hybridized)
The bonding description of phosphorus in PCl5 is
3d
Energy
P — Cl bonds
sp3d P atom (in PCl5 )
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10.7. The Lewis structure of CO2 is
O
C
O
The double bonds are each a π bond plus a σ bond. Hybrid orbitals are needed to describe the two σ bonds. This suggests sp hybridization. Each sp hybrid orbital on the carbon atom overlaps with a 2p orbital on one of the oxygen atoms to form a σ bond. The π bonds are formed by the overlap of a 2p orbital on the carbon atom with a 2p orbital on one of the oxygen atoms. Hybridization and bonding of the carbon atom are shown as follows:
C – O π bonds 2p 2p
2p
sp
sp
2s
Energy
σ bonds
1s
1s C atom (ground state)
1s
C atom (hybridized)
C atom (in CO2 )
10.8. The structural formulas for the isomers are as follows:
F N
N
N
F
F cis
N
F trans
These compounds exist as separate isomers with different properties. For these to interconvert, one end of the molecule would have to rotate with respect to the other end. This would require breaking the π bond and expending considerable energy.
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Chapter 10: Molecular Geometry and Chemical Bonding Theory
10.9. There are 2 x 6 = 12 electrons in C2. They occupy the orbitals as shown below.
σ
1s
σ1s *
σ 2s
π 2p
σ * 2s
The electron configuration is KK(σ2s)2(σ2s*)2(π2p)4 . There are no unpaired electrons; therefore, C2 is diamagnetic. There are eight bonding and four antibonding electrons. The bond order is 1/2 (8 − 4) = 2. 10.10. There are 6 + 8 = 14 electrons in CO. The orbital diagram is
σ
1s
σ1s *
σ2s
σ2s *
π 2p
σ 2p
The electron configuration is KK(σ2s)2(σ2s*)2(π2p)4(σ2p)2. There are ten bonding and four antibonding electrons. The bond order is 1/2(10 − 4) = 3. There are no unpaired electrons; hence, CO is diamagnetic.
■
ANSWERS TO CONCEPT CHECKS
10.1. The VSEPR model predicts that four electron pairs about any atom in a molecule will distribute themselves to give a tetrahedral arrangement. Any three of these electron pairs would have a trigonal pyramidal arrangement. The geometry of a molecule having a central atom with three atoms bonded to it would be trigonal pyramidal. 10.2. A molecule, AX3, could have one of three geometries: it could be trigonal planar, trigonal pyramidal, or Tshaped. Assuming the three groups attached to the central atom are alike, as indicated by the formula, the planar geometry should be symmetrical, so even if the A–X bonds are polar, their polarities would cancel to give a nonpolar molecule (dipole moment of zero). This would not be the case in the trigonal pyramidal geometry. In that situation, the bonds all point to one side of the molecule. It is possible for such a molecule to have a lone pair that points away from the bonds and whose polarity might fortuitously cancel the bond polarities, but an exact cancellation is not likely. In general, you should expect the trigonal pyramidal molecule to have a nonzero dipole moment, but a zero dipole is possible. The argument for the Tshaped geometry is similar to that for the trigonal pyramidal geometry. The bonds point in a plane, but toward one side of the molecule. Unless the sum of the bond polarities was fortuitously canceled by the polarities from the lone pairs, this geometry would have a nonzero dipole moment. This means that molecule Y is likely to be trigonal planar, but trigonal pyramidal or Tshaped geometries are possible. Molecule Z cannot have a trigonal planar geometry; it must be either trigonal pyramidal or Tshaped.
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10.3. Assuming there are no lone pairs, the atom has four electron pairs and, therefore, an octet of electrons about it. The single bond and the triple bond each require a sigma bond orbital for a total of two such orbitals. This suggests sp hybrids on the central atom.
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ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
10.1. The VSEPR model is used to predict the geometry of molecules. The electron pairs around an atom are assumed to arrange themselves to reduce electron repulsion. The molecular geometry is determined by the positions of the bonding electron pairs. 10.2. The arrangements are linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. 10.3. A lone pair is "larger" than a bonding pair; therefore, it will occupy an equatorial position, where it encounters less repulsion than if it were in an axial position. 10.4. The bonds could be polar, but if they are arranged symmetrically, the molecule will be nonpolar. The bond dipoles will cancel. 10.5. Nitrogen trifluoride has three N–F bonds arranged to form a trigonal pyramid. These bonds are polar and would give a polar molecule with partial negative charges on the fluorine atoms and a partial positive charge on the nitrogen atom. However, there is also a lone pair of electrons on nitrogen that is directed away from the bonds. The result is that the lone pair nearly cancels the polarity of the bonds and gives a molecule with a very small dipole moment. 10.6. Certain orbitals, such as p orbitals and hybrid orbitals, have lobes in given directions. Bonding to these orbitals is directional; that is, the bonding is in preferred directions. This explains why the bonding gives a particular molecular geometry. 10.7. The angle is 109.5°. 10.8. A sigma bond has a cylindrical shape about the bond axis. A pi bond has a distribution of electrons above and below the bond axis.
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10.9. In ethylene, C2H4, the changes on a given carbon atom may be described as follows:
π bonds
2p
2p
2p
sp2
sp2 2s
Energy
σ bonds
1s
1s C atom (ground state)
1s
C atom (hybridized)
C atom (in C2H4)
An sp2 hybrid orbital on one carbon atom overlaps a similar hybrid orbital on the other carbon atom to form a σ bond. The remaining hybrid orbitals on the two carbon atoms overlap 1s orbitals from the hydrogen atoms to form four C–H bonds. The unhybridized 2p orbital on one carbon atom overlaps the unhybridized 2p orbital on the other carbon atom to form a π bond. The σ and π bonds together constitute a double bond. 10.10. Both of the unhybridized 2p orbitals, one from each carbon atom, are perpendicular to their CH2 planes. When these orbitals overlap each other, they fix both planes in the same plane. The two ends of the molecule cannot twist around without breaking the π bond, which requires considerable energy. Therefore, it is possible to have stable molecules with the following structures:
X
X
H
H cis
H
X
X trans
H
Because these have the same molecular formulas, they are isomers. In this case, they are called cistrans isomers, or geometrical isomers.
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10.11. In a bonding orbital, the probability of finding electrons between the two nuclei is high. For this reason, the energy of the bonding orbital is lower than that of the separate atomic orbitals. In an antibonding orbital, the probability of finding electrons between the two nuclei is low. For this reason, the energy of the antibonding orbital is higher than that of the separate atomic orbitals. 10.12. The factors determining the strength of interaction of two atomic orbitals are (1) the energy difference between the interacting orbitals and (2) the magnitude of their overlap. 10.13. When two 2s orbitals overlap, they interact to form a bonding orbital, σ2s, and an antibonding orbital, σ2s*. The bonding orbital is at lower energy than the antibonding orbital. 10.14. When two 2p orbitals overlap along their axes, they interact to form one σ2p bonding orbital and one σ2p* antibonding orbital. When they overlap sideways, they form π2p and π2p* molecular orbitals. 10.15. A σ bonding orbital is formed by the overlap of the 1s orbital on the H atom with the 2p orbital on the F atom. This H–F orbital is made up primarily of the fluorine orbital. 10.16. The O3 molecule consists of a framework of localized orbitals and of delocalized pi molecular orbitals. The localized framework is formed from sp2 hybrid orbitals on each atom. Thus, an O–O bond is formed by the overlap of a hybrid orbital on the left O atom with a hybrid orbital on the center O atom. Another O–O bond is formed by the overlap of another hybrid orbital on the right O atom with a hybrid orbital on the center O atom. The remaining hybrid orbitals are occupied by lone pairs of electrons. Also, there is one unhybridized p orbital on each of the atoms. These p orbitals are perpendicular to the plane of the molecule and overlap sidewise to give three pi molecular orbitals that are delocalized. The two orbitals of lowest energy are occupied by pairs of electrons. 10.17. The answer is e, a little less than 109.5°. 10.18. The answer is d, square pyramidal. 10.19. The answer is b, H2S. 10.20. The answer is e, 5/2.
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ANSWERS TO CONCEPT EXPLORATIONS
10.21. a.
The Lewis structure does not obey the octet rule. There are ten electrons around the carbon atom. The carbon has a formal charge of 1−, and the singlebonded oxygen has a formal charge of 1−. It is difficult to describe the bonding in VB terms, since carbon cannot hybridize to form three σ bonds and two π bonds. Three σ bonds requires sp2 hybridization, leaving only one unhybridized 2p orbital. This is available to form only one π bond with the oxygen atoms, not two as required.
b.
The Lewis structure given is not the best. The following structure is a better one. 2
:O: .. ..O..
C
.. ..O..
This structure is better because all atoms obey the octet rule. Also, the formal charges are 1− on both the singlebonded oxygen atoms. c.
Yes, the description is incorrect. The left formula predicts a bent geometry, with angles around 120°. The right formula predicts a linear geometry, with angles around 180°.
d.
Both formulas are incorrect. Neither one is better than the other. They represent resonance structures of each other. The left structure will have a larger dipole moment than that of the right structure.
e.
A Lewis formula that gives a better description of CO2 is given below.
O
C
O
All atoms obey the octet rule, and there are no formal charges. 10.22. a.
The second structure from the left is the best description of SeO2. There is a double bond between the selenium atom and one of the oxygen atoms, sharing four electrons. The other oxygen atom is singly bonded to the selenium atom, sharing two electrons. There is also a lone pair on selenium, completing its octet.
b.
The selenium atom is sp2 hybridized, forming two σ bonds by overlap with the oxygen atom’s 2p orbitals. The third sp2 hybrid orbital holds the lone pair of electrons. The unhybridized 2p orbital overlaps with a 2p orbital on the oxygen atom, forming a double bond.
c.
The arrangement of electron pairs around the selenium atom is trigonal planar. The molecule has a bent geometry.
d.
The O–Se–O angle is slightly less than 120°. The lone pair of electrons are bulky, so the angle will be decreased.
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e.
The molecular geometry of H2Se is bent, the same as water. The selenium atom is sp3 hybridized.
H
■
345
Se
H
f.
The H–Se–H angle is less than the tetrahedral (sp3) angle of 109°, whereas the O–Se–O angle is less than the trigonal planar (sp2) angle of 120°. Therefore, the H–Se–H angle is smaller.
g.
Both molecules have dipole moments because they lack the proper molecular symmetry for the polar bonds to cancel each other.
ANSWERS TO CONCEPTUAL PROBLEMS
10.23. In order to solve this problem, draw the Lewis structure for each of the listed molecules. In each case, use your Lewis structure to determine the geometry, and match this geometry with the correct model. a.
SeO2 is angular and has an AX2 geometry. This is represented by model (ii).
b.
BeCl2 is linear and has an AX2 geometry. This goes with model (i).
c.
PBr3 is trigonal pyramidal and has an AX3 geometry. This goes with model (iv).
d.
BCl3 is trigonal planar and has an AX3 geometry. This is represented by model (iii).
10.24. First, determine the geometry about the central atom, carbon. There is a total of 16 valence electrons for the molecule. For each atom to obey the octet rule, eight electrons must be shared in bonds. This would require four bonding pairs, a single bond (one pair) between C and Cl, and a triple bond (three pairs) between C and N. The geometry about the central atom is AX2, which is linear. This is represented by model (a), on the left. 10.25. In a CH3CH3 molecule, each C atom has four electron pairs arranged tetrahedrally. Within this molecule, each CH3 considered as a separate group has a trigonal pyramidal geometry (with three C–H bonding pairs and a fourth pair from the C–C bond around the C atom). The :CH3 molecule retains this trigonal pyramidal geometry, having three bonding pairs and one lone pair around the C atom. The CH3 molecule, however, has only three electron pairs around the C atom. Initially, as the CH3 molecule breaks away from the ethane molecule, it has the trigonal pyramidal geometry it had in the ethane molecule. However, the repulsions of the bonding electron pairs on the CH3 molecule are no longer balanced by the fourth pair (from the C–C bond), so the molecule flattens out to form a trigonal planar geometry.
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10.26. The BF3 molecule starts out in a trigonal planar geometry with three bonding pairs of electrons distributed about the B atom in a plane and 120° apart. After the bond between B and N is formed, the geometry about the B atom is tetrahedral, with 109° bond angles. As the bond begins to form, the lone pair on N approaches the plane of the BF3 molecule and begins to interact with the bonding pairs. The repulsion between the electron pairs causes the bonding pairs to be pushed downward, forcing them closer together. 10.27. First, determine the geometry of each molecule using VSEPR theory. Then compare to the orbital pictures for the correct number of bonding and nonbonding orbitals. a.
BeF2 is an AX2 molecule, which has linear geometry. This is depicted by orbital drawing (i).
b.
SiF4 is an AX4 molecule, which has tetrahedral geometry. This is depicted by orbital drawing (iii).
c.
SeF4 is an AX4 molecule, which has seesaw geometry. This is depicted by orbital drawing (iv).
d.
RnF4 is an AX4 molecule, which has square planar geometry. This is depicted by orbital drawing (v).
10.28. The arrangement of electron pairs about this atom suggested by the two bonds and one lone pair is trigonal planar. You would expect sp2 hybrid orbitals for this atom (a total of three hybrid orbitals). Two of these hybrid orbitals would be used to form the two bonds; the third hybrid orbital would be used for the lone pair. 10.29. The reaction with Br2 indicates that C2H2Br2 has a double bond. There are three possible isomers of C2H2Br2 having double bonds:
Br
H C
Br
C
H
C Br
C
H C
H
H
II
I
Br
Br
Br
C III
H
Compounds II and III have dipole moments. The addition of Br2, with one Br going to each C atom, yields the following products:
Br
Br
Br
C
C
H
H
Br
Br
Br
Br
C
C
H
H
Br
Br
Br
H
C
C
Br
H
Br
IA
IIIA IIA Products IA and IIA are identical and arise from compounds I and II. Thus, the original two compounds, one not having a dipole moment, the other having a dipole moment, but both reacting with bromine to give the same product, are compounds I and II, respectively.
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10.30. A neutral molecule of the form XF4 could have four, five, or six electron pairs around X. With four bonding pairs and no lone pairs, the geometry would be tetrahedral. The molecule would be symmetrical and nonpolar. Similarly, with four bonding pairs and two lone pairs, the geometry would be square planar; the molecule would again be symmetrical and nonpolar. However, with four bonding pairs and one lone pair, the geometry would be seesaw; the molecule would be nonsymmetrical in shape and could be polar. This fits the description of a compound with dipole moment 0.63 D. To identify X, let’s look at the total number of valence electrons in XF4. Each F atom, with its bonding electrons, has an octet (8) of electrons. In addition, there is a lone pair, or two electrons. Thus, the total number of valence electrons is 4 x 8 + 2 = 34. Of these, 4 x 7 = 28 are from the F atoms, leaving 34 − 28 = 6 valence electrons coming from X. So X is a Group VIA element. X cannot be oxygen because a seesaw geometry would require sp3d hybrid orbitals, and oxygen does not have d orbitals to hybridize. Thus, X must be sulfur (S), selenium (Se), or tellurium (Te). 10.31. First, determine the geometry about atoms a, b, and c using VSEPR theory. Then compare to the geometry in each drawing. a.
The geometry about atom a (carbon) is AX4, which is tetrahedral geometry. This is depicted by drawing (ii).
b.
The geometry about atom b (carbon) is AX3, which is trigonal planar geometry. This is depicted by drawing (iv).
c.
The geometry about atom c (oxygen) is AX2, which is bent geometry. This is depicted by drawing (i).
10.32. The formate ion, HCO2−, is expected to have trigonal planar geometry by the VSEPR model. (Each resonance formula has one C=O bond, one C–O bond, and one C–H bond, giving a total of three groups about the C atom.) The VSEPR model predicts bond angles of 120°. However, a bond between the carbon atom and the oxygen atom has a bond order of 3/2 (resonance between a single and a double bond) and requires more room than a pure single bond. The repulsion between the two C–O bonds would be greater than the repulsion between a C–O bond and the C–H bond. Thus, you would predict that the O–C–O angle is slightly greater than 120°, whereas the O–C–H angle is slightly less than 120°.
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Chapter 10: Molecular Geometry and Chemical Bonding Theory
SOLUTIONS TO PRACTICE PROBLEMS
10.33. The number of electron pairs, number of lone pairs, and geometry are for the central atom. Number of Electron Pairs
ElectronDot Structure
Number of Lone Pairs
Geometry
F F
Si
F
4
0
Tetrahedral
4
2
Bent
3
0
Trigonal planar
4
1
Trigonal pyramidal
F S F
F
O C F
F
Cl Cl
P
Cl
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10.34. The number of electron pairs, number of lone pairs, and geometry are for the central atom. Number of Electron Pairs
ElectronDot Structure
Number of Lone Pairs
Geometry
Ge 3
1
Bent (angular)
4
1
Trigonal pyramidal
3
0
Bent (angular)
4
0
Tetrahedral
Cl
Cl F F
F
N S Cl
Cl O O
Xe
O
O
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349
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Chapter 10: Molecular Geometry and Chemical Bonding Theory
10.35. The number of electron pairs, number of lone pairs, and geometry are for the central atom. Number of Electron Pairs
ElectronDot Structure
Cl
P
4
1
4
0
Tetrahedral
2
0
Linear
4
1
Trigonal pyramidal
O 3
O O
Geometry

O O
Number of Lone Pairs
O
O
C
N
[
[S
H H
O

+ H
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351
10.36. The number of electron pairs, number of lone pairs, and geometry are for the central atom. Number of Electron Pairs
ElectronDot Structure .. N ..
N
Number of Lone Pairs
Geometry
.. N ..
2
0
Linear
H
4
0
Tetrahedral
4
1
Trigonal pyramidal
3
1
Bent (angular)
H H
B H .. :O:
.. :O ..
S ..
2
.. O: ..
.. N :O:
:O: ..
10.37. a.
CCl4 is tetrahedral, VSEPR predicts that the bond angles will be 109°, and you would expect them to be this angle.
b.
SCl2 is bent, and VSEPR theory predicts bond angles of 109°, but you would expect the Cl–S–Cl bond angle to be less.
c.
COCl2 is trigonal planar. VSEPR theory predicts 120° bond angles. The C–O bond is a double bond, so you would expect the Cl–C–O bond to be more than 120° and the Cl–C–Cl bond to be less than 120°.
d.
AsH3 is trigonal pyramidal, and VSEPR theory predicts bond angles of 109°, but you would expect the H–As–H bond angle to be less.
10.38. a.
NCl3 is trigonal pyramidal, and VSEPR theory predicts bond angles of 109°, but you would expect the Cl–N–Cl bond angle to be less.
b.
Both carbon atoms are trigonal planar. VSEPR theory predicts 120° bond angles. The C–C bond is a double bond, so you would expect the F–C–C bond to be more than 120° and the F–C–F bond to be less than 120°.
c.
SiF4 is tetrahedral, VSEPR predicts that the bond angles will be 109°, and you would expect them to be this angle.
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d.
OF2 is angular, and VSEPR theory predicts bond angles of 109°, but you would expect the F–O–F bond angle to be less.
10.39. The number of electron pairs, number of lone pairs, and geometry are for the central atom. ElectronDot Structure
Number of Electron Pairs
Number of Lone Pairs
Geometry
F F F
5
0
Trigonal bipyramidal
F
5
2
Tshaped
F
6
1
Square pyramidal
5
1
Seesaw
P F F
Br
F
F F F Br F
F Cl Cl S Cl Cl
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10.40. The number of electron pairs, number of lone pairs, and geometry are for the central atom. ElectronDot Structure
Number of Electron Pairs
Number of Lone Pairs
Geometry
F F
F
6
1
Square pyramidal
5
0
Trigonal bipyramidal
5
1
Seesaw
6
0
Octahedral
Cl F
F F F Sb
F
F F
F F Se F F F F
F Te F
F F
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10.41. The number of electron pairs, number of lone pairs, and geometry are for the central atom. ElectronDot Structure
Number of Electron Pairs
Number of Lone Pairs
Geometry
Cl Cl Cl
5
0
Trigonal bipyramidal
6
0
Octahedral
5
3
Linear
6
2
Square planar
Sn Cl Cl
F F
F P F
F F
.. :F ..
.. . . Cl ..
.. F: ..
F
F

I F
F
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10.42. The number of electron pairs, number of lone pairs, and geometry are for the central atom. ElectronDot Structure .. . . I ..
.. : Cl ..
Number of Electron Pairs .. Cl : ..
F
F Br F
Number of Lone Pairs
Geometry
5
3
Linear
6
2
Square planar
6
0
Octahedral
5
1
Seesaw
F +
F F
F Te F
F F
+
F F Cl
F
F 10.43. a.
Trigonal pyramidal and Tshaped. Trigonal planar would have a dipole moment of zero.
b.
Bent. Linear would have a dipole moment of zero.
10.44. a.
Trigonal pyramidal and Tshaped. Trigonal planar would have a dipole moment of zero.
b.
Seesaw. Square planar and tetrahedral would have dipole moments of zero.
10.45. a.
CS2 has a linear geometry and has no dipole moment.
b.
TeF2 has a bent geometry and has a dipole moment.
c.
SeCl4 has a seesaw geometry and has a dipole moment.
d.
XeF4 has a square planar geometry and has no dipole moment.
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10.46. a.
SeF2 has a bent, or angular, geometry and has a dipole moment.
b.
BeI2 has a linear geometry and has no dipole moment.
c.
TeF6 has an octahedral geometry and has no dipole moment.
d.
TeF4 has a seesaw geometry and has a dipole moment.
10.47. a.
SF2 is an angular model. The sulfur atom is sp3 hybridized.
b.
ClO3− is a trigonal pyramidal molecule. The chlorine atom is sp3 hybridized.
10.48. a.
AlCl3 is a trigonal planar molecule. The aluminum atom is sp2 hybridized.
b.
BF4− is a tetrahedral ion. The boron atom is sp3 hybridized.
10.49. a.
SeCl2 is a bent molecule. The Se atom is sp3 hybridized.
b.
NO2− is a bent ion. The N atom is sp2 hybridized.
c.
CO2 is a linear molecule. The C atom is sp hybridized.
d.
COF2 is a trigonal planar molecule. The C atom is sp2 hybridized.
10.50. a.
GeCl4 is a tetrahedral molecule. The Ge atom is sp3 hybridized.
b.
PBr3 is a trigonal pyramidal molecule. The P atom is sp3 hybridized.
c.
BeF2 is a linear molecule. The Be atom is sp hybridized.
d.
SO2 is a bent molecule. The S atom is sp2 hybridized.
10.51. a.
The Lewis structure is
Cl
Hg
Cl
The presence of two single bonds and no lone pairs suggests sp hybridization. Thus, an Hg atom with the configuration [Xe]4f145d106s2 is promoted to [Xe]4f145d106s16p1, then hybridized. An Hg–Cl bond is formed by overlapping an Hg hybrid orbital with a 3p orbital of Cl.
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b.
357
The Lewis structure is
Cl Cl
P
Cl
The presence of three single bonds and one lone pair suggests sp3 hybridization of the P atom. Three hybrid orbitals each overlap a 3p orbital of a Cl atom to form a P–Cl bond. The fourth hybrid orbital contains the lone pair. 10.52. a.
The Lewis structure is
F F
N
F
The presence of three single bonds and one lone pair suggests sp3 hybridization of the N atom. Three hybrid orbitals each overlap a 2p orbital of an F atom to form an N–F bond. The fourth hybrid orbital contains the lone pair. b.
The Lewis structure is
F F
Si
F
F The presence of four single bonds and no lone pairs suggests sp3 hybridization of the Si atom. Each hybrid orbital overlaps a 2p orbital of an F atom to form an Si–F bond. 10.53. a.
Xenon has eight valence electrons. Each F atom donates one electron to give a total of ten electrons, or five electron pairs, around the Xe atom. The hybridization is sp3d.
b.
Bromine has seven valence electrons. Each F atom donates one electron to give a total of twelve electrons, or six electron pairs, around the Br atom. The hybridization is sp3d2.
c.
Phosphorus has five valence electrons. The Cl atoms each donate one electron to give a total of ten electrons, or five electron pairs, around the P atom. The hybridization is sp3d.
d.
Chlorine has seven valence electrons, to which may be added one electron from each F atom minus one electron for the charge on the ion. This gives a total of ten electrons, or five electron pairs, around chlorine. The hybridization is sp3d.
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10.54. a.
Bromine has seven valence electrons, and each fluorine has seven valence electrons for a total of 28 valence electrons for the molecule. The bromine atom has a total of five electron pairs around it, giving sp3d hybridization.
b.
Tellurium has six valence electrons. The F atoms each have seven electron to give a total of 34 electrons for the molecule. The tellurium atom has five electron pairs around it, giving sp3d hybridization.
c.
Xenon has eight valence electrons. Each F atom has seven valence electrons to give a total of 36 electrons for the molecule. The xenon atom has six electron pairs around it, and the hybridization is sp3d2.
d.
Iodine has seven valence electrons, each F atom has seven valence electrons, and one electron to account for the charge on the ion gives 36 electrons for the ion. This gives a total of six electron pairs around the I atom. The hybridization is sp3d2.
10.55. The P atom in PCl6− has six single bonds around it and no lone pairs. This suggests sp3d2 hybridization. Each bond in this ion is a σ bond formed by overlap of an sp3d2 hybrid orbital on P with a 3p orbital on Cl. 10.56. The central I atom in I3− has two single bonds and three lone pairs around it. This suggests sp3d hybridization. Each I–I bond is formed by the overlap of an sp3d hybrid orbital on the central I with a 5p orbital from a terminal I. 10.57. a.
The structural formula of formaldehyde is
O C H
H
Because the C is bonded to three other atoms, it is assumed to be sp2 hybridized. One 2p orbital remains unhybridized. The carbon–hydrogen bonds are σ bonds formed by the overlap of an sp2 hybrid orbital on C with a 1s orbital on H. The remaining sp2 hybrid orbital on C overlaps with a 2p orbital on O to form a σ bond. The unhybridized 2p orbital on C overlaps with a parallel 2p orbital on O to form a π bond. Together, the σ and π bonds constitute a double bond. b.
The nitrogen atoms are sp hybridized. A σ bond is formed by the overlap of an sp hybrid orbital from each N. The remaining sp hybrid orbitals contain lone pairs of electrons. The two unhybridized 2p orbitals on one N overlap with the parallel unhybridized 2p orbitals on the other N to form two π bonds.
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10.58. a.
Each nitrogen atom is bonded to two other atoms and also has a lone pair of electrons in its valence shell. Therefore, the nitrogen atoms are expected to be sp2 hybridized. The N–H bonds are σ bonds formed by the overlap of an sp2 hybrid orbital of an N atom with a 1s orbital on an H atom. Each N also contains a lone pair of electrons in an sp2 hybrid orbital. The remaining sp2 hybrid orbitals, one on each N, overlap with each other to form a σ bond. The two 2p orbitals, one on each N, are oriented parallel to each other so that they may overlap to form a π bond.
b.
The carbon atom is bonded to two atoms and has no lone pairs; hence, it is expected to be sp hybridized. The C–H bond is a σ bond formed by the overlap of an sp hybrid orbital on carbon with a 1s orbital on hydrogen. The other sp hybrid orbital overlaps with an sp orbital on the nitrogen atom to form a σ bond. The remaining sp hybrid orbital on N contains a lone pair. There remain two unhybridized 2p orbitals on the carbon and two 2p orbitals on the nitrogen atom. One 2p orbital from the C and a parallel 2p orbital on the N overlap to form a π bond. The remaining two 2p orbitals overlap to form a second π bond between the carbon and nitrogen.
10.59. Each of the N atoms has a lone pair of electrons and is bonded to two atoms. The N atoms are sp2 hybridized. The two possible arrangements of the O atoms relative to one another are shown below. Because the π bond between the N atoms must be broken to interconvert these two forms, it is to be expected that the hyponitrite ion will exhibit cis–trans isomerism. .. :O: ..
..
N
N
:O .. :
2
2N ..
.. N :O: ..
:O .. : cis
trans
10.60. The C=C bond consists of a σ bond and a π bond. The π bond depends on the parallel orientation of the two p orbitals that overlap to form the bond. In order to interconvert the cis and trans forms, one end of the molecule would have to be rotated relative to the other end, thus changing the orientation of the p orbitals and breaking the bond. Because this requires considerable energy, the cis and trans forms are stable relative to interconversion.
H
H C
HOOC
H
C
COOH C
COOH
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HOOC
C H
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10.61. a.
Total electrons = 2 x 5 = 10. The electron configuration is KK(σ2s)2(σ2s*)2(π2p)2. Bond order = 1/2 (nb − na) = 1/2 (6 − 4) = 1 The B2 molecule is stable. It is paramagnetic because the two electrons in the π2p subshell occupy separate orbitals.
b.
Total electrons = (2 x 5) − 1 = 9. The electron configuration is KK(σ2s)2(σ2s*)2(π2p)1. Bond order = 1/2 (nb − na) = 1/2 (5 − 4) = 1/2 The B2+ molecule should be stable and is paramagnetic because there is one unpaired electron in the π2p subshell.
c.
Total electrons = (2 x 8) + 1 = 17. The electron configuration is KK(σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)3. Bond order = 1/2 (nb − na) = 1/2 (10 − 7) = 3/2 The O2− molecule should be stable and is paramagnetic because there is one unpaired electron in the π2p* subshell.
10.62. a.
Total electrons = (2 x 4) − 1 = 7. The electron configuration is KK(σ2s)2(σ2s*)1. Bond order = 1/2 (nb − na) = 1/2 (4 − 3) = 1/2 The Be2+ molecule is expected to be stable and paramagnetic.
b.
Total electrons = 2 x 10 = 20. The electron configuration is KK(σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)4(σ2p*)2. Bond order = 1/2 (nb − na) = 1/2 (10 − 10) = 0 The Ne2 molecule is expected to be unstable and diamagnetic.
c.
Total electrons = (2 x 5) + 1 = 11. The electron configuration is KK(σ2s)2(σ2s*)2(π2p)3. Bond order = 1/2 (nb − na) = 1/2 (7 − 4) = 3/2 The B2− molecule is expected to be stable. It is paramagnetic because there is one unpaired electron in the π2p subshell.
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10.63. Total electrons = 6 + 7 + 1 = 14. The electron configuration is KK(σ2s)2(σ2s*)2(π2p)4(σ2p)2. Bond order = 1/2 (nb − na) = 1/2 (10 − 4) = 3 The CN− ion is diamagnetic. 10.64. Total electrons = 5 + 7 = 12. The electron configuration is KK(σ2s)2(σ2s*)2(π2p)4. Bond order = 1/2 (nb − na) = 1/2 (8 − 4) = 2 The BN molecule is diamagnetic.
■
SOLUTIONS TO GENERAL PROBLEMS
10.65. The number of electron pairs, number of lone pairs, and geometry are for the central atom. Number of Electron Pairs
ElectronDot Structure
Sn
Cl
Cl
Number of Lone Pairs
Geometry
2
1
Bent
3
0
Trigonal planar
5
3
Linear
6
0
Octahedral
O C
[
Cl
Br
I
Cl
[
Br
Cl Cl
Cl P
Cl
Cl Cl
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Chapter 10: Molecular Geometry and Chemical Bonding Theory
10.66. ElectronDot Structure
H
Number of Electron Pairs
O
F
Geometry
4
2
Bent, or angular
4
0
Tetrahedral
5
0
Trigonal bypyramidal
5
1
Seesaw
+
H H
Number of Lone Pairs
N
H
H
F F P
F
F F + F F Cl
F
F
10.67. a.
F
Se
F
Bent
Cl Cl b.
C
Cl
Tetrahedral
H
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F F Se
Seesaw
F c.
F 2 F F
F
Sn
Octahedral
F
F F
d. 10.68.
Br Trigonal planar
B a.
Br
Br
H b.
As
H
H
Trigonal pyramidal
F
F Cl F
Square planar
F
c.
F F
F Te F
F
d.
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Square pyramidal
363
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Chapter 10: Molecular Geometry and Chemical Bonding Theory
10.69. H
H
C
C
H C
a.
a
H
b
c
O
Ca and Cb: Three electron pairs around them. They are sp2hybridized. Cc: Four electron pairs around it. It is sp 3 hybridized.
H
H
N
C
C
Both C atoms are bonded to two other atoms and have no lone pairs of electrons. They are sp hybridized.
N
b. 10.70. a. H
N
b.
H
H
N
N
C
C
Each N has three atoms and one lone pair of electrons. They are sp 3 hybridized.
H
N
Each N has a σ bond to one atom and a lone pair of electrons. They are sp hybridized.
10.71.
Br
Br
Br C
H
C
Br
C
has a net dipole
C
H
H
cis
Br
H
H C
H
has no net dipole. The two C–Br bond dipoles cancel and the two C–H bond dipoles cancel.
C Br
trans
10.72.
Cl
Cl H3N
Pt Cl zero
H3N
NH3 Identical bond dipoles cancel each other.
Pt H3N
Cl Bond dipoles cannot cancel. Dissimilar across from each other.
nonzero
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10.73. All four hydrogen atoms of H2C=C=CH2 cannot lie in the same plane because the second C=C bond forms perpendicular to the plane of the first C=C bond. By looking at Figure 10.26, you can see how this is so. The second C=C bond forms in the plane of the C–H bonds. Thus, the plane of the C–H bonds on the right side will be perpendicular to the plane of the C–H bonds on the left side. This is shown below using the π orbitals of the two C=C double bonds.
H H H
C
C
C H
10.74. All four hydrogen atoms of H2C=C=C=CH2 lie in the same plane because the second C=C bond forms perpendicular to the plane of the first C=C bond, and the third C=C bond forms perpendicular to the plane of the second C=C bond. Thus, the righthand CH2 is in the same plane as the lefthand H2C group.
H
C
C
H
C
C
H H
10.75. Total electrons = 2 + 1 − 1 = 2. The electron configuration is (σ1s)2. Bond order = 1/2 (nb − na) = 1/2 (2) = 1 The HeH+ ion is expected to be stable. 10.76. Total electrons = (2 x 2) − 1 = 3. The electron configuration is (σ1s)2(σ1s*)1. Bond order = 1/2 (nb − na) = 1/2 (2 − 1) = 1/2 The He2+ ion is expected to be stable. 10.77. Total electrons = (2 x 6) + 2 = 14. The electron configuration is KK(σ2s)2(σ2s*)2(π2p)4(σ2p)2. Bond order = 1/2 (nb − na) = 1/2 (10 − 4) = 3 10.78. Total electrons = (2 x 8) + 2 = 18. The electron configuration is KK(σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)4. Bond order = 1/2 (nb − na) = 1/2 (10 − 8) = 1
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10.79. The molecular orbital configuration of O2 is KK(σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2. O2+ has one electron less than O2. The difference is in the number of electrons in the π2p* antibonding orbital. This means that the bond order is larger for O2+ than for O2. O2: Bond order = 1/2 (nb − na) = 1/2 (10 − 6) = 2 O2+: Bond order = 1/2 (nb − na) = 1/2 (10 − 5) = 5/2 It is expected that the species with the higher bond order, O2+, has the shorter bond length. In O2−, there is one more electron than in O2. This additional electron occupies a π2p* orbital. Increasing the number of electrons in antibonding orbitals decreases the bond order; hence, O2− should have a longer bond length than O2. 10.80. The molecular orbital configuration of N2 is KK(σ2s)2(σ2s*)2(π2p)4(σ2p)2. N2+ has one electron less than N2. The difference is in the number of electrons in the σ2p bonding orbital. Because N2+ has fewer electrons in bonding orbitals than N2 and the same number in antibonding orbitals, N2+ has a lower bond order and a longer bond distance. In N2−, there is one more electron than in N2. This additional electron occupies a π2p* antibonding orbital. Because N2− has more electrons in antibonding orbitals and the same number in bonding orbitals, the bond order is lower; hence, N2− has a longer bond length than N2. 10.81. As shown in Figure 10.34, the occupation of molecular orbitals by the N2 valence electrons is KK(σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)0 To form the first excited state of N2, a σ2p electron is promoted to the π2p* orbital, giving KK(σ2s)2(σ2s*)2(π2p)4(σ2p)1(π2p*)1 The differences in properties are as follows: Magnetic character: The ground state is diamagnetic (all electrons paired), and the excited state is paramagnetic. Bond order: ground state order = 1/2 (8 − 2) = 3 excited state order = 1/2 (7 − 3) = 2 Bond dissociation energy: groundstate energy = 942 kJ; excitedstate energy is less than 942 kJ (excited state does not possess a stable triple bond like the ground state) Bond length: groundstate length = 110 pm; excitedstate length is more than 110 pm. 10.82. For N2, refer to Figure 10.34. Note that the energy of the highest occupied orbital of atomic nitrogen is greater than the energy of the highest occupied orbital of N2. Thus the outer electrons of atomic nitrogen are farther from the nucleus and have a smaller ionization energy than that of N2. The occupation of the molecular orbitals of O2 are almost the same as shown in Figure 10.34; the electron configuration for O2 given in Example 10.7 is below along with the electron configuration of atomic oxygen: O2: KK(σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2 O:
1s22s22p4
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Referring again to Figure 10.34 (after inserting two π2p* electrons for O2), we see that the energy of the highest occupied orbital of O2 is greater than the energy of the highest occupied orbital of atomic oxygen. Thus, the outer electrons of O2 are farther from the nucleus and have a smaller ionization energy than that of atomic oxygen. 10.83. A person’s two hands are similar but not identical. A person’s left hand is a mirror image of the other hand, yet the two hands are nonsuperimposible. The presence of handedness in molecules depends on the fact that atoms in the molecules occupy specific places in threedimensional space. 10.84. The chemical substance responsible for the flavor of spearmint is Lcarvone. The substance responsible for the flavor of caraway seeds is Dcarvone. These are two substances whose molecules differ only in being mirror images of one another. Nevertheless, they have strikingly different flavors: one is minty, the other a pungent aromatic. 10.85. Color vision is possible because three types of cone cells exist: one type absorbs light in the red region of the spectrum, another absorbs in the green region, and a third absorbs in the blue region. The chemical substance primarily responsible for human vision is 11cisretinal. In each of the different cone cells, 11cisretinal is attached to a different protein molecule, which affects the region of light that is absorbed by retinal. 10.86. When a photon of light in the ultraviolet region (near 180 nm) is absorbed by cis1,2dichloroethene, an electron in the π orbital undergoes a transition to the π* orbital. As a result, the double bond becomes a single bond, and rotation about the bond is now possible. The cis isomer tends to rotate into the trans isomer, which is more stable. The electron in the π* orbital undergoes a transition back to the π orbital; the single bond becomes a double bond again. The net result is the conversion of the cis isomer to the trans isomer. 10.87. Ozone in the stratosphere absorbs ultraviolet radiation between 200 and 300 nm, which is vitally important to life on earth. Radiation from the sum contains ultraviolet rays that are harmful to the DNA of biological organisms. Oxygen, O2, absorbs the most energetic of these ultraviolet rays in the earth’s upper atmosphere, but only ozone in the stratosphere absorbs the remaining ultraviolet radiation that is destructive to life on earth. 10.88. In 1974, Mario J. Molina and F. Sherwood Rowland expressed concern that chlorofluorocarbons (CFC’s), such as CClF3, and CCl2F2, would be a source of chlorine atoms that could catalyze the decomposition of ozone in the stratosphere, so that the ozone would be destroyed faster than it could be produced. The chlorofluorocarbons are relatively inert compounds used as refrigerants, spraycan propellants, and blowing agents (substances used to produce plastic foams). As a result of their inertness, however, CFC’s concentrate in the atmosphere, where they steadily rise into the stratosphere. Once they are in the stratosphere, ultraviolet light decomposes them to form chlorine atoms, which react with ozone to form ClO and O2. The ClO molecules react with oxygen atoms in the stratosphere to regenerate Cl atoms.
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Cl(g) + O3(g) → ClO(g) + O2(g) ClO(g) + O(g) → Cl(g) + O2(g) O3(g) + O(g) → 2O2(g) The net result is the decomposition of ozone to dioxygen. Chlorine atoms are consumed in the first step but regenerated in the second. Thus, chlorine atoms are not used up, so they function as a catalyst.
■
SOLUTIONS TO STRATEGY PROBLEMS
10.89. The electron dot formula for C22− is :C
C:
2
The valence bond description of the bonding is a triple bond, involving a σ bond formed by the overlap of a sp hybrid orbital on each carbon, and two π bonds formed by the overlap of the two unhybridized 2p orbitals on each carbon. Each atom obeys the octet rule, and there is a formal charge of −1 on each carbon. The molecular orbital description is based on a total of 14 electrons for the ion, with a configuration of KK(σ2s)2(σ2s*)2(π2p)4(σ2p)2. The bond order is 1/2(8 – 2) = 3. This agrees with the valence bond description. 10.90. The Lewis structures and formal charges are as follows.
0
0
1
1
B
F
B
F
4
4
3
3
B
F
B
F
2 B
2 F
The formula that best fits the molecule is in the upperright corner. The boron atom is singly bonded to the fluorine atom, and there is no formal charge on either atom. Also, the fluorine atom obeys the octet rule, but the boron atom does not, but this is not unusual. The valence bond description is a σ bond formed by the overlap of an sp hybrid orbital on boron with an sp3 orbital on fluorine. 10.91. In BF3NH3, both the boron atom and the nitrogen atom are in a tetrahedral geometry, with sp3 hybrid orbitals. In terms of valence bond theory, the bond between boron and fluorine is a σ bond, formed by the overlap of an sp3 orbital on boron with an sp3 orbital on nitrogen. For the reactant molecules, BF3 is trigonal planar with sp2 hybrid orbitals, and NH3 is trigonal pyramidal with sp3 hybrid orbitals.
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10.92. The bonding about the carbon atom is trigonal planar geometry, with the carbon having sp2 hybrid orbitals. The carbon–oxygen double bond is a σ bond, formed by the overlap of an sp2 orbital on carbon with a 2p orbital on oxygen, and a π bond, formed by the overlap of an unhybridized 2p orbital on carbon with another 2p orbital on oxygen. Another sp2 hybrid orbital on carbon overlaps with an sp2 orbital of the carbon in the ring. The third σ bond is another sp2 orbital on carbon overlapping with an sp3 orbital of the oxygen atom in the single bond. 10.93. In the molecule XF6, X must have six valence electrons. Therefore, XF4 has a total of 34 valence electrons. This gives four X–F single bonds and one lone pair around the X atom. The XF6 molecule has an octahedral geometry, with sp3d2 hybrid orbitals. The XF4 molecule has seesaw geometry, with sp3d hybrid orbitals. Molecules with seesaw geometries are polar, so the observed dipole moment of 0.632 D for XF4 is not unexpected. Some possibilities for X are sulfur (S), selenium (Se), and tellurium (Te). 10.94. The structures for the compounds are
F
F C
C
H
F
H C
A
C
C
H
H
F
H C
F
F
H
B
C
10.95. The Lewis formulas for the two isomers of N2H2 are
H N
N H
cis
N
N H
H
trans
The bonding descriptions for both structures are the same. All of the nitrogen–hydrogen bonds are σ bonds formed by the overlap of an sp2 orbital on nitrogen with a 1s orbital on hydrogen. The nitrogen–nitrogen double bond is a σ bond, formed by the overlap of an sp2 orbital on one nitrogen with an sp2 orbital on the other nitrogen, and a π bond, formed by the overlap of an unhybridized 2p orbital on one nitrogen with a 2p orbital on the other nitrogen. The double bond requires a lot of energy to break, allowing free rotation, so it doesn’t happen at normal temperatures. Thus, both isomers are stable. 10.96. The molecular orbital description of an O2 molecule is KK(σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2. In the π2p* orbital, both electrons must be in the same orbital for all electrons to be paired. The ground state would have one electron in each of the two π2p* orbitals. In both cases, the bond order is 2, but the excited state is diamagnetic, whereas the ground state is paramagnetic.
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10.97. If you construct molecular orbitals similar to oxygen, you get the following configuration for S2: [core](σ3s)2(σ3s*)2(π3p)4(σ3p)2(π3p*)2(σ3p*)0. This would predict a bond order of 2. If two electrons were added to give S22−, they would go into the π3p* orbital, and the bond order would be reduced to 1. Thus, the bond length should increase in this case. If two electrons were taken away to give S22+, they would be removed from the π3p* orbital, and the bond order would increase to 3. Thus, the bond length would decrease in this case. 10.98. The Lewis formula for formaldehyde is
O C H
H
Because the C is bonded to three other atoms, it is assumed to be sp2 hybridized. One 2p orbital remains unhybridized. The carbon–hydrogen bonds are σ bonds formed by the overlap of an sp2 hybrid orbital on C with a 1s orbital on H. The remaining sp2 hybrid orbital on C overlaps with a 2p orbital on O to form a σ bond. The unhybridized 2p orbital on C overlaps with a parallel 2p orbital on O to form the π bond. In molecular orbital theory, the carbon–oxygen double bond has a σ and a π molecular orbital that are occupied by electrons. There is also a π* orbital that is unoccupied in the ground state. The transition at 270 nm (UV region) is due to an electron moving from the occupied π orbital to the unoccupied π* orbital to give an excited state.
■
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 10.99. After assuming a 100.0g sample, convert to moles: 60.4 g Xe x
1 mol Xe = 0.46005 mol Xe 131.29 g Xe
22.1 g O x
1 mol O = 1.381 mol O 16.00 g O
17.5 g F x
1 mol F = 0.9215 mol F 18.99 g F
Divide by 0.460 : Xe:
0.460 1.381 0.9215 = 1; O: = 3.00; F: = 2.00 0.460 0.460 0.460
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The simplest formula is XeO3F2. This is also the molecular formula. The Lewis formula is
F
F O
O
O
Xe
or
O
O
Xe O
F
F
Number of electron pairs = 5, number of lone pairs = 0; hence, the geometry is trigonal bipyramidal. Because xenon has five single bonds, it will require five orbitals to describe the bonding. This suggests sp3d hybridization. 10.100. After assuming a 100.0g sample, convert to moles: 1 mol Xe = 0.4478 mol Xe 131.29 g Xe
58.8 g Xe x 7.2 g O x
1 mol O = 0.4500 mol O 16.00 g O
34.0 g F x
1 mol F = 1.7904 mol F 18.99 g F
Divide by 0.4478 : Xe:
0.4478 0.450 1.7904 = 1; O: = 1.00; F: = 4.00 0.4478 0.4478 0.4478
The simplest formula is XeOF4. This is also the molecular formula. The Lewis formula is
O F
F Xe
F
F
Number of electron pairs = 6, number of lone pairs = 1; hence, the geometry is square pyramidal. Because xenon has six electron pairs, it will require six orbitals to describe the bonding. This suggests sp3d2 hybridization.
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Chapter 10: Molecular Geometry and Chemical Bonding Theory
10.101. U(s) + ClFn
→
UF6 +
ClF(g)
mol UF6 = 3.53 g UF6 x mol ClF = n =
1 mol UF6 = 0.01003 mol UF6 352.07 g UF6
PV 2.50 atm x 0.343 L = = 0.3002 mol ClF RT 0.082057 L • atm/ (K • mol) x 348 K
0.010 mol UF6 = 0.060 mol F, and 0.030 mol ClF = 0.030 mol F; therefore, the total moles of F from ClFn = 0.090 mol F. Because mol ClFn must equal mol ClF, mol ClFn = 0.030 mol and n = 0.090 mol F ÷ 0.030 mol ClFn = 3. The Lewis formula is
F
Cl
F
F Number of electron pairs = 5, number of lone pairs = 2; hence, the geometry is Tshaped. Because chlorine has five electron pairs, it will require five orbitals to describe the bonding. This suggests sp3d hybridization. 10.102. Br2
+
excess F2
mol Br2 = n =
→
2BrFn
PV (748/ 760) atm x 0.423 L = = 0.01199 = 0.0120 mol Br2 RT 0.082057 L • atm/(K • mol) x 423 K
From the equation, 0.0120 mol Br2 must produce 0.0240 mol BrFn. Formula mass BrFn =
4.20 g = 175 g/mol 0.0240 mol
175 g/mol = 79.9 g Br/mol + 19.00 x n g F/mol n =
175  79.9 = 5.00 19.00
The Lewis formula is
F F
F Br
F
F
Number of electron pairs = 6, number of lone pairs = 1; hence, the geometry is square pyramidal. Because bromine has six electron pairs, it will require six orbitals to describe the bonding. This suggests sp3d2 hybridization.
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10.103. N2: Triple bond; bond length = 110 pm. Geometry is linear; sp hybrid orbitals are needed for one lone pair and one σ bond. N2F2: Double bond; bond length = 122 pm. Geometry is trigonal planar; sp2 hybrid orbitals are needed for one lone pair and two σ bonds. N2H4: Single bond; bond length = 145 pm. Geometry is tetrahedral; sp3 hybrid orbitals are needed for one lone pair and three σ bonds. 10.104. The bond length of C2 is close to that of C2H4, suggesting the same bond order for C2 as for C2H4, which has a C=C bond. Total electrons for C2 = 2 x 6 = 12. The molecular orbital configuration for C2 is KK(σ2s)2(σ2s*)2(π2p)4(σ2p)0. Bond order = 1/2(8 − 4) = 2. This agrees with the bondlength prediction. 10.105. HNO3 resonance formulas:
O
O H
O
H
N O
O
N O
The geometry around the nitrogen is trigonal planar; therefore, the hybridization is sp2. Formation reaction: H2(g) + 3O2(g) + N2(g) → 2HNO3(g) 2 x ΔHf° = [BE(H–H) + 3BE(O2) + BE(N2)] − [2BE(H–O) + 4BE(N–O) + 2BE(N=O)] = [(432 + 3 x 494 + 942) − (2 x 459 + 4 x 201 + 2 x 607) kJ/2 mol] ΔHf° = −80 kJ/2 mol = −40 kJ/mol
Resonance energy = −40 kJ − (−135 kJ) = 95 kJ
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10.106. This is the other resonance formula of benzene.
H
C
C C H
H
C
H
C C
H
H Geometry is trigonal planar; therefore, the hybridization is sp2. Formation reaction: 3H2(g) + 6C(graphite) → C6H6(g) [ + C(graphite) → C(g)] ΔHf° = [3BE(H–H) + 6 x ΔH C(g)] − [6BE(C–H) + 3BE(C–C) + 3BE(C=C)]
= [(3 x 432 + 6 x 715) − (6 x 411 + 3 x 346 + 3 x 602) kJ/mol] = +276 kJ/mol Resonance energy = +276 kJ − (−83 kJ) = 359 kJ
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CHAPTER 11
States of Matter; Liquids and Solids
■
SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 11.1. First, calculate the heat required to vaporize 1.00 kg of ammonia: 1.00 kg NH3 x
1 mol NH 3 1000 g x 17.03 g NH3 1 kg
x
23.4 kJ = 1374.04 kJ 1 mol
The amount of water at 0°C that can be frozen to ice at 0°C with this heat is 1374.04 kJ x
1 mol H 2 O 18.01 g H 2 O x = 4117.54g = 4.12 kg H2O 6.01 kJ 1 mol H 2 O
11.2. Use the twopoint form of the ClausiusClapeyron equation to calculate P2: ln
P2 26.8 x 103 J / mol ⎡ 1 1 ⎤ = ⎢ 8.31 J / (K • mol) ⎣ 319 K 308 K ⎥⎦ 760 mmHg ⎡ 1.12 x 104 ⎤ ⎥ = −0.36106 K ⎣ ⎦
= 3225 K ⎢ Converting to antilogs gives
P2 = antilog(−0.36106) = e−0.36106 = 0.69693 760 mmHg
P2 = 0.69693 x 760 mmHg = 529.6 = 5.30 x 102 mmHg 11.3. Use the twopoint form of the ClausiusClapeyron equation to solve for ΔHvap: ln
ΔH vap 757 mmHg 1 1 ⎡ ⎤ = ⎢ 522 mmHg 8.31 J/(K • mol) ⎣ 368 K 378 K ⎥⎦
0.37169 =
ΔH vap
⎡ 7.188 x 105 ⎤ ⎢ ⎥ K 8.31 J/(K • mol) ⎣ ⎦
ΔHvap = 4.296 x 104 J/mol (43.0 kJ/mol)
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Chapter 11: States of Matter; Liquids and Solids
11.4. a.
Liquefy methyl chloride by a sufficient increase in pressure below 144°C.
b.
Liquefy oxygen by compressing to 50 atm below −119°C.
a.
Propanol has a hydrogen atom bonded to an oxygen atom. Therefore, hydrogen bonding is expected. Because propanol is polar (from the O–H bond), we also expect dipoledipole forces. Weak London forces exist, too, because such forces exist between all molecules.
b.
Linear carbon dioxide is not polar, so only London forces exist among CO2 molecules.
c.
Bent sulfur dioxide is polar, so we expect dipoledipole forces; we also expect the usual London forces.
11.5.
11.6. The order of increasing vapor pressure is butane (C4H10), propane (C3H8), and ethane (C2H6). Because London forces tend to increase with increasing molecular mass, we would expect the molecule with the highest molecular mass to have the lowest vapor pressure. 11.7. Because ethanol has an H atom bonded to an O atom, strong hydrogen bonding exists in ethanol but not in methyl chloride. Hydrogen bonding explains the lower vapor pressure of ethanol compared to methyl chloride. 11.8. a.
Zinc, a metal, is a metallic solid.
b.
Sodium iodide, an ionic substance, exists as an ionic solid.
c.
Silicon carbide, a compound in which carbon and silicon might be expected to form covalent bonds to other carbon and silicon atoms, exists as a covalent network solid.
d.
Methane, at room temperature a gaseous molecular compound with covalent bonds, freezes as a molecular solid.
11.9. Only MgSO4 is an ionic solid; C2H5OH, CH4, and CH3Cl form molecular solids; thus MgSO4 should have the highest melting point. Of the molecular solids, CH4 has the lowest molecular mass (16.0 amu) and would be expected to have the lowest melting point. Both C2H5OH and CH3Cl have approximately the same molecular masses (46.0 amu vs. 50.5 amu), but C2H5OH exhibits strong hydrogen bonding and, therefore, would be expected to have the higher melting point. The order of increasing melting points is CH4, CH3Cl, C2H5OH, and MgSO4. 11.10. Each of the four corners of the cell contains one atom, which is shared by a total of four unit cells. Therefore, the corners contribute one whole atom. Atoms 1/4 atom = 4 corners x = 1 atom Unit cell 1 corner
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11.11. Use the edge length to calculate the volume of the unit cell. Then, use the density to determine the mass of one atom. Divide the molar mass by the mass of one atom. V = (3.509 x 10−10 m)3 = 4.321 x 10−29 m3 d =
0.534 g ⎡100 cm ⎤ x ⎢ 3 ⎥ 1 cm ⎣ 1m ⎦
Mass of 1 unit
3
= 5.34 x 105 g/m3
= d x V 5
= (5.34 x 10 g/m3) x (4.321 x 10−29 m3) = 2.3074 x 10−23 g There are two atoms in a bodycentered cubic unit cell; thus, the mass of one lithium atom is 1/2 x 2.3074 x 10−23 g = 1.1537 x 10−23 g The known atomic mass of lithium is 6.941 amu, so Avogadro's number is NA =
6.941 g/mol = 6.016 x 1023 = 6.02 x 1023 atoms/mol 1.1537 x 1023 g/atom
11.12. Use Avogadro's number to convert the molar mass of potassium to the mass per one atom. 39.0983 g K 1 mol K 6.4925 x 1023 g K x = 23 1 mol K 6.022 x 10 atoms 1 atom
There are two K atoms per unit cell; therefore, the mass per unit cell is 2 atoms 6.4925 x 1023 g K 1.2985 x 1022 g x = 1 unit cell 1 atom 1 unit cell
The density of 0.856 g/cm3 is equal to the mass of one unit cell divided by its unknown volume, V. After solving for V, determine the edge length from the cube root of the volume. 0.856 g/cm3 = V =
1.2985 x 1022 g = 1.517 x 10−22 cm3 (1.517 x 10−28 m3) 0.856 g/cm3
Edge length =
■
1.2985 x 1022 g V
3
1.517 x 1028 m3 = 5.333 x 10−10 = 5.33 x 10−10 m (533 pm)
ANSWERS TO CONCEPT CHECKS
11.1. a. (1)
At t = 0, since the system is not at equilibrium and there are no H2O molecules in the gaseous state, you would expect the rate of evaporation to exceed the rate of condensation. At t = 1, since evaporation has proceeded at a greater rate than condensation, there must now be fewer molecules in the liquid state, resulting in a lower level of H2O(l).
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(2)
At t = 1, since some of the H2O has gone into the vapor state, the vapor pressure must be higher.
(3)
At t = 1, since evaporation has occurred, there must be more molecules in the vapor state.
(4)
At t = 0, since the system is not at equilibrium, and there are no H2O molecules in the gaseous state, you would expect the rate of evaporation to exceed the rate of condensation.
(5)
t =1 b. (1)
Between t = 1 and t = 2, the system has still not reached equilibrium. Therefore, because the rate of evaporation continues to exceed the rate of condensation, you would expect the water level to be lower.
(2)
Prior to reaching equilibrium at t = 2, you would continue to observe a rate of evaporation greater than the rate of condensation, resulting in a higher vapor pressure than t = 1.
(3)
Since evaporation has been occurring at a greater rate than condensation between points t = 1 and t = 2, you would expect more molecules in the vapor state at t = 2.
(4)
When the system has reached equilibrium at t = 2, the rate of evaporation equals the rate of condensation.
(5)
t=2 11.2. You would have to cook the egg for a longer time. The reason is that, since the temperature is lower, it would take longer for the egg to become hard, a chemical process (which slows down at lower temperature). Copyright © Houghton Mifflin Company. All rights reserved.
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11.3. a.
If the first reaction occurred, the mixture of hydrogen and oxygen that resulted would form an explosive mixture. At 100°C, 2 moles of water make 61.2 L of gaseous water. In the incorrect reaction, 2 moles of water produce 91.8 L of gaseous products.
b.
Since you would be breaking strong chemical bonds and forming relatively weak bonds, the enthalpy for the first reaction (the wrong reaction) would be many times greater (more positive) than for the second reaction.
c.
Apply Hess’s law. The enthalpy for the wrong reaction would be equal to two times the negative of ΔHfo for H2O(l) plus the heat required to raise the temperature of 2 moles of water from 25°C to 100°C.
a.
First, consider the B balls (small). There are four atoms, each completely inside the cell. Thus, there are four B atoms per cell. Next, there are fourteen A atoms (large). Of these, eight are in corners, and contribute 1/8 to the cell. Six atoms are in faces, and contribute 1/2 to the cell. Thus, there are 8 x (1/8) + 6 x (1/2) = 4 A atoms per cell. The ratio of A atoms to B atoms is 4 to 4, or 1 to 1. Thus, the formula of the compound is AB.
b.
Since all of the B atoms are completely within the cell, the shape of the cell is determined by the A atoms only. It is a facecentered cubic unit cell.
11.4.
■
ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
11.1. The six different phase transitions, with examples in parentheses, are melting (snow melting), sublimation (dry ice subliming directly to carbon dioxide gas), freezing (water freezing), vaporization (water evaporating), condensation (dew forming on the ground), and gassolid condensation or deposition (frost forming on the ground). 11.2. Iodine can be purified by heating it in a beaker covered with a dish containing ice or ice water. Only pure iodine should sublime, crystallizing on the cold bottom surface of the dish above the iodine. The common impurities in iodine do not sublime nor do they vaporize significantly. 11.3. The vapor pressure of a liquid is the partial pressure of the vapor over the liquid, measured at equilibrium. In molecular terms, vapor pressure involves molecules of a liquid vaporizing from the liquid phase, colliding with any surface above the liquid, and exerting pressure on it. The equilibrium is a dynamic one because molecules of the liquid are continually leaving the liquid phase and returning to it from the vapor phase. 11.4. Steam at 100°C will melt more ice than the same weight of water at 100°C because it contains much more energy in the form of its heat of vaporization. It will transfer this energy to the ice and condense in doing so. The condensed steam and the water will both transfer heat to the ice as the temperature then drops.
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11.5. The heat of fusion is smaller than the heat of vaporization because melting requires only enough energy for molecules to escape from their sites in the crystal lattice, leaving other molecular attractions intact. In vaporization, sufficient energy must be added to break almost all molecular attractions and also to do the work of pushing back the atmosphere. 11.6. Evaporation leads to cooling of a liquid because the gaseous molecules require heat to evaporate; as they leave the other liquid molecules, they remove the heat energy required to vaporize them. This leaves less energy in the liquid, the temperature of which then drops. 11.7. As the temperature increases for a liquid and its vapor in the closed vessel, the two, which are separated by a meniscus, gradually become identical. The meniscus first becomes fuzzy and then disappears altogether as the temperature reaches the critical temperature. Above this temperature, only the vapor exists. 11.8. A permanent gas can be liquefied only by lowering the temperature below its critical temperature while compressing the gas. 11.9. The pressure in the cylinder of nitrogen at room temperature (above its critical temperature of −147°C) decreases continuously as gas is released because the number of molecules in the vapor phase, which governs the pressure, decreases continuously. The pressure in the cylinder of propane at room temperature (below its critical temperature) is constant because liquid propane and gaseous propane exist at equilibrium in the cylinder. The pressure will remain constant at the vapor pressure of propane until only gaseous propane remains. At that point, the pressure will decrease until all of the propane is gone. 11.10. The vapor pressure of a liquid depends on the intermolecular forces in the liquid phase since the ease with which a molecule leaves the liquid phase depends on how strongly it is attracted to the other molecules. If such molecules attract each other strongly, the vapor pressure will be relatively low; if they attract each other weakly, the vapor pressure will be relatively high. 11.11. Surface tension makes a liquid act as though it had a skin, because for an object to break through the surface, the surface area must increase. This requires energy, so there is some resistance to the object breaking through the surface. 11.12. London forces, also known as dispersion forces, originate between any two molecules that are weakly attracted to each other by means of small instantaneous dipoles that occur as a result of the varying positions of the electrons during their movement about their nuclei. 11.13. Hydrogen bonding is a weak to moderate attractive force that exists between a hydrogen atom covalently bonded to a very electronegative atom, X (N, O, or F), and a lone pair of electrons on another small, electronegative atom, Y. (X and Y may be the same or different elements.) Hydrogen bonding in water involves a hydrogen atom of one water molecule bonding to a lone pair of electrons on the oxygen atom of another water molecule.
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11.14. Molecular substances have relatively low melting points because the forces broken by melting are weak intermolecular attractions in the solid state, not strong bonding attractions. 11.15. A crystalline solid has a welldefined, orderly structure; an amorphous solid has a random arrangement of structural units. 11.16. In a facecentered cubic cell, there are atoms at the center of each face of the unit cell in addition to those at the corners. 11.17. The structure of thallium(I) iodide is a simple cubic lattice for both the metal ions and the anions. Thus, the structure consists of two interpenetrating cubic lattices of cation and anion. 11.18. The coordination number of Cs+ in CsCl is 8; the coordination number of Na+ in NaCl is 6; and the coordination number of Zn2+ in ZnS is 4. 11.19. Starting with the edge length of a cubic crystal, we can calculate the volume of a unit cell by cubing the edge length. Next, knowing the density of the crystalline solid, we can calculate the mass of the atoms in the unit cell. Then, the mass of the atoms in the unit cell is divided by the number of atoms in the unit cell, to give the mass of one atom. Dividing the mass of one mole of the crystal by the mass of one atom yields a value for Avogadro's number. 11.20. X rays can strike a crystal and be reflected at various angles; at most angles, the reflected waves will be out of phase and will interfere destructively. At certain angles, however, the reflected waves will be in phase and will interfere constructively, giving rise to a diffraction pattern. 11.21. The answer is e, cooling the H2O sample from 105°C to 84°C. 11.22. The answer is d, sublime. 11.23. The answer is c, London (dispersion) forces. 11.24. The answer is b, 6.40 x 107 pm3.
■
ANSWERS TO CONCEPT EXPLORATIONS
11.25. a.
In the container, A is in solid form, B is in liquid form, and C is in gaseous form.
b.
The greatest intermolecular attractions are in the solid (A), second is for the liquid (B), and the least intermolecular attractions are for the gas (C).
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c.
Since each of these substances is a monatomic nonmetal, the only forces present are London (dispersion) forces.
d.
The greatest atomic mass will be for the strongest intermolecular forces, which is A.
e.
The element with the normal boiling point of 2 K is the gas, C. A gas is a substance whose boiling point is below ambient temperature.
f.
The substance with a melting point of 50 K is A. Since it is a solid, its melting point must be above ambient temperature.
g.
The substance with a normal boiling point of 25 K is B. It cannot have a freezing point of 7 K, or it would be in solid form. In order for a substance to be a liquid, the freezing point must be below ambient temperature.
h.
As you began to heat the container to 20 K, the gas would expand and the piston would move up.
i.
There would not be much difference in the container and its contents at 20 K. The gas (C) would have expanded its volume in accordance with Charles’s law, and the new volume would be five times the volume at 4 K. The solid component (A) would appear unchanged. Since the temperature is nearing the boiling point of the liquid component (B), which is 25 K, some of B should appear in the vapor phase. In the drawing below, The gray is A, the black B, and the white C.
j.
At 30 K, the temperature is now above the boiling point of B, which is 25 K, so both B and C are now gases. A remains in the solid state.
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k.
383
When the temperature of the container reaches 60 K, component A has now melted and is in the liquid state. The gases (B and C) continue to expand according to Charles’s law.
11.26. Part 1 a.
Yes, it is possible to add heat to a pure substance and not observe a temperature change. This occurs during phase changes: melting (freezing) or boiling (condensing).
b.
On a molecular level, when heat is first added to a solid substance, it causes the particles that make up the substance to begin vibrating in their locus sites. This causes the temperature of the substance to increase, until the melting point is reached. At this point, enough energy has been added to the substance to overcome the intermolecular forces of attraction, and particles begin to escape into the liquid phase. This occurs at a constant temperature, until the entire sample has melted.
Part 2
bpt
Vapor Solid and liquid
Vapor Liquid and vapor
Liquid mpt 50
Solid Time (Heat added at constant rate)
a.
Heating curve for substance B
Temperature (°C)
Temperature (°C)
Heating curve for substance A
bpt
Solid and liquid Liquid and vapor Liquid
mpt 50
Solid Time (Heat added at constant rate)
No, substance B melted first. Since the melting point of substance B is lower, it requires less heat to reach the melting point. The curves for the solid region of each graph have about the same slope, so both solids warm at the same rate. Since it takes less heat, it also takes less time to reach the melting point for substance B, and it melts first.
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b.
The boiling points are not the same. The boiling point for substance B is lower than for substance A. Since the curves in the liquid portion of the graphs for both substances have the same slope, the boiling point for substance B is proportionately lower.
c.
Substance B reached the boiling point first. It requires less total heat to reach the boiling point, and thus less time.
d.
No, melting and vaporizing are not the same process, and they involve different forces. It takes more energy to transform a liquid to a vapor than to melt a solid, so the heat required to convert all of substance A to a gas at its boiling point is greater than the heat required to melt it. On the graphs, the horizontal line for liquidvapor equilibrium is longer than the horizontal line for solidliquid equilibrium.
■
ANSWERS TO CONCEPTUAL PROBLEMS
11.27. a.
The molecules of A would have the majority of molecules in the gas phase because at this temperature, liquid A has the higher fraction of molecules that have sufficient kinetic energy and can vaporize.
b.
The molecules with the strongest intermolecular attractions will have the fewest molecules in the gas phase because a higher kinetic energy is needed to overcome the attractions. Since molecules of C have the smallest fraction of molecules with sufficient kinetic energy, they have the strongest intermolecular attractions.
c.
The molecules with the strongest intermolecular attractions will have the lowest vapor pressure. Thus, molecules of C would have the lowest vapor pressure.
11.28. You will need to compare the heats of fusion and vaporization of substance X (ΔHfus = 9.0 kJ/mol and ΔHvap = 20.3 kJ/mol) with the values for water, which are ΔHfus = 6.01 kJ/mol and ΔHvap = 40.7 kJ/mol. Comparing values shows that ΔHfus is 1.5 times larger for substance X and that ΔHvap is 2.0 times larger for H2O. Heating the substance, or water, from −10°C to the boiling point is a threestep process. Step 1 is to heat the solid from −10°C to 0°C, the freezing point. The heat required for this step is equal to mass x specific heat capacity x temperature change. Step 2 is to melt the solid to liquid at 0°C. The heat required for this step is equal to moles x ΔHfus. Step 3 is to heat the liquid from 0°C to 100°C. The heat required for this step is equal to mass x specific heat capacity x temperature change. a.
Since the masses, heat capacities, and temperature changes for water and for substance X are all equal, the heat required for Step 1 and Step 3 are the same for both. Since ΔHfus is larger for substance X (per mole), Step 2 will require more heat for substance X and thus take longer. Therefore, H2O will reach the boiling point first.
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b.
To completely boil away the substance, an additional step is required. Step 4 is to boil the liquid to vapor at 100°C. The heat required for this step is equal to moles x ΔHvap. Since the ΔHvap values are much larger than the ΔHfus values, Step 4 will require much more heat than Step 2 for both substance X and H2O. Since ΔHvap is smaller for substance X (per mole), Step 4 will require less heat for substance X and thus take less time. The total heat required for the four steps is directly proportional to the time it would take to completely boil away the substance. Steps 1 and 3 are the same for both. Step 2 takes 1.5 times as long for substance X, but Step 4 takes 2.0 times as long for water. Since Step 4 requires the most heat, water will require more time to complete this step, so substance X will boil away first.
c.
The heating curves for substance X and for water are shown below.
Vapor
100
50
0 10
Heating curve f or X
Solid and liquid
Vapor
100 Liquid and vapor
Liquid
Solid Time (Heat added at constant rate)
Temperature (°C)
Temperature (°C)
Heating curve f or water
50
Solid and liquid
Liquid and vapor Liquid
0 10
Solid Time (Heat added at constant rate)
11.29. The water the farmers spray above and on their fruit is warmer than the temperature of the fruit on the trees. Therefore, as the temperature of the air drops, it absorbs heat from the water, converting it into ice, before absorbing any heat from the fruit. The heat released when the liquidtosolid phase change occurs prevents the fruit from freezing. 11.30. a.
Bottle A is probably mislabeled. If it is an ionic compound, the boiling point should be higher than 35°C. Most ionic compounds are solids with high melting points.
b.
The substance with the highest boiling point will have the strongest intermolecular attractions. Thus, the compound in bottle C has the strongest intermolecular attractions.
c.
The substance with the lowest boiling point will have the highest vapor pressure. Thus, the substance in bottle B will have the highest vapor pressure.
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11.31. a.
Considering the A atoms, there are nine A atoms per cell. Of these, eight are in corners and contribute 1/8 per cell, and one is completely inside the cell. Thus there are 8 x (1/8) + 1 x (1) = 2 A atoms per cell. Next, considering the B atoms, there are six per cell. Of these, four are in faces and contribute 1/2 per cell, and two are completely inside the cell. Thus, there are 4 x (1/2) + 2 x (1) = 4 B atoms per unit cell. The ratio of A atoms to B atoms is 2 to 4, or 1 to 2. Thus, the formula of the compound is AB2.
b.
The A atoms are in the arrangement of a bodycentered cell.
11.32. Consulting the phase diagram for water (Figure 11.11), you see that by increasing the pressure on solid ice at constant temperature, you will convert solid ice into liquid water. As the weights continue to exert pressure on the ice through the wire, the ice below the wire continues to melt. Above the wire, the pressure is released, and the water refreezes, reforming the ice. Thus the wire passed through the block but the block was not cut in half. 11.33. As the water evaporates, molecules with higher kinetic energy escape the liquid and leave behind molecules with lower energy. The result is a drop in temperature of the liquid. Since the cup is well insulated, the energy lost with the evaporated molecules is not rapidly replaced. 11.34. The heat released when the vapor of a substance condenses to liquid is equal to the negative (opposite) of the heat of vaporization for the substance. As a consequence of its strong intermolecular attractions, water has a larger heat of vaporization, so it releases more heat when condensing on the skin, causing a more severe burn. 11.35. This question can be answered by looking at the packing efficiencies of the three types of crystals: simple cubic, 52.4% bodycentered cubic, 68% and facecentered cubic, 74%. a.
The highest density would correspond to the type of crystal with the highest packing efficiency. Of the three types of crystals, the facecentered cubic has the highest packing efficiency (74%) and thus has the highest density.
b.
The most empty space corresponds to the type of crystal with the lowest packing efficiency. Of the three types of crystals, the simple cubic has the lowest packing efficiency (52.4%) and therefore the most empty space.
11.36. Since the liquids have comparable molar masses, differences in properties can be attributed to differences in intermolecular forces. The flask on the left has more molecules in the vapor phase than the flask on the right, and therefore, the substance in this flask has a higher vapor pressure. Since vapor pressure decreases as intermolecular forces increase, the intermolecular forces for the substance in the flask on the right are stronger than for the substance in the flask on the left. a.
Substance A has hydrogen bonding, whereas substance B does not. Hydrogen bonding is a stronger intermolecular force than a dipoledipole force. This implies that substance A, with the lower vapor pressure, is in the flask on the right.
b.
The amount of vapor present at 35°C (an increase of 15°C) would be greater in both flasks compared to the amount present at 20°C, but since the intermolecular forces of substance B are weaker, substance B will experience a larger increase.
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387
SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off.
11.37. a.
Vaporization
b.
Freezing of eggs and sublimation of ice
c.
Condensation
d.
Gassolid condensation, deposition
e.
Freezing
11.38. a.
Sublimation
b.
Vaporization
c.
Sublimation of the filament and gassolid condensation of the vapor
d.
Freezing
e.
Melting, fusion
11.39. Dropping a line from the intersection of a 350mmHg line with the diethyl ether curve in Figure 11.7 intersects the temperature axis at about 10°C. 11.40. Dropping a line from the intersection of a 250mmHg line with the carbon tetrachloride curve in Figure 11.7 intersects the temperature axis at about 40°C. 11.41. The total amount of energy provided by the heater in 4.54 min is 4.54 min x
3.48 J 60 s x = 947.95 J (0.94795 kJ) s 1 min
The heat of fusion per mole of I2 is 2 x 126.9 g I 2 0.9479 kJ 15.5 kJ x = 15.52 = mol I 2 15.5 g I 2 mol I 2
11.42. The total amount of energy provided by the heater in 6.92 min is 6.92 min x
4.66 J 60 s x = 1934.8 J (1.9348 kJ) 1s 1 min
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The heat of fusion per mole of Cd is 112.4 g Cd 6.07 kJ 1.9348 kJ x = 6.074 = mol Cd mol Cd 15.5 g Cd
11.43. The heat absorbed per 2.25 g of isopropyl alcohol, C3H8O, is 2.25 g C3H8O x
1 mol C3 H8 O 42.1 kJ x = 1.576 = 1.58 kJ 60.09 g C3 H8 O 1 mol C3 H8 O
11.44. For 39.3 g of butane, C4H10, the heat needed is 39.3 g C4H10 x
1 mol C4 H10 21.3 kJ x = 14.40 = 14.4 kJ 58.12 g C 4 H10 1 mol C 4 H10
11.45. Because all the heat released by freezing the water is used to evaporate the remaining water, you must first calculate the amount of heat released in the freezing: 9.31 g H 2 O x
mol H 2 O 6.01 kJ x = 3.10505 kJ 18.02 g H 2 O mol H 2 O
Finally, calculate the mass of H2O that was vaporized by the 3.10505 kJ of heat: 3.10505 kJ x
18.02 g H 2 O 1 mol H 2 O x = 1.246 = 1.25 g H2O 44.9 kJ 1 mol H 2 O
11.46. Enough ice must have been added so that the heat consumed in melting the ice is equal to the heat released in cooling the water from 21.0°C to 0.0°C. Heat released by cooling = [(33.6 g)(0.0°C − 21.0°C)(4.18 J/g•°C)] = −2.9494 x 103 J = −2.9494 kJ Convert this heat (2.9494 kJ) to the mass of ice melted: 2.9494 kJ x
1 mol H 2 O 18.02 g H 2 O x = 8.8433 = 8.84 g H2O 6.01 kJ 1 mol H 2 O
11.47. Calculate how much heat is released by cooling 64.3 g of H2O from 55°C to 15°C. ⎛ 4.18 J ⎞ 4 ⎟ = −1.07509 x 10 J = −10.7509 kJ • 1 g °C ⎝ ⎠
Heat released = (64.3 g)(15°C − 55°C) ⎜
The heat released is used first to melt the ice and then to warm the liquid from 0°C to 15°C. Let the mass of ice equal y grams. Then, for fusion, and for warming, we have Fusion: (y g H2O) x
1 mol H 2 O 6.01 kJ x = 0.3335 y kJ 18.02 g H 2 O 1 mol H 2 O
⎛ 4.18 J ⎞ ⎟ = 62.70 y J (0.06270 y kJ) ⎝ 1 g • °C ⎠
Warming: (y g H2O)(15°C − 0°C) ⎜
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Because the total heat required for melting and warming must equal the heat released by cooling, equate the two, and solve for y. 10.7509 kJ = 0.3335y kJ + 0.0627y kJ = y(0.3335 + 0.0627) kJ y = 10.7509 kJ ÷ 0.3962 kJ = 27.13 (grams)
Thus, 27 g of ice was added. 11.48. If all the steam condensed, the quantity of heat used to warm the water in the flask must equal the heat released by condensation and cooling of the steam. Thus, first find the quantity of heat used to warm the water in the flask. Heat required = (275 g)(83°C − 21°C)(4.18 J/(g•°C)) = 71.26 x 103 J (71.26 kJ) Now let y = the mass of the steam. Then find the quantity of heat released by condensation: y g H2O x
1 mol H 2 O  40.7 kJ x = −2.259 y kJ 18.02 g H 2 O 1 mol H 2 O
Now find the quantity of heat released by cooling y grams of H2O from 100°C to 83°C: Heat released. = y (83°C − 100°C) (4.18 J/(g•°C)) = −71.06y J (−0.07106y kJ) The total quantity of heat released in condensing and cooling the steam is equal in magnitude, but opposite in sign, to the quantity of heat required to warm the water in the flask: −71.26 kJ = −2.259y kJ + (−0.07106y) kJ y g steam =
71.26 kJ = 30.58 = 31 g (2.259 + 0.07106) kJ
Thus, 31 g of steam condensed. 11.49. At the normal boiling point, the vapor pressure of a liquid is 760.0 mmHg. Use the ClausiusClapeyron equation to find P2 when P1 = 760.0 mmHg, T1 = 334.85 K (61.7°C), and T2 = 309.35 K (36.2°C). Also use ΔHvap = 31.4 x 103 J/mol. ln
ΔH vap ⎛ 1 P2 1 ⎞ = ⎜ ⎟ R ⎝ T1 T2 ⎠ P1
ln
P2 31.4 x 103 J/mol ⎛ 1 1 ⎞ = ⎜ ⎟ = −0.93018 8.31 J/K • mol ⎝ 334.85 K 309.35 K ⎠ 760 mmHg
Taking antilogs of both sides gives P2 = e−0.93018 = 0.3944 760 mmHg
P2 = 0.3944 x 760 mmHg = 299.8 = 3.00 x 102 mmHg (300. mmHg)
11.50. At methanol's normal boiling point, its vapor pressure is 760.0 mmHg. Use the ClausiusClapeyron equation to calculate P2 when P1 = 760 mmHg, T1 = 338.2 K, and T2 = 295.2 K. Also use ΔHvap = 37.4 x 103 J/mol.
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ln
ΔH vap P2 = R P1
ln
P2 37.4 x 103 J/mol ⎛ 1 1 ⎞ = ⎜ ⎟ = −1.9384 8.31 J/K • mol ⎝ 338.2 K 295.2 K ⎠ 760 mmHg
⎛1 1 ⎞ ⎜ ⎟ T2 ⎠ ⎝ T1
Taking antilogs of both sides gives P2 = e−1.9384 = 0.14393 760 mmHg
P2 = 0.14393 x 760 mmHg = 109.38 = 109 mmHg
11.51. From the ClausiusClapeyron equation, ⎛ T2 T1 ⎞ ⎟ ⎝ T2  T1 ⎠
ΔHvap = R ⎜
⎡ P2 ⎤ ⎡ (553.2 K)(524.2 K) ⎤ ⎢ln ⎥ = [8.31 J/(K•mol)] ⎢ ⎥ ⎣ (553.2  524.2) K ⎦ ⎣ P1 ⎦
⎡ 760.0 mmHg ⎤ ⎢ln ⎥ ⎣ 400.0 mmHg ⎦
= 5.3336 x 104 J/ mol = 53.3 kJ/ mol 11.52. From the ClausiusClapeyron equation, ⎛ T2 T1 ⎞ ⎟ ⎝ T2  T1 ⎠
ΔHvap = R ⎜
⎡ P2 ⎤ ⎡ (319.7 K)(301.2 K) ⎤ ⎢ln ⎥ = [8.31 J/(K•mol)] ⎢ ⎥ ⎣ (319.7  301.2) K ⎦ ⎣ P1 ⎦
⎡ 760.0 mmHg ⎤ ⎢ln ⎥ ⎣ 400.0 mmHg ⎦
= 2.7762 x 104 J/ mol = 27.8 kJ/ mol 11.53. a.
At point A, the substance will be a gas.
b.
The substance will be a gas.
c.
The substance will be a liquid.
d.
No
11.54. a.
At point A, the substance will be a gas.
b.
The substance will be a solid.
c.
The substance will be a gas.
d.
No
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11.55. The phase diagram for oxygen is shown below. It is plotted from these points: triple point = −219°C, boiling point = −183°C, and critical point = −118°C. The graph is not drawn to scale. •
P ressure
50.1 atm
Critical point SOLID LIQU ID
1 atm •
1.10 mmHg
219°C
GA S
Triple point 183°C
118°C
Temperature
11.56. The phase diagram for argon below is plotted from these points: triple point = −189°C, boiling point = −186°C, and critical point = −122°C. The graph is not drawn to scale.
Pressure
48 atm
• Critical point
LIQUID 1 atm
0.68 mmHg
SOLID GAS • 189°C
Triple point 186°C
122°C
Temperature
11.57. Liquefied at 25°C: SO2 and C2H2. To liquefy CH4, lower its temperature below −82°C, and then compress it. To liquefy CO, lower its temperature below −140°C, and then compress it. 11.58. a.
If CF4 is in the tank, it's not in liquid form because the liquid phase cannot exist above −46°C.
b.
If C4H10 is in the tank, it's not in liquid form because 21°C is above its boiling point (1.0 atm).
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11.59. Br2 phase diagram: •
102 atm Pressure
Critical point SO LID
LIQUID GA S
44 mmHg
•
Triple point
7.3°C
315°C Temperature
a.
Circle "solid." The pressure of 40 mmHg is lower than the pressure at the triple point, so the liquid phase cannot exist.
b.
Circle "liquid." The pressure of 400 mmHg is above the triple point, so the gas will condense to a liquid.
11.60. Kr phase diagram:
Pressure
54 atm
133 mmHg
• Critical point SOLID
LIQUID GAS •
Triple point
169°C
63°C Temperature
a.
Circle "sublimes." The pressure of 130 mmHg is lower than the pressure at the triple point, so the liquid phase cannot exist.
b.
Circle "melts." The pressure of 760 mmHg is higher than the pressure at the triple point, so the solid melts to the liquid phase.
11.61. Yes, the heats of vaporization of 0.9, 5.6, and 20.4 kJ/mol (for H2, N2, and Cl2, respectively) increase in the order of the respective molecular masses of 2.016, 28.02, and 71.0. (London forces increase in order of increasing molecular mass.)
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11.62. The heats of vaporization of 1.8 kJ/mol for Ne and 6.8 kJ/mol for O2 increase in the order of the respective molecular masses of 20.1 and 32.0 because only London forces are involved. The heat of vaporization of 34.5 kJ/mol for methanol (molecular mass = 32.0) is higher than that of oxygen because of strong hydrogen bonding between H and O. 11.63. a.
London forces
b.
London and dipoledipole forces, hydrogen bonding
c.
London and dipoledipole forces
d.
London forces
11.64. Both c and d exhibit only London forces. Compound a is an ionic species. Thus, it has other forces present in addition to London forces. The substance depicted in b has a dipole moment as well as London forces. 11.65. The order is CCl4 < SiCl4 < GeCl4 (in order of increasing molecular mass). 11.66. The order is Ne < Kr < Xe (in order of increasing atomic mass). 11.67. CCl4 has the lowest vapor pressure because it has the largest molecular mass and the greatest London forces even though HCCl3 and H3CCl have dipoledipole interactions. 11.68. ClF has the smallest molecular mass and hence the smallest intermolecular forces and the highest vapor pressure. Thus, it should have the lowest boiling point. 11.69. The order of increasing vapor pressure is HOCH2CH2OH, FCH2CH2OH, FCH2CH2F. There is no hydrogen bonding in the third molecule; the second molecule can hydrogenbond at only one end; and the first molecule can hydrogenbond at both ends for the strongest interaction. 11.70. The order is HOCH2CH2CH2OH < CH3CH2CH2CH2OH < CH3CH2OCH2CH3. There is no hydrogen bonding in the third molecule; the second molecule can hydrogenbond at only one end; and the first molecule can hydrogenbond at both ends for the strongest interaction and lowest vapor pressure. 11.71. The order is CH4 < C2H6 < CH3OH < CH2OHCH2OH. The weakest forces are the London forces in CH4 and C2H6, which increase with molecular mass. The nextstrongest interaction is in CH3OH, which can hydrogenbond at only one end of the molecule. The strongest interaction is in the last molecule, which can hydrogenbond at both ends.
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11.72. The order is C2H6 < C3H8 < (CH3)3N < C4H9OH. The weakest forces are the London forces in C2H6, C3H8, and (CH3)3N, which increase with molecular mass. The strongest interaction is in C4H9OH, which can hydrogenbond from oxygen to hydrogen, a stronger force than in the other three molecules. 11.73. a.
Metallic
b.
Metallic
c.
Covalent network
d.
Molecular
e.
Ionic
11.74. a.
Ionic
b.
Ionic
c.
Molecular
d.
Molecular
e.
Molecular
11.75. a.
Metallic
b.
Covalent network (like diamond)
c.
Molecular
d.
Molecular
11.76. a.
Molecular
b.
Not molecular (ionic)
c.
Not molecular (metallic)
d.
Molecular
11.77. The order is (C2H5)2O < C4H9OH < KCl < CaO. Melting points increase in the order of attraction between molecules or ions in the solid state. Hydrogen bonding in C4H9OH causes it to melt at a higher temperature than (C2H5)2O. Both KCl and CaO are ionic solids with much stronger attraction than the organic molecules. In CaO, the higher charges cause the lattice energy to be higher than in KCl.
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11.78. The order is C2H6 < CH3OH < NaCl < Si. Melting points increase in the order of attraction between molecules or atoms. Hydrogen bonding in CH3OH causes it to melt at a higher temperature than C2H6. Because NaCl is an ionic solid, it melts at a higher temperature than either of the previous two. Silicon is a covalent network solid with the highest melting point. 11.79. a.
Lowmelting and brittle
b.
Highmelting, hard, and brittle
c.
Malleable and electrically conducting
d.
Hard and highmelting
11.80. a.
Metallic (from conductivity and luster)
b.
Covalent network (from highmelting, hard, and nonconducting liquid)
c.
Ionic (from high melting point and conducting liquid)
d.
Molecular (from low melting point and odor or vapor pressure at room temperature)
11.81. a.
LiCl
b.
SiC
c.
CHI3
d.
Co
11.82. a.
Pb
b.
CaCl2
c.
P4 S3
d.
BN
11.83. In a simple cubic lattice with one atom at each lattice point, there are atoms only at the corners of unit cells. Each corner is shared by eight unit cells, and there are eight corners per unit cell. Therefore, there is one atom per unit cell. 11.84. There are two atoms per unit cell, one from the corners and one atom at the center of the unit cell.
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11.85. Calculate the volume of the unit cell, change density to g/m3, and then convert volume to mass, using density: Volume = (2.866 x 10−10 m)3 = 2.354 x 20−29 m3 7.87 g ⎛ 100 cm ⎞ x ⎜ ⎟ 1 cm 3 ⎝ 1m ⎠
3
= 7.87 x 106 g/m3
Mass of one cell = (7.87 x 106 g/m3) x (2.354 x 10−29 m3) = 1.8526 x 10−22 g Because Fe is a bodycentered cubic cell, there are two Fe atoms in the cell, and Mass of one Fe atom = (1.8526 x 10−22 g) ÷ 2 = 9.263 x 10−23 g Using the molar mass to calculate the mass of one Fe atom, you find the agreement is good: 55.85 g Fe 1 mol Fe x = 9.2743 x 10−23 g/Fe atom 1 mol Fe 6.022 x 1023 Fe atoms
11.86. Calculate the volume of the unit cell, change density to g/m3, and then convert volume to mass, using density: Volume = (3.524 x 10−10 m)3 = 4.376 x 20−29 m3 8.91 g ⎛ 100 cm ⎞ x ⎜ ⎟ 1 cm3 ⎝ 1m ⎠
3
= 8.91 x 106 g/m3
Mass of one cell = (8.91 x 106 g/m3) x (4.376 x 10−29 m3) = 3.899 x 10−22 g Because Ni is a facecentered unit cell, there are four Ni atoms per cell, and Mass of one Ni atom = (3.899 x 10−22 g) ÷ 4 = 9.747 x 20−23 g Using the molar mass to calculate Avogadro's number, you obtain 58.70 g Ni 1 atom Ni x = 6.022 x 1023 atoms/mol 23 1 mol Ni 9.747 x 10 g Ni
11.87. There are four Cu atoms in the facecentered cubic structure, so the mass of one cell is 1 mol Cu 63.5 g Cu x = 4.218 x 10−22 g 6.022 x 1023 Cu atoms 1 mol Cu
4 Cu atoms x Cell volume =
4.218 x 1022 g = 4.723 x 10−23 cm3 8.93 g/cm3
All edges are the same length in a cubic cell, so the edge length, l, is l =
3
V =
3
4.723 x 1023 cm3 = 3.614 x 10−8
= 3.61 x 10−8 cm (361 pm)
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11.88. There are two Ba atoms in the facecentered cubic structure, so the mass of one cell is 1 mol Ba 137.33 g Ba x = 4.561 x 10−22 g 23 6.022 x 10 Ba atoms 1 mol Ba
2 Ba atoms x Cell volume =
4.561 x 1022 g = 1.2994 x 10−22 cm3 3 3.51 g/cm
All edges are the same length in a cubic cell, so the edge length, l, is l =
3
V =
3
1.2994 x 1022 cm3 = 5.0651 x 10−8
= 5.07 x 10−8 cm (507 pm) 11.89. Calculate the volume from the edge length of 407.9 pm (4.079 x 10−8 cm), and then use it to calculate the mass of the unit cell: Cell volume = (4.079 x 10−8 cm3 = 6.7869 x 10−23 cm3 Cell mass = (19.3 g/cm3)(6.7869 x 20−23 cm3) = 1.3098 x 10−21 g Calculate the mass of one gold atom: 1 Au atom x
1 mol Au 196.97 g Au x = 3.2708 x 10−22 g 23 6.022 x 10 Au atoms 1 mol Au
1.3098 x 1021 g 1 Au atom x 1 unit cell 3.2708 x 1022 g Au
=
4.004 Au atoms unit cell
Since there are four atoms per unit cell, it is a facecentered cubic. 11.90. Calculate the volume from the edge (288.5 pm = 2.885 x 10−8 cm). Use it to calculate the mass: Cell volume = (2.885 x 10−8 cm)3 = 2.401 x 10−23 cm3 Cell mass = (7.20 g/cm3)(2.401 x 10−23 cm3) = 1.729 x 10−22 g Calculate the mass of one chromium atom: 1 Cr atom x
1 mol Cr 51.996 g Cr x = 8.634 x 10−23 g Cr 23 6.022 x 10 Au atoms 1 mol Cr
2.002 Cr atoms 1.729 x 1022 g 1 Cr atom x = 1 unit cell 1 unit cell 8.634 x 1023 g Cr
Since there are two atoms per unit cell, it is a bodycentered cubic.
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11.91. Calculate the volume from the edge (316.5 pm = 3.165 x 10−8 cm). Use it to calculate the mass: Cell volume = (3.165 x 10−8 cm)3 = 3.1705 x 10−23 cm3 For a bodycentered cubic lattice, there are two atoms per cell, so their mass is 2 W atoms x Density =
1 mol W 183.8 g W x = 6.1043 x 10−22 g W 6.022 x 1023 W atoms 1 mol W
6.1043 x 1022 g W = 19.253 = 19.25 g/cm3 3.1705 x 1023 cm 3
11.92. Calculate the volume from the edge (495.0 pm = 4.950 x 10−8 cm). Use it to calculate the mass: Cell volume = (4.950 x 10−8 cm)3 = 1.2129 x 10−22 cm3 For a facecentered cubic lattice, there are four atoms per cell, so their mass is 4 Pb atoms x Density =
1 mol Pb 207.2 g Pb x = 1.3763 x 10−21 g Pb 6.022 x 10 23 Pb atoms 1 mol Pb
1.3763 x 1021 g Pb = 11.347 = 11.35 g/cm3 1.2129 x 1022 cm3
11.93. Use Avogadro's number to calculate the number of atoms in 1.74 g (= d x 1.000 cm3): 1.74 g Mg x
1 mol Mg 6.022 x 10 23 Mg atoms x 1 mol Mg 24.305 g Mg
= 4.311 x 1022 Mg atoms Because the space occupied by the Mg atoms = 0.741 cm3, each atom's volume is Volume 1 Mg atom = Volume =
0.741 cm3 = 1.719 x 10−23 cm3 4.311 x 1022 Mg atoms
4πr 3 3
so r =
3
3V = 4π
3
3 (1.719 x 1023 cm 3 ) = 1.601 x 10−8 = 1.60 x 10−8 cm 4π
= 1.60 x 102 pm
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11.94. Use Avogadro's number to calculate the number of atoms in 3.51 g (= d x 1.000 cm3): 1 mol Ba 6.022 x 10 23 Ba atoms x 1 mol Ba 137.33 g Ba
3.51 g Ba x
= 1.539 x 1022 Ba atoms Because the space occupied by the Ba atoms = 0.680 cm3, each atom's volume is Volume one Ba atom = Volume =
0.680 cm3 = 4.418 x 10−23 cm3 22 1.539 x 10 Ba atoms
4πr 3 3
so r =
■
3
3V = 4π
3
3 (4.418 x 1023 cm3 ) = 2.193 x 10−8 = 2.19 x 10−8 cm = 219 pm 4π
SOLUTIONS TO GENERAL PROBLEMS
11.95. Water vapor deposits directly to solid water (frost) without forming liquid water. After heating, most of the frost melted to liquid water, which then vaporized to water vapor. Some of the frost may have sublimed directly to water vapor. 11.96. Water vapor condenses directly to solid water (snow) in the upper atmosphere. After falling through the warm air mass, the snow melts to liquid water (rain). After falling on a sunny spot, the rain is vaporized to water vapor. 11.97. From Table 5.6, the vapor pressures are 18.7 mmHg at 21°C and 12.8 mmHg at 15°C. If the moisture did not begin to condense until the air had been cooled to 15°C, then the partial pressure of water in the air at 21°C must have been 12.8 mmHg. The relative humidity is Percent relative humidity =
12.8 mmHg x 100% = 68.44 = 68.4 percent 18.7 mmHg
11.98. The vapor pressure of water at 21°C is 18.7 mmHg (Table 5.6). Therefore, Percent relative humidity = 58 percent =
x mmHg x 100%; 18.7 mmHg
x = 10.8 mmHg The partial pressure of water in the air at 21°C is thus 10.8 mmHg. The water will begin to condense when the temperature drops to the temperature at which the vapor pressure of water is 10.8 mmHg. This temperature is between 12°C and 13°C (Table 5.5).
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11.99. After labeling the problem data as shown below, use the ClausiusClapeyron equation to obtain ΔHvap, which can then be used to calculate the boiling point. At T1 = 299.3 K, P1 = 100.0 mmHg; at T2 = 333.8 K, P2 = 400.0 mmHg ln
ΔH vap 400.0 mmHg ⎡ 333.8 K  299.3 K ⎤ = 8.31 J / (K • mol) ⎢⎣ 333.8 K x 299.3 K ⎥⎦ 100.0 mmHg
1.3862 = ΔHvap (4.1535 x 10−5 mol/J) ΔHvap = 33.37 x 103 J/mol (33.4 kJ/mol)
Now, use this value of ΔHvap and the following data to calculate the boiling point: At T1 = 299.3 K, P1 = 100.0 mmHg; at T2 (boiling pt.), P2 = 760 mmHg ln
33.4 x 103 J/mol ⎡ 1 1 ⎤ 760.0 mmHg = ⎢ ⎥ 8.314 J / (K • mol) ⎣ 299.3 K T2 ⎦ 100.0 mmHg
⎡
1 ⎤
1
2.0281 = 4.0144 x 103 K ⎢ ⎥ T2 ⎦ ⎣ 299.3 K 1 2.0281 1 = = 2.8359 x 10−3/K 299.3 K 4.0144 x 103 K T2
T2 = 352.6 = 353 K (80°C)
11.100. After labeling the problem data as shown below, use the ClausiusClapeyron equation to obtain ΔHvap, which can then be used to calculate the boiling point. At T1 = 293.2 K, P1 = 17.5 mmHg; at T2 = 353.2 K, P2 = 355.1 mmHg ln
ΔH vap 355.1 mmHg ⎡ 353.2 K  293.2 K ⎤ = 8.31 J / (K • mol) ⎢⎣ 353.2 K x 293.2 K ⎥⎦ 17.5 mmHg
3.0101 = ΔHvap (6.9687 x 10−5 mol/J) ΔHvap = 43.19 x 103 J/mol (43.2 kJ/mol)
Now, use this value of ΔHvap and the following data to calculate the boiling point: At T1 = 293.32 K, P1 = 17.5 mmHg; at T2 (boiling pt.), P2 = 760 mmHg ln
43.2 x 103 J/mol ⎡ 1 1 ⎤ 760.0 mmHg = ⎢ ⎥ 8.314 J / (K • mol) ⎣ 293.2 K T2 ⎦ 17.5 mmHg
⎡
1
1 ⎤
3.7711 = 5.1955 x 103 K ⎢ ⎥ T2 ⎦ ⎣ 293.2 K 1 3.7711 1 = = 2.6848 x 10−3/K 293.2 K 5.1955 x 103 K T2
T2 = 372.46 = 372 K (99°C)
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11.101. a.
As this gas is compressed at 20°C, it will condense into a liquid because 20°C is above the triple point but below the critical point.
b.
As this gas is compressed at −70°C, it will condense directly to the solid phase because the temperature of −70°C is below the triple point.
c.
As this gas is compressed at 40°C, it will not condense because 40°C is above the critical point.
11.102. a.
As I2 vapor is cooled at 120 atm, no change to a distinct liquid will be observed, but the I2 will condense to a solid phase at some definite temperature.
b.
As I2 vapor is cooled at 1 atm, the vapor will condense to a liquid and then freeze to the solid phase.
c.
As I2 vapor is cooled at 50 mmHg (below the triple point), the vapor will condense directly to the solid phase without going through the liquid phase.
11.103. In propanol, hydrogen bonding exists between the hydrogen of the OH group and the lone pair of electrons of oxygen of the OH group of an adjacent propanol molecule. For two adjacent propanol molecules, the hydrogen bond may be represented as follows: C3H7 –O–H•••O(H)C3H7 11.104. In hydrogen peroxide, hydrogen bonding exists between any hydrogen and the lone pair of electrons of oxygen of an adjacent hydrogen peroxide. For two adjacent hydrogen peroxide molecules, the hydrogen bond may be represented as follows: HO–O–H•••O(H)–OH 11.105. Ethylene glycol molecules are capable of hydrogen bonding to each other, whereas pentane molecules are not. The greater intermolecular forces in ethylene glycol are reflected in greater resistance to flow (viscosity) and high boiling point. 11.106. Pentylamine molecules are capable of hydrogen bonding to each other, but triethylamine molecules are not. The greater intermolecular forces in pentylamine cause a higher boiling point and greater resistance to flow. 11.107. Aluminum (Group IIIA) forms a metallic solid. Silicon (Group IVA) forms a covalent network solid. Phosphorus (Group VA) forms a molecular solid. Sulfur (Group VIA) forms a molecular (amorphous) solid. 11.108. AlF3 forms an ionic solid. SiF4, PF3, and SF4 form molecular solids.
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11.109. a.
Lower: KCl. The lattice energy should be lower for ions with a lower charge. A lower lattice energy implies a lower melting point.
b.
Lower: CCl4. Both are molecular solids, so the compound with the lower molecular mass should have weaker London forces and, therefore, the lower melting point.
c.
Lower: Zn. Melting points for Group IIB metals are lower than for metals near the middle of the transitionmetal series.
d.
Lower: C2H5Cl. Ethyl chloride cannot hydrogenbond, but acetic acid can. The compound with the weaker intermolecular forces has the lower melting point.
11.110. a.
Lower: C6H14. A molecular solid has a lower melting point than an ionic solid.
b.
Lower: 1propanol. The 1propanol can hydrogenbond at only one end; ethylene glycol can hydrogenbond at both ends, so its intermolecular forces are stronger.
c.
Lower: Na. Sodium is a metallic solid, but Si is a covalent network solid (high melting point).
d.
Lower: CH4. Both form molecular solids, but CH4 has a lower molecular mass with weaker London forces and, therefore, the lower melting point.
11.111. The facecentered cubic structure means one atom is at each lattice point. All edges are the same length in such a structure, so the volume is Volume = l3 = (3.839 x 10−8 cm)3 = 5.6579 x 10−23 cm3 Mass of unit cell = dV = (22.42 g/cm3)(5.6579 x 10−23 cm3) = 1.2685 x 10−21 g There are four atoms in a facecentered cubic cell, so Mass of one Ir atom = mass of unit cell ÷ 4 = (1.2685 x 10−21 g) ÷ 4 = 3.1712 x 10−22 g Molar mass of Ir = (3.1712 x 10−22 g/Ir atom) x (6.022 x 1023 Ir atoms/mol) = 190.96 = 191.0 g/mol (The atomic mass = 191.0 amu.) 11.112. The bodycentered cubic structure means that one atom is at each lattice point. All edges are the same length in such a structure, so the volume is Volume = l3 = (3.306 x 10−8 cm)3 = 3.6133 x 10−23 cm3 Mass of unit cell = dV = (16.69 g/cm3)(3.6133 x 10−23 cm3) = (6.0306 x 10−22 g) There are two atoms in a bodycentered cubic cell, so Mass of one Ta atom = mass of unit cell ÷ 2 = (6.0306 x 10−22 g) ÷ 2 = 3.0153 x 10−22 g Molar mass of Ta = (3.0153 x 10−22 g/Ta atom) x (6.022 x 1023 Ta atoms/ mol) = 181.58 = 181.6 g/mol (The atomic mass = 181.6 amu.)
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11.113. From Problem 11.87, the cell edge length (l) is 361.4 pm. There are four copper atom radii along the diagonal of a unitcell face. Because the diagonal square = l 2 + l 2 (Pythagorean theorem), 4r =
2 l2 =
2 (361.4 pm) = 127.7 = 128 pm 4
2 l, or r =
11.114. A bodycentered cubic cell has two atoms per cell. The mass of the unit cell is 2 Rb atoms x
1 mol Rb 85.468 g Rb x = 2.8385 x 10−22 g 23 6.022 x 10 Rb atoms 1 mol Rb
The volume of the unit cell is Volume = l =
3
2.8385 x 1022 g = 1.8528 x 10−22 cm3 1.532 g/cm3
Volume =
3
1.8528 x 1022 cm 3 = 5.7009 x 10−8 cm
Because the corner spheres touch the bodycentered sphere, the length of the body diagonal (diagonal passing through the center of the cell) must be four times the radius of the Rb atom. Also, from the geometry of a cube and the Pythagorean theorem, the square of the body diagonal equals l 2 + d 2 (l is the unitcell edge length, and d is the diagonal along a face of the unit cell). Because d 2 = l 2 + l 2, or d =
2 l, you can write
(Body diagonal)2 = l 2 + d 2 = l 2 + 2l 2 = 3l 2 Body diagonal =
3l =
3 (5.7009 x 10−8 cm) = 9.8742 x 10−8 cm
Radius of Rb atom = (9.8742 x 10−8 cm) ÷ 4 = 2.4686 x 10−8 cm (246.9 pm) 11.115. The body diagonal (diagonal passing through the center of the cell) is four times the radius, r, of a sphere. Also, from the geometry of a cube and the Pythagorean theorem, the body diagonal equals 3 l, where l is the edge length of the unit cell. Thus 4r =
3l
or l =
4r 3
Because the unit cell contains two spheres, the volume occupied by the spheres is Vspheres = 2 x
4 3 πr 3
and Vcell = l3 =
⎡ 4r ⎤ ⎢ ⎥ ⎣ 3⎦
3
=
64r 3 3 3
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Chapter 11: States of Matter; Liquids and Solids
Finally, to obtain the percent volume of the cell occupied, divide Vspheres by Vcell:
Percent V =
Vspheres Vcell
⎡ 4πr 3 ⎤ 2⎢ 3 ⎥ π 3 x 100% = ⎣ 3 ⎦ x 100% = x 100% 8 64r 3 3
= 68.01 = 68.0% 11.116. Because the spheres touch along the diagonal of a face, d, the radius, r, of the spheres is r = d/4 = l( 2 )/4
or l =
4r 2 ⎡ 4r ⎤ ⎥ ⎣ 2⎦
3
Vcell = l3 = ⎢
For a facecentered cubic structure, there are four spheres per cell, so the volume, Vspheres, occupied by the spheres is ⎡ 4πr 3 ⎤
Vspheres = 4 ⎢ ⎥ ⎣ 3 ⎦
Percent V =
Vspheres Vcell
⎡ 4πr 3 ⎤ 4 ⎢ ⎥ ⎣ 3 ⎦ x 100% = π 2 x 100% x 100% = 3 6 ⎡ 4r ⎤ ⎢ ⎥ ⎣ 2⎦
= 74.04 = 74 percent 11.117. a.
The boiling point increases as the size (number of electrons in the atom or molecule) increases. The London forces or dispersion forces increase.
b.
Hydrogen bonding occurs between the H–F molecules and is much stronger than the London forces.
c.
In addition to the dispersion forces, the hydrogen halides are polar (have dipole moments), so there are dipoledipole interactions.
11.118. a.
The boiling point increases as the size (number of electrons in the molecule) increases. The London forces or dispersion forces increase.
b.
Hydrogen bonding occurs among the NH3 molecules.
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c.
405
The nitrogen family compounds consist of polar molecules, whereas the carbon family molecules are nonpolar. Thus the nitrogen family has dipoledipole interactions as well as the contributions from the dispersion forces.
11.119. a.
Diamond and silicon carbide are covalent network solids with strong covalent bonds between all the atoms. Graphite is a layered structure, and the forces holding the layers together are weak dispersion forces.
b.
Silicon dioxide is a giant molecule with an infinite array of O–S–O bonds. Each silicon is bonded to four oxygen atoms in a covalent network solid. Carbon dioxide is a discrete, nonpolar molecule.
11.120. The first member atoms are very small and highly electronegative. Oxygen readily forms double bonds, whereas the larger atoms have difficulty getting close enough together to form multiple bonds. Hence, we have O2 vs. S8. Because oxygen has a higher attraction for hydrogen, the dipole moment is so large that it is possible to have hydrogen bonding in H2O, but there is no hydrogen bonding in H2S. 11.121. a.
CO2 consists of discrete nonpolar molecules that are held together in the solid by weak dispersion forces. SiO2 is a covalent network solid with all the atoms held together by strong covalent bonds.
b.
HF(l) has extensive hydrogen bonding among the molecules. HCl(l) boils much lower because it doesn't have hydrogen bonding.
c.
SiF4 is a larger molecule (it has more electrons than CF4), so it has stronger dispersion forces and a higher boiling point than CF4. Both molecules are tetrahadrally symmetrical and, therefore, nonpolar.
11.122.
H
H
Li+ H: a.
N
H
H
H
C
H
O
C
O
H
b.
LiH has the highest boiling point because of the ionic bonding and crystalline lattice of LiH.
c.
CH4 has the lowest boiling point because it is a small, nonpolar molecule. The only intermolecular forces are dispersion forces.
d.
NH3 is a polar molecule that also has hydrogen bonding between the molecules.
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11.123. In a tank of carbon dioxide under pressure (for example, in a CO2 fire extinguisher), the substance normally exists as the liquid in equilibrium with its gas phase. At 35°C (above 31°C), the two phases, gas and liquid, are replaced by a single fluid phase. Thus the carbon dioxide in the tank is above its critical temperature and pressure and exists as the supercritical fluid. 11.124. Under normal conditions, carbon dioxide is not a very good solvent for organic substances, but supercritical carbon dioxide readily dissolves many of these substances, including caffeine. It is nontoxic and nonflammable. It also has no effect on the stratospheric ozone layer, whereas methylene chloride does. Carbon dioxide does contribute to the greenhouse effect, but the gas once used can be recirculated for solvent use and not vented to the atmosphere. 11.125. Van der Waals forces exist between any two surfaces but are extremely weak unless relatively large areas of the two surfaces come quite close together. The toe of a gecko is covered with fine hairs, each hair having a thousand or so split ends. As the gecko walks across a surface, it presses these stalks of hairs against the surface. The intimate contact of a billion or so split ends of hairs with the surface results in a large attractive force, holding the gecko fast. This is not true for your finger, where the contact between the surfaces is much less. 11.126. As the gecko walks, its foot naturally bends in such a way that the hairs at the back edge of its toes disengage, row after row, until the toe is free. It is the mechanics of the gecko’s walk that allow it to connect to, and disengage easily from, a surface. 11.127. The nematic liquid crystals consist of rodlike, polar molecules in a phase that is intermediate in order between that of a liquid and that of a crystalline solid. The molecules tend to align or orient themselves in the same direction along their long axes, like matches in a matchbox, but they occupy random positions within the substance. Thus, nematic liquid crystals have orientation order but no position order. 11.128. The molecular order in a nematic liquid crystal, which results from weak intermolecular forces, is easily disrupted. And because these molecules are polar, they can be reoriented by the application of an electric field. A liquidcrystal display (LCD) uses this ease of molecular reorientation by an electric field to change small areas, or pixels, of the display from light to dark. An image is formed by turning various screen pixels on and others off by means of software. 11.129. The Bragg equation, nλ = 2d sin θ, relates the wavelength of x rays, λ, to the distance between atomic planes, d, and the angle of reflection, θ. Note that reflections occur at several angles, corresponding to different values of n (n = 1, 2, 3, …). 11.130. A molecular crystal has many different atomic planes, so it reflects an xray beam in many different directions. By analyzing the intensities and angular directions of the reflected beams, you can determine the exact positions of all the atoms in the unit cell of the crystal and therefore obtain the structure of the molecule.
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11.131. Water has several unusual properties that set it apart from other substances. For example, its solid phase, ice, is less dense than liquid water, whereas for most substances the solid phase is more dense than the liquid. In addition, water has an unusually large heat capacity. The unusual properties of water are largely linked to its ability to form hydrogen bonds. For example, ice is less dense than liquid water because ice has an open, hydrogenbonded structure. Each oxygen atom in the structure of ice is surrounded tetrahedrally by four hydrogen atoms: two that are close and covalently bonded to give the H2O molecule, and two others that are farther away and held by hydrogen bonds. The tetrahedral angles give rise to a threedimensional structure that contains open space. The solvent properties of water are also unusual. Water is both a polar substance and a hydrogenbonding molecule. As a result, water dissolves many substances, including ionic and polar compounds. These properties make water the premier solvent, biologically and industrially. 11.132. The fact that ice is less dense than water means that it forms on top of the liquid when freezing occurs. This has farreaching effects, both for weather and for aquatic animals. When ice forms on a body of water, it insulates the underlying water from the cold air and limits further freezing. Fish depend on this for winter survival. Consider what would happen to a lake if ice were more dense than water. The ice would freeze from the bottom of the lake upward. Without the insulating effect at the surface, the lake could well freeze solid, killing the fish. Spring thaw would be prolonged, because the insulating effect of the surface water would make it take much longer for the ice at the bottom of a lake to melt.
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SOLUTIONS TO STRATEGY PROBLEMS
11.133. PV (750.0/760 atm)(20.00 L) = = 0.820477 = 0.8205 mol N2 RT (0.082058 L • atm/K • mol)(293.15 K)
a.
n =
b.
353.6 x 10−3 g x
1 mol H 2 O = 0.0196277 = 0.01963 mol H2O 18.0153 g
The water evaporated and was pushed out of the flask by the nitrogen. c.
During the experiment, the total number of moles of gas that exited the flask is ntot = 0.820477 mol N2 + 0.0196277 mol H2O = 0.84010 mol
The partial pressure of nitrogen is PN2 = X N2 Ptot =
d.
0.820477 mol N 2 x 750 0 mmHg = 732.47 = 732.5 mmHg 0.84010 mol
The vapor pressure of water is PH2 O = Ptot − PN2 = 750.0 mmHg − 732.5 mmHg = 17.5 mmHg.
The appendix lists the vapor pressure of water at 20°C as 17.5 mmHg, so the calculated value agrees with this value.
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11.134. At 30.0°C, the vapor pressure of water is 31.8 mmHg. With a relative humidity of 80.0%, the number of moles of water vapor in a 1.00L flask of this air is n =
PV (0.80)(31.8/760 atm)(1.00 L) = = 0.0013456 mol RT (0.082058 L • atm/K • mol)(303.15 K)
The mass of this water is Mass = 0.0013456 mol x
18.0153 g = 0.02424 g H2O 1 mol H 2 O
At 5°C, the vapor pressure of water is 6.5 mmHg. The mass of water that still exists in the gas phase at this temperature is Mass =
(6.5/760 atm)(1.00 L) 18.0153 g x = 0.006750 g H2O (0.082058 L • atm/K • mol)(278.15 K) 1 mol H 2 O
The mass of liquid water that condensed out is Mass = 0.02424 g − 0.006750 g = 0.01749 = 0.0175 g 11.135. There are three triple points in the phase diagram for sulfur. The triple point at 95°C and 1 x 10−5 atm has the phases rhombic solid, monoclinic solid, and vapor in equilibrium. At 119°C and 6 x 10−5 atm, the phases present are monoclinic solid, liquid, and vapor. The triple point at 151°C and 1290 atm has the phases rhombic solid, monoclinic solid, and liquid in equilibrium. If you cooled liquid sulfur, the solid phase that would freeze out would be monoclinic. It is not the normal phase of solid sulfur, which is rhombic. The rhombic form of solid sulfur cannot form from the liquid except at very high pressures. Only the triple point at 151°C and 1290 atm has the two phases in equilibrium with each other. 11.136. Even though the liquid state is the stable state of water below 100°C, water vapor is present at all temperatures between the triple point and the critical point. At 20°C, liquid water will evaporate until the vapor pressure is reached. In an open container, the vapor pressure is never reached, so the water evaporates completely. The statement at the beginning of this problem is inaccurate because it describes a system undergoing change and not a system in a closed container at equilibrium. 11.137. a.
At room temperature, carbon dioxide is below its critical temperature, so it would be present in two different phases: liquid and gas. Thus, inside the tank you would expect to see a liquid phase separated from a gaseous phase by a meniscus.
b.
The pressure of carbon dioxide would remain constant at the vapor pressure at 20°C for the gas. It would remain the same as long as liquid carbon dioxide was present. As soon as all of the liquid carbon dioxide disappeared, the pressure of the gas would begin to drop, eventually reaching a gauge pressure of zero.
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11.138. a.
In 1pentanol, the intermolecular forces present are London forces, dipoledipole forces, and hydrogen bonding. In hexane, only London forces are present.
b.
Since boiling point increases with increasing intermolecular forces, 1pentanol would be expected to have the higher boiling point. Thus the boiling point of hexane would be 69°C, the lower value.
c.
Viscosity increases with increasing intermolecular forces. Since 1pentanol has the greater intermolecular forces, its viscosity should be 2.987 g/(cm.s), and hexane, with the weaker intermolecular forces, should have a viscosity of 0.313 g/(cm.s).
d.
The compound CH3CH2CH2CH2CH2Cl has London forces and dipoledipole forces, so its boiling point should be intermediate between those of hexane and 1pentanol.
11.139. a.
CH3CHO is a polar molecule, so it has London forces and dipoledipole forces present. CH3CH2CH3 is a nonpolar molecule and has only London forces present. CH3CH2OH is a polar molecule with London forces, dipoledipole forces, and hydrogen bonding.
b.
The heat of vaporization increases with increasing intermolecular forces. Since CH3CHO has stronger intermolecular forces than CH3CH2CH3 and weaker intermolecular forces than CH3CH2OH, you would expect it to have a heat of vaporization of 25.8 kJ/mol.
c.
Use the Clapeyron equation to find the vapor pressure at 15°C. ln
ΔH vap ⎛ 1 P2 1⎞ = ⎜ ⎟ R ⎝ T1 T2 ⎠ P1
ln
P2 25.8 x 103 J/mol ⎛ 1 1 ⎞ = ⎜ ⎟ = −0.2198 8.31 J/K • mol ⎝ 294 K 288 K ⎠ 760 mmHg
Taking antilogs of both sides gives P2 = e−0.2198 = 0.80260 760 mmHg
P2 = 0.80260 x 760 mmHg = 609.97 = 610. mmHg
11.140. Use the Clapeyron equation to find the vapor pressure at 15°C. ln
P2 32 x 103 J/mol ⎛ 1 1 ⎞ = ⎜ ⎟ = −2.4928 8.31 J/K • mol ⎝ 354 K 288 K ⎠ 760 mmHg
Taking antilogs of both sides gives P2 = e−2.4928 = 0.082673 760 mmHg
P2 = 0.082673 x 760 mmHg = 62.9 = 63 mmHg
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Chapter 11: States of Matter; Liquids and Solids
11.141. a.
The rhenium atoms (smaller spheres) are occupying each corner in the unit cell, each atom contributing 1/8 rhenium atom to the unit cell. This gives 1/8(8), or one, rhenium atom per unit cell. The oxygen atoms (larger spheres) are occupying each edge in the unit cell. There are twelve edges in the cell, and each atom on an edge contributes 1/4 oxygen atom to the unit cell. This gives 1/4(12), or three, oxygen atoms per unit cell.
b.
Each unit cell has one rhenium atom and three oxygen atoms for a formula unit of ReO3, which is red.
11.142.
■
a.
Since metal cations are smaller than nonmetal anions, the red spheres represent the metal ions.
b.
The red spheres occupy each corner and each face of the unit cell. This gives 1/8(8) + 1/2(6) = 4 metal atoms per unit cell. Each of the green spheres is internal and is totally within the unit cell. This gives 8 nonmetal atoms per unit cell.
c.
Since the unit cell has four metal atoms and eight nonmetal atoms, the simplest formula is MX2. The formula of the cation is M2+ and the formula of the anion is X−.
d.
An example of a compound with this formula is CaF2. It has the proper mole ratio and is an ionic compound. The red spheres would represent Ca2+.
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
11.143. Use the ideal gas law to calculate n, the number of moles of N2: N2: n =
PV (745/760 atm)(5.40 L) = = 0.2201 mol RT (0.082057 L • atm/K • mol)(293 K)
C3H8O: n = 0.6149 g C3H8O x X C3 H8 O =
1 mol C3 H8 O = 0.010232 mol 60.094 g C3 H8 O
0.010232 mol C3 H8 O = 0.04442 mole fraction (0.010232 mol + 0.2201 mol)
Partial P = 0.04442 x 745 mmHg = 33.093 mmHg = 33.1 mmHg Vapor pressure of C3H8O = 33.1 mmHg
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411
11.144. Use the ideal gas law to calculate n, the number of moles of N2: N2: n =
PV (768/760 atm)(6.35 L) = = 0.2624 mol RT (0.082057 L • atm/K • mol)(298 K)
C3H6O: n = 6.550 g C3H6O x X C3 H 6 O =
1 mol C3 H 6 O = 0.11283 mol 58.05 g C3 H 6 O
0.11283 mol C3 H 6 O = 0.3007 mole fraction (0.11283 mol + 0.2624 mol)
Partial P = 0.3007 x 768 mmHg = 230.9 mmHg = 231 mmHg Vapor pressure of C3H6O = 231 mmHg 11.145. Calculate the moles of HCN in 10.0 mL of the solution (density = 0.687 g HCN/mL HCN): 10.0 mL HCN x
0.687 g HCN 1 mol HCN x = 0.2541 mol HCN 1 mL HCN 27.03 g HCN
0.2541 mol HCN(l)
→
(ΔHf° = 108.9 kJ/mol)
0.2541 mol HCN(g) (ΔHf° = 135 kJ/mol)
ΔH° = 0.2541 mol x [135 kJ/mol − 108.9 kJ/mol] = 6.633 = 6.6 kJ
11.146. Calculate moles of CH3OH in the 20.0mL solution (density = 0.787 g CH3OH/mL CH3OH): 20.0 mL CH3OH x
0.787 g CH 3 OH 1 mol CH 3 OH x = 0.49126 mol CH3OH 1 mL CH 3 OH 32.04 g CH 3 OH
0.49126 mol CH3OH(l) (ΔHf° = −238.7 kJ/mol)
→0.49126 mol CH3OH(g)
(ΔHf° = −200.7 kJ/mol)
ΔH° = 0.49126 mol x [−200.7 kJ/mol − (−238.7 kJ/mol)] = 18.66 = 18.7 kJ
11.147. First, convert the mass to moles; then multiply by the standard heat of formation to obtain the heat absorbed in vaporizing this mass: 12.5 g P4 x
1 mol P4 = 0.1009 mol P4 123.88 g P4
0.1009 mol P4 x
95.4 J x (44.1°C − 25.0°C) = 183.86 J = 0.18386 kJ °C • mol P4
2.63 kJ/mol P4 x 0.1009 mol P4 = 0.26537 kJ Total heat = 0.26537 kJ + 0.18386 kJ = 0.44923 = 0.449 kJ = 449 J
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Chapter 11: States of Matter; Liquids and Solids
11.148. First, convert the mass to moles; then multiply by the standard heat of formation to obtain the heat absorbed in vaporizing this mass: 25.0 g Na x
1 mol Na = 1.0874 mol Na 22.99 g Na
1.0874 mol Na x
28.2 J °C • mol Na
x (97.8°C − 25.0°C) = 2232 J (2232 kJ)
2.60 kJ/mol Na x 1.0874 mol Na = 2.8273 kJ Total heat = 2.8273 kJ + 2.232 kJ = 5.0597 = 5.06 kJ 11.149. Use the ideal gas law to calculate the total number of moles of monomer and dimer: n =
PV (436/760)atm x 1.000 L = RT [0.082057 L • atm/(K • mol)] x 373.75 K
= 0.0187057 mol monomer + dimer (0.0187057 mol monomer + dimer) x
0.630 mol dimer = 0.01178 mol dimer 1 mol dimer + monomer
0.0187057 mol both − 0.01178 mol dimer = 0.00692 mol monomer Mass dimer = 0.01178 mol dimer x
120.1 g dimer = 1.414 g dimer 1 mol dimer
Mass monomer = 0.00692 mol monomer x Density =
60.05 g monomer = 0.4155 g monomer 1 mol monomer
1.414 g + 0.4155 g = 1.829 = 1.83 g/L vapor 1.000 L
11.150. Use the ideal gas law to calculate the total number of moles of monomer and dimer: n =
PV (146/760)atm x 1.000 L = = 0.006796 mol monomer + dimer RT [0.082057 L • atm/(K • mol)] x 344.45 K
Let Xd = mole fraction of dimer; then write one equation in one unknown based on 1.000 L of gas: 0.702 g both = 0.006796 mol Xd (120.1 g/mol) + 0.006796 (1 − Xd) (60.05 g/mol) 0.702 g = 0.8162 Xd + 0.4081 − 0.4081Xd Xd =
(0.702  0.4081) = 0.7201 = 0.720 0.4081
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CHAPTER 12
Solutions
■
SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 12.1. An example of a solid solution prepared from a liquid and a solid is a dental filling made of liquid mercury and solid silver. 12.2. The C4H9OH molecules will be more soluble in water because their –OH ends can form hydrogen bonds with water. 12.3. The Na+ ion has a larger energy of hydration because its ionic radius is smaller, giving Na+ a more concentrated electric field than K+. 12.4. Write Henry's law (S = kHP) for 159 mmHg (P2), and divide it by Henry's law for 1 atm, or 760 mmHg (P1). Then substitute the experimental values of P1, P2, and S1 to solve for S2.
S2 k P P = H 2 = 2 S1 kH P1 P1 Solving for S2 gives
S2 =
P2 S1 (159 mmHg)(0.0404 g O 2 /L) = = 8.452 x 10−3 = 8.45 x 10−3 g O2/L P1 760 mmHg
12.5. The mass of HCl in 20.2% HCl (0.202 = fraction of HCl) is 0.202 x 35.0 g = 7.070 = 7.07 g HCl The mass of H2O in 20.2% HCl is 35.0 g solution − 7.07 g HCl = 27.93 = 27.9 g H2O
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Chapter 12: Solutions
12.6. Calculate the moles of toluene using its molar mass of 92.14 g/mol: 35.6 g toluene x
1 mol toluene = 0.3863 mol toluene 92.14 g toluene
To calculate molality, divide the moles of toluene by the mass in kg of the solvent (C6H6):
Molality x
0.3863 mol toluene = 3.0904 = 3.09 m toluene 0.125 kg solvent
12.7. The number of moles of toluene = 0.3863 (previous exercise); the number of moles of benzene is 125 g benzene x
1 mol benzene = 1.6003 mol benzene 78.11 g benzene
The total number of moles is 1.6003 + 0.3863 = 1.9866, and the mole fractions are Mole fraction benzene x
1.6003 mol benzene = 0.80554 = 0.806 1.9866 mol
Mole fraction toluene x
0.3863 mol toluene = 0.1944 = 0.194 1.9866 mol
The sum of the mole fractions = 1.000. 12.8. This solution contains 0.120 moles of methanol dissolved in 1.00 kg of ethanol. The number of moles in 1.00 kg of ethanol is 1.00 x 103 g C2H5OH x
1 mol C2 H 5 OH = 21.706 mol C2H5OH 46.07 g C 2 H 5 OH
The total number of moles is 21.706 + 0.120 = 21.826, and the mole fractions are Mole fraction C2H5OH = Mole fraction CH3OH =
21.706 mol C 2 H 5OH = 0.994501 = 0.995 21.826 mol 0.120 mol CH 3OH = 0.005498 = 0.00550 21.826 mol
The sum of the mole fractions is 1.000. 12.9. One mole of solution contains 0.250 mol methanol and 0.750 mol ethanol. The mass of this amount of ethanol, the solvent, is 0.750 mol C2H5OH x
46.07 g C 2 H 5OH = 34.55 g C2H5OH (0.03455 kg) 1 mol C2 H 5 OH
The molality of methanol in the ethanol solvent is 0.250 mol CH 3OH = 7.2358 = 7.24 m CH 3OH 0.03455 kg C 2 H5 OH
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12.10. Assume an amount of solution contains 1 kg of water. The mass of urea in this mass is 3.42 mol urea x
60.05 g urea = 205.4 g urea 1 mol urea
The total mass of solution is 205.4 + 1000.0 g = 1205.4 g. The volume and molarity are Volume of solution = 1205.4 g x Molarity =
1 mL = 1153.49 mL = 1.15349 L 1.045 g
3.42 mol urea = 2.9649 mol/L = 2.96 M 1.15349 L solution
12.11. Assume a volume equal to 1.000 L of solution. Then Mass of solution = 1.029 g/mL x (1.000 x 103 mL) = 1029 g Mass of urea = 2.00 mol urea x
60.05 g urea = 120.1 g urea 1 mol urea
Mass of water = (1029 − 120.1) g = 908.9 g water (0.9089 kg) Molality =
2.00 mol urea = 2.2004 = 2.20 m urea 0.9089 kg solvent
12.12. Calculate the moles of naphthalene and the moles of chloroform: 0.515 g C10H8 x
1 mol C10 H8 = 0.004018 mol C10H8 128.17 g C10 H8
60.8 g CHCl3 x
1 mol CHCl3 = 0.50929 mol CHCl3 119.38 g CHCl3
The total number of moles is 0.004018 + 0.50929 = 0.5133 mol, and the mole fraction of chloroform is Mole fraction CHCl3 =
0.50929 mol CHCl3 = 0.9921 0.5133 mol
Mole fraction C10H8 =
0.004018 mol C10 H8 = 0.007828 0.5133 mol
The vaporpressure lowering is ΔP = P° X C10 H8 = (156 mmHg)(0.007828) = 1.221 = 1.22 mmHg
Use Raoult's law to calculate the vapor pressure of chloroform:
P = P° X CHCl3 = (156 mmHg)(0.9921) = 154.7 = 155 mmHg
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Chapter 12: Solutions
12.13. Solve for cm in the freezingpointdepression equation (ΔT = Kfcm; Kf in Table 12.3):
cm =
ΔT 0.150°C = = 0.8073 m 1.858°C/m Kf
Use the molal concentration to solve for the mass of ethylene glycol: 0.08073 mol glycol x 0.0378 kg solvent = 0.003051 mol glycol 1 kg solvent
0.003051 mol glycol x
62.1 g glycol = 1.894 x 10−1 = 1.89 x 10−1 g glycol 1 mol glycol
12.14. Calculate the moles of ascorbic acid (vitamin C) from the molality, and then divide the mass of 0.930 g by the number of moles to obtain the molar mass: 0.0555 mol vit. C x 0.0950 kg H2O = 0.005272 mol vit C 1 kg H 2 O 0.930 g vit. C = 176.4 = 176 g/mol 0.005272 mol vit. C
The molecular mass of ascorbic acid, or vitamin C, is 176 amu. 12.15. The molal concentration of white phosphorus is cm =
ΔT 0.159°C = = 0.06625 m 2.40°C/m Kb
The number of moles of white phosphorus (Px) present in this solution is 0.06625 mol Px x 0.0250 kg CS2 = 0.001656 mol Px 1 kg CS2
The molar mass of white phosphorus equals the mass divided by moles: 0.205 g ÷ 0.001656 mol = 123.77 = 124 g/mol Thus, the molecular mass of Px is 124 amu. The number of P atoms in the molecule of white phosphorus is obtained by dividing the molecular mass by the atomic mass of P: 123.77 amu Px = 3.9965 = 4.00 30.97 amu P Hence, the molecular formula is P4 (x = 4).
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417
12.16. The number of moles of sucrose is 5.0 g sucrose x
1 mol sucrose = 0.0146 mol sucrose 342.3 g sucrose
The molarity of the solution is 0.0146 mol sucrose = 0.146 M sucrose 0.100 L
The osmotic pressure, π, is equal to MRT and is calculated as follows: 0.146 mol sucrose 0.0821 L • atm x x 293 K = 3.51 = 3.5 atm 1L K • mol 12.17. The number of ions from each formula unit is i. Here, i = 1+2 = 3 The boilingpoint elevation is
ΔTb = Kbcm = 3 x
0.512°C x 0.050 m = 0.07681 = 0.077°C m
The boiling point of aqueous MgCl2 is 100.077°C. 12.18. AlCl3 would be most effective in coagulating colloidal sulfur because of the greater magnitude of charge on the Al ion (3+).
■
ANSWERS TO CONCEPT CHECKS
12.1. In each case, the component present in the greatest amount is the solvent. a.
Mo is the solute, and Cr is the solvent.
b.
MgCl2 is the solute, and water is the solvent.
c.
N2 and O2 are the solutes, and Ar is the solvent.
12.2. The two main factors to consider when trying to determine the solubility of an ionic compound in water are ionic size and lattice energy. In this case, the lattice energy for the two compounds is the same; you can discount its effects. Since a smaller cation will have a more concentrated electric field leading to a larger energy of hydration, you would expect that if AB2 has a smaller cation, then AB2 will have a greater energy of hydration and AB2 will be the more soluble compound. 12.3. As the altitude increases, the atmospheric pressure decreases, and thus the partial pressure of oxygen decreases. Above 3500 m, the partial pressure of oxygen in air has decreased to the point that not enough will dissolve in the water to sustain the fish.
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12.4. In order to boil at a lower temperature than water, the vapor pressure of the solution (water + liquid) must be greater than that of water. To make this solution, you must add a liquid that is both soluble in water and chemically similar to water. It must have a higher vapor pressure than water and a boiling point lower than 100°C. One possible liquid is ethanol, with a boiling point of 78.3°C (Table 12.3). 12.5. By the principle of osmosis, in a brine solution, water will flow out of the pickle (lower concentration of ions) into the brine (higher concentration of ions). If the pickles were stored in a water solution, the water (lower concentration of ions) would flow into the pickle (higher concentration of ions) and cause it to swell up and probably burst. 12.6. These are water solutions of identical volume (normal boiling point 100°C) containing different numbers of moles of solute. The boiling point of a solution can be determined by the formula ΔTb = iKbm. The solution with the largest ΔTb will have the highest boiling point. Since Kb is a constant, this will be the compound with the largest factor of i • m. Also, since the volume is constant, the factor reduces to i • moles. Ideally, all of the compounds will dissolve completely, so NaCl and KBr have i = 2, and Na2SO4 and MgCl2 have i = 3. This gives For NaCl, i • moles = 2 x 1.5 = 3.0 For Na2SO4, i • moles = 3 x 1.3 = 3.9 For MgCl2, i • moles = 3 x 2.0 = 6.0 For KBr, i • moles = 2 x 2.0 = 4.0 The result is given from highest boiling point to lowest boiling point: MgCl2 > KBr > Na2SO4 > NaCl 12.7. Iron(III) hydroxide is a hydrophobic colloid. As the colloid forms in water, an excess of iron(III) ion (Fe3+) is present on the surface, giving each crystal an excess positive charge. These positively charged crystals repel one another, so aggregation to larger particles of iron(III) hydroxide is prevented. When the electrodes are dipped into the colloidal solution, iron(III) hydroxide precipitates because electrons from the negative electrode neutralize the excess positive charge on the iron(III) hydroxide, allowing larger particles to form (precipitate).
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ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
12.1. An example of a gaseous solution is air, in which nitrogen (78%) acts as a solvent for a gas such as oxygen (21%). Recall that the solvent is the component that is present in greater amount. An example of a liquid solution containing a gas is any carbonated beverage in which water acts as the solvent for carbon dioxide gas. Ethanol in water is an example of a liquidliquid solution. An example of a solid solution is any goldsilver alloy.
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12.2. The two factors that explain differences in solubilities are (1) the natural tendency of substances to mix together, or the natural tendency of substances to become disordered, and (2) the relative forces of attraction between solute species (or solvent species) compared to that between the solute and solvent species. The strongest interactions are always achieved. 12.3. This is not a case of "like dissolves like." There are strong hydrogenbonding forces between the water molecules. For the octane to mix with water, hydrogen bonds must be broken and replaced by the much weaker London forces between water and octane. Thus, octane does not dissolve in water, because the maximum forces of attraction among molecules are obtained if it does not dissolve. 12.4. In most cases, the wide differences in solubility can be explained in terms of the different energies of attraction between ions in the crystal and between ions and water. Hydration energy is used to measure the attraction of ions for water molecules, and lattice energy is used to measure the attraction of positive ions for negative ions in the crystal lattice. An ionic substance is soluble when the hydration energy is much larger than the lattice energy. An ionic substance is insoluble when the lattice energy is much larger than the hydration energy. 12.5. A sodium chloride crystal dissolves in water because of two factors. The positive Na+ ion is strongly attracted to the oxygen (negative end of the water dipole) and dissolves as Na+(aq). The negative Cl− ion is strongly attracted to the hydrogens (positive end of the water dipole) and dissolves as Cl−(aq). 12.6. When the temperature (energy of a solution) is increased, the solubility of an ionic compound usually increases. A number of salts are exceptions to this rule, particularly a number of calcium salts such as calcium acetate, calcium sulfate, and calcium hydroxide (although the solubilities of calcium bromide, calcium chloride, calcium fluoride, and calcium iodide all increase with temperature). 12.7. Calcium chloride is an example of a salt that releases heat when it dissolves (exothermic heat of solution). Ammonium nitrate is an example of a salt that absorbs heat when it dissolves (endothermic heat of solution). 12.8. As the temperature of the solution was increased by heating, the concentration of the dissolved gas would decrease. 12.9. A carbonated beverage must be stored in a closed container because carbonated beverages must contain more carbon dioxide than is soluble in water at atmospheric pressure. It is possible to add carbon dioxide under pressure to a closed container before it is sealed and increase the solubility of carbon dioxide. This is an illustration of Le Châtelier's principle, which states that the equilibrium between gaseous and dissolved carbon dioxide is shifted in favor of the dissolved carbon dioxide by an increase in pressure.
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12.10. According to Le Châtelier's principle, a gas is more soluble in a liquid at higher pressures because, when the gas dissolves in the liquid, the system decreases in volume, tending to decrease the applied pressure. However, when a solid dissolves in a liquid, there is very little volume change. Thus, pressure has very little effect on the solubility of a solid in a liquid. 12.11. The four ways to express the concentration of a solute in a solution are (1) molarity, which is moles per liter; (2) mass percentage of solute, which is the percentage by mass of solute contained in a given mass of solution; (3) molality, which is the moles of solute per kilogram of solvent; and (4) mole fraction, which is the moles of the component substance divided by the total moles of solution. 12.12. The vapor pressure of the solvent of the dilute solution is larger than the vapor pressure of the solvent of the more concentrated solution (Raoult’s law). Thus, the more dilute solution loses solvent and becomes more concentrated while the solvent molecules in the vapor state condense into the concentrated solution, making it more dilute. After sufficient time has passed, the vapor pressure of the solvent in the closed container will reach a steady value (equilibrium), at which time the concentration of solute will be the same in the two solutions. 12.13. In fractional distillation, the vapor that first appears over a solution will have a greater mole fraction of the more volatile component. If a portion of this is vaporized and condensed, the liquid will be still richer in the more volatile component. After successive distillation stages, eventually the more volatile component will be obtained in pure form (Figure 12.20). 12.14. The boiling point of the solution is higher because the nonvolatile solute lowers the vapor pressure of the solvent. Thus, the temperature must be increased to a value greater than the boiling point of the pure solvent to achieve a vapor pressure equal to atmospheric pressure. 12.15. One application is the use of ethylene glycol in automobile radiators as antifreeze; the glycolwater mixture usually has a freezing point well below the average low temperature during the winter. A second application is spreading sodium chloride on icy roads in the winter to melt the ice. The ice usually melts, because at equilibrium, a concentrated solution of NaCl usually freezes at a temperature below that of the roads. 12.16. If a pressure greater than the osmotic pressure of the ocean water is applied, the natural osmotic flow can be reversed. Then, the water solvent flows from the ocean water through a membrane to a more dilute solution or to pure water, leaving behind the salt and other ionic compounds from the ocean in a more concentrated solution. 12.17. Part of the light from the sun is scattered in the direction of an observer by fine particles in the clouds (Tyndall effect) rather than being completely absorbed by the clouds. The scattered light becomes visible against the darker background of dense clouds. 12.18. Fog is an aerosol, whipped cream is a foam, mayonnaise is an emulsion, solid silver chloride dispersed in water is a sol, and fruit jelly is a gel.
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12.19. The polar –OH group on glycerol allows it to interact (hydrogenbond) with the polar water molecules, which means it is like water and, therefore, will dissolve in water. Benzene is a nonpolar molecule, which indicates that it is not “like” water and, therefore, will not dissolve in water. 12.20. Soap removes oil from a fabric by absorbing the oil into the hydrophobic centers of the soap micelles and off the surface of the fabric. Rinsing removes the micelles from contact with the fabric and leaves only water on the fabric, which can then be dried. 12.21. The answer is a, 6.50 x 10−2 mol. 12.22. The answer is c, 8.66 m. 12.23. The answer is e, 25.3 g/mol. 12.24. The answer is a, the solution would freeze at a lower temperature than pure water.
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ANSWERS TO CONCEPT EXPLORATIONS
12.25. a.
The lower boiling point will be for the solution with the smaller ΔTb and hence the lower particle concentration. For solutions with equal concentrations, the solution with the smallest i factor will have the lowest particle concentration. Substance AB breaks into ions and has an i factor of 2, whereas substance XY dissolves as a molecular compound and has an i factor of 1. Therefore, substance XY has the lower ΔTb and the lower boiling point.
b.
According to Raoult’s law, the vaporpressure lowering is proportional to the mole fraction of the solute multiplied by the i factor. Since the i factor for substance AB is 2, and the i factor for substance XY is 1, you would not expect the vapor pressures of the two solutions to be the same. The vaporpressure lowering is smaller for the substance with the smaller i factor. Therefore, the vapor pressure of a solution of XY will be higher than the vapor pressure of a solution of AB.
c.
To make the two solutions have the same boiling point, you would have to make the concentration of particles in the two solutions the same. This can be accomplished by doubling the concentration of XY, thereby making the factor i x m equal for both solutions.
d.
The boiling point of 250 mL of the AB(aq) solution will be the same as that of the original solution. The boiling point of a solution is independent of the volume of the sample and depends only on the concentration of solute particles.
e.
Freezingpoint depression depends on the concentration of the solute. As the solvent evaporates, the solute concentration increases, and the freezingpoint depression gets larger, resulting in a lower melting point. Thus, after the water evaporates, the melting point of the resulting solution will be lower than for the original solution.
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12.26. a.
Even though the two solutions were prepared using equal moles, they can have different freezing points if the two substances dissolve into a different number of solute particles. Since sample A freezes at −0.50°C and sample B freezes at −1.00°C, sample B must dissolve into twice as many particles as sample A.
b.
The sample with substance B still has the same freezing point as the original sample, so the difference is exactly the same.
c.
Since equal numbers of moles of A and B were used, both solutions have the same molality. Using substance A, with i = 1, you get m =
Kf
=
0.50°C = 0.269 = 0.27 m 1.858°C/m
d.
If another 1.0 kg of water were added to solution B, the molality would be reduced to onehalf the original molality, and ΔTf would also be reduced by onehalf. The new solution would have a freezing point of −0.50°C.
e.
To get a solution of A with a freezing point of −0.25°C, you would need the following molality:
m = f.
■
ΔT f
ΔT f Kf
=
0.25°C = 0.134 = 0.13 m 1.858°C/m
The boiling point of A will be lower than that of B, the vapor pressure of A will be higher than that of B, and the osmotic pressure of A will be lower than that of B.
ANSWERS TO CONCEPTUAL PROBLEMS
12.27. The amount of oxygen dissolved in water decreases as the temperature increases. Thus, at the lower temperatures, there is enough oxygen dissolved in the water to support both bass and trout. But as the temperature rises above 23°C, there is not enough dissolved oxygen in the warm water to support the trout, which need more O2 than bass. 12.28. The salt that would best accomplish the task would be the salt that lowers the freezing point of water the most. This, in turn, will be the salt with the largest i factor. Ideally, if each salt dissolved completely, KCl, MgSO4, and AgCl would have i = 2. Similarly, CaCl2 and PbS2 would have i = 3. Of the latter two salts, CaCl2 is more soluble than PbS2, so its i factor is closer to 3. Therefore, the salt with the largest i factor is CaCl2, so it would lower the freezing point of water the most and would best accomplish the task.
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12.29. The two main factors to consider when trying to determine the solubility of an ionic compound in water are ionic size and lattice energy. Ionic size is inversely related to the energy of hydration; the smaller the ion, the greater the energy of hydration. Keep in mind that the greater the energy of hydration, the more likely it is for a compound to dissolve. The amount of lattice energy is directly related to the solubility of the compound; the lower the lattice energy, the more likely it is for the compound to dissolve. Taking into account these factors, in order to increase the solubility of a compound you need to decrease the ionic size and decrease the lattice energy. Since the energy of hydration of the Y+ ion is greater than that of the X+ ion (making XZ less soluble), in order for XZ to be more soluble than YZ, the lattice energy must be less for the XZ compound. 12.30. a.
According to Raoult's law, the addition of a nonvolatile nonelectrolyte to a solvent will lower the vapor pressure of the solvent, so we would expect the vapor pressure of such a solution to be lower than that of the pure solvent (water in this case). When a volatile solute is added to a solvent, the vapor pressure of the solution is dependent on the mole fraction of the solute and solvent and on the vapor pressures of both the solute and the solvent. Since the solute is volatile (a high vapor pressure relative to water), the solution must have a higher vapor pressure than pure water.
b.
Keeping in mind that a solution will boil when the vapor pressure equals the pressure pushing on the surface of the solution, the solution with the lower vapor pressure will boil at a higher temperature. Thus, the solution with the nonvolatile solute will boil at the higher temperature.
12.31. Smoke particles carry a small net charge that prevents them from forming larger particles that would settle to the bottom of the smokestack. The charged smoke particles are neutralized by the current, which then allows them to aggregate into large particles. These large particles are too big to be carried out of the stack. 12.32. a.
Since beaker B contains more solute particles, according to Raoult's law, it will boil at a higher temperature than beaker A.
b.
More particles in solution lead to lower vapor pressure, which in turn lowers the freezing point of a solution. Since beaker B contains more solute particles, it will freeze at a lower temperature than beaker A.
c.
When separated by a semipermeable membrane, the solvent from the less concentrated solution flows into the more concentrated solution. Because of this, the water will flow from beaker A to beaker B, causing an increase in the concentration of NaCl in beaker A.
12.33. Vinegar is a solution of acetic acid (solute) and water (solvent). Because the salt concentration is higher outside the lettuce leaf than inside, water will pass out of the lettuce leaf into the dressing via osmosis. The result is that the lettuce will wilt.
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12.34. As a solution freezes, pure solvent forms without any of the solute present. This means that as ocean water freezes to make icebergs, it freezes as pure water without the salt present. 12.35. To answer these questions, it is necessary to determine the relative concentrations of solute particles in each beaker. With equal moles and equal volumes, the solution concentration will be directly proportional to the van’t Hoff factor, i, for that compound. Urea, a nonelectrolyte, has i = 1. NaCl, which is ionic, has i = 2, and CaCl2, which is also ionic, has i = 3. a.
The highest boiling point will correspond to the largest boilingpoint elevation, ΔTb. Since ΔTb = iKbm, the largest ΔTb will correspond to the highest value of i. Thus, the solution with CaCl2 (i = 3) will have the highest boiling point.
b.
The highest freezing point will correspond to the smallest freezingpoint depression, ΔTf. Since ΔTf = iKfm, the smallest ΔTf will correspond to the smallest value of i. Thus, the solution with urea (i = 1) will have the highest freezing point.
12.36. To answer these questions, it is necessary to determine the relative concentrations of solute particles in each beaker. In order to compare volumes, use the lines on the sides of the beakers. Beaker A has a concentration of five atoms per two volume units, 5/2 or 2.5/1. Beaker B has a concentration of ten atoms per two volume units, 10/2 or 5/1. Beaker C has a concentration of five atoms per volume unit, 5/1. Comparing the concentrations, beaker B = beaker C = 2 x beaker A.
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a.
Since PA = XAPA°, the solution with the highest vapor pressure is the solution with the highest mole fraction of solvent, or, conversely, the lowest concentration of solute particles. Thus, the solution in beaker A will have the highest vapor pressure.
b.
The lowest boiling point will correspond to the smallest boiling point increase, ΔTb. Since ΔTb = Kbm, the smallest ΔTb will correspond to the smallest concentration of solute particles (molality here). Thus, the solution in beaker A will have the lowest boiling point.
c.
For each solution to have the same freezing point, they all need the same concentration of solute particles. This can be accomplished by doubling the volume in beakers B and C.
SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. Atomic masses are rounded to two decimal places, except for that of hydrogen. 12.37. An example of a liquid solution prepared by dissolving a gas in a liquid is household ammonia, which consists of ammonia (NH3) gas dissolved in water. 12.38. An example of a solid solution prepared from two solids is almost any alloy, such as 18kt gold, which consists of 25 percent silver dissolved in 75% gold.
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12.39. Boric acid would be more soluble in ethanol because this acid is polar and is more soluble in a more polar solvent. It can also hydrogenbond to ethanol but not to benzene. 12.40. Naphthalene is more soluble in benzene because nonpolar naphthalene must break the strong hydrogen bonds between ethanol molecules and replace them with weaker London forces. 12.41. The order of increasing solubility is H2O < CH2OHCH2OH < C10H22. The solubility in nonpolar hexane increases with the decreasing polarity of the solute. 12.42. Acetic acid is more soluble because it is polar like ethanol and can form hydrogen bonds to it. Stearic acid also has a polar, Hbonding end, but it also has a long hydrocarbon part which is nonpolar and unlike ethanol. 12.43. The Al3+ ion has both a greater charge and a smaller ionic radius than Mg2+, so Al3+ should have a greater energy of hydration. 12.44. The F− ion has a smaller radius than the Cl− ion, so F− should have a greater hydration energy. 12.45. The order is Ba(IO3)2 < Sr(IO3)2 < Ca(IO3)2 < Mg(IO3)2. The iodate ion is fairly large, so the lattice energy for all these iodates should change to a smaller degree than the hydration energy of the cations. Therefore, solubility should increase with decreasing cation radius. 12.46. Fluorides (smaller anion): As the cation radius increases, the lattice energy of the fluoride salts decreases to a greater degree than the cation hydration energy decreases. Permanganates (larger anion): As the cation radius increases, the lattice energy of the permanganate salts decreases to a smaller degree than the cation hydration energy decreases. Therefore, the solubility of the fluorides increases with increasing cation radius, whereas the solubility of the permanganates decreases with increasing cation radius. 12.47. Using Henry's law, let S1 = the solubility at 1.00 atm (P1), and S2 = the solubility at 5.50 atm (P2).
S2 =
P2 S1 (5.50 atm)(0.161 g /100 mL) = 0.8855 = 0.886 g/100 mL = 1.00 atm P1
12.48. The partial pressure of N2 in air at 4.79 atm is 4.79 atm x 0.781 = 3.74099 atm Substituting this for P2 in S2 = S1 x P2/P1 (Henry's law), you obtain
S2 = (0.00175 g/100 mL) x
3.74099 atm = 0.006546 = 0.00655 g/100 mL 1.00 atm
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12.49. First, calculate the mass of KI in the solution; then calculate the mass of water needed. Mass KI = 72.5 g x
5.00 g KI = 3.6250 = 3.63 g KI 100 g solution
Mass H2O = 72.5 g soln − 3.6250 g KI = 68.875 = 68.9 g H2O Dissolve 3.63 g KI in 68.9 g of water. 12.50. First, calculate the mass of Na2SO4 in the solution; then calculate the mass of water needed. Mass of Na2SO4 = 455 g solution x
6.50 g Na 2SO 4 = 29.57 g Na2SO4 100 g solution
Mass of water = 455 g soln − 29.57 g Na2SO4 = 425.43 g water Dissolve 29.57 g Na2SO4 in 425.43 g of water. 12.51. Multiply the mass of KI by 100 g of solution per 5.00 g KI (reciprocal of percentage). 0.258 g KI x
100 g solution = 5.160 = 5.16 g solution 5.00 g KI
12.52. Multiply the mass of sodium sulfate by 100 g of solution per 6.50 g Na2SO4 (reciprocal of percentage). 1.50 g Na2SO4 x
100 g solution = 23.07 = 23.1 g solution 6.50 g Na 2SO 4
12.53. Convert mass of vanillin (C8H8O3, molar mass 152.14 g/mol) to moles, convert mg of ether to kg, and divide for molality. 0.0391 g vanillin x
1 mol vanillin = 2.570 x 10−4 mol vanillin 152.14 g vanillin
168.5 mg ether x 1 kg/106 mg = 168.5 x 10−6 kg ether Molality =
2.570 x 104 mol vanillin = 1.5252 = 1.53 m vanillin 168.5 x 106 kg ether
12.54. Convert mass of lauryl alcohol (LA, C12H25OH, molar mass 186.33 g/mol) to moles, convert g of ethanol to kg, and divide. 17.1 g LA x
1 mol LA = 9.177 x 10−2 mol LA 186.3 g LA
165 g ethanol x 1 kg/103 g = 165 x 10−3 kg ethanol Molality =
9.177 x 102 mol LA = 0.5561 = 0.556 m LA 165 x 103 kg ethanol
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12.55. Convert mass of fructose (C6H12O6, molar mass 180.16 g/mol) to moles, and then multiply by one kg H2O per 0.125 mol fructose (the reciprocal of molality). 1.75 g fructose x
1 kg H 2 O 1 mol fructose x = 0.07770 kg (77.7 g H2O) 0.125 mol fructose 180.16 g fructose
12.56. Multiply the kg of CHCl3 by the molality to get moles of caffeine; then convert moles to mass. 0.0946 mol caffeine 194.2 g caffeine x 1 mol caffeine 1 kg CHCl3
0.0450 kg CHCl3 x
= 0.8267 = 0.827 g caffeine 12.57. Convert masses to moles, and then calculate the mole fractions. 65.0 g alc. x
1 mol alc. = 1.0817 mol alc. 60.09 g alc.
35.0 g H2O x
1 mol H 2 O = 1.9422 = 1.94 mol H2O 18.02 g H 2 O mol alc. 1.0817 mol = = 0.3577 = 0.358 total mol 3.0239 mol
Mole fraction alc. = Mole fraction H2O =
mol H 2 O 1.9422 mol = = 0.6422 = 0.642 3.0239 mol total mol
12.58. Convert masses to moles, and then calculate the mole fractions. 2.50 x 103 g glycol x 2.00 x 103 g H2O x
1 mol glycol = 40.277 mol glycol 62.07 g glycol
1 mol H 2 O = 110.98 mol H2O 18.02 g H 2 O
Total moles = 40.277 + 110.98 = 151.26 mol Mole fraction H2O =
mol H 2 O 110.98 mol = 0.7336 = 0.734 = 151.26 mol total mol
Mole fraction glycerol =
mol glycerol 40.277 mol = = 0.2662 = 0.266 total mol 151.26 mol
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Chapter 12: Solutions
12.59. In the solution, for every 0.650 mol of NaClO there is 1.00 kg, or 1.00 x 103 g, of H2O, so 1.00 x 103 g H2O x
1 mol H 2 O = 55.49 mol H2O 18.02 g H 2 O
Total mol = 55.49 mol H2O + 0.650 mol NaClO = 56.14 mol Mole fraction NaClO =
mol NaClO 0.650 mol = 0.01157 = 0.0116 = 56.1 4 mol total mol
12.60. In the solution, for every 0.600 mol of H2O2, there is 1.00 kg, or 1.00 x 103 g, of H2O, so 1.00 x 103 g H2O x
1 mol H 2 O = 55.49 mol H2O 18.02 g H 2 O
Total mol = 55.49 mol H2O + 0.600 mol H2O2 = 56.093 mol Mole fraction H2O2 =
mol H 2 O 2 0.600 mol = = 0.01069 = 0.0107 56.093 mol mol solution
12.61. The total moles of solution = 3.31 mol H2O + 1.00 mol HCl = 4.31 mol Mole fraction HCl = 3.31 mol H2O x Molality =
1 mol HCl = 0.23201 = 0.232 4.31 mol
18.02 g H 2 O 1 kg H 2 O x = 5.965 x 10−2 kg H2O 1 mol H 2 O 103 g H 2 O
1.00 mol HCl = 16.76 = 16.8 m 5.965 x 102 kg H 2 O
12.62. The total moles of solution = 1.00 mol NH3 + 2.44 mol H2O = 3.44 mol Mole fraction NH3 = 2.44 mol H2O x Molality =
1 mol NH 3 = 0.2906 = 0.291 3.44 mol
18.02 g H 2 O 1 kg H 2 O x = 4.397 x 10−2 kg H2O 1 mol H 2 O 103 g H 2 O
1.00 mol NH 3 = 22.74 = 22.7 m 4.397 x 102 kg H 2 O
12.63. The mass of 1.000 L of solution is 1.022 kg. In the solution, there is 0.580 mol H2C2O4 (OA) for every 1.0000 kg of water. Convert this number of moles to mass. 0.580 mol OA x
90.04 g OA 1 kg OA = 0.05222 kg OA x 1 mol OA 103 g OA
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The total mass of the solution containing 1.000 kg H2O and 0.580 mol OA is calculated as follows: Mass = 1.0000 kg H2O + 0.05222 kg OA = 1.05222 kg Use this to relate the mass of 1.000 L (1.022 kg) of solution to the amount of solute: 1.022 kg solution x Molarity =
0.580 mol OA = 0.5633 mol OA 1.05222 kg solution
0.5633 mol OA = 0.5633 = 0.563 M 1.00 L solution
12.64. In the solution, there is 0.688 mol of citric acid (CA) for every 1.000 kg of water. Convert this number of moles to mass. 0.688 mol CA x
192.1 g CA 1 kg x = 0.1321 kg CA 1 mol CA 1000 g
The total mass of solution containing 1.000 kg H2O and 0.688 mol CA is Mass = 1.0000 kg H2O + 0.1321 kg CA = 1.1321 kg = 1132.1 g Now, use the density of the solution (1.049 g/mL) to convert this mass to liters of solution. Volume = 1132.1 g x
1 mL 1L x = 1.0792 L 1000 mL 1.049 g
Since there is 0.688 mol CA in this volume, you can now calculate the molarity of the solution. Molarity =
mol CA 0.688 mol = 0.63746 = 0.637 M = 1.0792 L Volume
12.65. In 1.000 L of vinegar, there is 0.763 mol of acetic acid. The total mass of the 1.000L solution is 1.004 kg. Start by calculating the mass of acetic acid (AA) in the solution. 0.763 mol AA x
60.05 g AA = 45.82 g AA (0.04582 kg AA) 1 mol AA
The mass of water may be found by difference: Mass H2O = 1.004 kg soln − 0.04582 kg AA = 0.9582 kg H2O Molality =
0.763 mol AA = 0.7962 = 0.796 m 0.9582 kg H 2 O
12.66. In 1.000 L of beverage, there is 0.265 mol of tartaric acid. The total mass of the 1.000L beverage is 1.016 kg. Start by calculating the mass of tartaric acid (TA) in the beverage. 0.265 mol TA x
150.1 g TA = 39.77 g TA (0.03977 kg TA) 1 mol TA
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The mass of water may be found by difference: Mass H2O = 1.016 kg soln − 0.03977 kg TA = 0.9762 kg H2O Molality =
0.265 mol TA = 0.2714 = 0.271 m 0.9762 kg H 2 O
12.67. To find the mole fraction of sucrose, first find the amounts of both sucrose (suc.) and water: 20.2 g sucrose x 70.1 g H2O x
1 mol sucrose = 0.05901 mol sucrose 342.30 g sucrose
1 mol H 2 O = 3.890 mol H2O 18.02 g H 2 O
0.05901 mol sucrose = 0.01494 (3.890 + 0.05901) mol
Xsucrose =
From Raoult's law, the vapor pressure (P) and lowering (ΔP) are
P = PH2 O X H2 O = PH2 O (1  X suc. ) = (42.2 mmHg)(1 − 0.01494) o
o
ΔP = PH2O X suc. o
= 41.569 = 41.6 mmHg = (42.2 mmHg)(0.01494) = 0.6306 = 0.631 mmHg
12.68. To find the mole fraction of benzene, first find the amounts of both benzene (ben.) and naphthalene (nap.): 1.20 g nap. x
1 mol nap. = 0.0093603 mol nap. 128.2 g nap.
25.6 g ben. x
1 mol ben. = 0.3277 mol ben. 78.11 g ben.
0.0093603 mol nap. = 0.02777 (0.3277 + 0.0093603) mol
Xnap. =
From Raoult's law, the vapor pressure (P) and lowering (ΔP) are P = Pben. (1  X nap. ) = (86.0 mmHg)(1 − 0.02777) = 83.61 = 83.6 mmHg o
ΔP = Pben. X nap. = (86.0 mmHg)(0.02777) = 2.388 = 2.39 mmHg o
12.69. Find the molality of glycerol (gly.) in the solution first: 0.150 g gly. x Molality =
1 mol gly. = 0.001628 mol gly. 92.095 g gly.
0.001628 mol gly. = 0.081437 m 0.0200 kg solvent
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Substitute Kb = 0.512°C/m and Kf = 1.858 °C/m (Table 12.3) into equations for ΔTb and ΔTf:
ΔTb = Kbm = 0.512°C/m x 0.081437 m = 0.04169°C Tb = 100.000 + 0.04169 = 100.04169 = 100.042°C
ΔTf = Kfm = 1.858°C/m x 0.081437 m = 0.15131°C Tf = 0.000 − 0.15131 = −0.15131 = −0.151°C 12.70. Find the molality of the solution first: 1 mol S8 = 0.003251 mol S8 256.52 g S8
0.834 g S8 x Molality =
0.003251 mol S8 = 0.03251 m 0.1000 kg solvent
Substitute Kb = 3.08°C/m and Kf = 3.59°C/m (Table 12.3) into equations for ΔTb and ΔTf:
ΔTb = Kbm =
3.08°C x 0.03251 m S8 = 0.1001°C m S8
Tb = 118.5°C + 0.1001°C = 118.600 = 118.6°C
ΔTf = Kfm =
3.59°C x` 0.03251 m S8 = 0.1167°C m S8
Tf = 16.60°C − 0.1167°C = 16.483 = 16.48°C 12.71. Calculate ΔTf, the freezingpoint depression, and, using Kf = 1.858°C/m (Table 12.3), the molality, cm.
ΔTf = 0.000°C − (−0.086°C) = 0.086°C cm =
ΔT f Kf
=
0.086°C = 4.628 x 10−2 = 4.6 x 10−2 m 1.858°C/m
12.72. Calculate ΔTf, the freezingpoint depression, and, using Kf = 1.858°C/m (Table 12.3), the molality, cm.
ΔTf = 0.000°C − (−0.085°C) = 0.085°C cm =
ΔT f Kf
=
0.085°C = 0.0457 m 1.858°C/m
Now, find the amount of urea in the solution from the definition of molality: Molurea = cm x (kg H2O) = 0.00457 mol urea x
0.0457 mol urea x 0.1000 kg H2O = 0.00457 mol 1 kg H 2 O
60.06 g urea = 0.274 = 0.27 g urea 1 mol urea
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Chapter 12: Solutions
12.73. Find the moles of unknown solute from the definition of molality: Molsolute = m x kg solvent = Molar mass =
0.0698 mol x 0.002135 kg solvent = 1.490 x 10−4 mol 1 kg solvent
0.0182 g = 122.1 = 122 g/mol 1.490 x 104 mol
The molecular mass is 122 amu. 12.74. Find the moles of the solute from the definition of molality: Molsolute = m x kg solvent = Molar mass =
0.0368 mol x 0.00831 kg solvent = 3.058 x 10−4 mol 1 kg solvent
0.0653 g = 213.5 = 214 g/mol 3.058 x 104 mol
The molecular mass is 214 amu. 12.75. Calculate ΔTf, the freezingpoint depression, and, using Kf = 1.858°C/m (Table 12.3), the molality, cm.
ΔTf = 26.84°C − 25.70°C = 1.14°C cm =
ΔT f Kf
=
1.14°C = 0.1425 m 8.00°C/m
Find the moles of solute by rearranging the definition of molality: Mol = m x kg solvent = Molar mass =
0.1425 mol x 103 x 10−6 kg solvent = 1.467 x 10−5 mol 1 kg solvent
2.39 x 103 g = 162.9 = 163 g/mol 1.467 x 105 mol
The molecular mass is 163 amu. 12.76. Calculate ΔTf, the freezingpoint depression, and, using Kf = 5.065°C/m (Table 12.3), the molality, cm.
ΔTf = 5.455°C − 4.880°C = 0.575°C cm =
ΔT f Kf
=
0.575°C = 0.1135 m 5.065°C/m
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Find the moles of solute by rearranging the definition of molality: Mol = m x kg solvent = Molar mass =
0.1135 mol x 0.1000 kg solvent = 0.01135 mol 1 kg solvent
2.500 g = 220.2 = 2.20 x 102 g/mol 0.01135 mol
The molecular mass is 2.20 x 102 amu. 12.77. Use the equation for osmotic pressure (π) to solve for the molarity of the solution. ⎛ 1 atm ⎞ 1.47 mmHg x ⎜ ⎟ π ⎝ 760 mmHg ⎠ M = = = 8.013 x 10−5 mol/L RT (0.0821 L • atm/K • mol)(21 + 273)K Now, find the number of moles in 106 mL (0.106 L) using the molarity. 0.106 L x
8.013 x 105 mol = 8.494 x 10−6 mol 1L
Molar mass =
0.582 g = 6.851 x 104 = 6.85 x 104 g/mol 6 8.494 x 10 mol
The molecular mass is 6.85 x 104 amu. 12.78. Use the equation for osmotic pressure (π) to solve for the molarity of the solution.
⎛ 1 atm ⎞ 3.70 mmHg x ⎜ ⎟ π ⎝ 760 mmHg ⎠ M = = = 1.989 x 10−4 mol/L RT (0.0821 L • atm/K • mol)(25 + 273)K Now, find the number of moles in 100.0 mL (0.1000 L) using the molarity. 0.100 L x
1.989 x 104 mol = 1.989 x 10−5 mol 1L
Molar mass =
0.0216 g = 1.0859 x 103 = 1.09 x 103 g/mol 1.989 x 105 mol
The molecular mass is 1.09 x 103 amu. 12.79. Begin by noting that i = 3. Then calculate ΔTf from the product of iKfcm: 3 x
1.858o C x 0.0075 m = 0.0418°C m
The freezing point = 0.000°C − 0.0418°C = −0.0418 = −0.042°C.
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Chapter 12: Solutions
12.80. Start by noting that i = 4. Then calculate ΔTf from the product of iKfcm: 4 x
1.858°C x 0.0091 m = 0.0676 °C m
The freezing point = 0.000°C − 0.0676°C = −0.0676 = −0.068°C. 12.81. Begin by calculating the molarity of Cr(NH3)5Cl3. 1.40 x 10−2 g Cr(NH3)5Cl3 x Molarity =
1 mol Cr(NH 3 )5 Cl3 = 5.749 x 10−5 mol Cr(NH3)5Cl3 243.5 g Cr(NH3 )5 Cl3
5.749 x 105 mol Cr(NH 3 )5 Cl3 = 0.002299 M 0.0250 L
Now find the hypothetical osmotic pressure, assuming Cr(NH3)5Cl3 does not ionize: π = MRT = (2.30 x 10−3 M) x
0.0821 L • atm 760 mmHg x 298 K x = 42.77 mmHg K • mol 1 atm
The measured osmotic pressure is greater than the hypothetical osmotic pressure. The number of ions formed per formula unit equals the ratio of the measured pressure to the hypothetical pressure:
i =
119 mmHg = 2.782 ≅ 3 ions/formula unit 42.77 mmHg
12.82. Begin by calculating the molality of the solution after rearranging ΔTb = Kbcm:
cm =
ΔTb 5°C = = 9.76 m 0.512°C/m Kb
The 9.76 m is the molality of the ions, not of the NaCl. The molality of the NaCl will be 9.76 m ÷ 2 = 4.88 m because i = 2. Thus, 4.88 mol of NaCl must be dissolved in the 1 kg of water. This is converted to mass using the molar mass of NaCl: 4.88 mol NaCl x
58.5 g NaCl = 285 = 3 x 102 g NaCl 1 mol NaCl
12.83. a.
Aerosol (liquid water in air)
b.
Sol [solid Mg(OH)2 in liquid water]
c.
Foam (air in liquid soap solution)
d.
Sol (solid silt in liquid water)
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435
12.84. a.
Aerosol (liquid water in air)
b.
Foam (air in liquid egg white)
c.
Aerosol (solid dust particles in air)
d.
Emulsion (liquid oil in liquid vinegar)
12.85. Because the As2S3 particles are negatively charged, the effective coagulation requires a highly charged cation, so Al2(SO4)3 is the best choice. 12.86. The K3PO4 contains highly charged PO43− ions that will be strongly attracted to the positively charged solution and will cause it to coagulate.
■
SOLUTIONS TO GENERAL PROBLEMS
12.87. Using Henry's law [S2 = S1 x (P2/P1), where P1 = 1.00 atm], find the solubility of each gas at P2, its partial pressure. For N2, P2 = 0.800 x 1.00 atm = 0.800 atm; for O2, P2 = 0.200 atm. N2: S1 x
P2 0.800 atm = (0.0175 g/L H2O) x = 0.01400 g/L H2O 1.00 atm P1
O2: S2 x
P2 0.200 atm = (0.0393 g/L H2O) x = 0.007860 g/L H2O 1.00 atm P1
In 1.00 L of the water, there are 0.0140 g of N2 and 0.00786 g of O2. If the water is heated to drive off both dissolved gases, the gas mixture that is expelled will contain 0.0140 g of N2 and 0.00786 g of O2. Convert both masses to moles using the molar masses: 0.0140 g N2 x 0.00786 g O2 x
1 mol N 2 = 4.998 x 10−4 mol 28.01 g N 2
1 mol O 2 = 2.456 x 10−4 mol 32.00 g O 2
Now calculate the mole fraction of each gas:
X N2 =
mol N 2 4.998 x 104 mol = = 0.67051 = 0.671 (4.998 + 2.456) x 104 mol total mol
X O2 = 1 − X N2 = 1 − 0.67051 = 0.32949 = 0.329
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Chapter 12: Solutions
12.88. Using Henry's law [S2 = S1 x (P2/P1), where P1 = 1.00 atm], find the solubility of each gas at P2, its partial pressure. For CH4, P2 = 0.900 x 1.00 atm = 0.900 atm; for C2H6, P2 = 0.100 x 1.00 atm = 0.100 atm. CH4: S1 x
P2 0.900 atm = (0.023 g/L H2O) x = 0.0207 g/L H2O 1.00 atm P1
C2H6: S1 x
P2 0.100 atm = 0.059 g/L H2O x = 0.00590 g/L H2O 1.00 atm P1
When 1.00 L of the water saturated with the gas mixture is heated, 0.0207 g of CH4 and 0.0059 g of C2H6 will be expelled. Convert these to moles using molar masses: 1 mol CH 4 = 1.291 x 10−3 mol 16.04 g CH 4
0.0207 g CH4 x 0.0059 g C2H6 x
1 mol C2 H 6 30.07 g C 2 H 6
= 1.962 x 10−4 mol
Now, find the mole fraction of each gas:
X CH4 =
mol CH 4 1.291 x 103 mol = = 0.86819 = 0.87 (1.291 + 0.196) x 103 mol total mol
X C2 H6 = 1 − X CH4 = 1 − 0.086819 = 0.13181 = 0.13 12.89. Assume a volume of 1.000 L whose mass is then 1.024 kg. Use the percent composition given to find the mass of each of the components of the solution. 8.50 kg NH 4 Cl = 0.08704 kg NH4Cl 100.00 kg soln
1.024 kg soln x
Mass of H2O = 1.024 kg soln − 0.08704 kg NH4Cl = 0.9370 kg H2O Convert mass of NH4Cl and water to moles: 87.04 g NH4Cl x 937.0 g H2O x
1 mol NH 4 Cl = 1.627 mol NH4Cl 53.49 g NH 4 Cl
1 mol H 2 O = 52.01 mol H2O 18.015 g H 2 O
Molarity =
mol NH 4 Cl 1.627 mol = 1.627 = 1.63 M = 1.00 L L solution
Molality =
mol NH 4 Cl 1.627 mol = = 1.736 = 1.74 m 0.9370 kg H 2O kg H 2 O
X NH4 Cl =
mol NH 4 Cl 1.627 mol = 0.03033 = 0.0303 = total moles (52.01 + 1.627) mol
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437
12.90. Assume a volume of 1.000 L whose mass is then 1.127 kg. Use the percent composition given to find the mass of each of the components of the solution. 1.127 kg soln x
27.0 kg LiCl = 0.3042 kg = 304.2 g LiCl 100.00 kg soln
Mass of H2O = 1.127 kg soln − 0.3042 kg LiCl = 0.8227 kg = 822.7 g H2O Convert mass of LiCl and water to moles: 304.2 g LiCl x
1 mol LiCl = 7.178 mol LiCl 42.39 g LiCl
822.7 g H2O x
1 mol H 2 O = 45.66 mol H2O 18.015 g H 2 O
Total moles = 7.178 + 45.66 = 52.838 mol The various quantities can now be calculated. Molarity =
mol LiCl 7.178 mol LiCl = = 7.178 = 7.18 M L soln 1.000 L soln
Molality =
mol LiCl 7.178 mol LiCl = = 8.7249 = 8.72 m kg H 2 O 0.8227 kg H 2 O
XLiCl =
mol LiCl 7.178 mol LiCl = 0.1358 = 0.136 = total moles 52.838 mol
12.91. In 1.00 mol of gas mixture, there are 0.51 mol of propane (pro.) and 0.49 mol of butane (but.). First, calculate the masses of these components. 0.51 mol pro. x
44.10 g pro. = 22.491 g pro. 1 mol pro.
0.49 mol but. x
58.12 g but. = 28.478 g but. 1 mol but.
The mass of 1.00 mol of gas mixture is the sum of the masses of the two components: 22.491 g pro. + 28.478 g but. = 50.969 g mixture Therefore, in 50.969 g of the mixture, there are 22.491 g of propane and 28.478 g of butane. For a sample with a mass of 55 g, 55 g mixture x
22.491 g pro. = 24.26 = 24 g pro. 50.969 g mixture
55 g mixture x
28.478 g but . = 30.73 = 31 g but. 50.969 g mixture
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Chapter 12: Solutions
12.92. In 1.000 mol of gas mixture, there are 0.036 mol O2, 0.056 mol N2, and 0.908 mol He. Calculate the masses of these components first. 0.036 mol O2 x
32.00 g O 2 = 1.15 g O2 1 mol O 2
0.056 mol N2 x
28.01 g N 2 1.57 g N2 1 mol N 2
0.908 mol He x
4.0026 g He = 3.634 g He 1 mol He
The mass of 1.000 mol of the gas mixture is the sum of the masses of the components: 1.15 g O2 + 1.57 g N2 + 3.634 g He = 6.35 g mixture For a sample with a mass of 8.32 g, 8.32 g mixture x
1.15 g O 2 = 1.50 = 1.5 g O2 6.35 g mixture
8.32 g mixture x
1.57 g N 2 = 2.05 = 2.1 g N2 6.35 g mixture
8.32 g mixture x
3.634 g He = 4.76 = 4.8 g He 6.35 g mixture
12.93. PED = P°ED(XED) = 173 mmHg (0.25) = 43.25 mmHg PPD = P°PD(XPD) = 127 mmHg (0.75) = 95.25 mmHg P = PED + PPD = 43.25 + 95.25 = 138.50 = 139 mmHg
12.94. PB = P°B(XB) = 75 mmHg (0.30) = 22.5 mmHg PT = P°T(XT) = 22 mmHg (0.70) = 15.4 mmHg P = PB + PT = 22.5 + 15.4 = 37.9 = 38 mmHg
12.95. Calculate the moles of KAl(SO4)2•12H2O using its molar mass of 474.4 g/mol, and use this to calculate the three concentrations. a.
The moles of KAl(SO4)2•12H2O are calculated below, using the abbreviation “Hyd” for the formula of KAl(SO4)2•12H2O. moles Hyd = 0.1186 g Hyd x
1 mol Hyd = 0.00025000 = 0.0002500 mol 474.4 g Hyd
Note that 1 mol of KAl(SO4)2•12H2O contains 1 mole of KAl(SO4)2, so the molarity of KAl(SO4)2 can be calculated using the moles of KAl(SO4)2•12H2O. 1 mol KAl(SO 4 ) 2 mol 0.0002500 mol Hyd = x = 0.0002500 M KAl(SO4)2 L 1.000 L soln 1 mol Hyd
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b.
439
The molarity of the SO42− ion will be twice that of the KAl(SO4)2. mol SO 24 2 mol SO 24 = 0.0002500 M KAl(SO4)2 x = 0.0005000 M L 1 mol KAl(SO 4 ) 2
c.
Since the density of the solution is 1.00 g/mL, the mass of 1.000 L of solution is 1000 g, or 1.000 kg. Since molality is moles per 1.000 kg of solvent, the molality of KAl(SO4)2 equals 0.0002500 mol divided by 1.000 kg or 0.0002500 m (the same as the molarity).
12.96. Calculate the moles of Al2(SO4)3•18H2O using its molar mass of 666.46 g/mol, and use this to calculate the three concentrations. a.
The moles of Al2(SO4)3•18H2O are calculated below, using the abbreviation “Hyd” for the formula of Al2(SO4)3•18H2O. moles of Hyd = 0.1593 g Hyd x
1 mol Hyd = 2.3902 x 10−4 mol 666.46 g Hyd
Note that 1 mole of Al2(SO4)3•18H2O contains 1 mole of Al2(SO4)3, so the molarity of Al2(SO4)3 can be calculated using the moles of Al2(SO4)3•18H2O. Molarity = b.
mol Hyd 2.3902 x 104 mol = = 2.3902 x 10−4 = 2.390 x 10−4 M L soln 1.000 L soln
The molarity of the SO42− ion will be three times that of the Al2(SO4)3: 2.3902 x 10−4 mol Al2(SO4)3 x
c.
3 mol SO 4 2− = 7.1707 x 10−4 = 7.171 x 10−4 M 1 mol Al2 (SO 4 )3
Since the density of the solution is 1.00 g/mL, the mass of 1.000 L of the solution is 1000. g. Of this, 0.1593 g is solute; however, since the compound is a hydrate, 0.07751 grams of this mass of solute is water. Therefore the mass of just the Al2(SO4)3 is 1000. − (0.1593 − 0.07751) = 999.91 g = 0.99991 kg H2O The molality of Al2(SO4)3 can now be calculated. Molality =
mol Al2 (SO 4 )3 2.3902 x 104 mol = 2.3904 x 10−4 = 2.39 x 10−4 M = 0.99991 kg kg H 2 O
12.97. In 1.00 kg of a saturated solution of urea, there are 0.41 kg of urea (a molecular solute) and 0.59 kg of water. First, convert the mass of urea to moles. 0.41 x 103 g urea x
1 mol urea = 6.826 mol urea 60.06 g urea
Then find the molality of the urea in the solution: Molality =
mol urea 6.826 mol urea x = 11.57 m kg H 2 O 0.59 kg
ΔTf = Kfcm = (1.858°C/m)(11.57 m) = 21.4°C
Tf = 0.0°C − 21.4°C = −21.4 = −21°C
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Chapter 12: Solutions
12.98. In 1.00 kg of a saturated solution of CaCl2, there are 0.32 kg of CaCl2 (ionic) and 0.68 kg of water. First, convert the mass of CaCl2 to moles. 0.32 x 103 g CaCl2 x
1 mol CaCl2 = 2.88 mol CaCl2 111.0 g CaCl2
Then find the molality of the CaCl2 in the solution: mol CaCl2 2.88 mol CaCl2 x = 4.24 m kg H 2 O 0.68 kg
Molality =
For ionic solutions, ΔTf = iKfcm
Calcium chloride dissolves to give three ions; thus i = 3, and ΔTf = (1.858°C/m)(3)(4.24 m) = 23.6°C
Tf = 0.0°C − 23.6°C = −24°C
12.99. M =
12.100. M =
π RT
π RT
=
7.7 atm = 0.302 = 0.30 mol/L (0.0821 L • atm/K • mol)(37 + 273)K
=
5.50 atm = 0.2248 = 0.225 mol/L (0.0821 L • atm/K • mol)(25 + 273)K
12.101. Consider the equation ΔTf = iKfcm. For CaCl2, i = 3; for glucose, i = 1. Because cm = 0.10 for both solutions, the product of i and cm will be larger for CaCl2, as will ΔTf. The solution of CaCl2 will thus have the lower freezing point. 12.102. Consider the equation ΔTb = iKbcm. For CaCl2, i = 3; for KCl it is 2. The product of i and cm will be larger for CaCl2, as will ΔTb. The solution of KCl will thus have the lower boiling point. 12.103. Assume there is 1.000 L of the solution, which will contain 18 mol H2SO4, molar mass 98.09 g/mol. The mass of the solution is Mass solution =
18 mol x 98.09 g/mol = 1802 g 0.98
Thus, the density of the solution is d =
1802 g = 1.802 = 1.8 g/mL 1000 mL
The mass of water in the solution is Mass H2O = 1802 g x 0.02 = 36.0 g = 0.0360 kg
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Thus, the molality of the solution is m =
18 mol H 2SO 4 = 500 = 5 x 102 m 0.0360 Kg
12.104. Assume there is 1.000 L of the solution, which will contain 15 mol H3PO4, molar mass 97.99 g/mol. The mass of the solution is Mass solution =
15 mol x 97.99 g/mol = 1729 g 0.85
Thus, the density of the solution is d =
1729 g = 1.729 = 1.7 g/mL 1000 mL
The mass of water in the solution is Mass H2O = 1729 g x 0.15 = 259 g = 0.259 kg Thus, the molality of the solution is m =
15 mol H3 PO 4 = 57.9 = 58 m 0.0259 Kg
12.105. Use the freezingpointdepression equation to find the molality of the solution. The freezing point of pure cyclohexane is 6.55°C, and Kf = 20.0°C/m. Thus m =
ΔT f
Kf
=
(6.55  5.28)°C = 0.06350 m 20.0°C/m
The moles of the compound are Moles of compound =
0.06350 mol x 0.00538 kg = 3.416 x 10−4 mol 1 kg solvent
The molar mass of the compound is Molar mass =
0.125 g = 365.8 g/mol 3.416 x 104 mol
The moles of the elements in 100 g of the compound are Moles of Mn = 28.17 g Mn x
1 mol = 0.51274 mol 54.94 g Mn
Moles of C = 30.80 g C x
1 mol = 2.5645 mol 12.01 g C
Moles of O = 41.03 g O x
1 mol = 2.5644 mol 16.00 g O
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Chapter 12: Solutions
This gives mole ratios of 1 mol Mn to 5 mol C to 5 mol O. Therefore, the empirical formula of the compound is MnC5O5. The mass of this formula unit is approximately 195 amu. Since the molar mass of the compound is 366. g/mol, the value of n is n =
366 g/mol = 1.88, or 2 195 g/unit
Therefore, the formula of the compound is Mn2C10O10. 12.106. Use the freezingpointdepression equation to find the molality of the solution. The freezing point of pure cyclohexane is 6.55°C, and Kf = 20.0°C/m. Thus m =
ΔT f
Kf
=
(6.55  5.23)°C = 0.06600 m 20.0°C/m
The moles of the compound are Moles of compound =
0.06600 mol x 0.00672 kg = 4.435 x 10−4 mol 1 kg solvent
The molar mass of the compound is Molar mass =
0.147 g = 331.4 g/mol 4.435 x 104 mol
The moles of the elements in 100 g of the compound are Moles of Co = 34.47 g Co x
1 mol = 0.58493 mol 58.93 g Co
Moles of C = 28.10 g C x
1 mol = 2.3397 mol 12.01 g C
Moles of O = 37.43 g O x
1 mol = 2.3394 mol 16.00 g O
This gives mole ratios of 1 mol Co to 4 mol C to 4 mol O. Therefore, the empirical formula of the compound is CoC4O4. The mass of this formula unit is approximately 171 amu. Since the molar mass of the compound is 331. g/mol, the value of n is n =
331 g/mol = 1.93, or 2 171 g/unit
Therefore, the formula of the compound is Co2C8O8.
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443
12.107. a.
Use the freezingpointdepression equation to find the molality of the solution. The freezing point of pure water is 0°C and Kf = 1.858°C/m. Thus m =
ΔT f
Kf
=
[0  (2.2)]°C = 1.184 m 1.858°C/m
The moles of the compound are Moles of compound =
1.184 mol x 0.100 kg = 0.1184 mol 1 kg solvent
The molar mass of the compound is Molar mass =
18.0 g = 152.0 g/mol 0.1184 mol
The moles of the elements in 100 g of the compound are Moles of C = 48.64 g C x Moles of H = 8.16 g H x Moles of O = 43.20 g O x
1 mol = 4.0500 mol 12.01 g C 1 mol = 8.095 mol 1.008 g H
1 mol = 2.7000 mol 16.00 g O
This gives mole ratios of 1.5 mol C to 3 mol H to 1 mol O. Multiplying by 2 gives ratios of 3 mol C to 6 mol H to 2 mol O. Therefore, the empirical formula of the compound is C3H6O2. The mass of this formula unit is approximately 74 amu. Since the molar mass of the compound is 152.0. g/mol, the value of n is n =
152.0 g/mol = 2.05, or 2 74 g/unit
Therefore, the formula of the compound is C6H12O4. b.
The molar mass of the compound is (to the nearest tenth of a gram) 6(12.01) + 12(1.008) + 4(16.00) = 148.156 = 148.2 g/mol
12.108. a.
Use the freezingpointdepression equation to find the molality of the solution. The freezing point of naphthalene is 80.0 °C and Kf = 6.8°C/m. Thus m =
ΔT f
Kf
=
(80.0  78.0)°C = 0.294 m 6.8°C/m
The moles of the compound are Moles of compound =
0.294 mol x 0.00750 kg = 2.21 x 10−3 mol 1 kg solvent
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Chapter 12: Solutions
The molar mass of the compound is Molar mass =
0.855 g = 387 g/mol 2.21 x 103 mol
The moles of the elements in 100 g of the compound are 1 mol = 3.2889 mol 12.01 g C
Moles of C = 39.50 g C x Moles of H = 2.21 g H x
1 mol = 2.192 mol 1.008 g H
Moles of Cl = 58.30 g Cl x
1 mol = 1.6446 mol 35.45 g Cl
This gives mole ratios of 2 mol C to 1.33 mol H to 1 mol Cl. Multiplying by 3 gives ratios of 6 mol C to 4 mol H to 3 mol Cl. Therefore, the empirical formula of the compound is C6H4Cl3. The mass of this formula unit is approximately 182.5 amu. Since the molar mass of the compound is 387. g/mol, the value of n is n =
387 g/mol = 2.12, or 2 182.5 g/unit
Therefore, the formula of the compound is C12H8Cl6. b.
The molar mass of the compound is (to the nearest tenth of a gram) 12(12.01) + 8(1.008) + 6(35.45) = 364.884 = 364.9 g/mol
12.109. Use the freezingpointdepression equation to find the molality of the solution. The freezing point of pure water is 0°C, and Kf = 1.86°C/m. Thus m =
ΔT f
Kf
=
[0  (2.3)]°C = 1.24 m 1.86°C/m
Since this is a relatively dilute solution, we can assume the molarity and molality of the fish blood are approximately equal. The calculated molarity is for the total number of particles, assuming they behave ideally.
π = MRT = 1.24 mol/L x 0.08206 L•atm/K•mol x 298.2 K = 30.3 = 30. atm 12.110. Use the osmoticpressure equation to find the molarity of the solution. M =
π RT
=
17 atm = 0.701 M 0.08206 L • atm / K • mol x 295.2 K
Since this is a relatively dilute solution, we can assume the molality and molarity of the solution are approximately equal and that the electrolyte (salt) is behaving ideally. The calculated temperature change is ΔT = Kfm = 1.86°C/m x 0.701 m = 1.30°C
Therefore, the solution will freeze at −1.3°C.
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445
12.111. Hemoglobin is the substance in red blood cells that carries oxygen. Hemoglobin is normally present in solution within the red blood cells. But in people with sicklecell anemia, the unoxygenated hemoglobin readily comes out of solution. It produces a fibrous precipitate that deforms the cell, giving it the characteristic sickle shape. Normal and sicklecell hemoglobins are almost exactly alike, except that each β chain of the hemoglobin responsible for sicklecell anemia differs from normal hemoglobin in one place. In this place, the normal hemoglobin has the group that helps confer water solubility on the molecule because of the polarity of the group and its ability to form hydrogen bonds. 12.112. In people with the disease, the red blood cells tend to become elongated (sickleshaped) when the concentration of oxygen (O2) is low, as it is in the venus blood supply. Once the red blood cells have sickled, they can no longer function in their normal capacity as oxygen carriers, and they often break apart. Moreover, the sickled cells clog capillaries, interfering with the blood supply to vital organs. 12.113. Like a soap ion, a phospholipid has a hydrophobic end (the two hydrocarbon groups) and a hydrophilic end (an ionic portion containing a phosphate group, –PO4−–). Like soap ions, phospholipid molecules in water tend to associate so that their hydrophilic (or ionic) heads dip into the water, with their hydrophobic (or hydrocarbon) tails pointing away. However, the hydrocarbon tails of a phospholipid molecule are too bulky to associate into micelles and instead form a bilayer, a layer two molecules thick. 12.114. Recently, chemists have prepared celllike structures (lipid vesicles) from phospholipids and have used them to carry out chemical reactions by combining the contents of two lipid vesicles, with diameters from about 50 nanometers to several micrometers. These lipid vesicles are the world’s smallest test tubes. In one experiment, two phospholipid vesicles, each about a micrometer in diameter and containing different reactant substances, were brought together under a special microscope. When the vesicles just touched, the researchers delivered an electrical pulse to them that opened a small pore in each vesicle. The two vesicles coalesced into one, and their contents reacted.
■
SOLUTIONS TO STRATEGY PROBLEMS
12.115. Moles solute = Molar mass =
ΔT f
Kf
x kg solvent =
1.34°C x 0.878 kg = 0.6332 mol 1.858°C/m
mass 79.3 g = = 125.2 = 125 g/mol moles 0.6332 mol
The molecular mass of the compound is 125 amu.
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Chapter 12: Solutions
12.116. First calculate the moles and molality of magnesium chloride. Use a molar mass of 95.21 g/mol. Moles = m =
6.69 g = 0.07026 mol 95.21 g/mol
mol solute 0.07026 mol = = 0.2888 m kg solvent 0.2433 kg
Next calculate the boilingpoint elevation. For magnesium chloride, MgCl2, i = 3. ΔTb = iKbm = (3) x (0.512°C/m) x (0.2888 m) = 0.4436 = 0.444°C.
Finally, the boiling point of the solution is Tb = Tb° + ΔTb = 100°C + 0.4436°C = 100.4436 = 100.444°C
12.117. S2 = S1 x
P2 24.6 atm = 179.3 = 179 g/L = 15.6 g/L x 2.14 atm P1
If the temperature of the liquid were increased, the solubility of the gas would decrease. 12.118. First calculate the mass of each component in a solution containing 1 mol. The molar mass of methanol (CH3OH) is 32.042 g/mol, and the molar mass of water is 18.015 g/mol. Mass CH3OH = 0.520 mol x Mass H2O = 0.480 mol x
32.042 g = 16.661 g 1 mol
18.015 g = 8.647 g 1 mol
The mass percentage water can now be calculated. Mass % H2O =
mass water 8.647 g 8.647 g = = = 0.3416 = 0.342 mass solution 8.647 g + 16.661 g 25.309 g
12.119. The molar mass of potassium chloride (KCl) is 74.55 g/mol. The mass required is (0.110 m) x (0.372 kg) x (74.55 g/mol) = 3.050 = 3.05 g 12.120. The molar mass of barium chloride (BaCl2) is 208.23 g/mol. 427 g x 0.0317 x
1 mol = 0.06500 = 0.0650 mol 208.23 g
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447
12.121. First calculate the moles of each substance. The molar mass of ethanol (C2H5OH) is 46.068 g/mol, and the molar mass of water is 18.015 g/mol. Moles of C2H5OH = 4.76 g x Moles of H2O = 50.0 g x
1 mol = 0.1033 mol 46.068 g
1 mol = 2.7754 mol 18.015 g
The mole fraction of ethanol can now be calculated. X C2 H5OH =
mol C2 H 5 OH 0.1033 mol = = 0.03589 = 0.0359 total mol 0.1033 mol + 2.775 mol
12.122. First, calculate the moles and molarity of the starch. Moles = 0.759 g x Molarity =
1 mol = 2.371 x 10−5 mol 3.20 x 104 g
2.371 x 105 mol = 2.117 x 10−4 M 0.112 L
The osmotic pressure can now be calculated.
π = MRT = 2.117 x 10−4 mol/L x 0.08206 L•atm/K•mol x 298.15 K x
760 torr 1 atm
= 3.937 = 3.94 torr 12.123. Assume 1 L of solution. The molar mass of magnesium chloride (MgCl2) is 95.21 g/mol. The mass of magnesium chloride, the mass of water, and the mass of the solution are Mass MgCl2 = 0.797 mol x
95.21 g = 75.882 g 1 mol
Mass solution = d x V = 1.108 g/mL x 1000 mL = 1108 g Mass water = 1108 g − 75.882 g = 1032.11 g = 1032.1 g = 1.0321 kg The molality can now be calculated. Molality =
0.797 mol = 0.7721 = 0.772 m 1.0321 kg
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Chapter 12: Solutions
12.124. First calculate the molarity of the solution. For CaCl2 use i = 3.
π
Molarity =
iRT
=
16 atm = 0.2180 M 3 x 0.08206 L • atm / K • mol x 298.15 K
Next, determine the molality of the solution. Assume 1 L of solution. The molar mass of calcium chloride (CaCl2) is 110.98 g/mol. The mass of calcium chloride, the mass of water, and the mass of the solution are Mass CaCl2 = 0.2180 mol x
110.98 g = 24.19 g 1 mol
Mass solution = d x V = 1.108 g/mL x 100 mL = 1108 g Mass H2O = 1108 g − 24.19 g = 1083.8 g = 1.0838 kg The molality of the solution is Molality =
0.2180 mol = 0.2011 m 1.0838 kg
For CaCl2, i = 3 The freezingpoint depression is ΔTf = iKfm = (3) x (1.858°C/m) x (0.2011 m) = 1.121 = 1.1°C
The freezing point of the solution can now be calculated. Tf = Tf° − ΔTf = 0°C − 1.121°C = −1.121 = −1.1°C
■
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
12.125. First, determine the initial moles of each ion present in the solution before any reaction has occurred. There are five ions present: Na+, 2 x 0.375 = 0.750 mol; CO32−, 0.375 mol; Ca2+, 0.125 mol; Ag+, 0.200 mol; and NO3−, 0.200 + 2 x 0.125 = 0.450 mol. Two precipitates form: CaCO3 and Ag2CO3. The balanced net ionic equations for their formation are Ca2+(aq) + CO32−(aq) +
2−
2Ag (aq) + CO3 (aq)
→
CaCO3(s)
→
Ag2CO3(s)
There is sufficient CO3 to precipitate all of the Ca2+ and all of the Ag+ ions. The moles of excess CO32− are calculated as follows: 0.375 − 0.125 − 1/2 x 0.200 = 0.150 mol remaining. Since the volume is 2.000 L, the molarities of the various ions are 2−
M of CO32− left = 0.150 mol ÷ 2.000 L = 0.0750 M M of NO3− left
= 0.450 mol ÷ 2.000 L = 0.225 M
M of Na+ left
= 0.750 mol ÷ 2.000 L = 0.375 M
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449
12.126. First, determine the initial moles of each ion present in the solution before any reaction has occurred. There are five ions present: Na+, 3 x 0.310 = 0.930 mol; PO43−, 0.310 mol; Ca2+, 0.100 mol; Ag+, 0.150 mol; and NO3−, 0.150 + 2 x 0.100 = 0.350 mol. Two precipitates form: Ca3(PO4)2 and Ag3PO4. The balanced net ionic equations for their formation are 3 Ca2+(aq) + 2 PO43−(aq)
→
Ca3(PO4)2(s)
3 Ag+(aq) + PO43−(aq)
→
Ag3PO4(s)
There is sufficient PO4 to precipitate all of the Ca2+ and all of the Ag+ ions. The moles of excess PO43− is calculated as follows: 0.310 − 2/3 x 0.100 − 1/3 x 0.150 = 0.1933 mol remaining. Since the volume is 4.000 L, the molarities of the various ions are 3−
M of PO43− left = 0.1933 mol ÷ 4.000 L = 0.04833 = 0.0483 M M of NO3− left
= 0.350 mol ÷ 4.000 L = 0.0875 M
M of Na+ left
= 0.930 mol ÷ 4.000 L = 0.2325 = 0.233 M
12.127. Na+(g) + Cl−(g)
→
ΔH = 787 kJ/mol
NaCl(s) +
−
NaCl(s)
→
Na (aq) + Cl (aq)
ΔH =
Na+(g) + Cl−(g)
→
Na+(aq) + Cl−(aq)
ΔH = 783 kJ/mol
+ 4 kJ/mol
The heat of hydration of Na+ is Na+(g) + Cl−(g) −
Na+(aq) + Cl−(aq)
→
−
ΔH = 783 kJ/mol
Cl (aq)
→
Cl (g)
ΔH = +338 kJ/mol
Na+(g)
→
Na+(aq)
ΔH = 445 kJ/mol
12.128. K+(g) + Cl−(g)
→
ΔH = 717 kJ/mol
KCl(s) +
−
KCl(s)
→
K (aq) + Cl aq)
ΔH =
K+(g) + Cl−(g)
→
K+(aq) + Cl−aq)
ΔH = 699 kJ/mol
+18 kJ/mol
The heat of hydration of K+ is K+(g) + Cl−(g)
→
K+(aq) + Cl−aq) −
ΔH = 699 kJ/mol
−
Cl (aq)
→
Cl g)
ΔH = +338 kJ/mol
K+(g)
→
K+(aq)
ΔH = 361 kJ/mol
Less heat is evolved in the hydration of K+(g) than in that of Na+(g) because smaller ions have larger hydration energies.
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Chapter 12: Solutions
1 mol = 0.060854 mol MgSO4•7H2O 246.5 g
12.129. 15.0 g MgSO4•7H2O x
7 mol H 2 O 18.0 g H 2 O x = 7.667 g H2O 1 mol hydrate 1 mol H 2 O
0.060854 mol MgSO4•7H2O x
kg H2O = (100.0 g H2O + 7.667 g H2O) x m =
0.060854 mol MgSO 4 0.10766 kg H 2 O
12.130. 15.0 g Na2CO3•10H2O x
= 0.5652 mol/kg = 0.565 m
1 mol = 0.05241 mol Na2CO3•10H2O, or Na2CO3 286.2 g
0.05241 mol Na2CO3•10H2O x
10 mol H 2 O 18.0 g H 2 O x = 9.434 g H2O 1 mol hydrate 1 mol H 2 O
kg H2O = (100 g H2O + 9.434 g H2O) x m =
100 g soln x
1 mol = 0.06007 mol CuSO4•5H2O, or CuSO4 249.7 g
1 mL soln 1L x = 0.085689 L 1.167 g soln 1000 mL
0.06007 mol CuSO 4 = 0.70102 = 0.701 mol/L 0.085689 L
12.132. 20.0 g Na2S2O3•5H2O x 100 g soln x M =
1 kg H 2 O = 0.10943 kg H2O 1000 g H 2 O
0.05241 mol Na 2CO3 = 0.4789 mol/kg = 0.479 m 0.10943 kg H 2 O
12.131. 15.0 g CuSO4•5H2O x
M =
1 kg H 2 O = 0.10766 kg H2O 1000 g H 2 O
1 mol = 0.08058 mol Na2S2O3•5H2O, or Na2S2O3 248.2 g
1 mL soln 1L = 0.08517 L x 1000 mL 1.174 g soln
0.08058 mol Na 2S2 O3 = = 0.9461 = 0.946 mol/L 0.08517 L
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12.133. 0.159°C x
m 0.08557 mol AA + H + = 0.08557 m = 1000 g (or 1000 mL) H 2 O 1.858°C
Note that 0.0830 mol AA + mol H+ = 0.08557 mol/L (AA + H+). Moles of H+ = 0.08557 − 0.0830 = 0.00257 mol Percent dissoc. =
12.134. 0.210°C x
0.00257 mol mol H + x 100% = x 100% = 3.09 = 3.1% 0.0830 mol mol AA
m 0.1130 mol FA + H + = 0.1130 m = 1000 g (or 1000 mL) H 2 O 1.858°C
Note that 0.109 mol FA + mol H+ = 0.113 mol/L (FA + H+). Moles of H+ = 0.113 − 0.109 = 0.0040 mol Percent dissoc. =
0.0040 mol mol H + x 100% = x 100% = 3.69 = 4 percent 0.109 mol mol FA
12.135. Calculate the empirical formula first, using the masses of C, O, and H in 1.000 g: 1.434 g CO2 x
12.01 g C = 0.39132 g C 44.01 g CO 2
0.783 g H2O x
2.016 g H = 0.08761 g H 18.016 g H 2 O
g O = 1.000 g − 0.39132 g − 0.08761 g = 0.5211 g O Mol C = 0.39132 g C x 1 mol/12.01 g = 0.03258 mol C (lowest integer = 1) Mol H = 0.08761 g H x 1 mol/1.008 g = 0.08691 mol H (lowest integer = 8/3) Mol O = 0.5211 g O x 1 mol/16.00 g = 0.03257 mol O (lowest integer = 1) Therefore, the empirical formula is C3H8O3. The formula mass from the freezing point is calculated by first finding the molality: 0.0894°C x (m/1.858°C) = 0.04811 = (0.04811 mol/1000 g H2O) (0.04811 mol/1000 g H2O) x 25.0 g H2O = 0.001203 mol (in 25.0 g H2O) Molar mass = Mm = 0.1107 g/0.001203 mol = 92.02 g/mol Because this is also the formula mass, the molecular formula is also C3H8O3.
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Chapter 12: Solutions
12.136. Calculate the empirical formula first, using the masses of C, O, and H in 1.000 g: 1.418 g CO2 x
12.01 g C = 0.3870 g C 44.01 g CO 2
0.871 g H2O x
2.016 g H = 0.09746 g H 18.016 g H 2 O
g O = 1.000 g − 0.3870 g − 0.09746 g = 0.5155 g O Mol C = 0.3870 g C x 1 mol/12.01 g = 0.0322 mol C (lowest integer = 1) Mol H = 0.09746 g H x 1 mol/1.008 g = 0.09668 mol H (lowest integer = 3) Mol O = 0.5155 g O x 1 mol/16.00 g = 0.03222 mol O (lowest integer = 1) The empirical formula is CH3O. For the freezingpoint calculation, the molality is calculated as follows: 0.0734°C x (m/1.858°C) = 0.039505 = (0.39505 mol/1000 g H2O) (0.39504 mol/1000 g H2O) x 45.0 g H2O = 0.001778 mol Molar mass = Mm = 0.1103 g/0.0017778 mol = 62.04 g/mol Because the formula mass of CH3O is 31.02, the molecular formula is C2H6O2.
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CHAPTER 13
Rates of Reaction
■
SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 13.1. Rate of formation of NO2F = Δ[NO2F]/Δt. Rate of reaction of NO2 = −Δ[NO2]/Δt. Divide each rate by the coefficient of the corresponding substance in the equation: 1/2
Δ[NO 2 F] Δ[NO 2 ] Δ[NO 2 F] Δ[NO 2 ] = −1/2 ; or = − Δt Δt Δt Δt
13.2. Rate = −
Δ[I ] Δ[0.00101 M  0.00169 M] = − = 1.13 x 10−4 = 1.1 x 10−4 M/s Δt 8.00 s  2.00 s
13.3. The order with respect to CO is zero and with respect to NO2 is 2. The overall order is 2, the sum of the exponents in the rate law. 13.4. By comparing Experiments 1 and 2, you see the rate is quadrupled when the [NO2] is doubled. Thus, the reaction is second order in NO2, and the rate law is Rate = k[NO2]2 The rate constant may be found by substituting experimental values into the ratelaw expression. Based on values from Experiment 1, k=
rate 7.1 x 10 mol/(L • s) = = 0.710 = 0.71 L/(mol•s) [NO 2 ]2 (0.010 mol/L) 2
13.5. a.
For [N2O5] after 6.00 x 102 s, use the firstorder rate law, and solve for the concentration at time t: ln
[N 2 O5 ]t = −(4.80 x 10−4 /s)(6.00 x 102 s) = −0.2880 2 [1.65 x 10 M]
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Chapter 13: Rates of Reaction
Take the antilog of both sides to get [N 2 O5 ]t = e−0.2880 = 0.74976 [1.65 x 102 M]
Hence, [N2O5]t = 1.65 x 10−2 M x 0.74976 = 0.01237 = 0.0124 mol/L b.
Substitute the values into the rate equation to get ln
10.0% = −(4.80 x 10−4/s) x t 100.0%
Taking the log on the left side gives ln(0.100) = −2.30258. Hence, −2.30258 = −(4.80 x 10−4 /s) x t Or, t=
2.30258 = 4797 = 4.80 x 103 s (80.0 min) 4.80 x 104 /s
13.6. Substitute k = 9.2/s into the equation relating k and t1/2. t1/2 =
0.693 0.693 = = 0.0753 = 0.075 s k 9.2/s
By definition, the halflife is the amount of time it takes to decrease the amount of substance present by onehalf. Thus, it takes 0.0753 s for concentration to decrease by 50% and another 0.0753 s for the concentration to decrease by 50% of the remaining 50% (to 25% left), for a total of 0.1506, or 0.151 s. 13.7. Solve for Ea by substituting the given values into the twotemperature Arrhenius equation: ln
Ea 2.14 x 102 1 ⎞ ⎛ 1 = ⎜ ⎟ 836 K ⎠ 8.31 J/(mol • K) ⎝ 759 K 1.05 x 103
3.0146 = Ea =
Ea (1.2135 x 10−4 /K) 8.31 J/(mol • K)
3.0146 x 8.31 J/mol = 2.064 x 105 = 2.06 x 105 J/mol 1.2135 x 104
Solve for the rate constant, k2, at 865 K by using the same equation and using Ea = 2.064 x 105 J/mol: ln
k2 2.064 x 105 J/mol ⎛ 1 1 ⎞ = ⎜ ⎟ = 0.99623 2 1/2 865 K ⎠ 8.31 J/(mol • K) ⎝ 836 K 2.14 x 10 /(M • s)
Taking antilogarithms, k2 = e0.99623 = 2.7080 2 1/2 2.14 x 10 /(M • s)
k2 = 2.7080 x (2.14 x 10−2)/(M1/2•s) = 5.7952 x 10−2 = 5.80 x 10−2 (M1/2•s)
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13.8. The net chemical equation is the overall sum of the two elementary reactions: H2O2 + I−
→
H2O2 + IO− → 2H2O2 →
H2O + IO− H2O + O2 + I− 2H2O + O2
The IO− is an intermediate; I− is a catalyst. Neither appears in the net equation. 13.9. The reaction is bimolecular because it is an elementary reaction that involves two molecules. 13.10. For NO2 + NO2 → N2O4, the rate law is Rate = k[NO2]2 (The rate must be proportional to the concentration of both reactant molecules.) 13.11. The first step is the slow, ratedetermining step. Therefore, the rate law predicted by the mechanism given is Rate = k1 [H2O2][I−] 13.12. According to the ratedetermining (slow) step, the rate law is Rate = k2[NO3][NO] Eliminate NO3 from the rate law by looking at the first step, which is fast and reaches equilibrium. At equilibrium, the forward rate and the reverse rate are equal. k1[NO][O2] = k−1[NO3] Therefore, [NO3] = (k1/k−1)[NO][O2], so Rate =
k 2 k1 [NO]2[O2] = k[NO]2[O2] k 1
where k2(k1/k−1) has been replaced by k, which represents the experimentally observed rate constant.
■
ANSWERS TO CONCEPT CHECKS
13.1. a.
Since the slope is steeper at point A, point A must be a faster instantaneous rate.
b.
Since the curve is not a flat line, the rate of reaction must be constantly changing over time. Therefore, the rate for the reaction cannot be constant at all points in time.
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Chapter 13: Rates of Reaction
13.2. a.
Keeping in mind that all reactant species must be present in some concentration for a reaction to occur, the reaction with [Q] = 0 is the slowest because no reaction occurs. The other two reactions are equal in rate because the reaction is zero order with respect to [Q]; as long as there is some amount of Q present, the reaction rate depends on the [R], which is constant in this case.
b.
Since [Q]o = 1, you can rewrite the rate law as follows: Rate = k[R]2.
a.
A possible rate law is Rate of aging = (diet)w(exercise)x(sex)y(occupation)z. Your rate law probably will be different; however, the general form should be the same.
b.
You will need a sample of people who have all the factors the same except one. For example, using the equation given in part a, you could determine the effect of diet if you had a sample of people who were the same sex, exercised the same amount, and had the same occupation. You would need to isolate each factor in this fashion to determine the exponent on each factor.
c.
The exponent on the smoking factor would be 2 since you see a fourfold rate increase: [2]2 = 4.
13.3.
13.4. The halflife of a firstorder reaction is constant over the course of the reaction. The halflife of a secondorder reaction depends on the initial concentration and becomes longer as time elapses. Thus, the reaction must be second order because the halflife increases from 20 s to 40 s after time has elapsed. 13.5. a.
Since the “hump” is larger, the A + B reaction has a higher activation energy.
b.
Since the activation energy is lower, the E + F reaction would have the larger rate constant. Keep in mind the inverse relationship between the activation energy, Ea, and the rate constant, k.
c.
Since, in both cases, energy per mole of the reactants is greater than that of the products, both reactions are exothermic.
a.
Her finding should increase the rate since the activation energy, Ea, is inversely related to the rate constant, k; a decrease in Ea results in an increase in the value of k.
b.
This is possible because the rate law does not have to reflect the overall stoichiometry of the reaction.
c.
No. Since the rate law is based on the slow step of the mechanism, it should be Rate = k[Y]2.
13.6.
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457
ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
13.1. The four variables that can affect rate are (1) the concentrations of the reactants, although in some cases a particular reactant's concentration does not affect the rate; (2) the presence and concentration of a catalyst; (3) the temperature of the reaction; and (4) the surface area of any solid reactant or solid catalyst. 13.2. The rate of reaction of HBr can be defined as the decrease in HBr concentration (or the increase in Br2 product formed) over the time interval, Δt: Rate = −1/4
Δ [Br2 ] Δ [Br2 ] Δ [HBr] Δ [HBr] = 1/2 or − =2 Δt Δt Δt Δt
13.3. Two physical properties used to determine the rate are color, or absorption of electromagnetic radiation, and pressure. If a reactant or product is colored, or absorbs a different type of electromagnetic radiation than the other species, then measurement of the change in color (change in absorption of electromagnetic radiation) may be used to determine the rate. If a gas reaction involves a change in the number of gaseous molecules, measurement of the pressure change may be used to determine the rate. 13.4. Use the general example of a rate law in Section 13.3, where A and B react to give D and E with C as a catalyst: Rate = k [A]m [B]n [C]p Note that the exponents m, n, and p are the orders of the individual reactants and catalyst. Assuming that m and n are positive numbers, the rate law predicts that increasing the concentrations of A and/or B will increase the rate. In addition, the rate will be increased by increasing the surface of the solid catalyst (making it as finely divided as possible, etc.). Finally, increasing the temperature will increase the rate constant, k, and increase the rate. 13.5. An example that illustrates that exponents have no relationship to coefficients is the reaction of nitrogen monoxide and hydrogen from Example 13.12 in the text. 2NO + 2H2 → N2 + 2H2O. The experimental rate law given there is Rate = k[NO]2[H2] Thus, the exponent for hydrogen is 1, not 2 like the coefficient, and the overall order is 3, not 4 like the sum of the coefficients. 13.6. The rate law for this reaction of iodide ion, arsenic acid, and hydrogen ion is Rate = k[I−][H3AsO4] [H+] The overall order is 1 + 1 + 1 = 3 (third order).
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13.7. Use m to symbolize the reaction order as is done in the text. Then, from the table for m and the change in rate in the text, m is 2 when the rate is quadrupled (increased fourfold). Using the equation in the text gives the same result: 2m = new rate/old rate = 4/1; thus, m = 2 13.8. Use m to symbolize the reaction order as is done in the text. The table for m and the change in rate in the text cannot be used in this case. When m = 0.5, the new rate should be found using the equation in the text: 20.50 =
2 = 1.41 = new rate/old rate
Thus, the new rate is 1.41 times the old rate. 13.9. Use the halflife concept to answer the question without an equation. If the halflife for the reaction of A(g) is 25 s, then the time for A(g) to decrease to 1/4 the initial value is two halflives, or 2 x 25 = 50 s. The time for A(g) to decrease to 1/8 the initial value is three halflives, or 3 x 25 = 75 s. 13.10. The halflife equation for a firstorder reaction is t1/2 = 0.693/k. The halflife for the reaction is constant, independent of the reactant concentration. For a secondorder reaction, the halflife equation is t1/2 = 1/(k[A]o), and it depends on the initial reactant concentration. 13.11. According to transitionstate theory, the two factors that determine whether a collision results in reaction or not are (1) the molecules must collide with the proper orientation to form the activated complex, and (2) the activated complex formed must have a kinetic energy greater than the activation energy. 13.12. The potentialenergy diagram for the exothermic reaction of A and B to give activated complex AB‡ and products C and D is given below. AB
‡
Energy
E a (forward) Ea (reverse) A + B Reactants ΔH C + D Products Progress of reaction
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13.13. The activated complex for the reaction of NO2 with NO3 to give NO, NO2, and O2 has the structure below (dashed lines represent bonds about to form or break): O–N–O–O–N(–O)2 13.14. The Arrhenius equation expressed with the base e is k = A eE /RT a
The A term is the frequency factor and is equal to the product of p and Z from collision theory. The term p is the fraction of collisions with properly oriented reactant molecules, and Z is the frequency of collisions. Thus, A is the number of collisions with the molecules properly oriented. The Ea term is the activation energy, the minimum energy of collision required for two molecules to react. The R term is the gas constant, and T is the absolute temperature. 13.15. In the reaction of NO2(g) with CO(g), an example of an intermediate is the temporary formation of NO3 from the reaction of two NO2 molecules in the first step: NO2 + NO2 → NO3 + NO NO3 + CO → NO2 + CO2 13.16. It is generally impossible to predict the rate law from the equation alone, because most reactions consist of several elementary steps whose combined result is summarized in the rate law. If these elementary steps are unknown, the rate law cannot be predicted. 13.17. The mechanism cannot be 2NO2Cl → 2NO2 + Cl2 because the reaction is first order in NO2Cl. The order in NO2Cl would have to reflect the total number of molecules (two) for the proposed mechanism, but it does not. 13.18. The characteristic of the ratedetermining step in a mechanism is that it is, relatively speaking, the slowest step of all the elementary steps (even though it may occur in seconds). Thus, the rate of disappearance of reactant(s) is limited by the rate of this step. 13.19. For the rate of decomposition of N2O4, Rate = k1[N2O4] For the rate of formation of N2O4, Rate = k−1[NO2]2 At equilibrium the rates are equal, so k1[N2O4] = k−1[NO2]2 [N2O4] = (k−1/k1) [NO2]2 13.20. A catalyst operates by providing a pathway (mechanism) that occurs faster than the uncatalyzed pathway (mechanism) of the reaction. The catalyst is not consumed because after reacting in an early step, it is regenerated in a later step.
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Chapter 13: Rates of Reaction
13.21. In physical adsorption, molecules adhere to a surface through weak intermediate forces, whereas in chemisorption, the molecules adhere to the surface by stronger chemical bonding. 13.22. In the first step of catalytic hydrogenation of ethylene, the ethylene and hydrogen molecules diffuse to the catalyst surface and undergo chemisorption. Then the pi electrons of ethylene form temporary bonds to the metal catalyst, and the hydrogen molecule breaks into two hydrogen atoms. The hydrogen atoms next migrate to an ethylene held in position on the metal catalyst surface, forming ethane. Finally, because it cannot bond to the catalyst, the ethane diffuses away from the surface. 13.23. The answer is c, [A] = 0.0250 M, [B] = 0.0400 M. 13.24. The answer is e, I and III only. 13.25. The answer is e; if the rate law for the reaction is Rate = k[E]2[F], then doubling the concentration of E will cause the rate to increase by a factor of 4 (4 times faster). 13.26. The answer is c, doubling the [C]o while keeping the [A]o and [B]o constant.
■
ANSWERS TO CONCEPT EXPLORATIONS
13.27. a.
Rate = k[A]2[B]
b.
The greatest reaction rate will be for the container with the largest value of [A]2[B]. For container W, this product is (2)2(3) = 12, for container X it is (2)2(4) = 16, for container Y it is (3)2(2) = 18, and for container Z it is (5)2(0) = 0. Therefore, container Y will have the greatest reaction rate.
c.
The slowest reaction rate will be for container Z because the concentration of one of the reactants (B) is zero.
d.
When the volume of the container is doubled from 1.0 L to 2.0 L, the concentrations of A and B are reduced by a factor of 2. This corresponds to a reduction in rate of (1/2)2(1/2) = 1/8. Thus, the reaction rate in the larger container is 1/8 the rate in the smaller container.
e.
Usually, reaction rates increase with an increase in temperature. At a higher temperature, molecules have a greater kinetic energy and collide with other molecules at a higher rate. Thus, the fraction of collisions with energy in excess of Ea is greater, and the reaction rate increases.
f.
To double the reaction rate, the product [A]2[B] must also double. This can be accomplished by doubling [B] while leaving [A] unchanged. Another possibility is to double [A] and reduce [B] by onehalf.
g.
The slowest rate of formation of C is for container Z, where the reaction rate is zero.
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h.
Since the reaction is second order in A and only first order in B, changes in the concentration of A will have the greater impact on the rate of reaction, so removing a unit of A will have the greater impact on the rate of reaction in container W.
i.
The reaction rates are −
Δ[A] 1 Δ[B] =− Δt 2 Δt
Thus, the rate of consumption of A is onehalf the rate of consumption of B. j.
Since [C] does not appear in the rate law, removing C from the container should have no effect on the reaction rate.
13.28. a.
Since the graph is not linear, there is no way to determine the reaction order by inspection. A secondorder reaction should exhibit linear behavior if a plot is made of 1/[A] vs time.
b.
The data seem to better support a firstorder reaction. As the time changes by 2 seconds, the concentration of A decreases by 102. This is the behavior expected for an exponential rate law like a firstorder reaction rate law, [A] = [A]0e−kt.
c.
The rate constant is k=
d.
0.693 ln 2 = = 1.980 = 1.98 /s 0.35 s t1/2
A possible mechanism with intermediates X and Y is A → X + Y (slow step) X + Y → B + C (fast step)
e.
The behavior of the rate constant k as a function of temperature is described by the Arrhenius equation. ln
E ⎛1 k2 1⎞ = a⎜ ⎟ R ⎝ T1 T2 ⎠ k1
As the temperature increases the rate constant also increases. More molecules collide with sufficient energy to overcome the potential energy barrier so the rate of reaction increases.
■
ANSWERS TO CONCEPTUAL PROBLEMS
13.29. a.
You can write the rate expression in terms of the depletion of A: Rate of depletion of A = −
Δ[A] Δt
or you can write the rate expression in terms of the formation of B: Rate of formation of B = +
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Δ[B] Δt
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b.
No. Consider the stoichiometry of the reaction, which indicates that the rate of depletion of A would be faster than the rate of formation of B; for every 3 moles of A that are consumed, 2 moles of B would be formed.
c.
Taking into account the stoichiometry of the reaction, the two rate expressions that would give an equal rate when calculated over the same time interval are Rate = −
Δ[A] Δ[B] = 3Δt 2 Δt
13.30. You cannot write the rate law for this reaction from the information given. The rate law can be determined only by experiment, not by the stoichiometry of the reaction. 13.31. a.
The rate law for a secondorder reaction is Rate = k[A]2.
b.
The faster reaction rate will correspond to the container with the higher concentration of A. Both containers contain the same number of A particles, but the volume of the container on the right is only onehalf the volume of the other. Therefore, the initial concentration of A in the container on the right is double the initial concentration of A in the left container. Thus, the reaction will be faster in the smaller container on the right.
c.
For a secondorder reaction, the relationship among the halflife, rate constant, and initial concentration of A is t1/2 =
1 k[A]o
Since the halflife is inversely proportional to the initial concentration, the shorter halflife will correspond to the higher initial concentration of A, which is in the right container. d.
The relative rates of the reactions can be determined as follows. Since the initial concentration of A in the container on the right is double the initial concentration of A in the left container, the ratio of the rate in the right container to that on the left is 2 ⎛ [A] k[A]o,R RateR = = ⎜⎜ o,R 2 k[A]o,L Rate L ⎝ [A]o,L
2
⎞ ⎛ 2[A]o,L ⎟⎟ = ⎜⎜ ⎠ ⎝ [A]o,L
2
⎞ 2 ⎟⎟ = 2 = 4 ⎠
Thus, the reaction rate in the right container is four times the reaction rate of the container on the left. e.
Since both containers start with the same number of A particles, and the reaction rate is faster in the right container, more A particles will have reacted in this container, so it will contain fewer atoms.
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13.32. a.
The reaction is first order with a halflife of 10 seconds. Starting with 16 AB particles in the container, after 10 seconds (one halflife), 8 of the particles will have reacted and 8 will remain unreacted. After 20 seconds (two halflives), 12 of the particles will have reacted and 4 will remain unreacted. The pictures are
t=0s
t = 10 s
t = 20 s
b.
If the halflife is the same for the secondorder reaction, the container t = 20 s will have more AB and fewer A and B particles than from part a.
c.
After 10 seconds (one halflife), the concentration of the particles is onehalf their initial value. If we call the particles A, then for the firstorder case, the relative reaction rates at the start and after 10 seconds are k[A]o,0 [A]o,0 Rate0 [A]o = = = =2 k[A]o,10 [A]o,10 Rate10 1/2[A]o
Thus, for the firstorder case, after 10 seconds the rate is onehalf the initial rate. d.
After 10 seconds (one halflife), the concentration of the particles is onehalf their initial value. If we call the particles A, then for the secondorder case, the relative reaction rates at the start and after 10 seconds are 2
2
2 ⎛ [A] ⎞ k[A]o,0 ⎛ [A]o ⎞ Rate0 2 = = ⎜⎜ o,0 ⎟⎟ = ⎜ ⎟ =2 =4 2 k[A]o,10 [A] 1/2[A] Rate10 o,10 o ⎠ ⎝ ⎝ ⎠
Thus, for the secondorder case, after 10 seconds, the rate is onefourth the initial rate. 13.33. a.
If the concentration is tripled but there is no effect on the rate, the order of the reaction must be zero. Thus, x = 0.
b.
If the concentration is doubled and the rate doubles, it is a firstorder reaction. Thus, x = 1.
c.
If the concentration is tripled and the rate goes up by a factor of 27, it is a thirdorder reaction. Thus, x = 3.
13.34. a.
E must be a product since its concentration increases with time. If E were a reactant, you would expect the concentration to decrease over time.
b.
The average rate is faster between points A and B because the slope of the curve is steeper in this region. Remember, the steeper the curve, the greater the rate of change.
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Chapter 13: Rates of Reaction
13.35. a.
The rate has a constant value in region C because the slope of the curve is constant (flat) in this region.
b.
The rate is the fastest in region A because the slope of the curve is steepest in this region.
13.36. Since NO2 is a product in the reaction, its concentration must increase with time. The only graph that has [NO2] increasing with time is curve C. 13.37. A number of answers will work as long as you match one of the existing concentration of A. For example: [A] = 2.0 M with [B] = 2.0 M, or [A] = 1.0 M with [B] = 2.0 M. 13.38. The Arrhenius equation is k = AeE /RT a
■
a.
When the temperature is decreased, the rate constant, k, will also decrease. When k decreases, the rate also decreases.
b.
When the activation energy is increased, the rate constant, k, also decreases. When k decreases, the rate also decreases.
c.
Since the activation energy is in the numerator and the temperature is in the denominator, you cannot predict the effect without knowing the magnitude of the changes.
SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 13.39. For the reaction 2NO2 → 2NO + O2, the rate of decomposition of NO2 and the rate of formation of O2 are, respectively, Rate = − Δ[NO2]/Δt Rate = Δ[O2]/Δt To relate the two rates, divide each rate by the coefficient of the corresponding substance in the chemical equation and equate them. −1/2
Δ[NO 2 ] Δ[O 2 ] = Δt Δt
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13.40. For the reaction H2 + I2 → 2HI, the rate of formation of HI and the rate of decomposition of H2 are, respectively, Rate = Δ[HI]/Δt Rate = − Δ[H2] /Δt To relate the two rates, divide each rate by the coefficient of the corresponding substance in the chemical equation and equate them. −
Δ[H 2 ] Δ[HI] = 1/2 Δt Δt
13.41. For the reaction 5Br− + BrO3− + 6H+ → 3Br2 + 3H2O, the rate of decomposition of Br− and the rate of decomposition of BrO3− are, respectively, Rate = − Δ[Br−]/Δt Rate = − Δ[BrO3−]/Δt To relate the two rates, divide each rate by the coefficient of the corresponding substance in the chemical equation and equate them. 1/5
Δ[BrO3 ] Δ[Br  ] = Δt Δt
13.42. For the reaction 3I− + H3AsO4 + 2H+ → I3− + H3AsO3 + H2O, the rate of decomposition of I− and the rate of formation of I3− are, respectively, Rate = − Δ[I−]/Δt Rate = Δ[I3−]/Δt To relate the two rates, divide each rate by the coefficient of the corresponding substance in the chemical equation and equate them. −1/3
Δ[I3 ] Δ[I ] = Δt Δt
13.43. Rate = −
Δ[NH 4 NO 2 ] [0.0432 M  0.500 M ] =− = 2.27 x 10−2 = 2.3 x 10−2 M/hr [3.00 hr  0.00 hr] Δt
13.44. Rate = −
Δ[FeCl3 ] [0.02638 M  0.3586 M ] =− = 0.002370 = 0.00237 M/min Δt [4.00 min  0.00 min]
13.45. Rate = −
Δ[Azo.] [0.0101 M  0.0150 M ] 1 min =− x = 1.16 x 10−5 = 1.2 x 10−5 M/s Δt 7.00 min  0.00 min 60 sec
13.46. Rate = −
Δ[NO 2 ] [0.1076 M  0.1103 M ] =− = 4.50 x 10−5 = 4.5 x 10−5 M/s 60.0 s  0.0 s Δt
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13.47. If the rate law is Rate = k[H2S][Cl2], the order with respect to H2S is 1 (first order), and the order with respect to Cl2 is also 1 (firstorder). The overall order is 1 + 1 = 2, secondorder. 13.48. If the rate law is Rate = k[NO]2[Cl2], the order with respect to NO is 2 (second order), and the order with respect to Cl2 is 1 (firstorder). The overall order is 2 + 1 = 3, thirdorder. 13.49. If the rate law is Rate = k[MnO4−][H2C2O4], the order with respect to MnO4− is 1 (first order), the order with respect to H2C2O4 is 1 (first order), and the order with respect to H+ is zero. The overall order is 2, second order. 13.50. If the rate law is rate = k[H2O2][Fe2+], the order with respect to H2O2 is 1 (first order), the order with respect to Fe2+ is 1 (first order), and the order with respect to H+ is zero. The overall order is 2, second order. 13.51. The reaction rate doubles when the concentration of CH3NNCH3 is doubled, so the reaction is first order in azomethane. The rate equation should have the form Rate = k[CH3NNCH3] Substituting values for the rate and concentration yields a value for k: k=
rate 2.8 x 106 M/s = = 2.47 x 10−4 = 2.5 x 10−4/s [Azo.] 1.13 x 102 M
13.52. The reaction rate doubles when the concentration of ethylene oxide is doubled, so the reaction is first order in ethylene oxide. The rate equation should have the form Rate = k[C2H4O] Substituting values for the rate and concentration yields a value for k: k=
rate 5.57 x 107 M/s = = 2.047 x 10−4 = 2.05 x 10−4/s 3 [Et. Ox.] 2.72 x 10 M
13.53. Doubling [NO] quadruples the rate, so the reaction is second order in NO. Doubling [H2] doubles the rate, so the reaction is first order in H2. The rate law should have the form Rate = k[NO]2[H2] Substituting values for the rate and concentrations yields a value for k: k=
rate 2.6 x 105 M/s = = 2.88 x 102 = 2.9 x 102/M2s 3 2 3 2 [NO] [H 2 ] [6.4 x 10 M ] [2.2 x 10 M ]
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13.54. Doubling [NO] quadruples the rate, so the reaction is second order in NO. Doubling [O2] doubles the rate, so the reaction is first order in O2. The rate law should have the form Rate = k[NO]2[O2] Substituting values for the rate and concentrations yields a value for k: k=
rate 0.0281 M/s = = 7.108 x 103 = 7.11 x 103/M2s [NO]2 [O 2 ] [0.0125 M ]2 [0.0253 M ]
13.55. By comparing Experiments 1 and 2, you see that tripling [ClO2] increases the rate ninefold; that is, 3m = 9, so m = 2 (and the reaction is second order in ClO2). From Experiments 2 and 3, you see that tripling [OH−] triples the rate, so the reaction is first order in OH−. The rate law is Rate = k[ClO2]2[OH−] Substituting values for the rate and concentrations yields a value for k: k=
rate 0.0248 M/s = = 2.29 x 102 = 2.3 x 102/M2s [ClO 2 ]2 [OH  ] [0.060 M ]2 [0.030 M ]
13.56. By comparing Experiments 2 and 3, you see that doubling [I−] doubles the rate, so the reaction is first order in I−. From Experiments 1 and 3, you see that doubling [ClO−] also doubles the rate, so the reaction is first order in ClO−. From Experiments 3 and 4, you see that doubling [OH−] halves the rate; that is, 2m = 1/2. Hence m = −1, and the rate is inversely proportional to the first power of OH−. The rate law is Rate = k[I−][ClO−]/[OH−] Substituting values for the rate and concentrations yields a value for k: k=
rate [OH  ] 6.1 x 102 M/s [0.010 M ] = = 6.10 = 6.1 /s [0.010 M ] [0.010 M ] [ClO ][I ]
13.57. Let [SO2Cl2]o = 0.0248 M and [SO2Cl2]t = the concentration after 2.0 hr. Substituting these and k = 2.2 x 10−5/s into the firstorder rate equation gives ln
[SO 2 Cl2 ]t 3600 s ⎞ ⎛ = − (2.2 x 10−5/s) ⎜ 2.0 hr x ⎟ = −0.1584 1 hr ⎠ [0.0248 M ] ⎝
Taking the antilog of both sides gives [SO 2 Cl2 ]t = e−0.1584 = 0.8535 [0.0248 M ]
Solving for [SO2Cl2]t gives [SO2Cl2]t = 0.8525 x [0.0248 M] = 0.02116 = 0.021 = 2.1 x 10−2 M
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Chapter 13: Rates of Reaction
13.58. Let [C3H6]o = 0.0226 M and [C3H6]t = the concentration after 899 s. Substituting these concentrations and k = 6.0 x 10−4/s into the firstorder rate equation gives ln
[C3 H 6 ]t = − (6.0 x 10−4/s)(899 s) = −0.5394 [0.0226 M ]
Taking the antilog of both sides gives [C3 H 6 ]t = e−0.5394 = 0.5830 [0.0226 M ]
Solving for [C3H6]t gives [C3H6]t = 0.5830 x [0.0226 M] = 0.0131 = 0.013 M 13.59. The secondorder integrated rate law is 1 1 = kt + [A] t [A] o
Using a rate constant value of 0.225 L/(mol•s) and an initial concentration of 0.293 mol/L, after 35.4 s the concentration of A is 1 1 = 0.225 L/(mol•s) x 35.4 s + 0.293 mol/L [A] t 1 = 7.965 L/mol + 3.412 L/mol = 11.377 L/mol [A] t
[A]t =
1 = 0.087889 = 0.08789 M 11.377 L/mol
13.60. The secondorder integrated rate law is 1 1 = kt + [A] t [A] o
Using a rate constant value of 0.169 L/(mol•s) and an initial concentration of 0.159 mol/L, the time it would take for [A] to decrease to 6.07 x 10−3 mol/L is 1 1 = 0.169 L/(mol•s) x t + 3 6.07 x 10 mol/L 0.159 mol/L
0.169 L/(mol•s) x t = t=
1 1 − = 158.45 L/mol 6.07 x 103 mol/L 0.159 mol/L
158.45 L/mol = 937.6 = 938 s 0.169 L/(mol • s)
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13.61. First, find the rate constant, k, by substituting experimental values into the firstorder rate equation. Let [Et. Cl.]o = 0.00100 M, [Et. Cl.]t = 0.00067 M, and t = 155 s. Solving for k yields ln
k=−
[0.00067 M]t [0.00100 M]o = 2.583 x 10−3/s 155 s
Now let [Et. Cl.]t = the concentration after 256 s, [Et. Cl.]o again = 0.00100 M, and use the value of k of 1.564 x 10−3/s to calculate [Et. Cl.]t. ln
[Et. Cl.]t = − (2.584 x 10−3/s)(256 s) = −0.6614 [0.00100 M]
Converting both sides to antilogs gives [Et. Cl.]t = 0.5161 [0.00100 M]
[Et. Cl.]t = 0.5161 x [0.00100 M] = 5.16 x 10−4 = 5.2 x 10−4 M 13.62. First, find the rate constant, k, by substituting experimental values into the firstorder rate equation. Let [Cyb.]o = 0.00150 M, [Cyb.]t = 0.00119 M, and t = 455 s. −k(455 s) = ln
[0.00119 M] = −0.23151 [0.00150 M]
Solving gives k = 5.088 x 10−4/s. Now let [Cyb.]t = the concentration after 750 s and [Cyb.]o = 0.00150 M, and calculate [Cyb.]t. ln
[Cyb.]t = − (5.088 x 10−4 /s)(750 s) = −0.3816 [0.00150 M]
Taking the antilog of both sides yields [Cyb.]t = 0.68275 [0.00150 M]
Solving for [Cyb.]t gives [Cyb.]t = 0.68275 x [0.00150 M] = 1.024 x 10−3 = 1.02 x 10−3 M 13.63. For a firstorder reaction, divide 0.693 by the rate constant to find the halflife: t1/2 = 0.693/(6.3 x 10−4/s) = 1.10 x 103 = 1.1 x 103 s (18.3 min) Now, use the halflife to determine the concentrations. When the concentration decreases to 50.0% of its initial value, this is equal to the halflife. t50.0% left = t1/2 = 1.10 x 103 s When the concentration decreases to 25.0% of its initial value, this is equal to two halflives. t25.0% left = t1/4 left = 2 x t1/2 = 2 x (1.10 x 103 s) = 2.20 x 103 s (37 min)
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Chapter 13: Rates of Reaction
13.64. For a firstorder reaction, divide 0.693 by the rate constant to find the halflife: t1/2 = 0.693/(6.2 x 10−4/min) = 1.117 x 103 = 1.1 x 103 min (18.6 hr) t25.0% left = t1/4 left = 2 x t1/2 = 2 x (1.117 x 103 min) = 2.235 x 103 = 2.2 x 103 min t12.5% left = t1/8 left = 3 x t1/2 = 3 x (1.117 x 103 min) = 3.353 x 103 = 3.4 x 103 min 13.65. For a firstorder reaction, divide 0.693 by the rate constant to find the halflife: t1/2 = 0.693/(2.0 x 10−6/s) = 3.465 x 105 s (96.25 or 96 hr) t25% left = t1/4 left = 2 x t1/2 = 2 x 96.25 hr = 192.5 = 1.9 x 102 hr t12.5% left = t1/8 left = 3 x t1/2 = 3 x 96.25 hr = 288.75 = 2.9 x 102 hr t6.25% left = t1/16 left = 4 x t1/2 = 4 x 96.25 hr = 385.0 = 3.9 x 102 hr t3.125% left = t1/32 left = 5 x t1/2 = 5 x 96.25 hr = 481.25 = 4.8 x 102 hr 13.66. For a firstorder reaction, divide 0.693 by the rate constant to find the halflife: t1/2 = 0.693/(1.27/s) = 0.5456 s t25% left = t1/4 left = 2 x t1/2 = 2 x 0.5456 s = 1.091 = 1.09 s t12.5% left = t1/8 left = 3 x t1/2 = 3 x 0.5456 s = 1.636 = 1.64 s t6.25% left = t1/16 left = 4 x t1/2 = 4 x 0.5456 s = 2.182 = 2.18 s t3.125% left = t1/32 left = 5 x t1/2 = 5 x 0.5456 s = 2.728 = 2.73 s 13.67. The halflife for a secondorder reaction is t1/2 =
1 k[A] o
Using a rate constant of 0.413 L/(mol•s) and an initial A concentration of 5.25 x 10−3 mol/L, the halflife is t1/2 =
1 = 461.2 = 461 s (0.413 L/(mol • s)(5.25 x 103 mol/L)
13.68. The rate constant for a secondorder reaction is related to the halflife by k=
1 t1/2 [A] o
Using a halflife of 425 s and an initial A concentration of 5.99 x 10−3 mol/L, the rate constant is k=
1 = 0.3928 = 0.393 L/(mol•s) (425 s)(5.99 x 103 mol/L)
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13.69. Use the firstorder rate equation, and solve for time, t. Let [Cr3+]o = 100.0%; then the concentration at time t, [Cr3+]t, = (100.0% − 85.0%, or 15.0%). Use k = 2.0 x 10−6/s. ln
[15.0] = −(2.0 x 10−6/s) t [100.0]
t=−
ln(0.150) = 9.48 x 105 s, or 2.6 x 102 hr 2.0 x 106 /s
13.70. Use the firstorder rate equation, and solve for time, t. Let [Fe3+]o = 100.0%; then the concentration at time t, [Fe3+]t, = (100.0% − 90.0%, or 10.0%). Use k = 1.27/s. ln
[ 10.0%] = −(1.27/s) t [ 100.0%]
t=−
ln(0.100) 1.813 = 1.81 s 1.27/s
13.71. The rate law for a zeroorder reaction is [A] = −kt + [A]o. Using a rate constant of 8.1 x 10−2 mol/(L•s) and an initial concentration of 0.10 M, the time it would take for the concentration to change to 1.0 x 10−2 M is 1.0 x 10−2 M = −8.1 x 10−2 mol/(L•s) x t + 0.10 M t=
0.10 M  1.0 x 102 M = 1.11 = 1.1 s 8.1 x 102 M/s
13.72. The rate law for a zeroorder reaction is [A] = −kt + [A]o. Using a time of 4.3 x 102 s to go from an initial concentration of 0.50 M to 0.25 M, the rate constant is 0.25 M = −k x 4.3 x 102 s + 0.50 M k=
0.50 M  0.25 M = 5.81 x 10−4 = 5.8 x 10−4 M/s 4.3 x 10 2 s
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Chapter 13: Rates of Reaction
13.73. For the firstorder plot, follow Figure 13.9, and plot ln [ClO2] versus the time in seconds. The data used for plotting are t, sec
[ClO2], M
ln [CIO2]
−7.647 0.00 4.77 x 10−4 1.00 4.31 x 10−4 −7.749 2.00 3.91 x 10−4 −7.846 −4 3.00 3.53 x 10 −7.949 5.00 2.89 x 10−4 −8.149 10.00 1.76 x 10−4 −8.645 30.00 2.4 x 10−5 −10.63 50.00 3.2 x 10−5 −12.65 The plot yields an approximate straight line, demonstrating the reaction is first order in [ClO2]. The slope of the line may be calculated from the difference between the last point and the first point: Slope =
[(12.65)  (7.647)] = −0.1001/s [50.00  0.00]s
Just as the slope, m, was obtained for the plot in Figure 13.9, you can also equate m to −k, and calculate k as follows: k = −slope = 0.1001 = 0.100/s 13.74. For the firstorder plot, follow Figure 13.9, and plot ln [MA], the methyl acetate concentration, versus time in minutes. This plot does not yield a straight line, so the reaction is not first order. For the secondorder plot, follow Figure 13.10, and plot 1/[MA] versus time in minutes. The data used for plotting are t, min
[MA], M
1/[MA]
0.00 0.01000 100.0 3.00 0.00740 135.1 4.00 0.00683 146.4 5.00 0.00634 157.7 10.00 0.00463 215.9 20.00 0.00304 328.9 30.00 0.00224 446.4 The plot requires a graph with too many lines to be reproduced here, but it yields a straight line, demonstrating that the reaction is second order in [MA]. The slope of the line may be calculated from the difference between the last point and the first point: Slope =
0.192 [446.4  100.0]/M 11.54 = = [30.00  0.00]min M •s M • min
In this case, the slope equals the rate constant, so k = 11.5/(M•m), or 0.192/(M•s).
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13.75. The potentialenergy diagram, not drawn to scale, is below. Because the activation energy for the forward reaction is +10 kJ, and ΔH° = −200 kJ, the activation energy for the reverse reaction is +210 kJ. AB‡
Energy
Ea = 10 kJ A + B Reactants
Ea = 210 kJ
ΔH = 200
C + D Products
Progress of reaction
13.76. The potentialenergy diagram, not drawn to scale, is below. Because the activation energy for the forward reaction is +251 kJ, and ΔH° is +167 kJ, the activation energy for the reverse reaction is +84 kJ. A‡
Ea = 84 kJ
Ea = 251 kJ Energy
B + C
Products
ΔH = 167 kJ
A Reactant Progress of reaction
13.77. Solve the twotemperature Arrhenius equation for Ea by substituting T1 = 308 K (from 35°C), k1 = 1.4 x 10−4/s, T2 = 318 K (from 45°C), and k2 = 5.0 x 10−4/s: ln
Ea 5.0 x 104 1 ⎞ ⎛ 1 = ⎜ ⎟ 4 318 K ⎠ 8.31 J/(mol • K) ⎝ 308 K 1.4 x 10
Rearranging Ea to the left side and calculating [1/308 − 1/318] give Ea =
8.31 J/K x ln(3.5714) = 1.037 x 105 = 1.0 x 105 J/mol 1.0209 x 104 /K
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Chapter 13: Rates of Reaction
To find the rate at 55°C (328 K), use the first equation. Let k2 in the numerator be the unknown and solve: ln
k2 1.037 x 105 J/mol ⎛ 1 1 ⎞ = ⎜ ⎟ = 2.468 4 328 K ⎠ 1.4 x 10 8.31 J/(mol • K) ⎝ 308 K
k2 = 11.80 1.4 x 104
k2 = 11.80 x (1.4 x 10−4/s) = 1.652 x 10−3 = 1.7 x 10−3/s 13.78. Solve the twotemperature Arrhenius equation for Ea by making these substitutions: T1 = 350 K, k1 = 9.3 x 10−6/s, T2 = 400 K, and k2 = 6.9 x 10−4/s. ln
Ea 6.9 x 104 /s 1 ⎞ ⎛ 1 = ⎜ ⎟ 6 400 K ⎠ 9.3 x 10 /s 8.31 J/(mol • K) ⎝ 350 K
Rearranging and solving for Ea give ⎛ 6.9 x 104 ⎞ 8.31 J/(mol • K) x ln ⎜ 6 ⎟ ⎝ 9.3 x 10 ⎠ = 1.002 x 105 = 1.00 x 105 J/mol (100. kJ/mol) Ea = 1 ⎞ ⎛ 1 ⎜ ⎟ 400 K ⎠ ⎝ 350 K
To find the rate constant at 435 K, use the first equation with k2 as the symbol for the rate constant at 475 K and Ea = 1.002 x 105 J/mol: ln
k2 1.002 x 105 J/mol ⎛ 1 1 ⎞ = ⎜ ⎟ = 6.732 6 435 K ⎠ 9.3 x 10 /s 8.31 J/(mol • K) ⎝ 350 K
Taking the antilog of both sides gives k2 = e6.732 = 839.05 6 9.3 x 10 /s
Solving for k2 gives k2 = 839.05 x (9.3 x 10−6/s) = 0.00780 = 0.0078 /s 13.79. Because the rate constant is proportional to the rate of a reaction, tripling the rate at 25°C means that the rate constant at 25°C is also tripled. Thus, k35 = 3k25, and the latter can be substituted for k25 in the Arrhenius equation: ln
3 k25 Ea 1 ⎞ ⎛ 1 = ⎜ ⎟ 308 K ⎠ k25 8.31 J/(mol • K) ⎝ 298 K
1.0986 = (1.311 x 10−5 mol/J) Ea Ea =
1.0986 = 8.379 x 104 J/mol = 83.8 kJ/mol 1.311 x 105 mol/J
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13.80. Because the rate constant is proportional to the rate of a reaction, quadrupling the rate at 25°C means that the rate constant at 25°C is also quadrupled. Thus, k35 = 4k25, and the latter can be substituted for k35 in the Arrhenius equation: ln
4 k25 Ea 1 ⎞ ⎛ 1 = ⎜ ⎟ 308 K ⎠ k25 8.31 J/(mol • K) ⎝ 298 K
1.3862 = (1.311 x 10−5 mol/J) Ea Ea =
1.3862 = 1.0572 x 105 J/mol = 106 kJ/mol 1.311 x 105 mol/J
13.81. For plotting ln k versus 1/T, the data below are used: k
ln k
1/T (1/K)
0.527 −0.64055 1.686 x 10−3 0.776 −0.25360 1.658 x 10−3 1.121 0.11422 1.631 x 10−3 1.607 0.47436 1.605 x 10−3 The plot yields an approximately straight line. The slope of the line is calculated from the difference between the last and the first points: Slope =
(0.47436)  (0.64055) = −13764 K [1.605 x 103  1.686 x 103 ]/K
Because the slope = −Ea/R, you can solve for Ea using R = 8.31 J/K:  Ea = −13764 K 8.31 J/ (K • mol)
Ea = 13764 K x 8.31 J/K = 1.143 x 105 J/mol, or 1.1 x 102 kJ/mol 13.82. For plotting ln k versus 1/T, the data below are used: ln k 1/T (K) k 2.69 x 10−3 −5.918 1.402 x 10−3 −3 6.21 x 10 −5.081 1.364 x 10−3 −2 1.40 x 10 −4.268 1.328 x 10−3 −2 3.93 x 10 −3.236 1.294 x 10−3 The plot yields an approximately straight line. The slope of the line may be calculated from the difference between the last point and the first point: Slope =
(3.236)  (5.918) = −2.483 x 104 K [1.294 x 103  1.402 x 103 ]/K
Because the slope = −Ea/R, you can solve for Ea using R = 8.31 J/K•mol:  Ea = 2.483 x 104 K 8.31 J/ (K • mol)
Ea = 2.483 x 104 K x 8.31 J/K • mol = 2.063 x 105 J/mol (2.1 x 102 kJ/mol)
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Chapter 13: Rates of Reaction
13.83. The NOCl2 is a reaction intermediate that is produced in the first reaction and consumed in the second. The overall reaction is the sum of the two elementary reactions: + Cl2 →
NOCl2
NOCl2+ NO →
2NOCl
2NO + Cl2 →
2NOCl
NO
13.84. The O atom is a reaction intermediate that is produced in the first reaction and consumed in the second. The overall reaction is the sum of the two elementary reactions: O3 O3
+O
2O3
→
O2 + O
→
2O2
→
3O2
13.85. a.
Bimolecular
b.
Bimolecular
c.
Unimolecular
d.
Termolecular
13.86. a.
Termolecular
b.
Bimolecular
c.
Bimolecular
d.
Unimolecular
13.87. a.
Only O3 occurs on the left side of the equation, so the rate law is Rate = k[O3]
b.
Both NOCl2 and NO occur on the left side of the equation, so the rate law is Rate = k[NOCl2][NO]
13.88. a.
Only CS2 occurs on the left side of the equation, so the rate law is Rate = k[CS2]
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b.
Both CH3Br and OH− occur on the left side of the equation, so the rate law is Rate = k[CH3Br][OH−]
13.89. Step 1 of the isomerization of cyclopropane, C3H6, is slow, so the rate law for the overall reaction will be the rate law for this step, with k1 = k, the overall rate constant: Rate = k[C3H6]2 13.90. Step 1 of the decomposition of NO2Cl is slow, so the rate law for the overall reaction will be the rate law for this step, with k1 = k, the overall rate constant: Rate = k[NO2Cl] 13.91. Step 2 of this reaction is slow, so the rate law for the overall reaction would appear to be the rate law for this step: Rate = k2[I]2[H2] However, the rate law includes an intermediate, the I atom, and cannot be used unless the intermediate is eliminated. This can be done only using an equation for Step 1. At equilibrium, you can write the following equality for Step 1: k1[I2] = k−1 [I]2 Rearranging and then substituting for the [I]2 term yield [I]2 = [I2]
k1 k1
Rate = k2(k1/k−1)[I2][H2] = k[I2][H2] (k = the overall rate constant) 13.92. Step 2 of this reaction is slow, so the rate law for the overall reaction would appear to be the rate law for this step: Rate = k2[O3][O] However, the rate law includes an intermediate, the O atom, and cannot be used unless the intermediate is eliminated. This can be done only using an equation for Step 1. At equilibrium, you can write the following equality for Step 1: k1[O3] = k−1 [O2][O] Rearranging and then substituting for the [O] term yield ⎡ k ⎤ [O ]
[O] = ⎢ 1 ⎥ 3 ⎣ k1 ⎦ [O 2 ] ⎡ k ⎤ [O ]
Rate = k2[O3] ⎢ 1 ⎥ 3 ⎣ k1 ⎦ [O 2 ] 2
2 [O ] ⎡ ⎤ Rate = k2 ⎢ k1 ⎥ [O3 ] = k 3 (k = overall rate constant)
⎣ k1 ⎦ [O 2 ]
[O 2 ]
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Chapter 13: Rates of Reaction
13.93. The Br− ion is the catalyst. It is consumed in the first step and regenerated in the second step. It speeds up the reaction by providing a pathway with a lower activation energy than that of a reaction pathway involving no Br−. The overall reaction is obtained by adding the two steps together: 2H2O2 → 2H2O + O2 Bromide ion is added to the mixture to give the catalytic activity, and BrO− is an intermediate. 13.94. The OH− ion is the catalyst. It is consumed in the first step and regenerated in the second step. It speeds up the reaction by providing a pathway with a lower activation energy than that of a reaction pathway involving no OH− ion. The overall reaction is obtained by adding the two steps together: NH2NO2 → N2O + H2O You could add NaOH to give the catalytic activity.
■
SOLUTIONS TO GENERAL PROBLEMS
13.95. All rates of reaction are calculated by dividing the decrease in concentration by the difference in times; hence, only the setup for the first rate (after 10 minutes) is given below. This setup is Rate (10 min) = −
1 min (1.29  1.50) x 102 M x = 3.5 x 10−6 M/s 60 s (10  0) min
A summary of the times and rates is given in the table. Time, min 10 20 30
Rate 3.50 x 10−6 = 3.5 x 10−6 M/s 3.17 x 10−6 = 3.2 x 10−6 M/s 2.50 x 10−6 = 2.5 x 10−6 M/s
13.96. All rates of reaction are calculated by dividing the decrease in concentration by the difference in times; only the setup for the first rate after 1.0 min is given below: Rate (1.0 min) = −
1 min (0.1076  0.1103) M x = 4.50 x 10−5 = 4.5 x 10−5 M/s 60 s (1.0  0) min
A summary of the times and rates is given in the table. Time, min 1.0 2.0 3.0
Rate −5
4.50 x 10 = 4.5 x 10−5 M/s 4.33 x 10−5 = 4.3 x 10−5 M/s 4.00 x 10−5 = 4.0 x 10−5 M/s
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13.97. The calculation of the average concentration and the division of the rate by the average concentration are the same for all three time intervals. Thus, only the setup for the first interval is given: k10 min =
rate 3.50 x 106 M/s = = 2.508 x 104 /s 2 avg. conc. ⎡ (1.50 + 1.29) x 10 M ⎤ ⎢ ⎥ 2 ⎣ ⎦
A summary of the times, rate constants, and average rate constant is given in the table. Time 10 min 20 min 30 min 
Rate 3.50 x 10−6 M/s 3.17 x 10−6 M/s 2.50 x 10−6 M/s  average k
k 2.508 x 10−4/s 2.652 x 10−4/s 2.439 x 10−4/s 2.533 x 10−4 = 2.5 x 10−4/s
13.98. The calculation of the average concentration and the division of the rate by the average concentration are the same for all three time intervals. Thus, only the setup for the first interval is given: k1.0 min =
4.50 x 105 M/s rate = = 3.79 x 10−3/(M•s) 2 avg. conc. ⎛ 0.1103 M + 0.1076 M ⎞ ⎜ ⎟ 2 ⎝ ⎠
A summary of the times, rate constants, and average rate constant is given in the table. Time
Rate 4.50 x 10−5 M/s 4.33 x 10−5 M/s 4.00 x 10−5 M/s  average k =
1.0 min 2.0 min 3.0 min 
k 3.79 x 10−3/(M•s) 3.83 x 10−3/(M•s) 3.71 x 10−3/(M•s) 3.78 x 10−3 = 3.8 x 10−3/(M•s)
13.99. Use the firstorder rate equation k = 1.26 x 10−4/s, the initial methyl acetate [MA]o = 100%, and [MA]t = (100% − 65%, or 35%). 35% 100% = 8.3319 x 103 = 8.33 x 103 s t=− 1.26 x 104 /s ln
13.100. Use the firstorder rate equation; substitute k = 4.3 x 10−5/s, the initial benzene diazonium chloride [BC]o = 100%, and [BC]t = (100% − 75%, or 25%). 25% 100% = 3.223 x 104 = 3.2 x 104 s t=− 4.3 x 105 /s ln
13.101. Use k = 1.26 x 10−4/s, and substitute into the t1/2 equation: t1/2 =
0.693 0.693 = = 5.500 x 103 = 5.50 x 103 s (1.53 hr) 1.26 x 104 /s k
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Chapter 13: Rates of Reaction
13.102. Use k = 4.3 x 20−5/s, and substitute into the t1/2 equation: t1/2 =
0.693 0.693 = = 1.61 x 104 = 1.6 x 104 s (4.5 hr) k 4.3 x 105 /s
13.103. First, find the rate constant from the firstorder rate equation, substituting the initial concentration of [comp.]o = 0.0350 M, and the [comp.]t = 0.0250 M. ln
0.0250 M = − k (65 s) 0.0350 M
Rearranging and solving for k give ⎛ 0.0250 M ⎞ ⎜ ln ⎟ 0.0350 M ⎠ k= ⎝ = 5.17 x 10−3/s 65 s
Now, arrange the firstorder rate equation to solve for [comp.]t; substitute the above value of k, again using [comp.]o = 0.0350 M. ln
[comp.]t = −(5.18 x 10−3/s)(88 s) = −0.4558 0.0350 M
Taking the antilog of both sides gives [comp.]t = e−0.4558 = 0.6339 0.0350 M
[comp]t = 0.6339 x [0.0350 M]o = 0.0221 = 0.022 M 13.104. First, find the rate constant from the rearranged firstorder rate equation, substituting the initial concentration of [comp.]o = 0.1180 M and the [comp.]t = 0.0950 M. Use k to find the fraction. ⎛ 0.0950 M ⎞ ⎜ ln ⎟ 0.1180 M ⎠ k= ⎝ = 4.169 x 10−2/min 5.2 min
ln
[comp.]t = −(4.169 x 10−3/s)(7.1 min) = −0.2960 0.1180 M
Taking the antilogarithm of both sides yields [comp.]t = 0.7437 = 0.74 (fraction remaining) 0.1180 M
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481
13.105. a.
The rate constant for a secondorder reaction is related to the halflife by k=
1 t1/2 [A] o
Using a halflife of 5.92 x 10−2 s and an initial A concentration of 0.50 mol/L, the rate constant is k= b.
1 = 33.78 = 34 L/(mol•s) (5.92 x 102 s)(0.50 mol/L)
The secondorder integrated rate law is 1 1 = kt + [A] t [A] o
Using an initial concentration of C4H8 (A) of 0.010 M, after 3.6 x 102 s, the concentration will be 1 1 = 33.78 L/(mol•s)(3.6 x 102 s) + 0.010 mol/L [A] t 1 = 12,162 L/mol + 100.0 L/mol = 12,262 L/mol [A] t
[A]t =
1 = 8.155 x 10−5 = 8.2 x 10−5 M 12,262 L/mol
13.106. a.
A plot of 1/[A] versus time is a straight line for a secondorder reaction.
b.
The secondorder integrated rate law is 1 1 = kt + [A] t [A] o
After 57 s, the concentration of A dropped 40% of its initial value, so [A]t = 0.60[A]o, or (0.60)(0.50 M) = 0.30 M. Using these values gives 1 1 = k x (57 s) + 0.30 mol/L 0.50 mol/L
k x (57 s) = k=
1 1 − = 1.333 L/mol 0.30 mol/L 0.50 mol/L
1.333 L/mol = 0.0233 = 0.023 L/(mol•s) 57 s
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Chapter 13: Rates of Reaction
13.107. The secondorder integrated rate law is 1 1 = kt + [A] t [A] o
The starting concentration of NO2 (A) is 0.050 M. Using a rate constant of 0.775 L/(mol•s), after 2.5 x 102 s the concentration of NO2 will be 1 1 = 0.775 L/(mol•s)(2.5 x 102 s) + [A] t 0.050 mol/L 1 = 193 L/mol + 20.0 L/mol = 213. L/mol [A] t
[A]t =
1 = 4.67 x 10−3 = 4.7 x 10−3 M 213 L/mol
The halflife is 1 k[A] o
t1/2 =
Using a rate constant of 0.775 L/(mol•s) and an initial A concentration of 0.10 mol/L, the halflife is t1/2 =
1 = 25.80 = 26 s (0.775 L/(mol • s)(0.050 mol/L)
13.108. The secondorder integrated rate law is 1 1 = kt + [A] t [A] o
The starting concentration is 0.10 M. Using a rate constant of 3.1 x 10−2 L/(mol•s), after 1.0 x 102 s the concentration will be 1 1 = 3.1 x 10−2 L/(mol•s)(1.0 x 102 s) + 0.10 mol/L [A] t 1 = 3.1 L/mol + 10.0 L/mol = 13.1. L/mol [A] t
[A]t =
1 = 0.0763 = 0.076 M 13.1 L/mol
The halflife is t1/2 =
1 k[A] o
Using a rate constant of 3.1 x 10−2 L/(mol•s) and an initial concentration of 0.10 mol/L, the halflife is t1/2 =
1 = 322.5 = 3.2 x 102 s (3.1 x 10 L/(mol • s)(0.10 mol/L) 2
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13.109. The ln[CH3NNCH3] and time data for the plot are tabulated below. ln[CH3NNCH3]
t min 0 10 20 30
−4.1997 −4.3505 −4.5098 −4.6564 4 .2
ln[CH3NHCH3]
4 .3 4 .4 4 .5 4 .6 4 .7 0
10
20
30
T im e (m in u te s)
The slope, m, is calculated from the graph: m=
(4.6564)  (4.1997) (30  0) min
= −0.01522 /min Because the slope also = −k, this gives k = −(−0.01522/min)(1 min/60 s) = 2.53 x 10−4 = 2.5 x 10−4/s
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Chapter 13: Rates of Reaction
13.110. The ln [H2O2] and time data for the plot are tabulated below. ln[H2O2]
t min 0.0 5.0 10.0 15.0
−2.3025 −2.5207 −2.7364 −2.9584
ln[H2O2]
2 .4
2 .6
2 .8
3 .0 0
5
10
15
T im e (m in u tes)
The slope, m, is calculated from the graph: m=
(2.9584)  (2.3025) (15.0  0) min
= −0.04372/min Because the slope also = −k, this gives k = −(−0.04372/min)(1 min/60 s) = 7.287 x 10−4 = 7.29 x 10−4/s 13.111. The rate law for a zeroorder reaction is [A] = −kt + [A]o. Using a rate constant of 3.7 x 10−6 mol/(L•s) and an initial concentration of 5.0 x 10−4 M, the time it would take for the concentration to drop to 5.0 x 10−5 M is 5.0 x 10−5 M = −8.1 x 10−2 mol/(L•s) x t + 5.0 x 10−4 M t=
(5.0 x 104 M)  (5.0 x 105 M ) = 121.6 = 1.2 x 102 s 3.7 x 106 M/s
The halflife for a zeroorder reaction is t1/2 =
[A]o 5.0 x 104 M = = 67.57 = 68 s 2k 2(3.7 x 106 M/s)
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13.112. The rate law for a zeroorder reaction is [A] = −kt + [A]o. If it takes 3.3 x 102 s for the initial concentration of B to go from 0.50 M to 0.20 M, the rate constant is 0.20 M = −k x 3.3 x 102 s + 0.50 M k=
(0.50 M )  (0.20 M ) = 9.09 x 10−4 = 9.1 x 10−4 mol/(L•s) 3.3 x 102 s
The halflife for a zeroorder reaction is t1/2 =
[A]o 0.50 M = = 275 = 2.8 x 102 s 2k 2(9.09 x 104 M/s)
13.113. Rearrange the twotemperature Arrhenius equation to solve for Ea in joules, using k1 = 0.498 M/s at T1 = 592 K (319°C) and k2 = 1.81 M/s at 627 K (354°C). Assume (1/T1 − 1/T2) has three significant figures. ⎛ 1.81 ⎞ (8.31 J/mol • K) ln ⎜ ⎟ ⎝ 0.498 ⎠ = 1.137 x 105 J/mol = 114 kJ/mol Ea = 1 ⎞ ⎛ 1 ⎜ ⎟ 627 K ⎠ ⎝ 592 K
To obtain A, rearrange the ln form of the onetemperature Arrhenius equation; substitute the value of Ea obtained above, and use k1 = 0.498 M/s at T1 = 592 K. ln A = ln0.498 +
1.137 x 105 J/mol = 22.420 8.31 J/(mol • K) x 592 K
A = 5.46 x 109 = 5 x 109 To obtain k at 420°C (693 K), also use the ln form of the onetemperature Arrhenius equation: ln k = 22.420 −
1.137 x 105 J/ mol = 2.671 8.31 J/(mol • K) x 693K
k = e2.671 = 14.45 = 14 M/s 13.114. Rearrange the twotemperature Arrhenius equation; solve for Ea using k1 = 8.7 x 10−4/(M•s) at T1 = 303 K (30°C) and k2 = 1.8 x 10−3 (M•s) at 313 K (40°C). ⎛ 1.8 x 103 /(M • s) ⎞ 8.31 J/(mol • K) x ln ⎜ ⎟ 4 ⎝ 8.7 x 10 /(M • s) ⎠ = 5.729 x 104 (57 kJ/mol) Ea = 1 ⎞ ⎛ 1 ⎜ ⎟ 313 K ⎠ ⎝ 303 K
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Chapter 13: Rates of Reaction
To obtain A, rearrange the ln form of the onetemperature Arrhenius equation; substitute the value of Ea obtained above and use k1 = 8.7 x 10−4/(M•s) at T1 = 303 K. ln A = ln[8.7 x 10−4] +
5.729 x 104 J/ mol = 15.70 8.31 J/ (mol • K) x 303K
A = e15.70 = 6.646 x 106 /(M•s) To obtain k at 45°C (318 K), also use the ln form of the onetemperature Arrhenius equation: ln k = 15.70 −
5.729 x 104 J/ mol = −5.970 8.31 J/ (mol • K) x 318 K
k = e−5.970 = 2.551 x 10−3 = 3 x 10−3/(M•s) 13.115. If the reaction occurs in one step, the coefficients of NO2 and CO in this elementary reaction are each one, so the rate law should be Rate = k[NO2][CO] 13.116. If the reaction occurs in one step, the coefficients of CH3Cl and OH− in this elementary reaction are each 1, so the rate law should be Rate = k[CH3Cl][OH−] 13.117. The slow step determines the observed rate, so the overall rate constant, k, should be equal to the rate constant for the first step, and the rate law should be Rate = k[NO2Br] 13.118. The slow step determines the observed rate, so the overall rate constant, k, should be equal to the rate constant for the first step, and the rate law should be Rate = k[(CH3)3CCl] 13.119. The slow step determines the observed rate; assuming k2 is the rate constant for the second step, the rate law would appear to be Rate = k2[NH3][HOCN] However, this rate law includes two intermediate substances that are neither reactants nor products. The rate law cannot be used unless both are eliminated. This can be done only using an equation from Step 1. At equilibrium in Step 1, you can write the following equality, assuming k1 and k−1 are the rate constants for the forward and back reactions, respectively: k1[NH4+][OCN−] = k−1[NH3][HOCN] Rearranging and then substituting for the [NH3][HOCN] product give [NH3][HOCN] = (k1/k−1)[NH4+][OCN−] Rate = k2(k1/k−1)[NH4+][OCN−] = k[NH4+][OCN−] (k = overall rate constant)
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13.120. The slow step determines the observed rate; assuming k2 is the rate constant for the second step and using [HA+] for [CH3C(OH+)CH3], the rate law would appear to be Rate = k2[HA+][H2O] However, this rate law includes the intermediates HA+ and H2O that are neither reactants nor products. The rate law cannot be used unless both are eliminated. This can be done only using an equation from Step 1. At equilibrium in Step 1, you can write the following equality, assuming k1 and k−1 are the rate constants for the forward and back reactions, respectively: k1[A][H3O+] = k−1[HA+][H2O] Rearranging and then substituting for the [HA+][H2O] product give Rate = k2(k1/k−1)[A][H3O+] = k[A][H3O+] (k = overall rate constant) 13.121. a.
The reaction is first order in O2 because the rate doubled with a doubling of the oxygen concentration. The reaction is second order in NO because the rate increased by a factor of 8 when both the NO and O2 concentrations were doubled. Rate = k[NO]2[O2]
b.
The initial rate of the reaction for Experiment 4 can be determined by first calculating the value of the rate constant using Experiment 1 for the data. 0.80 x 10−2 M/s = k[4.5 x 10−2 M]2[2.2 x 10−2 M] Solving for the rate constant gives k = 1.796 x 102 M−2s−1 Now, use the data in Experiment 4 and the rate constant to determine the initial rate of the reaction. Rate = 1.796 x 102 M−2s−1 [3.8 x 10−1 M]2[4.6 x 10−3 M] Rate = 0.119 = 0.12 mol/L•s
13.122. a.
From the information given, the rate law is Rate = k[CH3Cl][H2O]2 Using just the units for each term in the rate law will give the units for the rate constant. M/s = k(M)(M)2 Thus, the units for k must be k = M−2s−1
b.
Plugging into the rate equation gives 1.50 M/s = k[0.40 M][0.40 M]2 k = 23.43 = 23 M−2s−1
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Chapter 13: Rates of Reaction
13.123. a. Rate will decrease because OH− will react with H3O+ and lower its concentration.
i)
ii) Rate will decrease because the dilution with water will decrease both the H3O+ and the CH3CSNH2 concentration. b. i) The catalyst will provide another pathway, and k will increase because Ea will be smaller. ii) The rate constant changes with temperature, and it will decrease with a decrease in temperature, because fewer molecules will have enough energy to react. 13.124. a. The [H3O+] is increased, so the rate will increase.
i)
ii) The [H3O+] and [CH3COOCH3] concentrations will be decreased, so the rate of reaction will decrease. b. i)
The rate of reaction will increase, but the rate constant will remain the same.
ii)
The rate constant will increase as more molecules have enough energy to react.
13.125. a.
The diagram:
I
Energy
II Reactants Products Reaction progress b.
In the diagram, I represents the activation energy for the uncatalyzed reaction, and II represents the activation energy for the catalyzed reaction. A catalyst provides another pathway for a chemical reaction, and with a lower activation energy, more molecules have enough energy to react, so the reaction will be faster.
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13.126. a.
The diagram: A‡
Energy
Ea
Ea = 23 kJ
ΔH = 57 kJ
Reaction progress b.
Ea = Ear + ΔH Ea = 23 + 57 = 80 kJ
c.
The forward reaction will be more sensitive to a temperature change because it has the larger activation energy. When Ea is large, the ratio of the number of molecules that have enough energy to react is larger than when Ea is smaller.
13.127. a.
The rate of a chemical reaction is the change in the concentration of a reactant or product with time. For a reactant, −Δc/Δt or −d[c]/dt
b.
The rate changes because the concentration of the reactant has changed. Rate = k[A]m
c.
Rate = k[A]m[B]n The rate will equal k when the reactants all have 1.00 M concentrations, or if the reaction is zeroth order.
13.128. .Factors that affect the rates of reactions: i)
Concentrations of the reactants
ii)
Temperature
iii) Catalysts At higher concentrations, more molecules can undergo effective collisions and give more product per unit of time. The temperature affects the number of molecules that have enough energy to react. The higher the temperature, the larger the fraction of molecules that have enough energy to react.
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Chapter 13: Rates of Reaction
Catalysts provide another pathway for reaction that has a lower activation energy, so more molecules have the minimum energy to react. The value of the rate constant for a particular reaction depends on the temperature and the activation energy. The concentration of catalyst and of the solvent, if the reaction occurs in solution, can affect Ea. 13.129. The experiment involves using a laser, first to send a pump pulse to excite a reactant molecule to a higher energy state. Then, a weaker laser pulse or probe pulse of properly chosen wavelength is used to detect the molecule as it transforms to products. By varying the time between the pump pulse and the probe pulse, the experimenter can follow a molecule throughout the course of its reaction. The change of wavelength of the probe pulse tells the experimenter something about the change in character of the molecules as it reacts. 13.130. The diagram shows the cyclobutane molecule going first to an intermediate molecule (between the two peaks). In this intermediate molecule, one carboncarbon bond is broken. When the intermediate molecule goes over to products, another bond breaks, giving two ethylene molecules.
■
SOLUTIONS TO STRATEGY PROBLEMS
13.131. Use the Arrhenius equation. ln
k2 2.11 x 103 J/mol ⎛ 1 1 ⎞ = ⎜ ⎟ = 0.11008 8.314 J/K • mol ⎝ 483 K 611 K ⎠ 1.4 x 10 /M • min 5
k2 = (1.4 x 10−5 /M•min)e0.11008 = 1.562 x 10−5 = 1.56 x 10−5 /M•min 13.132. 0.693 ln2 = = 0.04076 = 0.0408 /min 17.0 min t1/2
a.
k=
b.
When 86.0% of the H2O2 is decomposed, the concentration remaining is 0.140[H2O2]o. Now use the firstorder rate law, rearranged to solve for t. 1 k
t = − ln c.
[A]t 0.14[A]o 1 =− ln = 48.23 = 48.2 min 0.04076 /min [A]o [A]o
[A] = [A]oe−kt = (0.100 M)e−(0.04076 /min)(15.0 min) = 0.05425 = 0.0543 M
13.133. k =
1 1 = = 0.1741 = 0.174 /M•s t1/2 [A]o (307 s)(1.87 x 102 M)
13.134. Rate = k[I]2[H2]
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13.135. First write the rate law in terms of the slow step. Rate = k2[NOCl2][NO] Next, set the forward rate equal to the reverse rate for the fast equilibrium. Ratef = kf[NO][Cl2] = Rater = kr[NOCl2] Rearrange this equation and solve for [NOCl2]. [NOCl2] =
kf [NO][Cl2] kr
Finally, substitute into the rate law above to eliminate [NOCl2]. This gives the following rate equation. ⎧
⎫
⎩ kr
⎭
Rate = k2 ⎨ kf [NO][Cl2 ]⎬ [NO] = k[NO]2[Cl2]
13.136. The rate law for this elementary is Rate = k[A]2[B] Rearrange this equation and solve for the rate constant k. k=
Rate 16 M/s = = 4.00 = 4.0 /M2•s 2 [A] [B] (2.0 M ) 2 (1.0 M )
13.137. You can set up an expression for the ratio of the two rates as follows. Use [B]2 = 2[B]1. 3
3
⎡ [B] ⎤ ⎡ 2[B]1 ⎤ Rate 2 k[A]2 2 [B]23 3 = = ⎢ 2⎥ = ⎢ ⎥ =2 =8 [B] [B] Rate1 k[A]12 [B]13 ⎣ 1⎦ ⎣ 1 ⎦
Therefore, the rate of the second run is eight times the rate of the first run. 13.138. a.
Since ΔH is positive, the reaction is endothermic.
b.
The activation energy for the reverse reaction is Ea,r = Ea,f − ΔH = 100 kJ − 38 kJ = 62 kJ
c.
If a catalyst were added to the reaction, the activation energy would be lowered.
13.139. a.
The overall reaction is 2H2O2 → 2H2O + O2
b.
The catalyst is I−, and the intermediate is IO−.
c.
No, the rate law can not be specified until the ratedetermining step is established.
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Chapter 13: Rates of Reaction
13.140. a.
The rate law for this reaction is Rate = k[A] x[B]y. To determine the value of x, find the ratio of the rates for Experiments 1 and 2. In these experiments, [B] is constant. This gives ⎡ [A] ⎤ Rate 2 k[A]2 x [B]2 y = = ⎢ 2⎥ x y k[A]1 [B]1 Rate1 ⎣ [A]1 ⎦ 3.0 x 103 M/s ⎛ 0.0300 M ⎞ =⎜ ⎟ 3 1.0 x 10 M/s ⎝ 0.0100 M ⎠
x
x
This reduces to 3 = 3x, or x = 1. To determine the value of y, find the ratio of the rates for Experiments 2 and 3. In these experiments, [A] is constant. This gives ⎡ [B] ⎤ Rate3 k[A]3 x [B]3 y = = ⎢ 3⎥ x y k[A]2 [B]2 Rate 2 ⎣ [B]2 ⎦ 2.7 x 102 M/s ⎛ 0.0300 M ⎞ =⎜ ⎟ 3.0 x 103 M/s ⎝ 0.0100 M ⎠
y
x
This reduces to 9 = 3y, or y = 2. The rate law is therefore Rate = k[A]x[B]y = k[A][B]2 b.
The rate constant can be calculated using the date from Experiment 1. k=
c.
■
1.0 x 103 M/s Rate = = 10.0 = 10. /M2•s 2 [A][B] (0.0100 M )(0.100 M ) 2
Rate = k[A][B]2 = (10. /M2•s)(0.200 M)(0.200 M)2 = 0.0800 = 0.080 M/s
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
13.141. The balanced equation is 2N2O5
→
4NO2 + O2.
Note that the number of moles of O2 formed will be onehalf the number of moles of N2O5 decomposed. Now use the integrated form of the firstorder rate law to calculate the fraction of the 1.00 mol N2O5 decomposing in 20.0 hr, or 1200 min. ln
[ N 2 O5 ]t = −kt = −(6.2 x 10−4/min)(1200 min) = −0.744 [ N 2 O5 ]o
[ N 2 O5 ]t = e−0.744 = 0.4752 [ N 2 O5 ]o
Fraction of N2O5 decomposed = 1.000 − 0.4752 = 0.5248 Since there is 1.00 mol of N2O5 present at the start, the moles of N2O5 decomposed is 0.5248 mol. Thus, Moles of O2 formed = (1 O2/2 N2O5 ) x 0.5248 mol N2O5 = 0.26237 mol O2
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Now, use the ideal gas law to calculate the volume of this number of moles of O2 gas at 45°C and 780 mmHg. V=
nRT (0.26237 mol) (0.08206 L • atm/ (K • mol)) (318 K) = = 6.758 = 6.8 L (770/ 760) atm P
13.142. The balanced equation is →
2H2O2
2H2O + O2.
Note that the moles of O2 formed will be equal to onehalf that of the moles of H2O2 decomposed. Now, use the integrated form of the firstorder rate law to calculate the fraction of the 1.00 mol H2O2 decomposing in 20.0 min, or 1200 s. ln
[ H 2 O 2 ]t = −kt = −(7.40 x 10−4/s)(1200 s) = −0.8880 [ H 2 O 2 ]o
[ H 2 O 2 ]t = e−0.8880 = 0.41147 [ H 2 O 2 ]o
Fraction of H2O2 decomposed = 1.00 − 0.41147 = 0.58852 Since there is 1.00 mol of H2O2 present at the start, the moles of H2O2 decomposed are 0.58852 mol. Thus, Moles of O2 formed = (1 O2/2 H2O2 ) x 0.58852 mol H2O2 = 0.29426 mol O2 Now, use the ideal gas law to calculate the volume of this number of moles of O2 gas at 25°C and 740 mmHg. V=
nRT (0.29426 mol) (0.08206 L • atm/(K • mol)) (298 K) = = 7.39 = 7.4 L (740/ 760) atm P
13.143. Using the firstorder rate law, the initial rate of decomposition is given by Rate = k[H2O2] = (7.40 x 10−4/s) x (1.50 M H2O2) = 1.110 x 10−3 M H2O2/s The heat liberated per second per mol H2O2 can be found by first calculating the standard enthalpy of the decomposition of 1 mol H2O2:
H2O2(aq) Hf °
=
191.2 kJ/mol
→
H2O(l)
+
285.8 kJ/mol
1/2 O2(g) 0 kJ/mol
For the reaction, the standard enthalpy change is ΔH° = −285.8 kJ/mol − (−191.2 kJ/mol) = −94.60 kJ/mol H2O2 The heat liberated per second is 1.110 x 103 mol H 2 O 2 94.60 kJ x x 2.00 L = 0.21001 = 0.210 kJ/s L•s mol H 2 O 2
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13.144. Using the secondorder rate law, the initial rate of reaction of NO2 is Rate = k[NO2]2 = 0.515 L/(mol•s) x [0.0250 M NO2]2 = 3.218 x 10−4 M NO2/s The heat liberated per second per mol of NO2 can be found by first calculating the standard enthalpy for the reaction of 1 mol NO2:
NO2(g) Hf °
=
+
33.10 kJ/mol
→
CO(g)
NO(g) 90.29 kJ/mol
110.5 kJ/mol
+
CO2(g) 393.5 kJ/mol
For the reaction, the change in standard enthalpy is ΔH° = 90.29 − 393.5 − (33.10) − (−110.5) = −225.81 kJ/mol NO2 The heat liberated per second is 3.218 x 104 mol NO 2 225.81 kJ x x 3.50 L = 0.2543 = 0.254 kJ/s L•s mol NO 2
13.145. Use the ideal gas law (P/RT = n/V) to calculate the mol/L of each gas: [O2] = [NO] =
(345/760) atm = 0.009039 mol O2/L 0.082057 L • atm/(K • mol) x 612 K (155/760) atm = 0.004061 mol NO/L 0.082057 L • atm/(K • mol) x 612 K
The rate of decrease of NO is ⎛ 4.061 x 103 mol ⎞ 1.16 x 105 L2 x ⎜ ⎟ 1L mol 2 • s ⎝ ⎠
2
x
9.039 x 103 mol = 1.729 x 10−12 mol/(L•s) 1L
The rate of decrease in atm/s is found by multiplying by RT: 0.082057 L • atm 1.729 x 1012 mol x x 612 K = 8.682 x 10−11 atm/s K • mol L•s
The rate of decrease in mmHg/s is 8.682 x 10−11 atm/s x (760 mmHg/atm) = 6.598 x 10−8 = 6.60 x 10−8 mmHg/s
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13.146. Use the ideal gas law (P/RT = n/V) to calculate the mol/L of each gas: [H2] = [NO] =
(324/760) atm = 0.004727 mol H2/L 0.082057 L • atm/(K • mol) x 1099 K (144/760) atm = 0.002101 mol NO/L 0.082057 L • atm/(K • mol) x 1099 K
The rate of NO decrease is twice the rate of H2 decrease, so 2
2x
⎛ 2.101 x 103 mol ⎞ 4.727 x 103 mol 1.10 x 107 L2 x⎜ = 4.5905 x 10−15 mol/(L•s) ⎟ x 2 1 L 1 L mol • s ⎝ ⎠
The rate of decrease in atm/s is found by multiplying by RT: 0.082057 L • atm 4.5905 x 1015 mol x x 1099 K = 4.139 x 10−13 atm/s K • mol L•s
The rate of decrease in mmHg/s is 4.139 x 10−13 atm/s x (760 mmHg/atm) = 3.146 x 10−10 = 3.15 x 10−10 mmHg/s
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CHAPTER 14
Chemical Equilibrium
■
SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 14.1. Use the table approach, and give the starting numbers, the change, and the equilibrium number of moles of each. H2O(g)
CO2(g)
1.00
1.00
0
0
Change
−x
−x
+x
+x
Equilibrium
(1.00 − x)
(1.00 − x)
x
x = 0.43
Amt. (mol)
CO(g)
Starting
+
+
H2(g)
Because we are given that x = 0.43 in the statement of the problem, we can use that to calculate the equilibrium amounts of the reactants and products: Equilibrium amount CO = 1.00 − 0.43 = 0.57 mol Equilibrium amount H2O = 1.00 − 0.43 = 0.57 mol Equilibrium amount CO2 = x = 0.43 mol Equilibrium amount H2 = x = 0.43 mol 14.2. For the equation 2NO2 + 7H2 → 2NH3 + 4H2O, the expression for the equilibrium constant, Kc, is Kc =
[ NH 3 ]2 [ H 2 O ]4 [ NO 2 ]2 [ H 2 ]7
Notice that each concentration term is raised to a power equal to its coefficient in the chemical equation. For the equation NO2 + 7/2H2 → NH3 + 2H2O, the expression for the equilibrium constant, Kc, is Kc =
[ NH 3 ][ H 2 O ]2 [ NO 2 ][ H 2 ]7/ 2
Note the correspondence between the power and coefficient for each molecule.
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14.3. The chemical equation for the reaction is CO(g) + H2O(g)
H2(g) + CO2(g)
The expression for the equilibrium constant for this reaction is Kc =
[ CO 2 ][H 2 ] [ CO ][ H 2 O ]
We obtain the concentration of each substance by dividing the moles of substance by its volume. The equilibrium concentrations are as follows: [CO] = 0.057 M, [H2O] = 0.057 M, [CO2] = 0.043 M, and [H2] = 0.043 M. Substituting these values into the equation for the equilibrium constant gives Kc =
(0.043)(0.043) = 0.569 = 0.57 (0.057)(0.057)
14.4. Use the table approach, and give the starting concentrations, the change, and the equilibrium concentration of each by dividing moles by volume in liters. Conc. (M) Starting Change Equilibrium
2H2S(g) 0.0100 −2x 0.0100 − 2x
2H2(g) 0 +2x (= 0.00285) 2x
+
S2(g) 0 +x x
Because the problem states that 0.00285 M H2 was formed, we can use the 0.00285 M to calculate the other concentrations. The S2 molarity should be onehalf that, or 0.001425 M, and the H2S molarity should be 0.0100 − 0.00285, or 0.00715 M. Substituting into the equilibrium expression gives Kc =
[ H 2 ]2 [S2 ] ( 0.00285) 2 ( 0.001425) = = 2.26 x 10−4 = 2.3 x 10−4 2 2 [ H 2S] ( 0.00715)
14.5. Use the expression that relates Kc to Kp: Kp = Kc(RT)Δn The Δn term is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. In this case, Δn = (2 − 1) = 1, and Kp is Kp = (3.26 x 10−2) (0.0821 x 464)1 = 1.241 = 1.24 14.6. For a heterogeneous equilibrium, the concentration terms for liquids and solids are omitted because such concentrations are constant at a given temperature and are incorporated into the measured value of Kc. For this case, Kc is defined as Kc =
[Ni(CO) 4 ] [CO]4
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14.7. Because the equilibrium constant is very large (> 104), the equilibrium mixture will contain mostly products. Rearrange the Kc expression to solve for [NO2]. [NO2] =
K c [O 2 ][NO]2 =
(4.0 x 1013 )(2.0 x 106 )(2.0 x 106 ) 2 = 1.78 x 10−2
= 1.8 x 10−2 = 0.018 M 14.8. First, divide moles by volume in liters to convert to molar concentrations, giving 0.00015 M CO2 and 0.010 M CO. Substitute these values into the reaction quotient and calculate Q. Q =
[CO]2 (0.010) 2 = = 0.666 = 0.67 [CO 2 ] (0.00015)
Because Q = 0.67 and is less than Kc, the reaction will go to the right, forming more CO. 14.9. Rearrange the Kc expression, and substitute for Kc (= 0.0415) and the given moles per 1.00 L to solve for moles per 1.00 L of PCl5. [ PCl3 ][ Cl2 ] ( 0.020 ) ( 0.020 ) = = 9.63 x 10−3 = 9.6 x 10−3 M 0.0415 Kc
[PCl5] =
Because the volume is 1.00 L, the moles of PCl5 = 0.0096 mol. 14.10. Use the table approach, and give the starting number, the change, and the equilibrium number of moles of each. Amt. (mol) Starting Change Equilibrium
H2(g) 0.500 −x 0.500 − x
+
I2(g) 0.500 −x 0.500 − x
2HI(g) 0 +2x 2x
Substitute the equilibrium concentrations into the expression for Kc (= 49.7). Kc =
[HI]2 (2 x) 2 (2 x) 2 ; 49.7 = = [H 2 ][I 2 ] (0.500  x)(0.500  x) (0.500  x) 2
Taking the square root of both sides of the righthand equation and solving for x gives ±7.05 =
2x , or ±7.05(0.500 − x) = 2x (0.500  x)
Using the positive root, x = 0.390. Using the negative root, x = 0.698 (this must be rejected because 0.698 is greater than the 0.500 starting number of moles). Substituting x = 0.390 mol into the last line of the table to solve for equilibrium concentrations gives these amounts: 0.11 mol H2, 0.11 mol I2, and 0.78 mol HI.
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14.11. Use the table approach for starting, change, and equilibrium concentrations of each species. Conc. (M) Starting Change Equilibrium
PCl5(g) 1.00 −x 1.00 − x
PCl3(g) 0 +x x
+
Cl2(g) 0 +x x
Substitute the equilibriumconcentration expressions from the table into the equilibrium equation, and solve for x using the quadratic formula. [PCl3][Cl2] = Kc x [PCl5] = 0.0211(1.00 − x) = x2 x2 + 0.0211x − 0.0211 = 0 x =
0.0211 ±
(0.0211) 2  4(0.0211) 0.0211 ± 0.2913 = 2 2
x = −0.1562 (impossible; reject), or x = 0.13509 = 0.135 M (logical) Solve for the equilibrium concentrations using x = 0.135 M: [PCl5] = 0.86 M, [Cl2] = 0.135 M, and [PCl3] = 0.135 M. 14.12. a.
Increasing the pressure will cause a net reaction to occur from right to left, and more CaCO3 will form.
b.
Increasing the concentration of hydrogen will cause a net reaction to occur from right to left, forming more Fe and H2O.
14.13. a.
Because there are equal numbers of moles of gas on both sides of the equation, increasing the pressure will not increase the amount of product.
b.
Because the reaction increases the number of moles of gas, increasing the pressure will decrease the amount of product.
c.
Because the reaction decreases the number of moles of gas, increasing the pressure will increase the amount of product.
14.14. Because this is an endothermic reaction and absorbs heat, high temperatures will be more favorable to the production of carbon monoxide. 14.15. Because this is an endothermic reaction and absorbs heat, high temperatures will give the best yield of carbon monoxide. Because the reaction increases the number of moles of gas, decreasing the pressure will also increase the yield.
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Chapter 14: Chemical Equilibrium
ANSWERS TO CONCEPT CHECKS
14.1. The statement that when reactant A decreases by an amount x, product C increases by amount x implies that A and C have the same coefficients. The statement that when reactant B decreases by an amount x, product C increases by amount 2x implies that the coefficient of C is twice that of B. Therefore, the coefficient of A is twice that of B. The simplest equation satisfying these conditions is 2A + B → 2C. 14.2. To answer this question, find the relationship between the two species present, using the B(g), with equilibriumconstant expression and its value. For the first reaction, A(g) K = 2, this becomes K = 2 =
[B] [A]
This reduces to [B] = 2[A]. This corresponds to the container that has twice as many balls of one color than of the other color, namely container IV. Here, the blue molecules are B (eight of them), and the red molecules are A (four of them). For the second reaction, X(g) K = 6 =
2Y(g), with K = 6, this becomes
[Y]2 [X]
This reduces to [Y]2 = 6[X]. This corresponds to container I, where there are six of each color molecule. Since there are the same numbers of each molecule, you cannot determine which color corresponds to which molecule. For the third reaction, 2C(g) K = 1 =
D(g), with K = 1, this becomes
[D] [C]2
which reduces to [C]2 = [D]. This corresponds to container II. The red balls (nine of them) correspond to molecule D, and the blue balls (three of them) correspond to molecule C. 14.3. The concentration of each substance initially doubles. This means that each concentration factor in the reaction quotient expression is double that in the initial equilibrium mixture. Because this expression contains [CO][H2]2 in the denominator, the denominator increases by a factor of 23. However, the numerator contains only [CH3OH], which merely doubles. Thus, the reaction quotient equals two divided by two cubed, or onequarter, times the equilibrium constant. To approach the equilibrium constant, the numerator of the reaction quotient must increase, and the denominator must decrease. This means that more CH3OH must be produced. The reaction goes from left to right.
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14.4. The equilibriumconstant expression is [C]/([A][B]). A new equilibrium is attained in which the equilibriumconstant expression is [C]´/([A]´[B]´) = [C]´/(2[A]2[B]), where primes indicate new equilibrium concentrations. The value of the equilibriumconstant expression, though, must remain fixed in value, so [C]´/(2[A]2[B]) equals [C]/([A][B]).This means that the new concentration of C, or [C]´, must be four times larger than the original equilibrium value. Thus, the concentration of C is quadrupled. 14.5. For an exothermic reaction, as the temperature is increased, the reaction shifts toward the reactant side to absorb the heat and counteract the temperature increase. This corresponds to the case where there are mostly reactant molecules and very few product molecules, namely container I. At lower temperatures, the reaction shifts toward the product side to release heat. This corresponds to the case where there is a larger ratio of product molecules to reactant molecules, namely container II.
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ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
14.1. A reasonable graph showing the decrease in concentration of N2O4(g) and the increase in concentration of NO2(g) is shown below.
Moles of substance
Moles N 2O 4
Equilibrium amounts
Moles NO 2
Time
At first, the concentration of N2O4 is large, and the rate of the forward reaction is large, but then as the concentration of N2O4 decreases, the rate of the forward reaction decreases. In contrast, the concentration of NO2 builds up from zero to a low concentration. Thus, the initial rate of the reverse reaction is zero, but it steadily increases as the concentration of NO2 increases. Eventually, the two rates become equal when the reaction reaches equilibrium. This is a dynamic equilibrium because both the forward and reverse reactions are occurring at all times, even though there is no net change in concentration at equilibrium.
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Chapter 14: Chemical Equilibrium
14.2. The 1.0 mol of H2(g) and 1.0 mol of I2(g) in the first mixture reach equilibrium when the amounts of reactants decrease to 0.50 mol each and when the amount of product increases to 1.0 mol. The total number of moles of the reactants at the start is 2.0 mol, which is the same number of moles as in the second mixture, the 2.0 mol of HI that is to be allowed to come to equilibrium. The second mixture should produce the same number of moles of H2, I2, and HI at equilibrium, because if the total number of moles is constant, it should not matter from which direction an equilibrium is approached. 14.3. The equilibrium constant for a gaseous reaction can be written using partial pressures instead of concentrations because all the reactants and products are in the same vessel. Therefore, at constant temperature, the pressure, P, is proportional to the concentration, n/V. (The ideal gas law says that P = [n/V]RT.) 14.4. The addition of reactions 1 and 2 yields reaction 3: (Reaction 1) (Reaction 2) (Reaction 3)
HCN
+
OH− H2O HCN
CN− H+ H+
+ + +
H2O OH− CN−
The rule states that if a given equation can be obtained from the sum of other equations, the equilibrium constant for the given equation equals the product of the other equilibrium constants. Thus, K for reaction 3 is K = K1 x K2 = (4.9 x 104) x (1.0 x 10−14) = 4.9 x 10−10 14.5. a.
Homogeneous equilibrium. All substances are gases and thus exist in one phase, a mixture of gases.
b.
Heterogeneous equilibrium. The two copper compounds are solids, but the other substances are gases. This fulfills the definition of a heterogeneous equilibrium.
c.
Homogeneous equilibrium. All substances are gases and thus exist in one phase, a mixture of gases.
d.
Heterogeneous equilibrium. The two coppercontaining substances are solids, but the other substances are gases. This fulfills the definition of a heterogeneous equilibrium.
14.6. Pure liquids and solids can be ignored in an equilibriumconstant expression because their concentrations do not change. (If a solid is present, it has not dissolved in any gas or solution present.) In effect, concentrations of liquids and solids are incorporated into the value of Kc, as discussed in the chapter.
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14.7. A qualitative interpretation of the equilibrium constant involves using the magnitude of the equilibrium constant to predict the relative amounts of reactants and products at equilibrium. If Kc is around 1, the equilibrium mixture contains appreciable amounts (same order of magnitude) of reactants and products. If Kc is large, the equilibrium mixture is mostly products. If Kc is small, the equilibrium mixture is mostly reactants. The type of reaction governs what "large" and "small" values are; but for some types of reactions, "large" might be no less than 102 to 104, whereas "small" might be no more than 10−4 to 10−2. 14.8. The reaction quotient, Qc, is an expression that has the same form as the equilibriumconstant expression but whose concentrations are not necessarily equilibrium concentrations. It is useful in determining whether a reaction mixture is at equilibrium or, if not, what direction the reaction will go as it approaches equilibrium. 14.9. The ways in which the equilibrium composition of a mixture can be altered are (1) changing concentrations by removing some of the products and/or some of the reactants, (2) changing the partial pressure of a gaseous reactant and/or a gaseous product, and (3) changing the temperature of the reaction mixture. (Adding a catalyst cannot alter the equilibrium concentration; it can affect only the rate of reaction.) 14.10. The role of the platinum is that of a catalyst; it provides conditions (a surface) suitable for speeding up the attainment of equilibrium. The platinum has no effect on the equilibrium composition of the mixture even though it greatly increases the rate of reaction. 14.11. In some cases, a catalyst can affect the product in a reaction because it affects only the rate of one reaction out of several reactions that are possible. If two reactions are possible, and the uncatalyzed rate of one is much slower but that is the only reaction that is catalyzed, then the products with and without a catalyst will be different. The Ostwald process is a good example. In the absence of a catalyst, NH3 burns in O2 to form only N2 and H2O, even though it is possible for NH3 to react to form NO and H2O. Ostwald found that adding a platinum catalyst favors the formation of the NO and H2O almost to the exclusion of the N2 and H2O. 14.12. Four ways in which the yield of ammonia can be improved are (1) removing the gaseous NH3 from the equilibrium by liquefying it, (2) increasing the nitrogen or hydrogen concentration, (3) increasing the total pressure on the mixture (the moles of gas decrease), and (4) lowering the temperature (ΔH° is negative, so heat is evolved). Each causes a shift to the right in accordance with Le Châtelier's principle. 14.13. The answer is d, 2.10 moles. 14.14. The answer is a, 1.24. 14.15. The answer is c, 0.157. 14.16. The answer is b, Qc = 6.33 x 10−4; the concentration of NOBr increases.
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Chapter 14: Chemical Equilibrium
ANSWERS TO CONCEPT EXPLORATIONS
14.17. Part I [E]2 [C][D]
a.
Kc =
b.
A possible set of concentrations that you may observe in an equilibrium mixture is [C] = 2.0 M, [D] = 1.0 M, and [E] = 2.0 M. Another possible set of concentrations is [C] = 0.50 M, [D] = 0.25 M, and [E] = 0.50 M.
c.
There are an infinite number of concentrations for C, D, and E possible at equilibrium. No matter what set of initial concentrations there are for C and D, the system will eventually reach equilibrium. There are no restrictions on the amounts of reactants required, as long as some of each of C and D are present.
d.
The concentration of C cannot decrease to zero. When equilibrium is reached, each compound in the chemical equation must be present. No compound can disappear completely. The concentration of C can be determined as follows. Kc =2.0 =
[E]2 (2x) 2 = [C][D] (1.0  x)(1.0  x)
After taking the square root of each side, this reduces to 2.0 =
2x (1.0  x)
Solving this equation for x gives x =
2.0 2.0 +
2.0
=
1.414 = 0.414 = 0.41 M 3.414
Part II a.
When A and B are mixed together, a reaction occurs and proceeds from left to right. The concentrations of A and B will decrease, and the concentrations of F and G will increase.
b.
When F and G are mixed together, a reaction occurs and proceeds from right to left. The concentrations of A and B will increase, and the concentrations of F and G will decrease.
c.
When A and F are mixed together, no reaction occurs.
d.
When B and G are mixed together, no reaction occurs.
e.
When just B is placed in a container, no reaction occurs.
f.
When just G is placed in a container, no reaction occurs.
In parts a and b, where a reaction occurs, the equilibrium composition is mostly A and B. The magnitude of the equilibrium constant (1.0 x 10−5) suggests the equilibrium lies far to the left.
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14.18.
■
a.
To study the equilibrium, create a saturated solution. Dissolve enough Mg(OH)2 so that after mixing well, there is still excess solid that settles to the bottom of the container. As long as there is solid present in the flask, equilibrium is established. This solution could then be filtered and titrated with standardized HCl solution to determine [OH−].
b.
Ksp = [Mg2+][OH−]2
c.
If more solid Mg(OH)2 is added to a saturated solution, it will settle to the bottom of the container and not dissolve. It will have no effect on the concentrations of Mg2+ and OH− ions.
d.
Yes, If a solution is prepared from solid Mg(OH)2, then [Mg2+] = 1/2[OH−].
e.
If OH− is added to the equilibrium mixture, the reaction will shift to the left, causing more solid Mg(OH)2 to form and reducing the Mg2+ ion concentration.
f.
If some, but not all, of the solid Mg(OH)2 is removed from the equilibrium mixture, it will have no effect on the concentrations of Mg2+ and OH− ions.
g.
If there is no solid Mg(OH)2 on the bottom of the container, it is not likely to be at equilibrium. You could add more solid Mg(OH)2 and stir until some excess solid settled to the bottom of the container. Alternatively, you could heat the solution to reduce the volume of water until a precipitate began to form. Once a solid is present, equilibrium has been reached.
h.
The system is at dynamic equilibrium. This means that individual ions can find themselves at one time part of the solution and at another time part of the undissolved solid. The system is constantly changing on a microscopic level.
ANSWERS TO CONCEPTUAL PROBLEMS
14.19. For each three moles of H2 that react, two moles of ammonia form. The mole ratio is 2 mol NH 3 3 mol H 2
If x mol H2 react, then the amount of ammonia that forms is x mol H2 x
2 mol NH 3 2x = mol NH3 3 3 mol H 2
14.20. The addition of a pure liquid does not affect an equilibrium. (The pure liquid does not appear in the equilibrium constant; in effect, the concentration of the liquid does not change.) Thus, the amount of water vapor does not change appreciably. (More precisely, the amount of vapor decreases slightly because the liquid takes up more room in the container.) If, instead, you add water vapor to the container, vapor condenses until the original vapor pressure is restored. Thus, the amount of liquid water in the container increases. 14.21. Hydrogen, H2, is the limiting reactant, so the maximum amount of CH3OH that could form is 1 mol. However, because the reaction comes to equilibrium before it can go to completion, less than 1 mol of CH3OH forms. The answer is a. Copyright © Houghton Mifflin Company. All rights reserved.
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Chapter 14: Chemical Equilibrium
14.22. The system must exist as an equilibrium mixture of all four substances. The reaction can be represented as Fe3O4(s) + 4H2(g)
3Fe(s) + 4H2O(g)
If you pass H2(g) over iron oxide, the reaction shifts to the right, and metallic iron and water vapor form. If, instead, you pass water vapor over metallic iron, the reaction shifts to the left, and iron oxide and H2 form. An excess of one reactant pushes the reaction in the opposite direction. 14.23. a.
The equilibriumconstant expression for this reaction is K =
[H 2 ][I 2 ] = 2.0 [HI]2
The equilibrium case is where the reaction quotient, Q, is equal to the equilibrium constant, K. This occurs in picture III, where Q = b.
(2)(4) = 2.0 = K (2) 2
For picture I, the reaction quotient is Q =
(2)(2) = 0.25 < K (4) 2
Since Q < K, the reaction shifts toward the product (H2 + I2) side. For picture II, the reaction quotient is Q =
(4)(4) = 1.0 < K (4) 2
Since Q < K, the reaction shifts toward the product (H2 + I2) side. 14.24. Equilibrium has been reached when the concentration of reactants and products is constant. The equilibrium region on the graph is where the lines flatten out, indicating that the concentrations of reactants and products are not changing. At equilibrium, the concentration of A is approaching 2.0 M, and the concentration of B is approaching 1.0 M. Substituting into the equilibriumconstant expression gives Kc =
[B] 1.0 = = 0.25 [A]2 (2.0) 2
14.25. a.
When the volume is doubled, the pressure is reduced by onehalf. A decrease in pressure results in the equilibrium shifting to the side of the reaction with the greater number of moles of gas. Note that in this reaction, there are 2 mol of gas reactants versus 1 mol of gas products. Since the pressure has been reduced, the reaction will shift to the left, toward the side of the reaction with the greater number of moles of gas.
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b.
507
Since the reaction has been shifted to the left, the concentrations of A and B will increase, while the concentration of C will decrease.
14.26. For an endothermic reaction, as the temperature lowers, the reaction shifts toward the reactant side, producing heat and counteracting the lowering of the temperature. This corresponds to container I, where there is a smaller ratio of product molecules to reactant molecules. In terms of the reaction quotient, Q, the lower temperature corresponds to the container with the smaller value of Q (container I). 14.27. The formation of SO3 can be represented by the following reaction: 2SO2(g) + O2(g)
2SO3(g)
The first condition mentioned is the oxygenenriched air. The higher concentration of O2 drives the reaction to the right, increasing the yield of SO3. Next, the reaction temperature is 420°C. Since the reaction is exothermic, the elevated temperature drives the reaction to the left, decreasing the yield of SO3. The vanadium(V) oxide catalyst affects the rate of the reaction, but not the amount of SO3 that forms at equilibrium. Finally, the SO3 that forms in the reaction is absorbed by concentrated sulfuric acid and removed from the system. This causes the reaction to shift to the right, forming more SO3. Another condition that can be explored is a change in pressure. Since there are more moles of gas on the left side, an increase in pressure (or decrease in volume) should cause more SO3 to form. 14.28. If you start with 1.00 mol CH3OH, there will be 0.100 mol CH3OH present at equilibrium. Assume the reaction could go to completion. Starting with 1.00 mol CO and 2.00 mol H2, stoichiometry shows you would obtain 1.00 mol CH3OH. Therefore, 1.00 mol CH3OH is chemically equivalent to 1.00 mol CO and 2.00 mol H2. At equilibrium, both starting with 1.00 mol CO and 2.00 mol H2 and starting with 1.00 mol CH3OH give the same equilibrium mixture, one containing 0.100 mol CH3OH.
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SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 14.29. Use the table approach, and give the starting, change, and equilibrium number of moles of each. Amt. (mol) Starting Change Equilibrium
PCl5(g) 2.500 −x 2.500 − x
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PCl3(g) 0 +x x = 0.338
+
Cl2(g) 0 +x x
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Chapter 14: Chemical Equilibrium
Since the equilibrium amount of PCl3 is given in the problem, this tells you x = 0.338. The equilibrium amounts for the other substances can now be determined. Equilibrium amount Cl2 = x = 0.338 mol Equilibrium amount PCl5 = 2.500 − x = 2.500 − 0.338 = 2.162 mol Therefore, the amounts of the substances in the equilibrium mixture are 2.162 mol PCl5, 0.338 mol PCl3, and 0.338 mol Cl2. 14.30. Because the amount of NO2 at the start is zero, the 0.90 mol of NO2 at equilibrium must also equal the change, x, in the moles of N2O3. Use the table approach, and insert the starting, change, and equilibrium number of moles. NO2(g) + NO(g) Amt. (mol) N2O3(g) Starting 3.00 0 0 Change −x +x +x Equilibrium 3.00 − x X = 0.90 x Since the equilibrium amount of NO2 is given in the problem, this tells you x = 0.90. The equilibrium amounts for the other substances can now be determined. Equilibrium amount NO = x = 0.90 mol Equilibrium amount N2O3 = 3.00 − x = 3.00 − 0.90 = 2.10 mol Therefore, the amounts of the substances in the equilibrium mixture are 2.10 mol N2O3, 0.90 mol NO2, and 0.90 mol NO. 14.31. Use the table approach, and give the starting, change, and equilibrium number of moles of each. 3H2(g) 2NH3(g) 1.800 0 −3x +2x 1.800 − 2x = 0.048 3x Since the equilibrium amount of NH3 is given in the problem, this tells you x = 0.024. The equilibrium amounts for the other substances can now be determined. Amt. (mol) Starting Change Equilibrium
N2(g) 0.600 −x 0.600 − x
+
Equilibrium amount N2 = 0.600 − 0.024 = 0.576 mol Equilibrium amount H2 = 1.800 − 3 x (0.024) = 1.728 mol Therefore, the amounts of the substances in the equilibrium mixture are 0.576 mol N2, 1.728 mol H2, and 0.048 mol NH3. 14.32. Use the table approach, and give the starting, change, and equilibrium number of moles of each. Amt. (mol) Starting Change Equilibrium
2NO(g) 0.0524 −2x 0.0524 − 2x
+
Br2(g) 0.0262 −x 0.0262 − x
2NOBr(g) 0 +2x 2x = 0.0311
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Since the equilibrium amount of NOBr is given in the problem, this tells you x = 0.01555. The equilibrium amounts for the other substances can now be determined. Equilibrium amount NO = 0.0524 − 2 x (0.01555) = 0.02130 = 0.0213 mol Equilibrium amount Br2 = 0.0262 − 0.01555 = 0.01065 = 0.0107 mol Therefore, the amounts of the substances in the equilibrium mixture are 0.0213 mol NO, 0.0107 mol Br2, and 0.0311 mol NOBr. 14.33. Use the table approach, and give the starting, change, and equilibrium number of moles of each. Amt. (mol) Starting Change Equilibrium
2SO2(g) 0.0400 −2x 0.0400 − 2x
+
O2(g) 0.0200 −x 0.0200 − x
2SO3(g) 0 +2x 2x (= 0.0296)
Therefore, x = 0.0148, and the amounts of substances at equilibrium are 0.0104 mol SO2, 0.0052 mol O2, and 0.0296 mol SO3. 14.34. Use the table approach, and give the starting, change, and equilibrium number of moles of each. Amt. (mol) Starting Change Equilibrium
CO(g) 0.1500 −x 0.1500 − x (= 0.1187)
+ 2H2(g) 0.3000 −2x 0.3000 − 2x
CH3OH(g) 0 +x x
Because 0.1500 − x = 0.1187, x = 0.0313. Therefore, the amounts of substances at equilibrium are 0.1187 mol CO, 0.2374 mol H2, and 0.0313 mol CH3OH. 14.35. a.
Kc =
[NO 2 ] [NO] [N 2 O3 ]
b.
Kc =
[H 2 ]2 [S2 ] [H 2S]2
c.
Kc =
[NO 2 ]2 [NO]2 [O 2 ]
d.
Kc =
[P(NH 2 )3 ] [HCl]3 [PCl3 ] [NH 3 ]3
Kc =
[N 2 H 4 ] [N 2 ] [H 2 ]2
14.36. a.
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Chapter 14: Chemical Equilibrium
b.
Kc =
[NO]2 [Cl2 ] [NOCl]2
c.
Kc =
[N 2 ] [H 2 O]2 [NO]2 [H 2 ]2
d.
Kc =
[Cl2 ] [H 2 O] [HCl]2 [O 2 ]1/2
14.37. The reaction is 2H2S(g) + 3O2(g)
2H2O(g) + 2SO2(g).
14.38. The reaction is C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(g).
14.39. When the reaction is halved, the equilibriumconstant expression is Kc =
[NO 2 ]2 [O 2 ]1/2 [N 2 O5 ]
When the reaction is then reversed, the equilibriumconstant expression becomes Kc =
[N 2 O5 ] [NO 2 ]2 [O 2 ]1/2
14.40. When the reaction is halved, the equilibriumconstant expression is Kc =
[NH 3 ]2 [O 2 ]5/2 [NO]2 [H 2 O]3
When the reaction is then reversed, the equilibriumconstant expression becomes Kc =
[NO]2 [H 2 O]3 [NH 3 ]2 [O 2 ]5/2
ZZX 2HI must be the 14.41. Because Kc = 1.84 for 2HI H2 + I2, the value of Kc for H2 + I2 YZZ reciprocal of Kc for the first reaction. Mathematically, this can be shown as follows:
Forward: Kc = Reverse: Kc =
[ H 2 ][ I 2 ] = 1.84 [ HI ]2 [ HI ]2 1 = [ H ][ I2 ] [ H 2 ][ I 2 ] 2 [ HI ]2
=
1 K c (f )
Thus, for the reverse reaction, Kc is calculated as follows: Kc = 1 ÷ 1.84 = 5.434 x 10−1 = 0.543
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14.42. Because Kc = 27.8 for CS2 + 4H2 CH4 + 2H2S, the value of Kc for 1/2CH4 + H2S 1/2CS2 + 2H2, the second (2) reaction, must be the reciprocal of the square root of Kc for the first (1) reaction. Mathematically, this can be shown as follows: Kc(1) =
[CH 4 ][H 2S]2 [CS2 ][H 2 ]4
Kc(2) =
⎡ [CH 4 ][H 2S]2 ⎤ [CS2 ]1/2 [H 2 ]2 = ⎢ 4 ⎥ [CH 4 ]1/2 [H 2S] ⎣ [CS2 ][H 2 ] ⎦
1/2
= [Kc(1)]−1/2
Thus, for the second (2) reaction, Kc is calculated as follows: Kc =
1 27.8
= 0.1896 = 0.190
14.43. First, calculate the molar concentration of each of the compounds in the equations: [H2] = 0.488 mol H2 ÷ 6.00 L = 0.08133 M [I2] = 0.206 mol I2 ÷ 6.00 L = 0.03433 M [HI] = 2.250 mol HI ÷ 6.00 L = 0.3750 M Now, substitute these into the Kc expression: Kc =
(0.3750) 2 = 50.36 = 50.4 (0.08133)(0.03433)
14.44. First, calculate the molar concentration of each of the compounds in the equilibrium: [PCl5] = 0.0126 mol PCl5 ÷ 4.00 L = 0.003150 M [PCl3] = 0.0148 mol PCl3 ÷ 4.00 L = 0.003700 M [Cl2] = 0.0870 mol Cl2 ÷ 4.00 L = 0.02175 M Now, substitute these into the expression for Kc: Kc =
0.003150 = 39.14 = 39.1 (0.003700)(0.02175)
14.45. Substitute the following concentrations into the Kc expression: [SO3] = 0.0296 mol ÷ 2.000 L = 0.01480 M [SO2] = 0.0104 mol ÷ 2.000 L = 0.005200 M [O2] = 0.0052 mol ÷ 2.000 L = 0.00260 M Kc =
( 0.01480 ) 2 = 3.11 x 103 = 3.1 x 103 ( 0.005200 ) 2 ( 0.00260 )
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14.46. Substitute the following concentrations into the Kc expression: [CH3OH] = 0.0313 mol ÷ 1.50 L = 0.02086 M [CO] = 0.1187 mol ÷ 1.50 L = 0.07913 M [H2] = 0.2374 mol ÷ 1.50 L = 0.1582 M Kc =
( 0.02086 ) ( 0.07913) ( 0.1582 ) 2
= 10.53 = 10.5
14.47. For each mole of NOBr that reacts, (1.000 − 0.094 = 0.906) mol remains. Starting with 2.00 mol NOBr, this means that 2 x 0.906 mol NOBr, or 1.812 mol NOBr, remains. Because the volume is 1.00 L, the concentration of NOBr at equilibrium is 1.812 M. Assemble a table of starting, change, and equilibrium concentrations: Conc. (M) Starting Change Equilibrium
2NOBr(g) 2.00 −2x 2.00 − 2x (= 1.812)
2NO(g) 0 +2x 2x
+
Br2(g) 0 +x x
Because 2.00 − 2x = 1.812, x = 0.094 M. Therefore, the equilibrium concentrations are [NOBr] = 1.812 M, [NO] = 0.188 M, and [Br2 ] = 0.094 M. Kc =
[NO]2 [Br2 ] (0.188) 2 (0.094) = = 1.01 x 10−3 = 1.0 x 10−3 [NOBr]2 (1.812) 2
14.48. A decomposition of 6.0% of NO2 means that 0.060 x 2.00 mol = 0.12 mol was decomposed, leaving 1.88 mol at equilibrium. Because the concentration at the start was 2.00 mol/80.0 L, or 0.0250 M, the concentration at equilibrium is 1.88 mol/80.0 L, or 0.0235 M. Assemble a table of starting, change, and equilibrium concentrations: Conc. (M) Starting Change Equilibrium
2NO2(g) 0.0250 −2x 0.0250 − 2x
2x
2NO(g) 0 +2x +2x (= 0.0235)
+
O2(g) 0 +x X
Because 0.0250 − 2x = 0.0235 M, x = 0.00075 M; thus, [NO2] = 0.0235 M, [NO] = 0.0015 M, and [O2] = 0.00075 M. The value of Kc is Kc =
( 0.0015) 2 ( 0.00075) = 3.055 x 10−6 = 3.1 x 10−6 2 ( 0.0235)
a.
Kp =
PHBr 2 PH2 PBr2
b.
Kp =
14.49.
PCH4 PH2S 2 PCS2 PH2 4
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c.
Kp =
d.
Kp =
513
PH2 O 2 PCl2 2 PHCl 4 PO2 PCH3OH PCO PH2 2
14.50. a.
Kp =
b.
Kp =
c.
Kp =
d.
Kp =
PNO2 2 PN2 O4
PNOBr 2 PNO 2 PBr2
PSO3 2 PSO2 2 PO2 PNO 4 PH2 O 6 PNH3 4 PO2 5
14.51. There are 3 mol of gaseous product for every 8 mol of gaseous reactant, so Δn = 3 − 5 = −2. Using this, calculate Kp from Kc: Kp = Kc(RT)Δn = 0.28 (0.0821 x 1173)−2 = 3.019 x 10−5 = 3.0 x 10−5 14.52. There is 1 mol of gaseous product for every 3 mol of gaseous reactant, so Δn = 3 − 1 = 2. Using this, calculate Kp from Kc: Kp = Kc(RT)Δn = (0.0952)(0.0821 x 500)2 = 160.4 = 160. 14.53. For each 1 mol of gaseous product, there are 1.5 mol of gaseous reactants; thus, Δn = 1 − 1.5 = −0.5. Using this, calculate Kc from Kp: Kc =
Kp (RT)Δn
=
6.55 = 6.55 x (0.0821 x 900)0.5 = 56.303 = 56.3 (0.0821 x 900)0.5
14.54. For each 1 mol of gaseous product, there is 0.5 mol of gaseous reactant; thus, Δn = 1 − 0.5 = 0.5. Using this, calculate Kc from Kp: Kc =
Kp (RT)
Δn
=
7.55 x 102 = 7.891 x 10−3 = 7.89 x 10−3 (0.0821 x 1115)0.5
14.55. a.
Kc =
[ CO ]2 [ CO 2 ]
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Chapter 14: Chemical Equilibrium
b.
Kc =
[ CO 2 ] [ CO]
c.
Kc =
[ CO 2 ] [SO 2 ] [ O 2 ]1/ 2
d.
Kc = [Pb2+] [I−]2
14.56. a.
Kc = [NH3] [HCl]
b.
Kc =
c.
Kc = [H2O] [CO2]
d.
Kc =
[N 2 O]2 [ CO 2 ] [N 2 ]2
1 [Fe3+ ] [ OH  ]3
14.57. a.
Not complete; Kc is very small (10−31), indicating very little reaction.
b.
Nearly complete; Kc is very large (1021), indicating nearly complete reaction.
14.58. a.
Nearly complete; Kc is very large (10113), indicating nearly complete reaction.
b.
Not complete; Kc is very small (10−17), indicating very little reaction.
14.59. Kc is extremely small, indicating very little reaction at room temperature. Because the decomposition of HF yields equal amounts of H2 and F2, at equilibrium [H2] = [F2]. So, for the decomposition of HF, Kc =
[H 2 ]2 [H 2 ] [F2 ] = [HF]2 [HF]2
[H2] = Kc1/2 [HF] = (1.0 x 10−95)1/2 (1.0 M) = 3.16 x 10−48 = 3.2 x 10−48 mol/L This result does agree with what is expected from the very small magnitude of Kc.
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515
14.60. Kc is extremely large, indicating nearly complete reaction at room temperature. The problem states that at equilibrium, [SO2] = [O2]. So, for the reaction of SO2, Kc =
[SO3 ]2 [SO3 ]2 = 2 [SO 2 ] [O 2 ] [SO 2 ]3
Solving for [SO2] yields ⎡ [SO ]2 ⎤
1/ 3
3 [SO2] = ⎢ ⎥ K c ⎣ ⎦
⎡ (1.0) 2
1/ 3
⎤
= ⎢ 35 ⎥ ⎣ 8.0 x 10 ⎦
= 1.07 x 10−12 = 1.1 x 10−12 M
This agrees with what we expect from the magnitude of Kc. 14.61. Calculate Q, the reaction quotient, and compare it to the equilibrium constant (Kc = 3.07 x 10−4). If Q is larger, the reaction will go to the left; if Q is smaller, the reaction will go to the right; if they are equal, the reaction is at equilibrium. In all cases, Q is found by combining these terms:
a.
Q =
[ NO ]2 [Br2 ] [ NOBr]2
Q =
(0.0162) 2 (0.0123) = 6.226 x 10−4 = 6.23 x 10−4 (0.0720) 2
Q > Kc. The reaction should go to the left. b.
Q =
(0.0159) 2 (0.0139) = 2.400 x 10−4 = 2.40 x 10−4 (0.121) 2
Q < Kc. The reaction should go to the right. c.
Q =
(0.0134) 2 (0.0181) = 3.063 x 10−4 = 3.06 x 10−4 (0.103) 2
Q = Kc. The reaction should be at equilibrium. d.
Q =
(0.0121) 2 (0.0105) = 6.900 x 10−4 = 6.90 x 10−4 (0.0472) 2
Q > Kc. The reaction should go to the left. 14.62. Calculate Q, the reaction quotient, and compare it to the equilibrium constant (Kc = 3.59). If Q is larger, the reaction will go to the left; if Q is smaller, the reaction will go to the right; if they are equal, the reaction is at equilibrium. In all cases, Q is found by combining these terms:
a.
Q =
[ CS2 ] [H 2 ]4 [CH 4 ] [ H 2S]2
Q =
(1.43)(1.12) 4 = 1.0249 = 1.02 (1.26)(1.32) 2
Q < Kc. The reaction should go to the right.
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Chapter 14: Chemical Equilibrium
b.
Q =
(1.15)(1.73) 4 = 3.566 = 3.57 (1.25)(1.52) 2
Q = Kc. The reaction should be at equilibrium. c.
Q =
(1.10)(1.20) 4 = 0.8825 = 0.883 (1.30)(1.41) 2
Q < Kc. The reaction should go to the right. d.
Q =
(1.23)(1.91) 4 = 5.1314 = 5.13 (1.56)(1.43) 2
Q > Kc. The reaction should go to the left. 14.63. Calculate Q, the reaction quotient, and compare it to the equilibrium constant. If Q is larger, the reaction will go to the left, and vice versa. Q is found by combining these terms: Q =
[ CH 3 OH ] (0.020) = = 20.0 (>10.5) (0.010)(0.010) 2 [CO][H 2 ]2
The reaction goes to the left. 14.64. Calculate Q, the reaction quotient, and compare it to the equilibrium constant. If Q is larger, the reaction will go to the left, and vice versa. Q is found by combining these terms: Q =
[SO3 ]2 ( 0.40 ) 2 = = 40.0 (>0.0417) 2 ( 0.20 ) 2 (0.10 ) [SO 2 ] [O 2 ]
The reaction goes to the left. 14.65. Substitute into the expression for Kc and solve for [COCl2]: Kc = 1.23 x 103 =
[COCl2 ] [COCl2 ] = [CO][Cl2 ] (0.012)(0.025)
[COCl2] = (1.23 x 103)(0.012)(0.025) = 0.369 = 0.37 M 14.66. Substitute into the expression for Kc and solve for [NO]: Kc = 0.0025 =
[NO]2 [NO]2 = [N 2 ][O 2 ] (0.031)(0.023)
[NO] = [(0.0025)(0.031)(0.023)]1/2 = 1.33 x 10−3 = 1.3 x 10−3 M
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517
14.67. Divide moles of substance by the volume 5.0 L to obtain concentration. The starting concentrations are 3.0 x 10−4 M for both [I2] and [Br2]. Assemble a table of starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
I2(g) 3.0 x 10−4 −x (3.0 x 10−4) − x
+
Br2(g) 3.0 x 10−4 −x (3.0 x 10−4) − x
2 IBr(g) 0 +2x 2x
Substituting into the equilibriumconstant expression gives Kc = 1.2 x 102 =
[IBr]2 (2 x) 2 = [I 2 ][Br2 ] (3.0 x 104  x)(3.0 x 104  x)
Taking the square root of both sides yields 10.95 =
( 2x ) ( 3.0 x 10 4  x )
Rearranging and simplifying the right side give ( 2x ) = (0.182 x) 10.95
(3.0 x 10−4 − x) = x = 2.53 x 10−4 M
Thus, [I2] = [Br2] = 4.70 x 10−5 = 4.7 x 10−5 M, and [IBr] = 5.06 x 10−4 = 5.1 x 10−4 M. 14.68. Divide moles of substance by the volume 8.00 L to obtain concentration. The starting concentrations are 0.09937 M for both [N2] and [O2]. Assemble a table of starting, change, and equilibrium concentrations. N2(g) 0.09937 −x 0.09937 − x
Conc. (M) Starting Change Equilibrium Kc = 0.0123 =
+
O2(g) 0.09937 −x 0.09937 − x
2NO(g) 0 +2x 2x
[NO]2 (2 x) 2 = [N 2 ][O 2 ] (0.09937  x)(0.09937  x)
Taking the square root of both sides gives 0.1109 =
( 2x ) ( 0.09937  x )
Rearranging and simplifying the right side yield (0.09937 − x) =
( 2x ) = (18.03 x) 0.1109
x = 5.222 x 10−3 M Thus [N2] = [O2] = 0.09414 = 0.0941 M, and [NO] = 1.044 x 10−2 = 1.04 x 10−2 M.
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Chapter 14: Chemical Equilibrium
14.69. Divide moles of substance by the volume 1.25 L to obtain concentration. The starting concentration is 1.00 M for [CO2], but the concentration of carbon, a solid, is omitted. Assemble a table of starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
CO2(g) 1.00 −x 1.00 − x
+
C(s)
2CO(g) 0 +2x 2x
Substituting into the equilibrium expression for Kc gives Kc = 14.0 =
[ CO ]2 ( 2x ) 2 = [ CO 2 ] (1.00  x )
Rearranging and solving for x yield 14.0 − 14.0x = 4x2 4x2 + 14.0x − 14.0 = 0 (quadratic equation) Using the solution to the quadratic equation gives x =
14.0 ±
(14.0) 2  4(4)(14.0) 2(4)
x = −4.31 (impossible; reject), or x = 0.8117 = 0.812 M (logical) Thus, [CO2] = 0.19 M, and [CO] = 1.62 M. 14.70. Divide moles of substance by the volume 4.0 L to obtain concentration. The starting concentrations are 0.10 M for both [PCl3] and [Cl2]. Assemble a table of starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
PCl3(g) 0.10 −x 0.10 − x
+
Cl2(g) 0.10 −x 0.10 − x
PCl5(g) 0 +x x
Substituting into the equilibrium expression for Kc gives Kc = 49 =
[ PCl5 ] x = ( 0.10  x ) ( 0.10  x ) [ PCl3 ][ Cl2 ]
Rearranging and solving for x yield 49(0.10 − x)2 = 49(x2 − 0.20x + 0.010) = x 49x2 − 10.8x + 0.49 = 0 (quadratic equation) Using the solution to the quadratic equation gives x =
10.8 ±
(10.8) 2  4(49)(0.49) 2(49)
x = 0.157 (impossible; reject), or x = 0.06389 = 0.064 M (logical)
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519
Thus, [PCl3] = [Cl2] = 0.036 M, and [PCl5] = 0.064 M; the vessel contains 0.14 mol PCl3, 0.14 mol Cl2, and 0.26 mol PCl5. 14.71. Divide moles of substance by the volume 10.00 L to obtain concentration. The starting concentrations are 0.1000 M for [CO] and 0.3000 M for [H2]. Assemble a table of starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
CO(g) 0.1000 −x 0.1000 − x
+
3H2(g) 0.3000 −3x 0.3000 − 3x
CH4(g) 0 +x x
+
H2O(g) 0 +x x
Substituting into the equilibrium expression for Kc gives Kc = 3.92 =
[ CH 4 ][ H 2 O ] x2 x2 = = [ CO ][ H 2 ]3 ( 0.1000  x )[ 3( 0.1000  x ) ]3 27 ( 0.1000  x ) 4
Multiplying both sides by 27 and taking the square root of both sides give 10.29 =
x (0.1000  x) 2
Or, 10.29x2 − 3.058x + 0.1029 = 0 Using the solution to the quadratic equation yields x =
3.058 ±
(3.058) 2  4(10.29)(0.1029) 2(10.29)
x = 0.2585 (can't be > 0.1000, so reject), or x = 0.03868 = 0.0387 M (use) Thus, [CO] = 0.0613 M, [H2] = 0.1839 M, [CH4] = 0.0387 M, and [H2O] = 0.0387 M. 14.72. Divide moles of substance by the volume 5.00 L to obtain concentration. The starting concentrations are 0.200 M for [N2] and 0.600 M for [H2]. Assemble a table of starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
N2(g) 0.200 −x 0.200 − x
+
3H2(g) 0.600 −3x 0.600 − 3x
2NH3(g) 0 +2x 2x
Substituting into the equilibrium expression for Kc gives Kc = 0.159 =
[NH 3 ]2 (2 x) 2 (2x ) 2 = = 3 3 (0.200  x)(0.600  3x) ( 0.200  x ) 33 ( 0.200  x )3 [N 2 ][H 2 ]
Combining the value of Kc and 33 from the righthand term gives (33)(0.159) =
(2x) 2 (0.200  x) 4
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Chapter 14: Chemical Equilibrium
Taking the square root of both sides gives 2.072 =
2x ( 0.200  x ) 2
0.08288 − 0.8288x + 2.072x2 = 2x 2.072x2 − 2.8288x + 0.08288 = 0 (quadratic equation) Using the solution to the quadratic equation yields x =
2.8288 ±
(2.8288) 2  4(2.072)(0.08288) 2(2.072)
x = 1.335 (larger than 0.200, so reject), or x = 0.02995 M (logical; use) Thus, [N2] = 0.02995 = 0.0300 M, [H2] = 0.08986 = 0.0899 M, and [NH3] = 0.05991 = 0.0599 M. 14.73. Forward direction 14.74. a.
Reverse direction
b.
Forward direction
14.75. a.
A pressure increase has no effect because the number of moles of reactants equals that of products.
b.
A pressure increase has no effect because the number of moles of reactants equals that of products.
c.
A pressure increase causes the reaction to go to the left because the number of moles of reactants is less than that of products.
14.76. a.
Pressure increase
b.
Pressure increase
c.
Pressure decrease
14.77. The fraction would not increase because an increase in temperature decreases the amounts of products of an exothermic reaction. 14.78. The decomposition would be favorable at high temperatures because an increase in temperature increases the amounts of products of an endothermic reaction.
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14.79. The value of ΔH° is calculated from the ΔHf° values below each substance in the reaction:
2NO2(g)
+
2(33.10)
7H2(g)
2NH3(g)
7(0)
2(45.9)
+
4H2O(g) 4(241.8)
ΔH° = −967.2 + (−91.8) − 66.20 = −1125.2 kJ/2 mol NO2 The equilibrium constant will decrease with temperature because raising the temperature of an exothermic reaction causes the reaction to go farther to the left. 14.80. The value of ΔH° is calculated from the ΔHf° values below each substance in the reaction:
CH4(g)
+
74.9
2H2S(g) 2(20)
CS2(g)
+
117
4H2(g) 4(0)
ΔH° = 117 − (−40) − (−74.9) = 231.9 = 232 kJ/mol CH4 The equilibrium constant will increase with temperature because raising the temperature of an endothermic reaction causes the reaction to go farther to the right. 14.81. Because the reaction is exothermic, the formation of products will be favored by low temperatures. Because there are more molecules of gaseous products than of gaseous reactants, the formation of products will be favored by low pressures. 14.82. Because the reaction is exothermic, the formation of products will be favored by low temperatures. Because there are more molecules of gaseous reactants than of gaseous products, the formation of products will be favored by high pressures.
■
SOLUTIONS TO GENERAL PROBLEMS
14.83. Substitute the concentrations into the equilibrium expression to calculate Kc. Kc =
[ CH 3 OH ] ( 0.015) = = 4.28 = 4.3 2 ( 0.096 ) ( 0.191) 2 [ CO ][ H 2 ]
14.84. Substitute the concentrations into the equilibrium expression to calculate Kc. Kc =
[SO3 ] ( 0.0160 ) = = 62.3 = 62 ( 0.0056 ) ( 0.0021)1/ 2 [SO 2 ][ O 2 ]1/2
14.85. Assume 100.00 g of gas: 90.55 g CO, and 9.45 g CO2. The moles of each are 90.55 g CO x
1 mol CO = 3.2328 mol COI 28.01 g CO
9.45 g CO2 x
1 mol CO 2 = 0.2147 mol CO2 44.01 g CO 2
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Chapter 14: Chemical Equilibrium
Total moles of gas = (3.2328 + 0.2147) mol = 3.4475 mol. Use the ideal gas law to convert to the volume of gaseous solution: V =
nRT (3.4475 mol)(0.082057 L • atm/K • mol)(850 + 273)K = = 317.69 L P 1.000 atm
The concentrations are [CO] = [CO2] =
3.2328 mol CO = 0.010176 M 317.69 L 0.2147 mol CO 2 317.69 L
= 6.758 x 10−4 M
Find Kc by substituting into the equilibrium expression: Kc =
[ CO ]2 ( 0.010176 ) 2 = = 0.1532 = 0.153 [ CO 2 ] ( 6.758 x 10 4 )
14.86. Assume 100.0 g of gas: 65.8 g NO2, and 34.2 g N2O4. The moles of each are 65.8 g NO2 x 34.2 g N2O4 x
1 mol NO 2 = 1.430 mol 46.01 g NO 2 1 mol N 2 O 4 92.01 g N 2 O 4
= 0.3717 mol
Total moles of gas = (1.430 + 0.3717) mol = 1.802 mol. Use the ideal gas law to convert to the volume of gaseous solution: V =
(1.802 mol)(0.082057 L • atm/K • mol)(25 + 273)K nRT = = 44.06 L 1.00 atm P
The concentrations are [NO2] = [N2O4] =
1.430 mol NO 2 = 0.03246 M 44.06 L 0.3717 mol N 2 O 4 = 0.008436 M 44.06 L
Find Kc by substituting into the equilibrium expression: Kc =
[NO 2 ]2 (0.03246) 2 = = 0.1248 = 0.125 M (0.008436) [N 2 O 4 ]
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14.87. After calculating the concentrations after mixing, calculate Q, the reaction quotient, and compare it with Kc. [N2] = [H2] = 1.00 mol ÷ 2.00 L = 0.500 M [NH3] = 2.00 mol ÷ 2.00 L = 1.00 M Q =
[NH 3 ]2 (1.00) 2 = = 16.00 = 16.0 (0.500)(0.500)3 [N 2 ][H 2 ]3
Because Q is greater than Kc, the reaction will go in the reverse direction (to the left) to reach equilibrium. 14.88. To calculate the concentrations after mixing, the volume can be assumed to be 1.00 L, symbolized as V. Because the volumes in the numerator and denominator cancel each other, they do not matter. Assume a 1.00L volume and calculate Q, the reaction quotient, and compare it with Kc. [I2] = [Br2] = 0.0010 mol/1.00 L [IBr] = 0.200 mol ÷ 1.00 L = 0.200 mol/1.00 L Q =
[ IBr ]2 ( 0.200 mol/1.00 L ) 2 = = 4.00 x 104 [ I 2 ][ Br2 ] ( 0.0010 mol/1.00 L ) ( 0.0010 mol/1.00 L )
Because Q is greater than Kc, the reaction will go in the reverse direction (to the left) to reach equilibrium. 14.89. To calculate the concentrations after mixing, assume the volume to be 1.00 L, symbolized as V. Because the volumes in the numerator and denominator cancel each other, they do not matter. Assume a 1.00L volume and calculate Q, the reaction quotient, and compare it with Kc. [CO] = [H2O] = [CO2 ] = [H2] = 1.00 mol/1.00 L Q =
[ CO 2 ][ H 2 ] (1.00 mol/1.00 L ) (1.00 mol/1.00 L ) = = 1.00 (1.000 mol/1.00 L ) (1.00 mol/1.00 L ) [ CO ][ H 2 O ]
Because Q is greater than Kc, the reaction will go in the reverse direction (left) to reach equilibrium. 14.90. After calculating the concentrations after mixing, calculate Q, the reaction quotient, and compare it with Kc. [CO] = 0.10 mol ÷ 2.0 L = 0.050 M [H2] = 0.20 mol ÷ 2.0 L = 0.10 M [CH3OH] = 0.50 mol ÷ 2.0 L = 0.25 M Q =
[ CH 3 OH ] ( 0.25) = = 5.00 x 102 = 5.0 x 102 2 ( 0.050 ) ( 0.10 ) 2 [ CO ][ H 2 ]
Because Q is greater than Kc, the reaction will go in the reverse direction (left) to reach equilibrium.
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Chapter 14: Chemical Equilibrium
14.91. Assemble a table of starting, change, and equilibrium concentrations, letting 2x = the change in [HBr]. Conc. (M) Starting Change Equilibrium
2HBr(g) 0.010 −2x 0.010 − 2x
Kc = 0.016 = 0.126 =
H2(g) 0 +x X
+
Br2(g) 0 +x x
[ H 2 ][ Br2 ] ( x)( x) = ( 0.010  2x ) 2 [ HBr]2
(x) ( 0.010  2x )
1.26 x 10−3 − (2.52 x 10−1)x = x x = (1.26 x 10−3) ÷ 1.252 = 1.006 x 10−3 = 1.0 x 10−3 M Therefore, [HBr] = 0.008 M, or 0.008 mol; [H2] = 0.0010 M, or 0.0010 mol; and [Br2] = 0.0010 M, or 0.0010 mol. 14.92. The starting concentration of [IBr] = 0.010 mol ÷ 1.0 L = 0.010 M. Assemble a table of starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
Kc = 0.026 = 0.161 −
I2(g) 0 +x x
2IBr(g) 0.010 −2x 0.010 − 2x
+
Br2(g) 0 +x x
[ I 2 ][ Br2 ] ( x)( x) = ( 0.010  2x ) 2 [ IBr]2
(x) ( 0.010  2x )
1.61 x 10−3 − (3.22 x 10−1)x = x x = (1.61 x 10−3) ÷ 1.322 = 1.217 x 10−3 = 1.2 x 10−3 M Therefore, [IBr] = 0.0076 M, or 0.008 mol; [I2] = 0.00121 M, or 0.0012 mol; and [Br2] = 0.00121 M, or 0.0012 mol. 14.93. The starting concentration of COCl2 = 1.00 mol ÷ 25.00 L = 0.0400 M. Assemble a table of starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
COCl2(g) 0.0400 −x 0.0400 − x
Cl2(g) 0 +x x
+
CO(g) 0 +x x
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525
Substituting into the equilibrium expression for Kc gives Kc = 8.05 x 10−4 =
[ CO ][ Cl2 ] ( x)( x) = ( 0.0400  x ) [ COCl2 ]
Rearranging and solving for x yield 3.22 x 10−5 − (8.05 x 10−4)x − x2 = 0 x2 + (8.05 x 10−4)x − 3.22 x 10−5 = 0 (quadratic equation) Using the solution to the quadratic equation gives x =
(8.05 x 104 ) ±
(8.05 x 104 ) 2  4(1)(3.22 x 105 ) 2(1)
x = −6.09 x 10−3 (impossible; reject), or x = 5.286 x 10−3 (logical; use) Percent dissoc. = (change ÷ starting) x 100% = (0.005286 ÷ 0.0400) x 100% = 13.21 = 13.2% 14.94. The starting concentration of N2O4 = 0.0300 mol ÷ 1.00 L = 0.0300 M. Assemble a table of starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
N2O4(g) 0.0300 −x 0.0300 − x
2NO2(g) 0 +2x 2x
Substituting into the equilibrium expression for Kc gives Kc = 0.125 =
[ NO 2 ]2 ( 2 x )2 = ( 0.0300  x ) [ N2O4 ]
Rearranging and solving for x yield 4x2 + 0.125x − 3.75 x 10−3 = 0 (quadratic equation) Using the solution to the quadratic equation gives x =
(0.125) ±
(0.125) 2  4(4)(3.75 x 103 ) 2(4)
x = −5.00 x 10−2 (impossible; reject), or x = 1.875 x 10−2 M (logical) Percent dissoc. = (change ÷ starting) x 100% = (0.01875 ÷ 0.0300) x 100% = 62.50 = 62.5%
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Chapter 14: Chemical Equilibrium
14.95. Using 1.00 mol/10.00 L, or 0.100 M, and 4.00 mol/10.00 L, or 0.400 M, for the respective starting concentrations for CO and H2, assemble a table of starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
CO(g) 0.100 −x 0.100 − x
3H2(g) 0.400 −3x 0.400 −3x
+
CH2(g) + 0 +x X
H2O(g) 0 +x x
Substituting into the equilibrium expression for Kc gives Kc = 3.92 = f(x) =
[ CH 4 ][ H 2 O ] x2 = [ CO ][ H 2 ]3 ( 0.100  x ) ( 0.400  3x )3
x2 ( 0.100  x ) ( 0.400  3x )3
Because Kc is > 1 (> 50% reaction), choose x = 0.05 (about half of CO reacting), and use that for the first entry in the table of x, f(x), and interpretations. x 0.05 0.06 0.055 0.0525 0.052
f(x) 3.20 8.45 5.18 4.07 3.87
Interpretation x > 0.05 x < 0.06 x < 0.055 x < 0.0525 (but close) f(x) of 3.87 ≅ 3.92
At equilibrium, concentrations and moles are CO: 0.048 M and 0.48 mol; H2: 0.244 M and 2.44 mol; CH4: 0.052 M and 0.52 mol; and H2O: 0.052 M and 0.52 mol. 14.96. Using 1.00 mol/10.00 L, or 0.100 M, and 4.00 mol/10.00 L, or 0.400 M, for the respective starting concentrations for N2 and H2, assemble a table of starting, change, and equilibrium concentrations. N2(g) 0.100 −x 0.100 − x
Conc. (M) Starting Change Equilibrium
+
3H2(g) 0.400 −3x 0.400 − 3x
2NH3(g) 0 +2x 2x
Substituting into the equilibrium expression for Kc gives Kc = 0.153 = f(x) =
[ NH 3 ]2 ( 2 x )2 = 3 ( 0.100  x ) ( 0.400  3x )3 [ N 2 ][ H 2 ]
4 x2 ( 0.100  x ) ( 0.400  3x )3
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527
Because Kc is < 1 (< 50% reaction), choose x = 0.010 (about 10% of N2 reacting), and use that for the first entry in the table of x, f(x), and interpretations. f(x) 0.0877 0.5089 0.237 0.136 0.1651 0.150
x 0.010 0.020 0.015 0.012 0.013 0.0125
Interpretation x too small x too big x still too big x < 0.153 (but close) f(x) of .1651 ≅ 0.153 best f(x)
At equilibrium, concentrations and moles are N2: 0.087 M and 0.87 mol; H2: 0.362 M and 3.62 mol; and NH3: 0.025 M and 0.25 mol. 14.97. The dissociation is endothermic. 14.98. The value of Kc and the ratio of products to reactants decrease with temperature, so the reaction must be exothermic. 14.99. For N2 + 3H2 Kp =
2NH3, Kp is defined in terms of pressures as
PNH3 2 PN2 PH2 3
But by the ideal gas law, where [ i ] = mol/L, Pi = (niRT)/V, or Pi = [ i ]RT Substituting the righthand equality into the Kp expression gives Kp =
[ NH 3 ]2 ( RT ) 2 [ NH 3 ]2 = (RT)−2 [ N 2 ]( RT ) [ H 2 ]3 ( RT )3 [ N 2 ] [ H 2 ]3
Kp = Kc(RT)−2, or Kc = Kp (RT)2 14.100. For COCl2 Kp =
CO + Cl2, Kp is defined in terms of pressure as
PCO PCl2 PCOCl2
Substituting [ i ]RT = P for each term as in the above problem yields Kp =
[ CO ]( RT )[ Cl2 ]( RT ) [ CO ][ Cl2 ] = (RT) [ COCl2 ]( RT ) [ COCl2 ]
Kp = KcRT, or Kc = Kp/(RT)
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Chapter 14: Chemical Equilibrium
14.101. a.
The change in the number of moles of gas for the reaction is Δn = 2 − 1 = 1. Using this, calculate Kp from Kc: Kp = Kc(RT)Δn = 0.153 (0.08206 x 1123)1 = 14.099 = 14.1
b.
Use the table approach, and give the starting, change, and equilibrium pressures in atm. Press. (atm) Starting Change Equilibrium
C(s)
+
CO2(g) 1.50 −x 1.50 − x
2CO(g) 0 +2x 2x
Substituting into the equilibriumconstant expression gives Kp = 14.10 =
2 PCO PCO2
=
(2x) 2 (1.50  x)
Rearranging and solving for x give a quadratic equation. 4x2 + 14.10x − 21.15 = 0 Using the quadratic formula gives x =
14.10 ±
(14.10) 2  (4) (4) (21.15) 2 (4)
x = 1.135 (positive root) Thus, at equilibrium, the pressures of CO and CO2 are PCO2 = 1.50 − 1.135 = 0.365 = 0.37 atm
PCO = 2x = 2(1.135) = 2.270 = 2.27 atm c.
Because the reaction is endothermic, the equilibrium will shift to the left, and the pressure of CO will decrease.
14.102. a.
The change in the number of moles of gas for the reaction is Δn = 2 − 1 = 1. Using this, calculate Kp from Kc: Kp = Kc(RT)Δn = 0.238 (0.08206 x 1173)1 = 22.909 = 22.9
b.
The total pressure of the system at equilibrium is 6.40 atm. This equals the sum of the pressures of CO and CO2. Thus, the pressure of CO2 can be expressed as PCO2 = 6.40 − PCO
Substituting into the equilibriumconstant expression gives Kp = 22.91 =
2 PCO PCO2
=
2 P CO 6.40  PCO
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Rearranging and solving for x (= PCO) give a quadratic equation. x2 + 22.91x − 146.6 = 0 Using the quadratic formula gives x =
 22.91 ±
(22.91) 2  (4) (1) (146.6) 2 (1)
x = 5.213 (positive root) Thus, at equilibrium, the pressures of CO and CO2 are PCO2 = 640 − 5.213 = 1.187 = 1.19 atm
PCO = x = 5.213 = 5.21 atm c.
Carbon doesn’t appear in the equilibrium expression, so it doesn’t have any effect on the position of the equilibrium.
14.103. a.
The molar mass of PCl5 is 208.22 g/mol. Thus, the initial concentration of PCl5 is 1 mol PCl5 208.22 g PCl5 5.0 L
35.8 g PCl5 x
= 0.0344 M
Use the table approach, and give the starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
PCl3(g) 0 x x
+
Cl2(g) 0 x x
PCl5(g) 0.0344 −x 0.0344 − x
Substituting into the equilibriumconstant expression gives Kc = 4.1 =
[PCl5 ] 0.0344  x = x2 [PCl3 ][Cl2 ]
Rearranging and solving for x gives a quadratic equation. 4.1x2 + x − 0.0344 = 0 Using the quadratic formula gives x =
1 ±
(1) 2  (4) (4.1) (0.0344) 2 (4.1)
x = 0.0306 (positive root) Thus, at equilibrium, [PCl3] = [Cl2] = x = 0.031 M. The concentration of PCl5 is [PCl5] = .0344 − x = 0.0344 − 0.0306 = 0.0038 = 0.004 M.
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Chapter 14: Chemical Equilibrium
b.
The fraction of PCl5 decomposed is Fraction decomposed =
c.
0.0306 M = 0.889 = 0.89 0.0344 M
There would be a greater pressure, so, in order to minimize the increase in pressure, less PCl5 would decompose.
14.104. a.
The total pressure at equilibrium is PNH3 + PH2S = 0.660 atm
Since the pressures of NH3 and H2S are equal, this gives PNH3 = PH2S =
0.660 atm = 0.330 atm 2
The equilibriumconstant expression is Kp = PNH x PH S = (0.330)2 = 0.1089 = 0.109 3
b.
2
Since PH S = 3 PNH3 , the equilibriumconstant expression becomes 2
Kp = PNH x PH S = 3 P 2NH = 0.1089 3
2
3
0.1089 = 0.1905 = 0.191 atm 3
PNH3 =
The partial pressure of H2S is PH2S = 3(0.1905) = 0.5715 = 0.572 atm
c.
Use the table approach, and give the starting, change, and equilibrium pressures in atm. NH4HS(s)
Press. (atm) Starting Change Equilibrium
NH3(g) 0.750 −x 0.750 − x
+
H2S(g) 0.500 −x 0.500 − x
Substituting into the equilibriumconstant expression gives Kp = PNH x PH S = (0.750 − x)(0.500 − x) = 0.1089 3
2
Rearranging and solving for x give a quadratic equation. x2 − 1.250x + 0.2661 = 0 Using the quadratic formula gives x =
 (1.250) ±
(1.250) 2  (4) (1) (0.2661) 2 (1)
x = 0.2721 (negative root)
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Thus, at equilibrium, the partial pressures of NH3 and H2S are PNH3 = 0.750 − 0.2721 = 0.4779 = 0.478 atm PH2S = 0.500 − 0.2721 = 0.2279 = 0.228 atm
For NH4HS, use the ideal gas equation to convert atm (= x) to moles. n =
PV (0.2721 atm) (1.00 L) = = 0.01113 = 0.0111 mol (0.08206 L • atm / K • mol) (298 K) RT
14.105. The initial moles of SbCl5 (molar mass 299.01 g/mol) are 1 mol SbCl5 = 0.2187 mol 299.01 g SbCl5
65.4 g SbCl5 x
The initial pressure of SbCl5 is P =
n RT (0.2187 mol) (0.08206 L • atm / K • mol) (468 K) = = 1.680 atm V 5.00 L
Use the table approach, and give the starting, change, and equilibrium pressures in atm. SbCl5(g) 1.680 −x (1.680)(0.642)
Press. (atm) Starting Change Equilibrium
SbCl3(g) 0 x (1.680)(0.358)
+
Cl2(g) 0 x (1.680)(0.358)
At equilibrium, 35.8% of the SbCl5 is decomposed, so x = (1.680)(0.358) in this table. The equilibriumconstant expression is Kp =
PSbC l3 • PC l2 PSbC l5
=
(1.680 x 0.358) 2 = 0.3354 = 0.335 1.680 x 0.642
14.106. Use the table approach, and give the starting, change, and equilibrium concentrations. The volume of the system is 1.00 L. Conc. (M) Starting Change Equilibrium
SO2(g) 0.0216 −x 0.0216 − 0.0175 = 0.00410
+
1/2O2(g) 0.0148 −1/2x 0.0148 − 1/2(0.0175) = 0.00605
Substituting into the equilibriumconstant expression gives Kc =
[SO3 ] 0.0175 = = 54.87 = 54.9 (0.00410)(0.00605)1/2 [SO 2 ][O 2 ]1/2
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SO3(g) 0 +x (= 0.0175) 0.0175
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Chapter 14: Chemical Equilibrium
14.107. a.
The initial concentration of SO2Cl2 (molar mass 134.97 g/mol) is 1 mol SO 2 Cl2 134.97 g SO 2 Cl2 1.00 L
8.25 g SO 2 Cl2 x
= 0.06112 M
Use the table approach, and give the starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
SO2Cl2(g) 0.06112 −x 0.06112 − x
SO2(g) 0 x x
+
Cl2(g) 0 x x
Substituting into the equilibriumconstant expression gives Kc =
[SO 2 ][Cl2 ] x2 = = 0.045 [SO 2 Cl2 ] (0.06112  x )
Rearranging and solving for x give a quadratic equation. x2 + 0.045x − 0.0027506 = 0 Using the quadratic formula gives x =
 (0.045) ±
(0.045) 2  (4) (1) (0.0027506) 2 (1)
x = 0.03456 (positive root) The concentrations at equilibrium are [SO2] = [Cl2] = x = 0.034 M. For SO2Cl2, [SO2Cl2] = 0.06112 − x = 0.06112 − 0.03456 = 0.02655 = 0.027 M. b.
The fraction of SO2Cl2 decomposed is Fraction decomposed =
c.
0.03456 M = 0.565 = 0.57 0.06112 M
This would shift the equilibrium to the left and decrease the fraction of SO2Cl2 that has decomposed.
14.108. a.
The initial concentration of COCl2 (molar mass 98.91 g/mol) is 1 mol COCl2 98.91 g COCl 2 1.00 L
6.55 g COCl 2 x
= 0.06622 M
Use the table approach, and give the starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
COCl2(g) 0.06622 −x 0.06622 − x
CO(g) 0 x x
+
Cl2(g) 0 x x
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533
Substituting into the equilibriumconstant expression gives Kc =
[CO][Cl2 ] x2 = = 0.0046 [COCl2 ] (0.06622  x )
Rearranging and solving for x give a quadratic equation. x2 + 0.0046x − 3.046 x 10−4 = 0 Using the quadratic formula gives x =
 (0.0046) ±
(0.0046) 2  (4) (1) (3.046 x 104 ) 2 (1)
x = 0.0153 (positive root) The concentrations at equilibrium are [CO] = [Cl2] = x = 0.015 M. For COCl2, [COCl2] = 0.06622 − x = 0.06622 − 0.0153 = 0.0509 = 0.051 M. b.
The fraction of COCl2 decomposed is 0.0153 M = 0.231 = 0.23 0.06622 M
Fraction decomposed = c.
This would shift the equilibrium to the left and decrease the fraction of COCl2 that has decomposed.
14.109. a.
First, determine the initial concentration of the dimer assuming complete reaction. The reaction can be described as 2A → D. Therefore, the initial concentration of dimer is onehalf of the concentration of monomer, or 2.0 x 10−4 M. Next, allow the dimer to dissociate into the monomer in equilibrium. Use the table approach, and give the starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
D(g) 2.0 x 10−4 −x 2.0 x 10−4 − x
Substituting into the equilibriumconstant expression gives Kc =
[A]2 (2x ) 2 1 = = = 3.125 x 10−5 3.2 x 104 [D] (2.0 x 104  x )
Rearranging and solving for x give a quadratic equation. 4x2 + (3.125 x 10−5) x − (6.250 x 10−9) = 0 Using the quadratic formula gives x =
 (3.125 x 105 ) ±
(3.125 x 105 ) 2  (4) (4) (6.250 x 109 ) 2 (4)
x = 3.58 x 10−5 (positive root)
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2A(g) 0 2x 2x
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Chapter 14: Chemical Equilibrium
Thus, the concentrations at equilibrium are [CH3COOH] = 2x = 2 (3.58 x 10−5) = 7.16 x 10−5 = 7.2 x 10−5 M [Dimer] = 2.0 x 10−4 − 3.58 x 10−5 = 1.64 x 10−4 = 1.6 x 10−4 M b.
Some hydrogen bonding can occur that results in a more stable system. The proposed structure of the dimer is O CH3
O
C
C O
c.
H
CH3
O
H
An increase in the temperature would facilitate bond breaking and would decrease the amount of dimer. We could also use a Le Châteliertype argument.
14.110. a.
First, determine the initial pressure of the dimer assuming complete reaction. The reaction can be described as 2A → D. Therefore, the initial pressure of the dimer is onehalf of the pressure of monomer, or 3.75 x 10−3 atm. Next, allow the dimer to dissociate into the monomer in equilibrium. Use the table approach, and give the starting, change, and equilibrium pressures. Press. (atm) Starting Change Equilibrium
D(g) 3.75 x 10−3 −x 3.75 x 10−3 − x
2A(g) 0 2x 2x
Substituting into the equilibriumconstant expression gives Kc =
P 2A PD
=
1 (2x ) 2 = = 7.692 x 10−4 3 3 1.3 x 10 (3.75 x 10  x )
Rearranging and solving for x give a quadratic equation. 4x2 + (7.692 x 10−4) x − (2.885 x 10−6) = 0 Using the quadratic formula gives x =
 (7.692 x 104 ) ±
(7.692 x 104 ) 2  (4) (4) (2.885 x 106 ) 2 (4)
x = 7.59 x 10−4 (positive root) Thus, the concentrations at equilibrium are [CH3COOH] = 2x = 2 (7.59 x 10−4) = 1.52 x 10−3 = 1.5 x 10−3 atm [Dimer] = 3.75 x 10−3 − 7.59 x 10−4 = 2.99 x 10−3 = 3.0 x 10−3 atm
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b.
Some hydrogen bonding can occur that results in a more stable system. The proposed structure of the dimer is O CH3
O
C
C O
c.
H
H
CH3
O
A decrease in the temperature would decrease bond breaking in the dimer, so this would decrease the amount of dimer.
14.111. The molar mass of Br2 is 159.82 g/mol. Thus, the initial concentration of Br2 is 1 mol Br2 159.82 g Br2 1.00 L
18.22 g Br2 x
= 0.1140 M
Use the table approach, and give the starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
2NO(g) + 0.112 −2x 0.112 − 2x = 0.02960
Br2(g) 0.1140 −x = 0.1140 − x 0.07280
2NOBr(g) 0 2x 2x (= 0.0824 M) = 0.0824
Substituting into the equilibriumconstant expression gives Kc =
[NOB r]2 (0.0824) 2 = = 106.4 = 1.1 x 102 2 [NO] [B r2 ] (0.02960) 2 (0.07280)
14.112. The molar mass of Br2 is 159.82 g/mol. Thus, the initial moles of Br2 are 1.52 g Br2 x
1 mol B r2 = 9.511 x 10−3 mol 159.82 g B r2
The initial pressures of NO and Br2 are PNO = P B r2 =
nRT (0.0322 mol) (0.08206 L • atm / K • mol) (298 K) = = 0.7874 atm V 1.00 L nRT (9.511 x 103 mol) (0.08206 L • atm / K • mol) (298 K) = = 0.2326 atm 1.00 L V
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Chapter 14: Chemical Equilibrium
Use the table approach, and give the starting, change, and equilibrium pressures. Press. (atm) Starting Change Equilibrium
2NO(g) 0.7874 −2x 0.7874 − 2x = 0.3494
+
Br2(g) 0.2326 −x 0.2326 − x = 0.0136
2NOBr(g) 0 2x 2x (= 0.438 atm) = 0.438
Substituting into the equilibriumconstant expression gives Kp =
P 2NOB r P
2 NO
P B r2
=
(0.438) 2 = 115.5 = 1.2 x 102 (0.3494) 2 (0.0136)
14.113. Chemists have shown that oscillation reactions occur by two different mechanisms, first by one, then by the other. These mechanisms are repeated in space or time, depending on the concentrations of intermediate substances. During the reaction, an indicator changes color depending on which mechanism is active. Although the complete set of elementary steps is complicated, the overall reaction occurs just as you would expect. The initial reactants continue to decrease over time, and the final products increase as the substances come to equilibrium. 14.114. An oscillating mechanism appears to be at work in the aggregation phase of cellular slime molds. During the initial stages of aggregation the amoebas form a distinctive pattern reminiscent of that seen by chemists in an oscillating reaction. One theory of the formation of fur patterns in animals, such as leopard’s spots, assumes that an oscillating reaction occurs in the skin of the embryo, which eventually generates the fur pattern you see in the adult. A similar theory has been used to explain striped patterns on seashells.
■
SOLUTIONS TO STRATEGY PROBLEMS
14.115. First write out the equilibrium expression Kp. Then write each of the partial pressures in terms of its mole fraction, Pi = XiPtot. Kp =
PCOF2 2 PCO2 PCF4
=
(X COF2 Ptot ) 2 (X CO2 Ptot ) (X CF4 Ptot )
=
X COF2 2 X CO2 X CF4
The mole fraction of CO2 is 1 − 0.40 − 0.20 = 0.40. Substitute the mole fractions and get Kp =
(0.20) 2 = 0.250 = 0.25 (0.40)(0.40)
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537
14.116. First write out the equilibrium expression Kp. Then write each of the partial pressures in terms of its moles, Pi = (RT/V)ni. Kp =
PNO 2 PN2 PO2
=
(n NO RT/V ) 2 (n NO ) 2 = (n N2 RT/V )(n O2 RT/V ) (n N2 )(n O2 )
Now calculate Q when equal moles of N2, O2, and NO are placed in the flask. Qp =
(n) 2 = 1.0 (n)(n)
Since Qp > Kp, the reaction is not at equilibrium and will undergo reaction to the left. 14.117. Let n be the initial moles of each substance, and set up the usual table. Amt. (mol) Starting Change Equilibrium
N2(g) n +x n+x
+
O2(g) n +x n+x
Substituting into the equilibriumconstant expression gives Kp = 0.0123 =
(n  2x) 2 (n + x)(n + x)
Take the square root of each side. 0.1109 =
(n  2x) (n + x )
Rearrange and solve for x. 0.1109(n + x) = n − 2x x =
1  0.1109 0.8890 n = n = 0.42119n = 0.421n 2 + 0.1109 2.1109
The equilibriumcomposition mixture is Moles of N2 = n + 0.4211n = 1.4211n Moles of O2 = = n + 0.421n = 1.4211n Moles of NO = n − 2(0.421n) = 0.1576n Total moles = 1.421n + 1.421n + 0.1576n = 2.9998n
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2NO(g) n −2x n − 2x
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Chapter 14: Chemical Equilibrium
The equilibriumcomposition mixture is best expressed in terms of the mole fractions of the substances, since the equilibrium mixture is independent of the initial moles, as long as it is the same for all three substances. X N2 =
1.4211n = 0.4737 = 0.474 2.9998n
X O2 =
1.4211n = 0.4737 = 0.474 2.9998n
XNO =
0.1576n = 0.05253 = 0.0525 2.9998n
14.118. Reverse the first equation (at 425°C) so that it is the same as the second equation (at 458°C). The equilibrium constant for the reverse reaction is K1 = 1/1.84 = 0.5434. The equilibrium constant at the higher temperature is K2 = 49.7. Since the equilibrium constant increases with increasing temperature, the reaction shifts to the right and is endothermic. Thus, increasing temperature favors the formation of HI. Since the change in the number of moles of gaseous substances, Δn, is zero for this reaction, increasing the pressure has no effect on the equilibrium. 14.119. This reaction requires the breaking of bonds, which requires energy and is an endothermic process. As you raise the temperature, the reaction shifts toward the formation of more NO2(g), so you would expect the reaction mixture to become more redbrown. 14.120. The initial concentration of PCl3 is 0.0800 M and of Cl2 is 0.120 M. Set up the usual table. Conc. (M) Starting Change Equilibrium
PCl3(g) 0.0800 −x 0.0800 − x
+
Cl2(g) 0.120 −x 0.120 − x
PCl5(g) 0 +x x
Substituting into the equilibriumconstant expression gives Kc = 25.6 =
[ PCl5 ] x = (0.0800  x)(0.120  x) [PCl3 ][Cl2 ]
Rearrange this expression into a quadratic equation in standard form. 25.6x2 − 6.12x + 0.2457 = 0 x =
 (6.12) ±
(6.12) 2  (4) (25.6) (0.2457) 6.12 ± 3.505 = 2 (25.6) 51.2
The positive root is 0.1879, which is not valid, so use the negative root, 0.05107 M. The equilibrium composition of the mixture is [PCl3] = 0.0800 M − x = 0.0800 − 0.05107 = 0.02893 = 0.0289 M [Cl2] = 0.120 − x = 0.120 − 0.05107 = 0.06893 = 0.069 M [PCl5] = x = 0.05107 = 0.0511 M
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539
14.121. a.
Removing some NH3 from the flask would cause the reaction to shift from left to right, increasing the amount of H2S produced.
b.
Adding some NH3 to the flask would cause the reaction to shift from right to left, decreasing the amount of H2S gas produced.
c.
Since NH4HS is a solid, removing some of it would have no effect on the equilibrium, nor would it have any effect on the amount of H2S produced.
d.
Increasing the pressure in the flask by adding helium gas would have no effect on the equilibrium, nor would it have any effect on the amount of H2S produced. Adding helium has no effect on the partial pressures of the gasses in the flask.
14.122. First calculate the initial concentrations of CO and Cl2 in the flask. [Cl2] = 8.00 g Cl2 x
1 mol 1 x = 0.01128 M 70.90 g 10.00 L
[CO] = 1/2[Cl2] = 0.005641 M Now set up the table. Conc. (M) Starting Change Equilibrium
CO(g) 0.005641 −x 0.005641 − x
+
Cl2(g) 0.01128 −x 0.01128 − x
COCl2(g) 0 +x x
Substituting into the equilibriumconstant expression gives Kc = 1.23 x 103 =
[ COCl2 ] x = (0.005641  x)(0.01128  x) [CO][Cl2 ]
Rearrange this expression into a quadratic equation in standard form. x2 − 0.01773x + 6.363 x 10−5 = 0 x =
 (0.01773) ±
(0.01773) 2  (4) (1) (6.363 x 105 ) 2 (1)
=
0.01773 ± 0.007735 2
The positive root is 0.01273, which is not valid, so use the negative root, 0.004997 M. The partial pressure of phosgene (COCl2) is calculated as follows. The temperature is 395°C, which is 668 K. P = (n/V)RT = (0.004997 M)(0.08206 L•atm/K•mol)(668 K) = 0.2739 = 0.274 atm 14.123. The reaction involves breaking of bonds, which requires energy and is endothermic. This means that the equilibrium constant should be larger at higher temperatures. This allows you to assign Kp = 0.001745 at 700°C and Kp = 0.01106 at 800°C.
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Chapter 14: Chemical Equilibrium
14.124. a.
The moles are equal to the mass, m, divided by the molar mass, Mm. This gives PV = nRT = (m/Mm)RT Rearrange this expression and solve for the mass of the gas. PVM m RT
m = b.
The mass of gas A and the mass of gas B are mA =
PAVM mA P VM mB = B mB RT RT
The total mass of the gas is m = mA + mB =
PAVM mA P VM V(PA M mA + PB M mB ) + B mB = RT RT RT
The density of the gas is d = c.
(P M + PB M mB ) m = A mA V RT
If A is NO2 and B is N2O4, then MmB = 2MmA. Also, PB = P − PA. Plug these relations into the result above to obtain dRT = PAMmA + (P − PA)(2MmA) = 2PMmA − PAMmA This equation can be solved for PA. PA = 2P −
d.
dRT M mA
The partial pressures of NO2 and N2O4 are PNO2 = 2(1.00 atm) −
(2.86 g/L)(0.082058 L • atm/K • mol)(308.15 K) 46.01 g/mol
= 0.4282 = 0.428 atm PN2 O4 = P − PNO2 = 1.00 atm − 0.4282 atm = 0.5718 = 0.572 atm
The equilibrium constant is Kp =
PN2 O4 PNO2 2
=
0.5718 = 3.118 = 3.12 (0.4282) 2
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SOLUTIONS TO CUMULATIVESKILLS PROBLEMS 2Sb(s) + 3H2S(g) [+3Pb2+ → 3PbS(s) + 6H+]:
14.125. For Sb2S3(s) + 3H2(g)
Starting M of H2(g) = 0.0100 mol ÷ 2.50 L = 0.00400 M H2 1.029 g PbS ÷ 239.26 g PbS/mol H2S = 4.3007 x 10−3 mol H2S 4.3007 x 10−3 mol H2S ÷ 2.50 L = [1.7203 x 10−3] = M of H2S Conc. (M) Starting Change Equilibrium
3H2(g) 0.00400 −0.0017203 0.0022797
+
Sb2S3(s)
3H2S(g) 0 +0.0017203 0.0017203
+ 2Sb(s)
Substituting into the equilibrium expression for Kc gives Kc =
[ H 2S]3 ( 0.0017203)3 = 0.4297 = 0.430 = [ H 2 ]3 ( 0.0022797 )3
LaOCl(s) + 2HCl(g) [+2Ag+ → 2AgCl(s) + 2H+]:
14.126. For LaCl3 + H2O
Starting M of H2O(g) = 0.0250 mol ÷ 1.25 L = 0.0200 M H2O 3.59 g AgCl ÷ 143.32 g AgCl/mol AgCl = 0.025048 mol AgCl 0.025048 mol AgCl ÷ 1.25 L = 0.020039 M Conc. (M) Starting Change Equilibrium Kc =
H2O(g) 0.0200 −0.0100195 0.0099805
+
LaCl3(s)
2HCl(g) 0 +0.020039 0.020039
[ HCl ]2 ( 0.020039 ) 2 = = 4.023 x 10−2 = 4.0 x 10−2 [ H2O ] ( 0.0099805)
14.127. for PCl5(g)
PCl3(g) + Cl2(g):
Starting M of PCl5 = 0.0100 mol ÷ 2.00 L = 0.00500 M Conc. (M) Starting Change Equilibrium
PCl5(g) 0.00500 −x 0.00500 − x
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PCl3(g) 0 +x x
+
Cl2(g) 0 +x x
+ LaOCl(s)
541
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Chapter 14: Chemical Equilibrium
Substituting into the equilibrium expression for Kc gives Kc = 4.15 x 10−2 =
[ PCl3 ][ Cl2 ] ( x)( x) = [ PCl5 ] ( 0.00500  x )
x2 + (4.15 x 10−2)x − 2.075 x 10−4 = 0 (quadratic) Solving the quadratic equation gives x = −4.60 x 10−2 (impossible) and x = 4.51 x 10−3 M (use). Total M of gas = 0.00451 + 0.00451 + 0.00049 = 0.009510 M P = (n/V)RT = (0.009510 M)[0.082057 L•atm/(K•mol)](523 K) = 0.4081 = 0.408 atm 14.128. For SbCl5(g)
SbCl3(g) + Cl2(g):
Starting M of SbCl5 = 0.0125 mol ÷ 3.50 L = 0.003571 M Conc. (M) Starting Change Equilibrium
SbCl5(g) 0.003571 −x 0.003571 − x
SbCl3(g) 0 +x x
+
Cl2(g) 0 +x x
Substituting into the equilibrium expression for Kc gives Kc = 2.50 x 10−2 =
[SbCl3 ][ Cl2 ] ( x)( x) = [SbCl5 ] ( 0.003751  x )
x2 + (2.50 x 10−2)x − 8.927 x 10−5 = 0 (quadratic) Solving the quadratic equation gives x = −2.817 x 10−2 (impossible) and x = 3.17 x 10−3 M (use). Total M of gas = 0.00317 + 0.00317 + 0.00040 = 0.006740 M P = (n/V)RT = (0.00674 M)[0.082057 L•atm/(K•mol)](521 K) = 0.2881 = 0.288 atm
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CHAPTER 15
Acids and Bases
■
SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 15.1. See labels below reaction:
H2CO3(aq)
+
acid
CN(aq)
HCN(aq)
base
+ HCO3 (aq)
acid
base
H2CO3 is the proton donor (BrønstedLowry acid) on the left, and HCN is the proton donor (BrønstedLowry acid) on the right. The CN− and HCO3− ions are proton acceptors (BrønstedLowry bases). HCN is the conjugate acid of CN−. 15.2. Part a involves molecules with all single bonds; part b does not, so bonds are drawn in.
F F
CH3 +
B
O H
F
CH3
B
O H
F
F Lewis acid
a.
F
Lewis base 2
O
O O
2
+
C O
b.
Lewis acid
Lewis base
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O
C O
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Chapter 15: Acids and Bases
15.3. The HC2H3O2 is a stronger acid than H2S, and HS− is a stronger base than the C2H3O2− ion. The equilibrium favors the weaker acid and weaker base; therefore, the reactants are favored. 15.4. a.
PH3
b.
HI
c.
H2SO3
d.
H3AsO4−
e.
HSO4−
15.5. A 0.125 M solution of Ba(OH)2, a strong base, ionizes completely to yield 0.125 M Ba2+ ion and 2 x 0.125 M, or 0.250 M, OH− ion. Use the Kw equation to calculate the [H3O+]. [H3O+] =
Kw 1.0 x 1014 = 4.00 x 10−14 = 4.0 x 10−14 M = (0.250) [OH ]
15.6. Use the Kw equation to calculate the [H3O+]. [H3O+] =
Kw 1.0 x 1014 = = 1.00 x 10−9 = 1.0 x 10−9 M 5 1.0 x 10 [OH ]
Since the [H3O+] concentration is less than 1.0 x 10−7, the solution is basic. 15.7. Calculate the negative log of the [H3O+]: pH = −log [H3O+] = −log (0.045) = 1.346 = 1.35 15.8. Calculate the pOH of 0.025 M OH−, and then subtract from 14.00 to find pH: pOH = −log [OH−] = −log (0.025) = 1.602 pH = 14.00 − 1.602 = 12.397 = 12.40 15.9. Because pH = 3.16, by definition log [H3O+] = −3.16. Enter this on the calculator and convert to the antilog (number) of −3.16. [H3O+] = antilog (−3.16) = 10−3.16 = 6.91 x 10−4 = 6.9 x 10−4 M 15.10. Find the pOH by subtracting the pOH from 14.00. Then enter −3.40 on the calculator to convert to the antilog (number) corresponding to −3.40. pOH = 14.00 − 10.6 = 3.40 [H3O+] = antilog (−3.40) = 10−3.40 = 3.98 x 10−4 = 4 x 10−4 M
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545
ANSWERS TO CONCEPT CHECKS
15.1. In any aqueous solution, you should consider the autoionization of water. And because we have a solution of a weak acid in water, you should also consider the equilibrium between this acid and water. Here are the two equilibria: H2O(l) + H2O(l)
H3O+(aq) + OH−(aq)
HCHO2(aq) + H2O(l)
CHO2−(aq) + H3O+(aq)
The species present in these equilibria are H2O(l), H3O+(aq), OH−(aq), HCHO2(aq), and CHO2−(aq). 15.2. The stronger acid gives up its proton more readily, and therefore, its conjugate base ion holds on to a proton less strongly. In other words, the stronger acid has the weaker conjugate base. Because formic acid is the stronger acid, the formate ion is the weaker base. Acetate ion is the stronger base. 15.3. Look at each solution, and determine whether it is acidic, basic, or neutral. In solution A, the numbers of H3O+ and OH− ions are equal, so the solution is neutral. For solution B, the number of H3O+ ions is greater than the number of OH− ions, so the solution is acidic. In solution C, the number of H3O+ ions is less than the number of OH− ions, so the solution is basic. Therefore, the ranking from most acidic to least acidic (most basic) is B > A > C. 15.4. In order to answer this problem qualitatively, it is essential that all the solutions have the same solute concentrations. Bases produce solutions of pH greater than 7, whereas acids produce solutions of pH less than 7. NH3 and NaOH are bases, and HCl and HC2H3O2 are acids. NaOH is a stronger base than NH3, so the NaOH solution would have the highest pH, followed by the NH3 solution. HC2H3O2 is a much weaker acid than HCl, so the HC2H3O2 solution would have a higher pH than the HCl solution. Therefore, the ranking from highest to lowest pH for solutions with the same solute concentrations is NaOH > NH3 > HC2H3O2 > HCl.
■
ANSWERS TO SELFASSESSMENT AND REVIEW QUESTIONS
15.1. You can classify these acids using the information in Section 15.1. Also recall that all diatomic acids of Group VIIA halides are strong except for HF. a.
Weak
b.
Weak
c.
Strong
d.
Strong
e.
Weak
f.
Weak
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Chapter 15: Acids and Bases
15.2. In Section 15.1, we are told that all neutralizations involving strong acids and bases evolve 55.90 kJ of heat per mole of H3O+. Thus, the thermochemical evidence for the Arrhenius concept is based on the fact that when 1 mol of any strong acid (1 mol H3O+) is neutralized by 1 mole of any strong base (1 mol OH−), the heat of neutralization is always the same (ΔH° = −55.90 kJ/mol). 15.3. A BrønstedLowry acid is a molecule or ion that donates an H+ ion (proton donor) to a base in a protontransfer reaction. A BrønstedLowry base is a molecule or ion that accepts an H+ ion (proton acceptor) from an acid in a protontransfer reaction. An example of an acidbase equation:
HF(aq) acid
+ NH3(aq)
→
base
NH4+(aq)
F (aq)
+
acid
base
15.4. The conjugate acid of a base is a species that differs from the base by only one H+. Consider the base, HSO3−. Its conjugate acid would be H2SO3 but not H2SO4. H2SO4 differs from HSO3− by one H and one O. 15.5. You can write the equations by considering that H2PO3− is both a BrønstedLowry acid and a BrønstedLowry base. The H2PO3− acts as a BrønstedLowry acid when it reacts with a base such as OH− : H2PO3−(aq) + OH−(aq) → HPO32−(aq) + H2O(l) The H2PO3− acts as a BrønstedLowry base when it reacts with an acid such as HCl: H2PO3−(aq) + HCl(aq) → H3PO3(aq) + Cl−(aq) 15.6. The BrønstedLowry concept enlarges on the Arrhenius concept in the following ways: (1) It expands the concept of a base to include any species that accepts protons, not just the OH− ion or compounds containing the OH− ion. (2) It enlarges the concepts of acids and bases to include ions as well as molecules. (3) It enables us to write acidbase reactions in nonaqueous solutions as well as in aqueous solutions, whereas the Arrhenius concept applies only to aqueous solutions. (4) It allows some species to be considered as acids or bases, depending on the other reactant with which they are mixed. 15.7. According to the Lewis concept, an acid is an electronpair acceptor and a base is an electronpair donor. An example is
Ag+(aq) + 2(:NH3) acid
→
Ag(NH3)2+(aq)
base
15.8. Recall that the weaker the acid, the stronger it holds on to its proton(s). Thus, if a reaction mixture consists of a stronger acid and base and a weaker acid and base, the weakeracid side will always be favored because the proton(s) will bond more strongly to the weaker acid.
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547
15.9. The two factors that determine the strength of an acid are (1) the polarity of the bond to which the H atom is attached, and (2) the strength of the bond, or how tightly the proton is held by the atom to which it is bonded. An increase in the polarity of the bond makes it easier to remove the proton, increasing the strength of the acid. An increase in the strength of the bond makes it more difficult to remove the proton, decreasing the strength of the acid. The strength of the bond depends in turn on the size of the atom, so larger atoms have weaker bonds, whereas smaller atoms have stronger bonds. 15.10. The selfionization of water is the reaction of two water molecules in which a proton is transferred from one molecule to the other to form H3O+ and OH− ions. At 25°C, the Kw expression is Kw = [H3O+][OH−] = 1.0 x 10−14. 15.11. The pH = −log [H3O+] of an aqueous solution. Measure pH by using electrodes and a pH meter or by interpolating the pH from the color changes of a series of acidbase indicators. 15.12. A solution of pH 4 has a [H3O+] = 1 x 10−4 M and is more acidic than a solution of pH 5, which has a [H3O+] = 1 x 10−5 M. 15.13. For a neutral solution, [H3O+] = [OH−]; thus, the [H3O+] of a neutral solution at 37°C is the square root of Kw at 37°C: [H3O+] =
2.5 x 1014 = 1.58 x 10−7 M
pH = −log (1.58 x 10−7) = 6.801 = 6.80 15.14. Because pH + pOH = pKw at any temperature, pH + pOH = −log (2.5 x 10−14) = 13.60 15.15. The answer is a, F−. 15.16. The answer is d, Fe3+. 15.17. The answer is d, 3.8 x 10−13 M. 15.18. The answer is b, 3.5 x 10−5 M.
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Chapter 15: Acids and Bases
■
ANSWERS TO CONCEPT EXPLORATIONS
15.19. a.
HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq)
b.
The pH of the solution is around 3. Since the volume of the solution is 1.0 L, the molarity and the number of moles are numerically equal, 0.0010 M. The first nonzero digit is three places to the right of the decimal, and this is the approximate pH.
c.
The H3O+ concentration and pH of the solution are 1 mol H 3O + 0.00100 mol HCl = 1.00 x 10−3 M x 1.0 L 1 mol HCl pH = −log(1.00 x 10−3) = 3.000 = 3.00
d.
Yes, there are OH− ions present in every solution that uses water as a solvent. The concentration of OH− can be determined using Kw. The OH− comes from the autoionization of water.
e.
If OH− were added, the H3O+ ion concentration would decrease. Some of the added OH− would react with the HCl present to form water, leaving less HCl remaining in the solution. Since the product [H3O+][OH−] is constant, as the OH− ion concentration increases, the H3O+ ion concentration must decrease.
f.
The pH of a sample removed from the original solution will have the same pH as the original solution. The pH does not depend on the volume of the solution being tested.
g.
The two different volumes have the same concentration of HCl, and therefore the same pH. They contain different numbers of moles because they contain different volumes. Since moles = M x V, different volumes of solution with the same concentration contain different moles of solute.
h.
If 1.0 L of pure water were added to the solution, the volume would double, and the solution would decrease in molarity by a factor of onehalf. Since the concentration of the solution is decreasing, the solution is less acidic and the pH is higher.
15.20. a.
The concentration of the two solutions is 0.10 M. The reactions are HA(aq) + H2O(l) → H3O+(aq) + A−(aq) HB(aq) + H2O(l) → H3O+(aq) + B−(aq)
b.
You could measure the pH of each solution. The pH of the strong acid solution will be lower than the pH of the weak acid solution.
c.
The strong acid will have a pH of 1.00. Since the HA solution has a pH of 3.7, it is the weaker of the two acids.
d.
[A−] = [H3O+] = 10−pH = 10−3.7 = 1.99 x 10−4 = 2.0 x 10−4 M
e.
If HB is a strong acid, then [H3O+] = [B−] = 0.10 M.
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549
f.
The greatest concentration should be for the H3O+. Since HB is an acid, [H3O+] > [OH−]. Also, since HB is a strong acid, the concentration of HB should be zero. The H3O+ and B− ions are formed from the acid in equal numbers, but since there is some H3O+ present in the water, it should have a slightly higher concentration than B−.
g.
Since HA is a weak acid, there should be more HA than either H3O+ or A− ions. Also, since HA is an acid, [H3O+] > [OH−].
h.
Adding 1.0 L of water to the HB solution would decrease the concentration of the acid by a factor of onehalf, thereby raising the pH of the solution. As long as the concentration changes, the pH will also change. As the concentration of the acid goes down, the pH goes up.
i.
A 200mL sample of the solution will have the same pH as the original solution. The pH depends on the concentration, not on the volume of the sample tested.
ANSWERS TO CONCEPTUAL PROBLEMS
15.21. It is not necessary to have the species NH4OH in order to have OH− in the solution. When ammonia reacts with water, hydroxide ion forms in the reaction. NH3(aq) + H2O(l)
NH4+(aq)
+ OH−(aq)
15.22. A reaction where HPO42− acts as an acid is HPO42−(aq) + OH−(aq)
PO43−(aq) + H2O(l)
A reaction where HPO42− acts as a base is HPO42−(aq) + H3O+(aq)
H2PO4−(aq) + H2O(l)
15.23. The hydroxide ion acts as a base and donates a pair of electrons on the O atom, forming a bond with CO2 to give HCO3−. 15.24. Nitrogen has a greater electronegativity than carbon. You would expect the H—O bond in the H—O—N group to be more polar (with the H atom having a positive partial charge) than the H—O bond in the H—O—C group. Thus, based on their structure, you would expect HNO2 to be the stronger acid. 15.25. When you lower the temperature of pure water, the value of Kw decreases. In pure water, the hydroniumion concentration equals the hydroxideion concentration, so Kw = [H3O+]2. When Kw decreases, the hydroniumion concentration decreases, and the corresponding pH increases. 15.26. Sodium hydroxide is a strong base, whereas ammonia is weak. As a strong base, NaOH exists in solution completely as ions, whereas NH3 exists in solution as an equilibrium in which only part of the NH3 has reacted to produce ions. Thus, a sodium hydroxide solution has a greater OH− concentration than the same concentration solution of NH3. At the same concentrations, the pH of the NaOH solution is greater (more basic) than that of the NH3 solution.
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Chapter 15: Acids and Bases
15.27. A strong monoprotic acid will dissolve completely in water to form hydronium ions (H3O+) and anions in equal number. None of the original monoprotic acid molecules will remain. This is represented by the picture in the middle. The picture on the left represents an undissociated acid. No ions are present in the solution. The picture on the right represents a weak acid where only some of the acid molecules have dissociated to form ions. 15.28. A weak acid will dissolve to form a solution that has some hydronium ions present, some anions present, and some undissociated acid molecules present. This is represented by beaker A. The solution in beaker B is a weak base, and the solution in beaker C is a strong acid.
■
SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 15.29. The reaction with labels of “acid” and “base” written below is as follows:
OH(aq) + HF(aq) base acid
F(aq) + H2O(l) base
acid
15.30. The reaction with labels of “acid” and “base” written below is as follows:
HOCl(aq) + acid
H2O(l) base
H3O+(aq) + OCl(aq) acid
base
15.31. a.
PO43−
b.
HS−
c.
NO2−
d.
HAsO42−
15.32. a.
SeO42−
b.
PH3
c.
S2−
d.
OCl−
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15.33. a.
HClO
b.
AsH4+
c.
H3PO4
d.
HTeO3−
15.34. a.
H2Se
b.
NH3
c.
HClO2
d.
N2H5+
15.35. Each equation is given below with the labels for acid and base:
HSO4(aq) a.
+
NH3 (aq)
acid
SO42(aq)
base
+ NH4+(aq) base
acid
The conjugate acidbase pairs are HSO4−, SO42−, and NH4+, NH3.
HPO42(aq) b.
H2PO4(aq) + NH3(aq)
+ NH4+(aq)
base
acid
base
acid
The conjugate acidbase pairs are H2PO4−, HPO42−, and NH4+, NH3.
Al(H2O)63+(aq) c.
+ H2O(l)
acid
Al(H2O)5(OH)2+(aq)
+ H3O+(aq)
acid
base
base
The conjugate acidbase pairs are Al(H2O)63+, Al(H2O)5(OH)2+, and H3O+, H2O.
d.
SO32(aq) base
+
NH4+(aq) acid
HSO3(aq) base
The conjugate acidbase pairs are HSO3−, SO32−, and NH4+, NH3. 15.36. Each equation is given below with the labels for acid and base:
HPO42(aq) + HCO3(aq) a.
acid
base
PO43(aq) + H2CO3(aq) base
acid
The conjugate acidbase pairs are HPO42−, PO43−, and H2CO3, HCO3−.
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+
NH3(aq) acid
551
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Chapter 15: Acids and Bases
b.
F(aq) + HSO4(aq)
HF(aq) + SO42(aq)
base
acid
acid
base
The conjugate acidbase pairs are HF, F−, and HSO4−, SO42−.
c.
HSO4(aq) + H2O(I )
SO42(aq) + H3 O+ (aq)
acid
base
base
acid
The conjugate acidbase pairs are HSO4−, SO42−, and H3O+, H2O.
d.
HS(aq) + CN(aq)
S2(aq) + HCN(aq)
acid
base
base
acid
The conjugate acidbase pairs are HS−, S2−, and HCN, CN−. 15.37. The reaction is
F
F +
F As
B
F
F
F Lewis base
F
F
F As
B
F
F
F
Lewis acid
15.38. The reaction is
Cl Cl Al
H +
N H
Cl
H
Lewis base
Lewis acid
Cl H Cl Al N H Cl H
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553
15.39. a.
The completed equation is AlCl3 + Cl−
AlCl4−
The Lewis formula representation is Cl
Cl Al
+
[
Cl
Cl
Al
Cl
[
Cl
Cl
Cl
AlCl3 is the electronpair acceptor and is the acid. Cl− is the electronpair donor and is the base. b.
The completed equation is I− + I2
I3−
The Lewis formula representation is
[
I
+
I
I
[
I
I
I
[
[
I2 is the electronpair acceptor and is the acid. The I− ion is the electronpair donor and is the base. 15.40. a.
The completed equation is GaBr3 + Br−
GaBr4−
The Lewis formula representation is Br
Br Ga
+
[
Br
Br
[
Br
Br
Ga
Br
Br
GaBr3 is the electronpair acceptor and is the acid. The Br− ion is the electronpair donor and is the base.
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Chapter 15: Acids and Bases
b.
The completed equation is BF3 + F−
BF4−
The Lewis formula representation is F
F
F
F
+
[
F
[
B
F
B
F
F
BF3 is the electronpair acceptor and is the acid. The F− ion is the electronpair donor and is the base. 15.41. a.
Each water molecule donates a pair of electrons to copper(II), making the water molecule a Lewis base and the Cu2+ ion a Lewis acid.
b.
The AsH3 donates a pair of electrons to the boron atom in BBr3, making AsH3 a Lewis base and the BBr3 molecule a Lewis acid.
15.42. a.
Each F− ion donates a pair of electrons to Be in BeF2, making F− a Lewis base and BeF2 a Lewis acid.
b.
Each Cl− ion donates a pair of electrons to SnCl4, making Cl− ion a Lewis base and SnCl4 a Lewis acid.
15.43. The equation is H2S + HOCH2CH2NH2 → HOCH2CH2NH3+ + HS−. The H2S is a Lewis acid, and HOCH2CH2NH2 is a Lewis base. The hydrogen ion from H2S accepts a pair of electrons from the N atom in HOCH2CH2NH2. 15.44. The equation is CaO + SO2 → CaSO3. The CaO is a Lewis base, and SO2 is a Lewis acid. The oxide ion from CaO donates a pair of electrons to the sulfur of SO2, forming the SO32− ion in CaSO3. 15.45. The reaction is HSO4− + ClO− → HClO + SO42−. According to Table 15.2, HClO is a weaker acid than HSO4−. Because the equilibrium for this type of reaction favors formation of the weaker acid (or weaker base), the reaction occurs to a significant extent. 15.46. The reaction is HCN + SO42− → CN− + HSO4−. According to Table 15.2, HCN is a weaker acid than HSO4−. Because the equilibrium for this type of reaction favors formation of the weaker acid (or weaker base), the reaction occurs in the opposite direction.
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15.47. a.
NH4+ is a weaker acid than H3PO4, so the lefthand species are favored at equilibrium.
b.
HCN is a weaker acid than H2S, so the lefthand species are favored at equilibrium.
c.
H2O is a weaker acid than HCO3−, so the righthand species are favored at equilibrium.
d.
H2O is a weaker acid than Al(H2O)63+, so the righthand species are favored at equilibrium.
15.48. a.
H2CO3 is a stronger acid than NH4+, so the lefthand species are favored at equilibrium.
b.
H2S is a weaker acid than H2CO3, so the lefthand species are favored at equilibrium.
c.
H2O is a weaker acid than HCN, so the lefthand species are favored at equilibrium.
d.
HCN is a stronger acid than HCO3−, so the lefthand species are favored at equilibrium.
15.49. Trichloroacetic acid is the stronger acid because, in general, the equilibrium favors the formation of the weaker acid, which is formic acid in this case. 15.50. The BF4− ion is the weaker base because the equilibrium favors the formation of the weaker base. 15.51. a.
H2S is stronger because acid strength decreases with increasing anion charge for polyprotic acid species.
b.
H2SO3 is stronger because, for a series of oxoacids, acid strength increases with increasing electronegativity.
c.
HBr is stronger because Br is more electronegative than Se. Within a period, acid strength increases as electronegativity increases.
d.
HIO4 is stronger because acid strength increases with the number of oxygen atoms bonded to the central atom.
e.
H2S is stronger because, within a group, acid strength increases with the increasing size of the central atom in binary acids.
15.52. a.
HBrO3 < HBrO4: Acid strength increases with the number of oxygen atoms bonded to the central atom.
b.
HCO3− < H2CO3: Acid strength decreases with increasing anion charge for polyprotic acid species.
c.
H2S < H2Te: Within a group, acid strength increases with increasing size of the central atom for binary acids.
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Chapter 15: Acids and Bases
d.
H2Se < HBr: Within a period, acid strength increases with increasing electronegativity of the central atom for binary acids.
e.
H3AsO4 < H3PO4: For a series of oxoacids, acid strength increases with increasing electronegativity.
15.53. a.
[H3O+] = 1.2 M [OH−] =
b.
[OH−] = 0.32 M [H3O+] =
c.
Kw 1.0 x 1014 = = 3.12 x 10−14 = 3.1 x 10−14 M 0.32 [OH ]
[OH−] = 2 x (0.085 M) = 0.170 M [H3O+] =
d.
Kw 1.0 x 1014 = = 8.33 x 10−15 = 8.3 x 10−15 M 1.2 [H 3O + ]
Kw 1.0 x 1014 = = 5.88 x 10−14 = 5.9 x 10−14 M 0.170 [OH ]
[H3O+] = 0.38 M [OH−] =
Kw 1.0 x 1014 = = 2.63 x 10−14 = 2.6 x 10−14 M [H 3O + ] 0.38
15.54. a.
[OH−] = 1.65 M [H3O+] =
b.
[OH−] = 2 x (0.35 M) = 0.70 M [H3O+] =
c.
Kw 1.0 x 1014 = = 1.42 x 10−14 = 1.4 x 10−14 M 0.70 [OH  ]
[H3O+] = 0.045 M [OH−] =
d.
Kw 1.00 x 1014 = = 6.060 x 10−15 = 6.06 x 10−15 M 1.65 [OH ]
Kw 1.0 x 1014 = = 2.22 x 10−13 = 2.2 x 10−13 M 0.045 [H 3O + ]
[H3O+] = 0.58 M [OH−] =
Kw 1.00 x 1014 = = 1.72 x 10−14 = 1.7 x 10−14 M + 0.58 [H 3O ]
15.55. The [H3O+] = 0.050 M (HCl is a strong acid); using Kw, the [OH−] = 2.0 x 10−13 M.
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15.56. The [H3O+] = 0.020 M (HNO3 is a strong acid); using Kw, the [OH−] = 5.0 x 10−13 M. 15.57. Because the Ba(OH)2 forms two OH− per formula unit, the [OH−] = 2 x 0.0085 = 0.017 M. [H3O+] =
Kw 1.0 x 1014 = = 5.88 x 10−13 = 5.9 x 10−13 M 0.017 [OH ]
15.58. Because Mg(OH)2 forms two OH− per formula unit, the [OH−] = 2 x 3.2 x 10−4 = 6.4 x 10−4 M. [H3O+] = Kw ÷ [OH−] = (1.0 x 10−14) ÷ (6.4 x 10−4) = 1.56 x 10−11 = 1.6 x 10−11 M 15.59 a.
5 x 10−6 M H3O+ > 1.0 x 10−7, so the solution is acidic.
b.
Use Kw to determine [H3O+]. [H3O+] =
Kw 1.00 x 1014 = = 2.0 x 10−6 = 2 x 10−6 M 5 x 109 [OH  ]
Since 2 x 10−6 M > 1.0 x 10−7, the solution is acidic. c.
When [OH−] = 1.0 x 10−7 M, [H3O+] = 1.0 x 10−7 M, and the solution is neutral.
d.
2 x 10−9 M H3O+ < 1.0 x 10−7, so the solution is basic.
15.60. a.
Use Kw to determine [H3O+]. [H3O+] =
Kw 1.00 x 1014 = = 5.0 x 10−4 = 5 x 10−4 M 2 x 1011 [OH  ]
Since 5 x 10−4 M > 1.0 x 10−7, the solution is acidic. b.
2 x 10−9 M < 1.0 x 10−7, so the solution is basic.
c.
Use Kw to determine [H3O+]. [H3O+] =
Kw 1.00 x 1014 = = 1.6 x 10−10 = 2 x 10−10 M 6 x 105 [OH  ]
Since 2 x 10−10 M < 1.0 x 10−7, the solution is basic. d.
6 x 10−3 M > 1.0 x 10−7 M, so the solution is acidic.
15.61. The [H3O+] calculated below is > 1.0 x 10−7 M, so the solution is acidic. [H3O+] =
Kw 1.00 x 1014 = = 6.66 x 10−6 = 6.7 x 10−6 M 9 1.5 x 10 [OH ]
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Chapter 15: Acids and Bases
15.62. The [H3O+] calculated below is < 1.0 x 10−7 M, so the solution is basic. [H3O+] =
Kw 1.00 x 1014 = = 1.19 x 10−10 = 1.2 x 10−10 M 5 8.4 x 10 [OH ]
15.63. a.
pH 4.6, acidic solution
b.
pH 7.0, neutral solution
c.
pH 1.6, acidic solution
d.
pH 10.5, basic solution
15.64. a.
pH 5.6, acidic solution
b.
pH 2.5, acidic solution
c.
pH 13.2, basic solution
d.
pH 8.3, basic solution
15.65. a.
Acidic (3.5 < 7.0)
b.
Neutral (7.0 = 7.0)
c.
Basic (9.0 > 7.0)
d.
Acidic (5.5 < 7.0)
15.66. a.
Basic (12.0 > 7.0)
b.
Neutral (7.0 = 7.0)
c.
Acidic (4.0 < 7.0)
d.
Acidic (5.7 < 7.0)
15.67. Record the same number of places after the decimal point in the pH as the number of significant figures in the [H3O+]. a.
−log (1.0 x 10−8) = 8.000 = 8.00
b.
−log (5.0 x 10−12) = 11.301 = 11.30
c.
−log (7.5 x 10−3) = 2.124 = 2.12
d.
−log (6.35 x 10−9) = 8.1972 = 8.197
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559
15.68. Record the same number of places after the decimal point in the pH as the number of significant figures in the [H3O+]. a.
−log (2.5 x 10−4) = 3.602 = 3.60
b.
−log (5.7 x 10−10) = 9.244 = 9.24
c.
−log (4.6 x 10−5) = 4.337 = 4.34
d.
−log (2.91 x 10−11) = 10.5361 = 10.536
15.69. Record the same number of places after the decimal point in the pH as the number of significant figures in the [H3O+]. −log (7.5 x 10−3) = 2.1249 = 2.12 15.70. Record the same number of places after the decimal point in the pH as the number of significant figures in the [H3O+]. −log (5.0 x 10−3) = 2.301 = 2.30 15.71. a.
pOH = −log (5.25 x 10−9) = 8.2798; pH = 14.00 − 8.2798 = 5.7202 = 5.72
b.
pOH = −log (8.3 x 10−3) = 2.0809; pH = 14.00 − 2.0809 = 11.91908 = 11.92
c.
pOH = −log (3.6 x 10−12) = 11.4436; pH = 14.00 − 11.4436 = 2.5563 = 2.56
d.
pOH = −log (2.1 x 10−8) = 7.6777; pH = 14.00 − 7.6777 = 6.322 = 6.32
15.72. a.
pOH = −log (6.74 x 10−11) = 10.1713; pH = 14.00 − 10.1713 = 3.828 = 3.83
b.
pOH = −log (5.8 x 10−5) = 4.236; pH = 14.00 − 4.236 = 9.763 = 9.76
c.
pOH = −log (3.4 x 10−10) = 9.468; pH = 14.00 − 9.468 = 4.531 = 4.53
d.
pOH = −log (7.1 x 10−4) = 3.148; pH = 14.00 − 3.148 = 10.851 = 10.85
15.73. First, convert the [OH−] to [H3O+] using the Kw equation. Then find the pH, recording the same number of places after the decimal point in the pH as the number of significant figures in the [H3O+]. [H3O+] = Kw ÷ [OH−] = (1.0 x 10−14) ÷ (0.0040) = 2.50 x 10−12 M pH = −log (2.50 x 10−12) = 11.602 = 11.60 15.74. First, convert the [OH−] to [H3O+] using the Kw equation. Then find the pH, recording the same number of places after the decimal point in the pH as the number of significant figures in the [H3O+]. [H3O+] = Kw ÷ [OH−] = (1.0 x 10−14) ÷ (0.050) = 2.00 x 10−13 M pH = −log (2.00 x 10−13) = 12.698 = 12.70
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Chapter 15: Acids and Bases
15.75. From the definition, pH = −log [H3O+] and −pH = log [H3O+], so enter the negative value of the pH on the calculator, and use the inverse and log keys (or 10x) key to find the antilog of –pH. log [H3O+] = −pH = −5.12 [H3O+] = antilog (−5.12) = 10−5.12 = 7.58 x 10−6 = 7.6 x 10−6 M 15.76. From the definition, pH = −log [H3O+] and −pH = log [H3O+], so enter the negative value of the pH on the calculator, and use the inverse and log keys (or 10x) key to find the antilog of –pH. log [H3O+] = −pH = −3.85 [H3O+] = antilog (−3.85) = 10−3.85 = 1.41 x 10−4 = 1.4 x 10−4 M 15.77. From the definition, pH = −log [H3O+] and −pH = log [H3O+], so enter the negative value of the pH on the calculator, and use the inverse and log keys (or 10x) key to find the antilog of −pH. Then use the Kw equation to calculate [OH−] from [H3O+]. log [H3O+] = −pH = −11.63 [H3O+] = antilog (−11.63) = 10−11.63 = 2.34 x 10−12 M [OH−] = Kw ÷ [H3O+] = (1.0 x 10−14) ÷ (2.34 x 10−12) = 4.27 x 10−3 = 4.3 x 10−3 M 15.78. From the definition, pH = −log [H3O+] and −pH = log [H3O+], so enter the negative value of the pH on the calculator, and use the inverse and log keys (or 10x) key to find the antilog of −pH. Then use the Kw equation to calculate [OH−] from [H3O+]. log [H3O+] = −pH = −9.61 [H3O+] = antilog (−9.61) = 10−9.61 = 2.45 x 10−10 M [OH−] = Kw ÷ [H3O+] = (1.0 x 10−14) ÷ (2.45 x 10−10) = 4.08 x 10−5 = 4.1 x 10−5 M 15.79. First, calculate the molarity of the OH− ion from the mass of NaOH. Next, convert the [OH−] to [H3O+] using the Kw equation. Then find the pH, recording the same number of places after the decimal point in the pH as the number of significant figures in the [H3O+]. 5.80 g NaOH 1 mol NaOH 0.1450 mol NaOH x = = 0.1450 M OH− 1.00 L 40.01 g NaOH 1.00 L
[H3O+] = Kw ÷ [OH−] = (1.0 x 10−14) ÷ (0.1450) = 6.896 x 10−14 M pH = −log [H3O+] = −log (6.896 x 10−14) = 13.1614 = 13.16
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561
15.80. First, calculate the molarity of the Ba(OH)2 from the mass of Ba(OH)2, and multiply the molarity of Ba(OH)2 by 2 to obtain [OH−]. Next, convert the [OH−] to [H3O+] using the Kw equation. Then find the pH, recording the same number of places after the decimal point in the pH as the number of significant figures in the [H3O+]. 6.78 g Ba(OH) 2 1 mol Ba(OH) 2 0.039556 mol Ba(OH) 2 x = 1.00 L 171.4 g Ba(OH) 2 1.00 L
[OH−] = 2 mol OH−/1 mol Ba(OH)2 = 2 x 0.039556 M Ba(OH)2 = 0.07911 M [H3O+] = Kw ÷ [OH−] = (1.0 x 10−14) ÷ (0.07912) = 1.264 x 10−13 M pH = −log [H3O+] = −log (1.264 x 10−13) = 12.8982 = 12.90 15.81. Figure 15.10 shows that the methylred indicator is yellow at pH values above about 5.5 (slightly past the midpoint of the range for methyl red). Bromthymol blue is yellow at pH values up to about 6.5 (slightly below the midpoint of the range for bromthymol blue). Therefore, the pH of the solution is between 5.5 and 6.5, and the solution is acidic. 15.82. Thymol blue is yellow at pH values above about 2.5. Bromphenol blue is yellow at pH values below about 3.5. Therefore, the pH of the aspirin solution must be in the range of 2.5 to 3.5. The solution is, of course, acidic.
■
SOLUTIONS TO GENERAL PROBLEMS
15.83. a.
BaO is a base; BaO + H2O → Ba2+ + 2OH−
b.
H2S is an acid; H2S + H2O → H3O+ + HS−
c.
CH3NH2 is a base; CH3NH2 + H2O → CH3NH3+ + OH−
d.
SO2 is an acid; SO2 + 2H2O → H3O+ + HSO3−
15.84. a.
P4O10 is an acid; P4O10 + 10H2O → 4H3O+ + 4H2PO4−
b.
K2O is a base; K2O + H2O → 2K+ + 2OH−
c.
N2H4 is a base; N2H4 + H2O → N2H5+ + OH−
d.
H2Se is an acid; H2Se + H2O → H3O+ + HSe−
15.85. a.
H2O2(aq) + S2−(aq) → HO2−(aq) + HS−(aq)
b.
HCO3−(aq) + OH−(aq) → CO32−(aq) + H2O(l)
c.
NH4+(aq) + CN−(aq) → NH3(aq) + HCN(aq)
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Chapter 15: Acids and Bases
d.
H2PO4−(aq) + OH−(aq) → HPO42−(aq) + H2O(l)
15.86. a.
H2O(l) + HCl(aq) → H3O+(aq) + Cl−(aq)
b.
HCO3−(aq) + HF(aq) → H2CO3(aq) + F−(aq)
c.
NH3(aq) + HBrO(aq) → NH4+(aq) + BrO−(aq)
d.
H2PO4−(aq) + H2SO3(aq) → H3PO4(aq) + HSO3−(aq)
15.87. The ClO− ion is a Brønsted base, and water is a Brønsted acid. The complete chemical equation is ClO−(aq) + H2O(l) HClO(aq) + OH−(aq). The equilibrium does not − favor the products because ClO is a weaker base than OH−. In Lewis language, a proton from H2O acts as a Lewis acid by accepting a pair of electrons on the oxygen of ClO−.
+
H b.
+
[
Cl O
[
a.
Cl O H
The NH2− ion is a Brønsted base, and NH4+ is a Brønsted acid. The complete chemical equation is NH4+ + NH2− 2NH3. The equilibrium favors the products because the reactants form the solvent, a weakly ionized molecule. In Lewis language, the proton from NH4+ acts as a Lewis acid by accepting a pair of electrons on the nitrogen of NH2−.
+ H H N H
+
N H
2 H N H
H
H
H
15.88. The HS− ion is a Brønsted base, and water is a Brønsted acid. The complete chemical equation is HS−(aq) + H2O(l) H2S(aq) + OH−(aq). The equilibrium does not − favor the products because the HS ion is a weaker base than OH−. In Lewis language, a proton from H2O acts as a Lewis acid by accepting a pair of electrons on the sulfur of HS−.
H O H
[
S H
H S
H + OH
The complete chemical equation is Fe3+(aq) + CN−(aq) → Fe(CN)2+. The reaction cannot be described in Brønsted language because no proton transfer occurs. In Lewis language, Fe3+ acts as a Lewis acid by sharing a pair of electrons on the carbon of CN−.
[
C
N
[
Fe 3+ +

[Fe
C
N
[
b.
+
[
a.
2+
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563
15.89. Table 15.2 shows that HNO2 is a stronger acid than HF. Because an acidbase reaction normally goes in the direction of the weaker acid, the reaction is more likely to go in the direction written HNO2 + F−
HF + NO2−
15.90. The HS− ion is acting as a base and will form its conjugate acid, H2S. Table 15.2 shows that H2S is a stronger acid than HCN. Because an acidbase reaction normally goes in the direction of the weaker acid, the reaction is more likely to go in the opposite direction: HS− + HCN
H2S + CN−
15.91. The order is H2S < H2Se < HBr. H2Se is stronger than H2S because, within a group, acid strength increases with increasing size of the central atom in binary acids. HBr is a strong acid, whereas the others are weak acids. 15.92. The order is HBrO < HBrO2 < HClO2. HClO2 is stronger than HBrO2 because, for oxoacids, acid strength increases with increasing electronegativity. HBrO2 is stronger than HBrO because acid strength increases with increasing oxidation number of Br. 15.93. The KOH is a strong base and is fully ionized in solution, so you can use its formula and molar concentration to determine the [OH−] of the solution. Therefore, the 0.25 M KOH contains 0.25 M OH−. The [H3O+] is obtained from the Kw expression: Kw = 1.0 x 10−14 = [H3O+] x 0.25 M OH− [H3O+] =
1.0 x 1014 = 4.00 x 10−14 = 4.0 x 10−14 M 0.25
15.94. The Sr(OH)2 is a strong base and is fully ionized in solution, so you can use its formula and molar concentration to determine the [OH−] of the solution. Therefore, the 0.35 M Sr(OH)2 contains 2 x 0.35 = 0.70 M OH−. The [H3O+] is obtained from the Kw expression: Kw = 1.0 x 10−14 = [H3O+] x 0.70 M OH− [H3O+] =
1.0 x 1014 = 1.42 x 10−14 = 1.4 x 10−14 M 0.70
15.95. Enter the H3O+ concentration of 1.5 x 10−3 into the calculator, press the log key, and press the sign key to change the negative log to a positive log. This follows the negativelog definition of pH. The number of decimal places of the pH should equal the significant figures in the H3O+. pH = − log [H3O+] = − log (1.5 x 10−3) = 2.823 = 2.82 15.96. Enter the H3O+ concentration of 2.5 x 10−2 into the calculator, press the log key, and press the sign key to change the negative log to a positive log. This follows the negativelog definition of pH. The number of decimal places of the pH should equal the significant figures in the H3O+. pH = − log [H3O+] = − log (2.5 x 10−2) = 1.602 = 1.60
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Chapter 15: Acids and Bases
15.97. Find the pOH from the pH using pH + pOH = 14.00. Then calculate the [OH−] from the pOH by entering the pOH into the calculator, pressing the sign key to change the positive log to a negative log, and finding the antilog. On some calculators, the antilog is found by using the inverse of the log; on other calculators, the antilog is found using the 10x key. The number of significant figures in the [OH−] should equal the number of decimal places in the pOH. pOH = 14.00 − 3.15 = 10.85 [OH−] = antilog (−10.85) = 10−10.85 = 1.412 x 10−11 = 1.4 x 10−11 M 15.98. Find the pOH from the pH using pH + pOH = 14.00. Then calculate the [OH−] from the pOH by entering the pOH into the calculator, pressing the sign key to change the positive log to a negative log, and finding the antilog. On some calculators, the antilog is found by using the inverse of the log; on other calculators, the antilog is found using the 10x key. The number of significant figures in the [OH−] should equal the number of decimal places in the pOH. pOH = 14.00 − 4.05 = 9.95 [OH−] = antilog (−9.95) = 10−9.95 = 1.12 x 10−10 = 1.1 x 10−10 M 15.99. a.
H2SO4(aq) + 2NaHCO3(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l) (molecular) H3O+(aq) + HCO3−(aq) → CO2(g) + 2H2O(l) (net ionic)
b.
The total moles of H3O+ from the H2SO4 is Moles of H3O+ =
2 mol H 3O + 0.437 mol H 2SO 4 x 0.02500 L x 1 mol H 2SO 4 1L
= 0.02185 mol H3O+ The moles of H3O+ that reacted with the NaOH is given by Moles of H3O+ =
0.108 mol NaOH x 0.0354 L = 0.003823 mol 1L
The moles of NaHCO3 present in the original sample is equal to the moles of H3O+ that reacted with the HCO3−, which is given by Total moles H3O+ − moles H3O+ reacted with the NaOH = moles HCO3− 0.02185 mol − 0.003823 mol = 0.01803 = 0.0180 mol NaHCO3
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c.
565
The mass of NaHCO3 (molar mass 84.01 g/mol) present in the original sample is 84.01 g NaHCO3 = 1.514 g 1 mol NaHCO3
0.01803 mol NaHCO3 x
Thus, the percent NaHCO3 in the original sample is given by Percent NaHCO3 =
1.514 g x 100% = 60.56 = 60.6% 2.500 g
The percent KCl in the original sample is Percent KCl = 100 − 60.56 = 39.44 = 39.4% 15.100. a.
CO32−(aq) + 2 H3O+(aq) → 3 H2O (l) + CO2(g)
b.
The total moles of H3O+ from the HCl is Moles of H3O+ =
0.798 mol HCl x 0.02500 L = 0.01995 mol 1L
The moles of H3O+ that reacted with the NaOH is given by Moles of H3O+ =
0.108 mol NaOH x 0.0287 L = 0.003100 mol 1L
The moles of Na2CO3 present in the original sample is equal to onehalf that of the moles of H3O+ that reacted with the CO32−, which is given by Moles Na2CO3 = 1/2(total moles H3O+ − moles H3O+ reacted with the NaOH) = 1/2(0.01995 mol − 0.003100 mol) = 0.008425 = 0.00843 mol c.
The mass of Na2CO3 (molar mass 105.99 g/mol) present in the original sample is 105.99 g Na 2 CO3 = 0.8929 g 1 mol Na 2 CO3
0.008425 mol Na2CO3 x
Thus, the percent Na2CO3 in the original sample is given by Percent Na2CO3 =
0.8929 g x 100% = 35.71 = 35.7% 2.500 g
The percent NaCl in the original sample is Percent NaCl = 100 − 35.71 = 64.28 = 64.3% H3O+(aq) + CO32−(aq)
15.101. HCO3−(aq) + H2O(l) HCO3−(aq) + H2O(l)
H2CO3(aq) + OH−(aq)
HCO3−(aq) + Na+(aq) + OH−(aq) HCO3−(aq) + H+(aq) + Cl−(aq)
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Na+(aq) + CO32−(aq) + H2O(l) H2O(l) + CO2(g) + Cl−(aq)
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Chapter 15: Acids and Bases
15.102. H2PO4− is amphiprotic or amphoteric. H2PO4−(aq) + H2O(l)
H3O+(aq) + HPO42−(aq)
H2PO4−(aq) + H2O(l)
H3PO4(aq) + OH−(aq)
H2PO4−(aq) + K+(aq) + OH−(aq)
K+(aq) + HPO42−(aq) + H2O(l)
H2PO4−(aq) + H+(aq) + NO3−(aq)
H3PO4(aq) + NO3−(aq)
15.103. CaH2(s) + 2H2O(l) → Ca(OH)2(s) + 2H2(g) The hydride ion is a stronger base because it took an H+ from water, leaving the OH− ion. Every time a strong base is added to water, it will react with the water, leaving the OH− as the product, so a strong base cannot exist in water. 15.104. Ca3N2(s) + 6H2O(l) → 3Ca(OH)2(s) + 2NH3(g) NaNH2(s) + H2O(l) → NaOH(aq) + NH3(g) N3− is a stronger base than NH2− because it has more negative charge, so it will have a greater attraction for H+. Also, we could consider that NH2− is a N3− that has already reacted with two H+'s. Stronger bases than OH− will produce OH− in aqueous solution. 15.105. H2F+ + F−
a.
2HF(l)
b.
NaF will be a base because F− is a conjugate base of HF.
c.
HClO4 + HF → H2F+ + ClO4− The conjugate acid is H2F+.
15.106. NH4+ + NH2−
a.
2NH3(l)
b.
NH2− is the conjugate base of NH3, so NaNH2 will be a base.
c.
NaNH2 + NH4Cl → 2NH3 + NaCl
15.107. The reaction of ammonia with water is given by NH4+(aq) + OH−(aq)
NH3(aq) + H2O(l)
The initial concentration of NH3 (molar mass 17.03 g/mol) is
1 mol NH 3 17.03 g NH 3 = 0.9982 M 0.2500 L
4.25 g NH 3 x Molarity =
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Since the NH3 is 0.42% reacted, the concentration of OH− is [OH−] = 0.9982 M x 0.0042 = 0.00419 M pOH = − log [OH−] = − log (0.00419) = 2.378 pH = 14 − pOH = 14 − 2.378 = 11.622 = 11.62 15.108. The reaction of C2H5NH2 with water is given by C2H5NH3+ (aq) + OH−(aq)
C2H5NH2(aq) + H2O(l)
The initial concentration of C2H5NH2 (molar mass 45.09 g/mol) is given by 1 mol C 2 H 5 NH 2 45.09 g C 2 H5 NH 2 = 0.1497 M 0.1000 L
0.675 g C 2 H 5 NH 2 x Molarity =
Since the C2H5NH2 is 0.98 percent reacted, the concentration of OH− is [OH−] = 0.1497 M x 0.0098 = 0.00147 M pOH = − log [OH−] = − log (0.00147) = 2.833 pH = 14 − pOH = 14 − 2.833 = 11.167 = 11.17 15.109. A bitter taste seems to be a common feature of a base. It is a fact that many medicinal substances are nitrogen bases, substances that organic chemists call amines, which are bases. 15.110. The nitrogen atom in an amine has a lone pair of electrons that can be donated to form a covalent bond, so an amine is a Lewis base. But like ammonia, an amine accepts a hydrogen ion from an acid to form an amine ion, so it is also a BrønstedLowry base.
CH3 H
N
+
CH3 H
+
H3O+
H
N
H
+
H2O
H
15.111. Such solutions work by chemically reacting with fat and with hair, the usual ingredients of a stopped drain. When fat reacts with a strong base such as sodium hydroxide, it forms a salt of a fatty acid, a product otherwise known as soap. Hair is a protein material, and in the presence of a strong base, the protein breaks up into the salts of its constituent amino acid salts. That solution washes easily down the drain.
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Chapter 15: Acids and Bases
15.112. Sodium hydroxide is produced commercially by sending a direct current through an aqueous solution of sodium chloride, a process referred to as electrolysis. Hydrogen gas is released at the negative pole, and chlorine gas is released at the positive pole. At the end of the electrolysis, the solution contains sodium hydroxide. The overall reaction is electrolysis 2NaCl(aq) + 2H2O(l) ⎯⎯⎯⎯⎯ → H2(g) + Cl2(g) + 2NaOH(aq)
■
SOLUTIONS TO STRATEGY PROBLEMS
15.113. Ethanol reacting as a BrønstedLowry acid with OH− is CH3CH2OH + OH−
CH3CH2O− + H2O
Ethanol reacting as a BrønstedLowry base with H3O+ is CH3CH2OH + H3O+
CH3CH2OH2+ + H2O
In both of these reactions, a proton is transferred from the acid to the base, which is in agreement with the BrønstedLowry definitions of an acid and a base. These reactions can also be considered Lewis acidbase reactions. In the first reaction, the oxygen atom of OH− can act as an electronpair donor, and the hydrogen of CH3CH2OH can act as an electronpair acceptor. In the second reaction, the oxygen of CH3CH2OH acts as the electronpair donor and the hydrogen of H3O+ as the electronpair acceptor. 15.114. The selfionization of ethanol is obtained by transferring a proton from one ethanol molecule to the other ethanol molecule. CH3CH2OH2+ + CH3CH2O−
CH3CH2OH + CH3CH2OH
In pure ethanol, [CH3CH2OH2+] = [CH3CH2O−], so setting the concentration equal to x gives Keth = 1.0 x 10−20 = [CH3CH2OH2+][CH3CH2O−] = x2
Solving for x gives 1.0 x 10−10 M for the concentration of each ion. 15.115. This is a Lewis acidbase reaction with N(CH3)3 the electronpair donor (Lewis base) and AlCl3 the electronpair acceptor (Lewis acid). The product is AlCl3–N(CH3)3.
Cl Cl
Al Cl
CH3 +
N CH3
CH3
Cl
Cl
CH3
Al
N
Cl
CH3
CH3
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15.116. a.
Fe3+ is a stronger Lewis acid than Fe2+. The higher the positive charge, the stronger the acid.
b.
BF3 is a stronger Lewis acid than BCl3. Fluorine is more electronegative than chlorine. Thus, the bonds in BF3 are more polar than the bonds in BCl3. Acid strength increases with increasing bond polarity.
15.117. a.
HBrO < HBrO2 < HBrO3. For a series of oxoacids with the same central atom, acid strength increases with increasing number of oxygen atoms bonded to the central atom.
b.
H2TeO3 < H2SeO3 < H2SO3. For a series of oxoacids having the same structure but different central atoms, acid strength increases with increasing electronegativity of the central atom. For elements in the same column on the periodic table, electronegativity decreases as you go down the column, so acid strength decreases as you go down the column.
c.
SbH3 < H2Te < HI. For acids with the structure HX, acid strength increases with increasing electronegativity of atom X. Going across a row of elements on the periodic table, the electronegativity increases, the bond polarity increases, and the acid strength increases.
d.
H2S < H2Se < HBr. For acids with the structure HX, acid strength increases with increasing electronegativity of atom X. In going down a column of the periodic table, the size of atom X increases, the bond strength decreases, and the acid strength increases. H2S and H2Se are in the same column, and Se has bigger atoms, so H2Se is a stronger acid than H2S. Also, going across a row of elements on the periodic table, the electronegativity increases, the bond polarity increases, and the acid strength increases. Se and Br are in the same row, and Br is more electronegative, so HBr is a stronger acid than H2Se.
e.
HBrO2 < HClO2 < HClO3. For a series of oxoacids having the same structure but different central atoms, acid strength increases with increasing electronegativity of the central atom. For elements in the same column on the periodic table, electronegativity decreases as you go down the column, so acid strength decreases as you go down the column. Thus, HClO2 is a stronger acid than HBrO2. Also, for a series of oxoacids with the same central atom, acid strength increases with increasing number of oxygen atoms bonded to the central atom. Thus, HClO3 is a stronger acid than HClO2.
15.118. The acids in each column have the same central atom, so the strongest acid in each column is the one with the most oxygen atoms. The acids in each row have the same structure but different central atoms. Thus, the strongest acid is the one with the most electronegative central atom, which is Cl. Overall, the strongest acid is HOClO2 (HClO3) and the weakest acid is HOI. 15.119. First, determine the H3O+ concentration in the solution with pH 1.831. [H3O+] = 10−pH = 10−1.831 = 0.014757 M Now, use the dilution formula, M1V1 = M2V2, to solve for the final volume. V2 =
M 1V1 (557 mL)(0.0300 M ) = = 1132.3 mL = 1.132 L = 1.13 L M2 (0.014757 M )
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Chapter 15: Acids and Bases
15.120. Water dissociates into ions according to H2O(l) → H+(aq) + OH−(aq) Kw = 1.00 x 10−14, so the concentration of each ion is 1.00 x 10−7 M. The mass of water in dissociated form is
1.00 L x
18.01 g 1.00 x 107 mol x = 1.801 x 10−6 = 1.80 x 10−6 g H2O 1L 1 mol H 2 O
The number of H+ ions in 1 L of pure water is 1.00 L x
1.00 x 107 mol 6.02 x 1023 H + ions x = 6.02 x 1016 H+ ions 1L 1 mol
15.121. In a solution with pH 2.00, the H3O+ ion concentration is 1.00 x 10−2 = 0.0100 M. Since HCl is a strong acid, this would also be the concentration of HCl. The mass of HCl required is 3.00 L x
0.0100 mol 36.458 g x = 1.0937 g HCl 1L 1 mol HCl
The solution is 37.2 mass percent HCl, so the mass of concentrated hydrochloric acid solution required is 1.097 g = 2.940 = 2.94 g concentrated HCl 0.372
15.122. In a solution of 1.00 x 10−7 M HCl, the amount of H+ ions from HCl is comparable to the number obtained from water itself. This is like a commonion problem, so set up a table and use Kw for the equilibrium. Conc. (M) From HCl From water Change Equilibrium
H3O+(aq) + 1.00 x 10−7 1.00 x 10−7 −x 2.00 x 10−7 − x
OH−(aq) 0 1.00 x 10−7 −x 1.00 x 10−7 − x
2H2O(l)
Kw = 1.00 x 10−14 = [H3O+][OH−] = (2.00 x 10−7 − x)(1.00 x 10−7 − x)
Rearrange this equation into standard quadratic form and use the quadratic equation to solve for x. x2 − 3.00 x 10−7x + 1.00 x 10−14 = 0 x =
=
(3.00 x 107 ) ±
(3.00 x 107 ) 2 (4)(1)(1.00 x 1014 ) 2(1)
3.00 x 107 ± 2.236 x 107 2
The positive root is 2.618 x 10−7, which is not allowed, so use the negative root, which is 0.382 x 10−7. Therefore, the concentration of H3O+ and the pH are [H3O+] = 2.00 x 10−7 − 0.382 x 10−7 = 1.618 x 10−7 = 1.62 x 10−7 M pH = −log(1.618 x 10−7) = 6.791 = 6.79
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571
SOLUTIONS TO CUMULATIVESKILLS PROBLEMS
15.123. For (HO)mYOn acids, acid strength increases with n regardless of the number of OH's. The structure of H3PO4 is (HO)3PO; because H3PO3 and H3PO4 have about the same acidity, H3PO3 must also have n = 1; thus, m = 2. This leaves one H, which must bond to phosphorus, giving a structure of (HO)2(O)PH. Assuming that only two H's react with NaOH, the mass of NaOH that reacts with 1.00 g of H3PO3 (PA) is calculated as follows: 1.00 g PA x
1 mol PA 2 mol NaOH 40.00 g NaOH x x 81.994 g PA 1 mol PA 1 mol NaOH
= 0.9756 = 0.976 g NaOH 15.124. For (HO)mYOn acids, acid strength increases with n regardless of the number of OH's. The structure of H3PO4 is (HO)3PO; because H3PO2 and H3PO4 have the same acidity, H3PO2 must also have n = 1; thus, m = 1. This leaves two H's, which must bond to phosphorus, giving a structure of (HO)(O)PH2. Assuming that only one H reacts with NaOH, the mass of NaOH that reacts with 1.00 g of H3PO2 (HA) is calculated as follows: 1.00 g HA x
1 mol HA 1 mol NaOH 40.00 g NaOH x x 1 mol HA 1 mol NaOH 65.994 g HA = 0.6061 = 0.606 g NaOH
15.125. BF3 acts as a Lewis acid, accepting an electron pair from NH3: BF3 + :NH3 → F3B:NH3 The NH3 acts as a Lewis base in donating an electron pair to BF3. When 10.0 g of each are mixed, the BF3 is the limiting reagent because it has the higher formula mass. The mass of BF3:NH3 formed is 10.0 g BF3 x
1 mol BF3 1 mol BF3 :NH 3 84.84 g BF3 :NH 3 x x 67.81 g BF3 1 mol BF3 1 mol BF3 :NH 3 = 12.51 = 12.5 g BF3:NH3
15.126. BF3 is the Lewis acid and accepts an electron pair from ether: BF3 + :OR2 → F3B:OR2 The ether (:OR2) acts as a Lewis base in donating an electron pair to BF3. When 10.0 g of BF3 and 20.0 g of ether are mixed, the BF3 is the limiting reagent because the formula masses are nearly equal. The mass of BF3:OR2 formed is 10.0 g BF3 x
1 mol BF3 1 mol BF3 :OR 2 141.93 g BF3 :OR 2 x x 67.81 g BF3 1 mol BF3 1 mol BF3 :OR 2
= 20.93 = 20.9 g BF3:OR2
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CHAPTER 16
AcidBase Equilibria
■
SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 16.1. Abbreviate the formula of lactic acid as HL. To solve, assemble a table of starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
H3O+ 0 +x x
HL + H2O 0.025 −x 0.025 − x
L− 0 +x x
+
Substituting into the equilibriumconstant equation gives Ka =
[H 3 O + ] [L ] (x) 2 = [HL] (0.025  x)
The value of x equals the value of the molarity of the H3O+ ion, which can be obtained from the pH: [H3O+] = antilog (−pH) = antilog (−2.75) = 0.00178 M Substitute this value for x into the equation to get Ka =
(x) 2 (0.00178) 2 = = 1.36 x 10−4 = 1.4 x 10−4 (0.025  x) (0.025  0.00178)
The degree of ionization is Degree of ionization =
0.00178 = 0.071 0.025
16.2. To solve, assemble a table of starting, change, and equilibrium concentrations. Use HAc as the symbol for acetic acid. Conc. (M) Starting Change Equilibrium
HAc + H2O 0.10 −x 0.10 − x
H3O+ 0 +x x
+
Ac− 0 +x x
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Now, substitute these concentrations and the value of Ka into the equilibriumconstant equation for acid ionization: Ka =
[H 3 O + ] [Ac ] (x) 2 = = 1.7 x 10−5 [HAc] (0.10  x)
Solve the equation for x, assuming x is much smaller than 0.10, so (0.10 − x) ≅ 0.10. (x) 2 ≅ 1.7 x 10−5 (0.10)
x2 = 1.7 x 10−5 x 0.10 = 1.7 x 10−6 x = 0.00130 M Check to make sure the assumption that (0.10 − x) ≅ 0.10 is valid: 0.10 − 0.00130 = 0.0987, = 0.10
(to two significant figures)
The concentrations of hydronium ion and acetate ion are [H3O+] = [Ac−] = x = 0.0013 = 1.3 x 10−3 M The pH of the solution is pH = − log [H3O+] = − log (0.00130) = 2.884 = 2.88 The degree of ionization is Degree of ionization =
0.00130 = 0.0130 = 0.013 0.10
16.3. Abbreviate the formula for pyruvic acid as HPy. To solve, assemble a table of starting, change, and equilibrium concentrations: Conc. (M) Starting Change Equilibrium
HPy + H2O 0.0030 −x 0.0030 − x
H3O+ 0 +x x
+
Py− 0 +x x
Substitute the equilibrium concentrations and the value of Ka into the equilibriumconstant expression to get Ka =
[H 3 O + ] [Py ] (x) 2 = = 1.4 x 10−4 (0.0030  x) [HPy]
Note that the concentration of acid divided by Ka is 0.0030/1.4 x 10−4 = 21, which is considerably smaller than 100. Thus, you can expect that x cannot be ignored compared with 0.0030. The quadratic formula must be used. Rearrange the preceding equation to put it into the form ax2 + bx + c = 0. x2 + 1.4 x 10−4 x − 4.20 x 10−7 = 0
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Chapter 16: AcidBase Equilibria
Substitute into the quadratic formula to get x =
1.4 x 104 ±
(1.4 x 104 ) 2 + 4(4.20 x 107 ) 2
=
1.4 x 104 ± 1.303 x 103 2
Using the positive root, x = [H3O+] = 5.81 x 10−4 M. Now you can calculate the pH. pH = − log [H3O+] = − log (5.81 x 10−4) = 3.2350 = 3.24 16.4. To solve, note that K a = 1.3 x 10−2 > K a = 6.3 x 10−8, and hence the second ionization and K a 1
2
2
can be ignored. Assemble a table of starting, change, and equilibrium concentrations. H3O+ 0 +x x
H2SO3 + H2O 0.25 −x 0.25 − x
Conc. (M) Starting Change Equilibrium
HSO3− 0 +x x
+
Substitute into the equilibriumconstant expression for the first ionization. K a1 =
[H 3 O + ] [HSO3 ] (x) 2 = = 1.3 x 10−2 = 0.013 (0.25  x) [H 2SO3 ]
This gives x2 + 0.013x − 0.00325 = 0. Note that the concentration of acid divided by Ka is 0.25/0.013 = 19, which is considerably smaller than 100. Thus, you can expect that x cannot be ignored compared with 0.25. Reorganize the above equilibriumconstant expression into the form ax2 + bx + c = 0, and substitute for a, b, and c in the quadratic formula. x =
0.013 ±
(0.013) 2 + 4(0.00325) 2
=
 0.013 ± 0.1147 2
Using the positive root, x = [H3O+] = 0.05087 M. pH = − log (0.05087) = 1.293 To calculate [SO32−], which will be represented by y, use the second ionization. Assume the starting concentrations of H3O+ and HSO3− are those from the first equilibrium. HSO3− + H2O 0.0508 −y 0.0508 − y
Conc. (M) Starting Change Equilibrium
H3O+ 0.0508 +y 0.0508 − y
+
SO32− 0 +y y
Now, substitute into the K a expression for the second ionization. 2
Ka2 =
[H 3 O + ] [SO32 ] (0.0508 + y )(y ) = = 6.3 x 10−8 [HSO3 ] (0.0508  y )
Assuming y is much smaller than 0.0508, note that the (0.0508 + y) cancels the (0.0508 − y) term, leaving y ≅ K a , or 2
y = [SO3 ] ≅ 6.3 x 10−8 M 2−
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575
16.5. Convert the pH first to pOH and then to [OH−]: pOH = 14.00 − pH = 14.00 − 9.84 = 4.16 [OH−] = antilog (−4.16) = 6.92 x 10−5 M Using the symbol Qu for quinine, assemble a table of starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
Qu + H2O 0.0015 −x 0.0015 − x
HQu+ 0 +x x
+
OH− 0 +x x
Note that x = 6.92 x 10−5. Substitute into the equilibriumconstant expression to get Kb =
[HQ + ] [OH  ] (x) 2 (6.92 x 105 ) 2 = = = 3.346 x 10−6 [Qu] (0.0015  x) (0.0015  6.92 x 105 )
= 3.3 x 10−6 16.6. Assemble a table of starting, change, and equilibrium concentrations. Conc. (M) Starting Change Equilibrium
NH3 + H2O 0.20 −x 0.20 − x
NH4+ 0 +x x
+
OH− 0 +x x
Assume x is small enough to ignore compared with 0.20. Substitute into the equilibriumconstant expression to get Kb =
[NH 4 + ] [OH  ] (x) 2 (x) 2 = ≅ = 1.8 x 10−5 (0.20  x) (0.20) [NH 3 ]
Solving for x gives x2 = (0.20) x 1.8 x 10−5 = 3.6 x 10−6 x = [OH−] ≅ 1.89 x 10−3 M (Note that x is negligible compared to 0.20.) Now calculate the hydroniumion concentration. [H3O+] =
Kw 1.0 x 1014 = = 5.29 x 10−12 = 5.3 x 10−12 M [OH ] 1.89 x 103
16.7. a.
Acidic. NH4NO3 is the salt of a weak base (NH4OH) and a strong acid (HNO3), so a solution of NH4NO3 is acidic because of the hydrolysis of NH4+.
b.
Neutral. KNO3 is the salt of a strong base (KOH) and a strong acid (HNO3), so a solution of NH4NO3 is neutral because none of the ions hydrolyze.
c.
Acidic. Al(NO3)3 is the salt of a weak base [Al(OH)3] and a strong acid (HNO3), so a solution of Al(NO3)3 is acidic because of the hydrolysis of Al3+.
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Chapter 16: AcidBase Equilibria
16.8. Calculate Kb of the F− ion from the Ka of its conjugate acid, HF:
a.
Kb = b.
Kw 1.0 x 1014 = = 1.47 x 10−11 = 1.5 x 10−11 Ka 6.8 x 104
Calculate Ka of C6H5NH3+ from the Kb of its conjugate base, C6H5NH2: Ka =
Kw 1.0 x 1014 = = 2.38 x 10−5 = 2.4 x 10−5 Kb 4.2 x 1010
16.9. Assemble the usual table, writing HBen for benzoic acid and Ben− for the benzoate ion. Let 0.015 − x equal the equilibrium concentration of the benzoate anion. Conc. (M) Starting Change Equilibrium
Ben− + H2O 0.015 −x 0.0015 − x
HBen 0 +x x
OH− 0 +x x
+
Calculate Kb for the benzoate ion from Ka for HBen Kb =
Kw 1.0 x 1014 = = 1.58 x 10−10 6.3 x 105 Ka
Substitute into the equilibriumconstant expression. Assume x is much smaller than 0.015. Kb =
(x) 2 (x) 2 [HBen] [OH  ] = ≅ = 1.58 x 10−10 (0.015  x) (0.015) [Ben ]
x = [HBen] = [OH−] ≅ 1.539 x 10−6 M
(x is negligible compared to 0.015.)
Thus, the concentration of benzoic acid in the solution is 1.5 x 10−6 M. The pH is pOH = − log [OH−] = − log (1.539 x 10−6) = 5.812 pH = 14.00 − 5.812 = 8.188 = 8.19 16.10. Assemble the usual table, using starting [H3O+] = 0.20 M from 0.20 M HCl and letting HFo symbolize HCHO2. Conc. (M) Starting Change Equilibrium
HFo + H2O 0.010 −x 0.10 − x
H3O+ 0.20 +x 0.20 + x
+
Fo− 0 +x x
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Assume x is negligible compared to 0.10 M and 0.20 M, and substitute into the equilibriumconstant expression to get Ka =
[H 3 O + ] [Fo  ] (0.20 + x)(x) (0.20)(x) = ≅ = 1.7 x 10−4 [HFo] (0.10  x) (0.10)
x = [Fo−] = 8.50 x 10−5 = 8.5 x 10−5 M The degree of ionization is Degree of ionization =
8.50 x 105 = 8.50 x 10−4 = 8.5 x 10−4 0.10
16.11. Assemble the usual table, using a starting [CHO2−] of 0.018 M from 0.018 M NaCHO2 and symbolizing HCHO2 as HFo and the CHO2− anion as Fo−. Conc. (M) Starting Change Equilibrium
HFo + H2O 0.025 −x 0.025 − x
H3O+ 0 +x x
+
Fo− 0.018 +x 0.018 + x
Substitute into the equilibriumconstant expression. Assume x is negligible compared to 0.025 M and 0.018 M. Ka =
[H 3 O + ] [Fo  ] (0.018 + x)(x) (0.018)(x) = ≅ = 1.7 x 10−4 [HFo] (0.025  x) (0.025)
x = [H3O+] ≅ 2.36 x 10−4 M The pH can now be calculated. pH = − log [H3O+] = − log (2.36 x 10−4) = 3.627 = 3.63 16.12. Let HOAc represent HC2H3O2 and OAc− represent C2H3O2−. The total volume of the buffer is Total volume = 30.0 mL + 70.0 mL = 100.0 mL = 0.1000 L The moles of HOAc and OAc− in the buffer are mol HOAc = 0.15 M x 0.0300 L = 0.00450 mol mol OAc− = 0.20 M x 0.0700 L = 0.0140 mol The concentrations of HOAc and OAc− in the buffer are [HOAc] = [OAc−] =
0.00450 mol = 0.0450 M 0.1000 L
0.0140 mol = 0.140 M 0.1000 L
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Chapter 16: AcidBase Equilibria
Now, assemble these starting concentrations into a table. Conc. (M) Starting Change Equilibrium
HOAc + H2O 0.0450 −x 0.0450 − x
H3O+ 0 +x x
+
OAc− 0.140 +x 0.140 + x
Substitute the equilibrium concentrations into the equilibriumconstant expression; then assume x is negligible compared to the starting concentrations of both HOAc and OAc−. Ka =
[H 3 O + ] [OAc ] (0.140 + x)(x) (0.140)(x) = ≅ = 1.7 x 10−5 [HOAc] (0.0450  x) (0.0450)
x = [H3O+] ≅ 5.46 x 10−6 M You can now calculate the pH. pH = − log [H3O+] = − log (5.46 x 10−6) = 5.262 = 5.26 16.13. First, do the stoichiometric calculation. From Exercise 16.11, [HFo] = 0.025 M and [Fo−] = 0.018 M. In 1 L of buffer, there are 0.025 mol HFo and 0.018 mol Fo−. The moles of OH− (equal to moles of NaOH) added are (0.10 M) x 0.0500 L = 0.00500 mol OH− The total volume of solution is Total volume = 1 L + 0.0500 L = 1.0500 L After reaction with the OH−, the moles of HFo and Fo− remaining in the solution are mol HFo = (0.025 − 0.00500) mol = 0.0200 mol mol Fo− = (0.018 + 0.00500) mol = 0.0230 mol The concentrations are [HFo] = [Fo−] =
0.0200 mol = 0.0190 M 1.0500 L
0.0230 mol = 0.0219 M 1.0500 L
Now account for the ionization of HFo to Fo− at equilibrium by assembling the usual table. Conc. (M) Starting Change Equilibrium
HFo + H2O 0.0190 −x 0.0190 − x
H3O+ 0 +x x
+
OAc− 0.0219 +x 0.0219 + x
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579
Assume x is negligible compared to 0.0190 M and 0.0219 M, and substitute into the equilibriumconstant expression to get Ka =
[H 3 O + ] [Fo  ] (0.0219 + x)(x) (0.0219)(x) = ≅ = 1.7 x 10−4 [HFo] (0.0190  x) (0.0190)
x = [H3O+] = 1.47 x 10−4 M Now, calculate the pH. pH = − log [H3O+] = − log (1.47 x 10−4) = 3.831 = 3.83 16.14. All the OH− reacts with the H3O+ from HCl. Calculate the stoichiometric amounts of OH− and H3O+. mol H3O+ = (0.10 mol/L) x 0.025 L = 0.0025 mol mol OH− = (0.10 mol NaOH/L) x 0.015 L = 0.0015 mol The total volume of solution is Total volume = 0.025 L + 0.015 L = 0.040 L Subtract the moles of OH− from the moles of H3O+, and divide by the total volume to get the concentration of H3O+. mol H3O+ left = (0.0025 − 0.0015) mol = 0.0010 mol [H3O+] =
0.0010 mol = 0.0250 M 0.040 L
Now calculate the pH. pH = − log [H3O+] = − log (0.0250) = 1.602 = 1.60 16.15. At the equivalence point, the solution will contain NaF. The molar amount of F− is equal to the molar amount of HF and is calculated as follows: (0.10 mol HF/L) x 0.025 L = 0.0025 mol F− The volume of 0.15 M NaOH added and the total volume of solution are calculated next. Volume NaOH =
M acidVacid (0.10 M )(25 mL) = = 16.6 mL 0.15 M M base
Total volume = 25 mL + 16.6 mL = 41.6 mL = 0.0416 L The concentration of F− at the equivalence point can now be calculated. [F−] =
0.0025 mol = 0.0600 M 0.0416 L
Next, consider the hydrolysis of F−. Start by calculating the hydrolysis constant of F− from the Ka of its conjugate acid, HF. Kb =
Kw Ka
=
1.0 x 1014 = 1.47 x 10−11 4 6.8 x 10
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Chapter 16: AcidBase Equilibria
Now, assemble the usual table of concentrations, assume x is negligible compared to 0.0600, and calculate [OH−]. Conc. (M) Starting Change Equilibrium
F + H2O 0.0600 −x 0.0600 − x
HF 0 +x x
OH− 0.0219 +x x
+
Substitute into the equilibriumconstant expression to get Kb =
[HF] [OH  ] (x) 2 (x) 2 = ≅ = 1.47 x 10−11 [F ] (0.0600  x) (0.0600)
x = [OH−] = 9.39 x 10−7 M Finally, calculate the pOH and then the pH. pOH = − log [OH−] = − log (9.39 x 10−7) = 6.026 pH = 14.00 − 6.026 = 7.974 = 7.97 16.16. a.
Assemble the usual table of concentrations. Conc. (M) Starting Change Equilibrium
NH3 + H2O 0.200 −x 0.200 − x
NH4+ 0 +x x
+
OH− 0 +x x
Now substitute into the equilibriumconstant expression, assume x is negligible compared to 0.200 M, and calculate [OH−]. Kb =
(x) 2 (x) 2 = = 1.8 x 10−5 (0.200  x) (0.200)
x = [OH−] = 1.897 x 10−3 M Finally, calculate the pOH and then the pH. pOH = − log [OH−] = − log (1.897 x 10−3) = 2.721 pH = 14.00 − 2.721 = 11.278 = 11.28 b.
At the halfway point in the titration of a weak base with a strong acid, pOH = pKb. pOH = − log (1.8 x 10−5) = 4.744 pH = 14.00 − pOH = 14.00 − 4.744 = 9.255 = 9.26
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c.
581
At the equivalence point, the solution will contain NH4Cl. The molar amount of NH4+ is equal to the molar amount of NH3 and is calculated as follows: (0.200 M) x 0.0800 L = 0.01600 mol NH4+ The volume of 0.100 M HCl added and the total volume of solution are calculated next. Volume HCl =
M baseVbase (0.200 M )(80.0 mL) = = 160.0 mL 0.100 M M acid
Total volume = 80.0 mL + 160.0 mL = 240.0 mL = 0.2400 L The concentration of NH4+ at the equivalence point can now be calculated. [NH4+] =
0.01600 mol = 0.06666 M 0.2400 L
Next, consider the hydrolysis of NH4+. Start by calculating the hydrolysis constant of NH4+ from the Kb of its conjugate base, NH3. Ka =
Kw Kb
=
1.0 x 1014 = 5.56 x 10−10 1.8 x 105
Now, assemble the usual table of concentrations, assume x is negligible compared to 0.06666, and calculate [H3O+]. Conc. (M) Starting Change Equilibrium
NH4+ + H2O 0.06666 −x 0.06666 − x
NH3 0 +x x
+
H3O+ 0 +x x
Substitute into the equilibriumconstant expression to get Ka =
[NH 3 ] [H 3 O + ] (x) 2 (x) 2 = ≅ = 5.56 x 10−10 + [NH 4 ] (0.06666  x) (0.06666)
x = [H3O+] = 6.08 x 10−6 M Finally, calculate the pH. pH = − log [H3O+] = − log (6.08 x 10−6) = 5.215 = 5.22 d.
Calculate the moles of acid added. (0.100 M) x 0.175 L = 0.01750 mol The moles of base added are the same as for part c, 0.01600 mol. The excess moles of acid and the total volume are Moles of acid remaining = 0.01750 mol − 0.01600 mol = 0.00150 mol V = 80.0 mL + 175 mL = 255.0 mL = 0.2550 L The [H3O+] concentration and the pH are [H3O+] =
0.00150 mol = 5.882 x 10−3 M 0.2550 L
pH = − log (5.882 x 10−3) = 2.230 = 2.23
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Chapter 16: AcidBase Equilibria
ANSWERS TO CONCEPT CHECKS
16.1. You would probably guess that the pH's of the acid solutions depend on their respective Ka's; the larger the Ka, the greater the acidity, or the lower the pH. We can put this on a firm basis by looking at the acidionization equilibrium. An acid, HA, ionizes in water as follows: HA(aq) + H2O(l)
H3O+(aq) + A−(aq)
The corresponding equilibrium constant, Ka, equals [H3O+][A−]/[HA]. When you start with the same concentration of HA, the concentration of HA in solution is essentially the same for each acid. Also, [H3O+] = [A−]. This means Ka is proportional to [H3O+]2, or pH = − log [H3O+] is proportional to − log Ka. Therefore, the larger the Ka, the lower the pH. As an example, compare two acids, one with Ka equal to 10−5 and the other with Ka equal to 10−4. The corresponding − log Ka values are 5 and 4, respectively. The second acid (the one with the greater Ka) would have the lower pH. If you look at Table 16.1, the acid with the largest Ka of those listed in the problem statement is HF. So the ranking from highest to lowest pH is HCN > HC2H3O2 > HNO2 > HF. 16.2. By examining the contents of the beaker, you see there are three different species present: the ions BH+ and OH− and unreacted B molecules. This represents a weak base. The ionization reaction for this base is B(aq) + H2O(l)
BH+(aq) + OH−(aq)
16.3. Ammonia, NH3, is a weak base; the other compounds are