Petrucci - General Chemistry Principles and Modern Applications 10th c2011 ANSWERS
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ANSWERSChapter 1 Answers Practice Examples 1a. 177 C 1b. −26.1 C 2a. 1.46 g/mL 2b. 2.987 cm 3a. 2.76 g/cm3 3b. The water level will remain unchanged. 4a. 1.8 kg ethanol 4b. 0.857 g/mL 5a. 21.3 5b. 1.1×106 6a. 26.7 6b. 15.6 Integrative Example A. 14% B. 4.239×1023 Na atoms Exercises 1. One theory is preferred over another if it can correctly predict a wider range of phenomena and if it has fewer assumptions. 3. For a given set of conditions, a cause, is expected to produce a certain result or effect. Although these cause-and-effect relationships may be difficult to unravel at times (“God is subtle”), they nevertheless do exist (“He is not malicious”). 5. The experiment should be carefully set up so as to create a controlled situation in which one can make careful observations after altering the experimental parameters, preferably one at a time. The results must be reproducible (to within experimental error) and, as more and more experiments are conducted, a pattern should begin to emerge, from which a comparison to the current theory can be made. 7a. Physical Copyright © 2011 Pearson Canada Inc. 1 7b. Chemical 7c. Chemical 7d. Physical 9a. Homogeneous mixture 9b. Heterogeneous mixture 9c. Heterogeneous mixture 9d. Substance 11a. If a magnet is drawn through the mixture, the iron filings will be attracted to the magnet and the wood will be left behind. 11b. When the glass-sucrose mixture is mixed with water, the sucrose will dissolve, whereas the glass will not. The water can then be boiled off to produce pure sucrose. 11c. Olive oil will float to the top of a container and can be separated from water, which is more dense. It would be best to use something with a narrow opening that has the ability to drain off the water layer at the bottom (i.e., buret). 11d. The gold flakes will settle to the bottom if the mixture is left undisturbed. The water then can be decanted (i.e., carefully poured off). 13a. 8.950 ×103 13b. 1.0700 ×104 13c. 2.40 ×10−2 13d. 4.7 ×10−3 13e. 9.383 ×102 13f. 2.75482 ×105 15a. 3.4 ×104 cm/s 15b. 6.378 ×103 km 15c. 7.4 × 10-11 m 15d. 4.6 ×105 17a. exact number 17b. measured quantity 17c. measured quantity 17d. measured quantity Copyright © 2011 Pearson Canada Inc. 2 19a. 3985 19b. 422.0 19c. 1.860 ×105 19d. 3.390 ×104 19e. 6.321×104 19f. 5.047 ×10−4 21a. 9.3 ×10−4 21b. 2.2 ×10−1 21c. 5.4 ×10−1 21d. 3.058 ×101 23a. 2.44 ×104 23b. 1.5 ×103 23c. 40.0 21d. 2.131×103 21e. 4.8 × 10-3 25a. 186.30 km/h 25b. 9.95 km/kg 27a. 127 mL 27b. 0.0158 L 27c. 0.981 L 27d. 2.65 ×106 cm3 29a. 174 cm 29b. 29 m 29c. 644 g 29d. 112 kg 29e. 7.00 dm3 Copyright © 2011 Pearson Canada Inc. 3 5 min 37.2 °M 47.790 g/mL 51. 3245µ g 33.2 ×103 g/cm 2 .47 acres 39.8 m Copyright © 2011 Pearson Canada Inc. 0. 9. 11% D .629 kg acetone 53. 0.9% B .2% A . 10. 28. 38. A temperature of −465 F cannot be achieved because it is below absolute zero.83 m/s 35c. 1. -59.52 ×103 mL 31. 0.04 ×104 g iron 57. 1. 1. iron bar < aluminum foil < water 59. 4 . 0.29f. 2. 3% F 65. 45a.7% C .07 kg fertilizer 55. 2.958 g/mL 49.4 ×1016 tons 72. high: 122 F 43. Low: 14 F .5 m 35a. 35.82 ×10−2 mm 61. 2. 1. 1. s 35b. 3. 9. 2.9 L blood 63.2 ×104 kg/m 2 41. 5. 48.10 ×103 g sucrose Integrative and Advanced Exercises 71.1°M 45b. 867 g/cm3 85b.119 mm 89. °C 76c. The answer is (e). a natural law.12 cm3 Self-Assessment Exercises 99.02 g/cm3 85c. 0. −19.74a.25 °C 76b. Copyright © 2011 Pearson Canada Inc. 11 g/mL 81.25 g/cm3 85a.3 g ethanol 88. 0. 160. Density g . %N = 58. 2. -101. The information provided in sketches (a) and (d) allows calculation of the density. 0. 102. 5 . 76. 100.869 g/cm3 85d. of the plastic material = 1. 5 mg 1m 2 1h 74b.98×104 kg sodium hypochlorite 84. The answer is (a).1 glasses of wine Feature Problems 94. 335 °C 80. 347. 4. 1.4 m 87.1 °C 76d. 1. 0.4 86.1×105 h 76a. The answer is (d) and (f). 101. The answer is (c). 0. The answer is (b). 6 . 109. The answer is (d). (c). The answer is (b). (e). 107. 104.103. Copyright © 2011 Pearson Canada Inc. No. (b).114 mm 111. The answer is (d). 108. 110. (a). (d). listed in order of increasing significant figures. and 50 Sn are all possible answers.39 g of unreacted O2 . Li + .61 g of MgO (product) and 3. 1 . U is an inner transition metal. 7. 117 50 Sn . 7b. 6. 3a.20 g Mg . In-115 6a. Kr is a nonmetal in group 18(8A). 4a.Chapter 2 Answers Practice Examples 1a.7 7a. Mg is a main-group metal in group 2(2A). As is a main-group metalloid in group 15(5A). 2b. 92. 0.9241 u. Na is a main-group metal in group 1(1A). A1 is a main-group metal in group 13(3A). 50 Sn . Ra 2+ .80 g oxygen. 116 50 2+ 118 2+ 119 2+ 120 2+ Sn 2+ . 109 47 Ag 3b. 50 Sn .5% lithium .87340 . Re is a transition metal in group 7(7B). 0. 248 g Cu Copyright © 2011 Pearson Canada Inc. S2 . Boron-11 5b. 8a. 16.529 g magnesium nitride 1b. S is a main-group nonmetal in group 16(6A) I is a main-group nonmetal in group 17(7A).08 u 6b. B is a metalloid in group 13(3A). F . 5a. 158 64 Gd .6. I . Si is a main-group metalloid in group 14(4A). 157.85 g magnesium 2a. Al 3+ . 28. 1. 9.5% lithium . 16.830885 4b. an actinide. 2 . 11% Copyright © 2011 Pearson Canada Inc. 0. 3. 7a.3% 9.46 1021 Pb atoms 9b.659 7c. These results are consistent with the Law of Multiple Proportions because the masses of hydrogen in the three compounds end up in a ratio of small whole numbers when the mass of nitrogen in all three compounds is normalized to a simple value (1. Compound A might be N2H6 and compound C.397 7b. The two samples of sodium chloride have the same %Na.14 105 atoms of 63Cu B. have reacted together to give two different compounds that have masses of oxygen that are in the ratio of small positive integers for a fixed amount of sulfur. The observations cited do not necessarily violate the law of conservation of mass.4477 106 servings . 62. the mass of oxygen in the second compound (SO3) relative to the mass of oxygen in the first compound (SO2) is in a ratio of 3:2.268 g oxygen 5. the mass of the ash thus is less than that of the match.422 g sulfur 13. 60. We would have to collect all reactants and all products and weigh them to determine if the law of conservation of mass is obeyed or violated.58×1022 206Pb atoms 9a. For a given mass of sulfur. 11. 9. The oxide formed when a match burns is a gas and will not remain with the solid product (the ash). These results are entirely consistent with the Law of Multiple Proportions because the same two elements.5% Integrative Example A. 15b. 0. sulfur and oxygen in this case. 0. These data support the law of conservation of mass. 15a. increasing the mass of the solid by an amount equal to the mass of the oxygen that has combined. 17.8b. 0. No. N2H4. 1. 3. The oxide formed when iron rusts is a solid and remains with the solid iron.000 g here). 2. Compare the mass before reaction (initial) with that after reaction (final). Exercises 1. 59 26 Fe 23d. express each in terms of e = 1. and he could have inferred the correct charge from these data. 23a. Use the mass of a proton plus that of an electron for the mass of a hydrogen atom. Determine the ratio of the mass of a hydrogen atom to that of an electron. 60 27 Co 23b. such as the charge on the species). The hydrogen ion is the lightest positive ion available. such as from the name or the symbol of the element involved). The mass-to-charge ratio for a positive particle is considerably larger than that for an electron. Does not characterize a specific nuclide. 3 . 21b. The values are consistent with the charge that Millikan found for that of the electron. and the mass number (or the number of neutrons). several possibilities exist. H+) and the electron. 32 15 P 23c. Calculate the charge on each drop.a Name Symbol number of protons number of electrons number of neutrons mass number sodium 23 11 Na 11 11 12 23 silicon 28 14 Si 14 14a 14 28 rubidium 85 37 Rb 37 37a 48 85 potassium 40 19 K 19 19 21 40 arsenic a 75 33 As 33a 33 42 75 neon 20 10 Ne2+ 10 8 10 20 bromine b 80 35 Br 35 35 45 80 lead b 208 82 82 82 126 208 Pb a This result assumes that a neutral atom is involved. b Copyright © 2011 Pearson Canada Inc. the number of electrons (or some way to obtain it. 21a. Insufficient data. Finally.6 10 19 C. express each in terms of 10 19 C.19. Use data in Table 2-1 to compare the mass-to-charge ratios those for the proton (a hydrogen ion. 226 88 Ra 25. The minimum information needed is the atomic number (or some way to obtain it. since they are all multiples of e . 18. K.936 u 31a. 39.4527 u.962 u Mass Number 49a. There are no chlorine atoms that have a mass of 35. 46 electrons. 60 27 Co3+ 33c. 24. 53 protons. and Rb. Copyright © 2011 Pearson Canada Inc. 2.27a.31 u 45.165216 31c. 108. 35 17 Cl 33b. Average atomic mass of germanium = 72. but particularly metals such as Na. 95 protons. The result is only approximately correct because the isotopic masses are given to only two significant figures. 43. 62 neutrons 27b. Other elements in group 16(6A) are similar to S: O. 70 72 74 76 10 20 Relative Number of atoms 30 40 49b. and Te.914071 31b.9 u 47. 70 neutrons. Se. 146 neutrons. 46 protons. 41. 95 electrons. 106.6 . 2. 51a. 4 . 54 electrons. 40. calcium 37. Ge 51b. 8.50146 33a.9919908 29. 124 50 Sn 2+ 35. Most of the elements in the periodic table are unlike S. 2. Answer (b). 1. 5. 3. 5 .3 10 21 Si atoms Copyright © 2011 Pearson Canada Inc.7 1014 Pb atoms / mL 65.811020 atoms Ag 55c.51 10 24 atoms Fe 55b.36% 77. Rb 51d.707 10 27 atoms Cr 57c. 74.6 u 82.7 1018 atoms 204 Pb 63a. 1. 110 210 Sn 2 Po 76. 2.154760 10 23 g F 1 atom F 59. Ti 2 . 72c. 9. 0. 119 55a. 6.45 106 mol Pb / L 63b. 200. 3 1015 g/cm3 72a. Integrative and Advanced Exercises 70. 8. 118 53b.347 mol Zn 57b. At 53a. 8. There is not enough information to determine the mass number.3 1010 g Au 57d.1 1013 atoms Na 57a. 3.51c.4 1022 Cu atoms 61. 234 Th 72b. 2. The answer is (d). (d) and (f) 106. 87. it can be as large as 3. Group 17 104c.01 grain.9% Cd. 159 ppm Rb Self-Assessment Exercises 97. which seems to be the limit of the readability of the balance. Group 18 104b.43 ×105 km3 93. which assumes that all of the error in the experiment is due to the (in)sensitivity of the balance. The answer is (a). 2. The difference between reactant mass and product mass (in grams) falls between the maximum error of a common modern laboratory balance and the minimum error of a good quality analytical balance. 100.86.01 grains and 3. Convert 0. Group 13 and Group 1 104d. The answer is (c).12 grains to grams. The answer is (d). 101. Copyright © 2011 Pearson Canada Inc. 99. 17. The sensitivity of Lavoisier’s balance can be as little as 0.1% Sn. alternatively. 102. 104a. The answer is (b). The answer is (b). The answer is (d).12 grains. The answer is (c). 107.0% Pb. Group 18 105. 32. Lavoisier’s results conform closely to the law of conservation of mass. 35 17 Cl 103. 98. 102 cm3 Feature Problems 89. 92. 108. The answer is (d). 50. 6 . 3%. 7 . 87Sr = 7. The reason for the imprecision is the low number of significant figures for 84Sr.5%.2 1014 atoms of Au Copyright © 2011 Pearson Canada Inc. Fe2O3 and Fe3O4 110.109. 86Sr = 9. 111. 1. sulfur hexafluoride. 23. chromium (III) chloride. 18. calcium bicarbonate or calcium hydrogen carbonate. The vapor will be detectable. 3a. nitrous acid. Li 2 O . each Hg has O. iron(II) sulfate.Chapter 3 Answers Practice Examples 1a.S. Al2S3 . 10a. 2. each Cr has an O. 7b. = +1 . 8b. 4.008% O . The empirical formula of the compound is C3H 7 O3 . CH2O. the oxidation state is 0. 4b. Copyright © 2011 Pearson Canada Inc. The empirical formula of the compound is C13 H16 O8 Cl12 .S. the O.S. the O. of C is 0. 8. each oxygen has an O. . Both (b) and (e) have the same empirical formula.180% H . = +6 .S.S. The molecular formula is C13 H16 O8 Cl12 . 5a.44×1016 O 2a.ions . 3.S. = 1/2 .S. C4 H 4S 7a. 6a. 175 mL C2 HBrClF3 4a. 5b. each chlorine has an O. Each S has an O. Mg3 N 2 . 9b.12×1015 NO 3. V2O3 . calcium hydride. 1 . 13.32% P . For an atom of a free element. silver(I) sulfide. Li3 N . iron(II) oxide. C7 H14O2 6b.9 g Br 3b. SnF2 . 3. cesium iodide.0 101 g MgCl2 1b. 9a.81% N . mercury(I) chloride. 41.681% C . that is. The molecular formula is C6 H14 O6 . = +1 . 8a. calcium fluoride. These two molecules have the same percent oxygen by mass. = +2 . 18.69 10 21 Au atoms 2b. copper(I) chloride. of Mn is +7. (a) CH3(CH2)3CH3. H 2SO 4 . ammonium nitrate. The compound is Mg3(PO4)2·5 H2O.4-dichlorobutane. 12b. (c) ICH2(CH2)6CH3. (a) . 12a. 15a. iron(III) sulfate. (a) 2-chloropropane. (a) alcohol. CH3CH2Cl 1c.10b. Mg3(PO4)2. (a) Molecules are isomers. (b) Not isomers. (b) Molecules are isomers. (b) 1. P4O10 1d. hypobromous acid. (b) CH3CO2H. 11a. (b) carboxylic acid. (b) 1-iodopropane. Al(NO3)3 . BF3 . (d) propene. B. silver(I) perchlorate. (c) chloro carboxylic acid. Exercises 1a. (d) bromoalkene. The compound is Cu(ClO4)2·6H2O. The compound is magnesium phosphate. OH Cl 15b. (d) O Integrative Examples A. (a) Not isomers. (c) O . 14b. K 2Cr2O7 . HCO2H Copyright © 2011 Pearson Canada Inc. 11b. (c) 3-methylbutanoic acid. phosphorus trichloride. (d) alkene. (c) carboxylic acid. P4O10 . 14a. (b) chloroalkane. (d) CH2(OH)(CH2)3CH3. 2 . 13a. HIO3 . CH3CH(OH)CH3 1e. (a) alkane. Cr(OH)3 . (c) 2-methyl propanoic acid. (b) HO OH . H2O2 1b. 13b. CaCl2 . The oxidation state of Cu is +2 and Cl is +7. (a) 2-propanol. 21 atoms 5b. 0. Uranium Copyright © 2011 Pearson Canada Inc.195 1025 F atoms 7a. False 17a. 1.800 H atoms/O atom 17c.587 mol N atoms 11c.25 mol N 2 O4 11b.H H H C 3. 3 . False 15d. 0. 11a.213u/C5 H11 NO2S molecule 7b.29 1022 atoms 5c.0 g N 2 O . (1e) HCO2H O C H 5a. (1d) CH3CH(OH)CH3 C C H H H H . (1b) CH3CH2Cl OH H OH H C Cl H H H C . 41 atoms 17b. True 15c.026 17d. 7c. 3 1022 Fe atoms 15a. 2. 1.055g C 7d. 1.206 mol N atoms 13.73 10 25 C atoms 9. False 15b. 2. 60. There are 11 moles of H atoms in each mole of C5 H11 NO 2S molecules. 0. The greatest number of N atoms is present in 50. answer (a). 149. 15.710 g CO 2 .497 % Fe 25c.7202 % Mg 27. 51a. 90. C16 H10 N 2O2 39. 9. 15. C19 H16 O4 33b. 3. C10 H8 . CH 4 N 47. titanium 41.046 %C . 8. 43b. 9. 2. Oxide with the largest %Cr will have the largest number of moles of Cr per mole of oxygen. also the largest mass of CO 2 . 7. SO3 (40.4566 %H . 25a.898 g H 2O . C4 H10O3 33a.05% S) and S2O (80.585% H 21. C14 H10 37.17e. 45. 49.1 g of Cu(UO2)2(PO4)2•8H2O 19.6349 %N .8634 %O . C = 4 in CH 4 Copyright © 2011 Pearson Canada Inc.49 % C . C4 H5 43c. 15. C8 H10 45. C9H10N4O2S2 35. thus.0 % S). 4 . 31. 29.88 %H 23. 64. 74. Arranged in order of increasing %Cr: CrO3 CrO 2 Cr2 O3 CrO .06 % Pb 25b. 894 u 43a. 1. The compound with the largest number of moles of C per mole of the compound will produce the largest amount of CO 2 and.495 % H . Fe = +6 in FeO 4 2 53. cesium sulfate 57e. calcium hydrogen sulfite 57j. CrO3 55a. potassium chromate 57d. nitric acid 57l. copper(II) hydroxide 57k. zinc sulfide 57c. O = 1 in BaO2 57a. iron(III) sulfate 57g. C = 0 in C2 H3O2 51e. 5 . magnesium hydrogen carbonate or magnesium bicarbonate 57h. O = 1 in Na 2 O 2 51d. O = 2 in OF2 55b. strontium oxide 57b. chromium(III) oxide 57f. Cr2 O3 . CrO2 . carbon disulfide Copyright © 2011 Pearson Canada Inc. ammonium hydrogen phosphate 57i. O = 1 in CsO2 2 55d. potassium perchlorate 57m. bromic acid 57n.51b. S = +4 in SF4 51c. O = +1 in O 2 F2 55c. phosphorous acid 59a. HBr(aq) 61i. Fe2O3 61e. C3S2 61f. Cl2O7 63d. sulfur hexafluoride 59f. hydroselenic acid 65d. HIO3 61j. Al2 SO 4 3 61b. silicon tetrafluoride 59c. PCl2 F3 63a. SiF4 61d. 6 . dinitrogen pentoxide 59e. Copyright © 2011 Pearson Canada Inc. Sr NO 2 2 61h. not ionic. S2 O8 2 65a. NH 4 2 Cr2 O 7 61c. TiCl4 63b.59b. diiodine hexachloride 61a. chlorous acid 65b. Fe 2 SO 4 3 63c. sulfurous acid 65c. chlorine pentafluoride 59d. oxygen difluoride. nitrous acid 67a. Co NO3 2 61g. pentanoic acid.0 mg Kr 99. Integrative and Advanced Exercises 83. 2-butanol. 2-chlorohexane.3 g CuSO 4 73. 22. CH3(CH2)3CHClCH3. 67c. CH3CH2CH2CH(CH3)CH2OH 79e.0 % . Molecular mass = 120.05 ppb 96. and (d) are structural isomers. It is not possible to have less than 1 molecule of S8. CH4 93. 81c. 426 yoctograms . MgCl2 6H 2 O 71. 81d.500 kg mass is required. 81b.0 %.6 u. 89. CH3(CH2)5CH3 79b. 2-methyl-1-propanol. methanol. (c). Molecular mass = 32. xenon difluoride. CH3CH2CO2H 79c. Molecules (a). ionic. 95. 79a. % H2O = 15. CH3CH(CH3)CH2OH. Molecular mass = 74. (b). 0.12 u. CH3OH.1 u.24 10 23 Li 6 atoms 86.9 g/mol 103. ZnSO4·7H2O Copyright © 2011 Pearson Canada Inc. 69. % H 2SO 4 = 85. 91. CuSiF6 6H2O 75. CH3CH2F 81a. 4. 26. 67d. ionic. 77. 1. not ionic.04 u. ammonium hydrogen phosphate. 7 .67b. copper (II) sulfite. Answer is (b). Molecular mass = 102. CH3(CH2)3CO2H. An additional 2. The answer is (d). 111c. 126. 5. I2Cl6 Feature Problems 111a. The answer is (c). % K = 4.45% O 130b. (ii) 53.8 1023 molecules per mole of oleic acid Self-Assessment Exercises 118.7 103 m2 113b.795% H. 8.106.15% . The answer is (a). 125. 120. 123. 111d. % Cu = 57.6% P2O5 . C12H8Cl6O 110.7% P2O5 . 113a.7-44. 2. 111b. 121. The answer is (c). 719. 4. % P = 4. The answer is (d). 124. The answer is (d). A “17. C13H18O2 Copyright © 2011 Pearson Canada Inc. 15. 119. 8 .5 nm 113c.5” fertilizer. The answer is (b). 127.67% C. The answer is (b). The answer is (c). It is impossible to make a “5-10-5” fertilizer if the only fertilizing components are (NH4)2HPO4 and KCl. C8H9NO2 130a.6-10. Na2SO3·7 H2O 128a.5 g 129. (i) 60. 122.46% 128b.37% . 75. 2 C7H6O2S(l) + 17 O2 (g) → 14 CO2(g) + 6 H2O(l) + 2 SO2(g) 3a.4 mg H 2 ( g ) 7b. 0. 5.15g CO 2 8a. 126 g H 2 5a. (a) 4 NH3(g) + 7 O2(g) → 4 NO2(g) + 6 H2O(g) (b) 6 NO2(g) + 8 NH3(g) → 7 N2(g) + 12 H2O(g) 2a. 18.63 mol Ag 4a.29 g Mg 3 N 2 4b. 0. 115g NaNO3 9b.1 mL Copyright © 2011 Pearson Canada Inc.64 mol O2 3b. 0. 3.9 g Na 2SO 4 ⋅10H 2 O 10a. 0. 50. (a) 2 H3PO4(aq) + 3 CaO(s) → Ca3(PO4)2(aq) + 3 H2O(l) (b) C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) 1b.Chapter 4 Answers Practice Examples 1a.50 g O 2 ( g ) 6a. 4 HgS(s) + 4 CaO(s) → 3 CaS(s) + CaSO4(s) + 4 Hg(l) 2b. 8.34 cm3 alloy 6b. 2.307 M 8b.122 M 11a.710 g NH 3 5b.524 M 9a. 0. 1 . 3. 0. 0.0675 M 10b.79 g Cu 7a. 0. 14b.7% 15a. 2. 6 S2 Cl2 + 16 NH 3 → N 4 S4 + 12 NH 4 Cl +S8 Copyright © 2011 Pearson Canada Inc. PCl3 + 3 H 2 O → H 3 PO3 + 3 HCl 3a. 2 SO3 → 2 SO 2 + O 2 1b. 936g PCl 3 12b. 4.8 g P4 13b.6 % yield . 19. (b) 25.8g CO 2 15b. 2 FeSO 4 → Fe 2 O3 + 2 SO 2 + 1 2 O 2 or 4 FeSO 4 → 2 Fe 2 O3 + 4 SO 2 + O 2 3c.0 g impure C6 H11OH 16a. (a) 30. 3 PbO + 2 NH 3 → 3 Pb + N 2 + 3 H 2 O 3b. Approximately 56 g of SiO2 and 30 g of KNO3 are needed. 3 NO 2 + H 2 O → 2 HNO3 + NO 1d. 1. 83. 57 g O 2 remaining 14a.7 g CH2O (g). 41.47 ×103 g HNO3 16b. 69.11b.86 kg POCl3 13a. 2 . 17a.96 g Ag2CrO4 12a. 83 wt% 17b. Cl2 O7 + H 2 O → 2 HClO 4 1c. 78. (c) 85.3% yield B.47% Integrative Example A.4 % Zn Exercises 1a. 3.0 g CH 2 O . 93. 284 g CaH 2 23. 2 N2H4(g) + N2O4(g) → 4 H2O(g) + 3 N2(g) 13. 3 . NH 4 NO3 ( s ) 9b. 0. 2 C3 H 7 CHOHCH(C2 H 5 )CH 2 OH + 23 O 2 →16 CO 2 +18 H 2 O 5a.3d. CH 3 CH(OH)COOH ( s ) + 3O 2 ( g ) → 3CO 2 ( g ) + 3H 2 O ( l ) ∆ → N2 O ( g ) + 2 H2 O ( g ) 9a.402 mol O 2 17b. 128 g KClO3 17c. 43.9 g KCl 19. 2 NO ( g ) + O 2 ( g ) → 2 NO 2 ( g ) 5c. 2 CH 4 ( g ) + 2 NH 3 ( g ) + 3O 2 ( g ) → 2 HCN ( g ) + 6 H 2 O ( g ) 11. 2 CH 3 CH(OH)CH 3 (l) + 9 O 2 ( g ) → 6 CO 2 ( g ) + 8 H 2 O ( l ) 7c. 12.7% Fe 2 O3 Copyright © 2011 Pearson Canada Inc. 79. 96. Ag 2 SO 4 ( aq ) + BaI 2 ( aq ) → BaSO 4 ( s ) + 2 AgI ( s ) 7a. 2 Mg ( s ) + O 2 ( g ) → 2 MgO ( s ) 5b. 2 C4 H10 (l) +13O 2 ( g ) → 8CO 2 ( g ) +10 H 2 O ( l ) 7b. 2 C2 H 6 ( g ) + 7 O 2 ( g ) → 4 CO 2 ( g ) + 6 H 2 O ( l ) 5d. 48.0 g Ag 2 CO3 21a. 2 Cr(s) + 3 O2(g) → 2 CrO3(s) 15. 785 g 17a.1 g H 2 O 21c.2 g H 2 21b. Na 2 CO3 ( aq ) + 2 HCl ( aq ) → 2 NaCl ( aq ) + H 2 O ( l ) + CO 2 ( g ) 9c. 408 M 31b. The ratio of the volume of the volumetric flask to that of the pipet would be 20:1. 4 . Al produces the largest amount of H 2 per gram of metal.0-mL flask and a 50. 0. or a 500.46 ×10−3 mol HNO3 L Copyright © 2011 Pearson Canada Inc.8% 27.0mL flask and a 25.0820 M 45. % by mass B10 H14 = 25.154 M 31c.1mL CH 3OH 37a.25. 59. 0. 0. 49a.177 g Na 2S 49b.03g H 2 29.753 M 33b.4 mL K 2 CrO 4 53. 1.00-mL pipet. 0.236 M 47. We could use a 100. 8.562 g Ag 2S 51. 0. 1. 31a.320 M CO(NH 2 ) 2 33c. Solution (d) is a 0. 4. 0. 0.206 M 35a.73g 35b.00-mL pipet. Molarity = 4.7 mmol C6 H12 O6 L 37b. 44.500 M KCl solution. 0. 1. a 1000. The 46% by mass sucrose solution is the more concentrated. 4.00-mL pipet. There are many combinations that could be used.53M 33a. 41.0-mL flask and a 5.7 ×10−3 mol C6 H12 O6 L 39. 43. 0386 g C2 H 6 87. a 100% yield is required. 87. A main criterion for choosing a synthesis reaction is how economically it can be run.8 g Ag 2 CrO 4 59.00 moles of NO (g) 65. Therefore.1% yield 75. 5 .91 M AlCl3 57. 86. 0. In the analysis of a compound.51 g CrSO 4 73a.624g Na 61. 1. 89a. 0. 2. 211. 1. 0. 10. leaving some Cu unreacted. 12. 143g Na 2 CS3 69a.80 mol CCl4 73b. 1. ∆ SiO2(s) + 2 C(s) → Si(s) + 2 CO(g) Si(s) + 2 Cl2(g) → SiCl4(l) SiCl4(l) + 2 H2(g) → Si(s. on the other hand.55 mol CCl2 F2 73c. it will be completely consumed. 10. 3. HNO3 (aq) is the limiting reactant. 67. it is essential that all of the material present be detected. 1. ultrapure) + 4 HCl(g) Copyright © 2011 Pearson Canada Inc.22 ×103 g CO 2 86. none of the material present in the sample can be lost during the analysis.0693mol AlCl3 55b. 0.5g NH 3 69b.34 kg AgNO3 per kg of I2 produced.55a.1 g excess Ca ( OH )2 71. 84.2649 M 63.6% 81. 91.8% CH 3CH 2 OH .96 ×104 g LiOH 96. Cu 5 (CrO 4 ) 2 (OH)6 5 CuSO 4 (aq) + 2 K 2 CrO 4 (aq) + 6 H 2 O (l) 116b.2 cm 2 106. 0. 90.89b. 119. 115. ↓ Cu 5 (CrO 4 ) 2 (OH)6 (s) + 2 K 2SO 4 (aq) + 3 H 2SO 4 (aq) 117. 144 g H 2 . CaCO 3 (s) → CaO(s) + CO 2 (g) ∆ 93b. 0. 4 SO 2 (g) + 2 Na 2 S(aq) + Na 2 CO 3 (aq) → CO 2 (g) + 3 Na 2 S 2 O 3 (aq) 95. 68.2 g Fe .95 % 116a. C3 H 4 O 4 (l) + 2 O 2 (g) → 3 CO 2 (g) + 2 H 2 O(l) 118. 0. 16. 143 mL 105. C 3 H 8 (g) + 3 H 2 O(g) → 3 CO(g) + 7 H 2 (g) 93d. 73.88 M CH 3CH 2 OH 101. 885 g C .33 % Integrative and Advanced Exercises ∆ 93a. 5. Both “a” and “b” are consistent with the stoichiometry of this equation.2% ( C2 H 5 )2 O . mass of excess Fe 2 O 3 = 2. The balanced equation for the reaction is: S 8 (s) + 4 Cl 2 (g) → 4 S 2 Cl 2 (l). Copyright © 2011 Pearson Canada Inc. 8.05 × 103 g Cl2 .1 g. 1. 9. 6 . 3 FeS + 5 O2 → Fe3O4 + 3 SO2 99. 2 ZnS(s) + 3 O 2 (g) → 2 ZnO(s) + 2 SO 2 (g) 93c.365g Zn 114. 5.850 g AgNO3 120.0% CaCO3 98. 142. The answer is (a). 137. 136. 2 C 3 H 6 (g) + 2 NH 3 (g) + 3 O 2 (g) → 2 C 3 H 3 N(l) + 6 H 2 O(l) 127b. A total of 374 mL may be prepared this way. The answer is (b).121. The answer is (a). The answer is (d). 2 C 8 H 18 + 25 O 2 → 16 CO 2 + 18 H 2 O 143b.25 g C3 N 3 (OH)3 123. Dolomite is the compound. 555 kg NH 3 Self-Assessment Exercises 135. 145. The answer is (b). 7 . 138. We must use all of the most concentrated solution and dilute this solution down using the next most concentrated solution. 1. 2 C8H18 + 25 O2 → 12 CO2 + 4 CO + 18 H2O 144. 143a. 141. Copyright © 2011 Pearson Canada Inc. 140. The answer is (c). The answer is (d). The answer is (d). The answer is (c). 139. 146. The answer is (d). 127a. .540 M Cl− 1b. Vanadium is oxidized. S ( s ) + 2 OH − ( aq ) + 2 OCl− ( aq ) → SO3 Copyright © 2011 Pearson Canada Inc. (a) 7. (a) Al3+ ( aq ) + 3 OH − ( aq ) → Al ( OH )3 ( s ) (b) No reaction occurs. 2− ( aq ) + H 2O(l) + 2 Cl − ( aq ) 1 .9 × 10-5 M F. manganese is reduced. (b) 3. 4b. 5a. 0. 3 UO 2+ ( aq ) + Cr2 O7 2− ( aq ) + 8 H + ( aq ) → 3 UO2 2+ ( aq ) + 2 Cr 3+ ( aq ) + 4 H 2O(l) 7a. Oxidation : Reduction: {Al ( s ) → Al ( aq ) + 3 e } × 2 {2 H ( aq ) + 2 e → H ( g )} × 3 − 3+ − + 2 Net equation : 2 Al ( s ) + 6 H + ( aq ) → 2 Al3+ ( aq ) + 3 H 2 ( g ) 5b. MnO 4 − ( aq ) + 8 H + ( aq ) + 5 Fe 2+ ( aq ) → Mn 2+ ( aq ) + 4 H 2 O(l) + 5 Fe3+ ( aq ) 6b. (a) Al3+ ( aq ) + PO 4 3− (b) Ba 2+ ( aq ) + SO 4 (c) Pb 2+ ( aq ) + CO3 ( aq ) → AlPO 4 ( s ) 2− ( aq ) → BaSO 4 ( s ) 2− ( aq ) → PbCO3 ( s ) 3a. (b) This is an oxidation–reduction reaction. + − NH 3 ( aq ) + HC3 H 5 O 2 ( aq ) → NH 4 ( aq ) + C3 H 5 O 2 ( aq ) 3b.1 kg CaF2 2a. The acid and base react to form a salt solution of ammonium propionate. (c) Pb 2+ ( aq ) + 2 I − ( aq ) → PbI 2 ( s ) 2b. (a) It is not an oxidation–reduction reaction.Chapter 5 Answers Practice Examples 1a. Oxidation : 2 Br − ( aq ) → Br2 ( l ) + 2 e − Reduction: Cl2 ( g ) + 2 e − → 2 Cl− ( aq ) Net equation : 2 Br − ( aq ) + Cl2 ( g ) → Br2 ( l ) + 2 Cl− ( aq ) 6a. CaCO3 ( s ) + 2 HC2 H 3 O 2 ( aq ) → CO 2 ( g ) + H 2 O ( l ) + Ca 2+ ( aq ) + 2 C 2 H 3 O 2 − ( aq ) 4a. 238 M K + 7b.4% Fe 10b. O has been reduced and thus. 0. Propionic acid: weak electrolyte 5c.03129 M Integrative Example A. The oxidation state of the element N decreases during this reaction. Nonelectrolyte.1019 M 9b.89% B. Ammonia: weak electrolyte 7a. Strong electrolyte 1d. O2(g) (oxidation state = 0) is the oxidizing agent. Strong electrolyte 1c. 0. thus. 65. 2 MnO 4 − ( aq ) + 3SO32− ( aq ) + H 2O(l) → 2 MnO 2 ( s ) + 3SO 4 2− ( aq ) + 2 OH − ( aq ) 8a.130 M 10a. Since the oxidation state of H is 0 in H2 (g) and is +1 in both NH3(g) and H2O(g). 9a.166 M Al3+ Copyright © 2011 Pearson Canada Inc. hydrogen is oxidized. Strong electrolyte 3. 5a. Au has been oxidized and. 0. 1. Weak electrolyte 1b.32% Exercises 1a. 0. Barium bromide: strong electrolyte 5b. 0. 0. 1e. A solution of both HCl and HC2H3O2 will yield similar results. meaning that NO2 (g) is reduced. Au(s) (oxidization state = 0). HCl is practically 100% dissociated into ions.334 M NO3− 7c. 49. The apparatus should light up brightly. 8b. The substance that is reduced is called the oxidizing agent. A substance that is oxidized is called a reducing agent.7b. is the reducing agent. 2 . 39 ×10−3 M K + 11c. Na 2 CO3 ( s ) dissolves. H 2 O ( l ) . BaCl2 ( aq ) + K 2SO 4 ( aq ) : Ba 2+ (aq) + SO 4 2− (aq) → BaSO 4 (s) 27a. 3. No reaction occurs (all are spectator ions). 4.1 mg K+ per mL gives the largest K+ of the three solutions.54 ×10−4 M Ca 2+ 11b. FeS ( s ) + 2 H + ( aq ) → H 2 S ( g ) + Fe 2+ ( aq ) Copyright © 2011 Pearson Canada Inc. OH − ( aq ) + HC2 H 3 O 2 ( aq ) → H 2 O ( l ) + C 2 H 3 O 2 − ( aq ) 27b. 15. 3. 25a. while Cu(NO3)2 (s) will dissolve. No reaction occurs.04 ×10-3 M OH 11a. 8. BaSO 4 ( s ) will form and MgSO4 will not precipitate.44 ×10−3 M Zn 2+ 13. 3 .732 M 19a. 0. This is the physical mixing of two acids. Mg ( NO3 )2 ( aq ) + NaOH ( aq ) : Mg 2+ ( aq ) + 2 OH − ( aq ) → Mg ( OH )2 ( s ) 25c. No reaction occurs.627 M Na + 9.7d.3 ×102 mg MgI 2 17. Pb 2+ ( aq ) + 2 Br − ( aq ) → PbBr2 ( s ) 19b. Sr ( NO3 )2 ( aq ) + K 2SO 4 ( aq ) : Sr 2+ ( aq ) + SO 4 2− ( aq ) → SrSO 4 ( s ) 25b. 0. 23c. but MgCO3 (s) will not dissolve (appreciably). 19c. 21b. 3Cu 2+ ( aq ) + 2 PO 43− ( aq ) → Cu 3 ( PO 4 )2 ( s ) 23a. Fe3+ ( aq ) + 3 OH − ( aq ) → Fe ( OH )3 ( s ) 21a. 23b. Cu 2+ ( aq ) + CO32− ( aq ) → CuCO3 ( s ) 21c. 27c. Add KCl(aq). The solution containing 8. Add K 2 SO 4 ( aq ) . 3. AgCl(s) will form. As a salt: NaHSO 4 ( aq ) → Na + ( aq ) + HSO 4 As an acid: HSO 4 − ( aq ) + ( aq ) 2− H 2 O ( l ) + SO 4 ( aq ) OH − ( aq ) → − 31. that of O is −2 . 33c. Fe 2+ ( aq ) + VO 4 3− ( aq ) + 6 H + ( aq ) → Fe3+ ( aq ) + VO 2+ ( aq ) + 3 H 2 O ( l ) 37d. and that of Mg is +2 on each side of this equation. On both sides of the equation the O. of H is +1 . Thus. 3 N 2 H 4 ( l ) + 2 BrO3 ( aq ) → 3 N 2 ( g ) + 2 Br − ( aq ) + 6 H 2 O ( l ) − 37c.+ 4 H2O(l) 3 Ag+(aq) + Cr(OH)3(s) + 5 OH- Copyright © 2011 Pearson Canada Inc. 33b. 4 . and that of Cr is +6 . 33d. This is not a redox equation.27d.S. NH3(aq). 33a. Al ( s ) + 4 OH − ( aq ) → Al ( OH )4 − ( aq ) + 3 e− 37a. This is a redox reaction. The O. HCO3 − ( aq ) + H + ( aq ) → "H 2 CO3 ( aq ) " → H 2 O ( l ) + CO 2 ( g ) 27e.ions necessary to form Mg(OH)2(s). 2 MnO 2 ( s ) + ClO3 ( aq ) + 2 OH − ( aq ) → 2MnO 4 ( aq ) + Cl− ( aq ) + H 2 O(l) − − 39b.S. that of C is +4 . This is a redox reaction. The O. The O. The O. this is not a redox equation. Use (b). of Ag is 0 on the left and +1 on the right side of this equation. Mg ( s ) + 2 H + ( aq ) → Mg 2+ ( aq ) + H 2 ( g ) 29. that of Ag is +1 . of N is +5 on the left and +4 on the right side of this equation. of Cl is 0 on the left and −1 on the right side of this equation. The O.S. 2SO3 2− ( aq ) + 6 H + ( aq ) + 4 e − → S2 O3 2− ( aq ) + 3 H 2 O(l) 35b. 6 ClO 2 (aq) + 6 OH − ( aq ) → 5ClO3 ( aq ) + Cl− ( aq ) + 3 H 2 O − 39d.S.S. of O is −2 . 35a. 3 Ag(s) + CrO42.S. 2 Fe ( OH )3 ( s ) + 3 OCl− ( aq ) + 4 OH − ( aq ) → 2FeO 4 2− ( aq ) + 3 Cl − ( aq ) + 5 H 2 O(l) 39c. 3 UO 2+ ( aq ) + 2 NO3− ( aq ) + 2 H + ( aq ) → 3 UO 2 2+ ( aq ) + 2 NO ( g ) + H 2 O ( l ) 39a. 10 I − ( aq ) + 2 MnO 4 − ( aq ) +16 H + ( aq ) → 5 I 2 ( s ) + 2 Mn 2+ ( aq ) + 8 H 2O ( l ) 37b. NH3 affords the OH. 2 NO3 ( aq ) +10 H + ( aq ) + 8 e − → N 2 O ( g ) + 5 H 2 O ( l ) − 35c. of Br is −1 on the left and 0 on the right side of this equation. SO3 2− ( aq ) is the reducing agent. 1. 108 ppm Mg Copyright © 2011 Pearson Canada Inc. 13. 3 Cl2 ( g ) + 6 OH − ( aq ) → 5 Cl− ( aq ) + ClO3 ( aq ) + 3 H 2 O(l) − 41b. 47b.(aq) + 4 OH.546 mL KOH solution 53. 5 NO 2 ( aq ) + 2 MnO 4 ( aq ) + 6 H + ( aq ) → 5 NO3 ( aq ) + 2 Mn 2+ ( aq ) + 3 H 2O ( l ) − − − 43b.(aq) → 5 MnO2 (s) + 2 H2O (l) 43c. Acidic 59. 10 NH3(g) +3 Cl2O(g) → 6 NH4Cl(s) + 2 N2(g) + 3 H2O(l) 47a. 3.3 mL NaOH ( aq ) soln 51. Fe ( CN )6 4− ( aq ) is the reducing agent. 49.0 g Na 2 C2 O 4 Integrative and Advanced Exercises 2− 71.41a. 2 S2 O 4 2− ( aq ) + H 2 O(l) → 2 HSO3 ( aq ) + S2 O3 − 2− ( aq ) 43a. 63.1230 M NaOH 55. 3 Ca 2 + (aq) + 2 HPO 4 (aq) → Ca 3 (PO 4 ) 2 (s) + 2 H + (aq) 74. 53. MnO 4 − ( aq ) is the oxidizing agent. 34 mL base 61.23% Fe 67.077 M NaOH 57. 0. 5 . 37. NO 2 ( g ) is the oxidizing agent. 3 Mn2+ (aq) + 2 MnO4. 47c. H 2 O 2 ( aq ) is the oxidizing agent. 0. Answer is (d). Cr2 O7 2− ( aq ) + 8 H + ( aq ) + 3 C2 H 5 OH → 2 Cr 3+ ( aq ) + 7 H 2 O ( l ) + 3 CH 3 CHO 45a. H 2 ( g ) is the reducing agent.968 ×10-2 M 65. 16 H2S(g) + 8 SO2(g) → 3 S8(s) + 16 H2O(g) 45c. CH4(g) + 4 NO(g) → 2 N2(g) + CO2(g) + 2 H2O(g) 45b. 75. 45. The answer is (c).03 95. 1. The answer is (d).6. 2I.00 L 89. 0. %Al(OH)3 = 78. 5.4. 3 Zn 2+ + PO34− → Zn 3 (PO 4 ) 2 (s) Copyright © 2011 Pearson Canada Inc. 44.+ Pb 2+ → PbI 2 (s) 107.4 g CaCO3 83. CaO(s) + H2O(l) → Ca2+(aq) + 2 OH−(aq) H2PO4−(aq) + 2 OH−(aq) → PO43−(aq) + 2 H2O(l) HPO4−(aq) + OH−(aq) → PO43−(aq) + H2O(l) 5 Ca2+(aq) + 3 PO43−(aq) + OH−(aq) → Ca5(PO4)3OH(s) 93b. CO 23 − + 2H + → H 2 O (l) + CO 2 (g) 108a.4346 % 93a. 105. 0. 91.0 ×102 g ClO 2 (g) 88a. x = 1.0% MnO 2 97. 106. 0.0874 L 80a. Before the breath test: 8×10-4 M. The answer is (a). 4 FeS2(s) + 15 O2(g) + 2 H2O(l) → 4 Fe3+(aq) + 8 SO42−(aq) + 4 H+(aq) 80b. 103.6 g Cl2 85. After the breath test 3 × 10−4 mol/L. 6 . Self-Assessment Exercises 102. 23. % Mg(OH)2 = 21. The answer is (b).9 g 88b. 104. 91.302 kg Feature Problems 94. Yes 114c. Cu 2+ + 2 OH − → Cu(OH) 2 (s) 108c. Loses electrons: NO 110. False 113d. 113a. No Copyright © 2011 Pearson Canada Inc. The answer is (b).108b. No 114b. Reducing agent: NO 109e. False 113b. False 113e. Species reduced: O2 109c. 112. Ni 2+ + CO32− → NiCO3 (s) 109a. Oxidizing agent: O2 109d. 111. Species oxidized: N in NO 109b. True 114a. Yes 114d. True 113c. The answer is (d). 7 . The answer is (a). Gains electrons: O2 109f. 0 g/mol .378 g O 2 7a. 5. 2. 24.8g/mol) the density of He is only about one seventh as much.16 g/L. 30. Thus.25 L O 2 g 10b.35 kg. 93mm glycerol 3a.Chapter 6 Answers Practice Examples 1a.11mL 6b. 0. When compared to the density of air under the same conditions (1.5atm Copyright © 2011 Pearson Canada Inc. 32. 86. 760.162 g/L .0 mmHg 2b.0 g/mol 9a. 464 K 5a. 0. 2. 1. 5. 8a. 150. 0. L NH 3 11a. 1.4 L NH 3 4b.mmHg 1b.6 g NaN3 9b. based on the “average molar mass of air”=28. It is not necessary to add a mass with the same cross sectional area. 35. 4a. 13atm 11b. 8b.59 × 1014 molecules N2 6a.13 g/cm 3 2a. 756. 1 .619 g Na(l) 10a. This answer is in good agreement with the molar mass of NO. helium is less dense (“lighter”) than air. 139 torr 3b.4 g/mol 7b. 1.11mol He 5b. 2. 52.0 mmHg.75 K 15a.17 atm 1d. C3H4.27 mmHg.12a. 661m/s . 1.27 L Copyright © 2011 Pearson Canada Inc. NH3 (g).968 atm 1b.1104 mol O2 15b.3 s bg 17b. 13a.03 kg cm-2 9a. B. Cl 2 g 2 Integrative Example A.90 102 g/mol 16b.47 atm CO 2 (g) .766 atm 1c. There are three possible Lewis structures. 2. 0. 76. Ar = 7. 7. 52. 1. 11.4 m benzene 5. C4H9NO3 Exercises 1a. O 2 pressure = 157 mmHg. 28. CO 2 = 0. 2.22 atm 3. 0.8 L 9b. N 2 = 584 mmHg. 0. 0. Cl bgg 17a. 2 .2 s 16a. 710 mm Hg 7. 14b.0348atm H 2 O(g) .00278 mol HCl 13b.382 L 14a. 12b. 1. 0. 23. 0. The density of CO2 is 1. 15. P4 47. no gas escapes. the bag is virtually leak proof. 24. 41.32 1020 molecules 21. 103 g/mol 37.1 10 11 Pa 33a. At the higher elevation of the mountains. Thus.89 g 27.18 g/L air 43b.1 107 L SO 2 Copyright © 2011 Pearson Canada Inc.132 g Ar 19a. 2. 31.56 L/mol 43a. 4.3 mg PH3 19b.4 L·mol-1 33b. 5. the balloon will not rise in air. 41. 7.30 L 25. 378 L O 2 49.80 g/L CO 2 . However. 3 . Since this density is greater than that of air. SF4 39.11. the gas inside the bag expands in the lower pressure until the bag is filled to near bursting.32 104 mL 29. 1. 255 K 17. 4. 225 K 13. the atmospheric pressure is lower than at the beach. 50. 3.6 atm 15.3 L·mol-1 35. 45. 55. 702 g Kr 31.8 g/mol. The formula is C 4 H 8 . as the volume has increased by 50%.05 83c. 0. 63.5% . Copyright © 2011 Pearson Canada Inc.0 atm 67. 65. 0. 4 .004 85. Pideal is off by 0.5 mmHg. 842 mmHg 61b. Pbenzene 89.83 u. 1. 2.35 atm.3atm .0 atm or +5.51 103 K 79.80 h 87a. Pideal 27. 17. 77.07 83b. 61a. 7.17 1021 J/molecule 81. 4. 751 mmHg 71.978 83d.2 atm or +8. 9.9% 73.37 L H 2 (g) 69. 326 m / s 75.8atm. 5. 1. Pideal 15. Pvdw 14.06 103 g Ne 59. 87b. 1.83 g/mol or 7. Pideal is off by 1. 87c. Pvdw = 18.6 atm .9% KClO3 53.00473 mol NO 2 83a. 73. Pvdw = 26.59 L gas 57.51. PAr 752 mmHg. The answer is (d).4 atm . Pideal is off by 1.5% .8 atm or +3. Situation (b) best represents the resulting mixture.7 L H 2 55. 10. 2.0% . 1. Pideal 19. 6.1atm. 13.39 atm 101. 19°C 118a. 3 121b. 97.25 atm 98.40 g/L 112b. Flask 2 n =0.92 103 119.609.046 104 115.42% by number 111c. 3. 31 mmHg 111a. 153 mmHg 102. 3.6 km 2 n 3 ab RT bP 2 n a V 0 V n 0 V P P P 121a. 5 .070×104 L 109. 2. O3 8.95 % H 2 O by mass 112a. 0. 0.37 L Copyright © 2011 Pearson Canada Inc. 542 L 99.42% H 2 O by volume 111b. 3. 482 m/s 118b.9% He 105. 23 L 108.89. 2.95 107 pm3 / He atom Integrative and Advanced Exercises 94.391. 7. Fu 1. Flask 1 n = 0. 19. 2. 1.9 atm 114. False 139b.3 L of CO remain. Just over 30 km. 134. The answer is (c). The answer is (d).Feature Problems 126. 141. bg 127c. 142. The number of atoms of X (oxygen) in each compound is : Nitryl Fluoride = 2 atoms of O. 143. True 139e. False 139d. 1. 546 K 136. False 140. The atomic mass of X is 16 u which corresponds to the element oxygen. 129. Magnesium will react with molecular nitrogen but not with Ar. 6 . The answer is (b). Copyright © 2011 Pearson Canada Inc. The answer is (d). bg 127b. 135. 138.50%. Thus. The answer is (c). The answer is (a).95 g/mol]. The answer is (c). Sulfuryl Fluoride = 2 atoms of O. bg 127a. Self-Assessment Exercises 133. Because of the presence of Ar(g) [39. The densities differ by 0. magnesium reacts with all the nitrogen in the mixture. The answer is (b). True 139c. The answer is (a). 127d. Nitrosyl Fluoride = 1 atom of O. the N 2 g [28.01 g/mol] from liquid air will have a greater density than N 2 g from nitrogen compounds. The N 2 g extracted from liquid air has some Ar(g) mixed in. 139a. Thionyl Fluoride = 1 atom of O. 145. 137. but leaves the relatively inert Ar(g) unreacted. 553 marks for O2. h is inversely proportional to D.146a. The height. and 25 for Ar. the larger the diameter of the tube. C3H8 has higher a and b values. 147. 7 . For every single mark representing CO2. 146c. 146b. the shorter the height of the liquid. Ne has higher a and b values. That is. Copyright © 2011 Pearson Canada Inc. C4H10 149. 148. Cl2 has higher a and b values. we need 2060 marks for N2. −2.9 C 3a. +1.83 ×103 kJ/mol 3b. 7a. -1367 kJ 11b.6 kJ 6a.0 ×102 g 2b.89 kJ 2a. 3. −124 kJ 9b. 1 . 4. −65. +14.11 kJ = +92. 60. 114 J 5b. 0.7 kJ 1b.70 ×102 J 6b. 32. −2006 kJ 10a. 33. +3. 179 J of work is done by the system to the surroundings. -1273 kJ/mol C6H12O6(s) Copyright © 2011 Pearson Canada Inc. 1.6 g C12 H 22 O11 7b. 37.31 kJ 8b.4 C 5a. −3.15 kJ heat evolved 8a.72 kg H2O 9a.Chapter 7 Answers Practice Examples 1a. 6.22 kJ bg b g 11a.26 kJ / C 4a. kJ/mol 4b. and its enthalpy change is minus two times the enthalpy of formation of NH 3 g : −2 × −46. 6 C ( graphite ) + 132 H 2 ( g ) + O 2 ( g ) + 12 N 2 ( g ) → C6 H13O 2 N ( s ) 10b.5 ×103 kJ/mole of mixture 12a. The specified reaction is twice the reverse of the formation reaction. 385 J g −1 C−1 for Zn 3b.905 J g −1 C−1 for Al 5. Integrative Example A. 1. 1. −6. +2.21 J/K 13. -125. 65.1 g of Ca(OH)2 and 66.3 g of H2O.45 ×103 kJ 17a.59 kJ 15b.49 × 105 kJ of heat evolved. 3. 0. 545 °C 7.65 ×103 kJ 15c. 0. -184 kJ/mol CH3OCH3(g) 13a.90 ×103 L H 2 O 19.0 C 9. 0.5 kJ/mol B. −112.8 kJ/mol Ag2CO3(s) formed. 15a.13 J g −1 C−1 for Pt 3c. -505.9 ×102 kJ 1b. Exercises 1a.20 × 105 kJ of heat energy 17c. 504 kg CH 4 17b. 2 .3 × 102 J mol−1 C−1 11.36 × 103 kJ heat Copyright © 2011 Pearson Canada Inc. 2. 2.12b. -177 kJ 3a. 2. 24.3 kJ/mol AgI(s) formed 13b. The contents of the vessel after completion of the reaction are 74. 1. Doubling the amount of KOH should give a temperature change known to two significant figures (1.6 g 31. 23. hence no work is done. 5.0 °C). 0.21a.2 × 102 g NH 4 Cl 25. -83 cal 49. -3. 51c.34 × 103 kJ 39a.7 ×102 kJ heat evolved . Work is done on the system by the surroundings (compression). When the Ne(g) sample expands into an evacuated vessel it does not push aside any matter.4 L atm 47b. 3 . 39. −2. -3. No pressure-volume work is done. Work is done on the system by the surroundings (compression).83 g H2O 33.34 ×103 kJ/mol C5 H10 O5 37b. The ∆T here is known to just one significant figure (0. 15 g 47a. −56 kJ/mol 27. 7.2. Copyright © 2011 Pearson Canada Inc. C5 H10 O5 ( g ) + 5O 2 ( g ) → 5CO 2 ( g ) + 5H 2 O ( l ) ∆H = .5 × 102 J 47c. 1.23 C 45.111 L 35. 51b. 29.72 kJ/ °C 43.98 kJ / o C 37a. 4. 4. 51a.6 °C) and using twenty times the mass of KOH should give a temperature change known to three significant figures (16. +18 kJ/mol 41. −5 ×101 kJ/mol 21b. 2.9 °C). Endothermic 39b. The temperature of the gas stays the same if the process is isothermal.53.5 kJ 73. ∆U for the gas must equal zero by definition (temperature is not changing). 59. the answer is (c). -545 J 63. −1366. −55.7 kJ 79b. 3. -562 J 55c. According to the First Law of Thermodynamics. ∆H = −290. +30. +202. 4 . -2012 kJ/mol 67.8 kJ 77. 0.74 kJ 69. or w = q. 65a. 0 55b. ∆H = −747. This situation is impossible. 57d. Yes 57b. kJ 71. 61.0 kJ/mol 83. −102. not w = -2q.08 kJ 57a. Yes 57c. −206.7 kJ 87. -2008 kJ/mol 65b.9 kJ/mol 89. ∆H = −217. ∆H° = -120.21 L 55a. ∆H = −35. kJ 79a. An ideal gas expanding isothermally means that ∆U = 0 = q + w.4 kJ 85. −55 kJ Copyright © 2011 Pearson Canada Inc.5 kJ 75. −1124 kJ 81. 91. 2.40×106 kJ 93. −424 kJ = ∆ H f HCOOH ( s ) Integrative and Advanced Exercises 96. 3.5 °C . This large a temperature rise is unlikely, as some of the kinetic energy will be converted into forms other than heat, such as sound and the fracturing of the object along with the surface it strikes. In addition, some heat energy would be transferred to the surface. 97. −1.95 ×103 kJ/mol C6 H8O7 103. The sewage gas. 107. 303g C6 H12 O6 109. The gas mixture contains 87.0% CH 4 and 13.0% C 2 H 6 , both by volume. 112. ΔH = −1.45 ×103 kJ/mol, 3 O 2 (g) + C2 H 6 O(g) → 2 CO 2 ( g ) + 3 H 2 O(l) . 113. 69 seconds 114. −12 J Feature Problems ∆ T (oC) 122a. The equivalence point occurs with 45.0 mL of 1.00 M NaOH(aq) [45.0 mmol NaOH] added and 15.0 mL of 1.00 M citric acid [15.0 mmol citric acid]. Plot of ∆ T versus Volume of NaOH 9 8 7 6 5 4 3 2 1 0 0 5 10 15 20 25 30 35 40 45 50 55 60 Volume of NaOH (mL) 122b. Heat is a product of the reaction, as are chemical species (products). Products are maximized at the exact stoichiometric proportions. Since each reaction mixture has the same volume, and thus about the same mass to heat, the temperature also is a maximum at this point. 122c. H 3C6 H 5O7 ( s ) + 3OH − ( aq ) → 3H 2 O(l) + C6 H 5O7 3− ( aq ) 125a. -147 J Copyright © 2011 Pearson Canada Inc. 5 125b. 2.50 Plot of Pressure Versus Volume Pressure (atm) 2.00 1.50 1.00 0.50 0.00 1.00 1.10 1.20 1.30 1.40 1.50 1.60 Volume (L) 1.70 1.80 1.90 2.00 125c. w = −152 J 125d. The maximum work is 209 J. The minimum work would be +152 J. 125e. ∆U = 0. Because ∆U = q + w = 0, q = −w. This means that −209 J corresponds to the maximum work of compression, and −152 J corresponds to the minimum work of compression. 125f. q/T = nR ln Vf /Vi . Yes, q/T is a state function. Self-Assessment Exercises 130. The answer is (b). 131. The answer is (c). 132. The answer is (d). 133. The answer is (a). 134. The answer is (b). 135. The answer is (a). 136. 52.7 °C 137a. 2 N2 + O2 → 2 N2O 137b. S + O2 + Cl2 → SO2Cl2 137c. 2 CH3CH2COOH + 7 O2 → 6 CO2 + 6 H2O 138. -218 kJ/mol 139. Enthalpy of formation for elements (even molecular ones, such as O2 or Cl2) is by convention set to 0. While it is possible for the enthalpy of formation of a compound to be near zero, it is unlikely. Copyright © 2011 Pearson Canada Inc. 6 140. From a theoretical standpoint, one can have a situation where the ΔU < 0, but there is enough work done on the system that makes ΔH > 0. 141. A gas stove works by combustion of a flammable fuel. Once shut off, the heat source instantly disappears. An electric stove works by the principle of heat conduction. Even after the electricity is shut off to the heating coil, it takes time for the coil to cool because of its heat capacity, and therefore it continues to supply heat to the pot. 142. The answer is (a). 143. The answer is (b). Copyright © 2011 Pearson Canada Inc. 7 Chapter 8 Answers Practice Examples 1a. 4.34 ×1014 Hz 1b. 3.28 m 2a. 520 kJ/mol 2b. 4.612 ×1014 Hz (orange) , 6.662 ×1014 Hz (indigo) . 3a. Yes, This is E9 for n = 9. 3b. n = 7 4a. 486.2 nm 4b. 121.6 nm (1216 angstroms) 5a. 80.13 nm 5b. Be3+ 6a. 1.21×10−43 m 6b. 3.96 ×104 m/s 7a. 6.4 × 10-43 m 7b. 1.3 m s-1 8a. 0.167 out of 1, or 16.7%. 8b. 100 and 200 pm. 9a. 0.52 nm 9b. 150. pm 10a. Yes 10b. l =1 and 2 11a. 3p 11b. n = 3; l = 0, 1, 2; ml = -2, -1, 0, 1, 2 12a. (3,2,-2,1) (3,1,-2, ½ ) (3,0,0, ½ ) m s = 1 is incorrect. The values of m s can only be + ½ or – ½. m l = -2 is incorrect. The values of m l can be +1,0,+1 when l=1 All quantum numbers are allowed. Copyright © 2011 Pearson Canada Inc. 1 12b. (2,3,0, ½ ) (1,0,0,- ½ ) (2,-1,-1, ½ ) l= 3 is incorrect. The value for l can not be larger than n. All quantum numbers are allowed. l= -1 is incorrect. The value for l can not be negative. (2,1,1,0) (1,1,0, ½ ) (3,-1,1, ½ ) (0,0,0, - ½ ) (2,1,2, ½ ) m s = 0 is incorrect. The values of m s can only be + ½ or – ½. l= 1 is incorrect. The value for l is 0 when n=1. l= -1 is incorrect. The value for l can not be negative. n= 0 is incorrect. The value for n can not be zero. m l = 2 is incorrect. The values of m l can be +1,0,+1 when l=1. 13a. (a) and (c) are equivalent. 13b. excited state of a neutral species 14a. Ti 14b. 1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 10 4 s 2 4 p 6 4d 10 5s 2 5 p5 . Each iodine atom has ten 3d electrons and one unpaired 5p electron. 15a. 15b. 16a. (a) Tin is in the 5th period, hence, five electronic shells are filled or partially filled; (b) The 3p subshell was filled with Ar; there are six 3p electrons in an atom of Sn; (c) The electron configuration of Sn is [Kr] 4d 10 5s 2 5 p 2 . There are no 5d electrons; (d) Both of the 5p electrons are unpaired, thus there are two unpaired electrons in a Sn atom. 16b. (a) The 3d subshell was filled at Zn, thus each Y atom has ten 3d electrons; (b) Ge is in the 4 p row; each germanium atom has two 4 p electrons; (c) We would expect each Au atom to have ten 5d electrons and one 6s electron. Thus each Au atom should have one unpaired electron. Integrative Example A. (a) 73.14 pm ; (b) 17.7 K . B. The possible combinations are 1s → np → nd, for example, 1s → 3p → 5d. The frequencies of these transitions are calculated as follows: 1s → 3p :ν =2.92 ×1015 Hz 3p → 5d :ν =2.34 ×1014 Hz Copyright © 2011 Pearson Canada Inc. 2 2. 4p → 3s. Exercises 1. 5d → 2p. 486. 3. 78. The difference between the sodium atoms is that the positions of the lines will be shifted to higher frequencies by 112.11 nm 9c. 7.28 nm .816 × 10-19 J. has the greatest energy per photon.60 ×10−19 J/photon 25b. 8. 410.3 min 9a. 2. n = 7. 3 . 434. UV radiation 25a.46 nm . 4p → 1s. 19a.1 × 10-5 nm radiation. 4. 397.16 nm . 6.68 nm 3a.6 kJ/mol 13. 6.9 ≈ 8 17.46 ×10−19 J/photon 19b. False 3c.26 nm .740 × 1014 s-1 15. Copyright © 2011 Pearson Canada Inc. 4. -1.The emission spectrum will have lines representing 5d → 4p. True 5. 5d → 3p. Choice (c). 3p → 1s. n = 10 11a. and 2p → 1s. True 3b. It will not display the photoelectric effect when exposed to infrared light. 23. 656. False 3d. 2.9050 ×1014 s −1 9b.90 ×10−18 J/photon 11b.1 × 103 nm has the least amount of energy per photon. 4.08 ×105 J/mol 21. 3p → 2s. Indium will display the photoelectric effect when exposed to ultraviolet light. 4p → 2s. The electron in the hydrogen atom does not orbit at a radius of 4. 2166 nm 29c.79 ×10−35 m . λ 17 cm 51. The distance of the electron from the nucleus. 43. which is far larger than the baseball’s wavelength. 0. 8.405 × 10-20 J 31d.053 ×10−20 J 29a. Be3+ cation 39. as originally proposed. n = 2 35a. Infrared radiation. 9. once its position is known at the present.0 55. Line A is for the transition n = 5 → n = 2. and the velocity of the electron in its orbit are also exactly known. Electrons 41.4 × 107 m s-1 49. n = 4. Bohr orbits.384 ×1014 s −1 29b. 31a. 4 . No 33. 45. −6.27a. are circular. All of these exactly known quantities can’t. 31c. while Line B is for the transition n = 4 → n = 1.00 Å. 1. 37b. while orbitals can be spherical. The diameter of a nucleus approximates 10−15 m.9 nm 27b. 35b.5 × 10-10 m 31b. or shaped like four tear drops meeting at their Copyright © 2011 Pearson Canada Inc. The Bohr model implies that the position of the electron is exactly known at any time in the future. ~1 × 10-13 m 47. Line A is for the transition n = 3 → n = 1. Hydrogen atom 37a. according to the Heisenberg uncertainty principle. be known with great precision simultaneously. -3. while Line B is for the transition n = 6 → n = 2. or shaped like two tear drops or two squashed spheres. 1. 1.55 nm 53. its energy. 4d 59c. Answer (c) is the only one that is correct. The electron in a Bohr orbit has a definite trajectory. Thus. does not have a well-known position or velocity. 106 pm 3 sinθ sinφ. 10 electrons 61d. 3d Copyright © 2011 Pearson Canada Inc. y j i Y(θφ) k l a. 3p 71b. The angular part of the 2py wave function is Y(θφ)py = 67. 2s 61a. while orbitals are three-dimensional regions of space in which there is a high probability of finding electrons. 59a.0 (that is. r = ao or r = 53 pm). however. Orbits and orbitals are similar in that the radius of a Bohr orbit is comparable to the average distance of the electron from the nucleus in the corresponding wave mechanical orbital. 2 electrons 61c. 71a. The electron in an orbital. 32 electrons 61e. 5p 59b. the probability of finding a 2py electron in the xz plane is zero. this means that the entire x z plane is a node. Bohr orbits are planar pathways. For all points in the 4π xz plane φ= 0. 65. 5 . A plot of radial probability distribution versus r/ao for a H1s orbital shows a maximum at 1.points. 5 electrons 63. and since the sine of 0° is zero. 57.m h g f positive phase x b c e negative phase d 69. 1 electron 61b. Its position and velocity are known at all times. 6f 73. Configuration (b) is correct for phosphorus. [Xe]6s24f145d106p4 85d. Ground state 83d. [Kr]5s24d105p2 85e.71c. five 75g. two 79d. 79a. Excited state 83c. 6 . 5dxyz 75a. 3 79b. ten 79c. fourteen 81a. 32 77. two 75c. [Xe]6s24f145d10 85b. Pb: [Xe] 4 f 14 5d 10 6s2 6p 2 81b. [Xe]6s24f145d3 85f. 114: [Rn] 5 f 14 6d 10 7 s 2 7 p 2 83a. three 75b. [Kr]5s24d105p5 Copyright © 2011 Pearson Canada Inc. two 75f. zero 75d. Excited state 83b. two 79e. fourteen 75e. Excited state 85a. [Ar]4s2 85c. 486.2203 ×1014 Hz . 121.2 m/s Feature Problems 114. The number of lines observed in the two spectra is not the same.66 ×105 m s −1 92. Green 96. 116a. rutherfordium 87b. 1875 nm.3 nm. 1. The slope is b = −3.0 ×1020 photons sec 93.5 nm.5 nm. carbon 87c. 6. They differ in that subshell energy levels are not degenerate and their radial wave functions no longer conform to the expressions in Table 8. Copyright © 2011 Pearson Canada Inc. 4.6 nm. 116c. Self-Assessment Exercises 122. Because the shortest wavelength of visible light is 390 nm. Not an element.8 ×10−3 J 102. 102. Plot ν on the vertical axis and 1/ n 2 on the horizontal axis.02 ×10−19 J 91c. 121. 91b. 101. 97. Atomic orbitals of multi-electron atoms resemble those of the H atom in having both angular and radial nodes.1 nm. 7 . m = 3 and n = 4.5 nm and 102. 66. 7 ×102 photons/s 105. 2. the photoelectric effect for mercury cannot be obtained with visible light. Integrative and Advanced Exercises 91a.2881 × 1015 Hz and the intercept is 8. 656. 116b.1.87a.2 nm. tellurium 87e. vanadium 87d. 123. Effective nuclear charge is the amount of positive charge from the nucleus that the valence shell of the electrons actually experiences. This amount is less than the actual nuclear charge, because electrons in other shells shield the full effect. 124. The px, py and pz orbitals are triply degenerate (the are the same energy), and they have the same shape. Their difference lies in their orientation with respect to the arbitrarily assigned x, y, and z axes of the atom. 125. The difference between the 2p and 3p orbitals is that the 2p orbital has only one node (n = 2-1) which is angular, whereas the 3p orbital has two nodes, angular and radial 126. The answer is (a). 127a. Velocity of the electromagnetic radiation is fixed at the speed of light in a vacuum. 127b. Wavelength is inversely proportional to frequency. 127c. Energy is directly proportional to frequency. 128. Sir James Jeans’s obtuse metaphor for the photoelectric effect points to the fact that it is a quantum-mechanical phenomenon. The photoelectric effect is a single photon-to-electron phenomenon; that is, a single photon that meets the minimum energy requirement can cause the ejection of an electron from the atom. If the photon is particularly energetic, the excess energy will not eject a second electron. Hence, you can’t kill two birds with one stone. Furthermore, the atom cannot accumulate the energy from multiple photon hits to eject an electron: only one hit of sufficient energy equals one and only one ejection. Therefore, you can’t kill a bird with multiple stones. Copyright © 2011 Pearson Canada Inc. 8 Chapter 9 Answers Practice Examples 1a. S 1b. Ca b b g b g g b g b g 2a. V 3+ 64 pm < Ti 2+ 86 pm < Ca 2+ 100 pm < Sr 2+ 113 pm < Br − 196 pm 2b. As 3a. K < Mg < S < Cl 3b. Sb 4a. Cl and Al are paramagnetic; K + , O 2− , and Zn are diamagnetic. 4b. Cr 2+ 5a. 280 K 5b. 570 K Integrative Example A. Based on rough approximations of the trends of data, the properties of francium can be approximated. Melting point: 22 °C, density: 2.75 g/cc, atomic radius: 4.25 Å. 200 5.0 M.P. density metallic radii 180 Melting Point (ºC) 4.5 4.0 140 3.5 120 3.0 100 2.5 80 2.0 60 1.5 40 1.0 20 Density (g/cc) or Atomic Radius (Å) 160 0.5 Li 0 0 Na 10 K 20 Rb 30 Fr Cs 40 50 60 70 80 0.0 90 Atomic # B. Element 168 should be a solid since the trend in boiling point and melting point would put the boiling point temperature above 298 K. The electronic configuration is [Unk]10s26h8. Exercises 1. 14 g g to 16 3 cm cm3 Copyright © 2011 Pearson Canada Inc. 1 3. A plot of density versus atomic number clearly shows that density is a periodic property for these two periods of main group elements. It rises, falls a bit, rises again, and falls back to the axis, in both cases. 5. Mendeleev arranged elements in the periodic table in order of increasing atomic weight. Atomic masses with non-integral values are permissible. Hence, there always is room for an added element between two elements that already are present in the table. On the other hand, Moseley arranged elements in order of increasing atomic number. Only integral (whole number) values of atomic number are permitted. Thus, when elements with all possible integral values in a certain range have been discovered, no new elements are possible in that range. 7a. 118 7b. 119 7c. A119 ≈ 298 u and A118 ≈ 295 u. 9a. Te 9b. K 9c. Cs 9d. N 9e. P 9f. Au 11. Sizes of atoms do not simply increase with atomic number is because electrons often are added successively to the same subshell. These electrons do not fully screen each other from the nuclear charge (they do not effectively get between each other and the nucleus). Consequently, as each electron is added to a subshell and the nuclear charge increases by one unit, all of the electrons in this subshell are drawn more closely into the nucleus, because of the ineffective shielding. 13a. B 13b. Te 15. Li + < Br < Se < I − 17. Fe 2+ and Co3+ , Sc3+ and Ca 2+ , F− and Al3+ , Zn 2+ and Cu + . 19. Ions can be isoelectronic without having noble-gas electron configurations. 21. Cs < Sr < As < S < F Copyright © 2011 Pearson Canada Inc. 2 23. For an ionization potential, negatively charged electron is being separated from a positively charged cation, a process that must always require energy, because unlike charges attract each other. 25. 9951.5 kJ/mol 27. Exothermic 29. In the case of Na + , the electron is being removed from a species that is left with a 2+ charge, while in the case of Ne, the electron is being removed from a species with a 1+ charge. The more highly charged the resulting species, the more difficult it is to produce it by removing an electron. 31a. Al < Si < S < Cl 31b. Al < Si < S < Cl 33. Fe 2+ 35a. Diamagnetic 35b. Paramagnetic 35c. Diamagnetic 35d. Diamagnetic 35e. Paramagnetic 35f. Diamagnetic 35g. Paramagnetic 37. All atoms with an odd number of electrons must be paramagnetic. There is no way to pair all of the electrons up if there is an odd number of electrons. Many atoms with an even number of electrons are diamagnetic, but some are paramagnetic. 39a. Elements that one would expect to exhibit the photoelectric effect with visible light should be ones that have a small value of their first ionization energy. Based on Figure 9-9, the alkali metals have the lowest first ionization potentials of these. Cs, Rb, and K are three suitable metals. Metals that would not exhibit the photoelectric effect with visible light are those that have high values of their first ionization energy. Again from Figure 9-9, Zn, Cd, and Hg seem to be three metals that would not exhibit the photoelectric effect with visible light. 39b. Rn 39c. +600 kJ/mol 39d. 5.7 g/cm3 41a. 5.4 g/cm3 Copyright © 2011 Pearson Canada Inc. 3 1155 = 0." 55. boiling point = 348 K 61a. Statement 3 45d. Since (0. B 62. "Gruppe V" or "Gruppe VI.011440 + (23. 53. so: 0. Ge 123 762 Al same same or smaller In larger smaller Se smaller larger BrCl: melting point = 226 K. Ga 2 O3 . B 61c. if we subtract 143 from 382 we get 239 u. 74. A 61d. ~210 K 45a.184 J/cal) = 0. which is within about 3% of the correct value. solving for atomic mass yields a value of 230 u. Ga4+ and Ge5+ Integrative and Advanced Exercises 51.5% Ga. Both (b) and (e) are compatible with the sketch. kJ/mol 57. Statement 1 45c.1155 J/goC . Statement 2 47. Statement 4 45e. Statement 1 45b. boiling point = 286 K ICl: melting point = 280 K. 4 . First we convert to J/goC: 0. 43. which is very close to the correct value of 238 u. Element Atomic Radius.967/atomic mass).0276×(4. Copyright © 2011 Pearson Canada Inc. Statement 2 45f. A 61b.41b.3734)(382) = 143 u. pm First Ionization Energy. 71a. 83. [Ne]4p1 : Zeff = 1. in the case of Mg. The more deeply an orbital penetrates (i. could represent the number of electrons left in the K shell after one K-shell electron has been ejected by a cathode ray. The answer is (a). 82. [Ne]3d1 : r 3d = 555 pm.e. As Copyright © 2011 Pearson Canada Inc. [Ne]3p1 : Zeff = 1. or [Kr] 5s24d105p6 . The answer is (d).65. [Ne]3d1 = 147 kJ mol-1 . an electron from a filled subshell is being removed. because in the case Mg. [Ne]3p1 =293 kJ mol-1. [Ne]4s1 = 188 kJ mol-1 . = 134 kJ mol-1. the electron configuration of Ar is achieved. The first ionization energy of Mg is higher than Na because.30 × 1015 Hz and b = 0. The answer is (a). I for F (1681) ≈ I for Ba (965) < I for Sc (2389) < I for Na (4562) < I for Mg (7733) 1 2 3 2 3 Feature Problems 70. 5 . It follows then that the 4s orbital will experience the greatest effective nuclear charge and that the Zeff values for the 3p and 4p orbitals should be larger than the Zeff for the 3d orbital. 78. the stable [Ar] configuration is being lost by removing an electron from the filled subshell. [Ne]3p1 : r 3p = 466 pm. the removing of one electron leads to the highly stable electron configuration of Ne. 71d. 79. Cs+: [Xe]. the closer the orbital is to the nucleus). A = 2. whereas in Na. [Ne]4s1 : r 4s = 823 pm . Thus.969. [Ne]4s1 : Zeff = 1. Self-Assessment Exercises 77. whereas in Na. Cs2+: [Kr] 5s24d105p5 84. 80. The constant b. 71b. __ __ __ __ 71c. the greater is the effective nuclear charge felt by the electrons in that orbital..51 . 81. E1 = E1 × N A = 8. [Ne]4p1 : r 4p = 950 pm.00 .022 × 1023 1 kJ × × =5249 kJ/mol 1 atom 1 mol 1000J 66. one can think of b as representing the screening afforded by the remaining electron in the K-shell.716 × 10-18 6. so it is probably the equivalent term in the Moseley equation. 85a. [Ne]4p1. The answer is (b).42. [Ne]3d1 : Zeff = 1. which is close to unity. The second ionization of Mg is lower than Na. The answer is (a).2881 × 1015 s-1). The value of A (calculated in this question) is close to the Rydberg frequency (3.28. The answer is (b). Carbon 86. whereas the S has one extra electron. Co/Ni. C 90b. Ba 91b. F 89b. Te/I and Th/Pa. Sc 89c. The exception is the case of S and P.85b. neutrons = 69 88c. True Copyright © 2011 Pearson Canada Inc. valence shell electrons = 4 89a. Rb 90c. Rb>Ca>Sc>Fe>Te>Br>O>F 93a. False 93b. where P has a slightly higher value. 88a. 3s electrons = 2 88e. At 91a. 5p electrons = 2 88f. Cl– 85d. S 91c. 6 . The pairs are Ar/Ca. 87. Si 90a. py and pz orbitals as in the case of P. The trends would generally follow higher first ionization energy values for a fuller subshell. protons = 50 88b. Carbon 85e. 4d electrons = 10 88d. This is because there is a slight energy advantage to having a half-filled subshell with an electron in each of px. F– 85c. Bi 92. False 94c. False 94b. True 93d. Ionization energy generally increases with Z for a given period. True 93e. These ionization energies are the reverse of electron affinities. 96. 7 . True 95.93c. Copyright © 2011 Pearson Canada Inc. True 94d. True 94a. . (a) : S C S : . ... (b) H C N: . :I N I: | :I: . The electrostatic potential map that corresponds to IF is the one with the most red in it. H H H H . (b) [Ba]2+ [: S :]2 . 3b. (c) [Li]+ [: O :]2 [Li]+ . . . N—H and PCl bonds are the most polar of the four bonds cited. Mg ...Chapter 10 Answers Practice Examples 1a. Ge . H O Cl : 3a. (c) : O C ( Cl :)2 ... . (a) N O .. . 5b. . The electrostatic potential map that corresponds to CH3OH is the one with the most red in it. . . .. (a) H C H O H 7a. . O C C H . (a) [: I :] [Ca]2+ [: I :] . ..... 2- 1 . K . O 6b. . Sn ... : Ne : 1b. . : Br Br : . (b) [Mg]2+ [: N :]3 [Mg]2+ [: N :]3 [Mg]2+ . (b) H H (c) O Copyright © 2011 Pearson Canada Inc. . ... . H N N H H C C H . 4a. H H C H ... H H . . (b) H H H N N H H . .. (a) [Na]+ [: S:]2 [Na]+ . .. . .. H .. .. The P — O bond is the most polar of the four bonds cited. [Tl]+ . 2b... 4b.. [: Br :] .. [: S:]2 2a. 5a. 6a. . which is the most electronegative atom in the molecule. Tetrahedral 11a. (b) N . formal charge on oxygen. (c) C N O Structure 1 C O Structure 2 N C O Structure 3 N O Cl is much poorer than the one derived in Example 10-8 because it has a positive 8a. O S O O 9a. Linear Copyright © 2011 Pearson Canada Inc. O S + O1 O3 O O O1 O2 N S O2 N + O1 O3 O2 N + O3 O1 9b.F F H B F H N O H F 7b. Linear 11b. (a) H . The first structure is the better of the two structures because it has no formal charges. and . Trigonal pyramidal 10b. O2 N O3 Resonance Hybrid 10a. 2 . H N C N H N C N H H 8b. 5 by the presence of two lone pairs on O. The H — C — N angles. 4 101 kJ/mol NH3 16a. From Table 10. The C — O — H bond angle is made somewhat smaller than 109.2. H 2 O 2 is polar. 486 kJ/mol 15b. linear. In H2O2. the molecular geometry around each O atom is bent. the length of a C — H bond is 110 pm. Since the geometries of the two molecules differ.5º. 13a. the H — C — H angle and the HCC angles all are very close to 109. the bond moments do not cancel. as are the HCO bond angles. A reasonable value is the average of the C — C and Br — Br bond lengths = 195pm. | — S C N | .5 . Endothermic Integrative Example A P Cl 165. The O — C — O bond angle and the O — C — C bond angles all are very close to 120 . it is a highly symmetrical molecule in which individual bond dipoles cancel out. the orbital overlap between P and the surrounding Cl atoms will be different and therefore the P–Cl bonds in these two compounds will also be slightly different.5 . 14a. 15a. Exothermic 16b. The H — N — H bond angle and the H — N — C bond angles are almost the tetrahedral angle of 109. Copyright © 2011 Pearson Canada Inc. Around the O there are two bonding pairs of electrons and two lone pairs.5 kJ/mol . . The length of a C — Br bond is not given in the table. PCl5 as the only nonpolar species. 13b. 3 . making its electron-group and molecular geometries trigonal planar. with a C — O — H bond angle of slightly less than 109. . The HCH bond angles are ~109. resulting in a tetrahedral electron-group geometry and a bent molecular shape around the O atom. made a bit smaller by the lone pair. 12b. 14b.5 . Three electron groups surround the right-hand C.H H C O H H 12a. (a)Formamide: H2N bond energy = 2129 kJ / mol . Rb 1f. Ga 1d. and its conversion endothermic. I I 3b. Formaldoxime: H2C OH . F Cl 3a. B.O C N H . Ge 1b. 4 . N 1c. Since BE of formamide is greater than that of formaldoxime. F F Copyright © 2011 Pearson Canada Inc. S 3c. As 1e. 124o O C H N H o (b) 109 H Exercises Kr 1a. it is more stable. bond energy = 2233 kJ/mol . 2- Li 5e. It is improperly written as a covalent Lewis structure. and only six electrons around the nitrogen.. [: Cl :] [Ca]2+ [: Cl :] .F N F F 3d. 11. SF6 . Cl Cl B Cl 5c. Te H 3e. (b) Neither C has an octet of electrons and the total number of valence electrons is incorrect. 13a. [Ba]2+ [: S :]2 Copyright © 2011 Pearson Canada Inc. 9b. Has two bonds (4 electrons) to the second hydrogen. . . . (c) The total number of valence electrons is incorrect. 5 . 7. H Cs Br 5a. NO 2 .. 13b. BF3 . The answer is (c). . 9a. Cs Cl Li O I Cl 5d.. The flaws with the other answers are as follows: (a) C does not have an octet of electrons. CaO is actually an ionic compound... H Sb H H 5b. . 5f.. 21a. H— 0 —C 0 C –1 =O –0 —O( 2) –1 C 0 —H(7) 0 side C(2) 0 central C +1 17b. such as oxygen in ozone. 17c. [Na]+ [: F :] . is that whereas the oxidation state of an element in its compounds is usually not zero.. [: Cl :] . . ... 6 . 13d.. . Another is that formal charges are used to decide between alternative Lewis structures. -2 21e. 15c. For instance.. -1 21c. . there are cases where atoms of the same type with the same oxidation state have different formal charges. . O 3 . 17a.. the oxidation state in a compound is invariant. its formal charge usually is.. 0 21d.. [Li]+ [: S:]2 [Li]+ . [: Cl :] [Sc]3+ [: Cl :] 15d. . Also. . .. though.. +1 21b.. 0 Copyright © 2011 Pearson Canada Inc. [Na]+ [: F :] 15a. [: I :] [Ca]2+ [: I :] . 13c.. . 15b. The most significant difference. [Li]+ [: O :]2 [Li]+ ...... while oxidation state is used in balancing equations and naming compounds. while the formal charge can change. . 19. we must conclude that structures (A) and (B) are equally plausible. Based on formal charge rules alone. H H N N H H H O Cl O 25b. 25a. H O O H 25d. 2- 2- O C 2- O O O O C O O O C O 27c. 7 . O N O O N O 27b. 2O O S O 27a.23. O HO S OH O 25e. 27d. O HO S OH 25c. H O O Copyright © 2011 Pearson Canada Inc. Group 16. Group 17 except fluorine. 35b. H 31b. Group 16 except oxygen. H Cl H C C C H H H H O 33a. O O 35c. 31a. 2S S 35a.H H H O H C C C C H H 29. 2- O O S O O . Cl O . 33b. 8 . . Copyright © 2011 Pearson Canada Inc. The molecular formulas for the compounds are SF4 and SiF4. [ O N O] - [O N O ] - 47. The electrostatic potential map on the right is for SiF4. Group 13. while H2C=O is represented on the right. 4% 39b.H H B H 35d. 45. 5% 39c. F2C=O is represented on the left. The molecule seems best represented as a resonance hybrid of (a) and (b). H . 33% 43. 9 . 60% 39d. Copyright © 2011 Pearson Canada Inc. C — H Br — H F — H Na — Cl K — F 39a. A 0 H B O 0 C O O H C O 51a. 37. 49. A 0 H B +1 O O C O H O C O C O O H O C O 51b. 10 .. O 51c.. SF4 and ICl3 require expanded octets. etc. Bent Copyright © 2011 Pearson Canada Inc. Paramagnetic 55d. Diamagnetic 55b. 59a. A 0 B O 0 0 O N N O +1 N O C -1 -1 O 0 -1 N O +1 N N O O O 51d. Linear 59b. Tetrahedral 59d. Linear 59c. Trigonal Planar 59e. 53b. H C 53a. H H H O O Cl O or Cl O 53c. Diamagnetic 55e.A +2 O F S B -1 O +1 O 0 F S O O . Paramagnetic 57. Paramagnetic 55c. 55a. The others do not. Diamagnetic 55f. O 65b. Planar 61c. Looking at the structures. F Cl F F 73b. Tetrahedral 63. O = C = O . linear. F Cl F 73a. the more acute the angle becomes. Copyright © 2011 Pearson Canada Inc. F 73c. Tetrahedral 67b. O 65c. F Cl . F F . Linear 69.61a. octahedral. Bent 61b. Tetrahedral 71. 11 . t-shaped. Tetrahedral 67c. CO 32 65a. Cl N O 67a. Cl C Cl . trigonal bipyramidal. Octahedral 67d. The more pairs there are. Linear 61d. square planar. . the molecular angle/shape depends on the number of valence electron pairs on the central atom. linear. trigonal planar. Octahedral 61e. trigonal bipyramidal. polar. polar. square pyramidal. Single bonds are generally longer than multiple bonds. nonpolar. polar. 81b. 233 pm Copyright © 2011 Pearson Canada Inc. Planar. 85. Octahedral. and Ob=C-Oc which 109o H O H 109o 120o N C C 109o H 120o 79. Tetrahedral. 12 . H H H Ob C C H Oc H Oa H H 77. 81c. Trigonal pyramidal. nonpolar. It cannot be linear. H H 81a. H C C C A m aximum of 5 atom s can be in the sam e p la n e H H 75. 83. Br2 possess the longest bond. 81e. Bent. 81d. 87a.5 except Oc-C-C. Bent. Ob=C-C. 81f. are 120° all angles ~109. F F . polar.F F F Cl 73d. octahedral. 00191. 166 kJ/mol 99. 448 kJ/mol Integrative and Advanced Exercises 101. 42. 172 pm 87c. 191 pm 89. 307 kJ 120. 144 pm 91.00199. C C H 109. Br = 0. -744 kJ/mol 104. EA for At : ~ -260 kJ/mol. Copyright © 2011 Pearson Canada Inc.0 g/mol . I = 0. 110.00192. H H H C C H. 13 . 149 pm 87d. Cl = 0. H f o = 99 kJ 97. Bent 113. EA ~ -260 kJ/mol. 113 kJ/mol 95a. H f = 39 kJ / mol o 95b. C3H6. H H N N N . C C H .00187. H 106. Neither is completely linear.87b. F = 0. 103 kJ 116. H N N N . Endothermic 93. The answer is (b).P P O O O P P P 127. 1. P P O O P O . 97.49 D 129b. EN = 0. O Cl O 140a. 91 kJ/mol 128b. 2O 140c. The P–O–P bond is not linear. C O O Copyright © 2011 Pearson Canada Inc. 137. 138. 23% 129a. Approximately 92. 136. The answer is (b).97 128c. The answer is (c). 135. The answer is (a). . 128a.6 Self-Assessment Exercises 134. 139. F P F F 140b. The answer is (c). 129c. 14 . The answer is (a). H 0 0 0 Copyright © 2011 Pearson Canada Inc. the molecular geometry is trigonal pyramidal. However. Tetrahedral 142. For both. A bent geometry is observed when an atom has two lone pairs and is bonded to two other atoms (AX2E2). H C H -1 S O +1 0 C H 0 S O +1 -1 15 . the bond angles will be approximately (usually smaller than) 109°. Bent 141b. H H C S O 147. Since there are 4 electron pairs around the central atom. 141a. since there are only three atoms bonding to the central atom. VSEPR theory is valence shell electron pair repulsion theory. Bond Bond Energy (kJ/mol) Bond Length (pm) C–H 414 110 C=O 736 120 C–C 347 154 C–Cl 339 178 144. 145.F F Br F F F 140d. the way to maximize the distance between them is to set up a tetrahedral electron group geometry. A pyramidal geometry is observed when an atom has one lone pair and is bonded to three other atoms (AX3E). 146. Trigonal pyramidal 141c. Bi 143. It is based on the premise that electron pairs assume orientations about an atom to minimize electron pair repulsions. The molecule is trigonal pyramidal. The carbon atoms have tetrahedral geometry. Thus. and z-directions of a Cartesian coordinate system. 2a. 3b. In both structures.) VSEPR theory begins with the Lewis structure and notes that there are three bond pairs and one lone pair attached to N. 4b. The three bonds between C and N consist of a sigma bond (sp on C with sp on N). See-saw molecular geometry. Each halffilled 2p orbital from N will overlap with one half-filled 5p orbital of an I. There are four bond pairs around the left-hand C. The three I atoms will lie in the same plane at the points of a triangle. the oxygen atom is bent. and two pi bonds (2p on C with 2p on N). sp3d hybridization. 4a. utilizing sp hybridization. Since VSEPR theory makes a prediction closer to the experimental bond angle of 107. structure(1) — — | N a N b — O — | | Na N b O — structure(2) .Chapter 11 Answers Practice Examples 1a. requiring sp 3 hybridization. there will be three N—I bonds. but bond angles are closer to 109. 1b. y . This produces a tetrahedral electron pair geometry and a trigonal pyramidal molecular shape with bond angles a bit less than the tetrahedral angle of 109. In structure (1) the N N bond results from the overlap of three pairs of half-filled orbitals: (1) spx on N b with 2 px on N a forming a bond. (We obtain the same molecular shape if N is assumed to be sp3 hybridized. it seems more appropriate in this case. The other C has two attached electron groups. There will be three N—H bonds. and possesses no lone pairs. 3a. and (3) 2 pz on N b Copyright © 2011 Pearson Canada Inc. The two C atoms join with a sigma bond: overlap of sp 2 on the lefthand C with sp on the right-hand C. The molecule is trigonal pyramid. and thus require sp 3 hybridization. Bent molecular geometry. The I atoms will be oriented in the same direction as the three 2p orbitals of N: toward the x . (2) 2 p y on Nb with 2 p y on Na forming a bond. the central N is attached to two other atoms.5 . Thus. The geometry of the molecule thus is linear and the hybridization on this central N is sp. There are three half-filled 2p orbitals on N. and one half-filled 5p orbital on I. Three of the bonds that form are C—H sigma bonds resulting from the overlap of a half-filled sp 3 hybrid orbital on C with a half-filled 1s orbital on H. with bond angles of approximately 90.5 because of the lone pair. with the N atom centered above them. the I—N—I angles will be approximately 90 (probably larger because the I atoms will repel each other). sp3 hybridization. requiring sp 3 hybridization. The central carbon is surrounded by three electron groups and is sp 2 hybridized. 2b. the tetrahedral bond angle. The left-most C and the right-most O are surrounded by four electron pairs. There are three half-filled orbitals on N and one half-filled orbital on each H. 1 . Each central atom is surrounded by four electron pairs. In structure (2) the N == O bond results from the overlap of two pairs of half-filled orbitals: (1) spy on N b with 2 p y on O forming a bond and (2) 2 pz on N b with 2 pz on O forming a bond. The N == N bond is a coordinate covalent bond. O.with 2 pz on N a also forming a bond. structure (1) is preferred. 2s 2s* 1s 1s* Ne2+ 1s 1s* 2s 2 s* 2 1s 1s* 2s 2s* N2 C2 + 2p 2p 2p 2p 2p 2p 2p* 2p* . BN Copyright © 2011 Pearson Canada Inc.5. 2p* 2p* . with a bond strength about half that of a hydrogen molecule. bond order = 2. bond order = 2. Based on formal charge arguments. CN+ 2s 2s 2p 2p 2p 2p . 2 . bond order = 3. The bond order in H 2 is ½. and (2) the overlap of the half-filled 2px orbital on N b with the half-filled 2px orbital on N a to form a bond. We would expect the ion H 2 to be stable. and is formed by two overlaps: (1) the overlap of the full spy orbital on N b with the empty 2py orbital on N a to form a bond.0.0.5. bond order = 0. 6b. 2p* 2p* . 5a. 53 kJ/mol Li 2 + 5b. 2p 2p 2p . and forms by the overlap of the full spx orbital on N b with the empty 2 p orbital on O. bond order = 2. 2s 2s 2p 7b. because the negative formal charge is on the more electronegative atom. The N—O bond is a coordinate covalent bond. 6a.0. 7a. The bond length increases as the bond order decreases. Longer bonds are weaker bonds. C. non-bonding -orbitals -molecular orbitals 3 .sp2 -bonds Copyright © 2011 Pearson Canada Inc. Atom Label sp hybrids 2p orbitals 0 A(N) _ _ 0 B(O) _ _ (1-) C(O) _ _ N O 8b. Atom Label sp hybrids np orbitals 2+ A(S) _ _ (3p) 0 B(O) _ _ (2p) (1-) C(O) _ _ (2p) (1-) D(O) _ _ (2p) O S O non-bonding -orbitals 120o O -molecular orbitals -framework 2 2 3 x sp .C. 2 F.sp -bonds 8a. O ~120o -framework 2 x sp2.2 F. (a) N C H N N H H H 1s H 1s N sp3 -system -system C N 1s C H sp2 H N sp2 C sp2 N sp3 (b) sp2 sp2 sp2C H N C N 1s N sp3 N 1s C N 1s H Antibonding moleclar orbitals Bonding moleclar orbitals One 2p electron from each of the atoms is placed in the ring (c) B. 4 . : Ca(2p) – N(2p) N-O: N(sp2) –O(sp3) Ca-Ca: Ca(sp2) –Ca(sp2) Ca-Cb: Cb(sp3) –Ca(sp2) O-H : H(1s) – O(sp3) Copyright © 2011 Pearson Canada Inc. (a) (b) O N N O H H Ca Ca H3Cb CbH3 Cb-H : H(1s) – Cb(sp3) Ca=N: Ca(sp2) –N(sp2) .Integrative Example H H N C N C H N A. 3c. molecular geometries are more directly obtained in valence-bond theory than in Lewis theory. each H—O bond results from the overlap of a 1s orbital on H with a 2p orbital on O. Lewis theory does not explain hindered rotation about double bonds. In valence-bond theory using simple atomic orbitals. sp 3d . CO32-. Third. 3d. The molecule is trigonal planar around N which is sp 2 hybridized. The lone pairs repel each other more than do the bond pairs. 7b.5º. C is the central atom. The electron-group geometry is tetrahedral. it is clear that a sigma bond must be stronger than a pi bond. indicating that B is sp 3 hybridized. 3b. 7a. 11a. 5 . N is the central atom.Exercises 1. B is the central atom. 7d. Each of the three sigma bonds are formed by the overlap of a 2 p orbital on F with one of the half-filled dsp 3 orbitals on Cl. Second. sp 3 . The electron-group geometry around Cl is tetrahedral. The molecule is linear and C is sp hybridized. 3a. indicating that Cl is sp 3 hybridized. Copyright © 2011 Pearson Canada Inc. 11d. The angle between sp3 orbitals is 109. each H—O bond results from the overlap of a 1s orbital on H with an sp3 orbital on O. and NO2 . 11b. 7c. The central atom is sp 2 hybridized in SO2. Lewis theory does not describe the shape of the water molecule. explaining the smaller than 109. 9. sp 11c. In VSEPR theory the H 2 O molecule is categorized as being of the AX2E2 type.5º tetrahedral bond angle. First. 11e. sp 2 . In valence-bond theory. Cl is the central atom.valence-bond theory clearly distinguishes between sigma and pi bonds. - 5. The two lone pairs occupy dsp 3 orbitals. The angle between 2p orbitals is 90 so this method initially predicts a 90 bond angle. for the orbitals overlap more effectively in a sigma bond (end-to-end) than they do in a pi bond (side-to-side). The hybridization is sp 3d . sp 3 d 2 . In valence-bond theory using hybrid orbitals. The bonds are: : O(2 p y )1 N( sp 2 )1 : N(sp 2 )1 Cl(3 pz )1 : O(2pz )1 N(2 pz )1 . all bond angles around C are 120º. : | N C—C N| The C—C bond is a bond. 17c. The four bonds in the molecules are: 2 : Cl 3 px —C sp 2 1 1 : O 2 p y —C sp 2 : O 2 pz — C 2 pz .group geometry at N is trigonal. . The hybridization for each C is sp. The C ==C bond is composed of one and one bond. 17a. The double bond is composed of one and one bond. The e. The four bonds are represented as follows.group geometry around C is trigonal planar. The geometry of Oa is tetrahedral. H O a N= O b . Neither linear nor planar. The hybridization for C is sp. 13d. The hybridization on C is sp 2 . The geometry at C is tetrahedral. H—C N| . The H—C bond is a bond.group geometry around N is triangular planar. 1 1 17d. Linear molecule.5. and the hybridization of C is sp 2 . N is sp2 hybridized.13a. 1 1 1 1 6 . Linear molecule. and each C N bond is composed of 1 and 2 bonds. C is sp 3 hybridized. and N is sp 2 hybridized. H Cl H C C C C Cl H H H Cl 15c. H O N O . Planar molecule. 15b. 13c.5 . : H 1s — O a sp 3 1 1 : O a sp 3 —N sp 2 1 : N sp 2 —O b 2 p y 1 1 1 : N 2 pz —O b 2 pz . 15a. The shape around the left-hand C is tetrahedral and that C has sp 3 hybridization. The e. The e. Cl—C—Cl bond angles are 109. and the O a — N — O b bond angle is120. 15d. Copyright © 2011 Pearson Canada Inc. and the C N bond is composed of 1 and 2 bonds. The O—N— C bond angle is about 120. All bonds are bonds except one of the bonds that comprise the C ==C bond. All single bonds in this structure are bonds. 13b. b g c h 1 Each C—Cl bond is represented by : C sp 3 — Cl 3 p 1 17b. Oa is sp3 hybridized and the H — O a — N bond angle is (at least close to) 109. O(sp3)) (C(sp2) .C(sp )) 3 4 C 6 C 3 O 1 H 6 1 (O(sp3) . 2 (S(sp ).S( 3p x)) (S(3p z). C1(2p)―O2(2p) . : Ca sp —H a 1s 1 1 : Cc sp 2 — H b 1s 1 : C b sp — Cc sp 2 : Ca sp —C b sp 1 1 : Ca 2 pz —C b 2 pz 1 1 : Cc sp 2 —O 2 py 1 1 1 1 : Ca 2 p y —C b 2 py 1 1 : Cc 2 pz —O 2 pz 1 1 1 σ: C3(sp2)―C4(sp3) 4 H 3 σ: C2(sp2)―C3(sp2) π.C(sp3)) (C(sp3) .O(2p)) + (C(2p) . C1-5.H(1s)) 3 (C(sp ) . | | Ha Ca Cb Cc O Hb 23. Furthermore. C2(2p)―C3(2p) CH3 C σ: C3(sp2)―H(1s) σ: C2(sp2)―C3(sp2) 5 H3C C 1 2 C O 1 H σ: C2(sp2)―O1(2p) O 2 σ: C1(sp2)―O2(2p) π. (O1. To C3). 7 .O(2p)) 5 3 2 (C(sp ) .O(sp3)) H 19.S (3 pz)) S 2 C e n tr a l S i s sp h y b r id i ze d an d the te rrmi n a l O an d S at o m s a r e u n hybridized O S 2 (S(sp )-O(2 pz)) 21a.H(1s)) 7 (C(sp3) . 21b. There are 8 atoms that are on the same plane 25. H att. σ: C2(sp2)―C5(sp3) Copyright © 2011 Pearson Canada Inc. O2.O H O C C O H 1 H O 2 6 1 2 5 C 5 3 1 O 3 2 4 H 4 C 2 7 2 H O H 5 4 (C(sp2) . two H atoms can also be added to this total. depending on the angle of rotation of the –CH3 groups (C4 and C5). there would have to be a region in a molecular orbital diagram where four bonding orbitals occur together in order of increasing energy.: 12s 1*2s 22s 2*2s 24 p 22 p 35f. No such region exists in any of the molecular orbital diagrams in Figure 11-25. 2 s * 33c. N 2– 2 bond order = 2 stable .27. 31. BN : 12s 1*2s 22s 2*2s 24 p Copyright © 2011 Pearson Canada Inc. NO+ : 12s 1*2s 22s 2*2s 22 p 24 p 35c. 2 p 35a. with no intervening antibonding orbitals. There is little concept of the relative energies of bonding in valence-bond theory. N 2– bond order = 2. CN.5 stable . The valence-bond method describes a covalent bond as the result of the overlap of atomic orbitals. NO : 12s 1*2s 22s 2*2s 22 p 24 p 2*1p 35b. CN+ : 12s 1*2s 22s 2*2s 24 p 35g. The molecular orbital bond does not have to be created from atomic orbitals (although it often is) and the orientations of atomic orbitals do not have to be manipulated to obtain the correct geometric shape. 33a. 8 . CN : 12s 1*2s 22s 2*2s 24 p 21 p 35e. Molecular orbital theory describes a bond as a region of space in a molecule where there is a good chance of finding the bonding electrons. CO : 12s 1*2s 22s 2*2s 22 p 24 p 35d. In molecular orbital theory. 1s 33d. bonds are ordered energetically. 29. In order to have a bond order higher than three. 1s 33b. C F σ*2p 2p π*2p π2p 2p σ2p 2s 39. 41.5 for N2+.N O N σ*2p σ*2p π*2p π*2p 2p N 2p 2p 2p π2p π2p σ2p 37a. Bond order is 3 for NO+. 37d. This creates a single delocalized structure for the C6 H 6 molecule. 43c. σ*2s σ2s 2s . NO+ is diamagnetic (all paired electrons). 37c. Copyright © 2011 Pearson Canada Inc. 2. Delocalized molecular orbitals required. Delocalized molecular orbitals required. N2+ is paramagnetic. 43a. This can be achieved by creating six molecular orbitals—three bonding and three antibonding—into which the 6 electrons are placed. 9 . 43b. The bonding requires that all six C—C bonds must be equivalent. The bond length in CF+ would be shorter. Delocalized molecular orbitals not required. N2+ has the greater bond length. σ2p 37b. because there is less electron density between the two nuclei. however.02 1020 electrons. there are no extra electrons in the lattice that can conduct an electric current.45a. Atomic number. If. Insulator 49d. No 51c. There are 22 electrons in Na2.02 1020 energy levels. Yes 53. 45d. Integrative and Advanced Exercises 58. Because metals occur in every period of the periodic table. Insulator 49c. These electrons are distributed in the molecular orbitals as Copyright © 2011 Pearson Canada Inc. has some minimal predictive value in determining metallic character. No 51f. The number of valence electrons has no bearing on the metallic character of an element. the silicon becomes contaminated with arsenic atoms. 7. 45c. 45b. 49a. 47. since atomic mass generally parallels atomic number for the elements. The answer for this part is much the same as the answer to part (a). Semiconductor 49e. Electrical conductor 49b. which provides the location of the element in the periodic table. In ultra pure crystalline silicon. 1090 nm. The arsenic atoms increase the conductivity of the solid by providing additional electrons that can carry a current after they are promoted into the conduction band by thermal excitation. No 51d. This is IR-radiation. No 51b. 10 . then there will be one additional electron added to the silicon crystal lattice for each arsenic atom that is introduced. 7. there is no particular relationship between the number of electron shells and the metallic behavior of an element. Insulator 51a. [Ne] 3s The half-filled 3s orbital on each Na overlaps with another to form a covalent bond. Electrical conductor 49f. Yes 51e. 55. For AsF5 the electron group geometry and molecular shape are trigonal bipyramidal. The Cl in ClF3 and As in AsF5 both have sp3d hybridization. O2 [ O O ] 2 O 64. For AsF6– the electron group geometry and the shape of this ion are octahedral. O2 60b. Thus Cl in ClF2+ has sp3 hybridization. we predict a single bond for Na2. Tetrahedral electron group geometry is associated with sp3 hybridization. the bonding in Na2 is very much like that in H2. H H C C N C C H 2p C H H . The bond order would be (4-0)/2 = 2. 11 . 1s 1s* 2s 2s* 2p 2p 2p* 2p* 3s Again. 1s 1s* 2s 2s* 2p 2p 2p* 2p* . O2 1s 1s* 2s 2s* 2p 2p . N [He] sp2 2p Number of -bonds = 3 This -bonding scheme produces three -bonds. [ O O ] . 2p* 2p* F Oa N O O F Oa N O F(2p) O ( sp 3 ) O ( sp 3 ) N(sp 2 ) a a sp 2 ) O(2p y ) N(2p ) O(2p ) z z O N O O F 66a. For ClF2+ the electron group geometry is tetrahedral and the ion is bent in shape. 70. For ClF3 the electron group geometry is trigonal bipyramidal and the molecule is T-shaped. O2 60a.5. bond order = 1.follows. The orbital diagrams for C and N are as follows. 68. Octahedral electron group geometry is associated with sp3d 2 hybridization. Trigonal bipyramidal electron group geometry is associated with sp3d hybridization. which is the hybridization adopted by As in AsF6–. 66b. which is identical to the number predicted by Lewis theory.0. C [He] sp2 antibonding molecular orbitals bonding molecular orbitals Copyright © 2011 Pearson Canada Inc. A Lewis-theory picture of the bonding would have the two Lewis symbols for two Na atoms uniting to form a bond: Na Na Na Na Thus. One possible configuration is 12s 1*s0 22s 20*s . Suppose two He atoms in the excited state 1s1 2s1 unite to form an He2 molecule. bond order =1. Resonance energy = –168 kJ. Feature Problems 80a. Hatomization = 5358 kJ (per mole of C6H6). 4. –205. Copyright © 2011 Pearson Canada Inc.9 kJ 80c.3 kJ 80d. because they both have a conjugated π system. –148. –117. : Cc (2p) – Ob (2p) Cc-Oa: Cc (sp2) –Oa (2p or sp3) Cd-Oa: Cd (sp3) –Oa (2p or sp3) CaN: Ca (sp) –N(sp) Two mutually perpendicular π -bonds: C(2p) – N(2p) Ca-Cb: Cb(sp3) –Ca(sp) Cd-Ce: Cd(sp3) –Ce(sp3) Cc-Cb: Cb(sp3) –Cc(sp2) 75a. The wavelength for both will be the same. Ob Ca CbH2 Cc Oa CdH2 CeH3 Cb-H . 12 .4 kJ 80b.9 amps 79.N 73. Ce-H : H(1s) – C(sp3) (all tetrahedral carbon uses sp3 hybrid orbitals) Cc=Ob: Cc (sp2) –Ob (2p or sp2) .00 watts 75b. Cd-H . 8. The answer is (c). The answer is (a). The answer is (d). 92. but the geometry is square pyramidal. 98. 96. 101. Therefore. and the sixth is a lone pair. 95. 103. The answer is (b). 102. 97. 99. The valence-bond method using pure s and p orbitals incorrectly predicts a trigonal pyramidal shape with 90° F—B—F bond angles. Copyright © 2011 Pearson Canada Inc. There are (a) 6 σ and (b) 2 π bonds. the hybridization is sp3d2. five are fluorine atoms. The answer is (c). The answer is (c). The answer is (c). C2+ 105. The answer is (c). B2– has 3 bonding electrons. 94. 104. 13 . C2 has the greater bond energy. 100. 2s 2s Self-Assessment Exercises 91. 93. All three are paramagnetic. so the B–B bond is the strongest. The answer is (a).σ*2p 1s π*2p π2p 2p σ2p 87. BrF5 has six constituents around it. 0 g/mol. Thus. their only intermolecular forces are London forces. Copyright © 2011 Pearson Canada Inc. the temperature drops. Once all of the water has been converted to H2O(s). the strong dipole–dipole forces developed between CH3NO2 molecules make the enthalpy of vaporization for CH3NO2 larger than that for C6H6.7 g/L 4a. When the temperature reaches the point on the vaporization curve. But the molecule is definitely polar. Vapor only 4b. When the temperature reaches the point on the fusion curve. therefore. CH3CN is polar and thus has the strongest intermolecular forces and should have the highest boiling point.9 mmHg 6a. water condenses at constant temperature (100C). The magnitude of the enthalpy of vaporization is strongly related to the strength of intermolecular forces: the stronger these forces. 5a. OD. 1b.0435 g H 2 O vapor. Moving from point R to P we begin with H2O(g) at high temperature (>100C). 42. OC. 1. CH 4 = 16. Once all of the water is in the liquid state. which is. C 6 H 6 = 78. Ne. the temperature of the sample decreases slightly until point P is reached. essentially non-polar. 7a. 121 mmHg 5b. of course. (CH3)2CO. the more endothermic the vaporization process. whose strength primarily depends on molar mass. He . 151 Torr 3b. 6b.0 g / mol. and also in order of increasing heat of vaporization. O2 . ice begins to form at constant temperature (0C).089 g liquid water .1 g / mol .0 g / mol. Cl2 . O3 . 0. -1025 J 3a. The substances are arranged in order of increasing molar mass: H 2 = 2. The last substance has a molar mass of 61.Chapter 12 Answers Practice Examples 1a. 1 .923 kJ 2b. which would produce intermolecular forces smaller than those of C 6 H 6 if CH 3 NO 2 were nonpolar. The first three substances all are nonpolar and. 0. (CH3)3CH < CH3CH2CH2CH3 < SO3 < C8H18 < C6H5CHO 2a. 0. 6. 669. 402 pm 11b. 51. Melting point higher than CaO: MgO. higher than H2O’s vapor pressure. B.21 g/cm3 12a. Lower melting point than KI: RbI or CsI. Mg 3 N 2 .1 mm Hg. 15. 2 . Ca 3 N 2 . 764 kJ/mol Integrative Example A. Ga 2 O 3 . additional pressure produces very little change in volume because liquids are not very compressible. A significant decrease in the volume occurs (10%) as ice is converted to liquid water. 6. Copyright © 2011 Pearson Canada Inc.8 mmHg. The vapor pressure for isooctane (using the Clausius-Clapeyron equation) is 43. Here. 3 L At Point R 7b.3 L at 100C 1/2 vap 8a.2 kJ/mol 12b.86 g / cm3 10b. EA 2 = +881 kJ . 0. At 25.0 °C. 524 pm 9b. melting begins.035 1023 atoms Al mol Al 11a. 8b. NaI 9a. 2. very little change occurs until the pressure reaches the point on the fusion curve OD.628 107 pm3 10a. After melting. the vapor pressure of water is 23.Since solids are not very compressible. like honey. 1b. No hydrogen bonds. London forces are not very important. in fact. 1d. molasses at low temperature is a very slow flowing liquid. Hydrogen bonding is weak . London forces are as strong. Molasses. Hydrogen bonding is most important interaction between HF molecules. 1e. Since both the silicone oil and the cloth or leather are composed of relatively nonpolar molecules. (c) < (b) < (d) < (a). on the other hand is polar and adheres very poorly to the silicone oil (actually. than it adheres to the cloth or leather. HH N HN N O 9. they attract each other. Viscosity generally increases as the temperature decreases. There are no dipole-dipole interactions. 3. Hence. This is because the oil is more nonpolar than is the cloth or the leather.” 15. H bonds are not important. Thus there is indeed a scientific basis for the expression “slower than molasses in January. O N H N H NH N N H 11. CH 3 OH 7. Dipole–dipole interactions should be relatively strong. Neither hydrogen bonds nor dipole–dipole attractions can occur. London forces are more important. CCl4 < CH3CH2OCH2CH3 < CH3OH Copyright © 2011 Pearson Canada Inc. is a very viscous liquid (high resistance to flow). 3 . Three water molecules. London forces are quite weak. the water is repelled by the oil). much more poorly. Water. 1c. Dipole–dipole interactions are important. 13. London forces between HCl molecules are expected to be relatively weak. water is repelled from the silicone-treated cloth or leather. No hydrogen bonds.Exercises 1a. 5. The coldest temperatures are generally in January (in the northern hemisphere). Thus. However. In order to vaporize water in the outer container. 120. 0. that same heat is liberated. 4 . Thus condensation is an exothermic process.88 L C6 H 6 (l) 23.5 kJ 25. Therefore. vaporization is an endothermic process).e. 110 C 29. if the evaporation occurs from an insulated container.7 C 35. 226 mmHg 31a.00 C 33. 221 L methane 27a. 6. 40. The intermolecular interactions in butanol are dominated by H-bonding. 31b. 100. 19. this energy is obtained from the surroundings. . When this vapor (steam) condenses on the outside walls of the inner container. 45 mmHg 27b. the only source of the needed energy is the liquid that is evaporating.17.5 atm. 21.907 mmHg 37. which is much stronger than the London dispersion forces dominant in pentane. The process of evaporation is endothermic. heat must be applied (i. If evaporation occurs from an uninsulated container. through the walls of the container. 8.. Copyright © 2011 Pearson Canada Inc. meaning it requires energy. the temperature of the liquid will decrease as the liquid evaporates. liquid cannot exist at this pressure.110 atm 53c. -776 kJ 47b. SO 2 . CO 2 . No 57a. NH 3 .117 atm 55a. No 55c.5 10 4 kJ 49. 53a. 33 C 43. Yes 55d. 49. The upper-right region of the phase diagram is the liquid region. The pressure then continues to drop. At a certain pressure. 0. 2.00 atm is far below 43 atm. 1. 293 K. 27 g steam Copyright © 2011 Pearson Canada Inc. As the phase diagram shows. 51a. with the entire sample being liquid while it does. and H 2 O can exist in liquid form at 20 C. Possible. 5 . the solid liquefies. with the gas becoming less dense as the pressure falls. while the lower-right region is the region of gas. initially nothing happens. 47a. Liquid and gas.0418 atm 53b.0981 atm 45. the lowest pressure at which liquid exists is at the triple point pressure. No 55b. 43 atm. 0. 51b. yes. Substances that can exist as a liquid at room temperature (about 20 C ) have critical temperature above 20 C . At this lower pressure the entire sample vaporizes.22 kg 57b.7 kJ/mol 41. 0. namely. Thus. 0. until point B is reached. and solid sublimes to gas instead. 0. As we move from point A to point B by lowering the pressure. Of the substances listed in Table 12. HCl.39. until another. 55e. lower pressure is reached. The pressure continues to drop. 51c.5. We expect forces in ionic compounds to increase as the sizes of ions become smaller and as ionic charges become greater. no first layer sphere is visible through the indentation. We expect the melting points to decrease in this series from NaF to NaI. An alternative explanation follows. . NaCl. one can look down into these indentations and see a sphere of the bottom (first) layer. The condition of supercooling is destroyed and the liquid reverts to the solid phase. causing it to freeze. 61. and NaI. 63b. and a hexagonal array of atoms. In one case. The liquid in the can is supercooled. The required heat is taken from the cooled liquid. As the forces between ions become stronger. The two different closest packing arrangements arise from two different ways of placing the third layer on top of these two. there are six spheres surrounding and touching any given sphere. Diamond 63a. In each layer of a closest packing arrangement of spheres. the two top unit cells in the diagram. gas bubbles released from the carbonated beverage serve as sites for the formation of ice crystals. 71a. When the can is opened. In the other case. 67. and thus the ionic forces become weaker. A second similar layer then is placed on top of this first layer so that its spheres fit into the indentations in the layer below. NaF 69.59. The half-filled sp 2 hybrid orbitals of the boron and nitrogen atoms overlap to form the bonding structure. In the series of compounds NaF. 65. The process of the gas coming out of solution is endothermic (heat is required). a higher temperature is required to melt the crystal. with its spheres fitting into the indentations of the layer below. 6 . Copyright © 2011 Pearson Canada Inc. sp 2 hybridization for each atom. NaBr. The 2 pz orbitals then overlap to form the bonding orbitals. the anions are progressively larger. This gives a total of four Ca 2+ ions per unit cell. Thus. since considerably more energy is released when it forms. Copyright © 2011 Pearson Canada Inc. 2 Si atoms per unit cell. Four formula units per unit cell of MgO. The small square outlined near the center of the figure drawn in part (a). RbCl(s). This gives a total of four O 2 ions per unit cell. There are two O2 ions totally contained within the unit cell. 15. The ratio of Ti 4+ ions to O 2 ions is 2 Ti 4+ ions per 4 O 2 ion: Ti 2 O4 or TiO 2 . CaF2 : There are eight Ca 2+ ions on the corners for a total of one corner ion per unit cell. -645 kJ/mol. TiO 2 . 1 diamond per unit cell. The ratio of Ca 2+ ions to F ions is 4 Ca 2+ ions per 8 F ions: Ca 4 F8 or CaF2 . Face centered cubic 85.25 g / cm3 75a. 81d. 79. KCl(s). each shared between two unit cells. 19. with NaCl(s) falling in the middle of the series. 81a. Face centered cubic 83c. wholly contained within the unit cell. each wholly contained within the unit cell. 3. 335 pm 75b.71b. the lattice energy for LiCl(s) should be the most exothermic and CsCl(s) the least in this series. The coordination number of Mg 2+ is 6 and that of O2 is 6 also. 9. 7 . for a total of two face atoms per unit cells. 2 circles per unit cell. 71c. 81c. There are six Ca 2+ ions on the faces for a total of three face ions per unit cell. There are eight Ti 4+ ions on the corners for a total of one Ti 4+ corner ion per unit cell. Lattice energies of a series such as LiCl(s). NaCl(s). Face centered cubic 83b.23 g cm-3 75c. 73. There are four O2 ions on the faces of the unit cell. The unit cell has one light-colored square per unit cell.62 1023 cm3 . There are eight F ions. A smaller cation will produce a more exothermic lattice energy. Thus.51 g/cm3 83a. Length = 424 pm . there are a total of two Ti 4+ ions per unit cell. 87. 81b. volume = 7. There is one Ti 4+ ion in the center.45 77. MgCl 2 is much more stable than MgCl. and CsCl(s) will vary approximately with the size of the cation. 0 100 Time(min) 200 109. As gas is released from the tank.84 kg N2 Temperature (K) 2300 300 107. 54. The final condition is a point on the vapor pressure curve at 30.021 J/m 2 P2 (15. maintaining a pressure in the tank equal to the vapor pressure of the substance at the temperature at which the cylinder is stored. Thus. The remaining gas will be quickly consumed. 1.0 Feature Problems 119. –0. 99a. but as a liquid in equilibrium with its vapor.971) 1 1 (14. the liquid will vaporize to replace it. and hence the reason for the warning. ln P R T .0 °C. less heat is needed to vaporize 1. the quantity of heat needed is 2257 J/g H2O. 104. for at the higher temperature the molecules of the liquid already are in rapid motion. 31 atm 99b.160) 120a. Some of this energy of motion or vibration will contribute to the energy needed to break the cohesive forces and vaporize the molecules.000 g of H2O at the higher temperature of 100.Integrative and Advanced Exercises 91. However. This makes sense. 87 C 103. 8 . This will continue until all of the liquid vaporizes. 16. At 100°C. 0. the situation of gas in equilibrium with liquid only applies to substances that have a critical temperature above room temperature (20 °C or 293 K).25 °C 100. °C.T + R ln T R (T2 T1 ) 1 1 2 1 120b. 169 K Copyright © 2011 Pearson Canada Inc. In many instances. the substance in the tank is not present as a gas only. 93. after which only gas will be present. 96. We would expect the % dimer to decrease with temperature.55) T2 (0.3% dimer . CH3CH2OH 134. 362 pm 139b. The anwer is (a). If too much is present initially. Copyright © 2011 Pearson Canada Inc. 133a. the liquid will all be converted to benzene(g) before Tc is reached. All the CCl4 will be in the vapor phase. 130. 141. The correct order of boiling points based on molar masses is: N2 < F2 < Ar < O3 < Cl2. 8. 129. 139d. The answer is (c). 4. and therefore only benzene(l) will be present at the time Tc is reached. Ne < C3H8 < CH3CH2OH < CH2OHCHOHCH2OH < KI < K2SO4 < MgO 136. 137. If there is too little benzene(l) in the sealed tube in Figure 12-22 initially. Refer to the photograph on page 518 of water boiling under a reduced pressure. 666 kJ 139a. the water may cool to 0 °C and ice may begin to form. 132. If the vapor is evacuated fast enough. The answer is (c). the liquid will expand and cause the benzene(l) to condense. The answer is (a). The answers are (d) and (f).74×107 pm3 139c. 135. 131. The answer is (d).2211022 g 139f. 138b. 9 . to supply the required ΔHvap.91 g/cm3 140. The answer is (b). 4 atoms/unit cell. 74% 139e.Self-Assessment Exercises 128. (CH3)2O 133c. 138a. 4. C10H22 133b. O3 is out of place. 2.13 Å 144. for example.142. 10 . 143. because it takes much more energy to overcome the covalent bonds in the solid (such as. Copyright © 2011 Pearson Canada Inc. The answer is (c). A network covalent solid will have a higher melting point. diamond) than to overcome ionic interactions. 8. Ppen = 127 mmHg .0 mmHg .3038 M .469. Oxalic acid should be the most readily soluble in water because it is polar and can form hydrogen bonds. 7a.89 atm 11b.927 . Ptot = 239 mmHg.2% ethanol by mass 1b.3246 m .005815 . 0. (c) 4.5 mmHg .857 atm 8b. Ptot = 64. (a) 0. Ptol = 13.798. 4b. Phex = 112 mmHg . 2a. 3. yb = 0. yhexane = 0. (b) 0. 8.531. 3a. 0.560 mL conc soln Copyright © 2011 Pearson Canada Inc. (a) 0. 6b. Suggestion 2: 46 g of NH4Cl crystallized.03593 2b.0105 atm 10a. 1 . (c) C17 H 20O 6 N 4 .Chapter 13 Answers Practice Examples 1a. 6.202. 10b.328 atm 6a. 0. yt = 0. KNO3 47% KClO 4 38% K 2SO 4 21% 5a. 0.122 m . (b) 3. (c) 0. (a) 0. 7b.73 M . Pbenz = 51. 16. ypentane = 0. 0.0 mmHg. 4a. I 2 should dissolve well in CCl 4 by simple mixing.24 103 g 9a. 11a.5 mmHg. Suggestion 1: 31 g NH 4 Cl crystallized . (b) 377 g/mol . 3b. Lower than 760.717 atm O 2 pressure 5b.60 104 g/mol 9b.34 m . 8a. the density of water is close to 1.00654 M 25. the mass in grams and the volume in mL are about equal. based on smallest lattice energy.85 M 21. 2 . 4. The reason for this assertion is that salicyl alcohol contains a benzene ring. 1.42 mmHg. For ethanol. 54. the density is about 0. As long as the final solution volume after mixing is close to the sum of the volumes for the two pure liquids. 113 mL conc.48 mmHg. (b) Benzoic acid and (d) 1Butanol are only slightly soluble in water. 0. 53. 3. 4.187 . 5.427 .3% H 2 O . on the other hand. For water. Exercises bg 1. the percent by volume of ethanol will have to be larger than its percent by mass.7 g NaBr/100 g solution.19 M .410 m 27. (c) Formic acid and (f) propylene glycol are soluble in water. NH 2 OH s should be the most soluble. 15.26 m . (b) 0. KF is probably the most water soluble. soln 23. 1. 0. and also can use its OH groups to hydrogen bond to water molecules.0 g/mL. (a) 0.Integrative Example A. C7 H16 = . C9 H20 = 0. B. 31a. Copyright © 2011 Pearson Canada Inc. Salicyl alcohol probably is moderately soluble in both benzene and water. It would only be true in those cases where the density of the other component is greater than the density of ethanol.386 . χphenol = 0.83 104 L sol'n 13. 21. (a) Iodoform and (e) chlorobenzene are insoluble in water. 7.8 g/mL.6 g HC2 H3O 2 17. This would not necessarily be true of other ethanol solutions.8 g I2 29. 9. which would make it soluble in benzene. 11. C8H18 = 0. 5.80 10-4 M 19.161. 29. 4 102 g CH 4 (natural gas) 47. 3. 3. 18. 1. 42.2 mmHg 55. 207 mL glycerol 37.00204 33b. the solution volume remains essentially constant as a gas dissolves in a liquid. 33a. Psol'n = 23. 0.7 mol% C8 H18 . 51. 4. 0. 1.66 m 41a.40 105 M 49. 22. 39.6 mol% C9 H 20 . 2.4 L solvent 65. Ptot = 56. the solution in the plant material moves across the semipermeable membrane in an attempt to dilute the salt solution.3 mmHg . 8.0101 35.8 105 g/mol 67. ye = 0. Unsaturated 41b.303 106 .32.68. Both the flowers and the cucumber contain ionic solutions (plant sap). ys = 0. 63. 4. leaving behind wilted flowers and shriveled pickles (wilted/shriveled plants have less water in their tissues).7 mol% C7 H16 . 57. 3 .9 mmHg 53.36 1016 atoms . Thus. This number is directly proportional to the mass of dissolved gas.0 g KClO 4 43. but both of these solutions are less concentrated than the salt solution.6 mmHg. 38. Pb 1.48 103 M 45. Pt = 16. Changes in concentrations in the solution result from changes in the number of dissolved gas molecules.29 103 mmHg 59.2 104 g/mol 61.31b. Pb = 40. 21 atm Copyright © 2011 Pearson Canada Inc. Because of the low density of molecules in the gaseous state. results in the formation of NH 4 C 2 H 3O 2 (aq). mole fraction of water = 0.56 103 m 2 100b. C 6 H 4 N 2 O 4 75. 73.372° C 81c. C 4 H 4S 77. The combination of NH 3 (aq) with HC 2 H 3O 2 (aq).62%. 682 g H 2 O 92. 4 . < –0.69.186° C 81b.99839.372° C 81d. The answer is (d). –0.C / m 71b. 85. 8. 120 g NaCl. –0.186° C 83. Volume % N 2 65. Copyright © 2011 Pearson Canada Inc. No.2 102 g/mol 71a.372° C 81f.34×10-7 m2 105. –0. 79. 20. –0.00161 .38% .558° C 81e. 4. –0. urea sucrose 0. 1 101 % palmitic acid 97. 100a. Cyclohexane is the better solvent for freezing-point depression determinations of molar mass.186° C 81g. which is a solution of the ions NH4+ and CH3COO-. volume % O 2 34. Integrative and Advanced Exercises 88. This solution of ions or strong electrolytes conducts a current very well. 1. 1. because a less concentrated solution will still give a substantial freezing-point depression. 1. –0.5 kg H 2 O 95.04 81a. 50 114b. 114c. CaCl 2 6 H 2 O deliquesces if the relative humidity is over 32%. 124.32 M Copyright © 2011 Pearson Canada Inc. 0.96 L Feature Problems 114a. close to the defined freezing point of 0. 0. 1. HCl = 0.60 C . 5 . 116a. 125. 121.12 and a boiling temperature of 110 C .0125 g / L . 115c. its saturated solution will have a low water . 0. The answer is (b).112 115a. Tf = 0. The composition of HCl(aq) changes as the solution boils in an open container because the vapor has a different composition than does the liquid. 122. which in turn will produce a low relative humidity. If the substance in the bottom of the desiccator has high water solubility. 111.16 M Na 126b. 90. the component with the lower boiling point is depleted as the solution boils. CaCl 2 6 H 2 O will deliquesce. Unsaturated 126a. fair agreement with 0. Thus. The answer is (a). The boiling point of the remaining solution must change as the vapor escapes due to changing composition. Using 0. 116b.52 C .04% 115b. The azeotrope occurs at the maximum of the curve: at HCl = 0. Self-Assessment Exercises 120.52 C .3 102 g CuSO4 5H 2O 109.73 atm . The answer is (c). HCl 0. 123. Thus.60 C . 114d. 5.107.92% NaCl (aq) and i = 2.0 gives Tf = 0. 0. The answer is (d). The answer is (d). 130.60 o C 127a. 135. 7. 4-a. 1-b. The magnitude of ΔHlattice is larger than the sum of the ΔHhydration of the individual ions. 75.242 127e. The answer is (e).80 M C3 H8O3 127b. 131.8% 128. 3-d. 0. 134. 136. The answer is (a). 133. 24.126c. Copyright © 2011 Pearson Canada Inc. The answer is (b).0 m H 2 O 127d. The answer is (b). The answer is (a). 2-d. 0. 34. 6 .4 M H 2 O 127c. 132.1 atm 126d. 8. The answer is (c). 46 105 M s 1 1b. 272 mmHg 7b. 4.28 103 M / min. Rate = k1k3 /k2 NO 2 F2 .56 103 mmHg. 2. agrees with the experimental rate law.1 min 7a. 8a.4 107 M min 1 5a. the reaction is first-order in [H2O2]. The first step is the slow step.9 107 M min 1 4a.62 104 M s 1 2a. 9a. First-order 3b.17 M . First-order. k = 6. 2b. Rate of reaction = k1 NO 2 . (a) 1.13. which is a bit faster than the average rate over the first 400 seconds.15 M determined in text Example 14-2b because the text used the initial rate of reaction. The two k values agree within the limits of the experimental error and thus. 2 10b.37 M . (b) 1. 311 K 10a. Zero-order. 3. 64. 1 . (b) 0.2% 6b.Chapter 14 Answers Practice Examples 1a. Substitute the provided values into text Equation 14. 2. k = 9. 25. 43 s 9b.9 107 mmHg . (2) The predicted rate law. Copyright © 2011 Pearson Canada Inc. 4. 6a. 1. 3a.3 104 M s 1 . 1. This value differs from the value of 2. (1) The steps of the mechanism add to produce the overall reaction.4 102 M 2 s 1 4b. (a) 3. 8b.93 103 s 1.0 M 5b. 1400.102 M 2 s 1 13. At low pressures ([A] ~ 0 and k -1 [A] k 2 hence k2 >> k-1[A]). (a) 44 kJ/mol .Integrative Example A. 15a. 3. 11b. First-order with respect to A. k -1 [A] k -1 B. k = 5.52 105 M/s 7b. True 15b. 0. 11c.1 10-4 Ms-1 1b. Third-order overall. mmHg 11a. second-order with respect to B. Rate = k [NO]2[Cl2].1 10-4 Ms-1 1c.4 min 7a.5 102 s 9a. 3. h . The rate of reaction is: Rate = k2[A*] = Exercises 1a. mmHg 9b. the denominator becomes ~ k-1[A] and the rate law k k [A]2 k k [A] = 2 1 is Rate = 2 1 . 3. 5. At high pressures ([A] is large k2 and k-1[A] >> K2). 4. 3.0 103 M s 1 5a. 9. 2 . (b) 20.575 M 5b. the denominator becomes ~ k2 and the rate law is k 2 k1 [A]2 Rate = = k1 [A]2 . Second-order with respect to [A].13% Copyright © 2011 Pearson Canada Inc. 1.3 10-4 Ms-1 3. 3000.70 M2 s1. 0. First-order with respect to [A]. 0. k 2 k1 [A]2 .357 M 7c. False 17a. +0. 0.010 M s 1 31b. 39b. 1. 37a. 0. 0. Set III 29. 0.3 L CO2 25a. 2.39 104 M / min 35.17b. k = 0. A second-order reaction's half-life varies inversely proportional to [A]0. 43.00193 M/s 19a. Set II 27b. 25b. 70 s 31a. Rate = k A 2 and k = 0. 0. 0. Second-order 41.089 M / min .6 min 23a. 20.020 L mol 1min 1 . 19 min 19b. +0. 1.0048 M s 1 31c. 218 min 23b. Rate = k[HI]2. Zero-order 37b. A zero-order reaction has a half life that varies proportionally to [A]0.00118 M-1s-1. that is. the half-life decreases. therefore. 72 s 39a. The reason for the Copyright © 2011 Pearson Canada Inc. The reaction is second-order in HI at 700 K. Set I 27c.148 M 27a. increasing [A]0 increases the half-life for the reaction.0034 M s 1 33.022 M / min . The virtual constancy of the rate constant throughout the time of the reaction confirms that the reaction is first-order. 3 . as [A]0 increases.86 103 s 1 25c.18 g A 21. by enabling an alternative mechanism.. even though the temperature does not.. The rate of a reaction depends on at least two factors other than the frequency of collisions. In a second-order reaction. Exothermic 51. The second factor is whether the molecules in a given collision are properly oriented for a successful reaction. Therefore. The lower activation energy of the alternative mechanism means that a larger fraction of molecules have sufficient energy to react. 4 . 45b. 45a.. Step 3 49d. the rate of reaction increases as the square of the [A]0. Two 49b. the fraction of those collisions that have sufficient energy to overcome the activation energy increases much more rapidly. Endothermic 49f. the rate of reaction will increase dramatically with temperature. Three 49c. 45c. 160 kJ/mol Copyright © 2011 Pearson Canada Inc. The first of these is whether each collision possesses sufficient energy to get over the energy barrier to products. Although the collision frequency increases relatively slowly with temperature. 49a. Potential Energy (kJ) 47a. Thus the rate increases. Step 1 49e. 63 kJ/mol A . The addition of a catalyst has the net effect of decreasing the activation energy of the overall reaction.difference is that a zero-order reaction has a constant rate of reaction (independent of [A]0).B Transiton State Ea(reverse) = 63 kJ Ea(forward) = 84 kJ H = +21 kJ Reactants A+B Products C+D Progress of Reaction 47b. 61. 59b. Although a catalyst is recovered unchanged from the reaction mixture. these negative catalysts are called inhibitors. Platinum is rather nonspecific.1 -11. the rate-determining step of the overall reaction.00325 1/T (K-1) 0. it does “take part in the reaction. An enzyme is quite specific. The molecularity is equal to the order of the overall reaction only if the elementary process in question is the slowest and. whereas enzymes are generally soluble in the reaction media (homogeneous). Since the activation energy for the depicted reaction is 209 kJ/mol. An excess of substrate must be present. The function of a catalyst is to change the mechanism of a reaction. Both platinum and an enzyme have a metal center that acts as the active site. 112 kJ mol-1 53c.8 kJ/mol 55b. 67. which is slow. 51 kJ/mol 57b. The second step. usually catalyzing only one reaction rather than all reactions of a given class.1 ln k -8.00345 0. Usually. we would not expect this reaction to follow the rule of thumb. platinum is not dissolved in the reaction solution (heterogeneous). Substituting for [N2O2] gives Copyright © 2011 Pearson Canada Inc. thus.3 102 s 55a. The new mechanism is one that has a different (lower) activation energy (and frequently a different A value). In addition. 34.Plot of ln k versus 1/T 0.1 53a.1 -10. however. y = -13520x + 37. 5 . 65.The rate of this rate- determining step is: Rate = k2 H2 N2O2 .1 53b. than the original reaction. is: H 2 + N 2 O 2 H 2 O + N 2 O . 63. the elementary process in question should be the only elementary step that influences the rate of the reaction.00365 -7.4 -12. Generally speaking.1 -9. 59a.” Some catalysts actually slow down the rate of a reaction. catalyzing many different reactions. 61 o C 57a. 2. Yes 95a. First-order 74b. Both reactions are first-order. 0. 1. 1. 94b. First step. k -1 90. Rate = 69.k2 k1 2 H 2 NO . 95c.29 min–1 74c.067 x 10-6 M-1s-1 91. rateoverall = k 2 [CHCl 3 ] 1 [Cl 2 (g)] . 0. 75. 0. 0. 95b.9235 M Copyright © 2011 Pearson Canada Inc. 6 .103 M/min 74d. The second step. 1. 0.015. No 94c.86 M 78.294 M/min. This k1 result conforms to the experimentally determined reaction order.0275 s-1 94a. 0. The reaction is first-order in H 2 and second-order in [NO].064 M/min 74e.9 102 min 1/2 k 87. {Units of k} M10 s 1 82. k 0. k1 Cl2 (g) + NO(g) NOCl(g) + Cl(g) k-1 k1 Cl(g) + NO(g) NOCl(g) k-1 71. k k S S d S1 : S2 k 2 S1 : S2 2 1 1 2 dt k 1 k 2 Integrative and Advanced Exercises 74a. min VN 2 . Reaction Rate (mM min ) 4.7 1. Time.25 -3.731 M N 2 O Feature Problems 96a.071 min 1 .3 32.75 Copyright © 2011 Pearson Canada Inc.4 37.7 -2 0.4 58. estimated half-life = 10. 0 Time (min) Plot of ln[C6H5N2Cl] versus time 10 20 30 ln[C6H5N2Cl] -2.5 48.0 0. 1. 7 . 96d.5 103 M min 1 96f. About 20 minutes.3 -3 -2 1. 96h.4 50.3 46.75 y = -0.3 3. -4.5 min.2. 96g. 5.0661x .7 96c.25 96i.3 2.8 19.1103 M min 1 96e.75 -3. Rate = k C6 H 5 N 2 Cl .8 min .7 2. mM 3 6 9 12 15 18 21 24 27 30 10. Calculated half-life = 9.3 58 47 time(min) 0 39 3 6 32 9 26 12 21 15 17 18 14 21 12 24 27 10 0 30 T(min) 3 3 3 3 3 3 3 3 3 3 [C6H5N2Cl](mM) 71 58 47 39 32 26 21 17 14 12 10 [C6H5N2Cl](mM) -13 -11 -8 -7 -6 -5 -4 -1 96b.66 -4.7 2. 0.3 44.0 1. k avg = 0.3 26.3 41. mL 0 0 C H 71 6 5 N 2 Cl .95d. 2 103 M 1 s 1 97d. This is exactly the same as the experimental rate law.0062 M 1 s 1 . 2. d C k1 A B .0661 min 1 2 97a. 45.33 M 111a. 110a.012 M 1 s 1 . 97b.0030 M 1 s 1 . second-order overall. 0. 0.312 M 110c. 97e. 102. The reaction is second-order. 8 . The answer is (c). 3. 60.0 min 108b. 1. 0. dt Copyright © 2011 Pearson Canada Inc. 103. 107. It is reasonable that the first step be slow since it involves two negatively charged species coming together. The mechanism is consistent with the stoichiometry. 106. The answer is (c).0 min 109.96j. first-order in I . The answer is (b). 0. First-order in S2 O8 .0×10-3 M·s-1 108a. The answers are (b) and (e).7 105 M s 1 97c. because the half-life doubles with each successive half-life period. and thus this should not be an easy or rapid process. 105. Self-Assessment Exercises 101. 6.0014 M 1 s 1 . The answer is (d). 3. This agrees with the observed reaction rate law. We know that like charges repel. 0. The answer is (a). 104. 0. 51 kJ/mol 97f.024 M/min 110b. The rate of the slow step of the mechanism is Rate1 = k1 S2 O 82 1 1 I . with the given proposed reaction mechanisms. the rate law for the product(s) is given as follows: d C dt k 2 k1 A B 2 k 1 k 2 A . However.111b. Adding the two reactions given. 115. This does not agree with the observed reaction rate law. we still get the same overall stoichiometry as part (a). 114. 112. The answer is (c) Copyright © 2011 Pearson Canada Inc. 9 . 113. The answer is (a). The answer is (d). The answer is (b). (a) Adding extra CaO(s) will have no direct effect on the position of equilibrium. Kc = 4b. 6b. that the NO2 concentration will still ultimately end up being higher than it was prior to pressurization. Decreasing the cylinder volume would have the initial effect of doubling both N 2 O 4 and NO 2 .7 ×10−6 3b. (b) The reaction will shift left. 6. ≈ 0. and smaller quantities of products than there were initially. Note.0098 M 2a. In order to reestablish equilibrium. The equilibrium system will shift right. 2. however. (c) Addition of solid CaCO 3 will not have an effect upon the position of equilibrium. 95 4a. A change in overall volume or total gas pressure will have no effect on the position of equilibrium. 8a. 5a. Cu 2+ = 1.Chapter 15 Answers Practice Examples x 1a. The net reaction will proceed to the right. Kc = 4 4 {P ( H 2 O )} [ H 2O] {P ( H 2 )}4 4 Kp = Kc . 1 . 7b. 1.3 ×102 2b. Greater 8b. Greater 9a. some NO2 will then be converted into N2O4.3 ×10−4 Copyright © 2011 Pearson Canada Inc. 7a. There will be greater quantities of reactants. Kp = . 8.22 1b. Ca 2+ 5 HPO 4 H+ 2− 3 4 [H2 ] .9 ×10−5 3a. 6a. 5b. 2 .5 ×10−4 . (b) 0.47 L . V 2+ = Cr 3+ = 0.47 L.481 atm 11b.2 g N 2 O 4 10a.113×10-6. K = ( ) ( ) c 2 S2 O8 2− Fe 2+ 3+ [ NO2 ] Kc = 2 [ NO] [O2 ] 2 3a. Cu ( s ) + 2 Ag ( aq ) Cu ( aq ) + 2 Ag ( s ) . V 3+ = Cr 2+ = 0. Exercises [ 1a. (b) ~8. 3b. 2 COF2 ( g ) CO 2 ( g ) + CF4 ( g ) . K= 1.20 − 0.0057 M.488 = 0. [F6P]eq = 1. B. F6P will decrease with an increase in temperature. K c = 2 Ag + + 2+ 2 1c.154 M. (a) The amount of N 2 O 4 will decrease. 25. 0.022 11a. 13b.7 ×10−3.262 mol HI 12b. Ag + = Fe 2+ = 0.716 atm 12a. Fe3+ = 1. (a) 8. Integrative Example A.71 M. c 10b. K p 0. 0. Kc = Zn 2+ Ag + 2 Copyright © 2011 Pearson Canada Inc. 13a. S2 O8 2− ( aq ) + 2 Fe ( aq ) 2 SO 4 2+ 2− 2 SO 4 2− Fe3+ aq + 2 Fe aq . K p = 3.0118 mol.9b. K c = CO 2 ][ CF4 ] [COF2 ] 2 Cu 2+ 1b.488 M. 0. 9.OH − 3c. K p = 0.429 9. NH 3 ( g ) + 7 4 NO 2 ( g ) + O 2 ( g ) 3 2 H 2O ( g ) 25b.7 ×10−16 13. Kc = CO 3 2 2− [ HF] 1/2 1/2 [ H 2 ] [ F2 ] 5a. 5c. K c = 1 [Cl2 ] [F2 ]3/2 1/2 7a. 3 . K = a(H2CO3)/a(CO2) .1×10−18 M 23.0012 7b.613 21. K c = [N 2 O]2 [N 2 ]2 [O 2 ] 5d. 0.0313 .28 × 10-3 . 17. 26. 0. 11. 0.55 × 105 7c.3 19b. K c = [ NH3 ] Kc = 3 [ N 2 ][ H 2 ] 2 5b.516 25a. 0. 5. K c = 1.03 ×1019 Copyright © 2011 Pearson Canada Inc. 5 ×1014 15. 9. 4.106 19a. 0. K = [ H 2CO3 ] c PCO2 P . n {CO} n {H 2 O} × V V 27a. 31b.601 atm total. The reaction will proceed to the right.82 < K c = 100 . Sketch (b) is the best representation because it contains numbers of SO2Cl2. Copyright © 2011 Pearson Canada Inc. 35g CH 3CO 2 H .46 atm 55.033 mol. Since V is present in both the denominator and the = n n{CO 2 } {H 2 } × V V numerator. K c = K p = 0. 31a. The mixture described cannot be maintained indefinitely. 749. 68g H 2 O . K c .41 mol. 32 g CH 3CO 2 C2 H 5 . the reaction is at equilibrium. n PCl5 = 0. 35. n PCl3 = 0. 41. Qc = 0. it can be stricken from the expression. 39b.659. Thus. The reaction will shift to the left. 39a.0 ×10−5 = = ( 0. Since Q = K. Amount H 2 = amount I 2 = 0. Q 4.0725 mol Cl 2 . the equilibrium system is disturbed by removing the products and the system will attempt to re-establish the equilibrium by shifting toward the right. CO ][ H 2 O ] [= [CO2 ][ H 2 ] 27b.529 atm O 2 .58 atm 45. Continuous removal of the product has the effect of decreasing the concentration of the products below their equilibrium values. to generate more products. The mixture is not at equilibrium. n Cl2 = x = 0. K = [aconitate ] .478 mol PCl 5 .0 mmHg 51.234 mol HI. 47.14 atm 53b. Amount HI = 0. 53a. 0.20 mol = amount Cl 2 . 4 .0 ×10−4 mol Cl2 37a. SO2. Amount PCl 3 = 0. 37b. 29. 33. 26 g C2 H 5OH . 49.031. 5.00128) [citrate] 0.623 mol PCl 3 . Amount PCl5 = 0. 32. and Cl2 molecules that are consistent with the Kc for the reaction. 30. 4. that is. Ptotal = 0. Increases 57c. More NO(g) formed. [Fe3+ ] 0. 87. 69.57a.049 M. 5 . Hb:O2 is reduced. no change in equilibrium position. 59b.0 81. No shift. 12. No change 57d.1×10−3 78. No change 59a. [Fe 2+ ] = = 0.070 moles H 2 O(g) . Shifts right. P{N2(g)} will have decreased when equilibrium is re-established. 91. Rate increases. No effect. 65b. 0. The pressure on N2O4 will initially increase as the crystal melts and then vaporizes.16 mol H 2 . Hb:O2 increases.6 atm Copyright © 2011 Pearson Canada Inc. 0.79 atm 76. its concentration will remain the same because it is a solid.085 moles CH 4 (g) . Integrative and Advanced Exercises 73. 24. 2. 0. Decreases 57b. but over time the new concentration decreases as the equilibrium is shifted toward NO2. toward products . 67. 61. 65c.34 ×10−4 87. 63c. Shifts right. 0.1 g/mol 85. [Ag + ] = 0. 63b. towards products . 0.15 M . 65a. 3. 63a. 74.12 mol CO 2 . 2. While the amount of calcium carbonate will decrease.111 M .177 82. Less Cl2 is made. The answer is (c). 108b. 108c.9 108a. 102. 109. 4. Self-Assessment Exercises 101. 111a. 106. 105. 0. 104. More Cl2 is produced. 108e. The answer is (a). 6 . 110. SO2 (g) will be less than SO2 (aq). C(aq) = 4×10-4 M. The answer is (d).0 94. C(or) = 6×10-4 M. 103. 108d. Less Cl2 is made. Less Cl2 is made. The answer is (b).80 × 10-8 97. The answer is (a). 0.232 moles Copyright © 2011 Pearson Canada Inc. 107. No change. The answer is (c).0578 moles 111b. 1. There will be much more product than reactant.Feature Problems 92. 4. OH = 4. M.9 10 8 5b.3 10 2 . 2. (c) Forward direction: HCl is the acid and C2H3O2 is the base. 1b. [H3O+ ] [HOOCCH 2COO ] 3. (b) Forward direction: HSO4 is the acid and NH3 is the base. OOCCH 2 COO = 2.4 103 .0 10 12 M. K a 2 = 5.0025 M.0 106 M.519 4b.28 8a. 2.Chapter 16 Answers Practice Examples 1a. (a) Forward direction: HF is the acid and H2O is the base. Copyright © 2011 Pearson Canada Inc.4 104 9a. Reverse direction: F is the base and H3O is the acid. 17% 8b.6 106 6a.1 L 3a. OH = 7. 1.3 105. 2. Reverse direction: SO42 is the base and NH4 is the acid. aq + H 2 O(l) H 2 CO3 (aq) CO 2 H 2 O aq + OH aq HCO3 aq + H 2O(l) PO 4 3 2a.11012. 11. 9b. 3. 11. 1.66 7a.739 5a. H 3O + = 1. 1.670 4a. 3b.7 102 M. NO 2 aq + H 3O + aq . 1 .29 7b. HNO 2 aq + H 2O(l) CO32 aq + H 3O + aq HCO3 aq + H 2O(l) HPO 4 2 aq + OH aq . Reverse direction: Cl is the base and HC2H3O2 is the acid. K a1 = 5. 13. I = H 3O + = 0. 2.8 6b. 2b. SO 4 2 0.68 5. 3. HOClO2 or ClO2(OH) has m = 2 and n = 1. (b) Ka = 10-2. HOClO3 or ClO3(OH) has m = 3 and n = 1. 8.08 13b. which is in good agreement with the accepted pKa = 7.21 M .3 108 K b = 1. (a) BF3 is a Lewis acid and NH 3 is a Lewis base. Codeine hydrochloride 12b. We expect Ka to be very large and in good agreement with the accepted Ka = -8. HOCl or Cl(OH) has m = 0 and n = 1.026 M . We expect Ka 10-7 or pKa 7. 11a. (c) basic. (a) Determine values of m and n in the formula EOm(OH)n. H 3O + = 0. Hydroxide ion and the chloride ion are the Lewis bases. H 3O + aq + HPO 4 2 aq H 2 PO 4 aq + H 2 O(l) OH aq + H3 PO 4 aq H 2 PO 4 aq + H 2 O(l) K a 2 = 6.014 M . b g 15b.92. HClO 4 and CH 2 FCOOH are the stronger acids in the pairs. pKa = -8. Ka = 10-pKa = 103. (b) H 2 O is a Lewis base and Cr 3+ is a Lewis acid.0060M. (c) OH O P H OH Copyright © 2011 Pearson Canada Inc. 10b. 11b.6 103 M 14a. pH 5. HSO 4 = 0. Ka = 10-pKa = 108 which turns out to be the case. SO 4 2 0. We expect Ka 10-2 or pKa 2. in good agreement with the pKa = 1. 14b.010M . HSO 4 = 0. Al OH 3 and SnCl 4 are the Lewis acids.10a. We expect Ka to be large and in good agreement with the accepted value of pKa = -3. Basic 13a. (b) neutral. (a) acidic.4 1012 12a.19 M. H 2SO 3 and CCl 2 FCH 2 COOH are the stronger acids in the pairs. H 3O + = 0. 2 . 15a.7 B. Integrative Example A.52. HOClO or ClO(OH) has m = 1 and n = 1. H 3O + = 4.00165 M.96 103 M 15. H 3O + = 2. acid base acid base HS (aq) + H 2 O(l) H 2S(aq) 3c. H 3O + = 5. 9b. acid base base acid 5.5 mL Copyright © 2011 Pearson Canada Inc. 11. Acid 1b. acid base acid base 2 HSO 4 (aq) + H 2 O(l) H 3O + (aq) + SO 4 (aq) 3b. Forward 7b. pH = 13. H 3O + = 0. Acid HOBr(aq) + H 2O(l) H 3O + (aq) + OBr (aq) 3a. Base 1d. H 3O + = 1. Answer (b). base acid acid + OH (aq) base C6 H 5 NH 3 (aq) + OH (aq) C6 H 5 NH 2 (aq) + H 2 O(l) 3d. OH = 0. 9d.7 10 11 M.8 104 M. 3 .0087 M. Reverse 9a.00426 M. OH = 1. 1.Exercises 1a.110 12 M. 9c. Reverse 7c. Base 1c. OH = 6.0 1014 M.1 10 12 M. Acid 1e.40. 13. 7a.3 10 12 M. 8. OH = 0. 47a. HSO 4 = 0.7 103 27.(produced in the 3rd ionization) compared to the HPO42-. [H 3O + ] 0.9 mL acid 19. S2 = 1 1019 M. pH 2. HS = 9.62 25.4 1013 M. 47b.20 21.74 M. 10. Copyright © 2011 Pearson Canada Inc. H 3O + 0.04 33. 31.17. there is little PO43. 49a. H 3O + = 0.073 M. 11.01.2 105 M.76 M. 53. 0. 0.14. pH = 1. HS = 8. 2.2 105 M.7 105 M. 4 .0053 39b. OH = 1. HS = 2. H 3O + = 9. which is invalid at the 13 and 42 percent levels of ionization seen here.7 105 M. 23a. Sketch (b) 39a.5 107 M. 34 mg of NaOH 37. pOH = 12. 0.011 M .0098 M. 2.0030 M 23b. We would not expect these ionizations to be correct because the calculated degree of ionization is based on the assumption that the [HC2H3O2]initial [HC2H3O2]initial .86.[HC2H3O2]equil. 0. S2 = 11019 M. SO 4 2 0.4 g HC7 H5O 2 29. there is little HPO42.. S2 = 11019 M. H3O + = 8.(produced in the 2nd ionization) compared to the H3O+ (produced in the 1st ionization). 45. 1.063 g vinegar 35a. H 3O + = 2. 11. Because H3PO4 is a weak acid. 47c.53% 41. In turn.5 107 M.23 35b.0098 M 43. and very little compared to the H3O+. 0. 49c. H 3CCH 2 CCl 2 COOH 69. C18 H 21O3 NH H 2O C18 H 21O3 N H3O . HOClO 67c.49b. 51a. The largest Kb belongs to (c) CH3CH2CH2NH2.95. SO 4 2 = 6. 2O 71c. 65a. O H O H H H . F F + B F F B O H F H O H 71b. H 3O + = 0. HI 67b. acid is H2O and 5 . HClO 3 65b. base is O2–. O C O O O + H + C H O 71a. The smallest Kb is that of (a) o-chloroaniline. HSO 4 = 9 105 M.0087 M. 9.6 55. acid is BF3 and .066M. base is H2O. SO 4 2 = 0.41 103 M. HSO 4 = 0. H 3O + = 1.6 104 M.084 M.0 C 20 H 24O 2 N 2 H 2 + OH pK b = 9. F O + O H Copyright © 2011 Pearson Canada Inc.8 Second ionization : C 20 H 24O 2 N 2 H + + H 2O 2 2+ 53. C 20 H 24 O 2 N 2 H + + OH C20 H 24 O 2 N 2 + H 2O First ionization : pK b1 = 6. H 2 O H H . HNO 2 65c. pK b 7. acid is CO2 and base is H2O. H 3 PO 4 67a. which is supplied by H3O+ F H H H H H C C O C C H H H 75c. . base is 75a. acid is B(OH)3. F F F B F H H B F H H H C C O C C H Acid: electron pair acceptor H H H H 6 .S + S S S O 2- O O 2- O O O S 71d. Lewis acid 73c. Lewis acid - H O O H + H O B O H H O B O O H H O H . H H Base: electron pair donor Copyright © 2011 Pearson Canada Inc. H N N H H H N N H H H H O H H H H H O The actual Lewis acid is H+. and base is S2–. acid is SO3 73a.. OH. H 75b. Lewis base 73b. 10 103 M 83a. 83d. 83b.73 g HC 2 H 3O 2 Copyright © 2011 Pearson Canada Inc. (e pair donor) (e pair acceptor) H O S O H O S O Integrative and Advanced Exercises 81a. 83e.pair acceptor) (e. 2. Not matched. 83c. Not matched. Other solutes don’t provide acidic solutions or lead to masses or volumes that are very difficult to measure. 7 . 0. Not matched. Not matched. 83g.096 g NH 4 Cl . 9. Not matched.1 matched. Base 81b. I I I Lewis acid Lewis base (e. Acid 82. Base 81c. 83f. 87. pH 4.77.8 matched. 94. pH 10. 5 105 88.6 matched. pH 2. 83h. Acid or Base 81d.pair donor) I I I O H H O H O H O S O Lewis base Lewis acid 79. Not matched. 83i. OH– > CH3NH2 > N2H4 > F–> HSO3– 99c. 37. 111.05 M Ba(OH)2 116.2 mL 114. The answer is (d).05 M H2SO4 < 0. [H3O+] = 2. The solution is basic. 118. 2.10 M HI < 0. The answer is (b). 8.58 115.05 M KC2H3O2 < 0.2 M.06 M NaOH < 0.98 102c. 117. Feature Problems 102a. 119.14×10-12 M. H2SO3 > HF >N2H5+ > CH3NH3+ > H2O 99b.05 M HC2H2ClO2 < 0. 0. The answer is (d).05 M NH3 < 0. 11. 110. 109. The answer is (c). 120. The answer is (c).70 113.23 Self-Assessment Exercises 106.0 M NaBr < 0.50 M HC2H3O2 < 0. 112. 0. O 99a. 8.2 Copyright © 2011 Pearson Canada Inc. [NH3] = 1. The answer is (a). 9. (i) to the right and (ii) to the left.95.50 M NH4Cl < 1.15 102b. The answer is (b). 108. The answer is (e). 2.16 H H O P O H 98. 8 . The answer is (a). 107. 79. (d) 11. The hydronium ion and the acetate ion react to form acetic acid.103 M.482 M. (a) 3. [H3O+ ] 0. 1 .238.Chapter 17 Answers Practice Examples 1a.70. (c) 7. 9a.08.09 5.15.20.18 9b. 15 g NaC 2 H 3O 2 3a. 3b. 4a.497 M. (b) 11.824. 30 drops 2a. (a) 12. 8b.02. Because a buffer contains approximately equal amounts of a weak base (NH3) and its conjugate acid (NH4+). (a) 2. 3.00.283 C Copyright © 2011 Pearson Canada Inc.150 M.53. 7b. 2b. to prepare a buffer we simply add an amount of HCl equal to approximately half the amount of NH3(aq) initially present. (c) 3. (a) 11.55.45 Integrative Examples A.74. 1.26. (c) 7. CHO 2 = 0.51 5a. 12. (d) 8.1 6a. This will be achieved if we add an amount of HCl equal to approximately half the original amount of acetate ion.00. 10. (b) 3. 0. 6b. (c) 3.2 10 4 M. This reacts with ammonia to form ammonium ion. 4. (d) 5.18. H 3O + 1. [H3O+] = 0. (a) 0.53. 8a. (b) 1. (b) 9.94 4b.018M. HF = 0. (b) 2. (c) 9.21. 1b. HCl dissociates essentially completely in water and serves as a source of hydronium ion. All that is necessary to form a buffer is to have approximately equal amounts of a weak acid and its conjugate base together in solution.785.3 mL 7a. pH = 5. 21 g 5b. [HF] = 0. 1. Not a buffer. 4.88 15a. 0. 0. 4.035 M 5b. 0. 5a. 3. No change. 11b.68 11a. 11e. just a change in total ionic strength.0892 M 3a. 80 g/mol . Buffer 11f. 4. The pH changes are not the same because there is an equilibrium system to be shifted in the first solution. whereas there is no equilibrium.B.05 3b. Not a buffer. 11c.41 g NH 4 2 SO 4 17. Buffer 11d.9 105 M 7.8 g NH 4 2 SO 4 15b. 9. 0.77 M 9a. 9. 8. 0. Exercises 1a.11013 M 1c. Kb = 1 × 10-6.0 105 M 1d. Not a buffer. Not a buffer. 13. 4. 2 .0892M 1b.64 9b.16 Copyright © 2011 Pearson Canada Inc. for the second solution. 1.0 104 M 5c. 48 23b. Thus. L. 25c. Thymolphthalein changes color in basic solution. the pH is probably about pH = 6. one adds only enough indicator to show a color change and not enough to affect solution acidity. An indicator is. Bromthymol changes color in neutral solution. Copyright © 2011 Pearson Canada Inc. L of solution. If chlorophenol red is orange.00 .00). 1. With sufficient dilution. If bromcresol green is green. 3 . 29a. The Henderson-Hasselbalch equation depends on the assumption that: NH3 1.01 9.28 10 5 M . 21b. 8. pH = pK a log [NH 3 ] 0. the pH of the solution changes sharply at a definite pH that is known prior to titration. pH = 3. Chlorophenol red changes color in acidic solution. in 1000. Thus.89 to pH = 5. and NH 3 = 1. 1.. after all. mL .89 . 21a.19.26 log 9.23 25a. 23a.1mL 1. a weak acid. Determining the pH of a solution is more difficult because the pH of the solution is not known precisely in advance. 3.00 0.128 M [NH 4 ] 25b. Use HCHO 2 and NaCHO 2 to prepare a buffer with pH = 3. Vacid = 100.100 mol of acid or base per liter of buffer solution. often less than 2.7 . 29b. several indicators—carefully chosen to span the entire range of 14 pH units—must be employed to narrow the pH to 1 pH unit or possibly lower. These concentrations are consistent with the assumption.2 106 M .0 . its pH will equal 7. and NH 4 = 1.50 .28 102 M .0720 M 9.e. the pH is probably about pH = 4.0 pH units.8 105 M NH 4 .4-Dinitrophenol changes color in acidic solution. the solution will become indistinguishable from pure water (i. the given quantities of NH 3 and NH 4 will not produce a solution with pH = 9. Its addition to a solution will affect the acidity of that solution. In an Acid–Base titration. 3. 27b. 0. Vsalt = 58 mL.20 10 3 M . Since each indicator only serves to fix the pH over a quite small region. Bromphenol blue changes color in acidic solution. if the solution is diluted to 1000. 2. NH 3 = 7.00 L.99 25d. Bromcresol green changes color in acidic solution. If the solution is diluted to 1. NH 3 = 7. and these two concentrations are not consistent with the assumption.52 23c.00 M HCl 27a. However. 3. 45a.87 41b. Copyright © 2011 Pearson Canada Inc. Yellow 31d. 2.346 39b. since the titrant has the same concentration in each case.63 39a. Yellow 33. Red 31f.034 41a. Thus. the amount of acid is the same in each case. Yellow 31c. This is a relatively good indicator. which will hydrolyze to produce a basic (alkaline) solution. 0. 2. 11. Red 31b. 4 . 1. and it is the same amount of base that reacts with a given amount (in moles) of acid.84. The approximate pKHIn = 3. The volume of acid and its molarity are the same. At the equivalence point in the titration of a strong acid with a strong base.51 43. 35. The volume of titrant needed to reach the equivalence point will also be the same in both cases. the solution contains ions that do not hydrolyze. Yellow 31e. But the equivalence point solution of the titration of a weak acid with a strong base contains the anion of a weak acid.31a.1508 M 37. Basic. 53. bromthymol blue 10 12 . H 3 PO 4 aq + CO3 2 H 2 PO 4 aq + CO3 HPO 4 2 H 2 PO 4 aq + HCO3 aq aq 2 HPO 4 2 aq + HCO3 aq aq PO 43 aq + H 2 O(l) aq + OH aq Copyright © 2011 Pearson Canada Inc.10 M NaOH added (mL) 49a. the anion of a very weak acid.45b. 5 . This alkalinity is created by the hydrolysis of the sulfide ion. See sketch in 49a.4 mL acid 47c.5 mL acid 12 10 pH 8 6 4 2 0 0 2 4 6 8 Volume of 0. 49b. 49c. Alizarin yellow. Thymol blue. See sketch in 49a. 11. 17.9 mL acid 47b. 55a. 17. 47a. 51. 18% 69a. No 61b. No 61c.00 1014 .3 g KH 2 PO 4 . No 61e. 2. No Integrative and Advanced Exercises 63a. 71. 1. 26 g Na 2 HPO4 12H 2 O .6 70b. 0. 0. Yes 61f.3 mL 64c.0038 M 59b.0 108.8% 63c. CO 32 is not a strong enough base to remove the third proton from H 3 PO 4 . 59a. 250 mL 66a. 6 .6 103. 57. bromthymol blue or phenol red.56 70a. 27. 9. HSO 4 (aq) OH – (aq) SO 42 –(aq) H 2 O(l) 63b. 64a. K a 2 = 1. Equation (1): K = 1. Yes 61d. pH = pKa + log (f /(1 – f )) 69b.55b. ~3 drops NH3 73a. Equation (2): K = 1.8 mL 64b. 3. 6. No 66b. Copyright © 2011 Pearson Canada Inc. K a1 = 1.49 M 61a.8 109 . Mixtures of this type are referred to as buffer solutions. HC 2 H 3 O 2 (aq) OH (aq) Therefore neither added strong acid nor added strong base alters the pH of the solution very much. 77a. Copyright © 2011 Pearson Canada Inc. 189 mL 77e. A buffer solution is able to react with small amounts of added acid or base. it reacts with formate ion. When strong acid is added. 77b. 77c. CHO 2 (aq) H 3 O (aq) HCHO 2 (aq) H 2 O Added strong base reacts with acetic H 2 O(l) C 2 H 3 O 2 (aq) acid. 7 .3% 81a. The extremely large size of each equilibrium constant indicates that each reaction goes essentially to completion. 8.73b. 119 mL 77d. 78 91c.4 83. Bromthymol blue or phenol red.8 82a. The H2PO4–/HPO42. 4.81b. 4. Copyright © 2011 Pearson Canada Inc. 8 . 3. 8.17 84. 5.2 89a. 12.system.40 81c.6 103 91b. 4. 82b. H 3O K1 CO 2 g Ca 2 K2 K3 K4 89b. 82c.57 91a. 7. where pH = pKa = 4.Feature Problems 92a. The two curves cross the point at which half of the total acetate is present as acetic acid and half is present as acetate ion. Copyright © 2011 Pearson Canada Inc. 92b.74 . f (H3PO4) f (H2PO4-) f (HPO42-) f (PO43-) 92c. 9 . This is the half equivalence point in a titration. 10 . C6H5NH2 + H3O+ → C6H5NH3+ + H2O 98c. HPO42. Use bromthymol blue. C6H5NH3+ + OH¯ → C6H5NH2 + H2O. H2PO4¯ + OH¯ → HPO42.Self-Assessment Exercises 98a. 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Vol of Titrant (mL) pH 99b.+ H2O. The pH at the equivalence point is 7. The pH at the equivalence point is ~5.+ H3O+ → H2PO4¯ + H2O pH 99a. CHO2¯ + H3O+ → HCHO2 + H2 98b. 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Vol of Titrant (mL) Copyright © 2011 Pearson Canada Inc. Use methyl red.3 for a 0. HCHO2 + OH¯ → CHO2¯ + H2O.1 M solution. 11. 3.20 100c. 8. The answer is (d).00 100d.1 101. 103.pH 99c. Copyright © 2011 Pearson Canada Inc. The answer is (b). Use phenolphthalein. Use bromthymol blue. The answer is (c). The pH for the first equivalence point (NaH2PO4– to Na2HPO42–) for a 0. 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Vol of Titrant (mL) pH 99d.1 M solution. The pH at the equivalence point is ~8. 4.7 for a 0. 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Vol of Titrant (mL) 100a. 11 . 102.12 100b.1 M solution is right around ~7. 32 106. 109. 12 . The answer is (b). 8. The answer is (b).104. pH < 7 110c. 108.75 107. The answer is (a). The answer is (b). 105. 10. pH > 7 110b. 110a. pH = 7 Copyright © 2011 Pearson Canada Inc. 8a. 3 ×10−7 2b.5 ×10−7 M. 1b. Cu ( OH )2 ( s ) Copyright © 2011 Pearson Canada Inc.55 mg 4a.7 ×10−8 M 3b. 0.02 M 7a. (b) Ksp = [Ca2+][HPO42−]. 3. Chromate.3 ×10−4 M 4b.9 ×10−9 3a. Yes 9a. 9. Yes 5b. Ba 2+ = 5. Cu 2+ ( aq ) + 2 OH − ( aq ) . (a) CaHPO 4 ( s ) 2a. No 8b. 10 drops 6a. No 6b. Ca 2+ ( aq ) + HPO 4 2− ( aq ) .8×10-21 M 5a.12% 7b.074M 9b. 0.7 ×10-3 M 10a. 1. (b) Ksp = [Ag+]3[PO43-]. 0. (a) Cu 2+ ( aq ) + 2 OH − ( aq ) → Cu ( OH )2 ( s ) .Chapter 18 Answers Practice Examples 1a. (c) ( aq ) + 2 OH − ( aq ) . (a) Ksp = [Mg2+][CO32-]. 1. 8. 0. (b) Cu ( NH 3 ) Cu ( OH )2 ( s ) + 4NH 3 ( aq ) 4 2+ Cu ( NH 3 ) Cu ( OH )2 ( s ) + 4 NH 3 ( aq ) 4 2+ ( aq ) + 2 OH − ( aq ) . 1 . s .7 Integrative Examples A. 4 × 10−6 M 13b. In Example 18-13 the expression for the solubility. K sp = Ag + SO 4 2 − 1b. 14b. K sp = PuO 2 CO3 Copyright © 2011 Pearson Canada Inc. Qspa = 0. Yes 11b.6 × 107 .1×10−4 > 6 ×10−30 = K spa . 87 ×106 g Ca ( NO3 )2 B. For all scenarios. Exercises 2− 1a. (c) ( aq ) + 4H3O+ ( aq ) 4 Zn ( H 2 O )4 2+ Zn ( OH ) ( s ) + 4 H 2 O ( l ) . (d) ( aq ) + 2 OH − ( aq ) 2 2+ Zn ( OH ) Zn ( OH )2 ( s ) + 2 OH − ( aq ) 4 2− ( aq ) .1000 M and K sp × K f = [NH 3 ] − 2 s K f is always 1. Qspa = 1. No 12a. We see that s will decrease as does Ksp .022 < 6 × 102 = K spa . 2. 14a. K sp = Ni 2+ PO 4 3 2 2+ 2− 1d. for Ag 2S. For FeS. No precipitate is remaining. Zn ( NH 3 ) 10b.20 M 13a. 0. (a) Zn 2+ ( aq ) + 4 NH 3 ( aq ) 4 2+ ( aq ) . (b) Zn ( NH 3 )4 2+ Zn ( H 2 O ) ( aq ) + 4NH 4 + ( aq ) . Ag2SO4. 11a. K sp = Ra 2+ IO3 2 3− 1c. 0. 2 . where NH 3 is the concentration of aqueous ammonia. of a silver halide in an aqueous ammonia solution.58 M 12b. the [ NH 3 ] stays fixed at 0. was given by: s . 5 ×10−5 21. The presence of KI in a solution produces a significant I − in the solution. Not as much AgI can dissolve in such a solution as in pure water. 27 ppm 27. 19. 4.5 × 10−9 5.6 ×10−11 3 2− 3b. 1.3 ×10−6 M 15c.7 ×10−4 M 15b. not as much AgI can dissolve as in pure water. since the ion product. Yes 29. The solubility of Ag 2 CrO 4 cannot be lowered to 5. 5. 25. 1.0×10-8 M by carefully adding Ag+(aq). again influencing the value of the ion product.079 M 23. K sp = Cr 3+ F− = 6. The solubility can be lowered to 5. K sp = Au 3+ C2 O 4 = 1×10−10 2 3c. Ag + I − . Ksp = Cd 2+ 3 PO 4 3d. Yes 9. 0.0 × 10−8 M with CrO 4 2− as the common ion. In similar fashion. pH > 8. 11 mg 11. Yes 31b.4 ×10−8 M 17. AgI < AgCN < AgIO 3 < Ag 2SO 4 < AgNO 2 .5 ×10−3 M 13. 3 . cannot exceed the value of K sp .32 31a. 7. AgNO 3 produces a significant Ag + in solution. 1.30 ×10−5 15a. 7.3 3a. Yes Copyright © 2011 Pearson Canada Inc.1 × 10−33 = 2. Ksp = Sr 2+ F − 2 3− 2 = 2. If the [H3O+] is maintained just a bit higher than 1. b g 61a. 0. AgI(s) 43b. No 33. Qspa = 1. 53. PbCl 2 will dissolve in the HCl(aq) up until the value of the solubility product is exceeded. Yes 43a. Yes 41a.60 49a. FeS will precipitate and Mn 2+ aq will remain in solution. CaC2O4 should precipitate. Copyright © 2011 Pearson Canada Inc. 41d. 3. Precipitation of AgI(s) should occur.8 ×10−5 M . 2. [Cl-] = 0.7 ×107 < 3 ×107 = K spa for MnS. 0.0 ×1030 55.052% 37. 57. It forms no such complex ion with nitrate ion. 2. [I-] = 4. Thus. Yes 45. Lead(II) ion forms a complex ion with chloride ion. 4 . Separation will be complete.31c. 35.5%. 43c. 1. Ag + = 3. 39a. AgI 41b. FeS (s). No 39b. MgCO3 (s).41 49b. Ca(OH)2 (s). Precipitation will not occur. 9.1×10−13 M . The formation of this complex ion decreases the concentrations of free Pb 2+ aq and − b g b g free Cl aq .16 M.7 ×10−4 M 41c.4 ×10−6 g 59.7×10-9 M.96 g 51. 47. 2. 5 ×10−4 M 85a. 2 g MnS/L 77a.8 ×10−5 M 89d. C2 H 3O 2 − = 0.7 ×10−4 M 89c. Yes 80.07 g AgCl . No 77c. if NH 3 is added first. Ag 2 SO 4 (s) + Ba 2 + (aq) + 2 Cl − (aq) → BaSO 4 (s) + 2 AgCl(s) 85b. 1. although Pb OH b g 2 will not darken with added NH 3 . 0. Integrative and Advanced Exercises 67. Ag 2SO 4 (s) = 0. The most important consequence would be the absence of a valid test for the presence or absence of Pb 2+ . 65. If Pb OH 2 b g does form. 1.61b. Yes 77b. 0. 3. 1. Feature Problems 87.2 ×10−16 82. 63. Presence of Ag + may be falsely concluded. 1. 5 .6 × 10–3 M 83. 1.059 M 89e.7 ×10−4 M 89b.012 M 74.63 g . In addition.0 ×10−4 89a.702 g . 1. 68% 69. (a) and (c) are the valid conclusions. 38% 71. mass dissolved Ag 2SO 4 = 0.875 g BaSO 4 .4 ×10−9 M Copyright © 2011 Pearson Canada Inc. PbCl 2 may form Pb ( OH )2 . 6. 0. it will be present with Hg 2 Cl 2 in the solid. 2.21M. The answer is NH3. 106. The answer is (d). The answer is (b). It will decrease the amount of precipitate. The answer is (b). The answer is (b). 102. 97. 100a. 100c.Self-Assessment Exercises 93. More soluble in basic solution. 98. 100e. The answer is (c). Solubility is independent of pH. The answers are (c) and (d). The answer is (d). 101. 94. More soluble in acidic solution. More soluble in acidic solution. CuCO3 is soluble in NH3(aq) and CuS(s) is not. Solubility is independent of pH. 104. The answer is (c). 6 . 95. 105. 108. 96. 107. The answer is (a). More soluble in acidic solution. Yes 103. 99. 100f. 100d. 100b. Copyright © 2011 Pearson Canada Inc. No precipitate will form. The answer is (a). 48 kJ mol1 6a. 5a.0 J mol1 K 1 2b.42 kJ/mol Copyright © 2011 Pearson Canada Inc. (a) 1. 1 . 70. 8. 5 109 Integrative Example A. 2a. (b) increases. K PSiCl4 PCl2 2 K p . (b) low temperatures. 312. 970 C 10b. 8a. (a) K = 6b. 1. (b) case 4. 402 J/mol 3a. (b) increases. 1b. (b) 1.7 105 .5 107 . (a) Uncertain.5 1017 8b. 4b. 10a.Chapter 19 Answers Practice Examples 1a. (b) K = HOCl H + Cl PCl2 . 83. (a) High temperatures. (a) Decreases. No appreciable forward reaction. 9a. Non-spontaneous 7b.4 J mol1 K 4a. (a) Case 2. 99. 607 K 9b.4 J mol1 K 1 3b. 2 [ Pb 2 ]3 pNO [ NO3 ]2 [ H ]8 7a. 1484 kJ 5b. NO2 will spontaneously convert into N2O4. B. Decrease 7a. Answer (b) is correct. Positive 7c.9 J mol 1K 1 . 167 JK -1mol-1 Exercises 1a. Uncertain 5d. Hydrogen bonding is more disrupted in water at 100 C than at 25 C. 5a. The first law of thermodynamics states that energy is neither created nor destroyed (thus. Uncertain 7e. Positive 7d. H vap 44. Negative 7b. Increase 5b. S vap = 118. Increase 3.0 kJ/mol . there is not as much energy needed to convert liquid to vapor (thus H vap has a smaller value at 100 C ). C6 H5CH3 13. 9b. Decrease 5c. A The reason why S vap vapor at 1 atm pressure (the case at both temperatures) has about the same entropy. 354 K 15. “The energy of the universe is constant”). On the other hand. and hence. Decrease 1b. 17a. 2 . 11. liquid water is more disordered (better able to disperse energy) at higher temperatures since more of the hydrogen bonds are disrupted by thermal motion. “the entropy of the universe increases toward a maximum”). Case 2 Copyright © 2011 Pearson Canada Inc. A consequence of the second law of thermodynamics is that entropy of the universe increases for all spontaneous processes (and therefore. Increase 1c. has a larger value at 25 C than at 100 C has to do with dispersion. Negative 9a. 21a. 37. this reaction will be more highly favored at low temperatures. The reaction does not proceed spontaneously in the forward direction. 1. 23. An exothermic reaction may not occur spontaneously if. and therefore G 0 . K = K p = K c RT 1 33b. the reaction is spontaneous. A reaction in which S 0 need not be spontaneous if that process also is endothermic. G o 11 kJmol-1 . K 14. Case 3 19. 21b. 474. goes to completion. the system becomes more ordered (concentrated) that is.6 kJ . 29a. 616 kJ 31c.4 kJ 25b.5 . where the T S term dominates the G expression. 285 J mol1 25a.17b. 3202 kJ 29b. 27c. 27a. The process is clearly spontaneous. 0. 33a.2 kJ. and hence feasible (at 25 C ). This is particularly true at low temperatures. reaches equilibrium. +4. H = 0 because the gases are ideal and thus there are no forces of attraction or repulsion between them. -574. K = K p = K c RT 2 35. goes to completion. 3 . Since the Grxn is negative.0545 kJ K1 31b. The gases are more dispersed when they are at a lower pressure and therefore S 0 . 3177 kJ 31a. 27b. No prediction 17c. K = K p = K c 33c. This is particularly true at a high temperature.3 kJ. 8 1013 Copyright © 2011 Pearson Canada Inc. at the same time. Because both the entropy and enthalpy changes are negative. 600 kJ. S o 0 . where the H term dominates the G expression. 2 105 57. Equation (b) is incorrect. P{O 2 g } = 3. +68. 41. +5. 0. the molecular oxygen was removed immediately after it was formed. 6. G o = 28. 23. 3.39a.4 kJ 49b. which causes the equilibrium to shift to the right continuously.9 kJ/mol 49c. 39c. not added to it. the side with molecular oxygen. The TSo term should be subtracted from the Ho term.2 55b.7 kJ/mol 51.40 kJ/mol 49d. High temperature. K = 8 104 .. G o = 32 kJ/mol . 43. 27 kJ/mol 55d. The reaction is spontaneous in the forward direction. 4 .659 47b. 215. 39b.6 kJ/mol 53a. +28.e.0 1021 atm 53b.36 103 K 59. 1. K = 4 105 . 45. 204. 49a.) In addition. 47a.467 kJ/mol 47c.0 kJ/mol . 55a. which favors the products (i.1kJ/mol . K = 2 1013 . The reaction will proceed to the right. The reaction is spontaneous in the forward direction. +91 J K mol kJ mol 55c. G o = 76.0 104 Copyright © 2011 Pearson Canada Inc. In the gas phase.7 kJmol-1 . 206.556 kJ/mol 93b. 5 .8mmHg Copyright © 2011 Pearson Canada Inc. 647 K 87. 23. 0.580 mol BrCl. 0. G o = 362.2 mg AgBr/L 91a. 4. In the liquid. Feature Problems 93a. 0. 329 K 65. False 72d. 297 K 86. True 72b. so do the translational. Integrative and Advanced Exercises 72a. 0.014 63b. The increase in translation and rotation on going from solid to liquid is much less than on going from liquid to gas.01 103 K 81.3 kJ/mol . 77. 91b. 3.1 kJ/mol 67b.210 mol Br2 . False 72c. G o 1. 650 K 63a. 69.61. and vibrational degrees of freedom. True 73. rotational. 2.1 kJ/mol 67a.0317 bar 93c. The reaction is not spontaneous. 0. all degrees of freedom are saturated. most of the vibrational degrees of freedom are saturated and only translational and rotational degrees of freedom can increase. The reaction is spontaneous.210 mol Cl2 .70 atm 83. +8. In the solid as the temperature increases. +152. therefore. When we combine two reactions and obtain the overall value of G o . where the value of G o for the plotted reaction becomes positive. T following the equation G o = ∆H T∆S. we subtract the values on the line labeled “ 2Zn + O2 2ZnO ” from those on the line labeled “ 2Mg + O2 2 MgO ”. 94c. Carbon would appear to be an excellent reducing agent. The slopes for these lines differ so markedly because these three reactions have quite different ∆S values . Temperature (°C) All three lines are straight-line plots of G o vs. because it will reduce virtually any metal oxide to its corresponding metal as long as the temperature chosen for the reaction is higher than the threshold temperature. 3. the value of G o for the decomposition becomes negative. The point where CO(g) would become less stable than 2C(s) and O2(g) looks to be below 1000 C (by extrapolating the line to lower temperatures). 97a.8 mmHg 94a. Thus. 23. Thus. 94f. This occurs above about 1850 C . 6 . The result for the overall G o will always be negative because every point on the “zinc” line is above the corresponding point on the “magnesium” line 94b. The slope of each line multiplied by minus one is equal to the ∆S for the oxide formation reaction. The “carbon” line has a negative slope. and the reaction becomes spontaneous. to reduce ZnO with elemental Mg. 5. it is not possible to decompose CO(g) to C(s) and O 2 g in a spontaneous reaction.4 JK -1mol-1 Copyright © 2011 Pearson Canada Inc. The decomposition of zinc oxide to its elements is the reverse of the plotted reaction. only at these elevated temperatures can ZnO be reduced by carbon. Based on this plot.93d. The “carbon” line is only below the “zinc” line at temperatures above about 1000 C . indicating that carbon monoxide becomes more stable as temperature rises.8 JK -1mol-1 97b. 94d. we subtract the value on the plot of the reaction that becomes a reduction from the value on the plot of the reaction that is an oxidation. Standard Free Energy Change (kJ) 0 500 1000 1500 2000 0 -100 1 -200 -300 2 -400 -500 3 -600 -700 94e. 2. Diagram (a). 109a. 7 . At room temperature and normal atmospheric pressure this process is spontaneous. 111. The reaction will be spontaneous at all temperatures.1kJmol-1 110c. 108a. The forward reaction should occur to a significant extent. Exothermic 110b. 107b. Because Gvap . The entropy of the universe is positive. not just for the system alone. 112. Equilibrium constant can be calculated from G o ( G o RT ln K ). 107a. 4. 108b. The answer is (d). The answers are (a) and (d). the vapor pressure is less than 1 atm at 298 K. 102. 107c. The answer is (a). No reaction is expected.11032 110d.6 oC 106b. consistent with o Tbp=57 C. 186. The answer is (c). 110a. 57o C 109b. The answer is (b). Copyright © 2011 Pearson Canada Inc. The melting point of carbon dioxide is expected to be very low.298 K >0. 104.8 kJmol-1 o 109c. 106a. Carbon dioxide is a gas at room temperature. G o = zero.Self-Assessment Exercises 101. Entropy change must be accessed for the system and its surroundings ( Suniv ). 105. to some extent. 113. The forward reaction should occur. and K permits equilibrium calculations for nonstandard conditions. 103. Sc s + 3Ag + aq Sc 3+ aq + 3Ag s 1b. Na(s) reacts with water to form H 2 g and NaOH(aq). Not be a feasible method. 0. Anode: Sn s Sn 2 + aq + 2e .48V 4b. Cathode: Ag + aq + e Ag s . Silver ion is one example. Copyright © 2011 Pearson Canada Inc. Overall: anode salt bridge Al cathode e- Ag NO3- K+ Al3+ Ag+ Al s + 3Ag + aq Al 3+ aq + 3Ag s .e. 1587 kJ 5b.. Overall: 2In(s) + 3Cd 2+ (aq) 2In 3+ (aq) + 3Cd(s) . 7a. 2a. Overall: Sn(s) + 2Ag + (aq) Sn 2+ (aq) + 2Ag(s) . 3a. The standard cell potential for reaction (2) is more positive than that for situation (1). 2b. +0. Thus. 0. Ecell 6b.74 V 4a. Anode: Al s Al 3+ aq + 3e .Chapter 20 Answers Practice Examples 1a. it should react with Sn(s) to form Sn 2 + aq spontaneously. Anode: In s In 3+ aq + 3e . The method is not feasible because another reaction occurs that has a more positive cell potential.460 V. if Sn 4 + aq is formed.229 V o = +0. +0. 7b. i. 1. . 6a. 1 .424V 5a. Also. Cathode: Cd 2+ aq 2e Cd(s) .587 V 3b. reaction (2) should occur preferentially. Cathode: Ag + aq 1e Ag(s) . + 3e.8a. The equilibrium constant's small size indicates that this reaction will not go to completion.4 g. +0. (c) 1303kJ/mol . 13a. The size of the equilibrium constant indicates that this reaction indeed will go to essentially 100% to completion. The largest positive cell potential will be obtained for the reaction involving the oxidation of o 2.329 V . I2 s and H 2 g . 0.000 V .755 V 5. 0.23V 11b. Yes 10b. 3. 0. 5.340 V E o 0.137 V B. 4OH-(aq). 0. 12b. (d) 13. Al(s) to Al3+(aq) and the reduction of Ag+(aq) to Ag(s). 2. 6.0150 11a. 1b.9 109 12a.89 amperes 13b.815 V 9b. +1. Exercises 1a. (a) Oxidation: Al(s) + 4OH-(aq) [Al(OH)4]. Reduction: O2(g) + 2H2O(l) + 4e. 9a. Ecell Copyright © 2011 Pearson Canada Inc. 2 . Overall: 4Al(s) + 4OH-(aq) +3O2(g) + 6H2O(l) 4[Al(OH)4]-(aq).23h Integrative Example A.31V 7a.017 V 10a. 0.800 V .476 V. +0.. Silver metal Ag + aq . 8b. (b) -2. 1.440 V E o 0. Significant extent. Ecell = 2. Yes 15b. Ni2+ 9b. Nonspontaneous 11c.103 V.42V . Ecell = +0.(aq) . Not to a significant extent. Large extent.828V o 23a. 2 Al s + 3 Sn 2+ aq 2 Al3+ aq + 3 Sn s . Fe 2+ aq + Ag + aq Fe3+ aq +Ag s . 15a. Ecell 0. Zn s + 2 H + aq Zn 2+ aq + H 2 g 17c. o 19d. Cd 11a. Copyright © 2011 Pearson Canada Inc.7b. o 23b. o = +1.539 V . 3Cr(s)+2Au 3+ (aq) 3Cr 2+ (aq)+2Au(s) . Ecell = 0.029 V . 3Ag s +NO3 aq +4 H + aq 3Ag + aq +NO g +2 H 2O(l) 17b. o 19c. Ecell o 19b. 13b. Spontaneous 13a. Significant extent. Spontaneous 11b. Does not react. 13e. Yes 15c. 3 . 19a. nonsponatneous. spontaneous. Significant extent. 13d. Ecell 9a. (cathode). The cell with the smallest positive cell potential will be obtained for the reaction involving the oxidation of Zn(s) to Zn2+(aq) (anode) and the reduction of Cu+2(aq) to Cu(s) o 1. No 17a. Ecell 0. 13c.029 V . H 2O(l) H + (aq)+OH . Nonspontaneous 11d.291V . 165 103 kJ 25b. H + aq will inevitably be on the left side of the reduction of an oxoanion because reduction is accompanied by not only a decrease in oxidation state. 0.249 V 37b. Ecell 0. These oxygen atoms appear on the right-hand side as H 2 O molecules. 5 × 10-6 M 37a. o 23d. 3.591V 33. 1. The hydrogens that are added to the right-hand side with the water molecules are then balanced with H + aq on the lefthand side.923 V 35. 25a. Ecell 0.1 102 kJ 27a.638 V 39a. This results in H 2 O(l) on the side that had H + aq ions (the left side in this case) and OH aq ions on the other side (the right side. nonspontaneous.5 109 27d. 39b.435V . To the left towards reactants.24 kJmol-1 27c. The reactions for which E depends on pH are those that contain either H + aq or OH aq in the balanced half-equation. nonspontaneous. Since K is very small the reaction will not go to completion. 31. 29a.435V . 0. +0. 268 kJ 25c. but also by the loss of oxygen atoms.o 23c. 1. 48.100 V 27b. it may be rebalanced by adding to each side OH aq ions equal in number to the H + aq originally present. No 29b. 1.) Copyright © 2011 Pearson Canada Inc. 39c. 3. If a half-reaction with H + aq ions present is transferred to basic solution. 4 . 229 V 53b. and hydroxide ion should be produced in the vicinity of the copper wire (pink color). 5 . Copyright © 2011 Pearson Canada Inc.025 V 47d.50 M. a more positive voltage will be created). 51b. which. while in the anode half-cell in the case at hand. Cell diagram: Cr s |Cr 2+ aq .039 V 47b.41a.Li + (aq) KOH(satd) MnO 2 (s).18 M. since Ecell is only slightly negative.07 104 C ). +1.65 M NaOH). by removing products as they are formed and replenishing reactants as they are consumed. +2.818 V 43b.237 M 47e. which serves as the cathode. Cell diagram: Li(s). with 1. 43a. 49. Sn 2+ = 0.00 g of metal is consumed in each cell. 0. is in fact.00 M KOH than with 0. (i. 1. 0.195 V 53a.65 M KOH.Mn(OH)3 (s) . 6 1038 M 41b.992 V 53c. [OH] = 1.e. 57. [OH] = 0.03 V.Cr 3+ aq ||Fe 2+ aq . 0.891V 55. Pb 2+ = 0. In the standard half-cell.177 V 47a. The forward reaction (dilution of H + ) should be even more spontaneous. We expect o (1. Decrease 47c. o 2. Calculate the quantity of charge transferred when 1. 1. Ecell = 0. Cl2(g) could be obtained by driving it to the product side with an external voltage or. 51a.65 M. +0. Oxidation of iron metal to Fe2+(aq) should be enhanced in the body of the nail (blue precipitate). 0. that Ecell the case.000 M NaOH) should be a little larger than Ecell (0. Ecell 59a. 45.Fe 3+ aq |Fe s .84 V. The reaction goes to completion.00 M. therefore the reaction will not be spontaneous under standard conditions. The aluminum-air cell has the highest value ( 1. the metal would develop a positive charge and no longer release electrons (or oxidize).9 g Ni 65a. 67a. 2 H 2 O(l) 2 H 2 g + O 2 g . A scratch tears the iron and exposes “fresh” metal. This change in polarity can be accomplished by hooking up the metal to an inert electrode in the ground and then applying a voltage across the two metals in such a way that the inert electrode becomes the anode and the metal that needs protecting becomes the cathode.62 g Zn 69b. There should be almost no blue precipitate. Ecell 69a.7642 A 73a. One way to retard oxidation of the metal would be to convert it into a cathode.229V . 63a.90 g Al 63c. 3. H 2 g and O 2 g . 1. Spontanteous 65c. Once transformed into a cathode. 59c. 0. 2. This way.229 V . We expect blue precipitate in the vicinity of the scratch. Zinc should protect the iron nail from corrosion. 0. 61. any oxidation that occurs will take place at the negatively charged inert electrode rather than the positively charged metal electrode. which makes it is more susceptible to corrosion. 65d. 67b. 37% Copyright © 2011 Pearson Canada Inc.59b. The pink color of OH should continue to form.2 min 71. Requires electrolysis with an applied voltage greater than +1. the zinc corrodes instead. 1079 C 71b. 11 g Ag 63d. 20. 6 . Copper 73b.3 g Zn 63b. 65b. Requires electrolysis with an applied voltage greater than +0. Requires electrolysis with an applied voltage greater than +0. o = 1.183V .236V . 102b.033 10 4 J .1 atm) . 0. eventually erode away. 2. 96a. G = 52. 100. The dissolved NaHCO3(s) serves as an electrolyte.8 kJ . It would also enhance the electrical contact between the two objects. Equation 2: 2Na (amalgam. E = 0. H o 26. 102c.66 1013 F 104b.Integrative and Advanced Exercises 75. Aluminum 6Ag(s) 2 Al3 (aq) 3 S2 (aq) 96b. 102d.61103 kWh 80. nonetheless.146 V 90. There are several chemicals involved: Al. K (at 50o C) 1. S o 189. 1.59 1014 .206 %) + 2H + (aq.26 V Feature Problems 102a. 2 Na(s) + 2H + (aq) 2 Na (1M) + H 2 (g. G1 8. 0.38 106 kJ 77b.1M) 2 Na + (1 M) + H 2 (g.61 .26 1014 C Copyright © 2011 Pearson Canada Inc.713V 104a. 0. 1 atm) . 2. E o Na + 1 M /Na s = 2. The standard hydrogen electrode is the anode. G2 36. 2. Ecell o = 2. Q 1.255 V 77a.77 1014 . H2O.9 104 M 82.3 kJ 87.2 JK -1 .713 V .156 104 J . 9. Although the aluminum plate will be consumed very slowly because silver tarnish is very thin.051% 92. K at 25o C 3.223 V. it will. and NaHCO3. Equation 1: Na s Na amalgam. 99. 2 Al(s) 3 Ag 2 S(s) 96c. 0. 127. 96d. 6.345 104 J . 7 . 2.206% . 41 105 K + ions 104d. The correct answer is (b). False 111b. False 111c. 119. S o .100. A net reaction occurs to the left. The correct answer is (a). Self-Assessment Exercises 111a. 107. The correct answer is (d). False 111g. electron flow from B to A Copyright © 2011 Pearson Canada Inc. 116. True 111d.NO-3 (1M) NO(g. 117. The correct answer is (a). Ecell 0. 8 . Ecell 1.000015 %) 105.1atm) Pt(s) . Zn(s) Zn 2+ (1M) H + (1M). The correct answer is (c). E Xo / X 0 V . 106. Fe(s)+Cu 2+ (1M) Fe 2+ (1M)+Cu(s) . E Xo / X > EYo2 /Y .719 V . 9. No 120b. 1. True 112. or reactions for which K>1.780V . H o and U o cannot be used alone to determine whether a particular electrochemical reaction will have a positive or negative value. 114. o 118. The correct answer is (d). Proceed similarly to the solution for 20. 1.5 107 (~ 0. 1. o 121a.82 120a. True 111e. Reactions with a positive cell potential are reactions for which G o 0 . 115.3 1011 ions 104e. 113. True 111f.104c. Zn(s)+Fe 2+ (0.646V . 122d. Sn 2+ (1M)+2Ag + (1M) Sn 4+ (aq)+2Ag(s) . O2(g) at anode and H2(g) and OH-(aq) at cathode. 122b.264V . Copyright © 2011 Pearson Canada Inc.0010M) Zn 2+ (0. O2(g) at anode and H2(g) and OH-(aq) at cathode. Cl2(g) at anode and Cu(s) at cathode. 9 . Ecell 0. electron flow from A to B. Ecell 0. Cl2(g) at anode and Ba(l) at cathode. 122a.o 121b. electron flow from A to B.10M)+Fe(s) . 122c. o 121c. 310-33) and therefore precipitation will occur. B2O3(s) + 3 CaF2(s) + 3 H2SO4(l) 2 BF3(g) + 3 CaSO4(s) + 3 H2O(g). The structure of the Y anion is linear: N C N 3a. 2 2 2b. a solution of BeCl2 4H 2 O is expected to be acidic. When BeCl2 4H 2 O is dissolved in water. 2 B OH 3 (s) B2 O3 (s) + 3H 2 O(g) . Hence. 1b. Integrative Example A. H2O(g). 2 NaCl aq + 2 H 2 O l electrolysis 2 NaOH aq + H 2 g + Cl2 g . When BeCl2 4H 2 O is heated. B. When CaCl2 6H 2 O is heated. 1 1 NaNO2 (s) + 3Na(s) 2Na 2O(s) + N 2 (g) . as discussed in Section 21-3. and [Be(H2O)3(OH)]+ and H3O+ ions are produced. 2 B2 O3 s + 3C s + 6 Cl 2 g 4 BCl3 g + 3CO 2 g . CaC 2 (s)+N 2 (g) CaCN 2 (s)+C(s) .10 H2O(s) + H2SO4(aq) 4 B(OH)3(s) + Na2SO4(aq) + 5 H2O(l). Compound X is likely calcium carbide and compound Y is calcium cyanamide. it decomposes to Be(OH)2(s). 2 NaCl aq + 2 H 2O l electrolysis 2 NaOH aq + H 2 g + Cl2 g . 2 B(OH)3(s) B2O3(s) + 3 H2O(l). Na 2 B4 O 7 10 H 2 O(s) + H 2SO 4 (l) 4 B OH 3 (s) + Na 2SO 4 (s) + 5 H 2 O(l) . Na 2 O(s) + O2 (g) Na 2 O 2 (s) . and HCl(g). Because of the high polarizing power of Be2+. Compound X is most likely Na2O and compound Y is Na2O2. BeCl2 4H 2 O comprises [Be(H2O)4]2+ and Cl− ions. 2 NaOH aq + 3 NO 2 g 2 NaNO3 aq + NO g + H 2 O(l) . 2 NaOH aq + SO 2 g Na 2SO3 aq + H 2 O(l) . 3b.Chapter 21 Answers Practice Examples 1a. it is much easier to drive off the Copyright © 2011 Pearson Canada Inc. boil Na 2S2 O3 aq . 4 BCl3 g + 3LiAlH 4 (s) 2 B2 H 6 g + 3LiCl(s) + 3AlCl3 (s) . The product [Al3+][OH-]3= 4. Na 2SO 3 aq + S(s) 2a.010-9 is much greater than KSP for Al(OH)3 (1.(aq). [Be(H2O)4]2+ ions react with water. Ca(s)+2C(s) CaC2 (s) . 1 . Al(NO3 )3 (aq) Al3+ (aq) + 3NO3.(aq). Because the charge density and polarizing power of Ca2+ is much less than that of Be2+. NaCN(aq) Na + (aq) + CN . CaCl2(s) and H2O(g) are produced. it is difficult to remove the coordinated water molecules by heating the solid and the acidity of the coordinated H2O molecules is enhanced. Na2B4O7. while the potassium flame test is violet. 11. 11. but Li 3 PO 4 is not very soluble. 7a.5% yield 9b. Na2O2(s) is thermodynamically stable with respect to Na2O(s) and O2(g) at 298 K. lithium gives a red color to a flame. 5. 9a. 2 Cs(s) + Cl2 (g) 2 CsCl(s) Na 2 O 2 (s) 1b. CaCO3 CaHPO4 Copyright © 2011 Pearson Canada Inc. the charge density of the Ca2+ ion is too low to affect the acidity of water molecules in the hydration sphere of the Ca2+ ion. the solution pH only depends on the number of electrons transferred. but no precipitate if the white solid is KCl.coordinated water molecules by heating the solid.87 1025. Exercises 1a. Hence. CO2 Ca(OH)2 H3PO4 HCl H2SO4 CaCl2 electrolysis Ca CaSO4 2 . However.61 7b. Both LiCl and KCl are soluble in water. Li CO s Li O s + CO g 2 3 2 2 1d. NH 3 is simply used during the Solvay process to produce the proper conditions for the desired reactions. Na 2 SO 4 s + 4 C s Na 2S s + 4 CO g 1e. As long as NaCl is in excess and the volume of the solution is nearly constant. K s + O 2 g KO 2 s 3. 2 Na(s) + O 2 (g) Δ 1c. Ca2+(aq) and Cl−(aq) are produced. a solution of CaCl2 6H 2 O has a neutral pH. Any net consumption of NH 3 is the result of unavoidable losses during production. 71. Based on molar solubilities: MgCO3 < Li2CO3 < Na2CO3. CaO 13. Hence the addition of K 3 PO 4 (aq) to a solution of the white solid will produce a precipitate if the white solid is LiCl. K 1. The best method is a flame test. When CaCl2 6H 2 O is dissolved in water. 19c. Ba s + Br2 l BaBr2 s 17c. Two H H H H H 23c. Equilibrium lies to the left. B4 H10 contains 22 valence electrons or 11 pairs. 19b. 17a. 23a. C C C C H H H H H 25a. 23b. Equilibrium lies to the right.electrolysis CaCl2 l Ca l + Cl2 g Ca OH 2 s + CO 2 g CaCO3 s + H 2O g CaCO3 s CaO s + CO 2 g Ca OH 2 s + H 2SO 4 aq CaSO 4 s + 2 H 2 O(l) 15. BeF2 (s) + Mg(s) Be(s) + MgF2 s 17b. 2BBr3(l) + 3H2(g) 2B(s) + 6HBr(g) 3 CO(g) + 2 B(s). Equilibrium lies slightly to the left. B2O3(s) + 3 C(s) Copyright © 2011 Pearson Canada Inc. 2 B(s) + 3 F2(g) 2 BF3(g) 25b. Mg2+ + 2 Cl(aq) Mg(s) + Cl2(g) (Overall reaction). leaving only one pair to bond the four B atoms together. BeSO4 will be the least stable with respect to decomposition. we expect Be2+ to be the most polarizing and thus. Clearly an electron deficient situation. A cation with a high polarizing power will polarize the SO42− ion and kinetically assist the decomposition to SO3. Because Be2+ has the largest charge density of the group 2 cations. The SO42− ion is a large polarizable ion. UO 2 s + 2 Ca s U s + 2 CaO s 17d. 3 . MgCO3 CaCO3 s MgO s + CaO s + 2 CO 2 g 17e. 2 H 3PO 4 aq + 3 CaO s Ca 3 PO 4 2 s + 3 H 2O l 19a. Charge is conserved. 21. Ten of these pairs could be allocated to form 10 B—H bonds.Ca OH 2 s + 2 HCl aq CaCl2 aq + 2H 2 O(l) . 2 Al2O3(s) + 3 Si(s) 41a. 2 Al s + 6 HCl aq 2 AlCl 3 aq + 3 H 2 g 27b. HCl(aq).5 to 8. 35. 3 SiO2(s) + 4 Al(s) CO2(g) + K2SiO3(s) 41b. Eventually. Al4C3(s) + 12 H2O(l) 3 CH4(g) + 4Al(OH)3(s) 43. Copyright © 2011 Pearson Canada Inc. aluminum is inert only when the medium to which it is exposed is neither highly acidic nor highly basic. 2 NaOH aq + 2 Al s + 6 H 2 O l 2 Na + aq + 2 Al OH 4 aq + 3 H 2 g 27c. In the sense that diamonds react imperceptibly slowly at room temperature (either with oxygen to form carbon dioxide. H H Ga H H 39. 25c. CS2 g + 3 Cl 2 g CCl4 l + S2 Cl 2 l . with the general formula Si n H 2 n +2 . Thus. Aluminum and its oxide are soluble in both acid and base. 2 CH 4 g + S8 g 2 CS2 g + 4 H 2S g . 2 KOH(aq) + 2 Al(s) + 6 H2O(l) 2 K[Al(OH)4](aq) + 3 H2(g). 2 B(s) + 3 N2O(g) 3 N2(g) + B2O3(s) 27a. crystallize 2 KAl(SO4)2(s). Al OH 3 s + HCO 3 aq . being a strong acid. K2CO3(s) + SiO2(s) 41c. or in its transformation to the more stable graphite). 33. A silanol is a compound in which one or more of the hydrogens of silane is replaced by an —OH group. 45. it is essentially true that “diamonds last forever. Al OH 4 aq + CO 2 aq can’t be used because it will dissolve the Al(OH)3(s). 4 CS2 g + 8 S2 Cl 2 g 4 CCl 4 l + 3 S8 (s) . A silane is a silicon-hydrogen compound.” However. with the elimination of a water molecule between every two silanol molecules. 2 Al s + 3 SO 4 2 2 Al3+ aq + 3 SO 2 aq + 6 H 2 O(l) aq +12 H + aq 29.5. H H Ga 37. 2 K[Al(OH)4](aq) + 4 H2SO4(aq) K2SO4(aq) + Al2(SO4)3(aq) + 8 H2O(l). the conversion to graphite occurs. Silicones are produced when silanols condense into chains. Al(s) is resistant to corrosion only over the pH range 4. 4 . at elevated temperatures. diamond will burn to form CO2(g) and thus the statement is false. Al 3+ aq + 3 HCO 3 aq Al OH 3 s + 3 CO 2 g 31. 59. 0. 62a. Copyright © 2011 Pearson Canada Inc. . then the product could be a mixture of magnesium nitride and magnesium oxide. + The Li ion will polarize the I3− to a significant extent and presumably assists the decomposition of I3− to I2 and I−. namely +4 and +3.332 g MgO 62b. 53. PbS s + 2 O 2 g PbSO 4 s 51a. all of the oxygen atoms in the mineral must be in the 2 oxidation state. SnCO 3 s Pb l + CO g 49c. we have -24 from the twelve oxygen atoms and +3 from the potassium and hydrogen atoms for a net number of -21 for the oxidation state. The product is 69% by mass MgO. 5 . Consequently. 62c. In oxygen-rich salts such as mica. 51c. we would expect that the silicon and the aluminum atoms would be in their highest possible oxidation states. The reaction is not spontaneous. I I I 58. The Li+ ion has a high charge density and significant polarizing power. Potassium is obviously in the +1 oxidation state. The reaction is not spontaneous. We still have three aluminum atoms and three silicon atoms to account for. Up to this point. respectively. Since the salt is neutral. This is precisely the total that is obtained if the silicon and aluminum atoms are in their highest possible oxidation states: (3 (+3) + 3 (+4) = +21). 51b. This scenario is not implausible since molecular nitrogen is readily available in the atmosphere. The formula is KAl2(OH)2(AlSi3O10). 49a. Since they are not segregated into O2 units in the formula. the empirical formula for white muscovite is consistent with the expected oxidation state for each element present.47.8 104. PbCl2 Integrative and Advanced Exercises.332 g MgO predicted. PbO s + 2 HNO3 aq Pb NO3 2 s + H 2 O(l) SnO s + CO2 g 49b. as are the hydrogen atoms in the hydroxyl groups. the oxidation numbers for the silicon and aluminum atoms must add up to +21. If the mass of product formed from the starting Mg differs from the 0. PbO s + C s 49d. The reaction will go to completion. 2 Fe 3+ aq + Sn 2 + aq 2 Fe 2 + aq + Sn 4 + aq 49e. K PO2 8. n = 3.03 V. Lattice energy of MgS = -3215 kJ. If the oxidation of C(s) to CO2(g) did not occur.4 103 M 73b.4 104 M 73c. 77.5°C) than K (773. 2. The carbonate ion concentration can readily be achieved by adding solid Na2CO3. 1.4) is not a feasible way of producing Li metal from LiCl. 67a. We would expect MgO(s) to have a larger value of lattice energy than MgS(s) because of the smaller interionic distance in MgO(s).87 V. and thus a reaction similar to (21. 4 Al(s) + 3 MnO2(s) 2 Al2O3(s) + 3 Mn(s). ΔH = +129 kJ.0 10 3 M 76. 48. 2.63a. 2.9°C). 7. 81. Cs has a lower boiling point (678. Ered = -3.605 V. 69. 70. ΔH = 277 kJ ΔG. -1. 19 g Pb(NO3 ) 2 73a. then the cell reaction would just be the reverse of the formation reaction of Al2O3 with n = 6 e-. 6. E = E o = -2.4 ppm Ca 2 67b. 6 . 64a. Mg(OH)2 should precipitate. Lattice energy of MgO = -3789 kJ. 66. Use the Ksp expression for CaCO3 to determine the needed [CO32–]. Copyright © 2011 Pearson Canada Inc. Feature Problems 78a. H = -1792 kJ. a reaction similar to (21. Since Li has a higher boiling point (1347°C) than K (773. 63b.2 102 g Na 2 CO3 68a. 64c.3 metric tons of coal .9°C).68 105 g Al 68b. The empirical formula is M3C60. 2 Al(s) + 3 MgO(s) Al2O3(s) + 3 Mg(s). 67d.4) is a feasible method of producing Cs metal from CsCl. 78b. Ered = -2. -852 kJ 64b.626 V .7 g CaO 67c. K 2CO 3 (aq)+Ba(OH)2 (aq) BaCO 3 (s)+2KOH(aq) 90b.aq) 90a. Ba(OH)2(aq) and H2O2(aq). Aluminum can be used for making products that last and for structural purposes. NaHCO 3 (s)+HCl(aq) NaCl(aq)+H 2O(l)+CO 2 (g) 90f. 86. CaCO 3 (s)+2HCl(aq) CaCl2 (aq)+H 2O(l)+CO 2 (g) 88c. 85a. The answer is (b). BCl3NH3(g) (an adduct) 85b. 2Na 2O 2 (s)+2CO 2 (g) 2Na 2CO 3 (s)+O 2 (g) 89a. Na 2SO 4 (s)+2HCl(g) 89c. The thermite reaction is evidence that aluminum will readily extract oxygen from Fe2O3.limited solubility) 88e. 2NaCl(s)+H 2SO 4 (concd. PbO 2 (s)+4HBr(aq) PbBr2 (s)+Br2 (l)+2H 2O(l) Copyright © 2011 Pearson Canada Inc. 88a. 2Na(s)+2H 2 O(l) 2NaOH(aq)+H 2 (g) followed by the reaction in Exercise 82(c). 7 . BaO(s)+H 2 O(l) Ba(OH)2 (aq.. Li2CO 3 (s) Li2O(s)+CO2 (g) 88b. SnO(s)+C(s) Sn(l)+CO(g) 90d. 2Al(s)+2NaOH(aq)+6H 2O(l) 2Na[Al(OH)4 ](aq)+2H 2 (g) 88d. 84. KO2(s) 85c. MgCO 3 (s)+2HCl(aq) MgCl2 (aq)+H 2 O(l)+CO 2 (g) 89b. CaF2 (s)+H 2SO 4 (concd. 87. Li2O(s) 85d. The answer is (f). aq) CaSO 4 (s)+2HF(g) 90e. Mg(HCO 3 )2 (aq) MgCO 3 (s)+H 2O(l)+CO2 (g) 90c. The answer is (c). because aluminum develops a coating of Al2O3 that protects the metal beneath it.Self-Assessment Exercises 83. SiF4 (g)+4Na(l) Si(s)+4NaF(s) 91.90g. CaCO3(s) 94b. H 2 SO 4 (aq)+BaO 2 (aq) H 2 O 2 (aq)+BaSO 4 (s) 93d. 8 . Al2O3(s) with Fe3+ and Ti4+ replacing some Al3+ in the crystal structure. No reaction. 2PbO(s)+Ca(OCl)2 (aq) CaCl2 (aq)+2PbO 2 (s) 94a. Pb(NO 3 )2 (aq)+2NaHCO 3 (aq) PbCO 3 (s)+2NaNO 3 (aq)+H 2 O(l)+CO 2 (g) 93b.27103 m3 Copyright © 2011 Pearson Canada Inc. 2B(OH)3 (s) B2O 3 (s)+3H 2O(g) 92b. CaSO 4 2H 2O(s)+(NH 4 ) 2CO3 (aq) CaCO3 (s)+(NH 4 ) 2SO 4 (aq)+2H 2O(l) 92a. CaSO42H2O(s) 94c. CaSO 4 2H 2O(s) CaSO 4 H 2O(s) + H 2 O(g) 2 2 93a. BaSO4(s) in water. 94d. 95. 1 3 92c. Li2 O(s)+(NH 4 )2 CO 3 (aq) Li2 CO 3 (s)+2NH 3 (g)+H 2 O(l) 93c. 1. Displacement reactions involve one element displacing another element from solution. and the others are molecular covalent compounds. 1. None of the halogens reacts with water to form H2(g).to NO3. Within the halogen group. Eo for the disproportionation reaction is negative. BeF2.Chapter 22 Answers Practice Examples 1a. Square pyramidal 9. 3. Bi2O3 is most basic. LiF. by oxygen in the air: 4 I − ( aq ) + O 2 ( g ) + 4 H + ( aq ) → 2 I 2 ( aq ) + 2 H 2 O(l) . CF4. which is yellow-brown in aqueous solution. The element that dissolves in the solution is more “active” than the element supplanted from solution.8 ×105 L air 7a. 5. The only halogen with sufficient oxidizing power to displace O2(g) from water is F2(g). the activity decreases from top to bottom. Iodide ion is slowly oxidized to iodine. The disproportionation of NO2. NF3. Trigonal pyramidal 7b. Copyright © 2011 Pearson Canada Inc. but not those above it. P4O6 is least basic (and is actually acidic). The bond energy of the noble gas fluorine is too small to offset the energy required to break the F—F bond. 4 ×10−15 % decomposed . BeF2 is a network covalent compound.7 ×10−48 2b. K p = 4 ×10−34. LiF is an ionic compound. o B.453 V 2a. 3 XeF4(aq) + 6 H2O(l) → 2 Xe(g) + 3/2 O2(g) + 12 HF(g) + XeO3(s) 11.622 V 1b. Thus. Ecell for the reaction is positive and therefore disproportionation of HNO2 to NO3. 15. BF3. Exercises 1.and NO is spontaneous under standard conditions. and Sb4O6 is amphoteric. Integrative Example A. 7. 0. 1 . 1. 13. Tetrahedral 7c. OF2. each halogen is able to displace the members of the group below it.and NO is not spontaneous. Keq = 2. T-shaped 21b. E°cell = -0. the reaction goes to completion.085 V 23b. 302 kJ / mol 37a. Keq = 3. we would expect the volume of hydrogen to be twice the volume of oxygen produced. 41. forms only weak hydrogen bonds. Keq = 9. 39d. producing a diamagnetic molecule. H 2 O forms much stronger hydrogen bonds. this method of displacement would not work for F2(g). 19. 2 . 2KClO 4 (s) → 2KClO 3 (s) + O 2 (g) 27.8 × 10-7. By the law of combining volumes. In this reaction. 37b. Since there is no chemical oxidizing agent that is stronger than F2(g). 33. 2HgO(s) → 2Hg(l) + O 2 (g) ∆ 25b. 39b. 2 moles of H 2 (g ) are produced for each mole of O 2 (g ) . 0. 2 ×106 kg F2 17b.17a. the disproportionation reaction will not occur spontaneously under standard conditions. 39c.1 × 10-65. Keq = 4. the reaction goes to completion. Square planar 23a.095 V.2 × 1041. leading to a higher boiling point. the reaction does not even come close to going to completion. while polar. the reaction does not even come close to going to completion. 3 ×10−5 mmHg 31. All electrons are paired in O3 . 21a. The electrolysis reaction is 2H 2O(l) electrolysis → 2H 2 (g) + O 2 (g) . Since the final cell voltage is negative. 5. 30 L Copyright © 2011 Pearson Canada Inc. ∆ 25a. H 2S.0 × 102. Square pyramidal 21c. CO 2 29. 39a. 6 Cl2(aq) + 6 H2O(l) → 2 ClO3-(aq) + 12 H+(aq) + 10 Cl-(aq).89 35. +5 53c. also are possible. FeS (s ) + 2 HCl (aq ) → FeCl2 (aq ) + H 2S (aq ) MnS(s).002% 53a. If the white solid is Na 2 SO 4 . S2 O3 2− 2− (aq ) → S (s ) + SO 2 (g ) + H 2 O(l) (aq ) + 2 H + (aq ) 47. potassium thiosulfate 43d. SO 2 (aq ) + MnO 2 (s ) 45d. CaSO3 (s ) + 2 HCl (aq ) → CaCl2 (aq ) + H 2 O(l) + SO 2 (g ) → Mn 2+ (aq ) + SO 4 45c. 1. +4 53b. zinc sulfide 43b. 3 . etc. 7. The decomposition of thiosulfate ion is more highly favored in an acidic solution. there will be no reaction with strong acids such as HCl. sulfur tetrafluoride 45a. Pt → 4NO(g) + 6H 2 O(g) 55b. upon addition of HCl(aq).rhombic) will form 2− → S (s ) + SO 2 (g ) + H 2 O(l) . HN3(aq) + HNO2(aq) → N2(g) + H2O(l) + N2O(g). 4NH 3 (g) + 5 O 2 (g) 55c. 2 NO (g ) + O 2 (g ) → 2 NO 2 (g ) . N 2 ( g ) + 3 H 2 ( g ) 850° C. S2 O3 (aq ) + 2 H + (aq ) 49. +4 2 NH 3 ( g ) 55a. HNO2(aq) + N2H5+(aq) → HN3(aq) + 2H2O(l) + H+(aq). -2 53d. ZnS(s). if the white solid is Na 2 S 2 O 3 . 57. 3 NO 2 (g ) + H 2O (l) → 2 HNO 3 (aq ) + NO (g ). SO 2 (g) will be liberated and a pale yellow precipitate of S(s. 6 × 109 kg N 2 O 4 ( g ) 61a. potassium hydrogen sulfite 43c.43a. 2 Na 2CO 3 (aq ) + 4 NO (g ) + O 2 (g ) → 4 NaNO 2 (aq ) + 2 CO 2 (g ) 59. By contrast. Copyright © 2011 Pearson Canada Inc. 45b.21 51. 2 NO 2 ( g ) 61b. 1. this molecule has the greatest mass percent hydrogen. ΔH° combustion = -1559.3 kJ. CaH 2 produces the greatest amount of H 2 per gram of solid. 2 Al (s ) + 6 HCl (aq ) → 2 AlCl3 (aq ) + 3 H 2 (g ) 73b.7 kJ. 3 CO (g ) + 7 H 2 (g ) → C 3H 8 (g ) + 3 H 2O (g ) ∆ 73c. 77. Nitrogen atom cannot bond to five fluorine atoms because. P O H H 65a. ΔH° combustion = -890. CaH 2 (s ) produces the most H2(g). The lone pair on the N atom causes the F—N—F bond angle to decrease from the ideal tetrahedral bond angle of 109° to 102. 75b. CH4(g) . 4 . C2H6(g). ΔH° combustion = -2877. calcium pyrophosphate or calcium diphosphate 65c. tetrapolyphosphoric acid 67.031 V 69a.61c.4 kJ. Since CH4 has the highest hydrogen atom to non-hydrogen atom ratio. H3PO4(aq) + 2 NH3(aq) → (NH4)2HPO4(aq) H H C H N N H H C H H H 63a. Copyright © 2011 Pearson Canada Inc. MnO 2 (s) + 2H 2 (g) → Mn(s) + 2H 2 O(g) 75a. as a second-row element. O Cl N O 63b. hydrogen phosphate ion 65b. C3H8(g). ΔH° combustion = -2219. 69b.5°. O H O 63c. 71. it cannot accommodate more than four electron pairs.9 kJ. The NF3 molecule is trigonal pyramidal. 73a. C4H10(g). A %BPL greater than 100% means that the material has a larger %P than does pure Ca3(PO4)2. 1 mol P4 O10 4 mol P 30. 100.23 g/cm3 94a. 95a.1.79. NaNO 2 decolorizes acidic solution of KMnO 4 according to the following balanced chemical equation: 2KMnO 4 + 3H 2 SO 4 + 5NaNO 2 → 5NaNO 3 + K 2 SO 4 + 2MnSO 4 + 3H 2 O.185 g Ca 3 (PO 4 ) 2 2 mol P 1 mol Ca 3 (PO 4 ) 2 Thus.185 will give the mass (or mass %) of BPL.974 g P × × = 0. NaNO 3 does not react with acidic solution of KMnO 4 . 283. H2O(l) is removed from the gas mixture. Because the 2px orbital on N is strongly bonding in the bent configuration.89 g P4 O10 1 mol P4 O10 1 mol Ca 3 (PO 4 ) 2 310. oxygen is converted to H2O by the reaction H2(g) + ½ O2(g) → H2O(l). The radiation is in the near ultraviolet. Solving for [H+] yields a value of 8 x 10-5 M which corresponds to a pH of 4.000 g P4 O10 × × 1 mol P4 O10 4 mol P × 283. The dehydrated zeolite has a very strong affinity for water molecules and thus. Integrative and Advanced Exercises 81.0 ×10−20 J/atom 89. 339 kJ/mol 94b. Thus. 15 g H 2SO 4 . the energy of the NH2− will be much lower for the bent configuration. 95c.18 g Ca 3 (PO 4 ) 2 × = 2. 92. 37 g Cl2 . multiplying the mass of P4O10 (283. 9. 3. ν =× 8. 85. The first four MO’s are fully occupied. In step (1). Copyright © 2011 Pearson Canada Inc. 2 NH 4 ClO 4 (s) → N 2 (g) + 4 H 2 O(g) + Cl 2 (g) + 2 O 2 (g) 92.436 will give the mass (or mass percent) of P.4% BPL 99. 0. 84. 95b. 5 .1716 M 86.50 1014 s −1 .89 g P4 O10 1 mol P4 O10 1 mol P multiplying the mass of P4O10 by 0. NH2− has 8 valence electrons. Step (2) ensures that unreacted hydrogen from step (1) is also converted to H2O(l). mass P = 1. The solution is still acidic.436 g P .000 g P4 O10 × mass Ca 3 (PO 4 ) 2 = 1.88) by 2. λ = 353 nm. 2NaI(s)+2H 2SO 4 (concd aq) → Na 2SO 4 (a)+2H 2O(g)+SO 2 (g)+I2 (g) Copyright © 2011 Pearson Canada Inc. 127a. 6 . 2 H 2 O(l) → O 2 (g) + 4 H + (aq) + 4 e − . [HOCl(aq)] − [Cl = (aq)] 0.187 V. This result supports the observation that Xe(g) does not react directly with O2(g) to form XeO3(g). 112. the formal charge on the S atom is zero and the sulfur–oxygen bonds are double bonds. The answer is (c). in the most important structure. 102. with a lone pair of electrons on the central atom. the central atom has a formal charge of +1.030 M and [Cl = 0. 126. ∆H fo = 639 kJ mol−1. The answer is (d). Oxidations: XeF4 (g) + 3 H 2 O → Xe(g) + 4 F− (aq) . 2 (aq)] 105. Cl2 (g)+2NaOH(aq) → NaCl(aq)+NaOCl(aq)+H 2O(l) 127b. 2 H+(aq) + ClO3-(aq) + 1 e. For O3. 122.70 V 107. H+(aq) + ClO2(aq) + 1 e. The answer is (b). The answer is (d). In each of these structures. 124. 125. The oxygen–oxygen bond order is between 1 and 2. Reduction: XeF4 (g) + 4 e − → 2 XeO3 (g) + 12 HF(aq) + 3/2 O 2 (g) + 2Xe(g) Net: 3 XeF4 (g) + 6 H 2 O(l) + = [H = (aq)] 104. First balance each equation and then consider how to manipulate and add the equations together to get the overall equation. thus. the reaction is also entropically unfavorable. The standard voltage for this half-reaction is given in Appendix D. The answer is (a) and (b). Both molecules are V shaped. Because the reaction converts 2. The formation reaction is very endothermic and thus energetically unfavorable. Feature Problems 110.→ HClO2(g). 109. Although many resonance structures can be drawn for SO2. Demonstrate that the three reactions result in the decomposition of water as the net reaction: 2 H 2 O → 2 H 2 + O 2 . 121. we expect ∆Sfo < 0.→ XeO3 (g) + 4 HF(aq) + 2 H + (aq) + 2 e − . Eo = 1.060 M . 0.→ ClO2(g) + H2O(l). The answer is (b). 123. Self-Assessment Exercises 120. The answer is (a).5 moles of gas into 1 mole of gas. the structure is a hybrid of two equivalent structures. (aq)+3H 2 O(l) → 6Cl. A number of other metals can be used. Other carbonates and acids can be used.38 V 134a. CaCO 3 (s)+2HCl(aq) → CaCl2 (aq)+H 2O(l)+CO 2 (g) . Cl2 (g)+2Br . LiH(s)+H 2O(l) → LiOH(aq)+H 2 (g) 129b. tellurous acid 132e.(aq)+IO3. 3NaBr(s)+H 3PO 4 (concd aq) → Na 3PO 4 +3HBr(g) 127e. 7 .(aq)+6H + (aq) . 5. 128b. AgAt 132b. NH 4 Cl(aq)+NaOH(aq) → NaCl(aq)+H 2O(l)+NH 3 (g) . Difficult to control reaction so that NO(g) is the only reduction product. 3Cl2 (g)+I. 3NO 2 (g)+H 2O(l) → 2HNO 3 (aq)+NO(g) 130. 128d.127c. 128e. 128c. A catalyst such as MnO2 is required. Cl2 (g)+2KI3 (aq) → 2KCl(aq)+3I2 (s) 127d. Zn(s)+2H + (aq) → Zn 2+ (aq)+H 2 (g) . 2KClO 3 (s) → 2KCl(s)+3O 2 (g) . 0. sodium perxenate 132c. SO3 134b. potassium perastatate 133. C(s)+H 2O(g) → CO(g)+H 2 (g) 129c.0 kg3 of seawater. Other ammonium salts and other bases work as well. 131. SO2 Copyright © 2011 Pearson Canada Inc. K2SeSO3 132f. MgPo 132d. 129a. 5HSO-3 (aq)+2MnO-4 (aq)+H + (aq) → 5SO 2-4 (aq)+2Mn 2+ (aq)+3H 2O(l) 128a.(aq)+Br2 (aq) . Electrolysis of H2O is an alternate method.(aq) → 2Cl. 132a. 3Cu(s)+8H+(aq)+2NO-3 (aq) → 3Cu 2+ (aq)+4H 2O(l)+2NO(g) . 8 . Cl2O7 134d. True Copyright © 2011 Pearson Canada Inc. True 135c. I2O5 135a. False 135b.134c. Sn 2+ . Air 2 Co (OH )3 (s ) → Co 2 O3 (s ) + 3H 2 O (g ) . and Cl−. (c) 2 HgO (s ) → 2 Hg (l ) + O 2 (g ) . Integrative Example A. Most compounds of main group Copyright © 2011 Pearson Canada Inc. The disproportionation reaction is 3 Ti2+ → 2 Ti3+ + Ti and E° = −1. PtCl42−. A given main-group metal typically displays just one oxidation state. Because E° < 0 the reaction is not spontaneous under standard conditions. usually equal to its family number in the periodic table. In practical terms. Exercises 1a. Sn 4+ . 1 . [Ar] 3d 4s 1e. -Eo for the couple must be > –0. and Zn(s) will do the job.13 V.041 V and < + 1. B. and Pb 2+ being major exceptions.Chapter 23Answers Practice Examples 1a. Main group metals do not form a wide variety of complex ions. (c) Mn 2+ (aq ) + 2H 2 O(l) → MnO 2 (s ) + 4H + (aq ) + 2e − . 2a. (a) 2 Cu 2S (s ) + 3 O 2 (g ) → 2 Cu 2 O (s ) + 2 SO 2 (g ) . NO3. Cr2+(aq). with Al3+ . (b) ∆ . PtCl42−. [Ar] 3d 4s 1d. [Ar] 3d 4s 1b. [Ar] 3d 4s 3. [Ar] 3d 4s 1f. an external voltage source would be required to make a significant amount of PtCl62− from V3+.(aq ) + 3V 3+ (aq ) + H 2 O(l) → NO (g ) + 3VO 2+ (aq ) + 2H + (aq ) Ecell o = +0. On the other hand. Fe(s). The formation of PtCl62− is favored by using a very low concentration of V2+ and very high concentrations of V3+. most transition metal ions form an extensive variety of complex ions.619V 2b. [Ar] 3d 4s 1c. (b) ∆ WO3 (s ) + 3 H 2 (g ) → W (s ) + 3 H 2 O (g ) . ∆ 1b. 2 V2+ + PtCl62− → 2 V3+ + PtCl42− + 2 Cl−.261 V. (a) 3 Si (s ) + 2 Cr2 O3 (s ) → 3 SiO 2 (s ) + 4 Cr (s ) . and Cl−. However. Ag (s ) + HCl (aq ) → no reaction 11d. As a result. the electron it is not added to the outermost shell. less strongly attracted to the positive charge of the nucleus in diffuse f-orbitals that are do not penetrate effectively and are very effectively shielded by the core electrons. Then there is no point of semistability until the remaining five d electrons are removed. which is farthest from the nucleus. 2 KOH (l ) + TiO 2 (s ) → K 2 TiO3 (s ) + H 2 O (g ) 13d. 9. 7. Removing one electron produces an electron configuration ( 3d 5 4 s1 ) with two half-filled subshells. removing two produces one with a half-filled and an empty subshell. many of the compounds of transition metal cations are colored. TiCl4 (g ) + 4 Na (l ) → Ti (s ) + 4NaCl (l ) ∆ 11b. they are removed much more readily. K 2 Cr2 O7 (aq ) + 2KOH (aq ) → 2 K 2 CrO 4 (aq ) + H 2 O(l) ∆ 11e. Thus it has small effect on the size of the atom.metals are colorless. many transition metals cations have one or more unpaired electrons and therefore are paramagnetic. Virtually every main-group metal cation has no unpaired electrons and hence is diamagnetic. as a result. is due to the larger size of lanthanide atoms. exceptions occur when the anion is colored. MnO 2 (s ) + 2 C (s ) → Mn (l ) + 2CO (g ) 13a. Cu (s ) + 2 H 2SO 4 (conc. On the other hand. Cr2 O3 (s ) + 2Al (s ) → 2 Cr (l ) + Al2 O3 (s ) 11c. The valence (outer shell) electrons of these larger atoms are further from the nucleus. Moreover. The greater ease of forming lanthanide cations compared to forming transition metal cations. which are one principal quantum number lower than the outermost shell. due to d-d electron transitions. Thus. when an electron and a proton are added to a transition metal atom to create the atom of next highest atomic number. ∆ 11a. this electron billows out and. 5. has a major influence on the size of the atom. the electron in the same subshell as the added electron offers little shielding. One possible explanation might be its 3d 5 4 s 2 electron configuration. aq ) → CuSO 4 (aq ) + SO 2 (g ) + 2 H 2 O(l) Copyright © 2011 Pearson Canada Inc. Manganese. The electron is added to the d-orbitals. On the other hand. 2 . When an electron and a proton are added to a main group element to create the element of next highest atomic number. 3 Fe 2+ (aq ) + MnO 4 − (aq ) + 2 H 2 O (l ) → 3 Fe3+ (aq ) + MnO 2 (s ) + 4 OH − ∆ 13c. Sc (OH )3 (s ) + 3H + (aq ) → Sc3+ (aq ) + 3 H 2 O(l) 13b. Also. The electron is added to the outermost shell (n value). it is the electron that influences the radius. the added electron is well shielded from the nucleus and hence it is only weakly attracted to the nucleus. H2(g) will do the job. Ecell o = +0. 21a.10 g Cr Copyright © 2011 Pearson Canada Inc. 0./Cr 3+ reduction potential = 1. We expect a positive slope with slight changes in the slope after the melting point (839 °C) and boiling point (1484 °C). 3 . 2.33 V and Cr 2+ /Cr reduction potential = -0.54V . 1.255 V.6 ×10−5 M 35. BaCO3 (s ) + 2 HCl (aq ) → BaCl2 (aq ) + H 2 O(l) + CO 2 (g ) 2 BaCl2 ( aq ) + K 2 Cr2 O 7 ( aq ) + 2 NaOH ( aq ) → 2 BaCrO 4 ( s ) + 2 KCl ( aq ) + 2 NaCl ( aq ) + H 2 O(l) ∆ ∆ 17. Cr 2 + (aq ) → Cr 3+ (aq ) + e − o = −0. The plot will be below the ΔG° line for 2 Mg(s) + O2(g) → 2 MgO(s) at all temperatures.424 V. Does not occur to a significant extent. Sn(s). Pb(s). The graph should be similar to that for 2 Mg(s) + O2(g) → 2 MgO(s). FeS (s ) + 2 HCl (aq ) → FeCl2 (aq ) + H 2S (g ) 4 Fe 2+ ( aq ) + O 2 ( g ) + 4H + ( aq ) → 4Fe3+ ( aq ) + 2 H 2 O(l) Fe3+ ( aq ) + 3 OH − ( aq ) → Fe ( OH )3 ( s ) 15b.704V . Does occur to a significant extent. HgS(s) + O2(g) → 4 Hg(l) + 3 CaS(s) + → Hg(l) + SO2(g).74 V./Cr 2+ . Pb 2+ ( aq ) + CrO 4 2− ( aq ) 31. Cr2 O7 2. E° = − 0. 4 HgS(s) + 4 CaO(s) CaSO4(s). 23a.229V . Ecell 23c. 19. 2CrO 4 2− ( aq ) + 2H + ( aq ) .90 V. Cr2 O7 2− ( aq ) + H 2 O(l) PbCrO 4 ( s ) .892 V and Cr 3+ /Cr . 25. Does occur to a significant extent.15a. VO 2+ ( aq ) + 2H + ( aq ) + e − → V 3+ ( aq ) + H 2O (l) 21b. Ecell o = +0. 23b. 27. The two unknown potentials are Cr2 O7 2.74 M 33b. 3 Zn (s ) + Cr2 O7 (aq ) +14 H + (aq ) → 3 Zn 2+ (aq ) + 2 Cr 3+ (aq ) + 7 H 2O(l) Zn (s ) + 2 Cr 3+ (aq ) → Zn 2+ (aq ) + 2 Cr 2+ (aq ) 4 Cr 2+ (aq ) + O 2 (g ) + 4 H + (aq ) → 4 Cr 3+ (aq ) + 2 H 2 O(l) 2− 33a. -Eo for the couple must be> –0. mainly owing to changes in entropy. 29.337 V and <+ 0. Table D-4 contains the following data: Cr 3+ /Cr 2+ reduction potential = -0. E° = 0. Oxoanions are better oxidizing agents in acidic solution because increasing the concentration of hydrogen ion favors formation → 2 Cr 3+ (aq) + 7H 2 O(l) . AuCl3 is not noted as being an insoluble chloride. which makes these compounds harder to dissolve. forms a very insoluble chloride. Copyright © 2011 Pearson Canada Inc. The trigonal bipyramidal shape of nickel and iron carbonyls does not fit well in crystalline lattices. the reflected light appears white. Spontaneous under these conditions. Notice of product. Compare this to octahedral complexes (Cr(CO)6 and others). The remaining reflected light is yellow in this case. Re(CO)557d. [AuCl4]– . The half-equation is: Cr2 O72. which probably adheres to the surface of the metal and prevents further reaction.24 V. Os(CO)5 57c. Fe3+(aq) + K4[F e(CN)6](aq) → K F e [F e(CN)6](s) + 3 K+(aq) 43a. For ZnO. No 45b. gold(III) forms a very stable complex ion with chloride ion. these five-coordinate carbonyls are liquids at room temperature. 0. Au + (aq ) + Fe 2 + (aq ) → Au (s ) + Fe 3+ (aq ) 43c. Mo(CO)6 57b.02 atm 51. Dichromate ion is the prevalent species in acidic solution.02 49b. E = 0. For CdS. Ag+. [II] [III] [II] 41. 39. Yes 47. AgCl. which are solids because they fit well into these lattices. 0. 0. 4 . [AgCl2]–. 57a.912 V 49a. the oxidation product. The blue light absorbed by CdS is subtracted from the white light incident on the surface of the solid. consequently. 2Cu 2+ (aq ) + SO 2 (g ) + 2 H 2 O(l) → 2 Cu + (aq ) + SO 4 2− (aq ) + 4 H + (aq ) 45a.(aq) + 14 H + (aq) + 6 e − 2− 2− also that CrO 4 is smaller than is Cr2 O7 . giving it a higher lattice energy in its compounds. Advanced and Integrative Exercises 53. λ = 479 nm (blue light) .37. Cu 2 + (aq ) + H 2 (g ) → Cu (s ) + 2H + (aq ) 43b. either because of its symmetry or repulsions with its neighbors and. λ = 413 nm (violet light) . First. This complex ion is much more stable than the corresponding dichloroargentate ion. When the violet light is subtracted from the white light incident on the ZnO surface. Second. 29. % Cr = 28. The % Ti by mass is 45%. If Δngas = 0. % Mn = 3. Spontaneous. E cell 64. If Δngas > 0.ions. then ΔS° ~ 0 and ΔG° is essentially independent of temperature (C(s) + O2(g) → CO2(g)). Dissolution: AgO(s) + 2 H + (aq) → Ag 2+ (aq) + H 2 O(l) 61a. Oxidation: 2 H 2 O(l) → O 2 (g) + 4 H + (aq) + 4 e − Reduction: Ag 2+ (aq) + e − → Ag + (aq) = +0. 82. NiC8H14O4N4 68b. Hg Hg O Mn Mn 71b.40 mL 67. O O O O O O O O 71c. hence the graph has a negative slope (2 C(s) + O2(g) → 2 CO (g)). Feature Problems 74a. hence the graph has a positive slope (2 CO(g) + O2(g) → 2 CO2 (g). 5 . Copyright © 2011 Pearson Canada Inc. The empirical formula is NiTi.52% 71a. then ΔS° > 0 and ΔG° will become more negative with increasing temperature. 68a. 65b.3% 65a. Os O O 73. 3.75 V .286%.2%. o 61b. then ΔS° < 0 and ΔG° will become more positive with increasing temperature. If Δngas < 0. This compound would be an ionic.57e. salt-like material consisting of Na+ and V(CO)6.1000 M MnO4– solution would be required. More of the 0. 74b. the plot of ΔG° for the net reaction as a function of temperature will be a straight line with a slope of − [{ΔS° (Rxn b)} − {ΔS° (Rxn c)}] (in kJ/K) and a y-intercept of [{ΔH° (Rxn b)} − {ΔH° (Rxn c)}] (in kJ). the plot of ΔG° vs. T for the reaction will follow the equation y = −0.176 x + 172.5 . PCO = 5.7 atm. Self-Assessment Exercises 79a. Impure iron (95% Fe). 79b. Iron-manganese alloy. 79c. Fe(CrO2)2 79d. An alloy of Zn and Cu with small amounts of Sn, Pb, and Fe. 79e. Mixture of HCl(aq) and HNO3(aq). 79f. Impure Cu(s) containing SO2(g). 79g. Iron alloy with varying quantities of metals such as Cr, Mn, and Ni, and a small and carefully controlled percentage of carbon. 80. The answers are (c), (f), and (g). 81. The answer is (b). 82. The answer is (d). 83. The answer is (a). 84. The answer is (d). 85. The answers are (c) and (e). 86a. CrO3(s) 86b. potassium manganate 86c. chromium carbonyl 86d. BaCr2O7 86e. lanthanum(III) sulfate nonahydrate 86f. Au(CN)3C3H2O 87a. 2Fe 2S3 (s) + 3O 2 (g) + 6H 2 O(l) → 4Fe(OH)3 (s) + 6S(s) 87b. 2Mn 2+ (aq) + 8H 2 O(l) + 5S2 O82- (aq) → 2MnO-4 (aq) + 16H + (aq) + 10SO 2-4 (aq) 87c. 4Ag(s) + 8CN - (aq) + O 2 (g) + 2H 2 O(l) → 4[Ag(CN 2 ]- (aq) + 4OH - (aq) Copyright © 2011 Pearson Canada Inc. 6 88. Atoms of Zn, Cd and Hg have configurations of 4s 2 3d 10 , 5s 2 4d 10 , and 6s 2 4 f 14 5d 10 , respectively. One the ns 2 electrons participate in bonding, and in this regard Zn, Cd, and Hg resemble the group 2 elements. 89. HNO3 is the oxidizing agent for oxidizing the metal to Au3+ but Au3+ must be stabilized in solution. In aqua regia, Cl- ions combine with Au3+ to form [AuCl4]-, which is stable in solution. 90. The Fe3+ ion forms a complex ion, Fe(H2O)63+, in aqueous solution. The complex behaves as a weak monoprotic acid in solution. Copyright © 2011 Pearson Canada Inc. 7 Chapter 24 Answers Practice Examples 1a. C.N. = 5 , O.S. = +2 . 1b. Fe CN 6 3 2a. K 2 PtCl6 2b. pentaamminethiocyanato-S-cobalt(III) chloride ox ox Cl Co NH3 Cl H3N NH3 ox NH3 Co Cl H3N Cl Cl Co NH3 Cl 3a. CO NC5H5 Cl Mo Cl OC NC5H5 3b. Cl Cl NC5H5 Cl Mo CO OC NC5H5 NC5H5 OC Mo CO Cl NC5H5 NC5H5 CO Cl Mo Cl OC NC5H5 NC5H5 CO Cl Mo CO Cl NC5H5 4a. Three 4b. There are three unpaired electrons in each case. 2 5a. Co CN 4 must be tetrahedral (3 unpaired electrons) and not octahedral (1 unpaired electron) because the magnetic behavior of a tetrahedral arrangement would agree with the experimental observations (3 unpaired electrons). 5b. The complex ion must be paramagnetic to the extent of one unpaired electron, regardless of the geometry the ligands adopt around the central metal ion. 6a. We are certain that Co H 2O 6 CoCl 2 4 2 is octahedral with a moderate field ligand. Tetrahedral has a weak field ligand. The relative values of ligand field splitting for the same ligand are t = 0.44 o . Thus, Co H 2O 6 2+ absorbs light of higher energy, blue or green light, leaving a light pink as the complementary color we observe. CoCl4 absorbs lower energy red light, leaving blue light to pass through and be seen. 2 6b. K 4 Fe CN 6 3 H 2O should appear yellow. Fe H 2O 6 NO 3 2 should be green. Copyright © 2011 Pearson Canada Inc. 1 Integrative Example A. (a) trans-chlorobis(ethylenediamine)nitrito-N-cobalt(III) nitrite (no reaction with AgNO3) + Cl H2 N H2 N NO2- Co H2N N H2 NO2 (b) trans-bis(ethylenediamine)dinitrito-N-cobalt(III) chloride (reacts with + NO2 H N H N Cl- Co HN NO2 N H AgNO3) (c) cis-bis(ethylenediamine)dinitrito-N-cobalt(III) chloride (reacts with AgNO3 and + H2 N NO2 H2 N Cl- Co H2N N H2 NO2 en) +3 NH3 H3N NH3 Pt H3N NH3 Cl B. [Pt(NH3)5Cl][Cl]3, . Exercises 1a. CrCl4 NH 3 2 , diamminetetrachlorochromate(III) ion. 1b. Fe CN 6 , hexacyanoferrate(III) ion. 3 1c. Cr en 3 Ni CN 4 , tris(ethylenediamine)chromium(III) tetracyanonickelate(II) ion. 3 2 3a. amminetetraaquahydroxocobalt(III) ion Copyright © 2011 Pearson Canada Inc. 2 7c. 7b. triamminetrinitrito-O-cobalt(III) 3c. tetraaquaplatinum(II) hexachloroplatinate(IV) 3d. S - H N O C N - 7a. Copyright © 2011 Pearson Canada Inc. silver(I) tetraiodomercurate(II) H 5a. O 5d. H O H H N C 5b. 3 . diaquadioxalatoferrate(III) ion 3e.3b. H H 5c. 11b.+ NH3 NH3 H3N Cr H3N OSO 3 NH3 9a. fac-triamminedichloronitro-O-cobalt(III) NH3 NH3 Cl Co NH3 ONO mer-triammine-cis-dichloronitro-O-cobalt(III) NH3 NH3 Cl Co ONO Cl Cl NH3 mer-triammine-trans-dichloronitro-O-cobalt(III) NH3 NH3 Cl Co Cl ONO 9c. Linear structures do not display cis-trans isomerism. Yes Copyright © 2011 Pearson Canada Inc. 4 . Square planar structures can show cis-trans isomerism. 13a. Tetrahedral structures do not display cis-trans isomerism. Three 13b. 3- O O C C O O O Co O C O C O O C O O O C 9b. NH3 11a. 11c. The Mo complex is paramagnetic. 25b. dark blue + 2 OH aq . The magnetic behavior cannot be used to determine whether the ammine complex of nickel(II) is octahedral or tetrahedral. Strong Field Weak Field eg (1 unpaired electron) t2g eg t2g (3 unpaired electrons) 21a.15. The trans-isomer is not optically active. light of certain wavelengths (corresponding to the energies absorbed) will no longer be present in the formerly white light. Three 23. The Co complex is diamagnetic. The energy spacing between these groups often corresponds to the energy of a photon of visible light. Co en 3 should have the largest overall Kf value. 17. 3+ 29. aq + 4 H 3O + aq 4 2+ 27. The cis-dichlorobis(ethylenediamine)cobalt(III) ion is optically active. 19. Cu 2+ aq + 2 OH aq 2 Cu NH 3 Cu OH 2 s + 4 NH 3 aq 4 Cu NH 3 4 2+ 2+ aq. the transition-metal complex will absorb light with energy corresponding to this spacing. If white light is incident on the complex ion. Thus. The octahedral and tetrahedral configurations have the same number of unpaired electrons. the light remaining after absorption will be missing some of its components. Zn OH 2 s + 4 NH 3 aq 4 2+ aq + 2 OH aq Cu OH s . Zn NH 3 25a. Cu H 2 O aq + 4 NH 4 + aq . Thus. First: Fe H 2 O 6 3+ Fe H 2O en aq + 2H 2O(l) aq + en aq 4 Second: Fe H 2 O 4 en Copyright © 2011 Pearson Canada Inc. The resulting light is colored. In crystal field theory. the five d orbitals of a central transition metal ion are split into two (or more) groups of different energies. 3+ 3+ aq + en aq Fe H 2O 2 en 2 3+ aq + 2H 2O(l) 5 . 21b. Third: Fe H 2 O 2 en 2 3+ aq + en aq Fe en 3 3+ aq + 2H 2 O l K f = 5. The successive acid ionizations of a complex ion such as [Fe(OH)6]3+ are more nearly equal in magnitude than those of an uncharged polyprotic acid such as H3PO4 principally because the complex ion has a positive charge. tetraammineplatinum(II) tetrachloroplatinate(II). K2[SbCl5].5 . With a reasonably high S2 O 32 . 39e. K overall = K sp K f 8. Although zinc(II) forms a soluble stable ammine complex ion. NH 3 aq cannot be used in the fixing of photographic film because of the relatively small value of Kf for Ag NH 3 2 aq .0 109 = 3 31a. This would produce a value of K = 8. 39d. Copyright © 2011 Pearson Canada Inc. its formation constant is not sufficiently large to dissolve highly insoluble ZnS. Al H 2O 6 3+ aq 35a. hexachloroplatinate(IV) ion. [Cu(NH3)4]2+. I− has a stronger tendency than does Cl− or NH3 for directing incoming ligands to the trans positions. sodium tetrachlorocuprate(II). successive ionizations should not show a great decrease in the magnitude of their ionization constants.6 107 . 39b. + 37. that dissolves the AgCl(s) that formed when Cl is low.0 106 which is nowhere large enough to drive the reaction to completion. [Pt(NH3)4][PtCl4]. [CoCl2(NH3)4]Cl. 39c. 33. it is sufficiently large to dissolve the moderately insoluble ZnCO 3 . this reaction will go essentially to completion. 31c. To make the cis isomer. Chloride ion forms a stable complex ion with silver(I) ion. The second proton is leaving a species which has one fewer positive charge but which is nonetheless positively charged. 45. Aluminum(III) forms a stable (and soluble) hydroxo complex but not a stable ammine complex. [PtCl6]2–. 40. Kf = 1. Integrative and Advanced Exercises 39a. tetraamminedichlorocobalt(III) chloride. tetraamminecopper(II) ion. potassium pentachloroantimonate(III). Since positive charges repel each other. 31b. Na2[CuCl4]. However. and so it is beneficial to convert K2[PtCl4] to K2[PtI4] before replacing ligands around Pt with NH3 molecules. ligands that show a strong tendency for directing incoming ligands to positions that are trans to themselves must be used. 6 . 35b. Copyright © 2011 Pearson Canada Inc. K[PtCl5(NH3)].142 V . 74. 52a.52 M. 52b. The answer is (d). E 0. which is one more than the actual number of geometric isomers found for this complex ion.59 V . The answer is (e). K+ occupies tetrahedral holes. each residing in a 3d orbital and would be sp3d2 hybridized. 4p. The negative cell potential indicates that the reaction indeed does not occur. [Co3 ] 2. [Pt(NH3)6]Cl4.2 10-28 M . An impossibly high concentration of H3O+ would be required. Ecell 0. The coordination compound is face-centered cubic. a trigonal prismatic structure cannot account for the optical isomerism that was observed for [Co(en)3]3+. M . 75. Feature Problems 64a. 4 Co3 (aq) 2 H2O(l) 4 Co2 (aq) 4 H (aq) O2 (g). 63. The answer is (a).99 V 57. Each Cr–NH3 coordinate covalent bond is a σ bond formed when a lone pair in an sp3 orbital on N is directed toward an empty sp3d2 orbital on Cr3+. E 0. while PtCl62. A trigonal prismatic structure predicts three geometric isomers for [CoCl2(NH3)4]+. The answer is (d). 9. In order from 0 NH3 ligands to 6 NH3 ligands: K2[PtCl6]. The answer is (c). 62. [PtCl(NH3)5]Cl3.02 47b. 52c. 72. Total [Cl–] = 0. Self-Assessment Exercises 70. The number of unpaired electrons predicted by valence bond theory would be the same as the number of unpaired electrons predicted by crystal field theory. [H3O+ ] 90. 54.47a. 64b. 2.occupies octahedral holes. [PtCl3(NH3)3]Cl. The hybrid orbitals would be hybrids of 4s. 51. The Cr3+ ion would have 3 unpaired electrons.0 10-4 M 47c. 73. PtCl4(NH3)2. The answer is (b). 7 . 71. and 3d (or 4d) orbitals. Barring any unusual stretching of the ethylenediamine ligand. [PtCl2(NH3)4]Cl2. no isomerism. K4[Cr(CN)6] Cl Cl 2- Pt Cl Cl 79a. Copyright © 2011 Pearson Canada Inc. 77b. [Pt(NO2)(NH3)3]+ 78c. no isomerism. 77c. sodiumhexanitrito-N-cobaltate(III). no isomerism 77d. The answer is (b). O + Cl H3N NH3 Cr H3N NH3 OH 79d. hexaamminechromium(III)hexacyanocobaltate(III). [CoCl(en)2(H2O)]2+ 78d. 78a.76. pentaamminebromocobalt(III)sulfate. 77a. 8 . [Ag(CN)2]78b. two optical isomers. tris(ethylenediamine)cobalt(III)chloride. Cl O O Cl Fe NH2 O H2N 79c. Cl H2 N Cl Fe H2N Cl Cl 79b. 82d. optical isomerism in the cis isomer. Two 80d. Copyright © 2011 Pearson Canada Inc. 82b. 82e. Two 81a. 82c. Linkage isomerism 81c. One 80b. No isomers. 83. one with the tridentate ligand occupying meridional positions and the other with the tridentate ligand occupying facial positions. Geometric isomerism 81e. Two 80c. Coordination isomerism 81b. Geometric isomerism 82a. Geometric isomerism (cis and trans) and optical isomerism in the cis isomer. No isomers. [Co(en)3]3+(aq) is yellow and [Co(H2O)6]3+(aq) is blue. 9 . No isomerism 81d.80a. Geometric isomerism (cis and trans). Two geometric isomers. and linkage isomerism in the thiocyanate ligand. 98 ×10−7 s −1 . (d) Possibly yes. (c) Radioactive.006001 u 6a. Integrative Example A. (b) 9.36 ×1018 atoms . 196 82 Pb + 0−1e → 3. 4. Exercises 1a. 75. (d) 2. 0 Cu →58 28 Ni + +1 β 4 124 124 1 53 I → 0+1 β + 124 52 Te . 2.14 B.40 ×1012 disintegrations / second . 234 94 230 Pu → 92 U + 42 He 1b. 0. 196 81 Tl + 0−1e → 196 80 Hg 0 C → 14 7 N + −1e Copyright © 2011 Pearson Canada Inc. 139 57 147 0 La +12 6 C → 63 Eu + 41 n 2b. 1 . 6b.Chapter 25 Answers Practice Examples 0 Pu → 241 95 Am + −1 β 1a. (c) 2. Positron emission by 17 F to produce 17 O and β − emission by 22 F to produce 22 Ne . (a) Stable.544 MeV 5b. 248 97 248 Bk → 98 Cf + −01e 1c. 0.7 ×103 y 4b. 14 6 196 81 Tl . (b) 4. (b) Radioactive.67 × 1040. (c) Yes. 58 29 2a. 3b. 121 51 Sb + 2 He → 53 I + 0 n . 3a. 241 94 1b.8 d 4a. 13 dis / min (per gram of C) 5a. (a) 9.36 ×1012 dis / s . (a) Yes. and 238. equals (4 × 51) + 2 . 222. 214 84 19b. The first of them. 32 17 4 W → 156 72 Hf + 2 He 38 Cl → 18 Ar + 0−1 β 214 Bi → 84 Po + 0−1 β 32 Cl → 16 S + 0+1 β 11a. 218. 37 Li + 11H → 84 Be + γ 1 11b. 2 .5 h Copyright © 2011 Pearson Canada Inc. 32 15 19c. Unk + 01n + 01n + 01n Po P Rn . 21.4 ×108 13 8 O. The mass numbers are separated from each other by 4 units. 4. 38 17 9c.235 Plot of Atomic Mass versus Atomic Number Atomic Mass (u) 230 225 220 215 210 205 82 80 84 5. 222 86 249 118 272 111 Rg → 5 42 He + 252 101 Md . 94 Be + 12 H → 10 5 B + 0n 11c. 226. and 214 84 Po . that is 4n + 2 . 230. 48 20 Ca + 249 98 Cf → 17. 58 26 302 Fe + 244 94 Pu → 120 Unk 19a. 214 83 9d. 87. In Figure 25–2. Co atoms 23. 234. 13. only the following mass numbers are represented: 206. 209 83 15. 60 27 28 12 Mg . 86 88 Atomic Number (Z) 90 92 7. 160 74 9b. 206. 80 35 Br . 210. 9a. 1 N + 10 n → 14 6 C + 1H 14 7 Bi + 64 28 Ni → 272 111 Rg + 10 n . where n = 51 . 214. An example is 16 O . 18 O 41c. 8.42 ×10−9 J 33b. 28.25. 8. which occurred about 3000 B. the rem provides a good idea of how much tissue damage a certain kind and quantity of radiation damage will do. 28. 20. 7 Li P should decay by β − emission. and hence often reacts in a similar fashion to calcium. 120 I should decay by β − emission. 50. The rem takes into account the quantity of biological damage done by a given dosage of radiation. 1. it will be retained in the body for a long time. For living tissue. 51. 184). P should decay by β + emission. 29. 7. 45. 3 . 3727 MeV 35. One reason why 90 Sr is hazardous is because strontium is in the same family of the periodic table as calcium. 82. The object is a bit more than 3000 years old. The most likely place for calcium to be incorporated into the body is in bones. 33 I should decay by β + emission. 142 days 27. where it resides for a long time. 50. But for nonliving materials. 29 43b. The rad is the dosage that places 0.805 MeV/nucleon 37. Strontium is expected to behave in a similar fashion. 7. 47. Mix a small amount of tritium with the H2 (g) and detect where the radioactivity appears with a Geiger counter.26 × 103 neutrons 41a. 20 Ne 41b. Thus. 4. 53. 20.06 MeV 39. 5. the rad is usually preferred. and thus is probably not from the pyramid era.37 mg 49.12 31. 82. 0. 134 43a. A “doubly magic” nuclide is one in which the atomic number is a magic number (2.03×109 y 33a. 114) and the number of neutrons also is a magic number (2. 126.C. 3.010 J of energy into each kilogram of irradiated matter. Copyright © 2011 Pearson Canada Inc. 73a. 59.0085 Plot of Packing Fraction versus Mass Number Packing Fraction 0. 2. The recovered sample will be radioactive. Only after some time has elapsed. A given anion does not remain associated with a particular cation. Thus. atoms multiplied by the very small decay constant is still larger than the product of the very small number of radon-222 atoms and its much larger decay constant. the larger number of radium-226. The graph and Fig. 4 . 2.2 × 107 kJ Feature Problems 0. In the early stages of the decay chain. 25-6 are almost exactly the inverse of one another. The rate of decay depends on both the half-life and the number of radioactive atoms present. 7. the ions are free to move throughout the solution. 29 y 67. with the maxima of one being the minima of the other.0 ×102 Ci 68.9 × 103 metric tons 62.0015 50 100 150 Mass Num ber (u) 72.0045 0.55.0 ×10−6 g 90Sr 63. 1.0065 0.0005 0 -0.0025 0. When NaCl(s) and NaNO 3 (s ) are dissolved in solution. does the rate of decay of radon-222 approach the rate at which it is formed from radium-226 and the amount of Copyright © 2011 Pearson Canada Inc. all the anions and cations are shuffled and some of the radioactive 24 Na will end up in the crystallized NaNO 3 . 232 90 Th+ 42 He → 243 97 232 92 Bk+2 01 n U+4 01 n 86. 85a.→ 212 86 Rn 85d. The answer is (c). 76 days 87a. 5 . The answer is (d). 375 days Copyright © 2011 Pearson Canada Inc. 81. 82. 80. The answer is (c).44 ppm 74c. 2. 174 days 87b. 267. The answer is (c). The answer is (d). 205 85 At → 205 84 Po+ +10 b 85c. 241 95 Am+ 42 He → 85f. 287 days 87c. The answer is (b). 214 88 Ra → 210 86 Rn+ 42 He 85b. the rate of decay of radon-222 exceeds its rate of formation.λd D = λp Po e p -λd D dt 73c. Beyond this point. The answer is (d). 21 H+ 21 H → 23 He+ 01 n 85e. 84. dD -λ t =λp P . 83. 87.5 u 74b.07 × 109 years Self-Assessment Exericises 78.95% 74d.radon-222 reaches a maximum. 212 87 Fr+e. 79. 73b. 2. 1 year 74a. The answer is (b). The answer is (b). 91. Copyright © 2011 Pearson Canada Inc. The answer is (a). 90.88. 89. 92. The answer is (c). The answer is (d). 6 . 6-dimethyloctane Copyright © 2011 Pearson Canada Inc. 3. 3. 2a.6. 1 .6-trimethyloctane 2b. C C C C C C C C CH3CH2CH2CH2CH2CH2CH3 C C C C C C C (CH3)2CHCH2CH2CH2CH3 C C C C C C CH3CH2CH(CH3)CH2CH2CH3 C C C C C C C C C (CH3)2CHCH2CHC(CH3)2 C C C C C C C (CH3)2CHCH(CH3)CH2CH3 C C C C C C C C C C (CH3)3CCH2CH2CH3 C C C C CH3CH2C(CH3) 2CH2CH3 C C C C C (CH3)3CCH(CH3)2 C C C C C C C CH(CH2CH3)3 1b.Chapter 26 Answers Practice Examples C C C C C C C CH3CH2CH2CH2CH2CH3 C C C C C C C (CH3)2CHCH2CH2CH3 C C C C C C C C C CH3CH2CH(CH3)CH2CH3 C C C C C C C (CH3)3CCH2CH3 (CH3)2CHCH(CH3)2 1a. CH3 H3C CH CH3 CH H3C 3b. All 2 . conformations have the same energy. Copyright © 2011 Pearson Canada Inc.3a. CH2 CH CH3 4a. 6b. 5a. (a) Achiral. (a) Achiral. 6a. (b) Chiral. (b) Chiral. 5b. Copyright © 2011 Pearson Canada Inc. 3 .4b. (c) Chiral. (c) Chiral. (b) R configuration. (c) R configuration. 8b. (c) (Z) stereoisomer. 8a. (a) Enantiomers. H3C H Cl CH2CH3 H3CH2C C H C H C . (E) stereoisomer CH2CH3 Cl H C C H (a) (Z) stereoisomer H3CH2C CH3 . (E) stereoisomer H3C C CH3 4 . (b) Enantiomers. (a) (E) stereoisomer. 9a.7a. (b) (Z) stereoisomer. H C H3CH2C Copyright © 2011 Pearson Canada Inc. (a) R configuration. H3CH2C C 9b. (E) stereoisomer H3CH2C CH3 C CH3 C H C H C (b) (Z) stereoisomer H3CH2C (c) (Z) stereoisomer . 7b. Ethers are relatively unreactive and the ether linkage is stable in the presence of most oxidizing and reducing agents. Compounds (a) and (b) are both ethers. as well as dilute acids and Copyright © 2011 Pearson Canada Inc. 5 .10a. CH3-CH2-CH2-OH (Compound A). CH3-CH2-CH2COO-CH3-CH2 (Compound C). B. Integrative Example A. 10b. CH3-CH2-CH2-COOH (Compound B). H H H Copyright © 2011 Pearson Canada Inc. NH2 H Cl H H C C C H 5a. C C C C C C C C C C 1b. on the other hand is an aldehyde. CH3 3a. Cl COOH Cl 3c. It will decolorize a Br2/CCl4 solution. Compound (c) is unsaturated secondary alcohol.alkalis. On treatment with chromic acid it will be oxidized to carboxylic acid. C C C C C C C C C C C C C C 1c. 6 . Compound (d). Exercises Br Br C 1a. H3C CH3 NO2 OH 3b. Both carbon atoms are sp 2 hybridized. H C O H C H H 5c. The C–C bond is between sp 3 orbitals on each C atom. The left-most C atom is sp 3 hybridized. Identical Copyright © 2011 Pearson Canada Inc.H H H O C C O H H H 5b. The C ≡ C triple bond is composed of one sigma bond formed by overlap of sp orbitals. Each carbon atom is sp 3 hybridized. All of the C – H bonds in the structure are sigma bonds between the 1s orbital of H and the sp 2 orbital of C. and two pi bonds. The righthand C – H bond is between the sp orbital on C and the 1s orbital on H. The C = C double bond is composed of a sigma bond between the sp 2 orbitals on each C atom and a pi bond between the 2 pz orbitals on the two C atoms. 7 . one from each C atom (that is a 2 p y 2 p y overlap and a 2 pz 2 pz overlap). and the C–H bonds to that C atom are between the sp 3 orbitals on C and the 1s orbital on H. H H H H H C C C H C H H H H 7b. The C–Cl bond is between the sp 2 orbital on C and the 3p orbital on Cl. H H C C C H H 9a. H C C Cl H H 7c. between the 1s orbital of H and the sp 3 orbital of C. All of the C – H bonds in the structure are sigma bonds. one from each C atom. The other two C atoms are sp hybridized. 7a. each formed by the overlap of two 2p orbitals. 15b. 21a. Ester group Copyright © 2011 Pearson Canada Inc. 19b. the carbonyl group is attached to two carbons. 19c. Aldehyde 17c. hydroxyl group and disubstituted aromatic ring. Carbon 2 is chiral. 15c. 17a. 13b. Phenol. Stereoisomers C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C 11. C C C C C C C C C C C C C C C C C 13a. Ketone 17d. An aldehyde has a carbonyl group with a hydrogen and a carbon group attached. Alkyl halide 17b. Constitutional isomers 9e. 15a. 8 . No relationship 9d. The essential difference is the presence of the — OH group in acetic acid.9b. The hydroxyl group. There are no chiral carbon atoms. In a ketone. Carbon 2 is chiral. There are no chiral carbon atoms. Constitutional isomers 9c. Carbon 2 is chiral. 19a. hydroxyl group. Carbon 2 is chiral. 21b. 13c. and phenyl group. Carboxylic acid. O O O O H OH O OH H H O H O O O O OH OH O O O OH H H O O O O O O HO O O O HO O O H O O O O H O H O O 27. 21d.21c. hydroxyl group. 1. OH 31a. Ketone and carboxylic acid group. O O H O O 25. 3.2-dimethylbutane 33b.2-dimethylcyclopropane Copyright © 2011 Pearson Canada Inc.3-dichloro-5-ethylheptane 33a. 2. 9 . Aldehyde. H OH OH O OH O 29. 2-methylpropene 33c.2-dimethylpropane 31c. and disubstituted aromatic ring.5-dimethyloctane 31b. 23. 2. 3. N. Sufficient 35f. HOOCCH 2 C ( OH )( COOH ) CH 2 COOH 39a. 3.N-diethylamine 39b. Insufficient 35e. 4-methyl-2-pentyne 33e.4-dimethyl-2-propyl-1-pentene 35a. N. Insufficient NO2 O2N CH3 NO2 37a. Sufficient 35c. Insufficient 35d.N-diethyl-N-methylamine Copyright © 2011 Pearson Canada Inc. N-cyclopentyl-N-ethylamine 39d. Insufficient 35b. 10 . p-nitroaniline 39c. 3.33d. O COCH3 OH 37b.4-dimethylhexane 33f. 37c. 47a. In the case of propene. 47b. there can be a double bond only between the central carbon atom Copyright © 2011 Pearson Canada Inc. 49. 43.41a. In the case of ethene there are only two carbon atoms between which there can be a double bond. 11 . 41b. 45. The Anti conformation is lowest in energy. (E) configuration 53a. Enantiomers 59e. 59a. (S)-3-bromo-2-methylpentane Copyright © 2011 Pearson Canada Inc.5-trihydroxybenzene 57.and a terminal carbon atom. The case of butene is different since 1-butene is distinct from 2butene. 12 . 53b. (Z) configuration 51c.3. Identical 59b. 1. 51a. Identical 59c. Isomers 59d. Identical 61a. Cl 55c. (E) configuration 51b. C CH 55a. Cl 55b. (E)-1-chloro-2-methylbut-1-ene 63c. 65b. 65e. (E)-4-(chloromethyl)-3. (S)-1. 65c. (R)-3-(bromomethyl)-1-chloropentan-3-ol 61d.2-dibromopentane 61c. (Z)-pent-2-ene 63b. (S)-1-bromopropan-2-ol 63a. Integrative and Advanced Exercises Copyright © 2011 Pearson Canada Inc. 13 . (Z)-5-bromo-3-(bromomethyl)-2-methylpent-2-enal 65a.7-dimethyloct-3-ene 63d.61b. 65d. 109. The correct answer is (b). C3H7OH 110c. C4 = sp3. The correct answer is (c). C4H8N2O2 84.79a. Copyright © 2011 Pearson Canada Inc. C6H12 110b. 3 CH 3CH= = CH 2 + 2 MnO 4 − + 4 H 2 O → 3CH 3CHOHCH 2 OH + 2 MnO 2 + 2 OH − 81. C2 = sp3. 14 . C6H5COOH 111a. The correct answer is (c). Ester. The correct answer is (e). Self-Assessment Exercises 101. 108. C1= sp2. 89b. 89a. C4H8 110a. 3 C6 H 5CH 2 OH + 2 Cr2 O7 2− + 16 H + → 3 C6 H 5COOH + 4 Cr 3+ + 11 H 2 O 79c. C3 = sp3. C6 H 5 NO 2 + 7 H + + 2 Fe → C6 H 5 NH 3+ + 2 H 2 O + 2 Fe3+ 79b. 102. arene. The unknown must be 1-butanol or butyraldehyde. 89c. The identity of the unknown can be established by treatment with a carboxylic acid. 103. amine(tertiary). N = sp3. Carbons 2 and 4 are chiral. 111b. 15 .111c. Copyright © 2011 Pearson Canada Inc. Chapter 27 Answers Practice Examples 1A (a) substitution (b) rearrangement (c) addition 1B (a) elimination (b) substitution (c) substitution 2A (a) (b) 2B (a) (b) 3A (a) CH3C≡C− + CH3Br ⎯ SN2→ CH3C≡CCH3 + Br− (b) no reaction is expected (c) CH3NH2 + (CH3)3CCl ⎯ SN1 → CH3NH2C(CH3)3 + Cl− Copyright © 2011 Pearson Canada Inc. + 1 . 3B (a) CN- SN2 mechanism H H3CH2C (b) - H3C H3C C (R) C δ+ NC δ− Cl Cl δ− CH3 NC + Cl- C (S) H CH CH 2 3 H CH2CH3 SN1 mechanism H O CH3 I H3C C CH2CH3 -I - H3C CH3CH2 H Start with pure (R) or (S) C H Topside attack OR bottomside attack H O OCH3 H3C C CH2CH3 H 50:50 mixture of (R) and (S) CH3 4A SN2 mechanism Copyright © 2011 Pearson Canada Inc. 2 . 5B 6A (a) Copyright © 2011 Pearson Canada Inc. 3 .4B SN1 mechanism 5A This is an SN2 reaction leading to the alkyl sulfide product. (b) 7A 7B Copyright © 2011 Pearson Canada Inc. 4 .(b) 6B (a) The reactions proceed via a SN2 mechanism. (a) Nucleophilic substitution is a fundamental class of substitution reaction in which an "electron rich" nucleophile selectively bonds with or attacks the positive or partially positive charge of an atom attached to a group or atom called the leaving group. (c) In an addition reaction. 5 . the positive or partially positive atom is referred to as an electrophile.Integrative Example 8A Alkanes are relatively unreactive. However. R-Br + OH− → R-OH + Br− (b) Electrophilic substitution reactions are chemical reactions in which an electrophile displaces another group. elimination will occur resulting in the formation of cyclohexene. 1. Subsequent repetition of these two steps will genererate the desired product. Such reaction will generate bromo-cyclohexane in the first step. Copyright © 2011 Pearson Canada Inc.3cyclohexadiene. a molecule adds across a double or triple bond in another molecule. In thepresence of a base. 8B Exercises 1. they will undergo bromination reaction in the presence of light. 6 . the carbon skeleton of a molecule is rearranged. atoms or groups that are bonded to adjacent atoms are eliminated as a small molecule.(d) In an elimination reaction. (e) When an organic compound undergoes a rearrangement reaction. Copyright © 2011 Pearson Canada Inc. 3. (a) substitution reaction (b) addition reaction (c) substitution reaction 5. (a) addition reaction (b) elimination reaction (c) substitution reaction 7. 7 . Copyright © 2011 Pearson Canada Inc.9. (d) Decreasing the concentration of NaOH by a factor of two will decrease the rate by a factor of 2. 11. (a) rate=k[CH3CH2CH2CH2Br][NaOH] (b) (c) Doubling the concentration of n-butyl bromide will also double the rate. The reaction proceeded via an SN1 mechanism. CH3CH2CH2OCH2CH3 + NaI CH3CH2ONa + CH3CH2CH2I leaving group nucleophile electrophile (b) The equilibrium favors the formation of the products CH3CH2NH3+ + KI NH3 + CH3CH2I electrophile nucleophile leaving group + K+ 15. In molecule (b). The reaction most likely proceeded via an SN1 mechanism.13. (a) E2 mechanism (b) SN2 mechanism (c) E2 mechanism Copyright © 2011 Pearson Canada Inc. H CH3 H3CH2CO (S and R) 21. 17. Molecule (a) reacts faster in the SN2 reaction than does molecule (b) because there is less steric hindrance. 19. 8 . the steric hindrance is associated with a methyl group being positioned in the axial position. (a) The equilibrium favors the formation of the products. 9 . (a) (b) (c) (d) There is no reaction.4-dimethyl-pent -2-ene.23. If no rearrangement occurs the expected product would be 4. 27. (a) (b) Copyright © 2011 Pearson Canada Inc. 29. 25. (a) (b) (c) (d) Copyright © 2011 Pearson Canada Inc.31. (a) (b) (c) 33. 10 . 11 .35. 37. Copyright © 2011 Pearson Canada Inc. 39. Copyright © 2011 Pearson Canada Inc. 12 . (a) (b) (c) 41. 47. During the polymerization reaction. we can only speak of average molecular weight.43. 49. Copyright © 2011 Pearson Canada Inc. its attack on alkane is purely statistical. There are six times as many 1° hydrogens and 2° hydrogens in the reactant molecule. 45. (a) (b) Whereas the flourine radical is so reactive. As a result. 13 . polymers of different chain-lengths are formed. (a) (b) (c) Copyright © 2011 Pearson Canada Inc. 53. Integrative and Advanced Exercises 55.51. 14 . 61. Copyright © 2011 Pearson Canada Inc. 15 . (a) (b) (c) (d) 59. (a) Compound (3) reacts with dilute HCl. (c) Compound (2) neutralizes dilute NaOH. 2-butanol. (a) (b) (c) no reaction (d) 63.57. (b) Compound (1) hydrolyzes. (e) Compound (2) and (3). 67. 16 . (c) Compound (2). (a) Compound (1) and (3). (b) Compound (2). Copyright © 2011 Pearson Canada Inc. (d) Compound (1) and (3).65. 1-chloro-2-methylbutane 1-chloro-3-methylbutane 2-chloro-3-methylbutane 2-chloro-2-methylbutane 31% 16% 31% 21% 73. 75. Copyright © 2011 Pearson Canada Inc.69. 71. 17 . one-step reactions that require an antiperiplanar conformation at the time of π-bond formation and β-bond breaking and do not involve carbocations. is a one-step process in which bond breaking and bond making occur simultaneously at a carbon atom with a suitable leaving group. on the other hand. (a) Nucleophilic substitution corresponds to a substitution (either SN1 or SN2) for aliphatic compounds. SN2 reaction. 18 . Self-Assessment Exercises 79. Copyright © 2011 Pearson Canada Inc. SN1 reaction involves the formation of carbocation. two or more atoms (molecules) combine to form a larger one. (b) (c) (d) Electrophilic aromatic substitution is typical for aromatic compounds (an atom is replaced by an electrophile) An addition reaction is the opposite of an elimination reaction.77. In an addition reaction. E1 reactions are unimolecular elimination reactions that proceed via carbocation intermediates. E2 reactions are bimolecular. (a) CN− (b) CH3I 83. the isopropoxide/ methyl iodide combination will be more ideal for SN2 (uncrowded alkyl halide). 19 . (a) (b) (c) (d) (e) Copyright © 2011 Pearson Canada Inc.81. Ether formation via the Williamson reaction proceeds via an SN2 mechanism. 85. As such. (b) K M = 0.300 m O O OH OH C C O O + H H2O O OH C O C O O + B. 0.Chapter 28 Answers Practice Examples 1a.01406 M .2 ×105 K + ions Copyright © 2011 Pearson Canada Inc. 2b. Pentapeptide sequence: Gly-Cys-Val-Phe-Tyr. 3 ×102 H 3O + ions 1b. (a) 4. dithreonylmethionine O O H2N CH C NH CH H CH2 OH O C NH CH H3C CH C OH CH3 1b. Hexapeptide sequence: Ser-Gly-Gly-Ala-Val-Trp. 2a. k= 2 Exercises 1a. Integrative Example A. 1 .8 ×105 s −1 . 1. Of the lipids listed in Table 27-2. 5b. phosphoric acid. Trinolein is unsaturated and an oil. Phospholipids are derived from glycerols. H C HO H H C OH H C OH CH2OH 15a. Polyunsaturated fatty acids are characterized by a large number of C = C double bonds in their hydrocarbon chain. ketotetrose. 7b. aldohexose. namely counterclockwise. 5 ×106 protein molecules 5a. Copyright © 2011 Pearson Canada Inc. 13a. myristate 7a. Both are triglycerides or glycerol esters. Polyunsaturated fatty acids are recommended in dietary programs since saturated fats are linked to a high incidence of heart disease. 15b. 2 .3. Soaps and phospholipids have hydrophilic heads and hydrophobic tails. Stearic acid has no C = C double bonds and therefore is not unsaturated. L-Glucose. D-Erythrulose. and a nitrogen containing base. Eleostearic acid has three C = C double bonds and thus is polyunsaturated. fatty acids. 13b. safflower oil has the highest percentage of unsaturated fatty acids. 11. glyceryl trioleate or triolein. C O H C O H C O H C O H C O H C O O C H H C OH HO C H H C OH HO C H H C OH HO C H H C OH HO C H HO C H H C OH H C OH HO C H H C OH HO C H C OH HO C H HO C H C OH CH2OH CH2OH H CH2OH CH2OH CH2OH H CH2OH 17a. Trilaurin is saturated and a fat. Soaps are salts of fatty acids. 17b. let alone polyunsaturated. A levorotatory compound is one that rotates the plane of polarized light to the left. CH 2 OHCHOHCH 2 OH and NaOOC ( CH 2 )14 CH 3 . A dextrorotatory compound rotates the plane of polarized light to the right. namely clockwise. 5c. 9. glyceryl palmitolinolenolaurate or glyceryl palmitoeleosterolaurate. 23d.S-configuration (rightmost structure). S-configuation (rightmost structure–top to bottom). 19. S. R. 0. A reducing sugar has a sufficient amount of the straight-chain form present in equilibrium with its cyclic form such that the sugar will reduce Cu 2+ aq to insoluble. (rightmost structure). 25c. b g bg 21. Enantiomers: S-configuration (leftmost structure). 3 . In organic nomenclature. Only free aldehyde groups are able to reduce the copper(II) ion down to copper (I). Diastereomers: S. such a mixture does not exhibit a net rotation of the plane of polarized light.397 g of Cu2O should be produced. Different molecules: different formulas 23c. The stereogenic center has an R-configuration if a curved arrow from the group of highest priority through to the one of lowest priority is drawn in a clockwise direction. 23b. 25b. this designation is given to a chiral carbon atom. R-config. red Cu 2 O s .17c. Copyright © 2011 Pearson Canada Inc. Since these two compounds rotate polarized light by the same amount but in opposite directions. 25a. 17d.R-configuration (leftmost structure–top to bottom). A racemic mixture has equal amounts of an optically active compound and its enantiomer.R configuration (leftmost structure–top to bottom). Diastereomers 23a. Diastereomers: R. with the elimination of a water molecule between them. + NH3 Cl CH2 − CH COOH 29a. to name a few.03 . O H2N CH CH3 31a. 27e. C NH O CH C OH CH2 SH Copyright © 2011 Pearson Canada Inc. For instance. The isoelectric point of glycine is pI = 6. and disulfide linkages. NH2 CH2 − CH COO Na + 29b. The peptide bond is the bond that forms between the carbonyl group of one amino acid and the amine group of another. glycine H 2 NCH 2 COOH is the simplest α -amino acid. For example. 27a.25d. NH3+ CH2 − CH COO 29c. A zwitterion is a form of an amino acid in which the amine group is protonated NH 3 + c − h i and the carboxyl group is deprotonated COO . b g d 27b. An α -amino acid has an amine group (NH2) bonded to the same carbon as the carboxyl group (COOH). 4 . 27c. The pH at which the zwitterion form of an amino acid predominates in solution is known as the isoelectric point. hydrogen bonding. the zwitterion form of glycine is + H 3 NCH 2 COO − . 27d. Tertiary structure describes how a coiled protein chain further interacts with itself to wrap into a cluster through a combination of salt linkages. The primary structure of an amino acid is the sequence of amino acids in the chain of the polypeptide. . Ser-Ala-Lys. There are 6 combinations: Ala-Ser-Ala-Ser. Ala-Ser-Ser-Ala. 35a. 39b. alanyl-seryl-glycyl-valyl-threonyl-leucine. H 33. 37b. H 2 NCH(CHOHCH 3 )COO − 37a. Ser-Ala-Ser-Ala. The tertiary structure of a protein refers to how different parts of the molecules interact with each other to maintain the overall shape of the protein macromolecule. 39a. 45. Ala-Ser-Gly-Val-Thr-Leu. or convoluted. + H 3 NCH(CHOHCH 3 )COO − 35c. Proline will not migrate very effectively under these conditions. Ser-Ala-Ala-Ser. or alanylserylglycylvalylthreonylleucine. Ala-Lys-Ser.H2N CH O O O C NH CH C NH CH C OH CHOH H3C CH CH3 CH3 31b. 41. The secondary structure describes how the protein chain is folded. + H 3 NCH(CHOHCH 3 )COOH 35b. There are 6 combinations: Lys-Ser-Ala. 5 . Not all proteins have a quaternary structure since many proteins have only one polypeptide chain. Ala-Ala-Ser-Ser. Ser-Ser-AlaAla. Ala-SerLys. Quaternary structure refers to how two or more protein molecules pack together into a larger protein complex. Lys-Ala-Ser. Aspartic acid will migrate toward the positively charged anode. Lysine will migrate toward the negatively charged cathode. coiled. CH3 H2N 43. Copyright © 2011 Pearson Canada Inc. R H R-alanine CO2H . Ser-Lys-Ala. One pyrimidine base is cytosine. CH3 O H N H H N N O N O O P OH OCH2 N O H N H O O OH O P OH N N OCH2 N O N H N O P OH N H N H OCH2 N O N O H H O N O OH N O H N H N H N OH Integrative and Advanced Exercises 52. 6 . Attached to each sugar is a purine or a pyrimidine base. 59. The sugars are deoxyribose in the case of DNA and ribose in the case of RNA. One example for the required sequence would be AGA CCA CAA CGA. 65a. In the case of RNA the other pyrimidine base is uracil. 9. Two major types of nucleic acids are DNA (deoxyribonucleic acid). The redundancy enables the organism to produce a given amino acid in several ways in case of transcription error. while for DNA the other pyrimidine base is thymine.47. Both of them contain phosphate groups. and RNA (ribonucleic acid).71 58.3 ×10−5 65b. 49. These phosphate groups alternate with sugars to form the backbone of the molecule. The purine bases are adenine and guanine. 2.9 ×103 g/mol . Nonapeptide: Arg-Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg. Met-Ser-Met-Met-Gly. The complementary sequence to AGC is TCG. 56.8 ×105 Copyright © 2011 Pearson Canada Inc. 62. 1. This is a minimum value because it is assumed that 1 Ag+ ion is all that is necessary to denature each protein molecule. 9. 75. Copyright © 2011 Pearson Canada Inc. The answer is (b). Saponification value = 189. Iodine number = 81. The answer is (d). 87. 66c.0. 7 . 83. 76. [B] = 2.65c. 84. The answer is (a). Self-Assessment Exercises 72. 85. Gly-Cys-Val-Phe-Tyr. 88. The answer is (d). 78. The iodine number for safflower oil may range between 150 to 144. The answer is (b). The answer is (e). The answer is (d). 129 g 80. 86. 73. The answer is (d). 79. 66b. The answer is (a). 77. Iodine number = 86. 82. The answer is (e). The saponification value may range between 211 and 179. The answer is (c).25 [A] Feature Problems 66a. 81. 74.6. The answer is (b). The answer is (b). Saponification value = 180. The answer is (c). RNA chain.
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