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188 CHAPTER 6 THERMOCHEMISTRY 34. a. ΔE = q + w = 23 J + 100. J = 77 J
b. w = PΔV = 1.90 atm(2.80 L 8.30 L) = 10.5 L atm × atmL
J3.101 = 1060 J
ΔE = q + w = 350. J + 1060 = 1410 J
c. w = PΔV = 1.00 atm(29.1 L11.2 L) = 17.9 L atm × atmL
J3.101 = 1810 J
ΔE = q + w = 1037 J 1810 J = 770 J 35. w = PΔV;; we need the final volume of the gas. Because T and n are constant, P1V1 = P2V2.
atm00.2
)atm0.15(L0.10P
PVV
2
112 = 75.0 L
w = PΔV = 2.00 atm(75.0 L 10.0 L) = 130. L atm × J1000
kJ1atmL
J3.101
= 13.2 kJ = work
36. w = 210. J = PΔV, 210 J = P(25 L 10. L), P = 14 atm 37. In this problem, q = w = 950. J.
950. J × J3.101
atmL1 = 9.38 L atm of work done by the gases
w = PΔV, 9.38 L atm = 760
.650 atm × (Vf 0.040 L), Vf 0.040 = 11.0 L, Vf = 11.0 L
38. ΔE = q + w, 102.5 J = 52.5 J + w, w = 155.0 J × J3.101
atmL1 = 1.530 L atm
w = PΔV, 1.530 L atm = 0.500 atm × ΔV, ΔV = 3.06 L ΔV = Vf – Vi, 3.06 L = 58.0 L Vi, Vi = 54.9 L = initial volume
39. q = molar heat capacity × mol × ΔT = molC
J8.20o × 39.1 mol × (38.0 0.0)°C = 30,900 J
= 30.9 kJ
w = PΔV = 1.00 atm × (998 L 876 L) = 122 L atm × = 12,400 J = 12.4 kJ
ΔE = q + w = 30.9 kJ + (12.4 kJ) = 18.5 kJ
atmLJ3.101
CHAPTER 6 THERMOCHEMISTRY 189 40. H2O(g) H2O(l); ΔE = q + w; q = 40.66 kJ; w = PΔV
Volume of 1 mol H2O(l) = 1.000 mol H2O(l) × g996.0
cm1mol
g02.18 3 = 18.1 cm3 = 18.1 mL
w = PΔV = 1.00 atm × (0.0181 L 30.6 L) = 30.6 L atm × = 3.10 × 103 J
= 3.10 kJ
ΔE = q + w = 40.66 kJ + 3.10 kJ = 37.56 kJ
Properties of Enthalpy 41. This is an endothermic reaction, so heat must be absorbed in order to convert reactants into
products. The high-temperature environment of internal combustion engines provides the heat.
42. One should try to cool the reaction mixture or provide some means of removing heat because
the reaction is very exothermic (heat is released). The H2SO4(aq) will get very hot and possibly boil unless cooling is provided.
43. a. Heat is absorbed from the water (it gets colder) as KBr dissolves, so this is an endothermic process. b. Heat is released as CH4 is burned, so this is an exothermic process. c. Heat is released to the water (it gets hot) as H2SO4 is added, so this is an exothermic
process. d. Heat must be added (absorbed) to boil water, so this is an endothermic process. 44. a. The combustion of gasoline releases heat, so this is an exothermic process. b. H2O(g) → H2O(l); heat is released when water vapor condenses, so this is an exothermic process. c. To convert a solid to a gas, heat must be absorbed, so this is an endothermic process. d. Heat must be added (absorbed) in order to break a bond, so this is an endothermic process. 45. 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) ΔH = −1652 kJ; note that 1652 kJ of heat is released when 4 mol Fe reacts with 3 mol O2 to produce 2 mol Fe2O3.
a. 4.00 mol Fe × = 1650 kJ; 1650 kJ of heat released
b. 1.00 mol Fe2O3 × = 826 kJ; 826 kJ of heat released
atmLJ3.101
Femol4kJ1652
32OFemol2kJ1652
CHAPTER 6 THERMOCHEMISTRY 191 50. H = E + PV; from this equation, H > E when V > 0, H < E when V < 0, and H
= E when V = 0. Concentrate on the moles of gaseous products versus the moles of gaseous reactants to predict V for a reaction.
a. There are 2 moles of gaseous reactants converting to 2 moles of gaseous products, so V = 0. For this reaction, H = E. b. There are 4 moles of gaseous reactants converting to 2 moles of gaseous products, so V < 0 and H < E. c. There are 9 moles of gaseous reactants converting to 10 moles of gaseous products, so V > 0 and H > E. Calorimetry and Heat Capacity 51. Specific heat capacity is defined as the amount of heat necessary to raise the temperature of
one gram of substance by one degree Celsius. Therefore, H2O(l) with the largest heat capacity value requires the largest amount of heat for this process. The amount of heat for H2O(l) is:
energy = s × m × ΔT = gCJ18.4
o × 25.0 g × (37.0°C 15.0°C) = 2.30 × 103 J
The largest temperature change when a certain amount of energy is added to a certain mass of substance will occur for the substance with the smallest specific heat capacity. This is Hg(l), and the temperature change for this process is:
ΔT = g.550
gCJ14.0
kJJ1000kJ7.10
msenergy
o
= 140°C
52. a. s = specific heat capacity = gK
J24.0gCJ24.0
o since ΔT(K) = ΔT(°C)
Energy = s × m × ΔT = gCJ24.0
o × 150.0 g × (298 K 273 K) = 9.0 × 102 J
b. Molar heat capacity = molCJ26
AgmolAgg9.107
gCJ24.0
oo
c. 1250 J = gCJ24.0
o × m × (15.2°C 12.0°C), m = 2.324.0
1250
= 1.6 × 103 g Ag
53. s = specific heat capacity = C)2.251.55(g00.5
J133Tm
qo
= 0.890 J/°C•g
From Table 6.1, the substance is solid aluminum.
192 CHAPTER 6 THERMOCHEMISTRY
54. s = C)0.205.53(g6.125
J585o
= 0.139 J/°C•g
Molar heat capacity = molC
J9.27Hgmol
g6.200gC
J139.0oo
55. | Heat loss by hot water | = | heat gain by cooler water |
The magnitudes of heat loss and heat gain are equal in calorimetry problems. The only difference is the sign (positive or negative). To avoid sign errors, keep all quantities positive and, if necessary, deduce the correct signs at the end of the problem. Water has a specific heat capacity = s = 4.18 J/°C•g = 4.18 J/K•g (ΔT in °C = ΔT in K).
Heat loss by hot water = s × m × ΔT = gK
J18.4 × 50.0 g × (330. K Tf)
Heat gain by cooler water = gK
J18.4 × 30.0 g × (Tf 280. K); heat loss = heat gain, so:
K
J209× (330. K Tf) =
KJ125× (Tf 280. K)
6.90 × 104 209Tf = 125Tf 3.50 × 104, 334Tf = 1.040 × 105, Tf = 311 K
Note that the final temperature is closer to the temperature of the more massive hot water, which is as it should be.
56. Heat loss by hot water = heat gain by cold water; keeping all quantities positive helps to
avoid sign errors:
gCJ18.4
o × mhot × (55.0°C 37.0°C) = gCJ18.4
o × 90.0 g × (37.0°C 22.0°C)
mhot = C18.0
C15.0g90.0o
o = 75.0 g hot water needed
57. Heat loss by Al + heat loss by Fe = heat gain by water; keeping all quantities positive to
avoid sign error:
gCJ89.0
o × 5.00 g Al × (100.0°C Tf) + gCJ45.0
o × 10.00 g Fe × (100.0 Tf)
= gCJ18.4
o × 97.3 g H2O × (Tf 22.0°C)
4.5(100.0 Tf) + 4.5(100.0 Tf) = 407(Tf 22.0), 450 (4.5)Tf + 450 (4.5)Tf = 407Tf 8950 416Tf = 9850, Tf = 23.7°C
CHAPTER 6 THERMOCHEMISTRY 193
58. Heat released to water = 5.0 g H2 × 2H gJ120. + 10. g methane ×
methanegJ50. = 1.10 × 103 J
Heat gain by water = 1.10 × 103 J = gCJ18.4
o × 50.0 g × T
T = 5.26°C, 5.26°C = Tf 25.0°C, Tf = 30.3°C 59. Heat gain by water = heat loss by metal = s × m × ΔT, where s = specific heat capacity.
Heat gain = gCJ18.4
o × 150.0 g × (18.3°C 15.0°C) = 2100 J
A common error in calorimetry problems is sign errors. Keeping all quantities positive helps to eliminate sign errors.
Heat loss = 2100 J = s × 150.0 g × (75.0°C 18.3°C), s = C7.56g0.150
J2100o
= 0.25 J/°C•g
60. Heat gain by water = heat loss by Cu; keeping all quantities positive helps to avoid sign errors:
gCJ18.4
o × mass × (24.9°C 22.3°C) = gCJ20.0
o × 110. g Cu × (82.4°C 24.9°C)
11(mass) = 1300, mass = 120 g H2O
61. 50.0 × 103 L × 0.100 mol/L = 5.00 × 103 mol of both AgNO3 and HCl are reacted. Thus 5.00 × 103 mol of AgCl will be produced because there is a 1 : 1 mole ratio between reactants.
Heat lost by chemicals = heat gained by solution
Heat gain = gCJ18.4
o × 100.0 g × (23.40 22.60)°C = 330 J
Heat loss = 330 J; this is the heat evolved (exothermic reaction) when 5.00 × 103 mol of AgCl is produced. So q = 330 J and ΔH (heat per mol AgCl formed) is negative with a value of:
ΔH = = 66 kJ/mol
Note: Sign errors are common with calorimetry problems. However, the correct sign for ΔH can be determined easily from the ΔT data;; i.e., if ΔT of the solution increases, then the reaction is exothermic because heat was released, and if ΔT of the solution decreases, then the reaction is endothermic because the reaction absorbed heat from the water. For calorimetry problems, keep all quantities positive until the end of the calculation and then decide the sign for ΔH. This will help eliminate sign errors.
J1000kJ1
mol1000.5J3303
194 CHAPTER 6 THERMOCHEMISTRY 62. NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)
We have a stoichiometric mixture. All of the NaOH and HCl will react.
0.10 L × Lmol0.1 = 0.10 mol of HCl is neutralized by 0.10 mol NaOH.
Heat lost by chemicals = heat gained by solution
Volume of solution = 100.0 + 100.0 = 200.0 mL
Heat gain =
mLg0.1mL0.200
gCJ18.4
o × (31.3 – 24.6)C = 5.6 × 103 J = 5.6 kJ
Heat loss = 5.6 kJ; this is the heat released by the neutralization of 0.10 mol HCl. Because the temperature increased, the sign for ΔH must be negative, i.e., the reaction is exothermic. For calorimetry problems, keep all quantities positive until the end of the calculation and then decide the sign for ΔH. This will help eliminate sign errors.
ΔH = mol10.0kJ6.5 = 56 kJ/mol
63. Heat lost by solution = heat gained by KBr; mass of solution = 125 g + 10.5 g = 136 g Note: Sign errors are common with calorimetry problems. However, the correct sign for
ΔH can easily be obtained from the ΔT data. When working calorimetry problems, keep all quantities positive (ignore signs). When finished, deduce the correct sign for ΔH. For this problem, T decreases as KBr dissolves, so ΔH is positive;; the dissolution of KBr is endothermic (absorbs heat).
Heat lost by solution = gCJ18.4
o × 136 g × (24.2°C 21.1°C) = 1800 J = heat gained by KBr
ΔH in units of J/g = KBrg5.10J1800 = 170 J/g
ΔH in units of kJ/mol = J1000
kJ1KBrmol
KBrg0.119KBrg
J170 = 20. kJ/mol
64. NH4NO3(s) NH4
+(aq) + NO3(aq) ΔH = ?;; mass of solution = 75.0 g + 1.60 g = 76.6 g
Heat lost by solution = heat gained as NH4NO3 dissolves. To help eliminate sign errors, we will keep all quantities positive (q and ΔT) and then deduce the correct sign for ΔH at the end of the problem. Here, because temperature decreases as NH4NO3 dissolves, heat is absorbed as NH4NO3 dissolves, so this is an endothermic process (ΔH is positive).
Heat lost by solution =
gCJ18.4
o × 76.6 g × (25.00 23.34)°C = 532 J = heat gained as NH4NO3 dissolves
CHAPTER 6 THERMOCHEMISTRY 195
ΔH = J1000
kJ1NONHmol
NONHg05.80NONHg60.1
J532
34
34
34 = 26.6 kJ/mol NH4NO3 dissolving
65. Because ΔH is exothermic, the temperature of the solution will increase as CaCl2(s)
dissolves. Keeping all quantities positive:
heat loss as CaCl2 dissolves = 11.0 g CaCl2 22
2
CaClmolkJ5.81
CaClg98.110CaClmol1
= 8.08 kJ
heat gained by solution = 8.08 × 103 J = gCJ18.4
o × (125 + 11.0) g × (Tf 25.0°C)
Tf 25.0°C = 13618.41008.8 3
= 14.2°C, Tf = 14.2°C + 25.0°C = 39.2°C
66. 0.1000 L × HClmol2releasedheatkJ118
LHClmol500.0
= 2.95 kJ of heat released if HCl limiting
0.3000 L ×
L)OH(Bamol100.0 2
2)OH(BamolreleasedheatkJ118
= 3.54 kJ heat released if Ba(OH)2 limiting
Because the HCl reagent produces the smaller amount of heat released, HCl is limiting and 2.95 kJ of heat are released by this reaction.
Heat gained by solution = 2.95 × 103 J = gCJ18.4
o × 400.0 g × ΔT
ΔT = 1.76°C = Tf Ti = Tf 25.0°C, Tf = 26.8°C
67. a. Heat gain by calorimeter = heat loss by CH4 = 6.79 g CH4 molkJ802
g04.16CHmol1 4
= 340. kJ
Heat capacity of calorimeter = C8.10
kJ.340o = 31.5 kJ/°C
b. Heat loss by C2H2 = heat gain by calorimeter = 16.9°C × C
kJ5.31o = 532 kJ
A bomb calorimeter is at constant volume, so the heat released/gained = qV = E:
ΔEcomb = 2222 HCmol
g04.26HCg6.12
kJ532
= 1.10 × 103 kJ/mol
68. First, we need to get the heat capacity of the calorimeter from the combustion of benzoic acid. Heat lost by combustion = heat gained by calorimeter.
Heat loss = 0.1584 g × g
kJ42.26 = 4.185 kJ
198 CHAPTER 6 THERMOCHEMISTRY 76. P4O10 → P4 + 5 O2 ΔH = (2967.3 kJ) 10 PCl3 + 5 O2 → 10 Cl3PO ΔH = 10(285.7 kJ) 6 PCl5 → 6 PCl3 + 6 Cl2 ΔH = 6(84.2 kJ) P4 + 6 Cl2 → 4 PCl3 ΔH = 1225.6 P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g) ΔH = 610.1 kJ
Standard Enthalpies of Formation 77. The change in enthalpy that accompanies the formation of 1 mole of a compound from its
elements, with all substances in their standard states, is the standard enthalpy of formation for a compound. The reactions that refer to H o
f are: Na(s) + 1/2 Cl2(g) → NaCl(s);; H2(g) + 1/2 O2(g) → H2O(l) 6 C(graphite, s) + 6 H2(g) + 3 O2(g) → C6H12O6(s)
Pb(s) + S(rhombic, s) + 2 O2(g) → PbSO4(s) 78. a. Aluminum oxide = Al2O3; 2 Al(s) + 3/2 O2(g) → Al2O3(s) b. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) c. NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq) d. 2 C(graphite, s) + 3/2 H2(g) + 1/2 Cl2(g) → C2H3Cl(g) e. C6H6(l) + 15/2 O2(g) → 6 CO2(g) + 3 H2O(l)
Note: ΔHcomb values assume 1 mole of compound combusted. f. NH4Br(s) → NH4
+(aq) + Br(aq) 79. In general, ΔH° = np products,fHΔ nr reactants,fΔH , and all elements in their standard
state have fHΔ = 0 by definition.
a. The balanced equation is 2 NH3(g) + 3 O2(g) + 2 CH4(g) → 2 HCN(g) + 6 H2O(g). ΔH° = (2 mol HCN × HCN,fHΔ + 6 mol H2O(g) × OH,f 2
HΔ )
(2 mol NH3 × 3NH,fHΔ + 2 mol CH4 × )H
4CH,f
ΔH° = [2(135.1) + 6(242)] [2(46) + 2(75)] = 940. kJ
CHAPTER 6 THERMOCHEMISTRY 199 b. Ca3(PO4)2(s) + 3 H2SO4(l) 3 CaSO4(s) + 2 H3PO4(l)
ΔH° =
molkJ1267)l(POHmol2
molkJ1433)s(CaSOmol3 434
molkJ814)l(SOHmol3
molkJ4126)s()PO(Camol1 42243
ΔH° = 6833 kJ (6568 kJ) = 265 kJ c. NH3(g) + HCl(g) → NH4Cl(s) ΔH° = (1 mol NH4Cl × ClNH,f 4
HΔ ) (1 mol NH3 × 3NH,fHΔ + 1 mol HCl × HCl,fHΔ )
ΔH° =
molkJ92mol1
molkJ46mol1
molkJ314mol1
ΔH° = 314 kJ + 138 kJ = 176 kJ 80. a. The balanced equation is C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g).
ΔH° =
molkJ278mol1
molkJ242mol3
molkJ5.393mol2
ΔH° = 1513 kJ (278 kJ) = 1235 kJ b. SiCl4(l) + 2 H2O(l) → SiO2(s) + 4 HCl(aq) Because HCl(aq) is H+(aq) + Cl(aq),
fHΔ = 0 167 = 167 kJ/mol.
ΔH° =
molkJ286mol2
molkJ687mol1
molkJ911mol1
molkJ167mol4
ΔH° = 1579 kJ (1259 kJ) = 320. kJ c. MgO(s) + H2O(l) → Mg(OH)2(s)
ΔH° =
molkJ286mol1
molkJ602mol1
molkJ925mol1
ΔH° = 925 kJ (888 kJ) = 37 kJ 81. a. 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g); ΔH° = np products,fHΔ nr reactants,fΔH
ΔH° =
molkJ46mol4
molkJ242mol6
molkJ.90mol4 = 908 kJ
200 CHAPTER 6 THERMOCHEMISTRY 2 NO(g) + O2(g) 2 NO2(g)
ΔH° =
molkJ.90mol2
molkJ34mol2 = 112 kJ
3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g)
ΔH° =
molkJ286mol1
molkJ34mol3
molkJ.90mol1
molkJ207mol2
140. kJ Note: All
fHΔ values are assumed ±1 kJ.
b. 12 NH3(g) + 15 O2(g) 12 NO(g) + 18 H2O(g) 12 NO(g) + 6 O2(g) 12 NO2(g) 12 NO2(g) + 4 H2O(l) 8 HNO3(aq) + 4 NO(g) 4 H2O(g) 4 H2O(l) 12 NH3(g) + 21 O2(g) 8 HNO3(aq) + 4 NO(g) + 14 H2O(g) The overall reaction is exothermic because each step is exothermic.
82. 4 Na(s) + O2(g) 2 Na2O(s) ΔH° = 2 mol
molkJ416 = 832 kJ
2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g)
ΔH° =
molkJ286mol2
molkJ.470mol2 = 368 kJ
2Na(s) + CO2(g) Na2O(s) + CO(g)
ΔH° =
molkJ5.393mol1
molkJ5.110mol1
molkJ416mol1 = 133 kJ
In Reactions 2 and 3, sodium metal reacts with the "extinguishing agent." Both reactions are exothermic, and each reaction produces a flammable gas, H2 and CO, respectively.
83. 3 Al(s) + 3 NH4ClO4(s) Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g)
ΔH° =
molkJ1676mol1
molkJ704mol1
molkJ.90mol3
molkJ242mol6
molkJ295mol3 = 2677 kJ
364 CHAPTER 10 LIQUIDS AND SOLIDS 95. 96. a. The plateau at the lowest temperature signifies the melting/freezing of the substance. Hence the freezing point is 20C. b. The higher temperature plateau signifies the boiling/condensation of the substance. The
temperature of this plateau is 120C. c. X(s) X(l) H = Hfusion; X(l) X(g) H = Hvaporization The heat of fusion and the heat of vaporization terms refer to enthalpy changes for the
specific phase changes illustrated in the equations above. In a heating curve, energy is applied at a steady rate. So the longer, higher temperature plateau has a larger enthalpy change associated with it as compared to the shorter plateau. The higher temperature plateau occurs when a liquid is converting to a gas, so the heat of vaporization is greater than the heat of fusion. This is always the case because significantly more intermolecular forces are broken when a substance boils than when a substance melts.
97. a. Many more intermolecular forces must be broken to convert a liquid to a gas as compared with converting a solid to a liquid. Because more intermolecular forces must be broken, much more energy is required to vaporize a liquid than is required to melt a solid. Therefore, Hvap is much larger than Hfus.
b. 1.00 g Na × NamolkJ60.2
g99.22Namol1
= 0.113 kJ = 113 J to melt 1.00 g Na
c. 1.00 g Na × NamolkJ0.97
g99.22Namol1
= 4.22 kJ = 4220 J to vaporize 1.00 g Na
d. This is the reverse process of that described in part c, so the energy change is the same quantity but opposite in sign. Therefore, q = 4220 J; i.e., 4220 of heat will be released.
CHAPTER 10 LIQUIDS AND SOLIDS 365
98. Melt: 8.25 g C6H6 × 66
66HCmol
kJ92.9g11.78HCmol1
= 1.05 kJ
Vaporize: 8.25 g C6H6 × 66
66HCmol
kJ7.30g11.78HCmol1
= 3.24 kJ
As is typical, the energy required to vaporize a certain quantity of substance is much larger than the energy required to melt the same quantity of substance. A lot more intermolecular forces must be broken to vaporize a substance as compared to melting a substance.
99. To calculate qtotal, break up the heating process into five steps. H2O(s, 20.°C) H2O(s, 0°C), ΔT = 20.°C; let sice = specific heat capacity of ice:
q1 = sice × m × ΔT = CgJ03.2
o × 5.00 × 102 g × 20.°C = 2.0 × 104 J = 20. kJ
H2O(s, 0°C) H2O(l, 0°C), q2 = 5.00 × 102 g H2O × mol
kJ02.6g02.18
mol1 = 167 kJ
H2O(l, 0°C) H2O(l, 100.°C), q3 = CgJ2.4
o × 5.00 × 102 g × 100.°C = 2.1 × 105J = 210 kJ
H2O(l, 100.°C) H2O(g, 100.°C), q4 = 5.00 × 102 g × mol
kJ7.40g02.18
mol1 = 1130 kJ
H2O(g, 100.°C) H2O(g, 250.°C), q5 = CgJ0.2
o × 5.00 × 102 g × 150.°C = 1.5 × 105 J
= 150 kJ
qtotal = q1 + q2 + q3 + q4 + q5 = 20. + 167 + 210 + 1130 + 150 = 1680 kJ 100. H2O(g, 125°C) H2O(g, 100.°C), q1 = 2.0 J/g•°C × 75.0 g × (−25°C) = 3800 J = −3.8 kJ
H2O(g, 100.°C) H2O(l, 100.°C), q2 = 75.0 g × mol
kJ7.40g02.18
mol1 = 169 kJ
H2O(l, 100.°C) H2O(l, 0°C), q3 = 4.2 J/g•°C × 75.0 g × (100.°C) = 32,000 J = 32 kJ To convert H2O(g) at 125°C to H2O(l) at 0°C requires (−3.8 kJ − 169 kJ − 32 kJ =) −205 kJ of heat removed. To convert from H2O(l) at 0°C to H2O(s) at 0°C requires:
q4 = 75.0 g × mol
kJ02.6g02.18
mol1 = −25.1 kJ
This amount of energy puts us over the −215 kJ limit (−205 kJ − 25.1 kJ = −230. kJ). There-fore, a mixture of H2O(s) and H2O(l) will be present at 0°C when 215 kJ of heat is removed from the gas sample.
CHAPTER 17 SPONTANEITY, ENTROPY, AND FREE ENERGY 663 30. 2 kJ AB B A B A 1 kJ AB B A A B 0 kJ AB A B A B Etotal = 0 kJ 2 kJ 4 kJ 1 kJ 1 kJ 2 kJ 2 kJ 3 kJ 3 kJ The most likely total energy is 2 kJ. 31. a. H2 at 100°C and 0.5 atm; higher temperature and lower pressure means greater volume
and hence larger positional probability. b. N2; N2 at STP has the greater volume because P is smaller and T is larger. c. H2O(l) has a larger positional probability than H2O(s). 32. Of the three phases (solid, liquid, and gas), solids are most ordered (have the smallest
positional probability) and gases are most disordered (have the largest positional probability). Thus a , b, and f (melting, sublimation, and boiling) involve an increase in the entropy of the system since going from a solid to a liquid or from a solid to a gas or from a liquid to a gas increases disorder (increases positional probability). For freezing (process c), a substance goes from the more disordered liquid state to the more ordered solid state; hence, entropy decreases. Process d (mixing) involves an increase in disorder (an increase in positional probability), while separation (phase e) increases order (decreases positional probability). So, of all the processes, a, b, d, and f result in an increase in the entropy of the system.
33. a. Boiling a liquid requires heat. Hence this is an endothermic process. All endothermic
processes decrease the entropy of the surroundings (ΔSsurr is negative). b. This is an exothermic process. Heat is released when gas molecules slow down enough to
form the solid. In exothermic processes, the entropy of the surroundings increases (ΔSsurr is positive).
34. a. ΔSsurr = K298
)kJ2221(T
HΔ = 7.45 kJ/K = 7.45 × 103 J/K
b. ΔSsurr = K298kJ112
THΔ = 0.376 kJ/K = 376 J/K
35. ΔG = ΔH TΔS;; when ΔG is negative, then the process will be spontaneous.
a. ΔG = ΔH TΔS = 25 × 103 J (300. K)(5.0 J/K) = 24,000 J; not spontaneous
b. ΔG = 25,000 J (300. K)(100. J/K) = 5000 J; spontaneous c. Without calculating ΔG, we know this reaction will be spontaneous at all temperatures.
ΔH is negative and ΔS is positive (TΔS < 0). ΔG will always be less than zero with these sign combinations for ΔH and ΔS.
d. ΔG = 1.0 × 104 J (200. K)( 40. J/K) = 2000 J; spontaneous
664 CHAPTER 17 SPONTANEITY, ENTROPY, AND FREE ENERGY 36. ΔG = ΔH TΔS;; a process is spontaneous when ΔG < 0. For the following, assume ΔH and
ΔS are temperature-independent. a. When ΔH and ΔS are both negative, ΔG will be negative below a certain temperature
where the favorable ΔH term dominates. When ΔG = 0, then ΔH = TΔS. Solving for this temperature:
T = K/J.60J000,18
SΔHΔ
= 3.0 × 102 K
At T < 3.0 × 102 K, this process will be spontaneous (ΔG < 0). b. When ΔH and ΔS are both positive, ΔG will be negative above a certain temperature
where the favorable ΔS term dominates.
T = K/J.60J000,18
SΔHΔ = 3.0 × 102 K
At T > 3.0 × 102 K, this process will be spontaneous (ΔG < 0). c. When ΔH is positive and ΔS is negative, this process can never be spontaneous at any
temperature because ΔG can never be negative. d. When ΔH is negative and ΔS is positive, this process is spontaneous at all temperatures
because ΔG will always be negative. 37. At the boiling point, ΔG = 0, so ΔH = TΔS.
ΔS = K)35273(
mol/kJ5.27THΔ
= 8.93 × 210 kJ/K•mol = 89.3 J/K•mol
38. At the boiling point, ΔG = 0, so ΔH = TΔS. T = molJ/K92.92J/mol1058.51
ΔSΔH 3
= 629.7 K
39. a. NH3(s) NH3(l);; ΔG = ΔH TΔS = 5650 J/mol 200. K (28.9 J/K•mol) ΔG = 5650 J/mol 5780 J/mol = 130 J/mol Yes, NH3 will melt because ΔG < 0 at this temperature.
b. At the melting point, ΔG = 0, so T = molJ/K28.9
J/mol5650ΔSΔH
= 196 K.
40. C2H5OH(l) C2H5OH(g); at the boiling point, G = 0 and Suniv = 0. For the vaporization process, S is a positive value, whereas H is a negative value. To calculate Ssys, we will determine Ssurr from H and the temperature; then Ssys = Ssurr for a system at equili-brium.
CHAPTER 17 SPONTANEITY, ENTROPY, AND FREE ENERGY 665
Ssurr = K351
J/mol1038.7T
H 3 = 110. J/K•mol
Ssys = Ssurr = (110.) = 110. J/K•mol Chemical Reactions: Entropy Changes and Free Energy 41. a. Decrease in positional probability;; ΔS° will be negative. There is only one way to
achieve 12 gas molecules all in one bulb, but there are many more ways to achieve the gas molecules equally distributed in each flask.
b. Decrease in positional probability; ΔS° is negative for the liquid to solid phase change. c. Decrease in positional probability; ΔS° is negative because the moles of gas decreased
when going from reactants to products (3 moles 0 moles). Changes in the moles of gas present as reactants are converted to products dictates predicting positional proba-bility. The gaseous phase always has the larger positional probability associated with it.
d. Increase in positional probability;; ΔS° is positive for the liquid to gas phase change. The
gas phase always has the larger positional probability. 42. a. Decrease in positional probability (Δn < 0);; ΔS°() b. Decrease in positional probability (Δn < 0);; ΔS°() c. Increase in positional probability;; ΔS°(+) d. Increase in positional probability;; ΔS°(+) 43. a. Cgraphite(s); diamond has a more ordered structure (has a smaller positional probability)
than graphite. b. C2H5OH(g); the gaseous state is more disordered (has a larger positional probability)
than the liquid state. c. CO2(g); the gaseous state is more disordered (has a larger positional probability) than the
solid state. 44. a. He (10 K); S = 0 at 0 K b. N2O; more complicated molecule, so has the larger positional probability. c. NH3(l); the liquid state is more disordered (has a larger positional probability) than the
solid state. 45. a. 2 H2S(g) + SO2(g) 3 Srhombic(s) + 2 H2O(g); because there are more molecules of
reactant gases than product molecules of gas (Δn = 2 3 < 0), ΔS° will be negative.
670 CHAPTER 17 SPONTANEITY, ENTROPY, AND FREE ENERGY b. Because ΔG° is positive, this reaction is not spontaneous at standard conditions and
298 K.
c. ΔG° = ΔH° TΔS°, ΔS° = K298
kJ53kJ.100T
GH oo = 0.16 kJ/K
We need to solve for the temperature when ΔG° = 0:
ΔG° = 0 = ΔH° TΔS°, ΔH° = TΔS°, T = kJ/K0.16kJ100.
SH
o
o
= 630 K
This reaction will be spontaneous (ΔG < 0) at T > 630 K, where the favorable entropy term will dominate.
62. a. ΔG° = 2(270. kJ) 2(502 kJ) = 464 kJ b. Because ΔG° is positive, this reaction is not spontaneous at standard conditions at 298 K. c. ΔG° = ΔH° TΔS°, ΔH° = ΔG° + TΔS° = 464 kJ + 298 K(0.179 kJ/K) = 517 kJ We need to solve for the temperature when ΔG° = 0:
ΔG° = 0 = ΔH° TΔS°, T = K/kJ179.0
kJ517SΔHΔ
o
o
= 2890 K
This reaction will be spontaneous at standard conditions (ΔG° < 0) when T > 2890 K. Here the favorable entropy term will dominate.
63. CH4(g) + CO2(g) CH3CO2H(l) ΔH° = 484 [75 + (393.5)] = 16 kJ;; ΔS° = 160. (186 + 214) = 240. J/K ΔG° = ΔH° TΔS° = 16 kJ (298 K)(0.240 kJ/K) = 56 kJ
At standard concentrations, this reaction is spontaneous only at temperatures below T = ΔH°/ΔS° = 67 K (where the favorable ΔH° term will dominate, giving a negative ΔG° value). This is not practical. Substances will be in condensed phases and rates will be very slow at this extremely low temperature.
CH3OH(g) + CO(g) CH3CO2H(l) ΔH° = 484 [110.5 + (201)] = 173 kJ;; ΔS° = 160. (198 + 240.) = 278 J/K
ΔG° = 173 kJ (298 K)(0.278 kJ/K) = 90. kJ
This reaction also has a favorable enthalpy and an unfavorable entropy term. But this reaction, at standard concentrations, is spontaneous at temperatures below T = ΔH°/ΔS° = 622 K (a much higher temperature than the first reaction). So the reaction of CH3OH and CO will be preferred at standard concentrations. It is spontaneous at high enough temperatures that the rates of reaction should be reasonable.
CHAPTER 17 SPONTANEITY, ENTROPY, AND FREE ENERGY 671 64. C2H4(g) + H2O(g) CH3CH2OH(l) ΔH° = 278 (52 242) = 88 kJ;; ΔS° = 161 (219 + 189) = 247 J/K
When ΔG° = 0, ΔH° = TΔS°, so T = J/K247
J1088SΔHΔ 3
o
o
= 360 K.
Since the signs of ΔH° and ΔS° are both negative, this reaction at standard concentrations will be spontaneous at temperatures below 360 K (where the favorable ΔH° term will dominate).
C2H6(g) + H2O(g) CH3CH2OH(l) + H2(g) ΔH° = 278 (84.7 242) = 49 kJ;; ΔS° = 131 + 161 (229.5 + 189) = 127 J/K
Both ΔH° and ΔS° have unfavorable signs. ΔG° can never be negative when ΔH° is positive and ΔS° is negative. So this reaction can never be spontaneous at standard conditions.
Thus the reaction C2H4(g) + H2O(g) C2H5OH(l) would be preferred at standard conditions.
Free Energy: Pressure Dependence and Equilibrium
65. ΔG = ΔG° + RT ln Q;; for this reaction: ΔG = ΔG° + RT ln3
22
ONO
ONO
PPPP
ΔG° = 1 mol(52 kJ/mol) + 1 mol(0) [1 mol(87 kJ/mol) + 1 mol(163 kJ/mol)] = 198 kJ
ΔG = 198 kJ + )10(1.00)10(1.00)10(1.00)10(1.00lnK)(298
J/kJ1000molJ/K8.3145
66
37
ΔG = 198 kJ + 9.69 kJ = 188 kJ 66. ΔG° = 3(0) + 2(229) [2(34) + 1(300.)] = 90. kJ
ΔG = ΔG° + RT ln 2S2H
O2H
SO2
2
PP
P
= 90. kJ +
)010.0()100.1()030.0(lnkJ
1000)298()3145.8(
4
2
ΔG = 90. kJ + 39.7 kJ = 50. kJ
67. ΔG = ΔG° + RT ln Q = ΔG° + RT ln 2NO
ON
2
42
PP
ΔG° = 1 mol(98 kJ/mol) 2 mol(52 kJ/mol) = 6 kJ a. These are standard conditions, so ΔG = ΔG° because Q = 1 and ln Q = 0. Because ΔG° is
negative, the forward reaction is spontaneous. The reaction shifts right to reach equili-brium.
672 CHAPTER 17 SPONTANEITY, ENTROPY, AND FREE ENERGY
b. ΔG = 6 × 103 J + 8.3145 J/K•mol (298 K) ln 2)21.0(50.0
ΔG = 6 × 103 J + 6.0 × 103 J = 0 Because ΔG = 0, this reaction is at equilibrium (no shift).
c. ΔG = 6 × 103 J + 8.3145 J/K•mol (298 K) ln 2)29.0(6.1
ΔG = 6 × 103 J + 7.3 × 103 J = 1.3 × 103 J = 1 × 103 J Because ΔG is positive, the reverse reaction is spontaneous, and the reaction shifts to the
left to reach equilibrium.
68. a. ΔG = ΔG° + RT ln 2HN
2NH
22
3
P P
P
; ΔG° = o
NH,f 3GΔ2 = 2(17) = 34 kJ
ΔG = 34 kJ + 3
2
.)200(.)200(.)50(ln
J/kJ1000K)(298mol)J/K(8.3145
ΔG =34 kJ 33 kJ = 67 kJ
b. ΔG = 34 kJ + 3
2
.)600(.)200(.)200(ln
J/kJ1000K)(298mol)J/K(8.3145
ΔG = 34 kJ 34.4 kJ = 68 kJ
69. NO(g) + O3(g) NO2(g) + O2(g); ΔG° = oreactantsf,r
oproductsf,p ΔGΣnΔGΣn
ΔG° = 1 mol(52 kJ/mol) [1 mol(87 kJ/mol) + 1 mol(163 kJ/mol)] = 198 kJ
ΔG° = RT ln K, K
K)mol(298J/K8.3145
J)101.98(expRTΔGexp
5o
e79.912 = 5.07 × 1034
Note: When determining exponents, we will round off after the calculation is complete. This helps eliminate excessive round off error.
70. ΔG° = 2 mol(229 kJ/mol) [2 mol(34 kJ/mol) + 1 mol(300. kJ/mol)] = 90. kJ
K)mol(298J/K8.3145
J)109.0(expRTΔGexpK
4o
e36.32 = 5.9 × 1015
ΔG° = ΔH° − TΔS°;; because there is a decrease in the number of moles of gaseous particles, ΔS° is negative. Because ΔG° is negative, ΔH° must be negative. The reaction will be spontaneous at low temperatures (the favorable ΔH° term dominates at low temperatures).