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).