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CHAPTER 6 
THERMODYNAMICS: THE FIRST LAW 
 
 
 
 
6.2 (a) The internal energy of an open system could be increased by (1) 
 adding matter to the system, (2) doing work on the system, and (3) adding 
 heat to the system. (b) Matter cannot be added to a closed system, so 
 only (2) doing work on the system and (3) adding heat to the system could 
 be used in increase the internal energy. (c) The internal energy of an 
 isolated system cannot be changed. 
 
6.4 (a) Given four cylinders operating at 60 cycles per minute and displacing 
 2.5 L each under a constant pressure of 1400 bar, the work done is given 
 by: 
 ( )
3
5
7
cycles Pa 0.001 m4 cylinders 60 1400 bar 10 2.5 L
min bar L
 = 8.4 10 J
⎛ ⎞⎛ ⎞ ⎛ ⎞⋅ ⋅ − ⋅ ⋅ ⋅⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
− ×
 
 (b) The expanding gas in the cylinder does work on the pistons (and to the 
 load connected to the pistons by the crankshaft of the engine). Therefore, 
 work done by the gas is negative. 
 
6.6 (a) The change in internal energy ∆U is given simply by summing the two 
 energy terms involved in this process. We must be careful, however, that 
 the signs on the energy changes are appropriate. In this case, internal 
 energy will be added to the gas sample by heating, but the expansion will 
 remove internal energy from the sample so it will be a negative number: 
 325 kJ 515 kJ 190 kJ∆ = − = −U
153 
 (b) If the heat added had exactly matched the amount of energy lost due 
 to the work of the gas (i.e., if q had been 235 kJ), then ∆U would have 
 been 0 and the temperature of the gas would not have changed. Because 
 less heat was added, however, and the gas was allowed to expand further, 
 the temperature of the gas would have had to decrease, and consequently 
 the pressure of the gas would be lower at the end. 
 
6.8 To calculate this, we use the relationship ∆U = q + w, which arises from 
 the first law of thermodynamics. Because the system releases heat, q will 
 be a negative number as will ∆U , because it is a decrease in internal 
 energy: 
 
125 kJ 346 kJ
221 kJ
− = − +
= +
w
w
 Because work is positive, the surroundings will do work on the system. 
 
6.10 If the heater operates as rated, then the total amount of heat transferred to 
 the cylinder will be 
 1 1 4(100 J s )(10 min)(60 s min ) 6.0 10 J or 60 kJ− −⋅ ⋅ = ×
 Work will be given by ext= − ∆w P V in this case because it is an expansion 
 against a constant opposing pressure: 
 (0.975 atm)(10.00 L 2.00 L) 7.80 L atm= − − = − ⋅w
 In order to combine the two terms to get the internal energy change, we 
 must first convert the units on the energy terms to the same values. A 
 handy trick to get the conversion factor for L atm⋅ to J or vice versa is to 
 make use of the equivalence of the ideal gas constant R in terms of 
 and J: L atm⋅
 
1 1
1 1
8.314 J K mol7.80 L atm 790 J or 0.790 kJ
0.082 06 L atm K mol
− −
− −
⎛ ⎞⋅ ⋅= − ⋅ = − −⎜ ⎟⋅ ⋅ ⋅⎝ ⎠
w 
 The internal energy change is then the sum of these two numbers: 
 60 kJ ( 0.790 kJ) 59 kJU q w∆ = + = + − ≅
154 
 The energy change due to the work term turns out to be negligible in this 
 problem. 
 
6.12 This is calculated from ∆ = +U q w where we know 892.4 kJ∆ = −U and 
 : 492 kJ= −w
 
892.4 kJ 492 kJ
400 kJ
− = −
= −
q
q
 4.00 × 102 kJ of heat are lost from the system. 
 
6.14 (a) true only if q = 0; (b) always true, volume is fixed and no work can be 
 done; (c) never true; (d) always true; (e) true if q = 0 
 
6.16 (a) The heat change will be made up of two terms: one term to raise the 
 temperature of the stainless steel and the other to raise the temperature of 
 the water: 
 1 1(450.0 g)(4.18 J ( C) g )(100.0 C 25.0 C)− −= ⋅ ° ⋅ ° −q °
° 1 1(500.0 g)(0.51 J ( C) g )(100 C 25 C)− −+ ⋅ ° ⋅ ° −
 2141 kJ 19 kJ 1.60 10 kJq = + = ×
 (b) The percentage of heat attributable to raising the temperature of the 
 water will be 
 141 kJ (100) 88.1%
160 kJ
⎛ ⎞ =⎜ ⎟⎝ ⎠
 
 (c) The use of the copper kettle is more efficient, as a larger percentage of 
 the heat goes into heating the water and not the container holding it. 
 
6.18 heat lost by metal = heat gained by water −
 
s
1 1
1 1
s
(20.0 g)(25.7 C 100.0 C)(C )
 (50.7 g)(4.18 J ( C) g )(25.7 C 22.0 C)
0.53 J ( C) g
− −
− −
° − °
= − ⋅ ° ⋅ ° − °
= ⋅ ° ⋅C
155 
6.20 
1
1
cal
1.236 g benzoic acid ( 3227 kJ mol benzoic acid)
122.12 g mol benzoic acid
2.345 C
13.93 kJ ( C)
−
−
⎛ ⎞− − ⋅⎜ ⎟⋅⎝ ⎠= °
= ⋅ °
C 
 
6.22 (a) Because the process is isothermal, 0∆ =U and .= −q w For a 
 reversible process, 
 2
1
ln= − Vw nRT
V
 
 n is obtained from the ideal gas law: 
 
1 1
1 1
(2.57 atm)(3.42 L) 0.359 mol
(0.082 06 L atm K mol )(298 K)
7.39(0.359 mol)(8.314 J K mol )(298 K) ln 685 J
3.42
685 J
− −
− −
= = =⋅ ⋅ ⋅
= − ⋅ ⋅ = −
= +
PVn
RT
w
q
 
 (b) For step 1, because the volume is constant, 0 and .= ∆ =w U q 
 In step 2, there is an irreversible expansion against a constant opposing 
 pressure, which is calculated from ex= − ∆w P V 
 The constant opposing pressure is given, and ∆V can be obtained from 
 final initial 7.39 L 3.42 L− = −V V
 
(1.19 atm)(7.39 L 3.42 L)
101.325 J4.72 L atm ( 4.72 L atm) 479 J
1 L atm
= − −
⎛ ⎞= − ⋅ = − ⋅ = −⎜ ⎟⋅⎝ ⎠
w
 
 The total work for part B is 0 J ( 479 J) 479 J+ − = − 
 
6.24 Molecules have a higher heat capacity than monatomic gases because they 
 are able to store energy in the form of rotations and vibrations, in addition 
 to simple translational kinetic energy, which is the only energy storage 
 mode available to monatomic gases. The heat capacity of C2H6 is larger 
 than that of CH4 because it is a more complicated molecule that has more 
156 
 possibilities for the molecule to rotate and for more bonds to vibrate than 
 does CH4. 
 
6.26 (a) The molar heat capacity of a monatomic ideal gas at constant pressure 
 is 5, 2C =P m R. The heat released will be given by 
 1 1 21
10.35 g (50 C 25 C)(20.8 J mol ( C) ) 2.7 10 J
20.18 g mol
− −
−
⎛ ⎞= ° − ° ⋅ ⋅ ° =⎜ ⎟⋅⎝ ⎠
q × 
 (b) Similarly, the molar heat capacity of a monatomic ideal gas at 
 constant volume is 3, 2C .=V m The heat released will be given by 
 1 1 21
10.35 g (50 C 25 C)(12.5 J mol ( C) ) 1.6 10 J
20.18 g mol
− −
−
⎛ ⎞= ° − ° ⋅ ⋅ ° =⎜ ⎟⋅⎝ ⎠
q × 
 
6.28 (a) BF3 is a polyatomic, nonlinear molecule. The contribution from 
 molecular motions will be 3R. 
 (b) N2O is a polyatomic, linear molecule. The contribution from 
 molecular motions will be 5/2 R. 
 (c) HCl is a linear molecule. The contribution from molecular motions 
 will be 5/2 R. 
 (d) SO2 is a polyatomic, nonlinear molecule. The contribution from 
 molecular motions will be 3R. 
 
6.30 No, the gasses do not have the same final temperature. The CH4 molecules 
have more internal vibrations in which to store energy than do N2 
molecules. As a result, it will require more energy to increase the 
temperature of CH4(g), and if the same amount of energy is supplied to 
both gasses, the temperature of N2 will be higher than that of CH4. 
 
6.32 Not counting internal vibrations, a nonlinear molecule will have a heat 
 capacity of CV,m = 3R. The temperatures at which the three vibrations of 
 SO2
become available are 1968 K, 1680 K, and 768 K. If a vibration is 
157 
 accessible, it will contribute a factor of R to the heat capacity. Therefore, 
 (a) CV,m = 6R; (b) CV,m = 4R; (c) CV,m = 3R 
 
6.34 (a) freezing fus∆ = −∆H H
 1fus
1
( 4.01 kJ) 5.09kJ mol
25.23 g
32.04 g mol
−
−
−∆ = − = ⋅⎛ ⎞⎜ ⎟⋅⎝ ⎠
H 
 (b) 1vap
1
37.5 kJ 31 kJ mol
95 g
78.11 g mol
−
−
∆ = = ⋅⎛ ⎞⎜ ⎟⋅⎝ ⎠
H 
 
6.36 This process is composed of two steps: raising the temperature of the 
 liquid water from 30°C to 100°C and then converting the liquid water to 
 steam at 100°C. 
 Step 1: 1 1(155 g)(4.18 J ( C) g )(100 C 30 C) 45.4 kJ− −∆ = ⋅ ° ⋅ ° − ° =H
 Step 2: 11
155 g (40.7 kJ mol ) 350 kJ
18.02 g mol
−
−
⎛ ⎞∆ = ⋅ =⎜ ⎟⋅⎝ ⎠
H 
 Total heat required = 45.4 kJ + 350 kJ = 395 kJ 
 
6.38 The heat lost by the metal must equal the heat gained by the water. Also, 
 the final temperature of the metal must be the same as that of the water. 
 We can set up the following relationship and solve for the specific heat 
 capacity of the metal: 
 metal water∆ = −∆H H
 metal(25.0 g)(Specific heat) (29.8 C 90.0 C)° − °
 1 1(50.0 g)(4.184 J ( C) g )(29.8 C 25.0 C)− −= − ⋅ ° ⋅ ° − ° 
 1 1metal(Specific heat) 0.667 J ( C) g
− −= ⋅ ° ⋅ 
 
6.40 (a) heat absorbed = (1.55 mol NO) 180.6 kJ 140 kJ
2 mol NO
⎛ ⎞ =⎜ ⎟⎝ ⎠
 
158 
 (b) 
2 1 1
1.00 atm 5.45 L 0.243 mol
0.082 06 L atm K mol 273 K− −
×= = =⋅ ⋅ ⋅ ×N
PVn
RT
 
 2
2
180.6 kJ(0.243mol N ) 43.9 kJ
mol N
⎛ ⎞ =⎜ ⎟⎝ ⎠
 
 (c) 
2
2
1
1
0.492 kJ 0.002 72 mol
180.6 kJ mol
(0.002 72 mol)(28.02 g mol ) 0.0762 g
−
−
= =⋅
= ⋅
N
N
n
n =
 
 
6.42 (a) 
3
1
1
4
(7.0 cm 6.0 cm 5.0 cm)(1.5 g cm C)heat ( 394 kJ mol C)
12.01 g mol C
1.0 10 kJ
−
−
−
× × ⋅= −⋅
= − ×
⋅ 
 (b) 
7
4
1 1
(1.0 10 J)mass of water 3.2 10 g
(4.18 J ( C) g )(100 C 25 C)− −
×= =⋅ ° ⋅ ° − ° × 
 
6.44 
2SO 1 1
(1.00 atm)(13.4 L) 0.598 mol
(0.082 06 L atm K mol )(273 K)− −
= = =⋅ ⋅ ⋅
PVn
RT
 
 
2
2
O 1
2
15.0 g O
0.469 mol
32.0 g mol O−
= =⋅n 
 The SO2 is the limiting reactant and can react with 0.299 mol of O2: 
 heat evolved 2
2
198 kJ(0.598 mol SO ) 59.2 kJ
2 mol SO
⎛ ⎞−= =⎜ ⎟⎝ ⎠
−
H w
 
 
6.46 65 kJ and 48 kJ. At constant pressure, .∆ = − = − ∆ = ∆ +U w U
 Therefore, 65 kJ ( 48 kJ) 17 kJ∆ = ∆ − = − − − = −H U w
 
6.48 Given the composition of “synthesis gas” it is easiest to first calculate the 
 enthalpy of combustion of 1.000 L of the gas. In 1.000 L of synthesis gas 
 there are 0.40 L of CO. The moles of CO present is given by: 
 
( ) ( )
( ) ( )CO L atmK×mol
1 atm 0.40 L
0.0164 mol of CO
0.0820574 298 K ⋅
⋅⋅= = =⋅ ⋅
P Vn
R T
 
159 
 The enthalpy of combustion of CO is calculated, using enthalpies of 
 formation and the balanced reaction 2CO(g) + O2(g) Æ 2CO2(g), to be 
 −566.0 kJ/mol. Therefore, if 1.000L of synthesis gas were to burn, 
 of energy would be released 
 due to the combustion of CO. The energies released due to the combustion 
 of H
( ) ( )0.0164 mol 566.0 kJ/mol 9.26 kJ⋅ − = −
2(g) and CH4(g) when 1.000 L of synthesis gas burns is calculated in a 
 similar way: 
 
( ) ( )
( ) ( )2H 2L atmK×mol
1 atm 0.25 L
0.0102 mol of H
0.0820574 298 K ⋅
⋅⋅= = =⋅ ⋅
P Vn
R T
 
 ( ) 120.0102 mol of H ( 571.66 kJ mol ) 5.84 kJ−⋅ − ⋅ = −
 
( ) ( )
( ) ( )4CH 4L atmK×mol
1 atm 0.25 L
0.0102 mol of CH
0.0820574 298 K ⋅
⋅⋅= = =⋅ ⋅
P Vn
R T
 
 ( ) 120.0102 mol of H ( 890 kJ mol ) 9.10 kJ−⋅ − ⋅ = −
 All together, when 1 L of synthesis gas burns, −24.2 kJ of energy is 
 released. 
 The amount of energy needed to heat 5.5 L of H2O(l) by 5oC is given by: 
 ( ) ( )1 o 1 o1000 g5.5 L 4.184 J g C 5 C 115 kJ
L
− −⎛ ⎞⋅ ⋅ ⋅ =⎜ ⎟⎝ ⎠ 
 Therefore, the volume of synthesis gas that must be burned to provide 
 this much energy is: 
 1
115 kJ 4.75 L
24.2 kJ L−
=⋅ 
 
6.50 (a) and (b) The balanced chemical equations and enthalpies of combustion 
 for methane, octane, and glucose are: 
 CH4(g) + 2 O2(g) Æ CO2(g) + 2 H2O(l) ∆H = −890 kJ/mol 
 C8H18 + 12 12 O2(g) Æ 8 CO2(g) + 9 H2O(l) ∆H = −5470 kJ/mol 
 C6H12O6(s) + 6 O2(g) Æ 6 CO2(g) + 6 H2O(l) ∆H = −890 kJ/mol 
 Dividing these combustion enthalpies by the moles of CO2(g) produced 
 we find: 
160 
 For combustion of CH4(g) : ∆H = −890 kJ/mol CO2(g) 
 For combustion of C8H18: ∆H = −684 kJ/mol CO2(g) 
 For combustion of C6H12O6(s): ∆H = −468 kJ/mol CO2(g) 
 (c) combustion of glucose produces more CO2 than does generating the 
 same amount of energy using octane. 
 
6.52 The combustion of the monoclinic sulfur is reversed and added to the 
 combustion reaction of rhombohedral sulfur: 
 2 2S(rhombic) O (g) SO (g)+ ⎯⎯→ 296.83 kJ∆ ° = −H 
 2 2 297.16 kJSO (g) S(monoclinic) O (g)⎯⎯→ + ∆ ° = +H 
 S(rhombic) S(monoclinic)⎯⎯→ 0.33 kJ∆ ° = +H 
 
6.54 The first reaction is reversed and added to the second: 
 2 22 NO(g) N (g) O (g)⎯⎯→ + 180.5 kJ∆ ° = −H 
 2 2 2N (g) 2 O (g) 2 NO (g)+ ⎯⎯→ 66.4 kJ∆ ° = +H 
 2 22 NO(g) O (g) 2 NO (g)+ ⎯⎯→
 180.5 kJ 66.4 kJ 114.1 kJ∆ ° = − + = −H
 
6.56 First, write the balanced equations for the known reactions: 
 4 2 2 2CH (g) 2 O (g) CO (g) 2 H O(l)+ ⎯⎯→ + 890 kJ∆ ° = −H 
 1 22CO(g) O (g) CO (g)+ ⎯⎯→ 2 283.0 kJ∆ ° = −H 
 The desired reaction can be obtained by multiplying the first reaction by 2 
 and adding it to the reverse of the second reaction, also multiplied by 2. 
 4 2 2 22 CH (g) 4 O (g) 2 CO (g) 4 H O(l)+ ⎯⎯→ + 1780 kJ∆ ° = −H 
 2 22 CO (g) 2 CO(g) O (g)⎯⎯→ + 566.0 kJ∆ ° = +H 
 4 2 22 CH (g) 3 O (g) 2 CO(g) 4 H O(l)+ ⎯⎯→ +
 1780 kJ 566.0 kJ 1214 kJ∆ ° = − + = −H
161 
6.58 f 2 f 2 f 2
f 3 f
1 1
1 1
1
(CaCl , aq) (H O, l) (CO , g)
[ (CaCO , s) 2( (HCl, aq)]
877.1 kJ mol ( 285.83 kJ mol ) ( 393.51 kJ mol )
[( 1206.9 kJ mol ) 2( 167.16 kJ mol )]
15.2 kJ mol
− −
− −
−
∆ ° = ∆ ° + ∆ ° + ∆ °
− ∆ ° + ∆ °
= − ⋅ + − ⋅ + − ⋅
− − ⋅ + − ⋅
= − ⋅
H H H H
H H
1−
 
6.60 The desired reaction is obtained adding reaction 1 to 6 times reaction 2, 3 
 times reaction 3, and 2 times the reverse of reaction 4: 
 3 22 Al(s) 6 HCl(aq) 2 AlCl (aq) 3 H (g)+ ⎯⎯→ + 1049 kJ∆ ° = −H 
 66[HCl(g) HCl(aq)]⎯⎯→ [ 74.8 kJ]∆ ° = −H 
 3[2 23[H (g) Cl (g) 2 HCl(g)]+ ⎯⎯→ 185 kJ]∆ ° = −H 
 23 32[AlCl (aq) AlCl (s)]⎯⎯→ [ 323 kJ]∆ ° = +H 
 2 3
2 Al(s) 3 Cl (g) 2 AlCl (s)
1049 kJ 6( 74.8 kJ) 3( 185 kJ) 2( 323 kJ) 1407 kJ
+ ⎯⎯→
∆ ° = − + − + − + + = −H
 
6.62 From Appendix 1f 32A, (O ) 142.7 kJ mol
−∆ ° = + ⋅H 
 2 2 3NO (g) O (g) NO(g) O (g)+ ⎯⎯→ + 200 kJ∆ ° = +H 
 33 2O (g) O (g)⎯⎯→ 2 142.7 kJ∆ ° = −H
 1 22 O (g) O(g)⎯⎯→ 249.2 kJ∆ ° = +H
 2NO (g) NO(g) O(g)⎯⎯→ + 306 kJ∆ ° = +H 
 
6.64 (a) 3 N 2 2 3O (g) H O(l) 2 HNO (aq) NO(g)+ ⎯⎯→ +
1−
O (s) 3 CaF (s) 2 BF (g) 3 CaO(s)+ ⎯⎯→ +
 
r f 3
f f 2 f 2
1 1
1
2 (HNO , aq)
 (NO, g [3 (NO , g) (H O, l)]
2( 207.36 kJ mol ) 90.25 kJ mol [3( 33.18 kJ mol )
 ( 285.83 kJ mol )]
138.18 kJ
− −
−
∆ ° = ∆ °
+ ∆ ° − ∆ ° + ∆ °
= − ⋅ + ⋅ − + ⋅
+ − ⋅
= −
H H
H H H
 (b) B 2 3 2 3
162 
 
r f 3 f
f 2 3 f 2
1 1
1 1
2 (BF , g) 3 (CaO, s)
 [ (B O , s) (CaF , s)]
2( 1137.0
kJ mol ) 3( 635.09 kJ mol )
 [ 1272.8 kJ mol 3( 1219.6 J mol )]
752.3 kJ
− −
− −
∆ ° = ∆ ° + ∆ °
− ∆ ° + ∆ °
= − ⋅ + − ⋅
− − ⋅ + − ⋅
= +
H H H
H H
 (c) 2 2H S(aq) 2 KOH(aq) K S(aq) 2 H O(l)+ ⎯⎯→ + 2
 
r f 2 f 2
f 2 f
1 1
1 1
(K S, aq) 2 (H O, l)
 [ (H S, aq) 2 (KOH, aq)]
417.5 kJ mol 2( 285.83 kJ mol )
 [ 39.7 kJ mol 2 ( 482.37 kJ mol )]
38.7 kJ
− −
− −
∆ ° = ∆ ° + ∆ °
− ∆ ° + ∆ °
= − ⋅ + − ⋅
− − ⋅ + − ⋅
= −
H H H
H H
 
6.66 (a) The reaction of interest is: 12 N2(g) + 1 12 H2(g) Æ NH3(l). The reaction 
 enthalpy may be calculated using individual bond enthalpies and the 
 enthalpy of vaporization of NH3. During the reaction, 12 of a mole of N2 
 bonds are broken as are 1 12 moles of H-H bonds while three N-H bonds 
 are formed. In addition, −23.4 kJ will be released due to condensation of 
 the product. Using the suggested data tables, the enthalpy of formation is 
 estimated to be: 
 
( ) ( )
( ) (
1 11 1
2 2
1 1
1
944 kJ mol 1 436 kJ mol
 3 388 kJ mol 23.4 kJ mol
61.4 kJ mol
− −
− −
−
+ ⋅ + ⋅
+ − ⋅ + − ⋅
= − ⋅
) 
 (b) The reaction of interest is 2 C(gr) + 3 H2(g) + 12 O2(g) Æ C2H6O(l). As 
 in part (a) above, the reaction enthalpy may be estimated using the bond 
 enthalpies, the enthalpy of sublimation of C(gr) and the enthalpy of 
 vaporization of the product. In this reaction, 3 moles of H-H bonds and 0.5 
 moles of O=O bonds, are broken while 1 mole of C-C bonds, 5 moles of 
 C-H bonds, 1 mole of O-H bonds, and 1 mole of C-O bonds are formed. 
 Furthermore, 717 kJ 1mol− are needed to vaporize the graphite and −43.5 
163 
 kJ are released when the product condenses into a liquid. Using the 
 suggested data tables, the enthalpy of formation is estimated to be: 
1mol−
 
( ) ( ) ( )
( ) ( ) (
( ) ( )
1 11
2
1 1
1 1
1
3 436 kJ mol 496 kJ mol 2 717 kJ mol
 348 kJ mol 5 412 kJ mol 463 kJ mol
 2 743 kJ mol 43.5 kJ mol
668 kJ mol
− −
− −
− −
−
+ ⋅ + ⋅ + ⋅
+ − ⋅ + − ⋅ + − ⋅
+ − ⋅ + − ⋅
= − ⋅
)
1
1
−
−
 
 (c) The reaction of interest is 3C(gr) + 3H2(g) + 12 ½O2(g) Æ C3H6O(l). As 
 in parts (a) and (b) above, 3 moles of H-H bonds and 0.5 moles of O-O 
 bonds are broken and three moles of C(gr) is vaporized while 2 moles of 
 C-C bonds, 1 mole of C=O bonds and 6 moles of C-H bonds are formed 
 and -21.9 kJ mol-1 are released when one mole of acetone condenses from 
 a gas to a liquid. Using the suggested data tables, the enthalpy of 
 formation is estimated to be: 
 
( ) ( ) ( )
( ) ( ) (
( )
-1 -1 -11
2
-1 -1 -1
-1
-1
3 436 kJ mol 496 kJ mol 3 717 kJ mol
 2 348 kJ mol 6 412 kJ mol 743 kJ mol
 29.1 kJ mol
233 kJ mol
+ ⋅ + ⋅ + ⋅
+ − ⋅ + − ⋅ + − ⋅
+ − ⋅
= − ⋅
) 
 
6.68 For the reaction 33AlCl (s) Al (g) 3 Cl (g)
+ −⎯⎯→ +
 
L f f 1 2 3
ea f 3
1 1 1
L
1 1 1
1
1
L
(Al, g) 3 (Cl, g) (Al) (Al) I (Al)
(Cl) (AlCl , s)
326 kJ mol 3(121 kJ mol ) 577 kJ mol
1817 kJ mol 2744 kJ mol 349 kJ mol
 704.2 kJ mol
6182 kJ mol
− − −
− − −
−
−
∆ = ∆ ° + ∆ ° + + +
− − ∆
∆ = ⋅ + ⋅ + ⋅
+ ⋅ + ⋅ − ⋅
+ ⋅
∆ = ⋅
H H H I I
E H
H
H
 
 
 
 
 
164 
6.70 The process can be broken down into the following steps: 
 1, kJ mol−∆ ° ⋅H 
 2Na (g) 2 Cl (g) NaCl (s)
+ −+ ⎯⎯→ 2524− *
 2(Na(s) Na(g)⎯⎯→ 107.32)+ 
 +494 Na(g) Na (g) e+⎯⎯→ + −
− +4560 2Na (g) Na (g) e+ +⎯⎯→ +
 +242 2Cl (g) 2 Cl(g)⎯⎯→
 2(2 (Cl(g) e Cl , g)− −+ ⎯⎯→ 349)− 
 2 2Na(s) 2 Cl (g) NaCl (s)+ ⎯⎯→ f 2289∆ ° = +H
 For comparison, the enthalpy of formation of 
 1NaCl(s) 411.15 kJ mol−= − ⋅ . Because the enthalpy of formation of 
 is a large positive number, would be considerably unstable. 
 The disproportionation reaction 
2NaCl 2NaCl
 2 22 NaCl (s) 2 NaCl(s) Cl (g)⎯⎯→ +
 would have an energy of 
 2 22(NaCl (s) Na(s) 2 Cl , g)⎯⎯→ + 12( 2289 kJ mol )−− ⋅
 1 222(Na(s) Cl (g) NaCl(s)+ ⎯⎯→ 12( 411.15 kJ mol )−− ⋅ 
 2 22NaCl (s) 2 NaCl(s) Cl (g)⎯⎯→ + 15400 kJ mol−− ⋅ 
 This would be very likely to take place, so NaCl2 would not be isolated. 
 
6.72 (a) break: 1 mol H—Cl bonds 1431 kJ mol−⋅ 
1 mol F—F bonds 1158 kJ mol−⋅ 
 form: 1 mol H—F bonds 1565 kJ mol−− ⋅ 
 1 mol Cl—F bonds 1256 kJ mol−− ⋅
 Total 1232 kJ mol−− ⋅ 
 
 
*From the assumption that is the same as that of MgCl ∆H L 2…
165 
 
 (b) break: 1 mol H—Cl bonds 1431 kJ mol−⋅ 
 1 mol C=C bonds 1612 kJ mol−⋅ 
 form: 1 mol C—C bonds 1348 kJ mol−− ⋅ 
 1 mol C—H bonds 1412 kJ mol−− ⋅ 
 1 mol C—Cl bonds 1338 kJ mol−− ⋅ 
 Total 155 kJ mol−− ⋅ 
 (c) break: 1 mol H—H bonds 1436 kJ mol−⋅ 
 1 mol C=C bonds 1612 kJ mol−⋅ 
 form: 1 mol C—C bonds 1348 kJ mol−− ⋅ 
 2 mol C—H bonds 12( 412) kJ mol−− ⋅ 
 Total 1124 kJ mol−− ⋅ 
 
6.74 (a) The Lewis structures for NO and NO2 show that the bond order in NO 
 is a double bond and that in NO2 is 1.5 on average. 
 
N
OO
N
O O NO NO2 N O
 
 The bond enthalpy of NO at 1632 kJ mol−⋅ is close to the N=O value listed 
 in the table, whereas the N—O bond enthalpy in NO2 of is 
 slightly greater than the average of a N—O single and N=O double bond: 
1469 kJ mol−⋅
 12 (630 kJ 210 kJ) 420 kJ+ = 
 The extra stability is due to resonance stabilization. 
 (b) The bond energies are the same for the two bonds, because the bonds 
 are equivalent due to resonance. 
166 
6.76 
N
N
N
N
N
N
 
 
 
 
 The reaction in question is 6 2N 3 N⎯⎯→ (g)
 For this to occur, we will need to break 3 N—N and 3 NRN bonds and 
 form 3 N—N triple bonds: 
 
1
1
3(163 kJ mol )
3(409 kJ mol )
−
−
⋅
⋅
 13(944 kJ mol )−− ⋅
 11116 kJ mol−− ⋅
 The reaction to form N2(g) is very exothermic and so the reaction is likely 
 to occur. Although N6 might be resonance-stabilized like benzene, it is 
 unlikely that resonance stabilization would overcome the tendency to form 
 the very strong N≡N triple bond; thus we do not expect N6 to be a stable 
 molecule. 
 
6.78 (a) The enthalpy of vaporization is the heat required for the conversion 
 at constant pressure. The value at 298.2 K 
 will be given by 
3 3CH OH(l) CH OH(g)⎯⎯→
 
vap at 298 K f 3 f 3
1 1
1
(CH OH, g) (CH OH, 1)
200.66 kJ mol ( 238.86 kJ mol )
38.20 kJ mol
− −
−
∆ ° = ∆ ° − ∆ °
= − ⋅ − − ⋅
= ⋅
H H H
 (b) In order to take into account the difference in temperature, we need to 
 use the heat capacities of the reactants and products in order to raise the 
 temperature of the system to 64.5°C. We can rewrite the reactions as 
 follows, to emphasize temperature: 
 3 at 298 K 3 at 298 KCH OH(l) CH OH(g)⎯⎯→ 38.20 kJ∆ ° =H 
167 
 3 at 298 K 3 at 337.6 KCH OH(l) CH OH(l)⎯⎯→
1 1
(1 mol)
 (337.8 K 298.2 K)
 (81.6 J mol K )
3.23 kJ
− −
∆ ° =
× −
× ⋅ ⋅
=
H 
 3 at 298 K 3 at 337.6 KCH OH(g) CH OH(g)⎯⎯→
1 1
(1 mol)
 (337.8 K 298.2 K)
 (43.89 J mol K )
1.74 kJ
− −
∆ ° =
× −
× ⋅ ⋅
=
H 
 To add these together to get the overall equation at 337.6 K, we must 
 reverse the second equation: 
 3 at 298 K 3 at 298 KCH OH(l) CH OH(g)⎯⎯→ 38.20 kJ∆ ° =H 
 3 at 337.8 K 3 at 298
KCH OH(l) CH OH(l)⎯⎯→ 3.23 kJ∆ ° = −H 
 3 at 298 K 3 at 337.8 KCH OH(g) CH OH(g)⎯⎯→ 1.74 kJ∆ ° =H 
 3 at 337.8 K 3 at 337.8 KCH OH(1) CH OH(g)⎯⎯→ 34.71 kJ°∆ =H
 (c) The value in the table is 35.3 1kJ mol−⋅ for the enthalpy of 
 vaporization of methanol. The value is close to that calculated as corrected 
 by heat capacities. At least part of the error can be attributed to the fact 
 that heat capacities are not strictly independent of temperature. 
 
6.80 The change in the reaction enthalpy is given by: 
 
2 1 2 1( ) ( ) ( ) where
 (prod.) (prod.)
∆ − ∆ = − ∆
∆ = ⋅ − ⋅∑ ∑
r r p
p p p
H T H T T T C
C n C n C
 given N2 and H2 are linear molecules, their heat capacities are both 
 approximately 72 R while that of NH3 is 4R. Therefore, for this reaction 
 the change in the reaction enthalpy is: 
 
7 7
2 1 2 2
1
( ) ( ) (500 K 300 K) 2(4 ) 3( )
 9.98 kJ mol−
∆ − ∆ = − − −⎡ ⎤⎣ ⎦
= − ⋅
r rH T H T R R R 
 
6.82 (a) The amount of heat lost by the piece of stainless steel must equal the 
 amount of heat absorbed by the water. The specific heat of stainless steel 
168 
 is found in Table 6.1. This problem is complicated because the water may 
 undergo one or more phase changes during this process. In order to answer 
 this question, we will need to analyze each step of the reaction to 
 determine if the steel can transfer enough heat to the water to cause the 
 given change. 
 For stainless steel, the enthalpy change will be given by 
 1 1 1155.7 g 0.51 J ( C) g 79 J ( C)− − −× ⋅ ° ⋅ × ∆ = ⋅ ° × ∆T T 
 For water, there are five separate processes that may be involved: (1) 
 heating solid water to 0°C, (2) converting the solid water to liquid at 
 0°C, (3) raising the temperature of the water from 0°C to 100°C, (4) 
 converting the liquid water to vapor at 100°C, (5) heating the vapor 
 above 100°C. 
 To heat the solid water from 24 C− ° to 0°C: 
 1 125.34 g 2.03 J ( C) g 24 C 1.23 kJ− −× ⋅ ° ⋅ × ° =
 For this process the corresponding decrease in temperature of the steel 
 would be 
 11.23 kJ 0.079 kJ ( C) 16 C−÷ ⋅ ° = °
 Temperature of the steel after heating the solid water to 0°C will be 
 475 C 16 C 459 C.° − ° = °
 Because the temperature of the steel is still above that of the water, there is 
 adequate heat for the next transformation to occur. 
 To convert all the solid water to liquid water at 0°C, we would require 
 121
2
25.34 g H O
(6.01 kJ mol ) 8.46 kJ
18.01 g mol H O
−
−
⎛ ⎞ ⋅ =⎜ ⎟⋅⎝ ⎠
 
 This change would require the temperature of the steel to decrease further 
 by 
 18.46 kJ 0.079 kJ ( C) 107 C−÷ ⋅ ° = ° 
169 
 The temperature of the steel after melting all of the solid water to 0°C will 
 be 459 The temperature of the steel is still above that 
 of the water, so more heat can be transferred to heat the liquid water. 
C 107 C 352 C.° − ° = °
 To heat the liquid water from 0°C to 100°C: 
 1 125.34 g 4.184 J ( C) g 100 C 10.6 kJ− −× ⋅ ° ⋅ × ° =
 This change would require the temperature of the steel to decrease further 
 by 
 110.6 kJ 0.079 kJ ( C) 134 C−÷ ⋅ ° = °
 The steel is still sufficiently hot enough to cause this transformation, so all 
 of the water will be heated to 100°C. The temperature of the steel after 
 melting all of the solid water to 0°C will be 352 . The 
 steel still has enough heat to convert at least some of the water from liquid 
 to vapor. 
C 134 C 218 C° − ° = °
 121
2
25.34 g H O
(40.7 kJ mol ) 57.3 kJ
18.01 g mol H O
−
−
⎛ ⎞ ⋅ =⎜ ⎟⋅⎝ ⎠
 
 157.3 kJ 0.079 kJ ( C) 725 C−÷ ⋅ ° = ° 
 The temperature of the steel is not sufficient to convert all the water to 
 vapor. The final temperature will then be 100°C, with some of the water 
 converted to vapor and some remaining in the liquid phase. 
 (b) To determine the amount of liquid and gaseous water present, we must 
 determine the amount of heat transferred to the water from the steel at 
 100°C. The steel should still be at 218°C when all the liquid water has 
 reached 100°C. The final temperature of the steel must also be 100°C, so 
 its change in temperature will be 118 C.− ° To go from 218°C to 100°C, 
 the steel will lose 
 10.079 kJ ( C) 118 C 9.3 kJ−⋅ ° × ° =
 We can then calculate the amount of water that can be converted to vapor 
 at 100°C by 9.3 kJ. 
170 
 11
9.3 kJ (18.02 g mol ) 4.1 g
40.7 kJ mol
−
−
⎛ ⎞ ⋅ =⎜ ⎟⋅⎝ ⎠
 
 4.1 g of water will be converted to vapor, leaving 
 of water in the liquid phase. 25.34 g 4.1 g 21.2 g− =
 
6.84 (a) First, we calculate the amount of heat needed to raise the temperature 
 of the water. Converting the temperatures from °F to °C: 
 59C ( F 32° = ° − °),
3−
 so 65°F corresponds to 18°C and 108°F corresponds to 
 42°C. 
 
1 3 1
1 1
(100 gal)(3.785 L gal )(1000 cm L )(1.00 g cm )
 (4.18 J ( C) g )(42 C 18 C)
= 38 MJ
− −
− −
∆ = ⋅ ⋅ ⋅
× ⋅ ° ⋅ ° − °
H
 The enthalpy of combustion of methane is 890 kJ mol− ⋅ (from Table 6.4). 
 The mass of methane required will be calculated as follows: 
 
3
1 2
4 41
38 10 kJ (16.04 g mol CH ) 6.8 10 g CH
890 kJ mol
−
−
⎛ ⎞× ⋅ = ×⎜ ⎟⋅⎝ ⎠
 
 (b) This quantity can be obtained from the ideal gas law. of 
 CH
26.8 10 g×
4 is 42 mol of CH4: 
 
1 1(42 mol)(0.082 06 L atm K mol )(298 K) 1030 L
1.00 atm
− −⋅ ⋅ ⋅= = =nRTV
P
 
 
6.86 (a) 1 1 15.50 L(1.00 mol)(8.314 J K mol )(298.2 K)ln
7.00 L
1.97 kJ
− −= − ⋅ ⋅
= −
w 
 (b) ext 1
743 Torr 101.325 J(15.5 L 7.00 L)
1 L atm760 Torr atm
0.842 kJ
−
⎛ ⎞ ⎛= − ∆ = − −⎜ ⎟ ⎜ ⋅⋅⎝ ⎠ ⎝
= −
w P V ⎞⎟⎠ 
 (c) The fact that the expansion occurs adiabatically—which means that 
 the system is isolated from its surroundings so that no heat is transferred—
 is important. It tells us that q = 0, and therefore 0.842 kJ.∆ = = −U w But 
171 
 will also be equal to ∆U = ∆nCv T (see equation 19) because U is a state 
 function. The heat capacity of an ideal gas is 12.5 1J K mol 1− −⋅ ⋅ (we will 
 assume it is constant over this temperature range). Therefore, J = 
 (1.00 mol) . The final temperature 
 will be 298.2 . 
0.842−
1 1(12.5 J K mol )( ). 67.4 K− −⋅ ⋅ ∆ ∆ = −T T
K 67.4K 230.6K− =
 
6.88 In order to solve this problem, we must be able to add reactions (a), (b), 
 and (c) to give the reaction desired. The most logical starting point for 
 solving this problem is to find the reactant or product that appears only 
 once in (a), (b), or (c), as this will determine how that equation must be 
 used. Because O2 only appears in (a), equation (a) must be multipled by 34 
 and left in its original direction in order to obtain the correct amount of 
 O2(g). 
 3 3 3 34 2 2 24 2 4 2CH (g) O (g) CO (g) H O(g)+ ⎯⎯→ + 34 ( 802 kJ)− 
 We now consider how to combine this equation with (b) and (c) to give 
 the desired overall reaction. Because H2O(g) appears in (c) and not in (b), 
 we must add equation (c) to the transformed (a) so as to obtain a total of 
 two H2O(g). To do this, we must reverse (c) and multiply it by 12 : 
 31 12 42 2 2 2CO(g) H (g) CH (g) H O(g)+ ⎯⎯→ + 1 2 12 ( 247 kJ)− 
 Equation (b) must be multiplied by 34 and left in its original direction in 
 order that the CO2(g) will be cancelled from the overall equation: 
 3 3 3 34 24 4 2 2CH (g) CO (g) CO(g) H (g)+ ⎯⎯→ + 2 34 ( 206 kJ)+ 
 The net result is 
 3 3 3 34 2 2 24 2 4 2CH (g) O (g) CO (g) H O(g)+ ⎯⎯→ + 34 ( 802 kJ)− 
 31 1 1 22 42 2 2 2CO(g) H (g) CH (g) H O(g)+ ⎯⎯→ + 12
( 247 kJ)− 
 3 3 3 34 2 24 4 2 2CH (g) CO (g) CO(g) H (g)+ ⎯⎯→ + 34 ( 206 kJ)+ 
 34 2 22CH (g) O (g) CO(g) 2 H O(g)+ ⎯⎯→ + 570 kJ∆ ° = −H 
 
172 
6.90 The formation reaction is 
 71 2 2 2 4 32 2N (g) H (g) 2 C(s) O (g) NH CH CO (s)+ + + ⎯⎯→ 2 
 The combustion reaction that we want is: 
 7 14 3 2 2 2 2 22 2NH CH CO (s) O (g) 2 CO (g) H O(l) N (g)+ ⎯⎯→ + + 
 Using Hess’ Law 
 
7
combustion f 2 f 22
f 4 3 2
1 17
2
1
1
2 (CO , g) (H O, l)
 (NH CH CO , s)
2 ( 393.15 kJ mol ) ( 285.83 kJ mol )
 ( 616.14 kJ mol )
117.28 kJ mol
− −
−
−
∆ ° = ∆ ° + ∆ °
− ∆ °
= − ⋅ + − ⋅
− − ⋅
= − ⋅
H H H
H
 
 
6.92 (a) The heat output of each sample is found using the heat capacity of the 
 calorimeter and the observed change in temperature of the calorimeter. For 
 Brand X, ∆T was 8.8oC and the energy content of 1.00 g is: 
 . For the ABC product, ∆T was 8.5( ) ( )o 18.8 C 600 J K 5300 J g−⋅ ⋅ = ⋅ 1− oC 
 and the energy content of 1.00 g is: ( ) ( )o 18.5 C 600 J K 5100 J g 1− −⋅ ⋅ = ⋅ . 
 (b) 30 g of Brand X cereal contains 
 ( )( )1 1 Calorie30 g 5.3 kJ g 38 Calories
4.184 kJ
− ⎛ ⎞⋅ =⎜ ⎟⎝ ⎠ while 30 g of ABC product 
 contains ( )( )1 1 Calorie30 g 5.1 kJ g 36 Calories
4.184 kJ
− ⎛ ⎞⋅ =⎜ ⎟⎝ ⎠ 
 
6.94 (a) and (b) 
 M c∆H density
 compound (g (k 1mol )−⋅ J mol )−⋅ 1 1kJ g−⋅ 3(g cm )−⋅ 1L mol−⋅ MJ 1L−⋅
 benzene 78.11 3268− 41.84− 0.8786 0.088 90 − 36.76
 ethanol 46.07 1368− 29.69− 0.7893 0.058 37 23.44−
 hexane 86.18 4163− 48.30− 0.6603 0.1305 31.89−
 octane 114.23 − 5471 47.89− 0.7025 0.1626 33.64−
173 
 (c) On a per-liter basis, the chemicals increase in cost in the following 
 order: ethanol < benzene ~ hexane < octane; (d) The choice of fuel is 
 not a simple matter. Although it would appear that benzene is the best as 
 far as heat production per liter is concerned, the cost of ethanol is roughly 
 two-thirds that of the cost of benzene, making the heat produced per dollar 
 about the same. Other issues such as cleanliness of burning, ecological 
 impact of the production of the fuel, capacity of fuel tanks (and how far a 
 car can drive on a tank of gas) then become important issues. 
 
6.96 In step 1, and w = 0. In step 2, 50 J∆ = =VU q 5 J= −q and 
 ; therefore, 50 J∆ = −U 50 J 5 J ; 45 J− = − + = −w w . Because 
 and P = 1.00 atm 
= − ∆w P V
 
1 1
1 1
45 J 0.082 06 L atm K mol 0.44 L
1.00 atm 8.314 J K mol
− −
− −
⎛ ⎞⎛ ⎞+ ⋅ ⋅ ⋅∆ = = +⎜ ⎟⎜ ⎟ ⋅ ⋅⎝ ⎠ ⎝ ⎠
V 
 The gas constants are used to convert the units of J to L atm⋅ . Because the 
 value ∆ is positive, this change will represent an expansion. V
 
6.98 The heat given off by the reaction that was absorbed by the calorimeter is 
 given by 
 1(488.1 J ( C) )(21.34 C 20.00 C) 654 J−∆ = − ⋅ ° ° − ° = −H
 This, however, is not all the heat produced, as the 50.0 mL of solution 
 resulting from mixing also absorbed some heat. If we assume that the 
 volume of NaOH and HCl are negligible compared to the volume of water 
 present and that the density of the solution is 1.00 1g mL−⋅ , then the 
 change in enthalpy of the solution is given by 
 1 1(21.34 C 20.00 C)(4.18 J ( C) g )(50.0 g) 280 J− −∆ = − ° − ° ⋅ ° ⋅ = −H
 The total heat given off will be 654 J ( 280 J) 934 J− + − = − 
 This heat is for the reaction of (0.700 M)(0.0250 L) = 0.0175 mol HCl, so 
 the amount of heat produced per mole of HCl will be given by 
174 
 1934 J 53.4 kJ mol HCl
0.0175 mol
−− = − ⋅ 
 
6.100 From Appendix 2A we can find the enthalpies of combustion of the gases 
 involved: 
 Compound Enthalpy of combustion 1(kJ mol )−⋅
 CH4(g), methane 890− 
 C2H6(g), ethane 1560− 
 C3H8(g), propane 2220− 
 C4H10(g), butane 2878− 
 (a) To calculate the mass of CO2 produced, we simply need to realize that 
 there will be one mole of CO2 produced per mole of carbon present in the 
 starting compound. So there will be one mole of CO2 produced per mole 
 of CH4, two per mole of C2H6, three per mole of C3H8, and four per mole 
 of C4H10. 
 Total moles of CO2 produced per minute 
 
2
4
4
2
2 6
2 6
2
3 8
3 8
1 mol CO 9.3 mol CH
1 mol CH
2 mol CO 3.1 mol C H
1 mol C H
3 mol CO 0.40 mol C H
1 mol C H
= ×
+ ×
+ ×
 
 
2
4 10
4 10
1
4 mol CO 0.20 mol C H
1 mol C H
17.5 mol min−
+ ×
= ⋅
 
 The mass of CO2 produced per minute will be 
 1 1 217.5 mol min 44.01 g mol 7.70 10 g min− −⋅ × ⋅ = × ⋅ 1−
 (b) The heat released per minute will be given by the enthalpies of 
 combustion multiplied by the number of moles of each gas combusted in 
 that time period: 
 heat released 14 4(9.3 mol CH )( 890 kJ mol CH )
−= − ⋅
175 
 
1
2 6 2 6
1
3 8
1
4 10 4 10
4 1
(3.1 mol C H )( 1560 kJ mol C H )
(0.40 mol C H )( 2220 kJ mol )
(0.20 mol C H )( 2878 kJ mol C H )
1.5 10 kJ min
−
−
−
−
+ − ⋅
+ − ⋅
+ − ⋅
= − × ⋅
 
6.102 The total amount of heat released by the reaction will be given by 
 15.270 kJ ( C) 1.140 C 6.008 kJ−− ⋅ ° × ° = −
 In this case, the heats of reaction for the two different outcomes will be 
 given simply by the values for and 
 . The number of moles of sulfur is 
 . We do not know the 
 relative amounts of SO
f∆ °H 12SO ( 296.83 kJ mol )−− ⋅
1
3SO ( 395.72 kJ mol )
−− ⋅
10.6192 g S(s) 32.06 kJ mol 0.01931 mol S(s)−÷ ⋅ =
2 versus SO3 formed, but these can be determined 
 using the following two relationships, which are required by the 
 stoichiometries of the reactions. 
 Let x = number of moles of SO2 formed 
 Let y = number of moles of SO3 formed 
 Then x + y = 0.01931 mol 
 1 1( 296.83 kJ mol ) ( 395.72 kJ mol ) 6.008 kJ− −− ⋅ + − ⋅ = −x y
 x = 0.01652 mol 
 y = 0.00279 mol 
 The ratio of SO2 to SO3 will be 0.01652 mol ÷ 0.00279 mol = 5.92 : 1. 
 
6.104 In order of increasing enthalpy of vaporization: N2, CH4, C6H6, H2O, 
 NaCl. A large enthalpy of vaporization indicates strong intermolecular 
 interactions in the liquid phase. The stronger these interactions the larger 
 the amount of energy needed to free a molecule into the gas phase. 
 Nitrogen, being a homonuclear diatomic molecule interacts through other 
 N2 molecules in the condensed phase through very weak London forces. 
 CH4, being a nonpolar molecule also interacts with other CH4 molecules 
 through weak London forces. However, the electron distribution along the 
176 
 C-H bonds is not uniform leading to stronger interactions than in N2. 
 C6H6 has no net dipole moment but, due to p bonding, above and below 
 the ring of carbons tends to be negative compared to the hydrogen atoms 
 along the edge of the ring giving rise to stronger bonding interactions than 
 in CH4 and N2. H2O is a highly polar molecule that can bind to other 
 water molecules through relatively strong hydrogen bonding interactions, 
 making its enthalpy of vaporization higher than that of the previous 
 molecules. Finally, NaCl forms an ionic solid with strong ionic bonds 
 extending throughout a three-dimensional crystalline lattice, requiring a 
 large amount of energy to free a NaCl molecule from the condensed 
 phase. 
 
6.106 The energy released by a falling body is given by: E = mgh where m is the 
 mass of the object, g is the acceleration due to free fall, and h is the height 
 though which the object falls. The energy
deposited into the water was 
 therefore, ( )( )( )215 kg 20 m 9.8067 m s 2940 J−= ⋅E = . If all of this 
 energy is deposited into the water upon impact, the change in the 
 temperature of the water would be: 
 (2500 g)(4.184 J⋅g−1⋅°C−1)∆T = 2940 J 
 ∆T = 0.28 °C 
 If the falling body does 900.0 J of work before impacting the water, only 
 2040 J will be deposited into the water, resulting in a temperature change 
 of: 
(2500 g)(4.184 J⋅g−1⋅°C−1)∆T = 2040 J 
 ∆T = 0.20 °C 
 
6.108 The amount of calcium chloride in the pipes is 
 or 1.25 × 10( )( )145.6 kJ pipe 60 pipes 2736 kg−⋅ = 4 moles of 
 CaCl2⋅6H2O. Given the enthalpy of fusion of CaCl2⋅6H2O, the amount of 
 energy released as the Ca recrystalizes is: 
177 
 . The mass of water that 
 may be heated by 10
( )( )1 427 kJ mol 1.25 10 mol 3.37 10 kJ−⋅ × = × 5
oC (from 15 oC to 25 oC) is: 
 ( )( )1 o 1 o 54.184 J g C 10 C 3.37 10 kJ
 8100 g or 8.1 kg
− −⋅ ⋅ = ×
=
m
m
 
178 
ch06_odd.pdf
CHAPTER 6 
THERMODYNAMICS: THE FIRST LAW 
 
 
 
 
6.1 (a) isolated; (b) closed; (c) isolated; (d) open; (e) closed; 
(f) open 
 
6.3 (a) Work is given by extw P V= − ∆ , The applied external pressure is 
known, but we must calculate the change in volume given the physical 
dimensions of the pump and the distance, d, the piston in the pump 
moves: 
 2 2
3
1 L(1.5 cm) (20 cm) 0.141 L
1000 cm
 is negative because the air in the pump is compressed to a smaller volume
 
 
 work is then:
101.325 J(2 atm)(-0.141 L) 28.7 J
L atm
V r d
V
w
π π ⎛ ⎞∆ = − = = −⎜ ⎟⎝ ⎠
= − =
∆
⋅
(b) Work on the air is positive by convention as work is done on the air, it 
is compressed. 
 
6.5 The change in internal energy U∆ is given simply by summing the two 
energy terms involved in this process. We must be careful, however, that 
the signs on the energy changes are appropriate. In this case, internal 
energy will be added to the gas sample by heating but the gas loses energy 
through expansion. Therefore the change in internal energy is: 
524 kJ 340 kJ 184 kJ∆ = − = +U 
 
SM-146 
 
6.7 (a) The internal energy increased by more than the amount of heat added. 
Therefore, the extra energy must have come from work done on the 
system. 
(b) 2982 J 492 J 4.90 10 J.w U q= ∆ − = − = + ×
 
6.9 To get the entire internal energy change, we must sum the changes due to 
heat and work. In this problem, 5500 kJ.q = + Work will be given by 
 because it is an expansion against a constant opposing 
pressure: 
extw P V= − ∆
1 1
750 Torr 1846 mL 345 mL 1.48 L atm
760 Torr atm 1000 mL L
w − −
⎛ ⎞ ⎛ ⎞−= − = − ⋅⎜ ⎟ ⎜ ⎟⋅ ⋅⎝ ⎠ ⎝ ⎠
 
To convert to J we use the equivalency of the ideal gas constants: 
1 1
2
1 1
8.314 J K mol(1.48L atm) 1.50 10 J
0.08206 L atm K mol
5500 kJ 0.150 kJ=5500 kJ
w
U q w
− −
− −
⎛ ⎞⋅ ⋅= − ⋅ = − ×⎜ ⎟⋅ ⋅ ⋅⎝ ⎠
∆ = + = −
 
The energy change due to the work term turns out to be negligible in this 
problem. 
 
6.11 Using where 2573 kJ andU q w U∆ = + ∆ = −
947 kJ 2573 kJ 947 kJ
1626 kJ
q w
w
= − − = − +
= −
 
1626 kJ of work can be done by the system on its surroundings. 
 
6.13 (a) true if no work is done; (b) always true; (c) always false; (d) true 
only if w = 0 (in which case ∆U = q = 0; (e) always true 
 
6.15 (a) The heat change will be made up of two terms: one term to raise the 
temperature of the copper and the other to raise the temperature of the 
water: 
SM-147 
 
1 1
1 1
(750.0 g) (4.18 J ( C) g ) (100.0 C 23.0 C)
(500.0 g) (0.38 J ( C) g ) (100 C 23 C)
242 kJ 14 kJ 256 kJ
q − −
− −
= ⋅ ° ⋅ ° −
+ ⋅ ° ⋅ ° −
= + =
°
° 
(b) The percentage of heat attributable to raising the temperature of water 
will be 
241 kJ (100) 94.3%
256 kJ
⎛ ⎞ =⎜ ⎟⎝ ⎠
 
 
6.17 heat lost by metal = 2 heat gained by water 
1 1
final
1 1
final
1 1
final final
final final
final final
(20.0 g) ( 100.0 C) (0.38 J ( C) g )
(50.7 g) (4.18 J ( C) g ) ( 22.0 C)
( 100.0 C) (7.6 J ( C) ) (212 J ( C) ) ( 22.0 C)
100.0 C 28( 22.0 C)
28 100.0 C 6
T
T
T T
T T
T T
− −
− −
− −
− ° ⋅ ° ⋅
= − ⋅ ° ⋅ − °
− ° ⋅ ° = − ⋅ ° − °
− ° = − − °
+ = ° +
final
final
16 C
29 716 C
25 C
T
T
°
= °
= °
 
 
6.19 1cal
22.5 kJ 14.8 kJ ( C)
23.97 C 22.45 C
C −= =° − ° ⋅ ° 
 
6.21 (a) The irreversible work of expansion against a constant opposing 
pressure is given by 
ex
1 1
(1.00 atm)(6.52 L 4.29 L)
2.23 L atm
2.23 L atm 101.325 J L atm 226 J
w P V
w
− −
= − ∆
= − −
= − ⋅
= − ⋅ × ⋅ ⋅ = −
 
(b) An isothermal expansion will be given by 
2
1
Vw nRT
V
= − 
n is calculated from the ideal gas law: 
SM-148 
 
1 1
1 1
(1.79 atm) (4.29 L) 0.307 mol
(0.082 06 L atm K mol ) (305 K)
6.52(0.307 mol) (8.314 J K mol ) (305 K) ln
4.29
326 J
PVn
RT
w
− −
− −
= = =⋅ ⋅ ⋅
= − ⋅ ⋅
= −
 
Note that the work done is greater when the process is carried out 
reversibly. 
 
6.23 The heat capacity increases with molecular complexity—as more 
atoms are present in the molecule, there are more possible bond vibrations 
that can absorb added energy. 
2NO .
 
6.25 (a) The molar heat capacity of a monatomic ideal gas at constant pressure 
is 5,m 2CP .R= The heat released will be given by 
1 1
1
5.025 g (25.0 C 97.6 C)(20.8 J mol ( C) ) 90.6 J
83.80 g mol
q − −−
⎛ ⎞= ° − ° ⋅ ⋅ °⎜ ⎟⋅⎝ ⎠
= − 
(b) Similarly, the molar heat capacity of a monatomic ideal gas at 
constant volume is 3,m 2CV .R= The heat released will be given by 
1 1
1
5.025 g (25.0 C 97.6 C)(12.5 J mol ( C) ) 54.4 J
83.80 g mol
q − −−
⎛ ⎞= ° − ° ⋅ ⋅ °⎜ ⎟⋅⎝ ⎠
= − 
 
6.27 (a) HCN is a linear molecule. The contribution from molecular motions 
will be 5/2 .R 
(b) is a polyatomic, nonlinear molecule. The contribution from 
molecular motions will be 3R. 
2 6C H
(c) Ar is a monoatomic ideal gas. The contribution from molecular 
motions to the heat capacity will be 3/2 R. 
(d) is a diatomic, linear molecule. The contribution from molecular 
motions will be 5/2 R. 
HBr
 
SM-149 
 
6.29 The strategy here is to determine the amount of energy per photon and the 
amount of energy needed to heat the water. Dividing the latter by the 
former will give the number of photons needed. Energy per photon is 
given by: 
34 8 -1
34 -1
3
6.626 10 J s (2.9979 10 m s ) 4.41 10 J photon
4.50 10 m
hcE λ
−
−
−
× ⋅ × ⋅= = = × ⋅× 
The energy needed to heat the water is: 
-1 o -1 o o 5350 g (4.184 J g C ) (100.0 C - 25.0 C) 1.10 10 J⋅ ⋅ = × 
The number of photons needed is therefore: 
5
27
23 -1
1.10 10 J 2.49 10 photons
4.41 10 J photon−
× = ×× ⋅ 
 
6.31 (a) Using the estimation that 3R = C: 
-1 -1
-1
-1 -1
3 (0.392 J K g )( )
3 63.6 g mol
0.392 J K g
This molar mass indicates that the atomic solid is Cu(s)
R C M
RM
= = ⋅ ⋅
= =⋅ ⋅ 
(b) From Example 5.3 we find that the density of a substance which form 
a face-centered cubic unit cell is given by: 3 / 2 3
4
8 A
Md
N r
= . Therefore, we 
expect the density of copper to be: 
-1
-3
3 / 2 23 -1 8 3
4 (63.55 g mol ) 8.90 g cm
8
(6.022 10 mol )(1.28 10 cm)
d −
⋅= =× × ⋅ 
 
6.33 (a) 1vap
1.93 kJ 8.21 kJ mol
0.235 mol
H −∆ = = ⋅ 
(b) 1vap
1
21.2 kJ 43.5 kJ mol
22.45 g
46.07 g mol
H −
−
∆ = = ⋅⎛ ⎞⎜ ⎟⋅⎝ ⎠
 
 
SM-150 
 
6.35 This process is composed of two steps: melting the ice at 0 and then 
raising the temperature of the liquid water from 0 C 
C°
to 25 C :° °
Step 1: 11
80.0 g (6.01 kJ mol ) 26.7 kJ
18.02 g mol
H −−
⎛ ⎞∆ = ⋅ =⎜ ⎟⋅⎝ ⎠
 
Step 2: 1 1(80.0 g) (4.18 J ( C) g ) (20.0 C 0.0 C) 6.69 kJH − −∆ = ⋅ ° ⋅ ° − ° =
Total heat required 26.7 kJ 6.69 kJ 33.4 kJ= + =
 
6.37 The heat gained by the water in the ice cube will be equal to the heat lost 
by the initial sample of hot water. The enthalpy change for the water in the 
ice cube will be composed of two terms: the heat to melt the ice at 0 
and the heat required to raise the ice from to the final temperature. 
C°
0 C°
heat (ice cube) 3 11
1 1
f
4 1
f
50.0 g (6.01 10 J mol )
18.02 g mol
(50.0 g) (4.184 J ( C) g ) ( 0 )
1.67 10 J (209 J ( C) ) ( 0 )
T
T
−
−
− −
−
⎛ ⎞= ×⎜ ⎟⋅⎝ ⎠
+ ⋅ ° ⋅
= × + ⋅ ° − °
⋅
− °
− °
 
heat (water) 1 1 f
3 1
f
(400 g) (4.184 J ( C) g ) ( 45 )
(1.67 10 J ( C) ) ( 45 )
T
T
− −
−
= ⋅ ° ⋅
= × ⋅ ° − °
Setting these equal: 
3 1 4 4 1
f f(1.67 10 J ( C) ) 7.5 10 J 1.67 10 J (209 J ( C) )T T
− −− × ⋅ ° + × = × + ⋅ ° 
Solving for f :T
4
f 3 1
5.8 10 J 31 C
1.88 10 J ( C)
T −
×= =× ⋅ ° ° 
 
6.39 (a) 1(1.25 mol) ( 358.8 kJ mol ) 448 kJH −∆ = + ⋅ =
(b) 21
197 g C 358.8 kJ 1.47 10 kJ
4 mol C12.01 g mol C
H −
⎛ ⎞ ⎛ ⎞∆ = = ×⎜ ⎟ ⎜ ⎟⋅ ⎝ ⎠⎝ ⎠
 
(c) 
2
1
CS
2
358.8 kJ mol )415 kJ ( )
4 mol CS
H n
−⎛ ⎞⋅∆ = = ⎜ ⎟⎝ ⎠
 
2CS 2
4.63 mol CSn = or 1 2(4.63 mol) (76.13 g mol ) 352 g CS−⋅ =
SM-151 
 
 
6.41 (a) 
3
7 330.48 cm(12 ft 12 ft 8 ft) 3.26 10 cm
1 ft
⎛ ⎞× × = ×⎜ ⎟⎝ ⎠
 
The heat capacity of air is 11.01 J ( C) mol 1− −⋅ ° ⋅ and the average molar mass 
of air is 128.97 g mol−⋅ (see Table 4.1). The density of air can be 
calculated from the ideal gas law: 
1 1 1
3
1.00 atm
(28.97 g mol ) (0.082 06 L atm K mol ) (277.6 K)
0.001 52 g cm
Pd
MRT
d
− − −
−
= = ⋅ ⋅ ⋅ ⋅
= ⋅
 
40 F 4.4 C, 78 F 26 C
26 C 4.4 C 22T
° = ° ° = °
∆ = ° − ° = ° 
The heat required is 
 (3.26 7 3 3 1 1
3
10 cm )(0.001 52 g cm ) (1.01 J ( C) mol ) (22 C)
1.1 10 kJ
− − −× ⋅ ⋅ ° ⋅ °
 = ×
The mass of octane required to produce this much heat will be given by 
3
1
1
1.1 10 kJ (114.22 g mol ) 23.0 g
5471 kJ mol
−
−
⎛ ⎞− × ⋅ =⎜ ⎟− ⋅⎝ ⎠
 
 3 1 1
1
5
(1.0 gal) (3.785 10 mL gal ) (0.70 g mL )
114.22 g mol
10 942 kJ 
2 mol octane
1.3 10 kJ
H
− −
−
⎛ ⎞× ⋅ ⋅∆ = ⎜ ⎟⋅⎝ ⎠
⎛ ⎞−× ⎜ ⎟⎝ ⎠
= − ×
(b) 
 
 
 
 
6.43 (a) 5kJ hr days kJ1250 1 150 1.88 10 
hr day year year
⎛ ⎞⎛ ⎞⎛ ⎞ = ×⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠ 
 
6
trips gal. L mL g150 0.40 3.785 1000 0.702 
year trip gal. L mL
1 mol kJ kJ5471 7.6 10
114.23 g mol year
⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠⎝⎝ ⎠⎝ ⎠⎝ ⎠
⎛ ⎞⎛ ⎞ = ×⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
⎞⎟⎠(b) 
 
 
 
SM-152 
 
6.45 From H U P V∆ = ∆ + ∆ at constant pressure, or .U H P V∆ = ∆ − ∆ 
Because we get 22 kJ,w P V= − ∆ = + 15 kJ 22 kJ 7 kJ.U− + = ∆ = + 
 
6.47 To determine the enthalpy of the reaction we must start with a balanced 
chemical reaction and determine the limiting reagent: 
2 22HCl(aq) + Zn(s) H (g) + ZnCl ( )aqU . 
-1
0.800 L 0.500 M HCl 0.400 mol HCl
8.5 g 0.130 mol Zn
65.37g mol
⋅ =
=⋅
 
Examining the reaction stoichiometry and the initial quantities of HCl and 
Zn, we note that Zn is the limiting reagent (0.260 mol of HCl is needed to 
completely react with 0.130 moles of Zn). The enthalpy of reaction may 
be obtained using tabulated enthalpies of formation: 
 
kJ kJ kJ153.89 2 167.16 2 167.16 0
mol mol mol
kJ153.89
mol
rH
⎛ ⎞ ⎛ ⎞
 ∆ =
 
 
− + − − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
This is the enthalpy per mole of Zinc consumed, Therefore, the energy 
released by the reaction of 8.5 g of Zinc is: 
( )kJ153.89 0.130 mol 20.0 kJ
mol
⎛ ⎞− =⎜ ⎟⎝ ⎠ − 
The change in the temperature of the water is then: 
 ( )o
o o o
J20000 J 4.184 800 g
C g
5.98 C and 25 C 5.98 C 31 Cf
T
T T
⎛ ⎞− = − ∆⎜ ⎟⎝ ⎠
∆ = = + = o
 
 
 
6.49 The enthalpy of reaction for the reaction 
4 C7H5N3O6(s) + 21 O2(g) → 28 CO2(g) + 10 H2O(g) + 6 N2(g) 
may be found using enthalpies of formation: 
kJ kJ kJ kJ28 393.51 10 241.82 4 67 1316.5
mol mol mol mol
⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − − − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 
SM-153 
 
This is the energy released per mole of reaction as written. One fourth of 
this amount of energy or kJ329.1
mol
− will be released per mole of TNT 
consumed. The energy density in kJ per gram may be found by dividing 
this amount of energy by the mass of one mole of TNT: 
kJ329.1 kJmol 14.5
g g227.11
mol
−
= − 
 
6.51 The combustion reaction of diamond is reversed and added to the 
combustion reaction of graphite to give the desired reaction: 
2 2C(gr) O (g) CO (g)+ ⎯⎯→ 393.51 kJH∆ ° = − 
2 2CO (g) C(dia) O (g)⎯⎯→ + 395.41 kJH∆ ° = + 
C(gr) C(dia)⎯⎯→ 1.90 kJH∆ ° = + 
 
6.53 The first reaction is doubled, reversed, and added to the second to give the 
desired total reaction: 
2 22[SO (g) S(s) O (g)]⎯⎯→ + (2)[ 296.83 kJ]+ 
 2 32S(s) 3 O (g) 2 SO (g)+ ⎯⎯→ 791.44 kJ− 
2 2 32 SO (g) O (g) 2 SO (g)+ ⎯⎯→ 
1 1(2) ( 296.83 kJ mol ) (791.44 kJ mol ) 197.78 kJH − −∆ ° = + ⋅ − ⋅ = − 
 
6.55 First, write the balanced equations for the reaction given: 
5
2 2 2 2 22C H (g) O (g) 2 CO (g) H O(l)+ ⎯⎯→ + 1300 kJH∆ ° = − 
 72 6 2 2 22C H (g) O (g) 2 CO (g) 3 H O(l)+ ⎯⎯→ + 1560 kJH∆ ° = − 
 12 2 22H (g) O (g) H O(l)+ ⎯⎯→ 286 kJH∆ ° = − 
The second equation is reversed and added to the first, plus two times the 
third: 
SM-154 
 
5
2 2 2 2 22C H (g) O (g) 2 CO (g) H O(l)+ ⎯⎯→ + 1300 kJH∆ ° = − 
 
2
7
2 2 6 22CO (g) 3 H O(l) C H (g) O (g)+ ⎯⎯→ + 2 1560 kJH∆ ° = + 
 12 2 222[H (g) O (g) H O(l)]+ ⎯⎯→ 2[ 286 kJ]H∆ ° = − 
2 2 2 2 6C H (g) 2 H (g) C H (g)
1300 kJ 1560 kJ 2 ( 286 kJ) 312 kJH
+ ⎯⎯→
∆ ° = − + + − = − 
 
6.57 
 
 
 
 
6.59 The desired reaction may be obtained by reversing the first reaction and 
multiplying it by 2, reversing the second reaction, and adding these to the 
third: 
 24 32[NH Cl(s) NH (g) HCl(g)]⎯⎯→ + [ 176.0 kJ]H∆ ° = + 
 3 22 NH (g) N (g) 3 H (g)⎯⎯→ + 2 92.22 kJH∆ ° = +
 2 2 2 4N (g) 4 H (g) Cl (g) 2 NH Cl(s)+ + ⎯⎯→ 628.86 kJH∆ ° = −
= +
 
 2 2H (g) Cl (g) 2 HCl(g)+ ⎯⎯→
 2( 176.0 kJ) 92.22 kJ 628.86 kJ 184.6 kJH∆ ° = + + − = −
 
6.61 From Appendix 2A,∆ ° f (NO) 90.25 kJH
The reaction we want is 
5
2 2 22N (g) O (g) N O (g)+ ⎯⎯→ 5 
Adding the first reaction to half of the second gives 
 2 22 NO(g) O (g) 2 NO (g)+ ⎯⎯→ 114.1 kJH∆ ° = − 
 12 2 222 NO (g) O (g) N O (g)+ ⎯⎯→ 5 55.1 kJH∆ ° = − 
 3 2 222 NO(g) O (g) N O (g)+ ⎯⎯→ 5 169.2 kJ− 
f 2 f 3 f 2 4
1 1
1
1
12 (H O, l) [4( [HNO , l]) 5 ( [N H , l])]
12( 285.83 kJ mol ) [4( 174.10 kJ mol )
5 ( 50.63 kJ mol )]
2480.41 kJ mol
H H H H
− −
−
−
∆ ° = ∆ ° − ∆ ° + ∆ °
= − ⋅ − − ⋅
+ + ⋅
= − ⋅
SM-155 
 
The
enthalpy of this reaction equals the enthalpy of formation of 
 minus twice the enthalpy of formation of NO, so we can write 2 5N O (g)
f 2 5
f 2 5
169.2 kJ (N O ) 2( 90.25 kJ)
(N O ) 11.3 kJ
H
H
− = ∆ ° − +
∆ ° = + 
 
6.63 The enthalpy of the reaction 
3 2 5PCl (l) Cl (g) PCl (s)+ ⎯⎯→ 124 kJH∆ ° = − 
is r f f(products) (reactants)H H H∆ ° = Σ ∆ ° − Σ ∆ °
f 5 f 3124 kJ (PCl , s) (PCl , l)H H− = ∆ ° − ∆ ° 
Remember that the standard enthalpy of formation of will be 0 by 
definition because this is an element in its reference state. From the 
Appendix we find that 
2Cl (g)
 
1
f 3
f 5
1
f 5
(PCl , l) 319.7 kJ mol
124 kJ (PCl ,s) ( 319.7 kJ)
(PCl ,s) 444 kJ mol
H
H
H
−
−
∆ ° = − ⋅
− = ∆ ° − −
∆ ° = − ⋅
 
6.65 (a) For , we want to find the enthalpy of the reaction 2H O(l)
1
2 2 22H (g) O (g) H O(l)+ ⎯⎯→ 
The enthalpy change can be estimated from bond enthalpies. We will need 
to put in to break the H—H bonds 
in
1(1 mol) (436 kJ mol )−⋅
1
2 21 mol H (g), ( mol)
1(496 kJ mol )−⋅ to break the O—O bonds in 
1
22 mol O (g) ; we will get back for the formation 
of O—H bonds. This will give 
(2 mol) 1(463kJ mol )−⋅
2 mol H∆ = 1242 kJ mol−− ⋅ . This value, 
however, will be to produce water in the gas phase. In order to get the 
value for the liquid, we will need to take into account the amount of heat 
given off when the gaseous water condenses to the liquid phase. This is 
 at 298 K: 144.0 kJ mol−⋅
SM-156 
 
vap
1
f,water(l) f,water(g)
1
242 kJ (1 mol) (44.0 kJ mol )
286 kJ mol
H H H −
−
∆ ° = ∆ ° − ∆ ° = − − ⋅
= − ⋅ 
(b) The calculation for methanol is done similarly: 
1
2 2 32C(gr) 2 H (g) O (g) CH OH(l)+ + ⎯⎯→ 
 H∆ for individual bond contributions: 
 atomize 1 m ol C(gr) 1(1 mol) (717 kJ mol )−⋅
 break 2 m H—H bonds ol 1(2 mol) (436 kJ mol )−⋅
 break 1 22 mol O bonds 
11
2( mol) (496 kJ mol )
−⋅ 
 form 3 m C—H bonds – ol 1(3 mol) (412 kJ mol )−⋅
 form 1 m C—O bonds – ol 1(1 mol) (360 kJ mol )−⋅
 form 1 m O—H bonds – ol 1(1 mol) (463 kJ mol )−⋅
 
 Total 222kJ− 
 
f, methanol(l) f, methanol(g) vap
1222 kJ (1 mol) (35.3 kJ mol )
257kJ
H H H
−
∆ ° = ∆ ° − ∆ °
= − − ⋅
= −
 
 
 
 (c) 6 C 2 6 6(gr) 3 H (g) C H (l)+ ⎯⎯→
Without resonance, we do the calculation considering benzene to have 
three double and three single C—C bonds: 
 atomize: 6 m ol C(gr) 1(6 mol) (717 kJ mol )−⋅
 break: 3 m H—H bonds ol 1(3 mol) (436 kJ mol )−⋅
 form: 3 m CRC bonds ol 1(3 mol) (612 kJ mol )−− ⋅
 form: 3 m C—C bonds ol 1(3 mol) (348 kJ mol )−− ⋅
 form: 6 m C—H bonds ol 1(6 mol) (412 kJ mol )−− ⋅
 __________________________________________________ 
 Total kJ258+ 
SM-157 
 
1
f,benzene(l) f,benzene(g) vap 258kJ (1 mol) (30.8 kJ mol )
227 kJ
H H H −∆ ° = ∆ ° = ∆ ° = + − ⋅
= + 
(d) 2 66 C(gr) 3 H (g) H (l)C+ ⎯⎯→ 6
1
With resonance, we repeat the calculation considering benzene to have six 
resonance-stabilized C—C bonds: 
 atomize: 6 m ol C(gr) 1(6 mol) (717 kJ mol )−⋅
break: H—H bonds 3 mol 1(3mol) (436 kJ mol )−⋅
form: C—C bonds, resonance 6 mol 1(6 mol) (518 kJ mol )−− ⋅
form: C—H bonds 6 mol 1(6 mol) (412 kJ mol )−− ⋅
________________________________________________________ 
Total 30 kJ+
1 1
f,benzene(l) f,benzene(g) vap 30 kJ mols (1 mol) (30.8 kJ mol )
1 kJ
H H H − −∆ ° = ∆ ° − ∆ ° = + ⋅ − ⋅
= −
 
6.67 For the reaction 22Na O(s) 2 Na (g) O (g)
+ −⎯⎯→ +
L f f 1
ea1 ea2 f 2
1 1 1
L
1 1
1
L
2 (Na, g) (O, g) 2 (Na)
(O) (O) (Na O(s)
2(107.32 kJ mol ) 249 kJ mol 2(494 kJ mol )
141 kJ mol 844 kJ mol 409 kJ mol
2564 kJ mol
H H H I
E E H
H
H
− − −
− − −
−
∆ = ∆ ° + ∆ ° +
− − − ∆
∆ = ⋅ + ⋅ + ⋅
− ⋅ + ⋅ + ⋅
∆ = ⋅
 
 
6.69 (a) ∆ =L f f 1
ea f
(Na, g) (Cl, g) (Na)
 of Cl (NaCl(s))
H H H I
E H
∆ ° + ∆ ° +
− − ∆
1−1 1 1
1
f
1
f
787 kJ mol 108 kJ mol 122 kJ mol 494 kJ mol
349 kJ mol (NaCl(s))
(NaCl(s)) 412 kJ mol
H
H
− − −
−
−
⋅ = ⋅ + ⋅ + ⋅
− ⋅ − ∆
∆ = − ⋅
 
 
 
 
 
 
 
SM-158 
 
(b) L f f 1
ea f
(K, g) (Br, g) (K)
 (Br) (KBr(s))
H H H I
E H
∆ = ∆ ° + ∆ ° +
− − ∆
1 1
L
1 1
1
89 kJ mol 97 kJ mol 418 kJ mol
325 kJ mol 394 kJ mol
673 kJ mol
H − −
− −
−
∆ = ⋅ + ⋅ + ⋅
− ⋅ + ⋅
= ⋅
1−
612) kJ mol
 
(c) L f f 1 ea f(Rb, g) (F, g) (Rb) (F) (RbF(s))H H H I E H∆ = ∆ ° + ∆ ° + − − ∆
 1 1f
1 1
1
f
774 kJ mol (Rb, g) 79 kJ mol
 402 kJ mol 328 kJ mol 558 kJ mol
(Rb, g) 63 kJ mol
H
H
− −
1− − −
−
⋅ = ∆ ° + ⋅
+ ⋅ − ⋅ + ⋅
∆ ° = ⋅
 
 
 
6.71 (a) break: 3(3 mol C=C bonds 1−⋅ 
 form: 6 mol C=C bonds 16(518) kJ mol−− ⋅ 
 ____________________________ 
 Total 11272 kJ mol−− ⋅ 
 (b) break: 4 mol C—H bonds 14(412) kJ mol−⋅ 
 4 mol Cl—Cl bonds 14(242) kJ mol−⋅ 
 form: 4 mol C—Cl bonds 14(338) kJ mol−− ⋅ 
 4 m H bol —Cl onds 14(431) kJ mol−− ⋅ 
 ______________________________ 
 Total 1460 kJ mol−− ⋅
(c) The number and types of bonds on both sides of the equations are 
equal, so we expect the enthalpy of the reaction to be essentially 0. 
 
6.73 (a) break: 1 mol N—N triple bonds 1(1 mol) (944 kJ mol )−⋅
 3 mol F—F bonds 1(3 mol) (158 kJ mol )−⋅
 form: 6 m ol N—F bonds 1(6 mol) ( 195 kJ mol )−− ⋅
 __________________________________________ 
 Total 248 kJ+ 
SM-159 
 
 (b) break: 1 m ol C=C bonds 1(1 mol) (612 kJ mol )−⋅
 1 mol O—H bonds 1(1 mol) (463 kJ mol )−⋅
 form: 1 mol C—C bonds 1(1 mol) (348 kJ mol )−− ⋅
 1 mol C—O bonds 1(1 mol) (360 kJ mol )−− ⋅
 1 mol C—H bonds 1(1 mol) (412 kJ mol )−− ⋅
 __________________________________________ 
 Total 45 kJ−
 (c) break: 1 m ol C—H bonds 1(1 mol) (412 kJ mol )−⋅
 1 mol Cl—Cl bonds 1(1 mol) (242 kJ mol )−⋅
 form: 1 m ol C—Cl bonds
 1 mol H—Cl bonds 1(1 mol) (431 kJ mol )−− ⋅
 _________________________________________ 
 Total 115 kJ− 
 
6.75 The value that we want is given simply by the difference between three 
isolated C=C bonds and three isolated single bonds, versus six 
resonance-stabilized bonds: 
C—C
3 C=C bonds 3 C—C bonds 3(348 kJ) 3(612 kJ) 2880 kJ+ = + = 
6 resonance-stabilized bonds 6(518 kJ) 3108 kJ= = 
As can be seen, the six resonance-stabilized bonds are more stable by 
ca. 228 kJ.
 
6.77 (a) The enthalpy of vaporization is the enthalpy change associated with 
the conversion at constant pressure. The value at 
298.2 K will be given by 
6 6 6 6C H (l) C H (g)⎯⎯→
vaporization at 298 K f 6 6 f 6 6
1 1
1
(C H , g) (C H , l)
82.93 kJ mol (49.0 kJ mol )
33.93 kJ mol
H H H
− −
−
∆ ° = ∆ ° − ∆ °
= ⋅ − ⋅
= ⋅
 
SM-160 
 
(b) In order to take into account the difference in temperature, we need to 
use the heat capacities of the reactants and products in order to raise the 
temperature of the system to We can rewrite the reactions as 
follows, to emphasize temperature, and then combine them according to 
Hess’s law: 
353.2 K.
6 6 at 298 K 6 6 at 298 KC H (l) C H (g)⎯⎯→ 33.93 kJH∆ ° = 
6 6 at 298 K 6 6 at 353.2 KC H (l) C H (l)⎯⎯→ 
1 1
(1 mol) (353.2 K 298.2 K)
(136.1 J mol K )
7.48 kJ
H
− −
∆ ° = −
⋅ ⋅
=
 
6 6 at 298 K
6 6 at 353.2 KC H (g) C H (g)⎯⎯→ 
1 1
(1 mol) (353.2 K 298.2 K)
(81.67 J mol K )
4.49 kJ
H
− −
∆ ° = −
⋅ ⋅
=
 
To add these together to get the overall equation at we must 
reverse the second equation: 
353.2 K,
6 6 at 298 K 6 6 at 298 KC H (l) C H (g)⎯⎯→ 33.93 kJH∆ ° = 
6 6 at 353.2 K 6 6 at 298 KC H (l) C H (l)⎯⎯→ 7.48 kJH∆ ° = − 
6 6 at 298 K 6 6 at 353.2 KC H (g) C H (g)⎯⎯→ 4.49 kJH∆ ° = 
6 6 at 353.2 K 6 6 at 353.2 KC H (l) C H (g)⎯⎯→ 30.94 kJH∆ ° = 
(c) The value in the table is 130.8 kJ mol−⋅ for the enthalpy of 
vaporization of benzene. The value is close to that calculated as corrected 
by heat capacities. At least part of the error can be attributed to the fact 
that heat capacities are not strictly constant with temperature. 
 
6.79
 at tem
 ∆ = −o o or,2 m,2 m,2
For the reaction: A + 2B 3C + D the molar enthalpy of reaction 
perature 2 is given by:
(products) (reactants)H H H
→
 
 
 
SM-161 
 
o o o o
m,2 m,2 m,2 m,2
o o
m,1 , 2 1 m,1 , 2 1
o o
m,1 , 2 1 m,1 , 2 1
o o o o
m,1 m,1 m,1 m,1
,
3 (C) (D) (A) 2 (B)
3[ (C) (C)( )] [ (D) (D)( )]
 [ (A) (A)( )] 2[ (B) (B)( )]
3 (C) (D) (A) 2 (B)
 +[3 (C)
p m p m
p m p m
p m p
H H H H
H C T T H C T T
H C T T H C T T
H H H H
C C
= + − −
= + − + + −
− + − − + −
= + − −
+ , , , 2 1
o
r,1 , , , , 2 1
(D) (A) 2 (B)]( )
[3 (C) (D) (A) 2 (B)]( )
m p m p m
p m p m p m p m
C C T T
H C C C C T T
− − −
= ∆ + + − − −
 
 
 
 
 
 
6.81 This process involves five separate steps: (1) raising the temperature of the 
ice from (2) melting the ice at 0.00 (3) raising the 
temperature of the liquid water from (4) vaporizing 
the water at 100.00 and (5) raising the temperature of the water vapor 
from1 
30.27 C to 0.00 C,− ° ° C,°
0.00 C to 100.00 C,° °
C,°
00.00 C to 150.35 C.° °
Step 1: 
 
1 1(27.96 g) (2.03 J ( C) g ) (0.00 C ( 30.27 C)) 1.72 kJH − −∆ = ⋅ ° ⋅ ° − − ° =
Step 2: 
 
 
1
1
27.96 g (6.01 kJ mol ) 9.32 kJ
18.02 g mol
H −−
⎛ ⎞∆ = ⋅ =⎜ ⎟⋅⎝ ⎠
Step 3: 
1 1(27.96 g) (4.18 J ( C) g ) (100.00 C 0.00 C) 11.7 kJH − − ∆ = ⋅ ° ⋅ ° − ° =
Step 4: 
 
1
1
27.96 g (40.7 kJ mol ) 63.1 kJ
18.02 g mol
H −−
⎛ ⎞ ⋅ =⎜ ⎟⋅⎝ ⎠ 
Step 5: 
 
∆ =
⋅ ° ⋅ ° − ° =1 1(27.96 g) (2.01 J ( C) g ) (150.35 C 100.00 C) 2.83 kJH − −∆ =
The total heat required 1.72 kJ 9.32 kJ 11.7 kJ 63.1 kJ 2.83 kJ
88.7 kJ
= + + + +
=
 
 
6.83 Appendix 2A provides us with the heat of formation of 
 mo and the heat capacities of 2I (g) at 298K ( 62.44 kJ+ ⋅ 1l )−
SM-162 
 
1 1 1 1
2 2I (g) (36.90 J K mol ) and I (s) (54.44 J K mol ).
− − − −⋅ ⋅ ⋅ ⋅ We can 
calculate the 0sub at 298K:H∆
2I (s) I (g)⎯⎯→ 2
0
T
1
 0 1sub 62.44 kJ molH
−∆ = + ⋅
We can calculate the enthalpy of fusion from the relationship 
0 0
sub fus vapH H H∆ = ∆ + ∆ 
but these values need to be at the same temperature. To correct the value 
for the fact that we want all the numbers for 298K, we need to alter the 
heat of vaporization, using the heat capacities for liquid and gaseous 
iodine. 
2 2I (l) at 184.3 C I (g) at 184.3 C° ⎯⎯→ ° 0 1vap 41.96 kJ molH −∆ = + ⋅
From Section 6.22, we find the following relationship 
0 0 0
r,2 r,1 , 2 1( )P mH H C T∆ = ∆ + ∆ − 
0 0 0 0
vap, 298K vap, 475.5K , 2 , 2 2 1
0 1
vap, 298K
1 1 1 1
1
( (I ,g) (I , l)) ( )
41.96 kJ mol
(36.90 J K mol 80.7 J K mol ) (298K 475.5K)
49.73kJ mol
P m P mH H C C T T
H −
− − − −
−
∆ = ∆ + − −
∆ = + ⋅
+ ⋅ ⋅ − ⋅ ⋅ −
= + ⋅
So, at 298K: 
1 0
fus
0 1
fus
62.44 kJ mol 49.73 kJ mol
12.71 kJ mol
H
H
− −
−
+ ⋅ = ∆ + ⋅
∆ = + ⋅ 
 
6.85 (a) 
0
1
2
3
4
5
6
7
0 2 4 6 8 1
Vf/Vi
w
(J
)
0
 
 
 
 
 
 
 
 
SM-163 
 
 
(b) The amount of work done is greater at the higher temperature. This 
can be seen from the equation: 
final
initial
ln
V
w nRT
V
= − 
The amount of work done is directly proportional to the temperature at 
which the expansion takes place. 
(c) The comparison requested is the comparison of the terms 
final
initial
ln
V
V
 
for the two processes. Even though in both cases the gas expands by 4 L, 
the relative amount of work done is different. We can get a numerical 
comparison by taking the ratio of this term for the two conditions: 
9.00 Lln
5.00 L 0.588 0.365
1.615.00 Lln
1.00 L
⎛ ⎞⎜ ⎟⎝ ⎠ = =⎛ ⎞⎜ ⎟⎝ ⎠
 
The second expansion by 4.00 L produces only about one third the amount 
of work that the first expansion did. 
 
6.87 First, we need to calculate how much energy from the sunshine will be 
hitting the surface of the ethanol, so we convert the rate 2 1kJ cm s :− −⋅ ⋅
2
2 1 4 2 1
4 2 1 2
1 m1 kJ m s 1 10 kJ cm s
100 cm
60 s(1 10 kJ cm s ) (50.0 cm ) 10 min 3 kJ
min
− − − − −
− − −
⎛ ⎞⋅ ⋅ = × ⋅ ⋅⎜ ⎟⎝ ⎠
⎛ ⎞× ⋅ ⋅ × =⎜ ⎟⎝ ⎠
 
The enthalpy of vaporization of ethanol is 143.5 kJ mol−⋅ (see Table 6.2). 
We will assume that the enthalpy of vaporization is approximately the 
same at ambient conditions as it would be at the boiling point of ethanol. 
1
1
3 kJ (46.07 g mol ) 3 g
43.5 kJ mol
−
−
⎛ ⎞ ⋅ =⎜ ⎟⋅⎝ ⎠
 
SM-164 
 
 
6.89 (a) 31 7 16 5 2 2 2 2 24 2C H NH (l) O (g) 6 CO (g) H O(l) N (g)+ ⎯⎯→ + + 2 
(b) 
 
2
2
12
CO 21
2
12
H O 21
2
6 mol CO0.1754 g aniline (28.01 g mol CO )
1 mol aniline93.12 g mol anline
0.4873 g CO (g)
3.5 mol H O0.1754 g aniline (18.02 g mol H O)
1 mol aniline93.12 g mol aniline
0.1188 g H O(l)
m
m
m
−
−
−
−
⎛ ⎞ ⎛ ⎞= ⋅⎜ ⎟ ⎜ ⎟⋅ ⎝ ⎠⎝ ⎠
=
⎛ ⎞ ⎛ ⎞= ⋅⎜ ⎟ ⎜ ⎟⋅ ⎝ ⎠⎝ ⎠
=
2
12
N 21
2
0.5 mol N0.1754 g aniline (28.02 g mol N )
1 mol aniline93.12 g mol aniline
0.026 39 g N (g)
−
−
⎛ ⎞ ⎛ ⎞= ⋅⎜ ⎟ ⎜ ⎟⋅ ⎝ ⎠⎝ ⎠
=
 
(c) 
2
31
24
O 1
2
mol O0.1754 g aniline
1 mol aniline93.12 g mol aniline
0.014 60 g O (g)
n −
⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⋅ ⎝ ⎠⎝ ⎠
=
 
1 1
2(0.014 60 mol O )(0.082 06 L atm K mol ) (296K)
0.355 L
0.999 atm
nRTP
V
− −⋅ ⋅ ⋅= =
=
 
 
6.91 (a) The reaction enthalpy is obtained by Hess’s law: 
r f f 2
1 1
r
1
r
(CO, g) (H O, g)
(1) ( 110.53 kJ mol ) (1) ( 241.82 kJ mol )
131.29 kJ mol
H H H
H
H
− −
−
∆ ° = ∆ ° − ∆ °
∆ ° = − ⋅ − − ⋅
∆ ° = + ⋅
 
endothermic 
(b) The number of moles of produced is obtained from the ideal gas 
law: 
2H
1
1 1
500 Torr (200 L)
760 Torr atm
4.74 mol
(0.082 06 L atm K mol ) (338 K)
PVn
RT
−
− −
⎛ ⎞⎜ ⎟⋅⎝ ⎠= = =⋅ ⋅ ⋅ 
The enthalpy change accompanying the production of this amount of 
hydrogen will be given by 
SM-165 
 
1(4.74 mol) (131.29 kJ mol ) 623 kJH −∆ = ⋅ = 
 
6.93 (a) The number of moles burned may be obtained by taking the difference 
in the number of moles of gas present in the tank before and after the drive 
using the ideal gas equation: 
1 2
1 2 1 2
-1 -1
( )
30.0 L(16.0 atm 4.0 atm)
(0.0820578 L atm K mol
14.7 mol
PV PV Vn n P P
RT RT RT
⎛ ⎞− = − = − ⎜ ⎟⎝ ⎠
⎛ ⎞= − ⎜ ⎟⋅ ⋅ ⋅⎝ ⎠
=
 
(b) From a table of enthalpies of formation,
the enthalpy of combustion of 
H2 is found to be -241.82 kJ�mol. The energy released was, therefore, 
(14.7 mol)( -241.82 kJ�mol) = -3560 kJ 
 
6.95 (a) First we must balance the chemical reaction: 
15
6 6 2 2 22C H (l) O (g) 6 CO (g) 3 H O(g)+ ⎯⎯→ + 
For 1 mol burned, the change in the number of moles of gas is 6 6C H (l)
(9.00 7.50) mol 2.50 mol n− = + = ∆ 
1 1 3( 2.50 mol) (8.314 J K mol ) (298 K) 6.19 10 J 6.19 kJ
nRTw P V P nRT
P
w − −
∆⎛ ⎞= − ∆ = − = − ∆⎜ ⎟⎝ ⎠
= − + ⋅ ⋅ = − × = −
 
1 1
c 6( 393.51 kJ mol ) 3( 241.82 kJ mol ) ( 49.0 kJ mol )
3135.5 kJ
H − −∆ = − ⋅ + − ⋅ − + ⋅
= −
1−(b) 
 
(c) ( 3135.5 6.19) kJ 3141.7 kJU H w∆ ° = ∆ ° + = − − = −
 
6.97 (a) The heat given off by the reaction, which was absorbed by the 
calorimeter, is given by 
1(525.0 J ( C) ) (20.0 C 18.6 C) 0.74 kJH −∆ = − ⋅ ° ° − ° = − 
This, however, is not all the heat produced, as the 100.0 mL of solution 
resulting from mixing also absorbed some heat. If we assume that the 
SM-166 
 
volume of NaOH and are negligible compared to the volume of 
water present and that the density of the solution is then the 
change in heat of the solution is given by 
3HNO
11.00 g mL ,−⋅
1 1(20.0 C 18.6 C) (4.18 J ( C) g ) (100.0 g) 0.59 kJH − −∆ = − ° − ° ⋅ ° ⋅ = − 
The total heat given off will be 0.74 kJ ( 0.59 kJ) 1.33 kJ− + − = − 
(b) This heat is for the reaction of (0.500 M) (0.0500L) = 
 so the amount of heat produced per mole of 
will be given by 
30.0250 mol HNO , 3HNO
11.33 kJ 53.2 kJ mol
0.0250 mol
−− = − ⋅ 
 
6.99 
(a) (1) (2) (3) 
 
(b) From bond enthalpies, each step is identical, as the number and types 
of bonds broken and formed are the same: 
 break: 1 mol C=C bonds 612 kJ 
 1 mol H—H bonds 436 kJ 
 form: 1 mol C—C bonds 348 kJ− 
 2 mol C—H bonds 2( 412 kJ)− 
 ____________________________________________ 
 Total: 124 kJ− 
The total energy change should be equal to the sum of the three steps or 
 3( 124 kJ) 372 kJ.− = −
(c) The Hess’s law calculation using standard enthalpies of formation is 
easily performed on the composite reaction: 
6 6 2 6 12
r f f
C H (l) 3 H (g) C H (l)
(products) (reactants)H H H
+ ⎯⎯→
∆ ° = Σ∆ ° − Σ∆ ° 
C
C C
C
H
H
C
C
H
H
H
H
H H
C C H
C
C
C
C
H
H
H
H
H
H
H
H
H
C CH
C
C
C
C
H
H
H
H
H
H
H
H
H
H
H
SM-167 
 
f f
1 1
1
(cyclohexane) (benzene)
156.4 kJ mol ( 49.0 kJ mol )
205.4 kJ mol
H H
− −
−
= ∆ ° − ∆ °
= − ⋅ − + ⋅
= − ⋅
 
(d) The hydrogenation of benzene is much less exothermic than predicted 
by bond enthalpy estimations. Part of this difference can be due to the 
inherent inaccuracy of using average values, but the difference is so large 
that this cannot be the complete explanation. As may be expected, the 
resonance energy of benzene makes it more stable than would be expected 
by treating it as a set of three isolated double and three isolated single 
bonds. The difference in these two values [ 205 kJ ( 372 kJ) 167 kJ]− − − = 
is a measure of how much more stable benzene is than the Kekulé 
structure would predict. 
 
6.101 (a) The combustion reaction is 
60 2 2C (s) 60 O (g) 60 CO (g)+ ⎯⎯→ 
The enthalpy of formation of will be given by 60C (s)
c f 2 f 60
1
f 60
1
f 60
60 (CO ,g) (C ,s)
25 937 kJ 60 mol ( 393.51 kJ mol ) (C ,s)
(C ,s) 2326 kJ mol
H H H
H
H
−
−
∆ ° = ∆ ° − ∆ °
− = × − ⋅ − ∆ °
∆ ° = + ⋅
 
(b) The bond enthalpy calculation is 
 60 C(gr) 60 C(g)⎯⎯→ 1)(60) ( 717 kJ mol )−+ ⋅
 Form 60 mol C—C bonds 160(348 kJ mol )−− ⋅
 Form 30 mol C=C bonds 130(612 kJ mol )−− ⋅
 60 60C (g) C (s)⎯⎯→ 233 kJ− 
 6060 C(gr) C (s)⎯⎯→ 3547 kJ+ 
(c) From the experimental data, the enthalpy of formation of shows 
that it is more stable by (
60C
3547 kJ 2326 kJ) 1221 kJ− = than predicted by 
the isolated bond model. 
(d) per carbon atom 1221 kJ 60 20 kJ÷ =
SM-168 
 
(e) per carbon atom 150 kJ 6 25 kJ÷ =
(f) Although the comparison of the stabilization of benzene with that of 
should be treated with caution, it does appear that there is slightly less 
stabilization per carbon atom in than in benzene. This fits with 
expectations, as the molecule is forced by its geometry to be curved. 
This means that the overlap of the p-orbitals, which gives rise to the 
delocalization that results in resonance, will not be as favourable as in the 
planar benzene molecule. Another perspective on this is obtained by 
nothing that the C atoms in are forced to be partially hybridized 
because they be rigorously planar as required by hybridization. 
60C
60C
60C
60C
3sp
2sp
 
6.103 The balanced combustion reactions are 
15 3 3
6 3 2 3 2 2 2 24 2
33 9 3
6 3 2 3 2 2 2 24 2
C H (NO ) (s) O (g) 6 CO (g) H O(l) N (g)
C H (NH ) (s) O (g) 6 CO (g) H O(l) N (g)
+ ⎯⎯→ + +
+ ⎯⎯→ + +
2
2
 
Because the fundamental structures of the two molecules are the same, we 
need only look at the differences between the two, which in this case are 
concerned with the groups attached to nitrogen. From the combustion 
equations we can see that the differences are (1) the consumption of 184 
more moles of and (2) the production of six more moles of 
for the combustion of aniline. Because the 
2O (g) 2H O(l)
f 2of O (g)H∆ ° is 0, the net 
difference will be the production of 6 more moles of 
 12H O(l) or 6 ( 285.83 kJ mol ) 1715.0 kJ.
−× − ⋅ = −
 
6.105 Start by calculating the amount of energy generated by the decomposition 
of lauric acid: 
 
12 24 2 2 2 2
-1 -1 -1
r
-1
C H O + 17 O 12 CO + 12 H O
H 12( 393.51 kJ mol ) 12( 285.83 kJ mol ) ( 774.6 kJ mol )
 -7377.48 kJ mol
→
∆ = − ⋅ + − ⋅ − − ⋅
= ⋅
SM-169 
 
if 15.0 g are consumed, then the energy released 
is: -1 -1
15.0 g-7377.48 kJ mol 552 kJ
200.32 g mol
⋅ =⋅ − 
The enthalpy for the decomposition of sucrose may be calculated given 
standard enthalpies of formation: 
12 22 11 2 2 2
-1 -1 -1
r
-1
C H O + 12 O 12 CO + 11 H O
H 12( 393.51 kJ mol ) 11( 285.83 kJ mol ) ( 2222 kJ mol )
 -5644.3 kJ mol
→
∆ = − ⋅ + − ⋅ − − ⋅
= ⋅
 
The amount of sucrose needed to produce the same amount of energy as 
15.0 g of lauric acid is: -1
-1
552 kJ 0.0979 mol 
5644.3 kJ mol
or (0.0979 mol)(342.3 g mol ) 33.5 g
k
−= =⋅
⋅ =
 
 
6.107 (a) Given the enthalpy of combustion, the enthalpy of formation of each 
compound may be found using: 
4 4 4 2 2 2
-1 -1
c f
-1 -1
f c
C H O (l) + 3 O (g) 4 CO (g) + 2 H O(l)
(X) 4( 393.51 kJ mol )+2(-285.83 kJ mol ) - (X)
(X) 4( 393.51 kJ mol )+2(-285.83 kJ mol ) - (X)
H H
H H
→
∆ = − ⋅ ⋅ ∆
∆ = − ⋅ ⋅ ∆
 
where X is either maleic or fumaric acid. Plugging in the appropriate 
enthalpies of combustion, the enthalpies of formation are found to be: 
The cis-trans isomerization reaction: maleic acid Æ fumaric acid has an 
enthalpy of 20.5 kJ⋅mol
-1 -1
f f(maleic acid) 790.5 kJ mol and (fumaric acid) 811.0 kJ molH H⊕∆ = − ⋅ ∆ = − ⋅
-1
(b) Fumaric acid has the lower enthalpy of formation. 
(c) Because one mole of gas is produced for each mole of acid consumed, 
work is done on the surroundings