Chapter 7 - Solutions
1. This exercise contains an error. It asks for a solution proving that Hvap in kilojoules
per mole is greater than the numerical value of Svap in joules per Kelvin per mole.
However, Hvap is smaller in every case for the conditions given.
Vaporization is a phase change; therefore, at the boiling point, the system is at
equilibrium. Hence Gvap = 0.
0 = G = Hvap – TbSvap
 H vap 

Tb = 
 S 
vap


All substances have a boiling-point temperature above 1 K; i.e., Tb > 1 K.
 H vap 

1 K < Tb = 
 S 
vap


 H vap 

1K< 
 S 
vap


Svap < Hvap
T 
T 
3. S = SNe + ΔS Fe = nCvln  2  + nCvln  2 
T1 
T1 
T 
T 
S = 0.345 J·K-1 = SNe + ΔS Fe = nCvln  2  + nCvln  2 
T1 
T1 
 288 
0.345 J·K-1 = S = (nNeCNe + nFeCFe)ln 

 273 
3
CNe =   R, CFe =
2
5
  R (no vibrational modes accessible at this temperature)
2
 288 
3
5
0.345 J·K-1 = S = (nNe   R + nFe   R)ln 

2
2
 273 
ntotal = nne + nFe =
(3.32 atm)(2.5 L)
PV
= 0.371 mol

RT (0.082057 L·atm·mol -1·K -1 )(273 K)
nFe = 0.371 mol – nNe
3
5
0.345 = [nNe   (8.3145 J·mol-1·K-1) + (0.371 – nNe)  
2
2
 288 
(8.3145 J·mol-1·K-1)]ln 

 273 
–1.2617 = –8.3145 nNe; nNe= 0.15 moles
nFe = 0.371 – 0.15 = 0.22 mol
checking answer
3
3
 288 
0.345 J·K-1 = S = [nNe   R + nFe   R] ln 

2
2
 273 
3
 288 
5
 288 
S = (0.15 mol)   Rln 
 + (0.22)   Rln 

2
 273 
2
 273 
S = 0.100 + 0.245 = 0.345 J·K-1
 1 kJ 
5. a. q = (power)(time) = (500 J·s-1)(240 s) 
 = 120 kJ
 1000 J 
 H vap 

Svap = 

T
 b 
 q 
Hvap =  
 n 
Sample
C2H5OH
C4H10
CH3OH
m
(g)
129.00
323.10
100.90
n
(mol)
2.80
5.56
3.15
Hvap
(kJ·mol-1)
43
22
38
Svap
(J·mol-1·K-1)
120
81
110
Boiling
Point
(K)
351.5
273.2
337.7
b. C2H5OH and CH3OH are more ordered liquids (not as random) than C4H10 is.
This is due to the hydrogen bonding in the alcohols.
7. Find G for rxn 1. G = H  TS
H = 278 kJ – 52 kJ – (242 kJ) = 88 kJ
S = 161 J·K-1 219 J·K-1 189 J·K-1= 248 J·K-1
 1 kJ 
G = 88 kJ – 298 K(248 J·K-1) 
 = 14 kJ
 1000 J 
(Rxn 1 is spontaneous under standard-state conditions and temperature 298 K.)
Find G for rxn 2. G = H  TS
H = 278 kJ – 0 kJ – (85 kJ) – (242 kJ) = +49 kJ
S = 161 J·K-1+ 131 J·K-1 230 J·K-1  189 J·K-1= 127 J·K-1
 1 kJ 
G = 49 kJ – 298 K(127 J·K-1) 
 = +87 kJ
 1000 J 
(Rxn 2 is nonspontaneous under standard-state conditions and temperature 298 K.)
Is there a temperature where rxn 2 is spontaneous?
Because H > 0 and S < 0, then G must be > 0, and the reaction is always
nonspontaneous.
Or, set G = 0 and solve for T, assuming H and S are temperature independent.
 H   49 kJ 
T= 
<0
=
 S   127 J·K -1 
There is no temperature at which Reaction 2 is spontaneous. Reaction 1 must be
used.
9. H2 (g) +
1
O2 (g)  H2O (l)
2
G = 237.25 kJ·mol-1
 70.9 kJ 2.0158 g H 2
H =
= –286 kJ·mol-1

0.50 g H 2
1 mol H 2
G = H  TS
 ( G   H )
S =
T
  (237 kJ·mol -1  (286 kJ·mol -1 )  10 3 J 
 
S = 

298 K
  1 kJ 


S = –164 J·K-1 per mole of H2
 1 mol H 2 

S (for 0.5 g H2) = 0.5 g H2  
2
.
0158
g
H
2 

-1
S (for 0.5 g H2) = 40.6 J·K
 -164 J·K -1 


 1 mol H 2 

