Chemistry I-H
Thermodynamic Problem Set #2
Solution Set
1.
Predict the sign for S for:
2.
Predict the sign for S for each of the following reaction:
(+)
(+)
B) ammonia vapor condensing;
D) tea dissolving in water.
(-)
(+)
o
A) CuSO4. 5 H2O(s)
B) 2 Cl(g) 

Cl2(g)
(+) [actually it is nearly zero]
CuSO4(s) + 5 H2O(liq)
(-) since ngas = -1
C) 2 H2(g) + O2(g) 
3.
A) a candle burning;
C) butter melting;
2 H2O(liq)
(-)
o
o
o
o
What is the significance of the " " symbol on the H , S , and G values?
means standard conditions [ = 298 K & 1 atm.]
4.
o
Using the table of thermodynamic values table, calculate the S for each of the following:
A) P4(s) + 6 Cl2(g)  4 PCl3(g)
So = [4(+7449)] - [(+66.90 J) + 6(+222.96 J)] = 2.839 x 104 J
B) Fe2O3(s) + 3 H2(g)

2 Fe(s)
+ 3 H2O(liq)
S = [2(+27.3 J) + 3(69.91 J) ] - [(87.40 J) + 3(+130.57 J)] = -214.78 J
o
C) 4 Al(s) + 3 MnO2(s)
 3 Mn(s) + 2 Al2O3(s)
S = [3(+7.59 J) + 2(23.15 J) ] - [4(6.77 J) + 3(+12.7 J)] = +3.89 J
o
5.
o
o
Calculate G at 25 C for which:
Go = Ho - TSo
o
o
o
o
A) H = -109.0 kJ and S = +27.8 J/K
B) H = +842 kJ and S = -116.2 J/K
Go = -109.0 kJ - 298(+0.0278 kJ)
Go = +842 kJ - 298(-0.1162 kJ)
= -117.3 kJ
o

4
= + 876.6 kJ
o
C) H = +8.29 x 10 J and S = +0.115 kJ/K
4
 Go = +8.29 x 10 J - 298(+115 J)
= +4.86 x 104 J
o
o
D) H = -39.3 kJ and S = -59.2 J/K
Go = -39.3 kJ - 298(-0.0592 kJ)
= -21.7 kJ
6.
For which of the reactions above would an increase in temperature result in a nonspontaneous
reaction becoming spontaneous? (C) since both are positive
7.
For which of the reactions above would an increase in temperature result in a spontaneous reaction
becoming nonspontaneous?
(D) since both are negative.
-28.
Calculate the following for the given reaction:
o
A) H
2 NO(g) + O2(g)  2 NO2(g)
o
o
B) S
= [2(240.45 j)] -[2(210.62) +(205.0)]
= 2(33.84 kJ) - 2(90.37 kJ)
= -145.34 joules
C) the temperature at which G = 0
T = H / S
= -113.06 kJ / -0.14534 kJ
= -113.06 kJ
9.
= 778 K
o
The normal boiling point of trichloromethane (CHCl3) is 61.0 C; its molar heat of
.
vaporization (Hvap) is 31.3 kJ/mole. The standard entropy of this liquid is 202.9 J/mole K.
o
o
Determine G at 61.0 C.
Go = Ho - TSo
= -31.3 kJ - (334 K)(0.2029 kJ) = -99.1 kJ
10. What is the normal freezing point for a substance if its heat of fusion is 34.5 kJ/mole and its entropy
.
in the solid phase is +56.7 J/mole K?
T = H / S = 34.5 kJ / 0.0567 kJ = 608 K
11.
A Which of the following quantities can be taken to be fairly independent of pressure?
A) H for a reaction; B) S for a reaction;
C) S for a substance;
D) G for a reaction.
12. Criticize each of the following statements:
o
A) When the value for G is positive, the reaction cannot occur.
The reaction can occur, but it needs an outside source of energy to sustain it. This
“outside source” is known as the activation energy.
o
B) When the values for Ho and S are both positive, the reaction is nonspontaneous at all
temperatures.

G = H - TS
At a high enough temperature, the reaction may be spontaneous because the TS
factor may offset the H factor.