When the reaction reaches equilibrium, ∆G^0 = 0
and Qc and Qp become Kc and Kp, respectively.
Thus,
0 = ∆G^0 + RT ln Kc and 0 = ∆G^0 + RT ln Kp
or
∆G^0 = -RT ln Kc and ∆G^0 = -RT ln Kp
(4.45)
or ∆G^0 = -2.303 RT log 10 Kc
and
∆G^0 = -2.303 RT log 10 Kp (4.46)
Problem 4.16 : State whether following
reactions are spontaneous or not. Further
state whether they are exothermic or
endothermic.
a. ∆H = -110 kJ and ∆S = +40 JK-1 at 400 K
b. ∆H = +50 kJ and ∆S = -130 JK-1 at 250 K
Solution :
a. ∆G = ∆H - T∆S
∆H = -110 kJ, ∆S = +40 J K-1
= +40 × 10-3 kJ K-1, T = 400 K
Therefore, ∆G = -110 kJ -400 K × 40
×10-3 kJ K-1
= -110 kJ - 16 kJ = -126 kJ
Since ∆G is negative, the reaction is
spontaneous. It is exothermic since ∆H is
negative
b. ∆H = +50 kJ, ∆S = -130 ×J K-1(^) = -130 ×10-3 kJ K-1 T = 250 K
∆G = +50 kJ - 250 K ×(-130 ×10-3 kJ K-1)
= 50 kJ + 32.5 kJ = +82.5 kJ
As ∆G is positive, the reaction is
nonspontaneous. It is endothermic since
∆H is positive.
Problem 4.17
For a certain reaction ∆H^0 is -224 kJ
and ∆S^0 is -153 J K-1. At what temperature
the change over from spontaneous to
nonspontaneous will occur?
Solution -
T =
∆H^0
∆S^0
∆H^0 = -224 kJ, ∆S^0 = -153 JK-1 = -0.153
kJ K-1
Therefore, T =
-224 kJ
-0.153 J K-1 = +1464 K
Since ∆H^0 and ∆S^0 are both
negative, the reaction is spontaneous at low
temperatures. A change over will occur at
1464 K. The reaction is spontaneous below
1464 K.
Problem 4.18
For the reaction,
CH 4 (g) + H 2 (g) C 2 H 6 (g),
Kp = 3.356 × 10^17
Calculate ∆G^0 for the reaction at 25^0 C.
Solution :
∆G^0 = -2.303 RT log 10 Kp
R = 8.314 J K-1mol-1, T = 298 K,
Kp = 3.356 × 10^17
∆G^0 = -2.303×8.314 × 298 ×
log 10 (3.356 × 10^17 )
= -2.303 × 8.314 J mol-1 × 298 × 17.526
= -100,000 J mol-1
= -100 kJ mol-1