Question

For an equilibrium reaction, AG = 0 kJ at 327.0 K. If the standard enthalpy of this reaction is -46.9 kJ, and the standard entropy is 107.0 J/K, what is the equilibrium constant? Multiple tries are permitted; however, 20% (1/5) point will be deducted for each incorrect response.

Answer 1

ΔG is 0 at 327 K.
So, reaction is at equilibrium at 327 K

Step 1:
ΔHo = -46.9 KJ
ΔSo = 107 J/K
= 0.107 KJ/K
T = 327 K

use:
ΔGo = ΔHo - T*ΔSo
ΔGo = -46.9 - 327.0 * 0.107
ΔGo = -81.889 KJ

Step 2:

T = 327 K
ΔGo = -81.889 KJ/mol
ΔGo = -81889 J/mol

use:
ΔGo = -R*T*ln Kc
-81889 = - 8.314*327.0* ln(Kc)
ln Kc = 30.1209
Kc = 1.206*10^13
Answer: 1.21*10^13