(1 /(2) N2(g) + (1) / (2)O2(g) → NO(g) ∆H° 89 KJ mol–. If the entropy change for the reaction above at 25°C is 11.8 J mol–, calculate the change in free energy. ∆G°, for the reaction at 25°C?
- 88.71 KJ
- 85.48 KJ
- -204.00 KJ
-
-3427.40 KJ
The correct answer is: D
Explanation
∆G = ∆H - T∆S; T = 2s° + 273 = 298 k∆G = 89 - 298 x 11.8 = -3427.4