(1 /(2) N_2(g) + (1) / (2)O_2(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

Correct answer is D

∆G = ∆H - T∆S; T = 2s° + 273 = 298 k
∆G = 89 - 298 x 11.8 = -3427.4

See all Chemistry questions & answers

See more JAMB questions & answers

(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?