(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?

(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?

A 88.71 KJ
B 85.48 KJ
C -204.00 KJ
D -3427.40 KJ

jamb 1997 chemistry

The correct answer is D. -3427.40 KJ

âG = âH - TâS; T = 2sÂ° + 273 = 298 k

âG = 89 - 298 x 11.8 = -3427.4

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