1 2 N2 g 1 2 O2 g NO g H 89 KJ mol- If the entr... - JAMB Chemistry 1997 Question
(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?
A
88.71 KJ
B
85.48 KJ
C
-204.00 KJ
D
-3427.40 KJ
correct option: d
∆G = ∆H - T∆S; T = 2s° + 273 = 298 k
∆G = 89 - 298 x 11.8 = -3427.4
∆G = 89 - 298 x 11.8 = -3427.4
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