Question

For the reaction at 298 K : 2A + B → C 

∆H = 400 J mol^-1 ∆S = 0.2 JK^-1 mol^-1 

Determine the temperature at which the reaction would be spontaneous.

Answer 1

Given, 

T = 298K 

∆H = 400 J mol^-1 

∆S = 0.2 JK^-1 mol 

∆G = ∆H – ∆S

if T = 2000K 

∆G = 400 – (0.2 x 2000) = 0

 if T > 2000 K 

∆G will be negative. 

The reaction would be spontaneous only beyond 2000 K.