For the reaction at 298 K, 2A + B → C ∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1

+1 vote
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asked Oct 6, 2017 in Chemistry by jisu zahaan (28,760 points) 28 438 1099
edited Oct 6, 2017 by jisu zahaan

For the reaction at 298 K, 

2A + B → C ∆H = 400 kJ mol–1 and

∆S = 0.2 kJ K–1 mol–1 

At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range?

1 Answer

+1 vote
answered Oct 6, 2017 by sforrest072 (157,439 points) 63 451 1294
selected Oct 6, 2017 by sanjeev
 
Best answer

From the expression, 

∆G = ∆H – T∆S 

Assuming the reaction at equilibrium, ∆T for the reaction would be: 

                            (∆G = 0 at equilibrium) 

T = 2000 K 

For the reaction to be spontaneous, ∆G must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K. 

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