At which temperature would a reaction with h -92 kj/mol, s -0.199 kj/(mol-k) be spontaneous? a.600k b.500k c.400k d.700k

The answer is C which is 400k.

Given information : H = -92 KJ/mol and S = -0.199 KJ/(mol.K)

At equilibrium G = 0

We have to find the Temperature at which reaction would be spontaneous.

For spontaneous reaction :

For non-spontaneous reaction :

We can find the temperature using the formula for Gibbs free energy which is:

Where, G = Gibbs free energy ,

H = Enthalpy

S = Entropy

T = Temperature

By plugging the value of G , H and S in the above formula we can find 'T'

Since reaction should be spontaneous that means should be negative , so the above formula can be written as :

On rearranging the above formula we get :

For the reaction to be spontaneous , T should be less than 462.3 K, so out of given option , C is correct which is 400 K.

108

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