Question

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

Answer 1

The answer is C which is 400k.

Answer 2

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 :
`\triangle G = negative (-)`

For non-spontaneous reaction :
`\triangle G = positive (+)`

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

`\triangle G = \bigtriangleup H - T\bigtriangleup S`

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'

`\triangle G = \bigtriangleup H - T\bigtriangleup S`

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

`\triangle G < \bigtriangleup H - T\bigtriangleup S`

On rearranging the above formula we get :

`0 < \bigtriangleup H - T\bigtriangleup S`

`T < (\bigtriangleup H)/(\bigtriangleup S)`

`T < (-92(KJ)/(mol))/(-0.199(KJ)/(mol.K))`

`T < ((-92)/(-0.199))* ((KJ)/(mol))* ((mol.K)/(KJ))`

`T < 462.3 K`

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.