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

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 :

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

$T < \frac{\bigtriangleup H}{\bigtriangleup S}$

$T < \frac{-92\frac{KJ}{mol}}{-0.199\frac{KJ}{mol.K}}$

$T < (\frac{-92}{-0.199})\times (\frac{KJ}{mol})\times (\frac{mol.K}{KJ})$

$T < 462.3 K$

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.

Answer 2

The answer is C which is 400k.