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 : [tex] \triangle G = negative (-)[/tex]
For non-spontaneous reaction : [tex] \triangle G = positive (+)[/tex]
We can find the temperature using the formula for Gibbs free energy which is:
[tex]\triangle G = \bigtriangleup H - T\bigtriangleup S[/tex]
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'
[tex]\triangle G = \bigtriangleup H - T\bigtriangleup S[/tex]
Since reaction should be spontaneous that means [tex]\triangle G[/tex] should be negative , so the above formula can be written as :
[tex]\triangle G < \bigtriangleup H - T\bigtriangleup S[/tex]
On rearranging the above formula we get :
[tex] 0 < \bigtriangleup H - T\bigtriangleup S [/tex]
[tex] T < \frac{\bigtriangleup H}{\bigtriangleup S} [/tex]
[tex] T < \frac{-92\frac{KJ}{mol}}{-0.199\frac{KJ}{mol.K}} [/tex]
[tex] T < (\frac{-92}{-0.199})\times (\frac{KJ}{mol})\times (\frac{mol.K}{KJ}) [/tex]
[tex] T < 462.3 K [/tex]
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.
