#### Question

For the reaction at 298 K,

2A + B → C

Δ*H *= 400 kJ mol^{–1} and Δ*S *= 0.2 kJ K^{–1} mol^{–1}

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

#### Solution 1

From the expression,

Δ*G* = Δ*H* – *T*Δ*S*

Assuming the reaction at equilibrium, Δ*T* for the reaction would be:

`T = (triangle h - triangle G)1/(triangle S)`

= `(triangle H)/ (triangle S) (triangle G = 0 " at equilibrium")`

` = (400 kJmol^(-1))/(0.2 kJ K^(-1) mol^(-1))`

*T* = 2000 K

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

#### Solution 2

As per the Gibbs Helmholtz equation:

ΔG = Δ H- TΔ S For ΔG=0 ; ΔH=TΔS or T=ΔH/ΔS

T = (400 KJ mol^{-1})/(0.2 KJ K^{-1} mol^{-1}) = 2000 k

Thus, reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature.