Question

A reversible reaction is carried out at 500 K where its equilibrium constant is unity. If $\Delta H$ at 500 K is $4.0kJ$, the value of ${\Delta S}^o$ is _______.

Answer 1

We know, $K=exp(-\Delta G^o/RT)$

Given, a reversible reaction is carried out at $500$ K for which, $K=1$

Hence, $\Delta G^o=0$

Now, $\Delta G^o=\Delta H^o-T\Delta S^o$

$\Delta S^o=\frac{\Delta H^o}{T}=\frac{4000}{500}=8J{K}^{-1}mo{l}^{-1}$

Hence, the correct answer is $8$.