Question

The standard Gibbs free energy $\Delta G^o$ is related to equilibrium constant $K_P$ as:

• A. $K_P=-RT\log \Delta G^o$
• B. $K_P={\left[e/RT\right]}^{\Delta G^o}$
• C. $K_P=-\Delta G^o/RT$
• D. $K_P=e^{-\Delta G^o/RT}$

Answer 1

The correct option is D $K_P=e^{-\Delta G^o/RT}$

For a reaction, the Gibb's free energy equation is :

$\Delta G=\Delta G^0+RTln{K}_P$

At equilibrium, $\Delta G=0$

$\therefore {\Delta G}^0=-RTln{K}_P$

$\therefore K_P=e^{-{\Delta G}^0/RT}$