Question

The correct relation between equilibrium constant $\left(K\right)$, standard free energy $\left(\Delta G^o\right)$ and temperature $\left(T\right)$ is:

• A. $\Delta G^o=RT\ln K$
• B. $K=e^{\Delta G^o/2.303RT}$
• C. $\Delta G^o=-RT\log K$
• D. $K={10}^{-\Delta G^o/2.303RT}$

Answer 1

The correct option is D $K={10}^{-\Delta G^o/2.303RT}$

Consider a reaction, $A+B\rightleftharpoons C+D$

$\Delta G={\Delta G}^0+RTlnQ$

for equilibrium, $\Delta G=0$

$\therefore 0={\Delta G}^0+RTlnK$

$\therefore {\Delta G}^0=-RTlnK$

$\therefore {\Delta G}^0=-2.303RTlogK$

i.e. $K={10}^{{-\Delta G}^0/2.303RT}$