Write the formula relating the equilibrium constant $K$ and the standard Gibbs free energy change $\Delta G^{\circ}$.

The equilibrium constant of a reaction is $0.008$ at $298 \ K$. The standard free energy change of the reaction at the same temperature is

If the equilibrium constant for a reaction is $10$,then the value of $\Delta G^o$ is ....... $(R = 8 \, J \, K^{-1} \, mol^{-1}, T = 300 \, K)$

The correct relationship between standard free energy change $(\Delta G^o)$ and equilibrium constant $(K)$ is:

The correct relation is:

The standard Gibbs free energy change $\Delta G^{\circ}$ at $25^{\circ} C$ for the dissociation of $N_2O_{4(g)}$ to $NO_{2(g)}$ is (given,equilibrium constant $K_{eq} = 0.15, R = 8.314 \ J \ K^{-1} \ mol^{-1}$) (in $kJ$)