The correct relation is:

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

Calculate $\Delta G^\circ$ for the reaction, $CH_4(g) + H_2(g) \rightarrow C_2H_6(g)$ at $298 \text{ K}$, given $K_p = 2 \times 10^{17}$ and $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$.

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 $kJ$)

For the reaction taking place at a certain temperature $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$,if the equilibrium pressure is $X \ bar$,then $\Delta_r G^o$ would be :-

For a reaction at $298 \ K$,the equilibrium constant is ${K_p} = 0.17 \times {10^{12}}$. Find the standard Gibbs free energy change $\Delta {G^\Theta }$. (Given: $R = 8.314 \ J \ mol^{-1} \ K^{-1}$)