Find the equilibrium constant for the reaction below at $25$ $^{\circ}C$:
$H_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons H_{2}O_{(g)}$ [ $\Delta_{f}G^{o} = -54.64 \ kcal$ ].

The equilibrium constant for the following reaction is $K_{p} = 3.44 \times 10^{24}$ at $25 ^{\circ}C$. Calculate the value of $\Delta_{f}G^{o}(SO_2)$. Given that the value of $\Delta_{f}G^{o}(SO_3)$ is $-88.52 \ kcal/mol$. The reaction is $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.

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: $R = 8.314 \ J \ mol^{-1} \ K^{-1}$)

For the reaction at $25\,^oC$,${N_2O_4}_{(g)} \rightleftharpoons 2NO_{2_{(g)}}$,if $\Delta G_f^o$ for $N_2O_4$ and $NO_2$ are $23.49 \, KCal$ and $12.39 \, KCal$ respectively,then $K_p$ for the reaction is: (in $, atm$)

Calculate the standard Gibbs energy change $(\Delta G^\circ)$ for a gaseous reaction at $298 \text{ K}$ if the equilibrium constant $K_p$ is $3.5 \times 10^{17}$ and the gas constant $R$ is $8.314 \text{ J K}^{-1} \text{mol}^{-1}$.

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