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

For the reaction $A \rightleftharpoons B$,find the value of $log_{10}K$. Given: $\Delta_rH^o_{298\,K} = -54.07\, kJ\, mol^{-1}$,$\Delta_rS^o_{298\,K} = 10\, J\, K^{-1}\, mol^{-1}$,$R = 8.314\, J\, K^{-1}\, mol^{-1}$,$2.303 \times 8.314 \times 298 = 5705$.

Describe the relationship between Gibbs energy change and chemical equilibrium.

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 $, atm$)

At $298 \ K$,for the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the $K_p$ value is $0.98$. Predict whether the reaction is spontaneous or not.

The value of $\log _{10} K$ for a reaction $A \rightleftharpoons B$ is
(Given : $\Delta _{r} H_{298 K}^{\circ} = -54.07 \ kJ \ mol^{-1}$,$\Delta _{r} S_{298 K}^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$; $2.303 \times 8.314 \times 298 = 5705$)