For an equilibrium reaction,if $\Delta G^{\circ} = 0$,the equilibrium constant $K$ is equal to:

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

Of the following reactions:
$(i) \, A \rightleftharpoons B, \Delta G^{\circ} = 250 \, kJ \, mol^{-1}$
$(ii) \, D \rightleftharpoons E, \Delta G^{\circ} = -100 \, kJ \, mol^{-1}$
$(iii) \, F \rightleftharpoons G, \Delta G^{\circ} = -150 \, kJ \, mol^{-1}$
$(iv) \, M \rightleftharpoons N, \Delta G^{\circ} = 150 \, kJ \, mol^{-1}$
The reaction with the largest equilibrium constant is:

Calculate $\Delta G^{\circ} \ (kJ/mol)$ at $127 \ ^{\circ}C$ for a reaction with $K_{equilibrium} = 10^5$.

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 standard free energy change $(\Delta G^{\circ})$ for $50\%$ dissociation of $N_2O_4$ into $NO_2$ at $27^{\circ}C$ and $1\,atm$ pressure is $-x\,J\,mol^{-1}$. The value of $x$ is $......$ (Nearest Integer)
[Given: $R = 8.31\,J\,K^{-1}\,mol^{-1}$,$\log 1.33 = 0.1239$,$\ln 10 = 2.3$]