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 $kJ$)

Calculate the standard Gibbs free energy change $\Delta G^o$ at $298 \ K$ for the conversion of oxygen to ozone,given by the reaction: $\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)}$. The equilibrium constant $K_p$ for this conversion is $3 \times 10^{-29}$.

At $227^{\circ} C$,dinitrogen tetraoxide is $60 \%$ dissociated. What is the standard free energy change at this temperature and at $1 \ atm$ pressure?

When a reaction is carried out at standard states,then at equilibrium:

For a hypothetical reversible reaction $\frac{1}{2} A_{2(g)} + \frac{3}{2} B_{2(g)} \rightarrow AB_{3(g)}$; $\Delta H = -20 \, kJ$. If the standard entropies of $A_2, B_2$ and $AB_3$ are $60, 40$ and $50 \, J K^{-1} mol^{-1}$ respectively,at what temperature $(K)$ will the reaction be in equilibrium?

In the equilibrium state,the value of $\Delta G$ is: