The correct relationship between the standard Gibbs free energy change $(\Delta G^o)$ and the equilibrium constant $(K_c)$ for a reaction is .......

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

At $298 \ K$, the equilibrium constant of the process $1.5 O_{2(g)} \rightleftharpoons O_{3(g)}$ is $3 \times 10^{-29}$. The standard free energy change (in $kJ \ mol^{-1}$) of the process is approximately ($R = 8.314 \ J \ mol^{-1} \ K^{-1}$; $\log 3 = 0.47$)

Consider the following reaction at $298 \ K$.
$\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)} ; K_{P} = 2.47 \times 10^{-29}$.
$\Delta_{r} G^{\ominus}$ for the reaction is $ . . . . . . \ kJ$. (Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

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 $3 \times 10^{-29}$.

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