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

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

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

At $298 \ K$,$\Delta_r G^{\ominus}$ for the following reaction is $165.469 \ kJ \ mol^{-1}$. What is the equilibrium constant for this reaction? $(R = 8.3 \ J \ mol^{-1} \ K^{-1})$
$\frac{3}{2} O_{2(g)} \longrightarrow O_{3(g)}$

If the equilibrium constant of a process is $3.8 \times 10^{-3}$ at $25^{\circ} C$, what is the standard free energy change of the process? $(R = 8.314 \ J \ mol^{-1} \ K^{-1}, \log 0.0038 = -2.42)$

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 $R = 8.314 \ J \ K^{-1} \ mol^{-1}$; $2.303 \times 8.314 \times 298 = 5705$)