Identify the relation between the standard Gibbs free energy change $\Delta G^{\circ}$ and the equilibrium constant $K_c$ for a chemical reaction.

The correct relationship between the standard free energy change in a reaction and the corresponding equilibrium constant $K_c$ 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? $(R = 8.3 \ J \ mol^{-1} \ K^{-1})$
$\frac{3}{2} O_{2(g)} \longrightarrow O_{3(g)}$

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

In a closed flask at $600 \text{ K}$, one mole of $X_2Y_4(g)$ attains equilibrium as given below: $X_2Y_4(g) \rightleftharpoons 2XY_2(g)$. At equilibrium, $75\%$ of $X_2Y_4(g)$ is dissociated and the total pressure is $1 \text{ atm}$. The magnitude of $\Delta_r G^\circ$ (in $\text{kJ mol}^{-1}$) at this temperature is . . . . . . .

For the reaction taking place at a certain temperature $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$,if the equilibrium pressure is $X \ bar$,then $\Delta_r G^o$ would be :-