$2O_{3(g)} \rightleftharpoons 3O_{2(g)}$
At $300 \ K$,ozone is $50\%$ dissociated. The standard free energy change at this temperature and $1 \ atm$ pressure is $(-) \dots \ J \ mol^{-1}$ (Nearest integer).
[Given: $\ln 1.35 = 0.3$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ ]

Consider the reaction $X \rightleftharpoons Y$ at $300 \text{ K}$. If $\Delta H^\circ$ and $K$ are $28.40 \text{ kJ mol}^{-1}$ and $1.8 \times 10^{-7}$ at the same temperature, then the magnitude of $\Delta S^\circ$ for the reaction in $\text{J K}^{-1} \text{ mol}^{-1}$ is . . . . . . . (Nearest integer) (Given: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$, $\ln 10 = 2.3$, $\log 3 = 0.48$, $\log 2 = 0.30$)

The equilibrium constant for a reaction is $100$. What will be the value of standard Gibbs energy change at $298 \text{ K}$? $(R = 8.314 \text{ J K}^{-1} \text{mol}^{-1})$

For the following reaction at $50^\circ$ $C$ and at $2 \text{ atm}$ pressure, $2N_2O_5(g) \rightleftharpoons 2N_2O_4(g) + O_2(g)$. $N_2O_5$ is $50\%$ dissociated. The magnitude of standard free energy change at this temperature is $x$. $x = . . . . . . \text{ J mol}^{-1}$.

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

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