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

The equilibrium constant for a reaction is $10$. What will be the value of $\Delta G^{\theta}$? $R = 8.314 \, J \, K^{-1} \, mol^{-1}, T = 300 \, K$

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: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$,$\ln 10 = 2.3$,$\log 3 = 0.48$,$\log 2 = 0.30$)

$2O_{3(g)} \rightleftharpoons 3O_{2(g)}$
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 $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ ]

At $320 \ K,$ a gas $A_2$ is $20 \%$ dissociated to $A_{(g)}.$ The standard free energy change at $320 \ K$ and $1 \ atm$ in $J \ mol^{-1}$ is approximately $(R = 8.314 \ J \ K^{-1} \ mol^{-1}; \ ln \ 2 = 0.693; \ ln \ 3 = 1.098).$

If $\Delta_fG^o [X_{(l)}] = -65 \, kcal \, mol^{-1}$ and $\Delta_fG^o [X_{(g)}] = -60.4 \, kcal \, mol^{-1},$ the vapour pressure of $X$ at $500 \, K$ would be about ...... $atm$.
Given: $R = 2 \, cal \, K^{-1} \, mol^{-1}$,$\ln \, a = 2.3 \, \log \, a$.