If the equilibrium constant for a reaction is $10$,then the value of $\Delta G^o$ is ....... $(R = 8 \, J \, K^{-1} \, mol^{-1}, T = 300 \, K)$

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

Calculate $\Delta G^{\circ} \ (kJ/mol)$ at $127 \ ^{\circ}C$ for a reaction with $K_{equilibrium} = 10^5$.

The $\Delta G^o$ for the reaction $X + Y \rightleftharpoons Z$ is $-4.606 \ kcal$. The value of the equilibrium constant of the reaction at $227 \ ^oC$ is $(R = 2.0 \ cal \ mol^{-1} K^{-1})$.

Assuming ideal behaviour,the magnitude of $\log \, K$ for the following reaction at $25^{\circ} C$ is $x \times 10^{-1}$. The value of $x$ is $......$. (Integer answer)
$3 HC \equiv CH_{(g)} \rightleftharpoons C_6H_{6(\ell)}$
[Given: $\Delta_f G^{\circ}(HC \equiv CH) = -2.04 \times 10^5 \, J \, mol^{-1}$
$\Delta_f G^{\circ}(C_6H_6) = -1.24 \times 10^5 \, J \, mol^{-1}; R = 8.314 \, J \, K^{-1} \, mol^{-1}$]

Calculate the standard Gibbs energy change $(\Delta G^\circ)$ for a gaseous reaction at $298 \text{ K}$ if the equilibrium constant $K_p$ is $3.5 \times 10^{17}$ and the gas constant $R$ is $8.314 \text{ J K}^{-1} \text{mol}^{-1}$.