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? $(R = 8.314 \ J \ mol^{-1} \ K^{-1}, \log 0.0038 = -2.42)$

For a chemical reaction, the standard Gibbs energy change, $\Delta G^{\circ}$ is $-7.64 \times 10^4 \ J \ mol^{-1}$. What is the value of equilibrium constant $(K)$?

$K_p$ for the conversion of oxygen to ozone at $400 \ K$ is $1.0 \times 10^{-30}$,its standard Gibbs energy change in $kJ \ mol^{-1}$ is approximately

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 $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ ]

The standard Gibbs free energy change $\Delta G^{\circ}$ at $25^{\circ} C$ for the dissociation of $N_2O_{4(g)}$ to $NO_{2(g)}$ is (given, equilibrium constant $K_{eq} = 0.15, R = 8.314 \ J \ K^{-1} \ mol^{-1}$) (in $kJ$)