The equilibrium constant for a reaction is $20$. What is the value of $\Delta G^{\circ}$ at $300 \ K$? (Given: $R = 8 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,$\ln(20) \approx 2.996$)

Find the equilibrium constant for the reaction below at $298 \ K$. $2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$ given that $\Delta H^o = 18.4512 \ kcal$ and $\Delta S^o = 29.16 \ cal/K$.

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

Calculate $\Delta G^\circ$ for the conversion of oxygen to ozone $\frac{3}{2} O_{2(g)} \to O_{3(g)}$ at $298 \ K$,if $K_p$ for this conversion is $2.47 \times 10^{-29}$.

When a reaction is carried out at standard states,then at equilibrium:

The standard free energy change $(\Delta G^{\circ})$ for $50\%$ dissociation of $N_2O_4$ into $NO_2$ at $27^{\circ}C$ and $1\,atm$ pressure is $-x\,J\,mol^{-1}$. The value of $x$ is $......$ (Nearest Integer)
[Given: $R = 8.31\,J\,K^{-1}\,mol^{-1}$,$\log 1.33 = 0.1239$,$\ln 10 = 2.3$]