The equilibrium constant of a reaction is $0.008$ at $298 \ K$. The standard free energy change of the reaction at the same temperature is

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: $R = 2 \, cal \, K^{-1} \, mol^{-1}$,$\ln \, a = 2.3 \, \log \, a$.

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

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: $R = 8.31\,J\,K^{-1}\,mol^{-1}$,$\log 1.33 = 0.1239$,$\ln 10 = 2.3$]

Find the equilibrium constant for the reaction below at $25$ $^{\circ}C$:
$H_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons H_{2}O_{(g)}$ [ $\Delta_{f}G^{o} = -54.64 \ kcal$ ].

The correct relation is: