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

At $300 \ K$,$\Delta_{r} G^{\circ}$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $-11.5 \ kJ \ mol^{-1}$. The equilibrium constant at $300 \ K$ is approximately $(R=8.314 \ J \ mol^{-1} \ K^{-1})$.

Consider the following reaction at $298 \ K$.
$\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)} ; K_{P} = 2.47 \times 10^{-29}$.
$\Delta_{r} G^{\ominus}$ for the reaction is $ . . . . . . \ kJ$. (Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

At $298 \ K$,if the standard Gibbs energy change $\Delta_r G^{\ominus}$ of a reaction is $-115 \ kJ \ mol^{-1}$,the value of $\log_{10} K_{p}$ will be $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$.

For the reaction $A_{(g)} \rightarrow B_{(g)},$ the value of the equilibrium constant at $300 \ K$ and $1 \ atm$ is equal to $100.0.$ The value of $\Delta_{r}G^{\circ}$ for the reaction at $300 \ K$ and $1 \ atm$ in $J \ mol^{-1}$ is $-xR,$ where $x$ is ........... (Rounded off to the nearest integer) ($R = 8.31 \ J \ mol^{-1} K^{-1}$ and $\ln 10 = 2.3$)

The standard Gibbs free energy change $\Delta G^o$ is related to the equilibrium constant $K_p$ as: