JEE Main 2023 — Chemical Equilibrium Question with Solution

The Gibbs energy change for a reaction at standard conditions, ${\Delta }_r{G}^{\theta }$, can be calculated using the equation:

$\Delta {\mathrm{G}}^{\theta }=-\mathrm{RT}\ln {\mathrm{K}}_{\mathrm{eq}}$

Where ${\mathrm{K}}_{\mathrm{eq}}$ is the equilibrium constant given by $\frac{{\mathrm{K}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{b}}}$. For the reaction:

$\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+\mathrm{D}$

The forward rate constant, $k_f$, is $10^3$, and the reverse rate constant, $k_r$, is $10^2$. Therefore,

${\mathrm{K}}_{\mathrm{eq}}=\frac{10^3}{10^2}=10$

Substitute the values into the Gibbs energy change formula:

$\Delta {\mathrm{G}}^{\theta }=-\mathrm{RT}\ln 10$

where $R=8.3\mathrm{J}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$ and $T=300\mathrm{K}$ (since the temperature is $27^{\circ }\mathrm{C}$ which converts to 300 K). The natural logarithm of 10 is approximately 2.3.

$\Delta {\mathrm{G}}^{\theta }=-(8.3\times 300\times 2.3)$

Calculating gives:

$\Delta {\mathrm{G}}^{\theta }=-5711\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$

Converting to kJ:

$\Delta {\mathrm{G}}^{\theta }=-5.7\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$

Rounding to the nearest integer, the standard Gibb's energy change is approximately $-6\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.