QJEEM23Chd91c: JEE Main 2023 Chemistry - Class 11 JEEM23Chd91c Solution: Consider the following reaction approaching equilibrium at 27 C …

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:

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

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 $R = 8.3 \, \mathrm{J \, K^{-1} \, mol^{-1}}$ 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:

$\Delta \mathrm{G}^\theta = -5711 \, \mathrm{J \, mol^{-1}}$

Converting to kJ:

$\Delta \mathrm{G}^\theta = -5.7 \, \mathrm{kJ \, mol^{-1}}$

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