PYQ NEET- Chemical Thermodynamics L-4

Question: Hydrolysis of sucrose is given by the following reaction.#

Sucrose $+\mathrm{H}2 \mathrm{O} \rightleftharpoons$ Glucose + Fructose If the equilibrium constant $\left(\mathrm{K}{\mathrm{c}}\right)$ is $2 \times 10^{13}$ at $800 \mathrm{~K}$, the value of $\Delta r G^{\Theta}$ at the same temperature will be :

A) $8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K} \times \operatorname{In}\left(2 \times 10^{13}\right)$

B) $8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K} \times \operatorname{In}\left(3 \times 10^{13}\right)$

C) $-8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K} \times \operatorname{In}\left(4 \times 10^{13}\right)$

D) -8.314 $\mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K} \times \operatorname{In}\left(2 \times 10^{13}\right)$

Answer: -8.314 $\mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K} \times \operatorname{In}\left(2 \times 10^{13}\right)$#

Solution:#

$$ \Delta G=\Delta G^{\circ}+\mathrm{RT} \ln \mathrm{Q} $$

At equilibrium $\Delta G=0, \mathrm{Q}=\mathrm{K}_{\mathrm{eq}}$

So, $\Delta_r G^{\circ}=-\mathrm{RT}$ in $\mathrm{K}_{\text {eq }}$ $$ \Delta_r G^{\circ}=-8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K} \times \ln \left(2 \times 10^{13}\right) $$