Standard Gibb energy chang |ChemistryEverydayLifeJEEMain2023

JEE MAIN 2023

29-1-23

Question

Consider the following reaction approaching equilibrium at 27ºC and 1 atm pressure The standard Gibb’s energy change (rG°) at 27ºC is (–) _______ kJ mol–1 (Nearest integer). (Given : R = 8.3 J K–1 mol–1 and ln 10 = 2.3)

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Solution

Sol. $$ \begin{aligned} & \because \Delta \mathrm{G}^{\circ}=\mathrm{RT} \text { In } \mathrm{K}_{\mathrm{eq}} \text { and } \mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{K}_{\mathrm{t}}}{\mathrm{~K}_{\mathrm{b}}} \\ & \therefore \quad \mathrm{~K}_{\mathrm{eq}}=\frac{10^3}{10^2}=10 \\ & \therefore \quad \Delta \mathrm{G}=-\mathrm{RT} \ln 10 \\ & \Rightarrow \quad-(8.3 \times 300 \times 2.3)=-5.7 \mathrm{~kJ} \mathrm{~mole}^{-1} \approx 6 \mathrm{~kJ} \mathrm{~mole}^{-1} \text { (nearest integer) } \\ & \text { Ans }=6 \end{aligned} $$

Question Tags

JEE Main

Chemistry

Easy

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