Chemistry · Chemical Kinetics

JEE Main 2025 — 24 January, Evening Shift — Question 44

Consider a complex reaction taking place in three steps with rate constants $\mathrm{k}_{1}, \mathrm{k}_{2}$ and $\mathrm{k}_{3}$ respectively. The overall rate constant $k$ is given by the expression $k=\sqrt{\frac{k_{1} k_{3}}{k_{2}}}$ . If the activation energies of the three steps are 60,30 and $10 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively, then the overall energy of activation in $\mathrm{kJ} \mathrm{mol}^{-1}$ is___ . (Nearest integer)

Answer: 20

Numerical answer — enter this value.

Step-by-step solution

$K=\sqrt{\frac{K_{1} K_{3}}{\mathrm{~K}_{2}}}$

A. $\mathrm{e}^{-\mathrm{Ea} / R T}=\sqrt{\frac{\mathrm{A}_{1} \mathrm{e}^{-\mathrm{Ea}_{1} / R T} \times \mathrm{A}_{3} \mathrm{e}^{-\mathrm{Ea}} / \mathrm{a}_{3} / \mathrm{RT}}{\mathrm{A}_{2} \mathrm{e}^{-\mathrm{Ea}_{2} / \mathrm{RT}}}}$

By comparinig exponential term

$\frac{E_{a}}{R T}=\frac{1}{2} \times\left(\frac{E_{a_{1}}}{R T}+\frac{E_{a_{3}}}{R T}-\frac{E_{a_{2}}}{R T}\right)$

$\mathrm{E}_{\mathrm{a}}=\left(\mathrm{E}_{\mathrm{a}_{1}}+\mathrm{E}_{\mathrm{a}_{3}}-\mathrm{E}_{\mathrm{a}_{2}}\right) / 2$

$\mathrm{E}_{\mathrm{a}}=(60+10-30) / 2=20 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Ans. 20

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Arrhenius Equation

Given below are two statements: In the light of the above statements, choose the correct answer from the options given below :

Answer: 20

Step-by-step solution

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