For a reaction taking place in three steps at same temperature, overall rate constant K=frac K 1 K 2 K 3 . If Ea 1 , Ea 2 and Ea 3 are 40,50 and 60 kJ / mol respectively, the overall Ea is kJ / mol .

Chemistry · Chemical Kinetics

JEE Main 2024 — 29 January, Shift 1 — Question 75

For a reaction taking place in three steps at same temperature, overall rate constant $K=\frac{\mathrm{K}_{1} \mathrm{~K}_{2}}{\mathrm{~K}_{3}}$ . If $\mathrm{Ea}_{1}, \mathrm{Ea}_{2}$ and $\mathrm{Ea}_{3}$ are 40,50 and $60 \mathrm{~kJ} / \mathrm{mol}$ respectively, the overall Ea is $\qquad$ $\mathrm{kJ} / \mathrm{mol}$ .

Answer: 30

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{K}=\frac{\mathrm{K}_{1} \cdot \mathrm{~K}_{2}}{\mathrm{~K}_{3}}=\frac{\mathrm{A}_{1} \cdot \mathrm{~A}_{2}}{\mathrm{~A}_{3}} \cdot \mathrm{e}^{-\frac{\left(\mathrm{E}_{\mathrm{a}_{1}}+\mathrm{E}_{\mathrm{a}_{2}}-\mathrm{E}_{\mathrm{a}_{3}}\right)}{\mathrm{RT}}}$

$A \cdot e^{-E_{a} / R T}=\frac{A_{1} A_{2}}{A_{3}} \cdot e^{-\frac{\left(E_{a_{1}}+E_{a_{2}}-E_{a_{3}}\right)}{R T}}$

$\mathrm{E}_{\mathrm{a}}=\mathrm{E}_{\mathrm{a}_{1}}+\mathrm{E}_{\mathrm{a}_{2}}-\mathrm{E}_{\mathrm{a}_{3}}=40+50-60=30 \mathrm{~kJ} /$ mole.

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Arrhenius Equation