Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below. A + B underset Step 3 stackrel Step 1 rightleftharpoons C xrightarrow Step 2 P Some details of the above reaction are listed below. Step Rate constant ( se c -1 ) Activation energy ( kJmo l -1 ) --- --- --- 1 k 1 300 2 k 2 200 3 k 3 E a 3 if the overall rate constant of the above transformation ( k ) is given as k =frac k 1 k 2 k 3 and the overall activation energy ( E a ) is 400 \!\! \!\! kJ \!\! \!\! mo l -1 , then the value of E a 3 is \!\! \!\! kJmo l -1 (nearest integer)

Chemistry · Chemical Kinetics

JEE Main 2024 — 4 April, Shift 1 — Question 80

Consider the following transformation involving first order elementary reaction in each step at constant temperature as

shown below. $\mathrm{A}+\mathrm{B} \underset{\text { Step } 3}{\stackrel{\text { Step } 1}{\rightleftharpoons}} \mathrm{C} \xrightarrow{\text { Step 2 }} \mathrm{P}$ Some details of the above reaction are listed below.

Step	Rate constant $\left( \text{se}{{\text{c}}^{-1}} \right)$	Activation energy $\left( \text{kJmo}{{\text{l}}^{-1}} \right)$
1	${{\text{k}}_{1}}$	300
2	${{\text{k}}_{2}}$	200
3	${{\text{k}}_{3}}$	$\text{E}{{\text{a}}_{3}}$

if the overall rate constant of the above transformation $\left( \text{k} \right)$ is given as $\text{k}=\frac{{{\text{k}}_{1}}{{\text{k}}_{2}}}{{{\text{k}}_{3}}}$ and the overall activation energy

$\left( {{\text{E}}_{\text{a}}} \right)$ is $400\text{ }\!\!~\!\!\text{ kJ }\!\!~\!\!\text{ mo}{{\text{l}}^{-1}}$ , then the value of $\text{E}{{\text{a}}_{3}}$ is $\text{ }\!\!~\!\!\text{ }$ $\text{kJmo}{{\text{l}}^{-1}}$ (nearest integer)

Answer: 100

Numerical answer — enter this value.

Step-by-step solution

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

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

$\mathrm{E}_{\mathrm{a}}=\mathrm{E}_{\mathrm{a}_{1}}+\mathrm{E}_{\mathrm{a}_{2}}-\mathrm{E}_{\mathrm{a}_{3}}$

$400=300+200-\mathrm{E}_{\mathrm{a}_{3}}$

$\mathrm{E}_{\mathrm{a}_{3}}=100 \mathrm{~kJ} / \mathrm{mole}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Arrhenius Equation