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JEE Main 202524 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2025 · 24 January · Evening Shift · Chemistry · Chemical Kinetics

Consider a complex reaction taking place in three steps with rate...

Consider a complex reaction taking place in three steps with rate constants k1,k2k_1, k_2 and k3k_3 respectively. The overall rate constant kk is given by the expression k=k1k3k2k=\sqrt{\frac{k_1 k_3}{k_2}}. If the activation energies of the three steps are 60,30 and 10 kJ mol−110 \mathrm{~kJ} \mathrm{~mol}{ }^{-1} respectively, then the overall energy of activation in kJmol−1\mathrm{kJ} \mathrm{mol}^{-1} is ……\ldots\ldots . (Nearest integer)

Correct answer

0

Step-by-step solution

k1,k2\mathrm{k}_1, \mathrm{k}_2 and k3\mathrm{k}_3 are given as rate constants of three steps of a complex reaction. Rate constant (k) of the overall reaction is given as k=k1k3k2k=\sqrt{\frac{k_1 k_3}{k_2}} Activation energies of the three steps are given as Ea1=60 kJ mol−1,Ea2=30 kJ mol−1,Ea3=10 kJ mol−1\begin{aligned} & \mathrm{E}_{\mathrm{a}_1}=60 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{E}_{\mathrm{a}_2}=30 \mathrm{~kJ} \mathrm{~mol}^{-1}, \\ & \mathrm{E}_{\mathrm{a}_3}=10 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} From Arrhenius equation, we know that k=Ae−Ea/RTT\mathrm{k}=\mathrm{Ae}^{-\mathrm{Ea} / \mathrm{RT} T} If EaE_a is the activation energy of the overall reaction, then Ea=12[Ea1+Ea3−Ea2]=12[60+10−30]=20 kJ mol−1\begin{aligned} & E_a=\frac{1}{2}\left[E_{a_1}+E_{a_3}-E_{a_2}\right] \\ & =\frac{1}{2}[60+10-30]=20 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}
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