Verified Solution • 2025–26 Session moderate

QJEEM25Chc520: JEE Main 2025 Chemistry - Class 12 JEEM25Chc520 Solution: Consider a complex reaction taking place in three steps with…

Consider a complex reaction taking place in three steps with rate constants k1,k2\mathrm{k}_1, \mathrm{k}_2 and k3\mathrm{k}_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 _________ . (Nearest integer)

Step-by-Step Verified Solution

amp;K=K1 K3 K2amp; A.e−Ea/RT=A1e−EEa1/RT×A3e−Ea3/RT A2e−EEa2/RT\begin{aligned}& \mathrm{K}=\sqrt{\frac{\mathrm{K}_1 \mathrm{~K}_3}{\mathrm{~K}_2}} \\& \mathrm{~A} . \mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}}=\sqrt{\frac{\mathrm{A}_1 \mathrm{e}^{-E \mathrm{Ea}_1 / \mathrm{RT}} \times \mathrm{A}_3 \mathrm{e}^{-E a_3 / \mathrm{RT}}}{\mathrm{~A}_2 \mathrm{e}^{-E \mathrm{Ea}_2 / \mathrm{RT}}}}\end{aligned}

By comparinig exponential term

amp;EaRT=12×(Ea1RT+Ea3RT−Ea2RT)amp;Ea=(Ea1+Ea3−Ea2)/2amp;Ea=(60+10−30)/2=20 kJ mol−1\begin{aligned}& \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) \\& E_a=\left(E_{a_1}+E_{a_3}-E_{a_2}\right) / 2 \\& E_a=(60+10-30) / 2=20 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}

Ans. 20

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring