Decomposition of a hydrocarbon follows the equation k = (5.5 × 10 11 s -1 ) e frac -28000 K T . The activation energy of reaction is \ \ \ \ kJ mol -1 . (Nearest Integer) Given : R =8.3 JK -1 mol -1

Chemistry · Chemical Kinetics

JEE Main 2026 — 6 April, Evening Shift — Question 64

Decomposition of a hydrocarbon follows the equation $\mathrm{k}=\left(5.5 \times 10^{11} \mathrm{~s}^{-1}\right) \mathrm{e}^{\frac{-28000 \mathrm{~K}}{\mathrm{~T}}}$ . The activation energy of reaction is $\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$ . (Nearest Integer) Given : $\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

Answer: 232

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{K}=\left(5.5 \times 10^{11} \mathrm{~S}^{-1}\right) \mathrm{e}^{-\frac{28000 \mathrm{~K}}{\mathrm{~T}}}$ $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$ $\frac{E_{a}}{R}=28000$ $\mathrm{E}_{\mathrm{a}}=\frac{28000 \times 8.3}{1000} \mathrm{~kJ} / \mathrm{mol}=232.4 \mathrm{~kJ} / \mathrm{mol}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Arrhenius Equation