The rate of a reaction doubles when its… - JEE Main 2013 Chemistry PYQ

JEE Main 2013 — Chemistry Physical Chemistry

2013mcqeasy

The rate of a reaction doubles when its temperature changes from $300K\mathrm{to}310K$ . Activation energy of such a reaction will be:

$(\text{R}={\text{8.314 JK}}^{-1}{mol}^{-1}\mathrm{and}log 2=\text{0.301})$

Official previous-year question

Held on 7 Apr 2013 · Verified 6 Jul 2026.

Options

A
$58.5\mathrm{kJ}{\mathrm{mol}}^{-1}$
B
$60.5\mathrm{kJ}{\mathrm{mol}}^{-1}$
C
$53.6\mathrm{kJ}\mathrm{mol}{}^{-1}$
D
$48.6\mathrm{kJ}{\mathrm{mol}}^{-1}$

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Solution

$\mathrm{log}\frac{{k}_{2}}{{k}_{1}}=-\frac{{E}_{a}}{2.303R}(\frac{1}{{T}_{2}}-\frac{1}{{T}_{1}})$

$\text{log}\frac{{\text{k}}_{2}}{{\text{k}}_{1}}=\frac{{\text{E}}_{\text{a}}}{2\text{.}303\text{R}}(\frac{{\text{T}}_{2}-{\text{T}}_{1}}{{\text{T}}_{1}{\text{T}}_{2}})$

$\text{log }2=\frac{{\text{E}}_{\text{a}}}{2\text{.}303\times 8\text{.}314}\times \frac{10}{300\times 310}$

${E}_{a}=\frac{0\text{.}301\times 2\text{.}303\times 8\text{.}314\times 300\times 310}{10}=53603.9J/\mathrm{mol}$

${E}_{a}={\text{53.6 kJ mol}}^{-1}$ .

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