QJEEM22Ch675f: JEE Main 2022 Chemistry - Class 12 JEEM22Ch675f Solution: The rate constant for a first order reaction is given by the…

Question

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The rate constant for a first order reaction is given by the following equation:

$\ln k = 33.24 - \frac{2.0 \times 10^{4} K}{T}$

The activation energy for the reaction is given by ____________ kJ mol^{$-$ 1}. (In nearest integer) (Given : R = 8.3 J K^{$-$ 1} mol^{$-$ 1})

Accepted Answer

\ln \mathrm{k}=\ln \mathrm{A}-\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{RT}}

Given:

\ln k=33.24-\frac{2.0 \times 10^4}{\mathrm{~T}}

\begin{aligned} ∴ on comparing \frac{E_{A}}{R} = 2.0 \times 10^{4} \\ ∴ E_{A} = 2.0 \times 10^{4} \times R \\ \Rightarrow E_{A} = 2.0 \times 10^{4} \times 8.3 J \\ \Rightarrow E_{A} = 16.6 \times 10^{4} J = 166 kJ \end{aligned}

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