The rate constant for a first order reaction is given by the following equation:
$\ln k = 33.24 - {{2.0 \times {{10}^4}\,K} \over T}$
The activation energy for the reaction is given by ____________ kJ mol$-$1. (In nearest integer) (Given : R = 8.3 J K$-$1 mol$-$1)
Answer (integer)
166
Solution
$\ln \mathrm{k}=\ln \mathrm{A}-\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{RT}}$<br/><br/>
Given: $\ln k=33.24-\frac{2.0 \times 10^4}{\mathrm{~T}}$<br/><br/>
$$
\begin{aligned}
&\therefore \text { on comparing } \frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{R}}=2.0 \times 10^4 \\\\
&\therefore \mathrm{E}_{\mathrm{A}}=2.0 \times 10^4 \times \mathrm{R} \\\\
&\Rightarrow \mathrm{E}_{\mathrm{A}}=2.0 \times 10^4 \times 8.3 \mathrm{~J} \\\\
&\Rightarrow \mathrm{E}_{\mathrm{A}}=16.6 \times 10^4 \mathrm{~J}=166 \mathrm{~kJ}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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