JEE MAIN - Chemistry (2022 - 27th June Morning Shift - No. 15)
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
166
Explanation
$\ln \mathrm{k}=\ln \mathrm{A}-\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{RT}}$
Given: $\ln k=33.24-\frac{2.0 \times 10^4}{\mathrm{~T}}$
$$ \begin{aligned} &\therefore \text { on comparing } \frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{R}}=2.0 \times 10^4 \\\\ &\therefore \mathrm{E}_{\mathrm{A}}=2.0 \times 10^4 \times \mathrm{R} \\\\ &\Rightarrow \mathrm{E}_{\mathrm{A}}=2.0 \times 10^4 \times 8.3 \mathrm{~J} \\\\ &\Rightarrow \mathrm{E}_{\mathrm{A}}=16.6 \times 10^4 \mathrm{~J}=166 \mathrm{~kJ} \end{aligned} $$
Given: $\ln k=33.24-\frac{2.0 \times 10^4}{\mathrm{~T}}$
$$ \begin{aligned} &\therefore \text { on comparing } \frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{R}}=2.0 \times 10^4 \\\\ &\therefore \mathrm{E}_{\mathrm{A}}=2.0 \times 10^4 \times \mathrm{R} \\\\ &\Rightarrow \mathrm{E}_{\mathrm{A}}=2.0 \times 10^4 \times 8.3 \mathrm{~J} \\\\ &\Rightarrow \mathrm{E}_{\mathrm{A}}=16.6 \times 10^4 \mathrm{~J}=166 \mathrm{~kJ} \end{aligned} $$
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