JEE MAIN - Chemistry (2022 - 28th July Evening Shift - No. 19)
For a reaction, given below is the graph of $$\ln k$$ vs $${1 \over T}$$. The activation energy for the reaction is equal to ____________ $$\mathrm{cal} \,\mathrm{mol}^{-1}$$. (nearest integer)
(Given : $$\mathrm{R}=2 \,\mathrm{cal} \,\mathrm{K}^{-1} \,\mathrm{~mol}^{-1}$$ )
_28th_July_Evening_Shift_en_19_1.png)
Answer
8
Explanation
$\begin{aligned}
&\mathrm{K}=\mathrm{Ae}^{-\mathrm{Ea} / \mathrm{RT}} \\\\
&\ln \mathrm{k}=\frac{-\mathrm{Ea}}{\mathrm{RT}}+\ln \mathrm{A} \\\\
&\text { Slope }=\frac{\mathrm{Ea}}{\mathrm{R}}=\frac{20}{5} \\\\
&\mathrm{E}_{\mathrm{a}}=4 \mathrm{R}=8 \,\mathrm{Cal} / \mathrm{mol}
\end{aligned}$
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