JEE MAIN - Chemistry (2025 - 4th April Evening Shift - No. 2)
Consider the following plots of $\log$ of rate constant $\mathrm{k}(\log \mathrm{k})$ vs $\frac{1}{\mathrm{~T}}$ for three different reactions. The correct order of activation energies of these reactions is :
_4th_April_Evening_Shift_en_2_1.png)
$\mathrm{Ea}_2>\mathrm{Ea}_1>\mathrm{Ea}_3$
$\mathrm{Ea}_1>\mathrm{Ea}_3>\mathrm{Ea}_2$
$\mathrm{Ea}_3>\mathrm{Ea}_2>\mathrm{Ea}_1$
$\mathrm{Ea}_1>\mathrm{Ea}_2>\mathrm{Ea}_3$
Explanation
$$\begin{aligned} & \mathrm{K}=\mathrm{A} \mathrm{e}^{-\mathrm{EaRT}} \\ & \operatorname{logk}=\log \mathrm{A}-\frac{\mathrm{Ea}}{2.303 \mathrm{RT}} \end{aligned}$$
For graph between logk with $\frac{1}{\mathrm{~T}}$
$$\mid \text { Slope of curve } \left\lvert\,=\frac{\mathrm{Ea}}{2.303 \mathrm{R}}\right.$$
From given graph
Magnitude of slope $\Rightarrow(2)>(1)>(3)$
Hence $\mathrm{Ea}_2>\mathrm{Ea}_1>\mathrm{Ea}_3$
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