Question:
If a reaction follows the Arrhenius equation, the plot lnk vs $\frac{1}{(\mathrm{RT})}$ gives straight line with a gradient $(-\mathrm{y})$ unit. The energy required to activate the reactant is:
Correct Option: , 2
Solution:
From Arrhenius equation,
$\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$
$\ln k=\ln A-\frac{E_{a}}{R T}$
slope $=-y$ (given)
$-y=-\mathrm{E}_{\mathrm{a}}$
$\Rightarrow \mathrm{E}_{\mathrm{a}}=y$
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