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If a reaction follows the Arrhenius equation, the plot $\ln k$
vs $\frac{1}{(\mathrm{RT})}$ gives straight line with a gradient $(-\mathrm{y})$ unit.
The energy required to activate the reactant is:
Options:
vs $\frac{1}{(\mathrm{RT})}$ gives straight line with a gradient $(-\mathrm{y})$ unit.
The energy required to activate the reactant is:
Solution:
1842 Upvotes
Verified Answer
The correct answer is:
y unit
From Arrhenius equation,
$k=A e^{-E_{n} / R T}$
$\ln k=\ln A-\frac{E_{a}}{R T}$
slope $=-y$ (given) $-y=-E_{s}$
$\Rightarrow \mathrm{E}_{\mathrm{a}}=y$
$k=A e^{-E_{n} / R T}$
$\ln k=\ln A-\frac{E_{a}}{R T}$
slope $=-y$ (given) $-y=-E_{s}$
$\Rightarrow \mathrm{E}_{\mathrm{a}}=y$
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