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The slope of the Arrhenius Plot $ln\ k\ v/s\ \frac{1}{T}$ the first-order reaction is $-5\times 10^{3}K.$ The value of $E_{a}$ of the reaction is. Choose the correct option for your answer.

$\left [ \text {Given R}=8.314\; \text{JK}^{-1}\text {mol}^{-1} \right ]$

Option: 1
$41.5\; \text {kJ mol}^{-1}$

Option: 2
$83.0\; \text {kJ mol}^{-1}$

Option: 3
$166\; \text {kJ mol}^{-1}$

Option: 4
$-83\; \text {kJ mol}^{-1}$

Answers (1)

best_answer

We know the Arrhenius equation

$\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$

$\ln \mathrm{k}=\ln \mathrm{A}+\ln \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$

$\mathrm{\ln k=\ln A-\frac{E_{a}}{R}\left(\frac{1}{T}\right)}$

The slope of $\left(\ln \mathrm{k\ v} / \mathrm{s}\ \frac{1}{\mathrm{T}}\right)$ of the above equation -

$\mathrm{m=-\frac{E_{a}}{R}}$

Given, m = $-5 \times 10^{3} \mathrm{~K}$ and $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

$\mathrm{-5 \times 10^{3}=-\frac{E_{a}}{8.314}}$

$\mathrm{E_{a}=5 \times 10^{3} \times 8.314 \mathrm{~J} / \mathrm{mol}}$

$\mathrm{E_{a} =41.57 \times 10^{3} \mathrm{~J} / \mathrm{mol}}$

$\mathrm{E_{a}\simeq 41.5 \mathrm{~kJ} / \mathrm{mol}}$

Therefore, the correct option is (1).

Posted by

Sanket Gandhi

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