The graph obtained between ln k (k= Rate constant) on y-axis and 1/T on x-axis is a straight line. The slope of it is -4 x10⁴ k. The activation energy of the reaction $(in kJ {mol}^{- 1})$ is $(R = 8.3 {JK}^{- 1} {mol}^{- 1})$

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166

332

382

765

answer is B.

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Detailed Solution

According to the Arrhenius equation:

$\ln k = - \frac{E_{a}}{RT} + \ln A \frac{E_{a}}{R} = 4 \times 10^{4} \frac{E_{a}}{8.3} = 4 \times 10^{4} E_{a} = 33.2 \times 10^{4} J / mol = 33.2 \times 10^{4} \times 10^{- 3} = 332 kJ / mol$

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