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The slope of Arrhenius Plot $\ln k$ vs. $\frac{1}{T}$ of first-order reaction is $-5 \times 10^3$ K. The value of $E_a$ of the reaction is: $\newline$ (Given $R = 8.314$ J K $^{-1}$ mol $^{-1})$

medium

ChemicalKinetics

2021

chemistry

$41.5$ kJ mol $^{-1}$

$83.0$ kJ mol $^{-1}$

Explanation

To solve this problem, we need to understand the relationship between the slope of an Arrhenius plot and the activation energy $(E_a). \newline$ The Arrhenius equation is given by: $\newline k = A \cdot e^{-\frac{E_a}{RT}} \newline$

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The slope of Arrhenius Plot ln⁡k \ln k lnk vs. 1T \frac{1}{T} T1​ of first-order reaction is −5×103 -5 \times 10^3 −5×103 K. The value of Ea E_a Ea​ of the reaction is: \newline (Given R=8.314 R = 8.314 R=8.314 J K−1^{-1}−1mol−1)^{-1})−1)

Explanation

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The slope of Arrhenius Plot $\ln k$ vs. $\frac{1}{T}$ of first-order reaction is $-5 \times 10^3$ K. The value of $E_a$ of the reaction is: $\newline$ (Given $R = 8.314$ J K $^{-1}$ mol $^{-1})$