The slope of Arrhenius Plot $\left(\text{In k v/s}\frac{1}{T}\right)$ of first order reaction is −5 x 10³ K. The value of E_a of the reaction is. Choose the correct option for your answer. [Given R = 8.314 JK⁻¹mol⁻¹]

The Arrhenius equation for a first‑order reaction is

$\ln \left(k\right)=-\frac{E}{a}\frac{1}{T}+\ln \left(A\right)$

When this equation is plotted as $\ln \left(k\right)$ versus $\frac{1}{T}$, the straight line has a slope

$m=-\frac{E}{a}$

We are given that the slope of the Arrhenius plot is $-5\times {10}^{3}K$. Taking the magnitude of the slope and solving for the activation energy:

$Ea=-m\times R$

Insert the numerical values (R = 8.314 J K⁻1 mol⁻1):

$Ea=5\times {10}^{3}K \times 8.314J K⁻1 \mathrm{mol}⁻1$

Calculating the product:

$5\times {10}^3 8.314=41570$ J mol⁻1

Convert joules to kilojoules (1 kJ = 1000 J):

$41570/1000=41.57$ kJ mol⁻1

Rounding to three significant figures gives

$41.5$ kJ mol⁻1

This matches the provided correct option.