The slope of Arrhenius Plot $\left(\text{In k v/s}\frac{1}{T}\right)$of first order reaction is –5 × 10³ K. The value of E_a of the reaction is. Choose the correct option for your answer.

[Given R = 8.314 JK^–1 mol^–1]

The correct option is 41.5 kJ mol⁻¹.

To find the activation energy ($E_a$) of the reaction, we can use the Arrhenius equation:
$k=A{e}^{-\frac{E_a}{RT}}$
Taking the natural logarithm ($\ln$) on both sides, we get:
$\ln k=\ln A-\frac{E_a}{RT}$
This equation can be rearranged into the slope-intercept form of a straight line, $y=mx+c$:
$\ln k=\left(-\frac{E_a}{R}\right)\left(\frac{1}{T}\right)+\ln A$
Here, plotting $\ln k$ on the y-axis against $\frac{1}{T}$ on the x-axis gives a straight line with a slope ($m$) equal to:
$\text{Slope}=-\frac{E_a}{R}$

According to the problem statement:
- Slope of the Arrhenius plot = $-5\times {10}^3\text{K}$
- Universal gas constant, $R=8.314{\text{J K}}^{-1}{\text{mol}}^{-1}$

Equating the formula for the slope to the given value:
$-\frac{E_a}{R}=-5\times {10}^3\text{K}$
Multiply both sides by -1 to solve for $E_a$:
$E_a=5\times {10}^3\text{K}\times R$
Substitute the value of $R$:
$E_a=5\times {10}^3\text{K}\times 8.314{\text{J K}}^{-1}{\text{mol}}^{-1}$
$E_a=41570{\text{J mol}}^{-1}$

To convert the activation energy from Joules ($\text{J}$) to kilojoules ($\text{kJ}$), divide the result by 1000:
$E_a=\frac{41570}{1000}{\text{kJ mol}}^{-1}\approx 41.5{\text{kJ mol}}^{-1}$

Therefore, the activation energy of the reaction is 41.5 kJ mol⁻¹.