The slope of Arrhenius Plot (ln k v/s 1/T) 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 = 8.314JK^-1mol^-1]

The correct option is 41.5 kJ mol⁻¹.

To find the activation energy ($E_a$) of the first-order reaction, we use the Arrhenius equation:

$k=A{e}^{-\frac{E_a}{RT}}$

Taking the natural logarithm (ln) on both sides of the equation, we get:

$\ln k=\ln A-\frac{E_a}{RT}$

We can rearrange this equation to highlight the relationship between $\ln k$ and $\frac{1}{T}$:

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

This is in the form of a straight-line equation, $y=mx+c$, where:
• $y=\ln k$
• $x=\frac{1}{T}$
• Slope, $m=-\frac{E_a}{R}$
• Intercept, $c=\ln A$

According to the problem, the slope of the Arrhenius plot ($\ln k$ versus $\frac{1}{T}$) is $-5\times {10}^3\text{K}$. Therefore, we can equate the theoretical slope to this given value:

$-\frac{E_a}{R}=-5\times {10}^3\text{K}$

Solving for $E_a$:

$E_a=5\times {10}^3\text{K}\times R$

Given the universal gas constant $R=8.314{\text{J K}}^{-1}{\text{mol}}^{-1}$, substitute its value into the equation:

$E_a=5\times {10}^3\text{K}\times 8.314{\text{J K}}^{-1}{\text{mol}}^{-1}$

$E_a=41570{\text{J mol}}^{-1}$

To convert the unit from Joules (J) to kilojoules (kJ), divide the result by 1000:

$E_a=\frac{41570}{1000}{\text{kJ mol}}^{-1}=41.57{\text{kJ mol}}^{-1}\approx 41.5{\text{kJ mol}}^{-1}$