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.

Step-by-Step Explanation:

According to the Arrhenius equation, the rate constant $k$ of a reaction depends on the absolute temperature $T$ as follows:
$k=A{e}^{-\frac{E_a}{RT}}$
where:
• $A$ is the pre-exponential factor,
• $E_a$ is the activation energy,
• $R$ is the gas constant, and
• $T$ is the temperature in Kelvin.

Taking the natural logarithm ($\ln$) on both sides of the equation yields:
$\ln k=\ln A-\frac{E_a}{RT}$

Rearranging the terms to match the equation of a straight line ($y=mx+c$):
$\ln k=\left(-\frac{E_a}{R}\right)·\frac{1}{T}+\ln A$

From this equation, a plot of $\ln k$ versus $\frac{1}{T}$ gives a straight line with:
$\text{Slope}\left(m\right)=-\frac{E_a}{R}$

We are given:
• $\text{Slope}=-5\times {10}^3\text{K}$
• $R=8.314{\text{J K}}^{-1}{\text{mol}}^{-1}$

Equating the theoretical slope to the given value:
$-\frac{E_a}{R}=-5\times {10}^3\text{K}$

Solving for the activation energy $E_a$:
$E_a=5\times {10}^3\text{K}\times 8.314{\text{J K}}^{-1}{\text{mol}}^{-1}$
$E_a=41570{\text{J mol}}^{-1}$

Converting the energy from Joules ($\text{J}$) to kilojoules ($\text{kJ}$):
$E_a=\frac{41570}{1000}{\text{kJ mol}}^{-1}=41.57{\text{kJ mol}}^{-1}\approx 41.5{\text{kJ mol}}^{-1}$

Therefore, the value of the activation energy $E_a$ is 41.5 kJ mol⁻¹.