Q25 / 90 · JEE Main 2022 (29 Jun Shift 2) · Chemistry

The equation $k = (6.5 \times 10^{12} s^{- 1}) e^{- 26000 K / T}$ is followed for the decomposition of compound $A$ . The activation energy for the reaction is ____ $KJ {mol}^{- 1}$ . [nearest integer]
(Given: $R = 8.314 J K^{- 1} {mol}^{- 1}$ )

Correct answer: 216

Solution

$K = {Ae}^{- E_{a} / RT} = (6.5 \times 10^{12} s^{- 1}) e^{- 26000 K / T}$

$\frac{Ea}{8.314} = 26000$

$Ea = 216.164 kJ / mol$ .

Hence,the activation energy of the given reaction is $216 KJ {mol}^{- 1_{}}$

Hence, the correct answer is 216.