Looking for classes? Ksquare Career Institute, Bengaluru →
Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature, T (K). $\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$ The energy of activation in kcal $mol^{-1}$ for the reaction is: [Given: k in $s^{-1}$, $R = 1.987$ cal $mol^{-1}\,K^{-1}$]
A
12.42
B
14.34
C
18.63
D
24.84
Detailed Solution
$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$
Comparing with $\ln k = \ln A - \frac{E_a}{RT}$:
$\frac{E_a}{R} = 1.25 \times 10^4$
$E_a = 1.25 \times 10^4 \times \frac{1.987}{1000}\ kcal\,mol^{-1}$
$E_a = 24.84\ kcal\,mol^{-1}$