Question

The rate constant for a first order reaction is given by the following equation:
$\ln k=33.24-\frac{2.0\times 10^{4}K}{T}$
The Activation energy for the reaction is given by _______$kJmo{l}^{-1}$. (In Nearest integer)
(Given: $R=8.3J{K}^{-1}mo{l}^{-1}$ )

Answer 1

Answer: 166

$\ln k=\ln A-\frac{E_A}{RT}$
Given: $\ln k=33.24-\frac{2.0\times 10^4}{T}$
$\therefore$ on comparing $\frac{E_A}{R}=2.0\times 10^4$
$\therefore E_A=2.0\times 10^4\times R$
$\Rightarrow E_A=2.0\times 10^4\times 8.3J$
$\Rightarrow E_A=16.6\times 10^{4}J=166kJ$