Question

The rate constant for a first order reaction is given by the following equation:
$\text{ln k}=33.34-\frac{2.0\times {10}^4}{T}K$
The activation energy for the reaciton is given by $kJ{\text{mol}}^{-1}$. (In nearest integer)
(Given: $R=8.3J{K}^{-1}mo{l}^{-1}$

Answer 1

$\text{ln k}=33.24-\frac{2.0\times {10}^4}{T}$
Comparing with
$\text{ln k}=\text{ln A}-\frac{E_a}{RT}$
$\therefore \frac{E_a}{R}=2\times {10}^4$
$E_a=2\times {10}^4\times 8.3J/\text{mol}$
$E_a=166kJ/\text{mol}$