Chemistry - Chemical Kinetics Question with Solution | TestHub

$\text{log}\text{k}=\text{log}\text{A}-\frac{{\text{E}}_{\text{a}}}{2.303\text{R}\text{T}}$...(1)

Also given$\text{log}\text{k}=6.0-\left(2000\right)\frac{1}{\text{T}}$…(2)

On comparing equations, (1) and (2)$\text{log}\text{A}=6.0\Rightarrow \text{A}={10}^6{\text{s}}^{-1}$

and$\frac{{\text{E}}_{\text{a}}}{2.303\text{R}}=2000;$

$\Rightarrow {\text{E}}_{\text{a}}=2000\times 2.303\times 8.314$

$=38.29\text{kJ}\text{mol}{}^{-1}$