Chemistry - Chemical Kinetics Question with Solution | TestHub

$\text{K}={\text{Ae}}^{-\text{Ea}/\text{(RT)}}$
Apply log on both sides:
$\text{ln k}=\text{ln A}-\frac{{\text{E}}_{\text{a}}}{\text{RT}}$
$\text{log}\text{k}=\text{log}\text{A}-\frac{{\text{E}}_{\text{a}}}{2.303\text{R}}\times \frac{1}{ \text{T}}$
$\text{y}=\text{mx}+\text{C}$
$\text{y}=\text{log}\text{k ,}\text{x}=\frac{1}{\text{T}},\text{Slope (m)}=\frac{-{\text{E}}_{\text{a}}}{2.303\text{R}}$