Chemistry - Chemical Kinetics Question with Solution | TestHub

$\mathrm{k}=\mathrm{A}{\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{R}\mathrm{T}}$
${\log }_{\mathrm{e}}\mathrm{k}={\log }_{\mathrm{e}}\mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}\mathrm{T}}$
${\log }_{10}\mathrm{k}={\log }_{10}\mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\times \frac{1}{\mathrm{T}}$
$\mathrm{y}=\mathrm{m}\mathrm{x}+\mathrm{C}$
Plot of log k against 1/T will give straight line with negative slope.