Arrhenius equation is $k=A e^{-E_a / R T}$
$\therefore \quad \ln \mathrm{k}=-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}+\ln \mathrm{A}$
$\therefore \quad \log _{10} \mathrm{k}=-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}} \frac{1}{\mathrm{~T}}+\log _{10} \mathrm{~A}$
Thus, the slope of the line is $-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}$
