The Arrhenius plots of two reactions, I and II are shown graphically
The graph suggests that:

Question

Professor PC Thomas Classes · Answer

The graphs are for logarithmic form of arrhenius equation for which we have the following relation,
$$
\begin{aligned}
& \ln \mathrm{k}=\ln \mathrm{A}-\left(\frac{\mathrm{E}}{\mathrm{R}}ight) \frac{1}{\mathrm{~T}} \
& 	ext { intercept }=\ln \mathrm{A} ; \
& 	ext { slope }=-\frac{\mathrm{E}}{\mathrm{R}} .
\end{aligned}
$$
From the given graphs, we can observe that,
$$
\begin{aligned}
& \ln \left(A_Iight)>\ln \left(A_{I I}ight) \Rightarrow A_I>A_{I I} \
& \left|\frac{-E_I}{R}ight|>\left|\frac{-E_{I I}}{R}ight| \Rightarrow E_I>E_{I I}
\end{aligned}
$$