Consider the following plots of log of rate constant $\mathrm{k}(\log \mathrm{k})$ vs $\frac{1}{\mathrm{~T}}$ for three different reactions. The correct order of activation energies of these reactions is

Question

Aicharya · Accepted Answer

$\begin{aligned}
& \mathrm{K}=\mathrm{A} \mathrm{e}^{-\mathrm{E} a \mathrm{RT}} \\
& \operatorname{logk}=\log \mathrm{A}-\frac{\mathrm{Ea}}{2.303 \mathrm{RT}}
\end{aligned}$
For graph between logk with $\frac{1}{\mathrm{~T}}$
$\mid \text { Slope of curve } \left\lvert\,=\frac{\mathrm{Ea}}{2.303 \mathrm{R}}\right.$
From given graph
Magnitude of slope $\Rightarrow(2) \gt (1) \gt (3)$
Hence $\mathrm{Ea}_2 \gt \mathrm{Ea}_1 \gt \mathrm{Ea}_3$

Consider the following plots of log of rate constant $\mathrm{k}(\log \mathrm{k})$ vs…

Solution