Find out the percentage of the reactant molecules crossing over the energy barrier at $325 \mathrm{~K}$.
Given: $\Delta H_{325 \mathrm{~K}}=0.12 \mathrm{kcal}$ $E_{\mathrm{a}(\mathrm{b})}=0.02 \mathrm{kcal}$

Question

Aicharya · Accepted Answer

$\Delta \mathrm{H}=\mathrm{E}_{\mathrm{a}(\mathrm{f})}-\mathrm{E}_{\mathrm{a}(\mathrm{b})}$
(activation energy of forward reactionaction energy of backward reaction)
$0.02+0.12=\mathrm{E}_{\mathrm{a}(\mathrm{f})}$
$\mathrm{E}_{\mathrm{a}(\mathrm{f})}=0.14 \mathrm{kcal} / \mathrm{mole} . \mathrm{k}=140 \mathrm{cal} / \mathrm{mol} . \mathrm{K}$
Only those molecules will cross the activation energy barrier which possess energy greater than $\mathrm{E}_{\mathrm{a}}$.
Formula:
Fraction of molecules with energy greater
$\operatorname{than} \mathrm{E}_{\mathrm{a}}=\mathrm{x}=\mathrm{e}^{-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}ight)}$
$\mathrm{x}=\mathrm{e}^{-\left(\frac{140}{2 \times 325}ight)}$ Where $\mathrm{R}=1.98 \mathrm{cal} / \mathrm{moleK}, \quad \mathrm{R} \approx$
$2 \mathrm{cal} / \mathrm{mol} . \mathrm{K}$
$\mathrm{x}=\mathrm{e}^{-0.2154}$
$\ln \mathrm{x}=0.2154$
$\ln \mathrm{x}=\frac{-0.2154}{2.303}=0.0935$
$\mathrm{x}=\operatorname{antilog}(0.0935)$
$\mathrm{x}=0.8063$
P ercentage $=80.63 \%$ ^

Find out the percentage of the reactant molecules crossing over the energy barrier at $325 \mathrm{~K}$.…

Solution