Question

What percentage of reactant molecules will crossover the energy barrier at $325$ $K$? Heat of reaction is $0.12$ $kcal$ and activation energy of backward reaction is $0.02$ $kcal$.

Answer 1

Activation energy of forward reaction $=0.12-0.02=0.10$ $kcal$
Fraction of molecules which are active or which crossover the energy barrier $\left(\frac{k}{A}\right)=e^{-E/RT}$
${\log }_e\left(\frac{k}{A}\right)=-\frac{E}{RT}$
$\frac{k}{A}=antilog\left[\frac{-E}{2.303RT}\right]$
$=antilog\left[\frac{-0.10\times 1000}{2.303\times 2\times 325}\right]$
$=antilog[-0.06680]=0.8574$
$\therefore$ Percentage of reactant molecules crossing over the barrier $=0.8574\times 100=85.74$