What is the percentage of the reactant molecules crossing over the energy barrier at 325K (nearest integer) given that $∆$H_325K = 0.12Kcal , E_a (backward) = 0.02 Kcal ?

$\begin{array}{l}\mathrm{\Delta H}={\mathrm{Ea}}_{\left(\mathrm{f}\right)}-{\mathrm{Ea}}_{\left(\mathrm{b}\right)};0.12={\mathrm{Ea}}_{\left(\mathrm{f}\right)}-0.02\\ {\mathrm{Ea}}_{\left(\mathrm{f}\right)}=140\mathrm{cal}\end{array}$
  Ea​=x=e^{−(Ea​​/RT)}

x=e−(140) ​/ (2×325) 

 R=1.98cal/molek,R≈2cal/mol.k

x=e^−0.2154

ln x=0.2154

ln x=2.303−0.2154​=0.0935

x=antilog(0.0935)

x=0.8063

Percentage=80.63%

% of reactant crossing over the energy barrier