Find  out  percentage  of the reactant   molecules crossing  over the  energy barrier at 325K. given that delta H₃₂₅= 0.12 kcal , E_a(b)  0.02kcal       

Dear Student,

Fraction of molecules having energy higher than activation energy = $e^{\frac{-E_a}{RT}}$
Ea = activation energy = 0.12+0.02 kcal = 0.14*4184 J = 585.8 J
R = 8.314 J/K.mol
T = temperature = 325 K
fraction = $e^{\frac{-E_a}{RT}}=e^{\frac{-585.8}{8.314*325}}$ = 80.5
Fraction of molecules = 0.85
% of molecules = 85 %

Regards