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

Find out the percentage of the reactant molecules crosisng over the energy barrier at 325 K . Given: Delta H_(325 K) = 0.12 kcal , E_(a(b)) = 0.02 kcal

Activation energy of the reaction AtoB+38KCal is 20KCal. Then activation energy of reaction BtoA bill be :-

Given that the enthalpy change of a reaction at 325 K is 0.12 kcal and energy of activation for the backward reaction is 0.02 kcal, calculate the percentage of the reactant molecules, crossing over the energy barrier.