Question

An enzyme catalyzes a reaction such that the rate constant is 10^6 times larger than that for the same reaction in solution at 37 degrees C. Use TST to predict the change in the barrier height (in kJ/mol) for the reaction.

Answer 1

According to Arrhenius equation.
Let k' be the rate constant of the enzyme-catalyzed reaction and k that of the uncatalyzed reaction:
k = A∙e{ -Ea/(R∙T) }
k' = A∙e{ -Ea'/(R∙T) }

Consider the quotient of the rate constants, such that unknown frequency factor cancels out:
k'/k = A∙e{ -Ea/(R∙T) } / A∙e{ Ea'/(R∙T) }
<=>
k'/k = e{ -Ea'/(R∙T) }∙e{ Ea/(R∙T) }
<=>
k'/k = e{ (Ea-Ea')/(R∙T) }

That relation can be solved for difference of the activation energies:
∆Ea = Ea - Ea' = R∙T∙ln(k'/k)
= 8.3145J/molK ∙ 310K ∙ ln(10⁶)
= 35.6 kJ/mol

The enzyme lowers activation energy of reaction by about 35.6kJ/mol