Question

An uncatalyzed reaction has a rate of 4.8 x 10-7 s–1 at room temperature (25°C). When an enzyme is added the rate is 3.3 x 104 s–1. If the height of the activation barrier for the uncatalyzed rate is 28.2 kcal/mol, what is the height of the activation barrier for the catalyzed rate? Report your answer in terms of kcal/mol to the nearest tenths. Also, assume the pre-exponential terms for the uncatalyzed and catalyzed reactions are the same.

Answer 1

Arrhenus equation k = A e^-Ea/RT

In can be written as

In k1 = In A - Ea1/RT         for uncatalysed      --- eq (1)

In k2 = In A - Ea2/RT         for catalysed          ---- eq(2)

Eq(2) - Eq(1) gives,

Ink2 - Ink1 =   (-Ea2/RT) + (Ea1/RT)

In(k2/k1) = [Ea1-Ea2]/RT

[Ea1-Ea2] = RT In(k2/k1)

Ea2 = Ea1 - { RT In(k2/k1)}

      = 28.2 x 10³ - { 1.987 cal/mol/K x 298 K In (3.3 x 10⁴/ 4.8 x 10^-7) }

     = 13424 cal/mol

      = 13.4 kcal/mol

Ea2 = 13.4 kcal/mol

Therefore,

activation barrier for the catalyzed rate = 13.4 kcal/mol