Suppose that a catalyst lowers the activation barrier of a reaction from 125 kj/mol to 57 kj/mol. by what factor would you expect the reaction rate to increase at 25 âc? (assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.)

Question

Milo Davila

Chemistry

28 January, 13:50

Suppose that a catalyst lowers the activation barrier of a reaction from 125 kj/mol to 57 kj/mol. by what factor would you expect the reaction rate to increase at 25 âc? (assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.)

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Answers (1)

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Malia Hamilton · Answer 1

We can begin tackling this problem by using the Arrhenius equation which is as follows:

k = Ae^ (-Ea/RT)

k = rate of reaction

A = frequency factor

Ea = activation energy

R = 0.008314 kJ/Kmol

T = 298.15 K

We can solve for the rate using each set of conditions, and then compare the rate of the two reactions:

k₁ = Ae^ (-125 / (0.008314) (298.15))

k₂ = Ae^ (-57 / (0.008314) (298.15)

We can now find the ratio of the rates to determine the factor by which the rate increases:

k₂/k₁ = [e^ (-57 / (0.008314) (298.15)) ]/[e^ (-125 / (0.008314) (298.15)) ]

k₂/k₁ = 8.20 x 10¹¹

The rate of reaction increases by a factor of 8.20 x 10¹¹ by using a catalyst to lower the activation energy.