Question

Suppose that a catalyst lowers the activation barrier of a reaction from 122 kJ/mol to 57...

Suppose that a catalyst lowers the activation barrier of a reaction from 122 kJ/mol to 57 kJ/mol .

By what factor would you expect the reaction rate to increase at 25 ∘C? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.)

Express your answer using two significant figures.

Homework Answers

Answer #1

we know that

K = Ae^(-Ea/RT)

ln K = lnA - (Ea/RT)

now

at two different conditions

ln K1 = lnA - (Ea1/RT)

lnA = lnK1 + (Ea1/RT)

and

ln K2 = ln A - (Ea2/RT)

ln A = lnK2 + (Ea2/RT)

so

we get


lnK1 + (Ea1/RT) = lnK2 + (Ea2/RT)

ln K2 - ln K1 = (Ea1/RT) - (Ea2/RT)

ln (K2/K1) = (Ea1 - Ea2)/RT

given

Ea1 = 122

Ea2 = 57

T = 25 C = 298 kelvin

so

ln (K2/K1) = (122-57) x 1000 / 8.314 x 298

(K2/K1) = 2.4767 x 10^11

so

the rate rate will increase by a factor of 2.4767 x 10^11

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