Question

The activation barrier for an uncatalyzed reaction is estimated to be 15.3 kcal/mol. The activation barrier for the catalyzed reaction is estimated to be 8.7 kcal/mol. How many times faster is the catalyzed rate versus the uncatalyzed rate? In other words, by what factor/coefficient do you have to multiply the uncatalyzed rate to equal the catalyzed rate? Assume the temperature is 298 K, and enter your answer to the nearest ones.

Answer 1

Given data: Let,

For Uncatalysed reaction rate constant be K1 and Activation energy be E1 and E1 = 15.3 kcal/mol

For catalysed reaction rate constant be K2 and activation energy be E2 and E2 = 8.7 kcal/mol.

The rate constant given by Arrhenius equation is, K = A x e^{-(E / RT)} .

Hence for catalysed reaction, K1 = A x e^{-(E1 / RT)}...........(I)

and for uncatalysed reaction,  K2 = A x e^{-(E2 / RT)}...........(II)

By Eq (I) / (II) we get

K1/K2 = e^{-(E1 / RT)} / e^{-(E2 / RT)}

K1/K2 = e(^E2-E1/RT)

K1/K2 = e^{[(8.7 - 15.3) x 1000 / (1.987 x 298)]}

K1/K2 = 1.44 x 10^-5.

K1 = K2 x 1.44 x 10^-5.

So, K2 = [1/(1.44 x 10^-5) x K1

K2 = 6.9145 x 10⁴ x K1

Rate of Catalysed reaction = 6.9145 x 10⁴ x Rate of Uncatalysed reaction.

i.e. Catalysed reaction is 6.9145 x 10⁴ = 69148 times faster that Uncatalysed one.