Question

Suppose that a catalyst lowers the activation barrier of a reaction from 128 kJ mol−1kJ mol−1 to 55 kJ mol−1kJ mol−1 .

By what factor would you expect the reaction rate to increase at 25 ∘C∘C?

Answer 1

Rate constant of the reaction when activation energy is 128 kJ/mol is given by (Arrhenius equation):

Where, A is frequency factor, Ea is activation energy and T is temperature. R is gas constant.

Similarly, Rate constant of the reaction when activation energy is 55 kJ/mol is given by (Arrhenius equation):

K₂/K₁ is given by:

Where:

• Ea₁ = 128
• Ea₂ = 55
• R = 8.314 x 10^-3 = 0.008314
• T = 25^oC = 298 K

• That is rate constant becomes 6.25 x 10¹² times that of initial value.
• Since rate constant is directly proportional to the rate of reaction, rate of reaction also become 6.25 x 10¹² times that of initial value of rate (without catalyst).