Question

Some reaction has an activation energy of 44 kJ/mol and a rate constant of 4.0 × 10₋4 s-1 at 37 oC. In the presence of a catalyst, the same reaction has an activation energy of 9.1 kJ/mol. Calculate the rate constant (s-1) of the catalyzed reaction at 37 oC. Enter your answer as an integer.

Answer 1

We have Arrhenius equation in the form log k = log A - ( Ea / 2.303 RT)

Where k is rate constant of reaction, A is frequency factor , Ea is energy of activation and T is temperature of reaction.

Rearranging above equation , we get ( Ea / 2.303 RT) = log A/ k

We have, k = 4.0 x 10 ^-04 s ^-1 , Ea = 44 kJ / mol = 44000 J / mol, T = 37 +273 = 310 K

Hence, ( 44000 J / mol / 2.303 x 8.314 J / K mol x 310 K ) = log A / 4.0 x 10 ^-04 s ^-1

7.413 = log A / 4.0 x 10 ^-04 s ^-1

A / 4.0 x 10 ^-04 s ^-1 = 10 ^7.413 = 2.587 x 10 ⁷

A = 2.587 x 10 ⁷ x 4.0 x 10 ^-04 s ^-1

A = 10350.0 s ^-1

Calculation for k

We have, log k = log A - ( Ea / 2.303 RT)

We have, A = 10350.0 s ^-1 , Ea = 9.1 kJ / mol =9100 J / mol, T = 37 +273 = 310 K

log k = log 10350.0 s ^-1 - ( 9100 J / mol / 2.303 x 8.314 J / K mol x 310 K )

log k = 4.015 s ^-1 - 1.533

log k = 2.482  s ^-1

k= 10 ^2.482 = 303.3 s ^-1 = 3.033 x 10 ² s ^-1