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.
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
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