Question

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the rate constant for the reaction increases by a factor of 2.50 ✕ 103 as compared with the uncatalyzed reaction. Assuming the frequency factor A is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.

Answer 1

Use the Arrhenius equation.
ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)
I would use k1 = 1 and k2 = 2.5E3. Solve for Ea

Alternatively, you may use the 50 as Ea for the uncatalyzed reaction and calculate k1. Then multiply k1 by 2.50E3, use that for k2 and solve for the new Ea. The first way means solving that equation just once and is a shortcut. Try it both ways to prove it to yourself.
Post your work if you get stuck.