Question

An uncatalyzed reaction has a rate of 0.0000048 sec–1 at room temperature (25 °C). When an enzyme is added the rate is 32955 sec–1. If the height of the activation barrier for the uncatalyzed rate is 27.9 kcal/mol, what is the height of the activation barrier for the catalyzed rate? Report your answer in terms of kcal/mol to the nearest tenths. Also, assume the pre-exponential terms for the uncatalyzed and catalyzed reactions are the same.

Answer 1

I'm not quite sure if the height of activation barrier is the same of activation energy Ea. I'm looking in different sources and I can't fin another equation or formula refering to height of activation, so, I'll use instead that formula to give you the idea of how to do it. If we have to use another formula then put it, and I'll do it for you.

the general equation for the rate would be:

K = Ae^-Ea/RT

for the uncatalized reaction:

K1 = Ae^-Ea1/RT

For the catalized reaction:

K2 = Ae^-Ea2/RT

so:

K1/K2 = A/A e^Ea2-Ea1/RT

ln K1/K2 = Ea2 - Ea1/RT

ln K1/K2 x RT + Ea1 = Ea2

Ea2 = ln (0.0000048/32955) x 8.3143 x 298 + (27.9 x 4.184)

Ea2 = 60.62 kJ/mol

Ea2 = 60.64 / 4.184

Ea2 = 14.48 kcal/mol