The rate constants of some reactions double with every 10 degree rise in temperature. Assume that a reaction takes place at 336K and 346K. What must the activation energy be for the rate constant to double as described?© BrainMass Inc. brainmass.com October 24, 2018, 11:17 pm ad1c9bdddf
Okay, I am attaching the worked problem in both a Word ...
The activation energy is determined so that the rate constant to the reaction will be doubled from 336K to 346K.
Determining Activation Energy from Rate Constant
Here is the problem I have to solve:
If 10 g of pure carbonic anhydrase catalyzes the hydration of 0.30 g of carbon dioxide in 1 min at 37 degrees C under optimal conditions, what is the turnover number (kcat) of carbonic anhydrase (in units of min-1)?
I have already calculated the turnover number, which is 2.0 x 107 min-1, but I'm kind of lost with two other questions.
(b) From the answer in (a) calculate the activation energy of the enzyme catalyzed reaction (in kJ/mol)
(c) If carbonic anhydrase provides a rate enhancement of 107, what is the activation energy for the uncatalyzed reaction?