For a simple reaction, S 1 P, k1 is the rate of the forward reaction and k-1 is the rate of the backwards reaction. The k1 in the absence of any enzyme was 0.01/s. The k1 in the presence of enzyme was 5,000,000/s. Calculate the rate enhancement (k1 catalyzed divided by k1 uncatalyzed) provided by the enzyme. If the equilibrium constant in the absence of enzyme is 1 x 104, what is the equilibrium constant (Keq) in the presence of enzyme?
For a simple reaction, S 1 P, k1 is the rate of the forward reaction and k-1 is the rate of the backwards reaction. The k1 in the absence of any enzyme was 0.01/s. The k1 in the presence of enzyme was 5,000,000/s. Calculate the rate enhancement (k1 catalyzed divided by k1 uncatalyzed) provided by the enzyme. If the equilibrium constant in the absence of enzyme is 1 x 104, what is the equilibrium constant (Keq) in the presence of enzyme?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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For a simple reaction, S 1 P, k1 is the rate of the forward reaction and k-1 is the rate of the backwards reaction. The k1 in the absence of any enzyme was 0.01/s. The k1 in the presence of enzyme was 5,000,000/s.
Calculate the rate enhancement (k1 catalyzed divided by k1 uncatalyzed) provided by the enzyme.
If the equilibrium constant in the absence of enzyme is 1 x 104, what is the equilibrium constant (Keq) in the presence of enzyme?
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