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: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter11: Chemical Kinetics: Rates Of Reactions
Section11.7: Reaction Mechanisms
Problem 11.12E
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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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