[S] >> [Etotal]. The rate of formation of ES equals its rate of breakdown. The rate of breakdown of ES to E + P is much faster than the rate of breakdown of ES back to E + S. KM = [E][S]/[ES] The concentration of ES is small relative to the concentration of Etotal. [Etotal] >> [Efree]. The rate of reaction is equal to k₂[ES]. The velocity of the catalyzed reaction is equal to k₁ [E][S].
[S] >> [Etotal]. The rate of formation of ES equals its rate of breakdown. The rate of breakdown of ES to E + P is much faster than the rate of breakdown of ES back to E + S. KM = [E][S]/[ES] The concentration of ES is small relative to the concentration of Etotal. [Etotal] >> [Efree]. The rate of reaction is equal to k₂[ES]. The velocity of the catalyzed reaction is equal to k₁ [E][S].
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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Which of the following are not assumptions made in the derivation of the Michaelis-Menten equation for a one-substrate, one-product reaction? (Put a check by the inappropriate assumptions.)
![[S] >> [Etotal].
The rate of formation of ES
equals its rate of breakdown.
The rate of breakdown of ES to
E + P is much faster than the
rate of breakdown of ES back to
E + S.
KM = [E][S]/[ES]
The concentration of ES is small
relative to the concentration of
Etotal.
[Etotal] >> [Efree ].
The rate of reaction is equal to
k₂[ES].
The velocity of the catalyzed
reaction is equal to k₁[E][S].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39da671c-e2d4-421a-b335-55bc5205deec%2Fbd50b195-b33b-49b4-a138-f9b8ed74f1ed%2Fmmt0zye_processed.jpeg&w=3840&q=75)
Transcribed Image Text:[S] >> [Etotal].
The rate of formation of ES
equals its rate of breakdown.
The rate of breakdown of ES to
E + P is much faster than the
rate of breakdown of ES back to
E + S.
KM = [E][S]/[ES]
The concentration of ES is small
relative to the concentration of
Etotal.
[Etotal] >> [Efree ].
The rate of reaction is equal to
k₂[ES].
The velocity of the catalyzed
reaction is equal to k₁[E][S].
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