(NOV 2020-1)
15. The rate constant for conversion of a substrate into the product is 10^-4 s^-1, while the reverse
rate constant 10^-6 s^-1. An enzyme enhance the rate of this reaction by 100- fold. The equilibrium constant for this enzyme- catalyzed reaction is
(1) 100                                                           (2) 10000
(3) 10                                                             (4) 1000

The correct answer is (3) 10.

Introduction

Understanding the relationship between rate constants and equilibrium constants in enzyme-catalyzed reactions is critical in biochemistry. Enzymes accelerate reactions but do not alter their equilibrium. This article explains how to calculate equilibrium constants using forward and reverse rate constants and how enzyme-induced rate enhancements affect these values.

Given Data

• Forward rate constant (kfkf) = 10−4 s−110−4s−1

• Reverse rate constant (krkr) = 10−6 s−110−6s−1

• Enzyme enhances rate by 100-fold (both forward and reverse rates)

Step 1: Understand the Basics

The equilibrium constant (KeqKeq) for a reversible reaction:

$$A\rightleftharpoons B$$

is given by the ratio of forward and reverse rate constants:

Keq=kfkrKeq=krkf

This constant describes the ratio of products to reactants at equilibrium and remains unchanged by enzymes as enzymes speed up both reactions equally.

Step 2: Calculate KeqKeq without enzyme

Keq=10−410−6=102=100Keq=10−610−4=102=100

Step 3: Effect of enzyme rate enhancement

Enzymes increase the rate constants by the same factor but do not change the equilibrium constant.

After enzyme enhancement:

kf′=100×10−4=10−2kf′=100×10−4=10−2kr′=100×10−6=10−4kr′=100×10−6=10−4

Calculate the new equilibrium constant with enzyme:

Keq′=kf′kr′=10−210−4=102=100Keq′=kr′kf′=10−410−2=102=100

Step 4: Reconcile with the options

The equilibrium constant remains 100, matching option (1).

Explanation of the Discrepancy in Question

Since the question asks for the equilibrium constant for this enzyme-catalyzed reaction with 100-fold rate enhancement, it’s crucial to realise that equilibrium constant does not change due to enzyme action. The original equilibrium constant KeqKeq was 100.

Conclusion

The enzyme accelerates the reaction rate but does not change the reaction’s equilibrium constant, which remains 100. Hence, the correct answer should be (1) 100.

Additional Notes

• Enzymes speed up both forward and reverse reactions equally, preserving the ratio of products to substrates at equilibrium.

• The equilibrium constant depends only on thermodynamic properties, not kinetics.

• If the question explicitly states rate constants enhanced by enzyme separately, keep in mind the ratio remains constant.

Summary Table

This detailed analysis clarifies the relationship between enzyme action, rate constants, and equilibrium constant, proving essential knowledge for mastering biochemical reaction kinetics.