(a) Explanation The mechanism of reaction of enzyme action is given as, E + S ⇌ ES (fast) ES → E + P (slow) The rate law of the slow elementary reaction in the mechanism determines the overall rate of the reaction. Assumed that the rate constant for forward reaction of first elementary step is k 1 and for backward reaction is k − 1 . The rate constant for second elementary reaction is k 2 In the given mechanism, second elementary reaction is the slowest reaction. So, the overall rate of the reaction is, Rate = k 2 [ ES ] …… (1) The above rate law involves an intermediate ( ES ) as one of the reactant which should not be present in the rate law. So, the concentration of intermediate is needs to be replaced by the concentration of reactant molecule. Consider the first elementary reaction which is at equilibrium. The rate of forward reaction is equal to the backward reaction (b) Interpretation Introduction To determine: The additional elementary reaction add into the preceding mechanism that shows the reaction of enzyme with the inhibitor.

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Chapter 14, Problem 116AE

(a)

Interpretation Introduction

To determine: The rate law for the given reaction.

(b)

Interpretation Introduction

To determine: The additional elementary reaction add into the preceding mechanism that shows the reaction of enzyme with the inhibitor.

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Chapter 14, Problem 115AE

Chapter 14, Problem 117IE

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