   Chapter 14, Problem 84GQ

Chapter
Section
Textbook Problem

Many biochemical reactions are catalyzed by acids. A typical mechanism consistent with the experimental results (in which HA is the acid and X is the reactant) is Step 1: Fast, reversible: HA ⇄ H + +A − Step 2: Fast, reversible: X+H + ⇄ XH + Step 3: Slow XH + → products What rate law is derived from this mechanism? What is the order of the reaction with respect to HA? How would doubling the concentration of HA affect the reaction?

Interpretation Introduction

Interpretation:

The rate law derived from the given mechanism, order of the reaction respect to HA has to be given and how doubling the concentration of HA affect the reaction should be explain.

Concept introduction:

Rate law or rate equation: Rate law:

It is generally the rate equation that consists of the reaction rate with the concentration or the pressures of the reactants and constant parameters.

aA + bBxXRate of reaction = k [A]m[B]n

Order of a reaction:  The order of each reactant in a reaction is represented by the exponential term of the respective reactant present in the rate law and the overall order of the reaction is the sum of all the exponents of all reactants present in the chemical reaction. The order of the reaction is directly proportional to the concentration of the reactants.

Rate constant, k: The rate constant for a chemical reaction is the proportionality term in the chemical reaction rate law which gives the relationship between the rate and the concentration of the reactant present in the chemical reaction.

Molecularity: It is defined as the number of reacting species involved in a chemical reaction.

Explanation

The rate law is generally obtained by considering the reactant involved in the reaction. It is generally the rate equation that consists of the reaction rate with the concentration or the pressures of the reactants and constant parameters.

The rate law for the given reaction is Rate=k[X][HA]

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