   Chapter 12, Problem 53E

Chapter
Section
Textbook Problem

# Consider the following initial rate data for the decomposition of compound AB to give A and B: [AB]o (mol/L) Initial Rate (mol/L · s) 0.200 3.20 × 10–3 0.400 1.28 ×10–2 0.600 2.88 ×10–2 Determine the half-life for the decomposition reaction initially having 1.00 M AB present.

Interpretation Introduction

Interpretation:

The half-life time of the given reaction has to be found.

Concept introduction:

Rate law: Rate law is the equation that relates the concentration of the reactant to the rate of the reaction.  Consider the reaction where the reactant A and B is giving product P .

A+BP

The rate law of the reaction can be expressed as given below.

Rate[A][B]orRate=k[A][B]

Where,

k is the rate constant of the reaction.

Half- life (t1/2) : It is the time taken to reduce the concentration of the reactant to half of its initial value.  Half-life of a second order reaction is calculated using the equation given below.

t1/2=1k[AB]°

Where,

k is the rate constant of the reaction.

[AB]° is the initial concentration of the reactant.

Explanation

To find the order of the given reaction

Given,

The concentration of the reactant AB in the first experiment = 0.200 mol L-1

The rate of the first experiment =3.20×103molL-1s-1

The concentration of the reactant AB in the second experiment = 0.400mol L-1

The rate of the second experiment =1.28×102molL-1s-1

The concentration of the reactant in the second experiment is twice of the concentration of the reactants in the first experiment.  It can be observed that, when concentration is doubled, the rate has increased to a factor of 4.

The rate of the second experiment =4×The rate of the first experiment =4×3.20×103 =1

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