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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

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BuyFindarrow_forward

Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

Derive expressions for the half-life of zero-, first-, and second-order reactions using the integrated rate law for each order. How does each half-life depend on concentration? If the half- life for a reaction is 20. seconds, what would be the second half-Life assuming the reaction is either zero, first, or second order?

Interpretation Introduction

Interpretation: The expression for the half-life of a zero-, first- and a second order reaction using the integrated rate law for each order is to be stated. The second half life for a reaction with the given first half-life value is to be calculated.

Concept introduction: The time taken by the reactants to decrease to half of their initial concentration is known as the half-life of the reaction.

To determine: The expression for the half-life of a zero-, first- and a second order reaction using the integrated rate law for each order.

Explanation

The integrated rate law for a zero order reaction is [A]=kt+[A]ο .

The time taken by the reactants to decrease to half of their initial concentration is known as the half-life of the reaction.

Therefore, at t12 , the concentration becomes half.

The integrated rate law at t12 is,

[A]ο2=kt12+[A]ο

Simplify the above expression,

kt12=[A]ο[A]ο2t12=[A]ο2k

The integrated rate law for a first order reaction is ln[A]=kt+ln[A]ο .

The time taken by the reactants to decrease to half of their initial concentration is known as the half-life of the reaction.

Therefore, at t12 , the concentration becomes half.

The integrated rate law at t12 is,

ln[A]ο2=kt12+ln[A]ο

Simplify the above expression,

kt12=ln[A]οln[A]ο2t12=ln(2)kt12=0.693k

The integrated rate law for a second order reaction is 1[A]=kt+1[A]ο .

The time taken by the reactants to decrease to half of their initial concentration is known as the half-life of the reaction.

Therefore, at t12 , the concentration becomes half

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