Student Solutions Manual to accompany Atkins' Physical Chemistry 11th  edition
Student Solutions Manual to accompany Atkins' Physical Chemistry 11th edition
11th Edition
ISBN: 9780198807773
Author: ATKINS
Publisher: Oxford University Press
Question
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Chapter 17, Problem 17.3IA

(a)

Interpretation Introduction

Interpretation:

The rate law of the base-catalysed reaction in which AH goes to products according to the following scheme has to be deduced.

  AH+Bka'kaBH++A-A-+AHkbproduct(rate-determining)

(a)

Expert Solution
Check Mark

Explanation of Solution

Rate-determining step is the slow step of the reaction.  The rate law for the slow reaction is

  υ=kb[A-][AH]

A- is the intermediate.  So,

  d[A-]dt=0ka[AH][B]-ka'[BH+][A-]-kb[A-][AH]=0ka'[BH+][A-]+kb[A-][AH]=ka[AH][B][A-](ka'[BH+]+kb[AH])=ka[AH][B][A-]=ka[AH][B]ka'[BH+]+kb[AH]

If the rate of conversion of intermediates to reactants is greater than rate of conversion of intermediate to products, then pre-equilibrium will be present.

  ka'[BH+]>>kb[AH][A-]willbechangedaska[AH][B]ka'[BH+]

The rate law can be calculated as

  υ=kbka[AH][B]ka'[BH+][AH]υ=kbka[AH]2[B]ka'[BH+]υ=kbkaka'[AH]2[B][BH+]

The equilibrium constant for the fast reaction is

  K=kaka'

The rate law of base-catalysed reaction is

  υ=kbK[AH]2[B][BH+]

(b)

Interpretation Introduction

Interpretation:

The rate law of the acid-catalysed reaction in which HA goes to products according to the following scheme has to be deduced.

  HA+H+ka'kaHAH+HAH++BkbBH++AH(rate-determining)

(b)

Expert Solution
Check Mark

Explanation of Solution

Rate-determining step is the slow step of the reaction.  The rate law for the slow reaction is

  υ=kb[HAH+][B]

HAH+ is the intermediate.  So,

  d[HAH+]dt=0ka[HA][H+]-ka'[HAH+]-kb[HAH+][B]=0ka'[HAH+]+kb[HAH+][B]=ka[HA][H+][HAH+](ka'+kb[B])=ka[HA][H+][HAH+]=ka[HA][H+]ka'+kb[B]

If the rate of conversion of intermediates to reactants is greater than rate of conversion of intermediate to products, then pre-equilibrium will be present.

  ka'>>kb[B][HAH+]willbechangedaska[HA][H+]ka'

The rate law can be calculated as

  υ=kbka[HA][H+]ka'[B]υ=kbka[HA][H+][B]ka'

The equilibrium constant for the fast reaction is

  K=kaka'

The rate law of acid-catalysed reaction is

  υ=kbK[HA][H+][B]

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Chapter 17 Solutions

Student Solutions Manual to accompany Atkins' Physical Chemistry 11th edition

Ch. 17 - Prob. 17A.2DQCh. 17 - Prob. 17A.3DQCh. 17 - Prob. 17A.4DQCh. 17 - Prob. 17A.1AECh. 17 - Prob. 17A.1BECh. 17 - Prob. 17A.2AECh. 17 - Prob. 17A.2BECh. 17 - Prob. 17A.3AECh. 17 - Prob. 17A.3BECh. 17 - Prob. 17A.4AECh. 17 - Prob. 17A.4BECh. 17 - Prob. 17A.5AECh. 17 - Prob. 17A.5BECh. 17 - Prob. 17A.6AECh. 17 - Prob. 17A.6BECh. 17 - Prob. 17A.7AECh. 17 - Prob. 17A.7BECh. 17 - Prob. 17A.9AECh. 17 - Prob. 17A.9BECh. 17 - Prob. 17A.1PCh. 17 - Prob. 17A.2PCh. 17 - Prob. 17A.3PCh. 17 - Prob. 17B.1DQCh. 17 - Prob. 17B.2DQCh. 17 - Prob. 17B.3DQCh. 17 - Prob. 17B.1BECh. 17 - Prob. 17B.2AECh. 17 - Prob. 17B.2BECh. 17 - Prob. 17B.3AECh. 17 - Prob. 17B.3BECh. 17 - Prob. 17B.4AECh. 17 - Prob. 17B.4BECh. 17 - Prob. 17B.5AECh. 17 - Prob. 17B.5BECh. 17 - Prob. 17B.6BECh. 17 - Prob. 17B.3PCh. 17 - Prob. 17B.4PCh. 17 - Prob. 17B.5PCh. 17 - Prob. 17B.6PCh. 17 - Prob. 17B.7PCh. 17 - Prob. 17B.8PCh. 17 - Prob. 17B.9PCh. 17 - Prob. 17B.10PCh. 17 - Prob. 17B.11PCh. 17 - Prob. 17B.12PCh. 17 - Prob. 17B.14PCh. 17 - Prob. 17B.15PCh. 17 - Prob. 17B.16PCh. 17 - Prob. 17B.17PCh. 17 - Prob. 17B.18PCh. 17 - Prob. 17C.1DQCh. 17 - Prob. 17C.2DQCh. 17 - Prob. 17C.1BECh. 17 - Prob. 17C.2AECh. 17 - Prob. 17C.2BECh. 17 - Prob. 17C.6PCh. 17 - Prob. 17D.1DQCh. 17 - Prob. 17D.1AECh. 17 - Prob. 17D.1BECh. 17 - Prob. 17D.2AECh. 17 - Prob. 17D.2BECh. 17 - Prob. 17D.3AECh. 17 - Prob. 17D.3BECh. 17 - Prob. 17D.4AECh. 17 - Prob. 17D.4BECh. 17 - Prob. 17D.5BECh. 17 - Prob. 17D.1PCh. 17 - Prob. 17D.3PCh. 17 - Prob. 17D.4PCh. 17 - Prob. 17D.5PCh. 17 - Prob. 17D.6PCh. 17 - Prob. 17E.1DQCh. 17 - Prob. 17E.2DQCh. 17 - Prob. 17E.3DQCh. 17 - Prob. 17E.4DQCh. 17 - Prob. 17E.5DQCh. 17 - Prob. 17E.6DQCh. 17 - Prob. 17E.1AECh. 17 - Prob. 17E.1BECh. 17 - Prob. 17E.2AECh. 17 - Prob. 17E.2BECh. 17 - Prob. 17E.3AECh. 17 - Prob. 17E.3BECh. 17 - Prob. 17E.4PCh. 17 - Prob. 17F.1DQCh. 17 - Prob. 17F.3DQCh. 17 - Prob. 17F.4DQCh. 17 - Prob. 17F.1AECh. 17 - Prob. 17F.1BECh. 17 - Prob. 17F.2AECh. 17 - Prob. 17F.2BECh. 17 - Prob. 17F.3AECh. 17 - Prob. 17F.3BECh. 17 - Prob. 17F.4AECh. 17 - Prob. 17F.4BECh. 17 - Prob. 17F.2PCh. 17 - Prob. 17F.3PCh. 17 - Prob. 17F.4PCh. 17 - Prob. 17F.6PCh. 17 - Prob. 17F.7PCh. 17 - Prob. 17G.1AECh. 17 - Prob. 17G.1BECh. 17 - Prob. 17G.2AECh. 17 - Prob. 17G.2BECh. 17 - Prob. 17G.3AECh. 17 - Prob. 17G.3BECh. 17 - Prob. 17G.1PCh. 17 - Prob. 17G.2PCh. 17 - Prob. 17G.7PCh. 17 - Prob. 17.3IACh. 17 - Prob. 17.6IACh. 17 - Prob. 17.7IA
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