GENERAL CHEMISTRY ACHIEVE ACCESS W/BOOK
GENERAL CHEMISTRY ACHIEVE ACCESS W/BOOK
4th Edition
ISBN: 9781319405212
Author: McQuarrie
Publisher: MAC HIGHER
Question
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Chapter 17, Problem 17.20P
Interpretation Introduction

Interpretation:

The rate law and the value of the rate constant for formation of hydrogen ion have to be calculated.

Concept Introduction:

Rate law:  The mathematical equation that relates the concentrations of the reactants to

the rate of reaction is called the rate law of a reaction.  An example of rate law of thermal decomposition of N2O5 is given by

    rate of reaction = k[N2O5]x

The proportionality constant, k, in a rate law is called the rate constant of the reaction.

Expert Solution & Answer
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Explanation of Solution

The order of reaction with respect to CH3COCH3(aq), Br2(aq) and H+(aq) are calculated as,

    (rate of reaction)0 for run 1(rate of reaction)0 for run 2=k[CH3COCH3(aq)]0x[Br2(aq)]oy[H+(aq)]ozfor run1k[CH3COCH3(aq)]0x[Br2(aq)]oy[H+(aq)]ozfor run24.0 ×10-37.0×10-3=(1.00)x(1.00)y(1.00)z(1.75)x(1.00)y(1.00)z(0.571)=(0.571)xx=1OrderofreactionwithrespecttoCH3COCH3(aq)isone(rate of reaction)0 for run 2(rate of reaction)0 for run 3=k[CH3COCH3(aq)]0x[Br2(aq)]oy[H+(aq)]ozfor run2k[CH3COCH3(aq)]0x[Br2(aq)]oy[H+(aq)]ozfor run37.0 ×10-39.8×10-3=(1.75)(1.00)y(1.00)z(1.75)(1.40)y(1.00)z(0.714)=(0.714)yy=1OrderofreactionwithrespecttoBr2(aq)isone(rate of reaction)0 for run 2(rate of reaction)0 for run 3=k[CH3COCH3(aq)]0x[Br2(aq)]oy[H+(aq)]ozfor run2k[CH3COCH3(aq)]0x[Br2(aq)]oy[H+(aq)]ozfor run37.0 ×10-39.8×10-3=(1.75)(1.00)(1.00)z(1.75)(1.40)(1.00)z(0.714)(0.714)=(1.00)zz=1OrderofreactionwithrespecttoH+(aq)isoneOverallorderofreaction=1+1+1=3

The rate law of reaction and rate constant are given by

    k=(rateofreaction)o[CH3COCH3(aq)]0[Br2(aq)]o[H+(aq)]o=4.0 ×10-3(1.00)(1.00)(1.00)=4.0 ×10-3M-1s-1Ratelaw:rate of reaction =4.0 ×10-3M-1s-1[CH3COCH3][H+][Br2]

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Chapter 17 Solutions

GENERAL CHEMISTRY ACHIEVE ACCESS W/BOOK

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