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 17.75P
Interpretation Introduction

Interpretation:

The rate law and the value of the rate constant for formation of OI 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 OCl(aq), I(aq) and OH(aq) are calculated as,

    (rate of reaction)0 for run 1(rate of reaction)0 for run 2=k[OCl-(aq)]0x[ I-(aq)]oy[OH-(aq)]ozfor run1k[OCl-(aq)]0x[ I-(aq)]oy[OH-(aq)]ozfor run23.06 ×10-45.44×10-4=(1.62×10-3)x(1.62×10-3)y(0.52)z(1.62×10-3)x(2.88×10-3)y(0.52)z(0.563)=(0.563)yy=1OrderofreactionwithrespecttoI-(aq)isone(rate of reaction)0 for run 2(rate of reaction)0 for run 4=k[OCl-(aq)]0x[ I-(aq)]oy[OH-(aq)]ozfor run2k[OCl-(aq)]0x[ I-(aq)]oy[OH-(aq)]ozfor run45.44×10-43.11×10-4=(1.62×10-3)x(2.88×10-3)y(0.52)z(1.62×10-3)x(2.88×10-3)y(0.91)z(1.749)=(0.571)zz=1OrderofreactionwithrespecttoOH-(aq)isone(rate of reaction)0 for run 2(rate of reaction)0 for run 3=k[OCl-(aq)]0x[ I-(aq)]oy[OH-(aq)]ozfor run2k[OCl-(aq)]0x[ I-(aq)]oy[OH-(aq)]ozfor run35.44×10-43.16×10-4=(1.62×10-3)x(2.88×10-3)(0.52)(2.71×10-3)x(1.62×10-3)(0.84)(1.65)=(0.60)xx=1OrderofreactionwithrespecttoOCl-(aq)isone

The rate law of reaction and rate constant are given by

    k=(rateofreaction)o[OCl-(aq)]0[ I-(aq)]o[OH-(aq)]o=5.44 ×10-4(1.62×10-3)(1.62×10-3)(0.52)=4.0 ×102M-1s-1Ratelaw:rate of reaction =4.0 ×102M-1s-1[OCl-(aq)]0[ I-(aq)]o[OH-(aq)]o

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Chapter 17 Solutions

GENERAL CHEMISTRY ACHIEVE ACCESS W/BOOK

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