Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 13, Problem 13.51QA
Interpretation Introduction

To find:

The rate law for the reaction 2BrOgBr2g+O2(g) under each of the given reaction conditions.

Expert Solution & Answer
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Answer to Problem 13.51QA

Solution:

a) The rate doubles when [BrO] doubles; Rate=k[BrO]

b) The rate quadruples when [BrO] doubles; Rate=k[BrO]2

c) The rate is halved when [BrO] is halved; Rate=k[BrO]

d) The rate is unchanged when [BrO] doubles; Rate=k

Explanation of Solution

1) Concept:

Rate law is the equation, which determines the relation between the rate of reaction and the concentration of reactant.

2BrOgBr2g+O2(g)

Consider the general rate law for this reaction, which is

Rate=k[BrO]n

where k is the rate constant and n is the order in BrO.

2) Calculation:

a) The rate doubles when [BrO] doubles.

Let the initial rate be Rate 1, and when the concentration of BrO doubles, the rate also doubles. Consider it as Rate 2.

i.e., Rate 2 = 2 (Rate 1) and [BrO] = 2 [BrO]

So the rate law for both will be

Rate 1=k[BrO]n

Rate 2=k2[BrO]n

Take the ratio of Rate 2 to Rate 1,

Rate 2Rate 1=k2[BrO]nk [BrO]n

Plug the value of Rate 2=2 (Rate 1),

Therefore,2 Rate 1Rate 1=k2[BrO]nk [BrO]n

2=2n [BrO]n [BrO]n

2=2n

n = 1

That means reaction is first order in BrO. Hence the rate law can be written as

Rate=k[BrO]

b) The rate quadruples when [BrO] doubles.

Let the initial rate be Rate 1m, and when the concentration of BrO doubles, the rate quadruples. Consider it as Rate 2.

i.e., Rate 2 = 4 (Rate 1) when [BrO] = 2 [BrO]

So the rate law for both will be

Rate 1=k[BrO]n

Rate 2=k2[BrO]n

Take the ratio of Rate 2 to Rate 1.

Rate 2Rate 1=k2[BrO]nk [BrO]n

Plug the value of Rate 2=4 (Rate 1).

Therefore,4 Rate 1Rate 1=k2[BrO]nk [BrO]n

4=2n [BrO]n [BrO]n

4=2n

n =2

i.e., reaction is second order in BrO, and the rate law will be

Rate=k[BrO]2

c) The rate is halved when [BrO] is halved.

Let the initial rate be Rate 1, and when the concentration of BrO halves, the rate also halves. Consider it as Rate 2.

i.e., Rate 2 = 1/2 (Rate 1) when [BrO] = 1/2 [BrO]

so the rate law for both will be

Rate 1=k[BrO]n

Rate 2=k1/2[BrO]n

Take the ratio of Rate 2 to Rate 1.

Rate 2Rate 1=k1/2 [BrO]nk [BrO]n

Plug the value of Rate 2=12Rate 1.

Therefore,

1/2 Rate 1Rate 1=k1/2[BrO]nk [BrO]n

1/2=1/2n [BrO]n [BrO]n

1/2= 1/22

n =1

Hence, reaction is first order in  BrO, and rate law is

Rate=k[BrO]

d) The rate is unchanged when [BrO] doubles.

When rate the concentration of BrO doubles, the rate remain unchanged,

i.e., [BrO] =2 [BrO] , Rate 2 = Rate 1

So the rate law for both will be

Rate 1=k[BrO]n

Rate 2=k2[BrO]n

Take the ratio of Rate 2 to Rate 1.

Rate 2Rate 1=k2[BrO]nk [BrO]n

Since, Rate 2 = Rate 1, equation becomes

Rate 1Rate 1=k2[BrO]nk [BrO]n

0=2n BrOn BrOn

0=2n

n = 0

Hence, the reaction is zero order in BrO.

The rate law will be

Rate=kBrOo

Rate=k

Conclusion:

The effect of change in concentration of reactants on rate depends on the order of reaction with respect to that reactant.

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Chapter 13 Solutions

Chemistry: An Atoms-Focused Approach

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