The rate law for the reaction 2 B r O g → B r 2 g + O 2 ( g ) under each of the given reaction conditions.

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Answer 1

Question

Chapter 13, Problem 13.51QA

Interpretation Introduction

To find:

The rate law for the reaction $2BrOg→Br2g+O2(g)$ under each of the given reaction conditions.

Expert Solution & Answer

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.

$2BrOg→Br2g+O2(g)$

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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Answer 2

Question

Chapter 13, Problem 13.51QA

Interpretation Introduction

To find:

The rate law for the reaction $2BrOg→Br2g+O2(g)$ under each of the given reaction conditions.

Expert Solution & Answer

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.

$2BrOg→Br2g+O2(g)$

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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Answer 3

Question

Chapter 13, Problem 13.51QA

Interpretation Introduction

To find:

The rate law for the reaction $2BrOg→Br2g+O2(g)$ under each of the given reaction conditions.

Expert Solution & Answer

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.

$2BrOg→Br2g+O2(g)$

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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Answer 4

Question

Chapter 13, Problem 13.51QA

Interpretation Introduction

To find:

The rate law for the reaction $2BrOg→Br2g+O2(g)$ under each of the given reaction conditions.

Expert Solution & Answer

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.

$2BrOg→Br2g+O2(g)$

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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Answer 5

Chapter 13, Problem 13.51QA

Interpretation Introduction

To find:

The rate law for the reaction $2BrOg→Br2g+O2(g)$ under each of the given reaction conditions.

Expert Solution & Answer

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.

$2BrOg→Br2g+O2(g)$

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.

Answer 6

Chapter 13, Problem 13.51QA

Interpretation Introduction

To find:

The rate law for the reaction $2BrOg→Br2g+O2(g)$ under each of the given reaction conditions.

Expert Solution & Answer

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.

$2BrOg→Br2g+O2(g)$

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.

Answer 7

Chapter 13, Problem 13.51QA

Interpretation Introduction

To find:

The rate law for the reaction $2BrOg→Br2g+O2(g)$ under each of the given reaction conditions.

Answer 8

Expert Solution & Answer

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; $Rate=k$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

1) Concept:

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

$2BrOg→Br2g+O2(g)$

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.

Answer 12

Ch. 13 - Prob. 13.1VP Ch. 13 - Prob. 13.2VP Ch. 13 - Prob. 13.3VP Ch. 13 - Prob. 13.4VP Ch. 13 - Prob. 13.5VP Ch. 13 - Prob. 13.6VP Ch. 13 - Prob. 13.7VP Ch. 13 - Prob. 13.8VP Ch. 13 - Prob. 13.9VP Ch. 13 - Prob. 13.10VP

The rate law for the reaction 2 B r O g → B r 2 g + O 2 ( g ) under each of the given reaction conditions.

Answer to Problem 13.51QA

Explanation of Solution

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