The rate of the decomposition of hydrogen peroxide, H 2 O 2 , depends on the concentration of iodide ion present. The rate of decomposition was measured at constant temperature and pressure for various concentrations of H 2 O 2 and of KI. The data appear below. Determine the order of reaction for each substance, write the rate law, and evaluate the rate constant. Rate [H 2 OJ [Kll (mL min-’) (mol L ’) (mol L ’) 0.090 0.15 0.033 0.178 0.30 0.033 0.184 0.15 0.066

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Answer 1

Textbook Question

Chapter 11, Problem 11.30PAE

The rate of the decomposition of hydrogen peroxide, H₂O₂, depends on the concentration of iodide ion present. The rate of decomposition was measured at constant temperature and pressure for various concentrations of H₂O₂and of KI. The data appear below. Determine the order of reaction for each substance, write the rate law, and evaluate the rate constant.

Rate	[H₂OJ	[Kll
(mL min-’)	(mol L ’)	(mol L ’)
0.090	0.15	0.033
0.178	0.30	0.033
0.184	0.15	0.066

Expert Solution

Interpretation Introduction

To determine:

Order of reaction of each substance, rate law and rate constant

Explanation of Solution

Let following chemical reaction

aA+bB→cC+dD

Rate law for this reaction can be written as

R=K[A]x[B]y.

A=H2O2

B=KI.

The rate law for the given reaction can be written as

R=K[H2O2]x[KI]y

You have given three set of data. Put each in the above equation

R=0.090mLmin−1

1mL=10−3L

So,

R=0.090×10−3molL−1min−1

[H2O2]=0.15molL−1.[KI]=0.033molL−1

Putting these values on equation, you get

(0.09×10−3molL−1min−1)=K(0.15molL −1)x(0.033molL −1)y⋯⋯(1)

R=0.178molL−1min−1=0.178×10−3molLmin−1

[H2O2]=0.30molL−1[KI]=0.033molL−1

Putting these value in equation you, get

(0.178×10−3molL−1min−1)=K(0.30molL −1)x.(0.033molL −1)y⋯⋯(2)

R=0.184molL−1min−1=0.184×10−3molL−1min−1

[H2O2]=0.15molL−1[KI]=0.066molL−1

Putting these value in equation you, get

0.184×10−1molL−1min−1=K(0.15molL −1)x(0.66molL −1)y⋯⋯(3)

(1)÷(2)

(0.09× 10 −3)molL−1min−10.178×10−3molL−1min−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.30mol L −1 )x( 0.033mol L −1 )y

11.98=1 ( 2 )x2x=1.982x=(2)1x=1

(1)÷(3)

(0.09× 10 −3)Lmin−1(0.178× 10 −3)Lmin−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.15mol L −1 )x( 0.066mol L −1 )y

11.98=1 ( 2 )y(2)y=1.98(2)y=(2)1y=1

Hence

x=1,y=1.

Hence overall order of reaction =1+1=2

Rate law can be written as

R=K[H2O2][KI]

Putting the value R,

[H2O2],[KI], to find the value of K.

(0.09×10−3)L−min−1=K(0.15molL−1)(0.033molL−1).

K=0.01818mol−1Lmin−1K=18.18mol−1mLmin−1.

Hence rate law can be written as

R=18.18mol−1mLmin−1[H2O2][KI]

Conclusion

It is the initial rates experiment method.

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Answer 2

Textbook Question

Chapter 11, Problem 11.30PAE

The rate of the decomposition of hydrogen peroxide, H₂O₂, depends on the concentration of iodide ion present. The rate of decomposition was measured at constant temperature and pressure for various concentrations of H₂O₂and of KI. The data appear below. Determine the order of reaction for each substance, write the rate law, and evaluate the rate constant.

Rate	[H₂OJ	[Kll
(mL min-’)	(mol L ’)	(mol L ’)
0.090	0.15	0.033
0.178	0.30	0.033
0.184	0.15	0.066

Expert Solution

Interpretation Introduction

To determine:

Order of reaction of each substance, rate law and rate constant

Explanation of Solution

Let following chemical reaction

aA+bB→cC+dD

Rate law for this reaction can be written as

R=K[A]x[B]y.

A=H2O2

B=KI.

The rate law for the given reaction can be written as

R=K[H2O2]x[KI]y

You have given three set of data. Put each in the above equation

R=0.090mLmin−1

1mL=10−3L

So,

R=0.090×10−3molL−1min−1

[H2O2]=0.15molL−1.[KI]=0.033molL−1

Putting these values on equation, you get

(0.09×10−3molL−1min−1)=K(0.15molL −1)x(0.033molL −1)y⋯⋯(1)

R=0.178molL−1min−1=0.178×10−3molLmin−1

[H2O2]=0.30molL−1[KI]=0.033molL−1

Putting these value in equation you, get

(0.178×10−3molL−1min−1)=K(0.30molL −1)x.(0.033molL −1)y⋯⋯(2)

R=0.184molL−1min−1=0.184×10−3molL−1min−1

[H2O2]=0.15molL−1[KI]=0.066molL−1

Putting these value in equation you, get

0.184×10−1molL−1min−1=K(0.15molL −1)x(0.66molL −1)y⋯⋯(3)

(1)÷(2)

(0.09× 10 −3)molL−1min−10.178×10−3molL−1min−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.30mol L −1 )x( 0.033mol L −1 )y

11.98=1 ( 2 )x2x=1.982x=(2)1x=1

(1)÷(3)

(0.09× 10 −3)Lmin−1(0.178× 10 −3)Lmin−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.15mol L −1 )x( 0.066mol L −1 )y

11.98=1 ( 2 )y(2)y=1.98(2)y=(2)1y=1

Hence

x=1,y=1.

Hence overall order of reaction =1+1=2

Rate law can be written as

R=K[H2O2][KI]

Putting the value R,

[H2O2],[KI], to find the value of K.

(0.09×10−3)L−min−1=K(0.15molL−1)(0.033molL−1).

K=0.01818mol−1Lmin−1K=18.18mol−1mLmin−1.

Hence rate law can be written as

R=18.18mol−1mLmin−1[H2O2][KI]

Conclusion

It is the initial rates experiment method.

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Answer 3

Textbook Question

Chapter 11, Problem 11.30PAE

The rate of the decomposition of hydrogen peroxide, H₂O₂, depends on the concentration of iodide ion present. The rate of decomposition was measured at constant temperature and pressure for various concentrations of H₂O₂and of KI. The data appear below. Determine the order of reaction for each substance, write the rate law, and evaluate the rate constant.

Rate	[H₂OJ	[Kll
(mL min-’)	(mol L ’)	(mol L ’)
0.090	0.15	0.033
0.178	0.30	0.033
0.184	0.15	0.066

Expert Solution

Interpretation Introduction

To determine:

Order of reaction of each substance, rate law and rate constant

Explanation of Solution

Let following chemical reaction

aA+bB→cC+dD

Rate law for this reaction can be written as

R=K[A]x[B]y.

A=H2O2

B=KI.

The rate law for the given reaction can be written as

R=K[H2O2]x[KI]y

You have given three set of data. Put each in the above equation

R=0.090mLmin−1

1mL=10−3L

So,

R=0.090×10−3molL−1min−1

[H2O2]=0.15molL−1.[KI]=0.033molL−1

Putting these values on equation, you get

(0.09×10−3molL−1min−1)=K(0.15molL −1)x(0.033molL −1)y⋯⋯(1)

R=0.178molL−1min−1=0.178×10−3molLmin−1

[H2O2]=0.30molL−1[KI]=0.033molL−1

Putting these value in equation you, get

(0.178×10−3molL−1min−1)=K(0.30molL −1)x.(0.033molL −1)y⋯⋯(2)

R=0.184molL−1min−1=0.184×10−3molL−1min−1

[H2O2]=0.15molL−1[KI]=0.066molL−1

Putting these value in equation you, get

0.184×10−1molL−1min−1=K(0.15molL −1)x(0.66molL −1)y⋯⋯(3)

(1)÷(2)

(0.09× 10 −3)molL−1min−10.178×10−3molL−1min−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.30mol L −1 )x( 0.033mol L −1 )y

11.98=1 ( 2 )x2x=1.982x=(2)1x=1

(1)÷(3)

(0.09× 10 −3)Lmin−1(0.178× 10 −3)Lmin−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.15mol L −1 )x( 0.066mol L −1 )y

11.98=1 ( 2 )y(2)y=1.98(2)y=(2)1y=1

Hence

x=1,y=1.

Hence overall order of reaction =1+1=2

Rate law can be written as

R=K[H2O2][KI]

Putting the value R,

[H2O2],[KI], to find the value of K.

(0.09×10−3)L−min−1=K(0.15molL−1)(0.033molL−1).

K=0.01818mol−1Lmin−1K=18.18mol−1mLmin−1.

Hence rate law can be written as

R=18.18mol−1mLmin−1[H2O2][KI]

Conclusion

It is the initial rates experiment method.

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Answer 4

Textbook Question

Chapter 11, Problem 11.30PAE

The rate of the decomposition of hydrogen peroxide, H₂O₂, depends on the concentration of iodide ion present. The rate of decomposition was measured at constant temperature and pressure for various concentrations of H₂O₂and of KI. The data appear below. Determine the order of reaction for each substance, write the rate law, and evaluate the rate constant.

Rate	[H₂OJ	[Kll
(mL min-’)	(mol L ’)	(mol L ’)
0.090	0.15	0.033
0.178	0.30	0.033
0.184	0.15	0.066

Expert Solution

Interpretation Introduction

To determine:

Order of reaction of each substance, rate law and rate constant

Explanation of Solution

Let following chemical reaction

aA+bB→cC+dD

Rate law for this reaction can be written as

R=K[A]x[B]y.

A=H2O2

B=KI.

The rate law for the given reaction can be written as

R=K[H2O2]x[KI]y

You have given three set of data. Put each in the above equation

R=0.090mLmin−1

1mL=10−3L

So,

R=0.090×10−3molL−1min−1

[H2O2]=0.15molL−1.[KI]=0.033molL−1

Putting these values on equation, you get

(0.09×10−3molL−1min−1)=K(0.15molL −1)x(0.033molL −1)y⋯⋯(1)

R=0.178molL−1min−1=0.178×10−3molLmin−1

[H2O2]=0.30molL−1[KI]=0.033molL−1

Putting these value in equation you, get

(0.178×10−3molL−1min−1)=K(0.30molL −1)x.(0.033molL −1)y⋯⋯(2)

R=0.184molL−1min−1=0.184×10−3molL−1min−1

[H2O2]=0.15molL−1[KI]=0.066molL−1

Putting these value in equation you, get

0.184×10−1molL−1min−1=K(0.15molL −1)x(0.66molL −1)y⋯⋯(3)

(1)÷(2)

(0.09× 10 −3)molL−1min−10.178×10−3molL−1min−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.30mol L −1 )x( 0.033mol L −1 )y

11.98=1 ( 2 )x2x=1.982x=(2)1x=1

(1)÷(3)

(0.09× 10 −3)Lmin−1(0.178× 10 −3)Lmin−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.15mol L −1 )x( 0.066mol L −1 )y

11.98=1 ( 2 )y(2)y=1.98(2)y=(2)1y=1

Hence

x=1,y=1.

Hence overall order of reaction =1+1=2

Rate law can be written as

R=K[H2O2][KI]

Putting the value R,

[H2O2],[KI], to find the value of K.

(0.09×10−3)L−min−1=K(0.15molL−1)(0.033molL−1).

K=0.01818mol−1Lmin−1K=18.18mol−1mLmin−1.

Hence rate law can be written as

R=18.18mol−1mLmin−1[H2O2][KI]

Conclusion

It is the initial rates experiment method.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

Interpretation Introduction

To determine:

Order of reaction of each substance, rate law and rate constant

Explanation of Solution

Let following chemical reaction

aA+bB→cC+dD

Rate law for this reaction can be written as

R=K[A]x[B]y.

A=H2O2

B=KI.

The rate law for the given reaction can be written as

R=K[H2O2]x[KI]y

You have given three set of data. Put each in the above equation

R=0.090mLmin−1

1mL=10−3L

So,

R=0.090×10−3molL−1min−1

[H2O2]=0.15molL−1.[KI]=0.033molL−1

Putting these values on equation, you get

(0.09×10−3molL−1min−1)=K(0.15molL −1)x(0.033molL −1)y⋯⋯(1)

R=0.178molL−1min−1=0.178×10−3molLmin−1

[H2O2]=0.30molL−1[KI]=0.033molL−1

Putting these value in equation you, get

(0.178×10−3molL−1min−1)=K(0.30molL −1)x.(0.033molL −1)y⋯⋯(2)

R=0.184molL−1min−1=0.184×10−3molL−1min−1

[H2O2]=0.15molL−1[KI]=0.066molL−1

Putting these value in equation you, get

0.184×10−1molL−1min−1=K(0.15molL −1)x(0.66molL −1)y⋯⋯(3)

(1)÷(2)

(0.09× 10 −3)molL−1min−10.178×10−3molL−1min−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.30mol L −1 )x( 0.033mol L −1 )y

11.98=1 ( 2 )x2x=1.982x=(2)1x=1

(1)÷(3)

(0.09× 10 −3)Lmin−1(0.178× 10 −3)Lmin−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.15mol L −1 )x( 0.066mol L −1 )y

11.98=1 ( 2 )y(2)y=1.98(2)y=(2)1y=1

Hence

x=1,y=1.

Hence overall order of reaction =1+1=2

Rate law can be written as

R=K[H2O2][KI]

Putting the value R,

[H2O2],[KI], to find the value of K.

(0.09×10−3)L−min−1=K(0.15molL−1)(0.033molL−1).

K=0.01818mol−1Lmin−1K=18.18mol−1mLmin−1.

Hence rate law can be written as

R=18.18mol−1mLmin−1[H2O2][KI]

Conclusion

It is the initial rates experiment method.

Answer 8

Expert Solution

Interpretation Introduction

To determine:

Order of reaction of each substance, rate law and rate constant

Explanation of Solution

Let following chemical reaction

aA+bB→cC+dD

Rate law for this reaction can be written as

R=K[A]x[B]y.

A=H2O2

B=KI.

The rate law for the given reaction can be written as

R=K[H2O2]x[KI]y

You have given three set of data. Put each in the above equation

R=0.090mLmin−1

1mL=10−3L

So,

R=0.090×10−3molL−1min−1

[H2O2]=0.15molL−1.[KI]=0.033molL−1

Putting these values on equation, you get

(0.09×10−3molL−1min−1)=K(0.15molL −1)x(0.033molL −1)y⋯⋯(1)

R=0.178molL−1min−1=0.178×10−3molLmin−1

[H2O2]=0.30molL−1[KI]=0.033molL−1

Putting these value in equation you, get

(0.178×10−3molL−1min−1)=K(0.30molL −1)x.(0.033molL −1)y⋯⋯(2)

R=0.184molL−1min−1=0.184×10−3molL−1min−1

[H2O2]=0.15molL−1[KI]=0.066molL−1

Putting these value in equation you, get

0.184×10−1molL−1min−1=K(0.15molL −1)x(0.66molL −1)y⋯⋯(3)

(1)÷(2)

(0.09× 10 −3)molL−1min−10.178×10−3molL−1min−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.30mol L −1 )x( 0.033mol L −1 )y

11.98=1 ( 2 )x2x=1.982x=(2)1x=1

(1)÷(3)

(0.09× 10 −3)Lmin−1(0.178× 10 −3)Lmin−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.15mol L −1 )x( 0.066mol L −1 )y

11.98=1 ( 2 )y(2)y=1.98(2)y=(2)1y=1

Hence

x=1,y=1.

Hence overall order of reaction =1+1=2

Rate law can be written as

R=K[H2O2][KI]

Putting the value R,

[H2O2],[KI], to find the value of K.

(0.09×10−3)L−min−1=K(0.15molL−1)(0.033molL−1).

K=0.01818mol−1Lmin−1K=18.18mol−1mLmin−1.

Hence rate law can be written as

R=18.18mol−1mLmin−1[H2O2][KI]

Conclusion

It is the initial rates experiment method.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

To determine:

Order of reaction of each substance, rate law and rate constant

Answer 13

Explanation of Solution

Let following chemical reaction

aA+bB→cC+dD

Rate law for this reaction can be written as

R=K[A]x[B]y.

A=H2O2

B=KI.

The rate law for the given reaction can be written as

R=K[H2O2]x[KI]y

You have given three set of data. Put each in the above equation

R=0.090mLmin−1

1mL=10−3L

So,

R=0.090×10−3molL−1min−1

[H2O2]=0.15molL−1.[KI]=0.033molL−1

Putting these values on equation, you get

(0.09×10−3molL−1min−1)=K(0.15molL −1)x(0.033molL −1)y⋯⋯(1)

R=0.178molL−1min−1=0.178×10−3molLmin−1

[H2O2]=0.30molL−1[KI]=0.033molL−1

Putting these value in equation you, get

(0.178×10−3molL−1min−1)=K(0.30molL −1)x.(0.033molL −1)y⋯⋯(2)

R=0.184molL−1min−1=0.184×10−3molL−1min−1

[H2O2]=0.15molL−1[KI]=0.066molL−1

Putting these value in equation you, get

0.184×10−1molL−1min−1=K(0.15molL −1)x(0.66molL −1)y⋯⋯(3)

(1)÷(2)

(0.09× 10 −3)molL−1min−10.178×10−3molL−1min−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.30mol L −1 )x( 0.033mol L −1 )y

11.98=1 ( 2 )x2x=1.982x=(2)1x=1

(1)÷(3)

(0.09× 10 −3)Lmin−1(0.178× 10 −3)Lmin−1=K( 0.15mol L −1 )x( 0.033mol L −1 )yK( 0.15mol L −1 )x( 0.066mol L −1 )y

11.98=1 ( 2 )y(2)y=1.98(2)y=(2)1y=1

Hence

x=1,y=1.

Hence overall order of reaction =1+1=2

Rate law can be written as

R=K[H2O2][KI]

Putting the value R,

[H2O2],[KI], to find the value of K.

(0.09×10−3)L−min−1=K(0.15molL−1)(0.033molL−1).

K=0.01818mol−1Lmin−1K=18.18mol−1mLmin−1.

Hence rate law can be written as

R=18.18mol−1mLmin−1[H2O2][KI]

Answer 14

Conclusion

It is the initial rates experiment method.

Answer 15

Ch. 11 - Prob. 1CO Ch. 11 - . define the rate of a chemical reaction and...Ch. 11 - Prob. 3CO Ch. 11 - Prob. 4CO Ch. 11 - . explain the difference between elementary...Ch. 11 - . find the rate law predicted for a particular...Ch. 11 - . use a molecular perspective to explain the...Ch. 11 - Prob. 8CO Ch. 11 - . explain the role of a catalyst in the design of...Ch. 11 - Prob. 11.1PAE

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