Mastering Chemistry Assignment

9/25/16, 4:46 PMChapter 14

Page 1 of 13https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4636745

Chapter 14 Due: 11:59pm on Sunday, September 25, 2016

To understand how points are awarded, read the Grading Policy for this assignment.

Rate Law

Learning Goal:

To understand reaction order and rate constants.

For the general equation , the rate law is expressed as follows:

where and indicate the order of the reaction with respect to each reactant and must be determined experimentally and is the rate constant, which is specific to each reaction.

Order

For a particular reaction, , the rate law was experimentally determined to be

A. This equation is zero order with respect to . Therefore, changing the concentration of has no effect on the rate because will always equal 1.

B. This equation is first order with respect to . This means that if the concentration of is doubled, the rate will double. If is reduced by half, the rate will be halved. If is tripled, the rate will triple, and so on.

C. This equation is second order with respect to . This means that if the concentration of is doubled, the rate will quadruple. If is tripled, the rate will increase by a factor of 9, and so on.

.

Overall reaction order and rate-constant units

The sum of the individual orders gives the overall reaction order. The example equation above is third order overall because .

For the units of rate to come out to be , the units of the rate constant for third-order reactions must be since

For a second-order reaction, the rate constant has units of because . In a first-order reaction, the rate constant has the units because .

Analyzing a specific reaction

Consider the following reaction:

Part A

aA + bB→cC + dD

rate = k[A [B]m ]n

m n k

aA + bB + cC→dD

rate = k[A [B [C = k[B][C]0 ]1 ]2 ]2

A A [A]0

B B [B] [B]

C C

[C]

0 + 1 + 2 = 3

M /s ⋅M −2 s−1

M /s = ( ⋅ )( )M −2 s−1 M 3

⋅M −1 s−1 M /s = ( ⋅ )( )M −1 s−1 M 2 s−1 M /s = ( )( )s−1 M 1

2Mg + →2MgO, rate = k[Mg][O2 O2 ] 2

9/25/16, 4:46 PMChapter 14

Page 2 of 13https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4636745

What is the overall reaction order?

Express your answer as an integer.

You did not open hints for this part.

ANSWER:

Part B

What are the units of the rate constant for this reaction?

You did not open hints for this part.

ANSWER:

Part C

What would happen to the rate if were doubled?

You did not open hints for this part.

ANSWER:

Part D

What would happen to the rate if were doubled?

You did not open hints for this part.

ANSWER:

k

M ⋅ s−1

s−1

⋅M −1 s−1

⋅M −2 s−1

⋅M −3 s−1

[Mg]

The rate would

stay the same.

double.

triple.

quadruple.

[ ]O2

9/25/16, 4:46 PMChapter 14

Page 3 of 13https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4636745

Reaction Order

For the reaction , the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected:

Trial

( )

( )

( ) Initial Rate

( )

1 0.10 0.10 0.10

2 0.10 0.10 0.30

3 0.20 0.10 0.10

4 0.20 0.20 0.10

Part A

What is the reaction order with respect to ?

You did not open hints for this part.

ANSWER:

Part B

What is the reaction order with respect to ?

You did not open hints for this part.

ANSWER:

Part C

What is the reaction order with respect to ?

You did not open hints for this part.

ANSWER:

The rate would

stay the same.

double.

triple.

quadruple.

A + B + C→D + E

[A] M

[B] M

[C] M M /s

3.0 × 10−5

9.0 × 10−5

1.2 × 10−4

1.2 × 10−4

A

B

C

9/25/16, 4:46 PMChapter 14

Page 4 of 13https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4636745

Part D

This question will be shown after you complete previous question(s).

Reaction Rates

To measure the speed of a car, we use miles per hour ( or ). To measure the rate of a reaction we use molar concentration per second ( ).

Part A

Imagine that you took a road trip. Based on the information in the table, what was the average speed of your car? Time Mile marker

3:00 pm 13

8:00 pm 294

Express your answer in miles per hour.

You did not open hints for this part.

ANSWER:

Now consider the following reaction and data:

Time ( )

concentration ( )

5 1.38

15 1.65

Part B

What is the average rate of formation of ?https://penstrokeswriters.com/order/order

Express your answer in molar concentration per second.

You did not open hints for this part.

ANSWER:

Part C

This question will be shown after you complete previous question(s).

miles/h mph M /s

mph

+ 2ICl→2HCl +H2 I2

s I2

M

I2

M /s

9/25/16, 4:46 PMChapter 14

Page 5 of 13https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4636745

Part D

This question will be shown after you complete previous question(s).

Introduction to Reaction Rates

Learning Goal:

To calculate average and relative reaction rates.

Reaction rate can be defined either as the increase in the concentration of a product per unit time or as the decrease of the concentration of a reactant per unit time. By definition, reaction rate is a positive quantity.

In the reaction , for example, is being produced twice as fast as is consumed and thus

Each rate can be expressed as the change in concentration over the change in time, :

Part A

Consider the reaction

Using the information in the following table, calculate the average rate of formation of between 10.0 and 40.0 .

Time ( ) 0 10.0 20.0 30.0 40.0 50.0

( ) 0 1.40×10−3 4.40×10−3 6.20×10−3 7.40×10−3 8.00×10−3

Express your answer numerically in molar per second.

You did not open hints for this part.

ANSWER:

Part B

This question will be shown after you complete previous question(s).

Part C

Consider the reaction

The average rate of consumption of is 1.10×10−4 over the first two minutes. What is the average rate of formation of during the same time interval?

Express your answer numerically in molar per second to three significant figures.

X→2Y Y X

rate of X = (rate of Y)12 ∆t

= ( )−∆[X] ∆t

1 2

∆[Y] ∆t

2 P → + 3 OH3 O4 P2 O5 H2

P2 O5 s

s

[ ]P2 O5 M

Rate of formation of = P2 O5 M /s

5B (aq) + Br (aq) + 6 (aq)→3B (aq) + 3 O(l)r− O−3 H + r2 H2

Br− M/s Br2

9/25/16, 4:46 PMChapter 14

Page 6 of 13https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4636745

You did not open hints for this part.

ANSWER:

Part D

This question will be shown after you complete previous question(s).

Experimental Determination of a Rate Law

Consider the reaction

whose rate at 25 was measured using three different sets of initial concentrations as listed in the following table:

Trial

( )

( ) Rate

( )

1 0.50 0.050 1.5×10−2

2 0.50 0.100 3.0×10−2

3 1.00 0.050 6.0×10−2

Part A

What is the rate law for this reaction?

Express the rate law symbolically in terms of , , and .

You did not open hints for this part.

ANSWER:

Part B

This question will be shown after you complete previous question(s).

Half-life (kinetics) for First Order Reactions

The integrated rate law allows chemists to predict the reactant concentration after a certain amount of time, or the time it would take for a certain concentration to be reached. The integrated rate law for a first-order reaction is:

Rate of formation of = Br2 M /s

A + 2B ⇌ C C∘

[A] M

[B] M M /s

k [A] [B]

rate =

[A] = [A]0 e −kt

[A] 0

9/25/16, 4:46 PMChapter 14

Page 7 of 13https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4636745

Now say we are particularly interested in the time it would take for the concentration to become one-half of its inital value. Then we could substitute for and rearrange the equation to:

This equation caculates the time required for the reactant concentration to drop to half its initial value. In other words, it calculates the half-life.

Half-life equation for first-order reactions:

where is the half-life in seconds , and is the rate constant in inverse seconds .

Part A

What is the half-life of a first-order reaction with a rate constant of 9.00×10−4 ?

You did not open hints for this part.

ANSWER:

Part B

What is the rate constant of a first-order reaction that takes 7.10 minutes for the reactant concentration to drop to half of its initial value?

You did not open hints for this part.

ANSWER:

Part C

A certain first-order reaction has a rate constant of 5.70×10−3 . How long will it take for the reactant concentration to drop to of its initial value?

You did not open hints for this part.

ANSWER:

Half-life for First and Second Order Reactions

The half-life of a reaction, , is the time it takes for the reactant concentration to decrease by half. For example, after one half-life the concentration falls from the initial concentration to , after a second half-life to , after a third half-life to , and so on. on.

For a first-order reaction, the half-life is constant. It depends only on the rate constant and not on the reactant concentration. It is expressed as

For a second-order reaction, the half-life depends on the rate constant and the concentration of the reactant and so is expressed as

[A] 0 2

[A] = t1/2

0.693 k

= t1/2 0.693

k t1/2 (s) k ( )s−1

s−1

s

min−1

s−1 18

s

t1/2 [A] [A]0 [A /2]0 [A /4]0 [A /8]0

k

=t1/2 0.693

k

=1/2 1

9/25/16, 4:46 PMChapter 14

Page 8 of 13https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4636745

Part A

A certain first-order reaction ( ) has a rate constant of 1.00×10−2 at 45 . How many minutes does it take for the concentration of the reactant, , to drop to 6.25 of the original concentration?

Express your answer numerically in minutes.

You did not open hints for this part.

ANSWER:

Part B

A certain second-order reaction ( ) has a rate constant of 1.45×10−3 at 27 and an initial half-life of 278 . What is the concentration of the reactant after one half-life?

Express the molar concentration numerically.

You did not open hints for this part.

ANSWER:

Problem 14.6

The following diagrams represent mixtures of and . These two substances react as follows:

It has been determined experimentally that the rate is second order in and first order in .

Part A

Based on this fact, which of the following mixtures will have the fastest initial rate?

=t1/2 1

k[A]0

A→products s−1 C∘ [A] %

min

B→products ⋅M−1 s−1 C∘ s B

M

NO(g) (g)O2 2 NO(g) + (g) → 2 N (g)O2 O2

NO O2

9/25/16, 4:46 PMChapter 14

Page 9 of 13https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4636745

ANSWER:

Problem 14.9

The following diagram shows the reaction profile of a reaction.

Part A

Label the components indicated by the boxes.

Label the diagram by dragging the labels to the appropriate targets.

ANSWER:

Problem 14.8

The mixture (1).

The mixture (2).

The mixture (3).

9/25/16, 4:46 PMChapter 14

Page 10 of 13https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4636745

Part A

Given the following diagrams at = 0 and = 30, what is the half-life of the reaction if it follows first-order kinetics?

ANSWER:

Part B

After four half-life periods for a first-order reaction, what fraction of reactant remains?

ANSWER:

Problem 14.13

Part A

Based on the reaction profile above, how many intermediates are formed in the reaction ?

ANSWER:

t t

= τ1/2 min

= [A]t A0

A → C

9/25/16, 4:46 PMChapter 14

Page 11 of 13https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4636745

Part B

How many transition states are there?

ANSWER:

Part C

Which step is the fastest?

ANSWER:

Part D

Is the reaction exothermic or endothermic?

ANSWER:

Problem 14.25

Part A

Consider the combustion of . If hydrogen is burning at the rate of 0.52 , what is the rate of consumption of oxygen?

ANSWER:

Part B

What is the rate of formation of water vapor?

ANSWER:

Part C

= intermediates k

= transition states n

The first step.

The second step.

A → C

The reaction is endothermic.

The reaction is exothermic.

(g) : 2 (g) + (g)→2 O(g)H2 H2 O2 H2 mol/s

= ∆[ ]/∆tO2 mol/s

= ∆[ O]/∆tH2 mol/s

2NO(g) + (g)→2NOCl(g)Cl2 NO torr/min

9/25/16, 4:46 PMChapter 14

Page 12 of 13https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4636745

The reaction is carried out in a closed vessel. If the partial pressure of is decreasing at the rate of 53 , what is the rate of change of the total pressure of the vessel?

ANSWER:

Problem 14.58

For the elementary process the activation energy and overall are 154 , and 136 , respectively.

Part A

What is the activation energy for the reverse reaction?

Express your answer using two significant figures.

ANSWER:

Problem 14.105

Americium-241 is used in smoke detectors. It has a first order rate constant for radioactive decay of . By contrast, iodine-125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of 0.011 .

Part A

What is the half-life of americium-241?

Express your answer using two significant figures.

ANSWER:

Part B

What is the half-life of iodine-125?

Express your answer using two significant figures.

ANSWER:

Part C

Which one decays at a faster rate?

ANSWER:

2NO(g) + (g)→2NOCl(g)Cl2 NO torr/min

= ∆ /∆tPtotal torr/min

(g)→ (g) + (g)N2 O5 NO2 NO3 ( )Ea ∆E kJ/mol kJ/mol

= Ea kJ/mol

k = 1.6 × 10−3 yr−1 k = day−1

= t1/2 yr

= t1/2 days

9/25/16, 4:46 PMChapter 14

Page 13 of 13https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4636745

Part D

How much of a 1.00 sample of americium remains after 3 half-lives?

Express your answer using two significant figures.

ANSWER:

Part E

How much of a 1.00 sample of iodine remains after 3 half-lives?

Express your answer using two significant figures.

ANSWER:

Part F

How much of a 1.00 sample of americium remains after 4 days?

Express your answer using two significant figures.

ANSWER:

Part G

How much of a 1.00 sample of iodine remains after 4 days?

Express your answer using three significant figures.

ANSWER:

Score Summary: Your score on this assignment is 0.0%. You received 0 out of a possible total of 10 points.

americium-241

iodine-125

mg

= m mg

mg

= m mg

mg

= m mg

mg

= m mg