Lecture13 Chemical Kinetics

1

Chemical Kinetics

2

Chapter Goals

1. The Rate of a ReactionFactors That Affect Reaction Rates

2. Nature of the Reactants3. Concentrations of the Reactants: The

Rate-Law Expression4. Concentration Versus Time: The

Integrated Rate Equation5. Collision Theory of Reaction Rates

2

3

Chapter Goals

6. Transition State Theory7. Reaction Mechanisms and the Rate-Law

Expression8. Temperature: The Arrhenius Equation9. Catalysts

4

The Rate of a Reaction

• Kinetics is the study of rates of chemicalreactions and the mechanisms by whichthey occur.

• The reaction rate is the increase inconcentration of a product per unit time ordecrease in concentration of a reactantper unit time.

• A reaction mechanism is the series ofmolecular steps by which a reactionoccurs.

3

5

The Rate of a Reaction• Thermodynamics (Chapter 15) determines if a reaction

can occur.• Kinetics (Chapter 16) determines how quickly a reaction

occurs.– Some reactions that are thermodynamically

feasible are kinetically so slow as to beimperceptible.

OUSINSTANTANE

kJ-79=GOHOH+H

SLOWVERY

kJ396GCOOC

o2982

-aq

+aq

o298g2g2diamond

l

6

The Rate of Reaction

• Consider the hypothetical reaction,aA(g) + bB(g) cC(g) + dD(g)

• equimolar amounts of reactants, A and B,will be consumed while products, C and D,will be formed as indicated in this graph:

4

7

0

0.2

0.4

0.6

0.8

1

1.2

0 50 100

150

200

250

300

350

Time

Con

cent

ratio

ns o

fR

eact

ants

& P

rodu

cts

[A] & [B][C] & [D]

• [A] is the symbol for the concentration of A in M ( mol/L).• Note that the reaction does not go entirely to completion.

– The [A] and [B] > 0 plus the [C] and [D] < 1.

8


• Reaction rates are the rates at which reactantsdisappear or products appear.

• This movie is an illustration of a reaction rate.

5

9


• Mathematically, the rate of a reaction canbe written as:

tdD+

tcC+or

tbB-

taA-=Rate

10

The Rate of Reaction• The rate of a simple one-step reaction is directly

proportional to the concentration of the reactingsubstance.

• [A] is the concentration of A in molarity ormoles/L.

• k is the specific rate constant.– k is an important quantity in this chapter.

Ak=RateorARate

C+BA (g)(g)(g)

6

11


• For a simple expression like Rate = k[A]– If the initial concentration of A is doubled, the initial

rate of reaction is doubled.• If the reaction is proceeding twice as fast, the amount of

time required for the reaction to reach equilibrium wouldbe:

A. The same as the initial time.B. Twice the initial time.C. Half the initial time.

• If the initial concentration of A is halved theinitial rate of reaction is cut in half.

12


• If more than one reactant molecule appearsin the equation for a one-step reaction, wecan experimentally determine that thereaction rate is proportional to the molarconcentration of the reactant raised to thepower of the number of molecules involved inthe reaction.

22

ggg

Xk=RateorXRate

Z+YX2

7

13


• Rate Law Expressions must be determinedexperimentally.

– The rate law cannot be determined from thebalanced chemical equation.

– This is a trap for new students of kinetics.• The balanced reactions will not work because most

chemical reactions are not one-step reactions.

• Other names for rate law expressions are:1. rate laws2. rate equations3. rate expressions

14


• Important terminology for kinetics.• The order of a reaction can be expressed in

terms of either:1 each reactant in the reaction or2 the overall reaction. Order for the overall reaction is the sum of the orders

for each reactant in the reaction.

• For example:

overall.orderfirstandONinorderfirstisreactionThis

ONk=Rate

O+NO4ON2

52

52

g2g2g52

8

15


• A second example is:

overall.orderfirstand,OHinorderzero

CBr,CHinorderfirstisreactionThis]CBrCHk[=Rate

BrCOHCHOHCBrCH

-33

33

-aqaq33

-aqaq33

16


• A final example of the order of a reaction is:

ALLYEXPERIMENTDETERMINEDARESEXPRESSIONRATEALLREMEMBER,

overallorderthirdand,OinorderfirstNO,inordersecondisreactionThis

Ok[NO]=RateNO2O+NO2

2

2

2

g2g2g

9

17


• Given the following one step reaction and itsrate-law expression.– Remember, the rate expression would have to be

experimentally determined.

• Because it is a second order rate-lawexpression:– If the [A] is doubled the rate of the reaction will

increase by a factor of 4. 22 = 4– If the [A] is halved the rate of the reaction will

decrease by a factor of 4. (1/2)2 = 1/4

2ggg

Ak=Rate

CBA2

18

Factors That Affect ReactionRates• There are several factors that can

influence the rate of a reaction:1. The nature of the reactants.2. The concentration of the reactants.3. The temperature of the reaction.4. The presence of a catalyst.• We will look at each factor individually.

10

19

Nature of Reactants

• This is a very broad category that encompassesthe different reacting properties of substances.

• For example sodium reacts with waterexplosively at room temperature to liberatehydrogen and form sodium hydroxide.

burns.andignitesHThereaction.rapidand violentaisThis

HNaOH2OH2Na2

2

g2aq2s

20

Nature of Reactants

• Calcium reacts with water only slowly atroom temperature to liberate hydrogenand form calcium hydroxide.

reaction.slowratheraisThis

HOHCaOH2Ca g2aq22s

11

21

Nature of Reactants

• The reaction of magnesium with water atroom temperature is so slow that that theevolution of hydrogen is not perceptible tothe human eye.

reactionNoOHMg 2s

22

Nature of Reactants

• However, Mg reacts with steam rapidly toliberate H2 and form magnesium oxide.

• The differences in the rate of these threereactions can be attributed to the changing“nature of the reactants”.

g2sC100

)g(2s HMgOOHMg o

12

23

Concentrations of Reactants:The Rate-Law Expression• This movie illustrates how changing the

concentration of reactants affects the rate.

24

Concentrations of Reactants:The Rate-Law Expression• This is a simplified representation of the

effect of different numbers of molecules inthe same volume.– The increase in the molecule numbers is

indicative of an increase in concentration.A(g) + B (g) Products

A B

A B

A BB

A B

A BA BA B

4 different possibleA-B collisions



13

25

Concentrations of Reactants:The Rate-Law Expression

• Example 16-1: The following rate data were obtained at25oC for the following reaction. What are the rate-lawexpression and the specific rate-constant for thisreaction?

2 A(g) + B(g) 3 C(g)

ExperimentNumber

Initial [A](M)

Initial [B](M)

Initial rate offormation of

C (M/s)1 0.10 0.10 2.0 x 10-4

2 0.20 0.30 4.0 x 10-4

3 0.10 0.20 2.0 x 10-4

26


.Bignorecancan weThus,k[A]=RateorBAk=Rate

012

constant.remainsrateinitial that theNotice2.byincreases[B] that theandconstantis[A]the

thatsee we3and1sexperimentcompare weIfBAk=Rate

:form theofbemustlawrateThe

0 xx

y

yx

y

14

27


You do it!

xx

reaction?for thiskofunitsand value theisWhat

overall.order1andAtorespectorder with1isreactionThis

k[A]=Rateork[A]=Rate122

2.byincreasesrate theand2byincreases[A] that theNotice2.and1sexperimentcompareNext,

st

st

1

28


[A]10 x2.0=Rateas writtenbecanlawrate theThus

10 x2.010.010 x2.0=k

1experimentfrom[A]andRateof values theUsingA

Rate=k

law.rate thefromkof value thefindcanWe

s13-

s13-s

4-

M

M

15

29

Concentrations of Reactants:The Rate-Law Expression• Example 16-2: The following data were obtained

for the following reaction at 25oC. What are therate-law expression and the specific rate constantfor the reaction?

2 A(g) + B(g) + 2 C(g) 3 D(g) + 2 E(g)

Experiment

Initial [A](M)

Initial [B](M)

Initial[C](M)

Initial rateof

formationof D (M/s)

1 0.20 0.10 0.10 2.0 x 10-4

2 0.20 0.30 0.20 6.0 x 10-4

3 0.20 0.10 0.30 2.0 x 10-4

4 0.60 0.30 0.40 1.8 x 10-3

30


yxzyx

z z

BAk=RateorCBAk=Rate

013

constant.remainsratebut the3byincreasesCTheconstant.remainBandA thatNotice

3.and1sexperimentCompare

16

31


BAk=RateorBAk=Rate

133

3.byincreasesrate theand3byincreasesBTheconstant.remainsAThe

2.and1sexperimentcompareNext,

1 xx

y y

32


overall.order2andB,respect toorder with1A,respect toorder with1isreactionThis

BAk=RateorBAk=Rate

133

3.byincreasesrate theand3byincreasesATheconstant.remainsBThe

4.and2sexperimentcompareNext,

ndst

st

11

xx

17

33


BA100.1=Rateas writtenbecanlaw-rate theThus,

100.10.100.20

100.2BA

Rate=k

4.or3,2,1,experimentfromdata theuseCank.ofunitsand value thedetermineFinally,

s12

s12

s4

M

M

M

MM

34

Concentrations of Reactants:The Rate-Law Expression• Example 16-3: consider a chemical reaction

between compounds A and B that is first orderwith respect to A, first order with respect to B,and second order overall. From the informationgiven below, fill in the blanks.

You do it!ExperimentInitial Rate

(M/s)Initial [A]

(M)Initial [B]

(M)1 4.0 x 10-3 0.20 0.050

2 1.6 x 10-2 ? 0.050

3 3.2 x 10-2 0.40 ?

18

35


BA40.0Rate theThus40.0

0.0500.20100.4

BARate=k

k.of value thedeterminecan we1experimentFromBAk=Rate

s1

s1

s3

M

M

M

MM

36


M

MMM

80.0]A[050.0s0.40s106.1]A[

k[B]Rate[A]

2experimentfromdataandkof value theUse

1-1

1-2

19

37


M

MMM

20.0[B]40.0s0.40s102.3[B]

k[A]R[B]

determinecan we3experimentfromSimilarly,

1-1

1-2

38

Concentration vs. Time: TheIntegrated Rate Equation• The integrated rate equation relates time

and concentration for chemical andnuclear reactions.– From the integrated rate equation we can

predict the amount of product that is producedin a given amount of time.

• Initially we will look at the integrated rateequation for first order reactions.These reactions are 1st order in the reactant

and 1st order overall.

20

39

Concentration vs. Time: TheIntegrated Rate Equation• An example of a reaction that is 1st order

in the reactant and 1st order overall is:a A products

This is a common reaction type for manychemical reactions and all simpleradioactive decays.

• Two examples of this type are:2 N2O5(g) 2 N2O4(g) + O2(g)

238U 234Th + 4He

40

Concentration vs. Time: TheIntegrated Rate Equation

where:[A]0= mol/L of A at time t=0. [A] = mol/L of A at time t.k = specific rate constant. t = time elapsed since

beginning of reaction.a = stoichiometric coefficient of A in balanced overall

equation.

• The integrated rate equation for first order reactionsis:

k taAAln 0

21

41


• Solve the first order integrated rate equation for t.

• Define the half-life, t1/2, of a reactant as the timerequired for half of the reactant to be consumed,or the time at which [A]=1/2[A]0.

AAln

ka1t 0

42

Concentration vs. Time: TheIntegrated Rate Equation• At time t = t1/2, the expression becomes:

ka693.0t

2lnka

1t

A1/2Aln

ka1t

1/2

1/2

0

01/2

22

43


• Example 16-4: Cyclopropane, ananesthetic, decomposes to propeneaccording to the following equation.

The reaction is first order in cyclopropanewith k = 9.2 s-1 at 10000C. Calculate the halflife of cyclopropane at 10000C.

CH2 CH2CH2

CH2CH3CH

(g) (g)

s075.0s2.9

693.0k693.0t 1-1/2

44

Concentration vs. Time: TheIntegrated Rate Equation• Example 16-5: Refer to Example 16-4.

How much of a 3.0 g sample ofcyclopropane remains after 0.50 seconds?– The integrated rate laws can be used for any

unit that represents moles or concentration.– In this example we will use grams rather than

mol/L.

23

45


remains1%org03.0eA

5.310.16.4Aln6.4Aln10.1

s50.0s2.9Aln-3.0lnk tAlnAln

.logarithmsoflaws theusemustWe

reactionfor thisk tk taAAln

3.5-

1-

0

0

46

Concentration vs. Time: TheIntegrated Rate Equation• Example 16-6: The half-life for the

following first order reaction is 688 hoursat 10000C. Calculate the specific rateconstant, k, at 10000C and the amount ofa 3.0 g sample of CS2 that remains after48 hours.

CS2(g) CS(g) + S(g)

You do it!

24

47


1-

1/2

1/2

hr00101.0hr688

0.693k

t0.693k

k693.0t

1.areactionFor this

48


hr)48)(hr00101.0(Aln-ln(3.0)

k tAlnAlnk tAAln

1-

00

unreacted97%org9.2g86.2eA

1.0521.10)--(0.048Aln0.048Aln-1.10

hr)48)(hr00101.0(Aln-ln(3.0)

k tAlnAlnk tAAln

1.052

1-

00

25

49

Concentration vs. Time: TheIntegrated Rate Equation• For reactions that are second order with respect

to a particular reactant and second orderoverall, the rate equation is:

• Where:[A]0= mol/L of A at time t=0. [A] = mol/L of A at time t.

k = specific rate constant. t = time elapsed sincebeginning of reaction.

a = stoichiometric coefficient of A in balanced overall equation.

k taA1

A1

0

50

Concentration vs. Time: TheIntegrated Rate Equation• Second order reactions also have a half-life.

– Using the second order integrated rate-law as astarting point.

• At the half-life, t1/2 [A] = 1/2[A]0.

0

1/200

Aofrdenominatocommonahaswhich

k taA1

A2/11

1/20

1/200

k taA1

ork taA1

A2

26

51


• If we solve for t1/2:

• Note that the half-life of a second orderreaction depends on the initialconcentration of A.

01/2 Aka1t

52

Concentration vs. Time: TheIntegrated Rate Equation• Example 16-7: Acetaldehyde, CH3CHO, undergoes

gas phase thermal decomposition to methane andcarbon monoxide.

The rate-law expression is Rate = k[CH3CHO]2,and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) Whatis the half-life of CH3CHO if 0.10 mole isinjected into a 1.0 L vessel at 527oC?

CH CHO CH + CO3 g 4 g g

27

53


tk A

hr

hr

1/2

-1

1

12 0 10 010

5 0 10

0

2 1

2

. .

.

M M

54

Concentration vs. Time: TheIntegrated Rate Equation• (b) How many moles of CH3CHO remain after

200 hours?

1 1

1 1010

2 0 10 200

1 10 4 0

0

2 1

1 1

A Ak t

A hr hr

A

-1

..

.

MM

M M

1 14 114

0 071

0 071 mol

11A

A

A

mol = 1.0 L x 0.071 molL

M

MM.

? .

28

55

Concentration vs. Time: TheIntegrated Rate Equation• (c) What percent of the CH3CHO remains

after 200 hours?

reacted29%andunreacted%71

%100mol0.10mol0.071=unreacted%

56

Concentration vs. Time: TheIntegrated Rate Equation• Example 16-8: Refer to Example 16-7. (a)

What is the half-life of CH3CHO if 0.10mole is injected into a 10.0 L vessel at527oC?– Note that the vessel size is increased by a

factor of 10 which decreases theconcentration by a factor of 10!

You do it!

29

57


tk A

hr

hrnote the time has increased by 10over Example 16 - 7:

1/2

-1

1

12 0 10 0 010

5 0 10

0

2 1

3

. .

.

M M

58

Concentration vs. Time: TheIntegrated Rate Equation• (b) How many moles of CH3CHO remain

after 200 hours?You do it!

30

59


1 1

1 10 010

2 0 10 200

1 100 4 0

0

2 1

1 1

A Ak t

A hr hr

A

-1

..

.

MM

M M

1 104 1104

0 0096

0 096 mol

11A

A

A

mol = 10.0 L x 0.0096 molL

M

MM.

? .

60

Concentration vs. Time: TheIntegrated Rate Equation• (c) What percent of the CH3CHO remains

after 200 hours?You do it!

31

61


% unreacted = 0.096 mol0.100 mol

unreacted & 4% reacted

100%

96%

62

Concentration vs. Time: TheIntegrated Rate Equation• Let us now summarize the results from the last

two examples.

InitialMoles

CH3CHO

[CH3CHO]0

(M)

[CH3CHO]

(M)

Moles ofCH3CH

O at 200hr.

%CH3CHOremainin

gEx. 16-

70.10 0.10 0.071 0.071 71%

Ex. 16-8

0.010 0.010 0.0096 0.096 96%

32

63

Enrichment - Derivation ofIntegrated Rate Equations• For the first order reaction

a A productsthe rate can be written as:

tA

a1-=Rate

64

Enrichment - Derivation ofIntegrated Rate Equations• For a first-order reaction, the rate is proportional

to the first power of [A].

-1a

At

k A

33

65

Enrichment - Derivation ofIntegrated Rate Equations• In calculus, the rate is defined as the

infinitesimal change of concentration d[A] in aninfinitesimally short time dt as the derivative of[A] with respect to time.

-1a

At

k Add

66

Enrichment - Derivation ofIntegrated Rate Equations• Rearrange the equation so that all of the [A]

terms are on the left and all of the t terms are onthe right.

- AA

a k td d

34

67

Enrichment - Derivation ofIntegrated Rate Equations• Express the equation in integral form.

- A

Aa k t

A

A td d0 0

68

Enrichment - Derivation ofIntegrated Rate Equations• This equation can be evaluated as:

-ln A a k t or

-ln A A a k t - a k 0which becomes-ln A A a k t

t0t

t

t

0

0

0

ln

ln

35

69

Enrichment - Derivation ofIntegrated Rate Equations• Which rearranges to the integrated first order

rate equation.

k taAAln

t

0

70

Enrichment - Derivation ofIntegrated Rate Equations• Derive the rate equation for a reaction that is

second order in reactant A and second orderoverall.

• The rate equation is:

2Ak ta

A

dd

36

71

Enrichment - Derivation ofIntegrated Rate Equations• Separate the variables so that the A terms are

on the left and the t terms on the right.

tkAaA

2 dd

72

Enrichment - Derivation ofIntegrated Rate Equations• Then integrate the equation over the limits as for

the first order reaction.

t

0

A

A2 tka

AA

0

dd

37

73

Enrichment - Derivation ofIntegrated Rate Equations• Which integrates the second order integrated

rate equation.

k taA1

A1

0

74

Enrichment - Derivation ofIntegrated Rate Equations• For a zero order reaction the rate expression is:

k ta

A

dd

38

75

Enrichment - Derivation ofIntegrated Rate Equations• Which rearranges to:

tkaA dd

76

Enrichment - Derivation ofIntegrated Rate Equations• Then we integrate as for the other two cases:

t

0

A

A

tkaA0

dd

39

77

Enrichment - Derivation ofIntegrated Rate Equations• Which gives the zeroeth order integrated rate

equation.

k ta-AAor

k t-aAA

0

0

78

Enrichment - Rate Equations toDetermine Reaction Order• Plots of the integrated rate equations can help us

determine the order of a reaction.• If the first-order integrated rate equation is

rearranged.– This law of logarithms, ln (x/y) = ln x - ln y, was applied

to the first-order integrated rate-equation.

0

0

Alnk taAlnor

k taAlnAln

40

79

Enrichment - Rate Equations toDetermine Reaction Order• The equation for a straight line is:

• Compare this equation to the rearranged firstorder rate-law.

bmy x

80

Enrichment - Rate Equationsto Determine Reaction Order

bmy x

• Now we can interpret the parts of the equationas follows:– y can be identified with ln[A] and plotted on the y-axis.– m can be identified with –ak and is the slope of the

line.– x can be identified with t and plotted on the x-axis.– b can be identified with ln[A]0 and is the y-intercept.

0Alnk taAln

41

81

Enrichment - Rate Equations toDetermine Reaction Order• Example 16-9: Concentration-versus-time data

for the thermal decomposition of ethyl bromideare given in the table below. Use the followinggraphs of the data to determine the rate of thereaction and the value of the rate constant.

700KatHBrHCBrHC gg42g52

82

Enrichment - Rate Equations toDetermine Reaction Order

Time(min) 0 1 2 3 4 5

[C2H5Br] 1.00 0.82 0.67 0.55 0.45 0.37

ln [C2H5Br] 0.00 -0.20 -0.40 -0.60 -0.80 -0.99

1/[C2H5Br] 1.0 1.2 1.5 1.8 2.2 2.7

42

83

Enrichment - Rate Equationsto Determine Reaction Order• We will make three different graphs of the

data.1 Plot the [C2H5Br] (y-axis) vs. time (x-axis)

– If the plot is linear then the reaction is zeroorder with respect to [C2H5Br].

2 Plot the ln [C2H5Br] (y-axis) vs. time (x-axis)– If the plot is linear then the reaction is first order

with respect to [C2H5Br].3 Plot 1/ [C2H5Br] (y-axis) vs. time (x-axis)

– If the plot is linear then the reaction is secondorder with respect to [C2H5Br].

84

Enrichment - Rate Equationsto Determine Reaction Order• Plot of [C2H5Br] versus time.

– Is it linear or not?[C2H5Br] vs. time

00.20.40.60.8

11.2

0 1 2 3 4 5

Time (min)

[C2H

5B

r]

43

85

Enrichment - Rate Equationsto Determine Reaction Order• Plot of ln [C2H5Br] versus time.

– Is it linear or not?

ln [C2H5Br] vs. time

-1.2-1

-0.8-0.6-0.4-0.2

00 1 2 3 4 5

Time (min)

ln [C

2H5B

r]

86


• Plot of 1/[C2H5Br] versus time.– Is it linear or not?

1/[C2H5Br] vs. time

0

1

2

3

0 1 2 3 4 5Time (min)

1/[C

2H5B

r]

44

87

Enrichment - Rate Equationsto Determine Reaction Order• Note that the only graph which is linear is the plot of

ln[C2H5Br] vs. time.– Thus this is a first order reaction with respect

to [C2H5Br].• Next, we will determine the value of the rate constant

from the slope of the line on the graph of ln[C2H5Br] vs.time.– Remember slope = y2-y1/x2-x1.

1-

12

12

min20.0min3

60.0slope

min14)20.0(80.0

x-xy-yslope

88

Enrichment - Rate Equations toDetermine Reaction Order• From the equation for a first order reaction we

know that the slope = -a k.– In this reaction a = 1.

.min0.20kconstantrate theThus-k-0.20slope

1-

45

89

Enrichment - Rate Equationsto Determine Reaction Order• The integrated rate equation for a reaction that is

second order in reactant A and second orderoverall.

• This equation can be rearranged to: k ta

A1

A1

0

0A1k ta

A1

90

Enrichment - Rate Equationsto Determine Reaction Order• Compare the equation for a straight line and the

second order rate-law expression.

• Now we can interpret the parts of the equationas follows:– y can be identified with 1/[A] and plotted on the y-axis.– m can be identified with a k and is the slope of the

line.– x can be identified with t and plotted on the x-axis– b can be identified with 1/[A]0 and is the y-intercept.

bmy x

0A1k ta

A1

46

91

Enrichment - Rate Equationsto Determine Reaction Order• Example 16-10: Concentration-versus-

time data for the decomposition of nitrogendioxide are given in the table below. Usethe graphs to determine the rate of thereaction and the value of the rate constant

500KatONO2NO2 g2gg2

92

Enrichment - Rate Equationsto Determine Reaction OrderTime(min) 0 1 2 3 4 5[NO2] 1.0 0.53 0.36 0.27 0.22 0.18

ln [NO2] 0.0 -0.63 -1.0 -1.3 -1.5 -1.7

1/[NO2] 1.0 1.9 2.8 3.7 4.6 5.5

47

93

Enrichment - Rate Equationsto Determine Reaction Order• Once again, we will make three different graphs

of the data.1. Plot [NO2] (y-axis) vs. time (x-axis).• If the plot is linear then the reaction is zero order with

respect to NO2.2. Plot ln [NO2] (y-axis) vs. time (x-axis).• If the plot is linear then the reaction is first order with

respect to NO2.3. Plot 1/ [NO2] (y-axis) vs. time (x-axis).• If the plot is linear then the reaction is second order

with respect to NO2.

94

Enrichment - Rate Equationsto Determine Reaction Order• Plot of [NO2] versus time.


[NO2] vs. time

00.20.40.60.8

11.2

0 1 2 3 4 5

Time (min)

[NO

2]

48

95

Enrichment -Rate Equationsto Determine Reaction Order• Plot of ln [NO2] versus time.

– Is it linear or not?ln [NO2] vs. time

-2

-1.5

-1

-0.5

00 1 2 3 4 5

Time (min)

ln [N

O2]

96

Enrichment - Rate Equationsto Determine Reaction Order• Plot of 1/[NO2] versus time.


1/[NO2] vs.time

0123456

0 1 2 3 4 5

Time (min)

1/[N

O2]

49

97

Enrichment - Rate Equationsto Determine Reaction Order• Note that the only graph which is linear is

the plot of 1/[NO2] vs. time.• Thus this is a second order reaction with

respect to [NO2].• Next, we will determine the value of the

rate constant from the slope of the line onthe graph of 1/[NO2] vs. time.

98


• From the equation for a second order reactionwe know that the slope = a k– In this reaction a = 2.

1-1 min0.45kconstantrate theThusk20.90slope

M

min1

1

1

12

12

90.0min4

60.3slope

min15)90.1(50.5

x-xy-yslope

MM

M

50

99

Collision Theory ofReaction Rates• Three basic events must occur for a

reaction to occur the atoms, molecules orions must:

1. Collide.2. Collide with enough energy to break and

form bonds.3. Collide with the proper orientation for a

reaction to occur.

100

Collision Theory ofReaction Rates• One method to increase the number of collisions

and the energy necessary to break and reformbonds is to heat the molecules.

• As an example, look at the reaction of methaneand oxygen:

• We must start the reaction with a match.– This provides the initial energy necessary to break

the first few bonds.– Afterwards the reaction is self-sustaining.

kJ891OHCOOCH (g)22(g)2(g)4(g)

51

101

Collision Theory ofReaction Rates• Illustrate the proper orientation of

molecules that is necessary for thisreaction.

X2(g) + Y2(g)2 XY(g)

• Some possible ineffective collisions are :

X

XY Y Y

YX X X X Y Y

102

Collision Theory ofReaction Rates• An example of an effective collision is:

X Y

X Y

X Y

X Y

X Y+

X Y

52

103

Transition State Theory• Transition state theory postulates that

reactants form a high energy intermediate,the transition state, which then falls apart intothe products.

• For a reaction to occur, the reactants mustacquire sufficient energy to form thetransition state.– This energy is called the activation energy or Ea.

• Look at a mechanical analog for activationenergy

104

Transition State Theory

Epot = mg h

Cross sectionof mountain

BoulderEactivation

h

h2

h1

Epot=mgh2

Epot=mgh1

Height

53

105


PotentialEnergy

Reaction Coordinate

X2 + Y2

2 XY

Eactivation - a kinetic quantity

E Ha thermodynamicquantity

Representation of a chemical reaction.

106


54

107


• The relationship between the activationenergy for forward and reverse reactionsis– Forward reaction = Ea

– Reverse reaction = Ea + E– difference = E

108


• The distribution of molecules possessingdifferent energies at a given temperature isrepresented in this figure.

55

109

Reaction Mechanisms and theRate-Law Expression• Use the experimental rate-law to postulate a

molecular mechanism.• The slowest step in a reaction mechanism is the

rate determining step.

110

Reaction Mechanisms and theRate-Law Expression• Use the experimental rate-law to

postulate a mechanism.• The slowest step in a reaction mechanism

is the rate determining step.• Consider the iodide ion catalyzed

decomposition of hydrogen peroxide towater and oxygen.

g22I

22 O+OH2OH2-

56

111

Reaction Mechanisms and theRate-Law Expression• This reaction is known to be first order in H2O2 ,

first order in I- , and second order overall.• The mechanism for this reaction is thought to be:

-22

2222

-2222

-2

--22

IOHk=RlawratealExperiment

O+OH2OH2reactionOverall

I+O+OHOH+IOstepFast

OH+IOI+OHstepSlow

112

Reaction Mechanisms and theRate-Law Expression• Important notes about this reaction:1. One hydrogen peroxide molecule and one

iodide ion are involved in the rate determiningstep.

2. The iodide ion catalyst is consumed in step 1and produced in step 2 in equal amounts.

3. Hypoiodite ion has been detected in thereaction mixture as a short-lived reactionintermediate.

57

113

Reaction Mechanisms and theRate-Law Expression• Ozone, O3, reacts very rapidly with nitrogen

oxide, NO, in a reaction that is first order in eachreactant and second order overall.

NOOk=Rateislaw-ratealExperiment

O+NONO+O

3

g2g2gg3

114

Reaction Mechanisms and theRate-Law Expression• One possible mechanism is:

223

223

33

O+NONO+OreactionOverall

O+NONO+OstepFastO+NONO+OstepSlow

58

115

Reaction Mechanisms and theRate-Law Expression• A mechanism that is inconsistent with the

rate-law expression is:

correct.becannotmechanism thisproveswhich

Ok=Rateismechanism thisfromlaw-rateTheONONO+OreactionOverall

NONO+OstepFastO+OOstepSlow

3

223

2

23

116

Reaction Mechanisms and theRate-Law Expression• Experimentally determined reaction

orders indicate the number of moleculesinvolved in:

1. the slow step only or2. the slow step and the equilibrium steps

preceding the slow step.

59

117

Temperature:The Arrhenius Equation• Svante Arrhenius developed this

relationship among (1) the temperature(T), (2) the activation energy (Ea), and (3)the specific rate constant (k).

k = Aeor

ln k = ln A - ERT

-E RT

a

a

118

Temperature:The Arrhenius Equation• This movie illustrates the effect of temperature

on a reaction.

60

119

Temperature:The Arrhenius Equation• If the Arrhenius equation is written for two

temperatures, T2 and T1 with T2 >T1.

ln k ln A - ERT

and

ln k ln A - ERT

1a

1

2a

2

120

Temperature:The Arrhenius Equation1. Subtract one equation from the other.

ln k k A - ln A - ERT

ERT

ln k k ERT

- ERT

2 1a

2

a

1

2 1a

1

a

2

ln ln

ln

61

121

Temperature:The Arrhenius Equation2. Rearrange and solve for ln k2/k1.

ln kk

ER T T

or

ln kk

ER

T - TT T

2

1

a

1 2

2

1

a 2 1

2 1

1 1

122

Temperature:The Arrhenius Equation• Consider the rate of a reaction for which Ea=50

kJ/mol, at 20oC (293 K) and at 30oC (303 K).– How much do the two rates differ?

ln kk

ER

T - TT T

ln kk 8.314

K

ln kkkk

e

2

1

a 2 1

2 1

2

1

Jmol

JK mol

2

1

2

1

0.677

50 000 303 293303 293

0 677

197 2

,

.

.

62

123

Temperature:The Arrhenius Equation• For reactions that have an Ea50 kJ/mol, the

rate approximately doubles for a 100C rise intemperature, near room temperature.

• Consider:2 ICl(g) + H2(g) I2(g) + 2 HCl(g)

• The rate-law expression is known to beR=k[ICl][H2].

At 230 C, k = 0.163 s

At 240 C, k = 0.348 sk approximately doubles

0 -1 -1

0 -1 -1

M

M

124

Catalysts

• Catalysts change reaction rates by providing analternative reaction pathway with a differentactivation energy.

63

125

Catalysts

• Homogeneous catalysts exist in same phase asthe reactants.

• Heterogeneous catalysts exist in differentphases than the reactants.– Catalysts are often solids.

126

Catalysts

• Examples of commercial catalyst systemsinclude:

systemconvertercatalyticAutomobile

ONNO2

CO2O+CO2

OH18CO16O25+HC

g2g2PtandNiO

g

g2PtandNiO

g2g

g2g2PtandNiO

g2g188

64

127

Catalysts

• This movie shows catalytic converterchemistry on the Molecular Scale

128

Catalysts

• A second example of a catalytic system is:

npreparatioacidSulfuric

SO2OSO2 g3NiO/PtorOV

g2g252

65

129

Catalysts

• A third examples of a catalytic system is:

ProcessHaber

NH2H3N g3OFeorFe

g2g232

130

Catalysts• Look at the catalytic oxidation of CO to CO2• Overall reaction

2 CO(g)+ O2(g)2CO2(g)

• AbsorptionCO(g)CO(surface) + O2(g)

O2(g)O2(surface)

• ActivationO2(surface) O(surface)

• ReactionCO(surface) +O(surface)CO2(surface)

• DesorptionCO2(surface)CO2(g)

66

131

Synthesis Question

• The Chernobyl nuclear reactor accidentoccurred in 1986. At the time that thereactor exploded some 2.4 MCi ofradioactive 137Cs was released into theatmosphere. The half-life of 137Cs is 30.1years. In what year will the amount of137Cs released from Chernobyl finallydecrease to 100 Ci? A Ci is a unit ofradioactivity called the Curie.

132

Synthesis Question

Ci100reachesChernobylatemittedCsfromityradioactiv the when24264401986

years440 years439 y0230.0

10.1t

t y0230.010.1

t y0230.0Ci100

Ci102.4ln

decayeradioactivfor this1aandk taAAln

Ci102.4 MCi2.4

y0230.0 y1.30

693.0t693.0k

k693.0t

137

1-

1-

1-6

0

6

1-

21

21

67

133

Group Question

• 99mTc has a half-life of 6.02 hours and isoften used in nuclear medical diagnostictests. Patients are injected withapproximately 10 mCi of 99mTc that is thendirected to specific sites in the patient’sbody to detect gallstones, brain tumorsand function, and other medicalconditions. How long will the patient havea higher than normal radioactivity levelafter they have been injected with 10 mCiof 99mTc?

16 Chemical Kinetics

Lecture13 Chemical Kinetics

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