CHAPTER 14 Chemical Kinetics. 2 Chapter Goals 1. The Rate of a Reaction Factors That Affect Reaction Rates 2. Nature of the Reactants 3. Concentrations

CHAPTER 14

Chemical Kinetics

2

Chapter Goals

1. The Rate of a Reaction

Factors That Affect Reaction Rates

2. Nature of the Reactants

3. Concentrations of the Reactants: The Rate-Law Expression

4. Concentration Versus Time: The Integrated Rate Equation

5. Collision Theory of Reaction Rates

3

Chapter Goals

6. Transition State Theory

7. Reaction Mechanisms and the Rate-Law Expression

8. Temperature: The Arrhenius Equation

9. Catalysts

4

The Rate of a Reaction

Kinetics is the study of rates of chemical reactions and the mechanisms by which they occur.

The reaction rate is the increase in concentration of a product per unit time or decrease in concentration of a reactant per unit time.

A reaction mechanism is the series of molecular steps by which a reaction occurs.

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 be imperceptible.

OUSINSTANTANE

kJ -79=G OHOH+H

SLOW VERY

kJ 396G COO C

o2982

-aq

+aq

o298g2g2diamond

l

6

The Rate of Reaction

Consider the hypothetical reaction,

A(g) + B(g) C(g) + D(g)

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

7

0

0.2

0.4

0.6

0.8

1

1.2

0 50 100

150

200

250

300

350

Time

Con

cent

rati

ons

of

Rea

ctan

ts &

Pro

duct

s

[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 reactants disappear or products appear.

This movie is an illustration of a reaction rate.

9


Mathematically, the rate of a reaction can be written as:

t d

D+

t c

C+or

t b

B-

t a

A-= Rate

10


The rate of a simple one-step reaction is directly proportional to the concentration of the reacting substance.

[A] is the concentration of A in molarity or moles/L. k is the specific rate constant.

k is an important quantity in this chapter.

Ak = Rateor ARate

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

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 would be:

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 the initial rate of reaction is cut in half.

12


If more than one reactant molecule appears in the equation for a one-step reaction, we can experimentally determine that the reaction rate is proportional to the molar concentration of the reactant raised to the power of the number of molecules involved in the reaction.

22

ggg

Xk = Rateor XRate

Z+ YX 2

13


Rate Law Expressions must be determined experimentally.

The rate law cannot be determined from the balanced 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 laws

2. rate equations

3. rate expressions

14

The Rate of Reaction 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.order first and

ONin order first isreaction This

ONk= Rate

O + NO4ON 2

52

52

g2g2g52

15


A second example is:

overall.order first and ,OHin order zero

CBr,CHin order first isreaction This

]CBrCHk[= Rate

BrCOHCHOHCBrCH

-

33

33

-aqaq33

-aqaq33

16


A final example of the order of a reaction is:

ALLYEXPERIMENT DETERMINED

ARE SEXPRESSION RATE ALL REMEMBER,

overallorder third and ,Oin order first

NO,in order second isreaction This

isreaction This Ok[NO]=Rate

NO 2O+NO 2

2

22

g2g2g

17

The Rate of Reaction Given the following one step reaction and its rate-law expression.

Remember, the rate expression would have to be experimentally determined.

Because it is a second order rate-law expression: 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

CBA 2

18

Factors That Affect Reaction Rates

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.

19

Nature of Reactants

This is a very broad category that encompasses the different reacting properties of substances.

For example sodium reacts with water explosively at room temperature to liberate hydrogen and form sodium hydroxide.

burns. and ignites H The

reaction. rapid and violent a is This

HNaOH 2OH 2Na 2

2

g2aq2s

20

Nature of Reactants

• Calcium reacts with water only slowly at room temperature to liberate hydrogen and form calcium hydroxide.

reaction. slowrather a is This

HOHCaOH 2Ca g2aq22s

21

Nature of Reactants

• The reaction of magnesium with water at room temperature is so slow that that the evolution of hydrogen is not perceptible to the human eye.

reaction No OH Mg 2s

22

Nature of Reactants

However, Mg reacts with steam rapidly to liberate H2 and form magnesium oxide.

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

g2sC100

2s HMgOOHMg o

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 in the 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 B BA B

A BA BA B

4 different possible A-B collisions



25


Example 16-1: The following rate data were obtained at 25oC for the following reaction. What are the rate-law expression and the specific rate-constant for this reaction?

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

Experiment

Number

Initial [A]

(M)

Initial [B]

(M)

Initial rate of formation 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


.Bignorecan can weThus,

k[A]=Rateor BAk=Rate

012

constant. remains rate initial that theNotice

2.by increases [B] that theandconstant is [A] the

thatsee we3 and 1 sexperiment compare weIf

BAk=Rate

:form theof bemust law rate The

0 xx

y

yx

y

27


You do it!

xx

reaction? for this

k of units and value theisWhat

overall.order 1 andA to

respect order with 1 isreaction This

k[A]=Rateor k[A]= Rate

122

2.by increases rate theand

2by increases [A] that theNotice

2. and 1 sexperiment compare Next,

st

st

1

28


[A] 10 x 2.0=Rate

as written becan law rate theThus

10 x 2.0 10.0

10 x 2.0=k

1 experiment from [A] and Rate of values theUsing

A

Rate=k

law.rate thefromk of value thefindcan We

s13-

s13-s

4-

M

M

29


Example 16-2: The following data were obtained for the following reaction at 25oC. What are the rate-law expression and the specific rate constant for 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 rate of

formation of 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=Rateor CBAk=Rate

013

constant. remains rate but the 3by increases C The

constant.remain B and A that Notice

3. and 1 sexperiment Compare

31


BAk=Rateor BAk=Rate

133

3.by increases rate theand 3by increases B The

constant. remains A The


1 xx

y y

32


overall.order 2 and B, respect to order with1

A, respect to order with 1 isreaction This

BAk=Rateor BAk= Rate

133

3.by increases rate theand 3by increases A The

constant. remains B The


ndst

st

11

xx

33


BA100.1 = Rate

as written becan law-rate theThus,

100.1

0.10 0.20

100.2

BA

Rate=k

4.or 3, 2, 1, experiment from data theuseCan

k. of units and value thedetermine Finally,

s12

s12

s4

M

M

M

MM

34


Example 16-3: consider a chemical reaction between compounds A and B that is first order with respect to A, first order with respect to B, and second order overall. From the information given below, fill in the blanks.

You do it!Experiment

Initial 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 ?

35


BA40.0 Rate theThus

40.0

0.050 0.20

100.4

BA

Rate=k

k. of value thedeterminecan we1 experiment From

BAk=Rate

s 1

s 1

s3

M

M

M

MM

36


M

MM

M

80.0]A[

050.0s0.40

s106.1]A[

k[B]

Rate[A]

2 experiment from data andk of value theUse

1-1

1-2

37


M

MM

M

20.0[B]

40.0s0.40

s102.3[B]

k[A]

R[B]

determinecan we3 experiment from Similarly,

1-1

1-2

38

Concentration vs. Time: The Integrated Rate Equation The integrated rate equation relates time and

concentration for chemical and nuclear reactions. From the integrated rate equation we can predict

the amount of product that is produced in a given amount of time.

Initially we will look at the integrated rate equation for first order reactions.These reactions are 1st order in the reactant and 1st

order overall.

39

Concentration vs. Time: The Integrated 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 many chemical reactions and all simple radioactive decays.

Two examples of this type are:

2 N2O5(g) 2 N2O4(g) + O2(g)

238U 234Th + 4He

40

Concentration vs. Time: The Integrated 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 reactions is:

k t aA

Aln 0

41


Solve the first order integrated rate equation for t.

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

A

Aln

k a

1t 0

42


At time t = t1/2, the expression becomes:

k a

693.0t

2lnk a

1t

A1/2

Aln

k a

1t

1/2

1/2

0

01/2

43


Example 16-4: Cyclopropane, an anesthetic, decomposes to propene according to the following equation.

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

CH2 CH2

CH2CH2CH3

CH

(g) (g)

s 075.0s 2.9

693.0

k

693.0t

1-1/2

44


Example 16-5: Refer to Example 16-4. How much of a 3.0 g sample of cyclopropane 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.

45


remains 1%or g 03.0eA

5.31.16.4Aln

6.4Aln 1.1

s 50.0s 2.9Aln -3.0ln

k t AlnAln

.logarithms of laws theusemust We

reaction for thisk t k t aA

Aln

3.5-

1-

0

0

46


Example 16-6: The half-life for the following first order reaction is 688 hours at 10000C. Calculate the specific rate constant, k, at 10000C and the amount of a 3.0 g sample of CS2 that remains after 48 hours.

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

You do it!

47


1-

1/2

1/2

hr 00101.0hr 688

0.693k

t

0.693k

k

693.0t

1. areaction For this

48


hr) 48)(hr 00101.0(Aln-ln(3.0)

k tAlnAlnk tA

Aln

1-

00

unreacted 97%or g 9.2g86.2eA

1.0521.1)--(0.048Aln

0.048Aln-1.1

hr) 48)(hr 00101.0(Aln-ln(3.0)

k tAlnAlnk tA

Aln

1.052

1-

00

49


For reactions that are second order with respect to a particular reactant and second order overall, 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 since beginning of reaction.

a = stoichiometric coefficient of A in balanced overall equation.

k t aA

1

A

1

0

50


Second order reactions also have a half-life. Using the second order integrated rate-law as a starting

point.

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

0

1/200

A ofr denominatocommon a haswhich

k t aA

1

A2/1

1

1/20

1/200

k t aA

1

or k t aA

1

A

2

51


If we solve for t1/2:

Note that the half-life of a second order reaction depends on the initial concentration of A.

01/2 Ak a

1t

52


Example 16-7: Acetaldehyde, CH3CHO, undergoes gas phase thermal decomposition to methane and carbon monoxide.

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

CH CHO CH + CO3 g 4 g g

53


tk A

hr

hr

1/2

-1

1

1

2 0 10 010

5 0 10

0

2 1

2

. .

.

M M

54


(b) How many moles of CH3CHO remain after 200 hours?

1 1

1 1010

2 0 10 200

110 4 0

0

2 1

1 1

A Ak t

A hr hr

A

-1

..

.

MM

M M

114

1

14

0 071

0 071 mol

11A

A

A

mol = 1.0 L x 0.071 molL

M

M

M.

? .

55


(c) What percent of the CH3CHO remains after 200 hours?

reacted 29% and unreacted %71

%100mol 0.10

mol 0.071= unreacted %

56


Example 16-8: Refer to Example 16-7. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 10.0 L vessel at 527oC? Note that the vessel size is increased by a factor

of 10 which decreases the concentration by a factor of 10!

You do it!

57


tk A

hr

hr

note the time has increased by 10

over Example 16 - 7:

1/2

-1

1

1

2 0 10 0 010

5 0 10

0

2 1

3

. .

.

M M

58


(b) How many moles of CH3CHO remain after 200 hours?

You do it!

59


1 1

1 10 010

2 0 10 200

1100 4 0

0

2 1

1 1

A Ak t

A hr hr

A

-1

..

.

MM

M M

1104

1

104

0 0096

0 096 mol

11A

A

A

mol = 10.0 L x 0.0096 molL

M

M

M.

? .

60


(c) What percent of the CH3CHO remains after 200 hours?

You do it!

61


% unreacted =0.096 mol0.100 mol

unreacted & 4% reacted

100%

96%

62


Let us now summarize the results from the last two examples.

Initial Moles

CH3CHO

[CH3CHO]0

(M)

[CH3CHO]

(M)

Moles of CH3CHO at 200 hr.

% CH3CHO remaining

Ex. 16-7 0.10 0.10 0.071 0.071 71%

Ex. 16-8 0.010 0.010 0.0096 0.096 96%

63

Enrichment - Derivation of Integrated Rate Equations

For the first order reaction

a A products

the rate can be written as:

t

A

a

1-=Rate

64


For a first-order reaction, the rate is proportional to the first power of [A].

-1a

At

k A

65


In calculus, the rate is defined as the infinitesimal change of concentration d[A] in an infinitesimally short time dt as the derivative of [A] with respect to time.

-1a

At

k Ad

d

66


Rearrange the equation so that all of the [A] terms are on the left and all of the t terms are on the right.

-A

Aa k t

dd

67


Express the equation in integral form.

-

AA

a k tA

A tdd

0 0

68


This equation can be evaluated as:

-ln A a k t or

-ln A A a k t - a k 0

which becomes

-ln A A a k t

t0t

t

t

0

0

0

ln

ln

69


Which rearranges to the integrated first order rate equation.

k t aA

Aln

t

0

70


Derive the rate equation for a reaction that is second order in reactant A and second order overall.

The rate equation is:

2Ak t a

A

d

d

71


Separate the variables so that the A terms are on the left and the t terms on the right.

tk A a

A2 d

d

72


Then integrate the equation over the limits as for the first order reaction.

t

0

A

A2 tk a

A

A

0

dd

73


Which integrates the second order integrated rate equation.

k t aA

1

A

1

0

74


For a zero order reaction the rate expression is:

k

t a

A

d

d

75


Which rearranges to:

tk aA dd

76


Then we integrate as for the other two cases:

t

0

A

A

tk aA0

dd

77


Which gives the zeroeth order integrated rate equation.

k t a-AA

or

k t -aAA

0

0

78

Enrichment - Rate Equations to Determine 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 t aAln

or

k t aAlnAln

79


The equation for a straight line is:

Compare this equation to the rearranged first order rate-law.

bmy x

80


b m y x

Now we can interpret the parts of the equation as 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 t aAln

81


Example 16-9: Concentration-versus-time data for the thermal decomposition of ethyl bromide are given in the table below. Use the following graphs of the data to determine the rate of the reaction and the value of the rate constant.

700Kat HBrHCBrHC gg42g52

82


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

83


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 zero order

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 second

order with respect to [C2H5Br].

84

Enrichment - Rate Equations to Determine Reaction Order Plot of [C2H5Br] versus time.

Is it linear or not?

[C2H5Br] vs. time

0

0.20.4

0.60.8

11.2

0 1 2 3 4 5

Time (min)

[C2

H5

Br]

85


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

0

0 1 2 3 4 5

Time (min)

ln [

C2H

5B

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 5

Time (min)

1/[C

2H5B

r]

87

Enrichment - Rate Equations to 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

min 20.0min 3

60.0slope

min 14

)20.0(80.0

x-x

y-y slope

88


From the equation for a first order reaction we know that the slope = -a k. In this reaction a = 1.

.min 0.20kconstant rate theThus

-k-0.20slope1-

89


The integrated rate equation for a reaction that is second order in reactant A and second order overall.

This equation can be rearranged to:

k t aA

1

A

1

0

0A

1k t a

A

1

90


Compare the equation for a straight line and the second order rate-law expression.

Now we can interpret the parts of the equation as 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.

b m y x

0A

1k t a

A

1

91


Example 16-10: Concentration-versus-time data for the decomposition of nitrogen dioxide are given in the table below. Use the graphs to determine the rate of the reaction and the value of the rate constant

500Kat ONO 2NO 2 g2gg2

92


Time

(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

93


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


Plot of [NO2] versus time. Is it linear or not?

[NO2] vs. time

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5

Time (min)

[NO

2]

95

Enrichment -Rate Equations to Determine Reaction Order

Plot of ln [NO2] versus time. Is it linear or not?

ln [NO2] vs. time

-2

-1.5

-1

-0.5

0

0 1 2 3 4 5

Time (min)

ln [

NO

2]

96


Plot of 1/[NO2] versus time. Is it linear or not?

1/[NO2] vs.time

01

23

45

6

0 1 2 3 4 5

Time (min)

1/[

NO

2]

97


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 on the graph of 1/[NO2] vs. time.

98


From the equation for a first order reaction we know that the slope = a k In this reaction a = 2.

1-1 min 0.45kconstant rate theThus

k 20.90slope

M

min 1

1

1

12

12

90.0min 4

60.3slope

min 15

)90.1(50.5

x-x

y-y slope

MM

M

99

Collision Theory of Reaction Rates

Three basic events must occur for a reaction to occur the atoms, molecules or ions 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


One method to increase the number of collisions and the energy necessary to break and reform bonds is to heat the molecules.

As an example, look at the reaction of methane and 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.

kJ 891 OH CO O CH (g)22(g)2(g)4(g)

101


Illustrate the proper orientation of molecules that is necessary for this reaction.

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

Some possible ineffective collisions are :

X

X

Y YY

Y

X X X X Y Y

102


An example of an effective collision is:

X Y

X Y

X Y

X Y

X Y +

X Y

103


This picture illustrates effective and ineffective molecular collisions.

104

Transition State Theory

Transition state theory postulates that reactants form a high energy intermediate, the transition state, which then falls apart into the products.

For a reaction to occur, the reactants must acquire sufficient energy to form the transition state. This energy is called the activation energy or Ea.

Look at a mechanical analog for activation energy

105


Epot = mgh

Cross section of mountain

Boulder

Eactivation

h

h2

h1

Epot=mgh2

Epot=mgh1

Height

106


PotentialEnergy

Reaction Coordinate

X2 + Y2

2 XY

Eactivation - a kinetic quantity

E Ha thermodynamic quantity

Representation of a chemical reaction.

107


108


The relationship between the activation energy for forward and reverse reactions is Forward reaction = Ea

Reverse reaction = Ea + E difference = E

109


The distribution of molecules possessing different energies at a given temperature is represented in this figure.

110

Reaction Mechanisms and the Rate-Law Expression

Use the experimental rate-law to postulate a molecular mechanism.

The slowest step in a reaction mechanism is the rate determining step.

111


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 to water and oxygen.

g22I

22 O + OH 2 OH 2-

112


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=R law rate alExperiment

O+OH 2OH 2reaction Overall

I+O+OHOH+ IO stepFast

OH+IOI+OH step Slow

113


Important notes about this reaction:1. One hydrogen peroxide molecule and one

iodide ion are involved in the rate determining step.

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

3. Hypoiodite ion has been detected in the reaction mixture as a short-lived reaction intermediate.

114


Ozone, O3, reacts very rapidly with nitrogen oxide, NO, in a reaction that is first order in each reactant and second order overall.

NOOk=Rate is law-rate alExperiment

O+NONO+O

3

g2g2gg3

115


One possible mechanism is:

223

223

33

O+NONO+Oreaction Overall

O+NONO+O stepFast

O+NONO+O step Slow

116


A mechanism that is inconsistent with the rate-law expression is:

correct. becannot mechanism thisproveswhich

Ok=Rate is mechanism thisfrom law-rate The

ONONO+Oreaction Overall

NONO+O stepFast

O+OO step Slow

3

223

2

23

117


Experimentally determined reaction orders indicate the number of molecules involved in:

1. the slow step only or

2. the slow step and the equilibrium steps preceding the slow step.

118

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 = Ae

or

ln k = ln A -ERT

-E RT

a

a

119


This movie illustrates the effect of temperature on a reaction.

120


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 -E

RT

1a

1

2a

2

121


1. Subtract one equation from the other.

ln k k A - ln A -E

RTERT

ln k kERT

-E

RT

2 1a

2

a

1

2 1a

1

a

2

ln ln

ln

122


2. 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

123


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?

lnkk

ER

T - TT T

lnkk

8.314

K

lnkk

kk

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

,

.

.

124

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

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

Consider:

2 ICl(g) + H2(g) I2(g) + 2 HCl(g) The rate-law expression is known to be R=k[ICl][H2].

At 230 C, k = 0.163 s

At 240 C, k = 0.348 s

k approximately doubles

0 -1 -1

0 -1 -1

M

M

125

Catalysts

Catalysts change reaction rates by providing an alternative reaction pathway with a different activation energy.

126

Catalysts

Homogeneous catalysts exist in same phase as the reactants.

Heterogeneous catalysts exist in different phases than the reactants. Catalysts are often solids.

127

Catalysts

Examples of commercial catalyst systems include:

systemconverter catalytic Automobile

ONNO 2

CO 2O+CO 2

OH 18CO16O 25+HC

g2g2Pt and NiO

g

g2Pt and NiO

g2g

g2g2Pt and NiO

g2g188

128

Catalysts

This movie shows catalytic converter chemistry on the Molecular Scale

129

Catalysts

A second example of a catalytic system is:

npreparatio acid Sulfuric

SO 2OSO 2 g3NiO/Ptor OV

g2g252

130

Catalysts

A third examples of a catalytic system is:

ProcessHaber

NH 2H 3 N g3OFeor Fe

g2g232

131

Catalysts Look at the catalytic oxidation of CO to CO2

Overall reaction

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

Absorption

CO(g) CO(surface) + O2(g)

O2(g) O2(surface)

Activation

O2(surface) O(surface)

Reaction

CO(surface) +O(surface) CO2(surface) Desorption

CO2(surface) CO2(g)

132

Synthesis Question

The Chernobyl nuclear reactor accident occurred in 1986. At the time that the reactor exploded some 2.4 MCi of radioactive 137Cs was released into the atmosphere. The half-life of 137Cs is 30.1 years. In what year will the amount of 137Cs released from Chernobyl finally decrease to 100 Ci? A Ci is a unit of radioactivity called the Curie.

133

Synthesis Question

Ci 100 reaches Chernobylat emitted Csfrom

ity radioactiv the when2426 440 1986

years440 years439 y0230.0

10.1t

t y0230.010.1

t y0230.0Ci 100

Ci 102.4ln

decay eradioactiv for this 1a andk t a A

Aln

Ci 102.4 MCi2.4

y0230.0 y1.30

693.0

t

693.0k

k

693.0t

137

1-

1-

1-6

0

6

1-

21

21

134

Group Question

99mTc has a half-life of 6.02 hours and is often used in nuclear medical diagnostic tests. Patients are injected with approximately 10 mCi of 99mTc that is then directed to specific sites in the patient’s body to detect gallstones, brain tumors and function, and other medical conditions. How long will the patient have a higher than normal radioactivity level after they have been injected with 10 mCi of 99mTc?

135

End of Chapter 16

A great deal of chemistry is a competition between Thermodynamics (Chapter 15) and Kinetics (Chapter 16).

Documents

CHAPTER 14 Chemical Kinetics. 2 Chapter Goals 1. The Rate of a Reaction Factors That Affect Reaction Rates 2. Nature of the Reactants 3. Concentrations