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Chapter 14 Chemical Kinetics

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Title: Chapter 14 Chemical Kinetics


1
Chapter 14Chemical Kinetics
2
14.1 Factors that Affect Reaction Rates
3
Kinetics
  • In kinetics we study the rate at which a chemical
    process occurs.
  • Besides information about the speed at which
    reactions occur, kinetics also sheds light on the
    reaction mechanism (exactly how the reaction
    occurs).

4
Factors That Affect Reaction Rates
  • Physical State of the Reactants
  • In order to react, molecules must come in contact
    with each other.
  • The more homogeneous the mixture of reactants,
    the faster the molecules can react.

5
Factors That Affect Reaction Rates
  • Concentration of Reactants
  • As the concentration of reactants increases, so
    does the likelihood that reactant molecules will
    collide.

6
Factors That Affect Reaction Rates
  • Temperature
  • At higher temperatures, reactant molecules have
    more kinetic energy, move faster, and collide
    more often and with greater energy.

7
Factors That Affect Reaction Rates
  • Presence of a Catalyst
  • Catalysts speed up reactions by changing the
    mechanism of the reaction.
  • Catalysts are not consumed during the course of
    the reaction.
  • P.575 GIST How does increasing the partial
    pressures of the reactive components of a gaseous
    mixture affect the rate at which the components
    react with one another?

8
14.2 Reaction Rates
9
Reaction Rates
  • Rates of reactions can be determined by
    monitoring the change in concentration of either
    reactants or products as a function of time.

10
Sample Exercise 14.1 Calculating an Average Rate
of Reaction
From the data given in the caption of Figure
14.3, calculate the average rate at which A
disappears over the time interval from 20 s to 40
s.
11
Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
  • In this reaction, the concentration of butyl
    chloride, C4H9Cl, was measured at various times.

12
Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
  • The average rate of the reaction over each
    interval is the change in concentration divided
    by the change in time

p.577 GIST Why do the rates of reactions
decrease as concentrations decrease?
13
Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
  • Note that the average rate decreases as the
    reaction proceeds.
  • This is because as the reaction goes forward,
    there are fewer collisions between reactant
    molecules.

14
Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
  • A plot of C4H9Cl vs. time for this reaction
    yields a curve like this.
  • The slope of a line tangent to the curve at any
    point is the instantaneous rate at that time.

p.578 GIST This figure shows two triangles used
to determine the slope of the curve at two
different times. How do you determine how large
to draw the triangle when determining the slope
of a curve at a particular point?
15
Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
  • All reactions slow down over time.
  • Therefore, the best indicator of the rate of a
    reaction is the instantaneous rate near the
    beginning of the reaction.

16
Sample Exercise 14.2 Calculating an Instantaneous
Rate of Reaction
Using Figure 14.4, calculate the instantaneous
rate of disappearance of C4H9Cl at t 0 (the
initial rate).
17
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18
Reaction Rates and Stoichiometry
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
  • In this reaction, the ratio of C4H9Cl to C4H9OH
    is 11.
  • Thus, the rate of disappearance of C4H9Cl is the
    same as the rate of appearance of C4H9OH.

19
Reaction Rates and Stoichiometry
  • What if the ratio is not 11?

2 HI(g) ??? H2(g) I2(g)
  • In such a case,

20
Reaction Rates and Stoichiometry
  • To generalize, then, for the reaction

21
Sample Exercise 14.3
  • How is the rate at which ozone disappears related
    to the rate at which oxygen appears in the
    reaction 2O3(g) ? 3O2(g)?
  • If the rate at which O2 appears is 6.0 x 10-5 M/s
    at a particular instant, at what rate is O3
    disappearing at the same time?

22
Practice
  • The decomposition of N2O5 proceeds according to
    the following equation
  • 2N2O5(g) ? 4NO2(g) O2(g)
  • If the rate of decomposition of N2O5 at a
    particular instant in a reaction vessel is 4.2 x
    10-7 M/s, what is the rate of appearance of NO2?
    Of O2?

23
14.3 Concentration and Rate
24
Concentration and Rate
  • One can gain information about the rate of a
    reaction by seeing how the rate changes with
    changes in concentration.

25
Concentration and Rate
  • If we compare Experiments 1 and 2, we see that
    when NH4 doubles, the initial rate doubles.

26
Concentration and Rate
  • Likewise, when we compare Experiments 5 and 6,
    we see that when NO2- doubles, the initial rate
    doubles.

27
Concentration and Rate
  • This means
  • Rate ? NH4
  • Rate ? NO2-
  • Rate ? NH4 NO2-
  • which, when written as an equation, becomes
  • Rate k NH4 NO2-
  • This equation is called the rate law, and k is
    the rate constant.

Therefore,
28
Rate Laws
  • A rate law shows the relationship between the
    reaction rate and the concentrations of
    reactants.
  • The exponents tell the order of the reaction with
    respect to each reactant.
  • Since the rate law is
  • Rate k NH4 NO2-
  • the reaction is
  • First-order in NH4 and
  • First-order in NO2-.

29
Rate Laws
  • Rate k NH4 NO2-
  • The overall reaction order can be found by adding
    the exponents on the reactants in the rate law.
  • This reaction is second-order overall.

30
14.3 GIST
  • What is a rate law?
  • What is the name of the quantity k in any rate
    law?
  • The experimentally determined rate law for the
    reaction 2NO 2H2 ? N2 2H2O is rate
    kNO2H2. What are the reaction orders in this
    rate law? Does doubling the concentration of NO
    have the same effect on rate as doubling the
    concentration of H2?

31
Sample Exercise 14.4
  • Consider a reaction A B ? C for which rate
    kAB2. Each of the boxes represent a reaction
    mixture (A is red, B is purple). Rank these
    mixtures in order of increasing rate of reaction.

32
Practice
  • Assuming the rate kAB, rank the mixtures in
    order of increasing rate.

33
Units of Rate Constant
  • Units of rate (units of rate constant)(units of
    concentration)x So,
  • Units of rate constant units of rate / (units
    of concentration)x
  • Units of k MEMORIZE
  • Zero order M/s
  • First order s-1
  • Second order M-1s-1
  • Third order M-2s-1

34
Sample Exercise 14.5 Determining Reaction Order
and Units of Rate Constants
(a) What are the overall reaction orders for the
reactions described in Equations 14.9 and 14.10?
(b) What are the units of the rate constant for
the rate law in Equation 14.9?
14.9 2 N2O5 ? 4 NO2 O2 Rate kN2O5
14.10 CHCl3 Cl2 ? CCl4 HCl Rate
kCHCl3Cl21/2
35
14.11 H2 I2 ? 2HI Rate kH2I2
36
Method of Initial Rates
  • Used to find the form of the rate law
  • Write the general form of the rate law, then
    determine order by doing the following.
  • Choose one reactant to start with
  • Find two experiments where the concentration of
    that reactant changes but all other reactants
    stay the same
  • Write the rate laws for both experiments
  • Divide the two rate laws
  • Use log rules to solve for the order
  • Follow the same technique for other reactants

37
Example
  • Choose one reactant to start with
  • Find two experiments where the concentration of
    that reactant changes but all other reactants
    stay the same

38
Example
39
Sample Exercise 14.6 Determining a Rate Law from
Initial Rate Data
40
Sample Exercise 14.6 Determining a Rate Law from
Initial Rate Data
41
14.4 The Change of Concentration With Time
42
Integrated Rate Laws
  • Using calculus to integrate the rate law for a
    first-order process gives us

Where
A0 is the initial concentration of A, and At
is the concentration of A at some time, t, during
the course of the reaction.
43
Integrated Rate Laws
  • Manipulating this equation produces

ln At - ln A0 - kt
ln At - kt ln A0
which is in the form
y mx b
44
First Order
  • lnA -kt lnA0
  • can calculate the concentration at any time
  • can be determined by a graph
  • where
  • y lnA
  • x t
  • m -k
  • b lnA0

lnA vs. t is a straight line
45
First Order
  • half-life
  • time is takes for a reactant concentration to
    reach half of its original concentration
  • does not depend on the concentration

46
First-Order Processes
ln At -kt ln A0
  • Therefore, if a reaction is first-order, a plot
    of ln A vs. t will yield a straight line, and
    the slope of the line will be -k.

47
First-Order Processes
  • Consider the process in which methyl isonitrile
    is converted to acetonitrile.

48
First-Order Processes
  • This data was collected for this reaction at
    198.9 C.

49
First-Order Processes
  • When ln P is plotted as a function of time, a
    straight line results.
  • Therefore,
  • The process is first-order.
  • k is the negative of the slope 5.1 ? 10-5 s-1.
  • P.587 GIST what do the y-intercepts in the above
    graphs represent?

50
Sample Exercise 14.7 Using the Integrated
First-Order Rate Law
The decomposition of a certain insecticide in
water follows first-order kinetics with a rate
constant of 1.45 yr1 at 12 C. A quantity of
this insecticide is washed into a lake on June 1,
leading to a concentration of 5.0 107 g/cm3.
Assume that the average temperature of the lake
is 12 C. (a) What is the concentration of the
insecticide on June 1 of the following year? (b)
How long will it take for the concentration of
the insecticide to decrease to 3.0 107 g/cm3?
51
Sample Exercise 14.7 Using the Integrated
First-Order Rate Law
52
Second-Order Processes
  • Similarly, integrating the rate law for a
    process that is second-order in reactant A, we get

also in the form
y mx b
53
Second-Order Processes
  • So if a process is second-order in A, a plot of
    vs. t will yield a straight line, and the
    slope of that line is k.

54
Second-Order Processes
The decomposition of NO2 at 300C is described by
the equation
and yields data comparable to this
Time (s) NO2, M
0.0 0.01000
50.0 0.00787
100.0 0.00649
200.0 0.00481
300.0 0.00380
55
Second-Order Processes
  • Plotting ln NO2 vs. t yields the graph at the
    right.
  • The plot is not a straight line, so the process
    is not first-order in A.

Time (s) NO2, M ln NO2
0.0 0.01000 -4.610
50.0 0.00787 -4.845
100.0 0.00649 -5.038
200.0 0.00481 -5.337
300.0 0.00380 -5.573
56
Second-Order Processes
  • Graphing ln vs. t, however, gives this
    plot.
  • Because this is a straight line, the process is
    second-order in A.

Time (s) NO2, M 1/NO2
0.0 0.01000 100
50.0 0.00787 127
100.0 0.00649 154
200.0 0.00481 208
300.0 0.00380 263
57
Second Order
  • 1/A kt 1/A0
  • 1/A vs. t is a straight line with a slope of k

58
Sample Exercise 14.8 Determining Reaction Order
from the Integrated Rate Law
59
Sample Exercise 14.8 Determining Reaction Order
from the Integrated Rate Law
60
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61
Half-Life
  • Half-life is defined as the time required for
    one-half of a reactant to react.
  • Because A at t1/2 is one-half of the original
    A,
  • At 0.5 A0.

62
Half-Life
  • For a first-order process, this becomes

ln 0.5 -kt1/2
-0.693 -kt1/2
NOTE For a first-order process, then, the
half-life does not depend on A0.
63
Half-Life
  • For a second-order process,

64
Second Order
  • half-life does depend on the concentration for
    second order

65
14.5 GIST
  • If a solution containing 10.0 g of a substance
    reacts by first-order kinetics, how many grams
    remain after 3 half-lives?
  • How does the half-life of a second-order reaction
    change as the reaction proceeds?

66
Sample Exercise 14.9 Determining the Half-Life of
a First-Order Reaction
The reaction of C4H9Cl with water is a
first-order reaction. Figure 14.4 shows how the
concentration of C4H9Cl changes with time at a
particular temperature. (a) From that graph,
estimate the half-life for this reaction. (b) Use
the half-life from (a) to calculate the rate
constant.
67
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68
Zero Order
  • half-life does depend on the concentration

69
Example
  • A certain first order reaction has t1/2 of 20.0
    minutes
  • Find k
  • How much time is required for this reaction to be
    75 complete?

70
14.5 Temperature and Rate
71
Temperature and Rate
  • Generally, as temperature increases, so does the
    reaction rate.
  • This is because k is temperature dependent.

72
The Collision Model
  • In a chemical reaction, bonds are broken and new
    bonds are formed.
  • Molecules can only react if they collide with
    each other.

73
The Collision Model
  • Furthermore, molecules must collide with the
    correct orientation and with enough energy to
    cause bond breakage and formation.

74
Activation Energy
  • In other words, there is a minimum amount of
    energy required for reaction the activation
    energy, Ea.
  • Just as a ball cannot get over a hill if it does
    not roll up the hill with enough energy, a
    reaction cannot occur unless the molecules
    possess sufficient energy to get over the
    activation energy barrier.

75
Reaction Coordinate Diagrams
  • It is helpful to visualize energy changes
    throughout a process on a reaction coordinate
    diagram like this one for the rearrangement of
    methyl isonitrile.

76
Reaction Coordinate Diagrams
  • The diagram shows the energy of the reactants and
    products (and, therefore, ?E).
  • The high point on the diagram is the transition
    state.
  • The species present at the transition state is
    called the activated complex.
  • The energy gap between the reactants and the
    activated complex is the activation energy
    barrier.

77
MaxwellBoltzmann Distributions
  • Temperature is defined as a measure of the
    average kinetic energy of the molecules in a
    sample.
  • At any temperature there is a wide distribution
    of kinetic energies.

78
MaxwellBoltzmann Distributions
  • As the temperature increases, the curve flattens
    and broadens.
  • Thus at higher temperatures, a larger population
    of molecules has higher energy.

79
MaxwellBoltzmann Distributions
  • If the dotted line represents the activation
    energy, then as the temperature increases, so
    does the fraction of molecules that can overcome
    the activation energy barrier.
  • As a result, the reaction rate increases.

80
MaxwellBoltzmann Distributions
  • This fraction of molecules can be found through
    the expression
  • where R is the gas constant and T is the Kelvin
    temperature.

f e
81
Arrhenius Equation
  • Svante Arrhenius developed a mathematical
    relationship between k and Ea
  • k A e
  • where A is the frequency factor, a number that
    represents the likelihood that collisions would
    occur with the proper orientation for reaction.

82
Arrhenius Equation
  • Taking the natural logarithm of both sides, the
    equation becomes
  • ln k - ( ) ln A

y m x b
Therefore, if k is determined experimentally at
several temperatures, Ea can be calculated from
the slope of a plot of ln k vs. .
83
14.5 GIST
  • What is the central idea of the collision model?
  • Why isnt collision frequency the only factor
    affecting a reaction rate?

84
Sample Exercise 14.10 Relating Energy Profiles to
Activation Energies and Speed of Reaction
85
Sample Exercise 14.11 Determining the Energy of
Activation
86
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87
Sample Exercise 14.11 Determining the Energy of
Activation
88
14.6 Reaction Mechanisms
89
Reaction Mechanisms
  • The sequence of events that describes the actual
    process by which reactants become products is
    called the reaction mechanism.

90
Reaction Mechanisms
  • Reactions may occur all at once or through
    several discrete steps.
  • Each of these processes is known as an elementary
    reaction or elementary process.
  • Rate laws can be determined directly from
    elementary reactions.
  • The rate law for the overall reaction is
    determined from the slowest elementary step in
    the mechanism

91
Reaction Mechanisms
  • Example
  • Overall NO2(g) CO(g) ? NO(g) CO2(g)
  • 2 Steps NO2(g) NO2(g) ? NO3(g) NO(g)
  • NO3(g) CO(g) ? NO2(g) CO2(g)

92
Reaction Mechanisms
  • intermediate
  • species that is formed and consumed in the steps
    so does not appear in overall reaction
  • appears as a product in one reaction and then
    reactant in following reaction
  • catalyst
  • species used to speed up the reaction
  • is not in overall reaction
  • appears as reactant and then as a product
  • Intermediates and catalysts do not appear in the
    rate law.

93
Reaction Mechanisms
  • The molecularity of a process tells how many
    molecules are involved in the process.

94
How do you know if the proposed mechanism is
correct?
  • 1. The series of elementary steps must add up to
    give overall balanced equation
  • 2. must agree with the experimentally determined
    rate law

95
p.598 GIST
  • What is the molecularity of this elementary
    reaction?
  • NO(g) Cl2(g) ? NOCl(g) Cl(g)

96
Sample Exercise 14.12 Determining Molecularity
and Identifying Intermediates
97
Sample Exercise 14.12 Determining Molecularity
and Identifying Intermediates
98
Sample Exercise 14.13 Predicting Rate Law for an
Elementary Reaction
99
Multistep Mechanisms
  • In a multistep process, one of the steps will be
    slower than all others.
  • The overall reaction cannot occur faster than
    this slowest, rate-determining step.

100
p.601 GIST
  • Why cant the rate law for a reaction generally
    be deduced from the balanced equation for the
    reaction?

101
Slow Initial Step
NO2 (g) CO (g) ??? NO (g) CO2 (g)
  • The rate law for this reaction is found
    experimentally to be
  • Rate k NO22
  • CO is necessary for this reaction to occur, but
    the rate of the reaction does not depend on its
    concentration.
  • This suggests the reaction occurs in two steps.

102
Slow Initial Step
  • A proposed mechanism for this reaction is
  • Step 1 NO2 NO2 ??? NO3 NO (slow)
  • Step 2 NO3 CO ??? NO2 CO2 (fast)
  • The NO3 intermediate is consumed in the second
    step.
  • As CO is not involved in the slow,
    rate-determining step, it does not appear in the
    rate law.

103
Sample Exercise 14.14 Determining the Rate Law
for a Multistep Mechanism
104
The reaction is believed to occur in two
steps The experimental rate law is rate
kO3NO2. What can you say about the relative
rates of the two steps of the mechanism?
105
Fast Initial Step
2 NO (g) Br2 (g) ??? 2 NOBr (g)
  • The rate law for this reaction is found to be
  • Rate k NO2 Br2
  • Because termolecular processes are rare, this
    rate law suggests a two-step mechanism.

106
Fast Initial Step
  • A proposed mechanism is

Step 2 NOBr2 NO ??? 2 NOBr (slow)
Step 1 includes the forward and reverse reactions.
107
Fast Initial Step
  • The rate of the overall reaction depends upon the
    rate of the slow step.
  • The rate law for that step would be
  • Rate k2 NOBr2 NO
  • But how can we find NOBr2?

108
Fast Initial Step
  • NOBr2 can react two ways
  • With NO to form NOBr
  • By decomposition to reform NO and Br2
  • The reactants and products of the first step are
    in equilibrium with each other.
  • Therefore,
  • Ratef Rater

109
Fast Initial Step
  • Because Ratef Rater ,
  • k1 NO Br2 k-1 NOBr2
  • Solving for NOBr2 gives us

110
Fast Initial Step
  • Substituting this expression for NOBr2 in the
    rate law for the rate-determining step gives

k NO2 Br2
111
Sample Exercise 14.15 Deriving the Rate Law for a
Mechanism with a Fast Initial Step
112
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113
14.7 Catalysis
114
Catalysts
  • Catalysts increase the rate of a reaction by
    decreasing the activation energy of the reaction,
    more collisions have the minimum Ea.
  • Catalysts change the mechanism by which the
    process occurs.
  • is never consumed in the reaction so it is not
    considered a reactant

115
Catalysts
  • One way a catalyst can speed up a reaction is by
    holding the reactants together and helping bonds
    to break.

116
Catalysis
  • heterogeneous
  • in different phases
  • homogeneous
  • same phase
  • enzyme
  • catalysts in body

117
Enzymes
  • Enzymes are catalysts in biological systems.
  • The substrate fits into the active site of the
    enzyme much like a key fits into a lock.

118
14.7 GIST
  • How does a catalyst increase the rate of a
    reaction?
  • How does a homogeneous catalyst compare with a
    heterogeneous one regarding the ease of recovery
    of the catalyst from the reaction mixture?
  • What names are given to the following aspects of
    enzyme catalysis
  • The place on the enzyme where catalysis occurs
  • The substances that undergo catalysis

119
Sample Integrative Exercise Putting Concepts
Together
Formic acid (HCOOH) decomposes in the gas phase
at elevated temperatures as follows HCOOH(g) ?
CO2(g) H2(g) The uncatalyzed decomposition
reaction is determined to be first order. A graph
of the partial pressure of HCOOH versus time for
decomposition at 838 K is shown as the red curve
in Figure 14.28. When a small amount of solid ZnO
is added to the reaction chamber, the partial
pressure of acid versus time varies as shown by
the blue curve in Figure 14.28. (a) Estimate the
half-life and first-order rate constant for
formic acid decomposition. (b) What can you
conclude from the effect of added ZnO on the
decomposition of formic acid? (c) The progress of
the reaction was followed by measuring the
partial pressure of formic acid vapor at selected
times. Suppose that, instead, we had plotted the
concentration of formic acid in units of mol/L.
What effect would this have had on the calculated
value of k? (d) The pressure of formic acid vapor
at the start of the reaction is 3.00 102 torr.
Assuming constant temperature and ideal-gas
behavior, what is the pressure in the system at
the end of the reaction? If the volume of the
reaction chamber is 436 cm3, how many moles of
gas occupy the reaction chamber at the end of the
reaction? (e) The standard heat of formation of
formic acid vapor is ?Hf 378.6 kJ/mol.
Calculate ?H for the overall reaction. If the
activation energy (Ea) for the reaction is 184
kJ/mol, sketch an approximate energy profile for
the reaction, and label Ea, ?H, and the
transition state.
120
Figure 14.28
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