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Chapter 14 Chemical Kinetics

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Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 14 Chemical Kinetics John D. Bookstaver – PowerPoint PPT presentation

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Title: Chapter 14 Chemical Kinetics


1
Chapter 14Chemical Kinetics
Chemistry, The Central Science, 10th
edition Theodore L. Brown H. Eugene LeMay, Jr.
and Bruce E. Bursten
  • John D. Bookstaver
  • St. Charles Community College
  • St. Peters, MO
  • ? 2006, Prentice Hall, Inc.

2
March 4
  • Reaction rates. Factors that affect the rate of a
    reaction.
  • Determining the instantaneous rate from a graph.
  • Stoichiometry and reaction rates
  • Homework p 617 q 1to3
  • and up to what we get to for 11 to 15 , 17 to
    31 odd

3
Kinetics
  • Is the branch of chemistry concerned with the
    rate of chemical reactions and the mechanism by
    which occur
  • Besides information about the speed at which
    reactions occur, kinetics also sheds light on the
    reaction mechanism (exactly how the reaction
    occurs).

4
Collision Theory
  • In order for two particles to react chemically
    they must collide. Not only they must collide,
    but it must be an effective collision. That is,
    they must have the correct amount of energy and
    collide with the proper orientation in space.
  • Any factor which increases the likehood that they
    will collide will increase the rate of the
    chemical reaction.

5
Factors That Affect Reaction Rates (bonds must
break)
  • Surface area/contact area (opportunity for
    collisions).
  • Concentration (increase frequency)
  • Temperature (increase frequency)
  • Catalyst ( Effective collisions)
  • Nature of reactants (effective collisions)

6
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.
  • For solid reactants increasing the surface area
    increases the rate of reaction.

7
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.

8
Factors That Affect Reaction Rates
  • Concentration of Reactants
  • As the concentration of reactants increases, so
    does the likelihood that reactant molecules will
    collide.

9
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.

10
Reaction Rates
  • Speed of a reaction is measured by the change in
    concentration with time.
  • For a reaction A ? B
  • Suppose A reacts to form B. Let us begin with
    1.00 mol A.

11
Reaction Rates
  • Rates of reactions can be determined by
    monitoring the change in concentration of either
    reactants or products as a function of time.

12
  • At t 0 (time zero) there is 1.00 mol A (100 red
    spheres) and no B present.
  • At t 20 min, there is 0.54 mol A and 0.46 mol
    B.
  • At t 40 min, there is 0.30 mol A and 0.70 mol
    B.
  • Calculating,

13
  • For the reaction A ? B there are two ways of
    measuring rate
  • the speed at which the products appear (i.e.
    change in moles of B per unit time), or
  • the speed at which the reactants disappear (i.e.
    the change in moles of A per unit time).

14
  • By convention rates are always positive
    quantities. Because A is decreasing with time, we
    use a negative sign to convert the negative
    change into a positive rate.
  • Because one molecule of A is consumed for every
    molecule of B that forms the average rate of
    disappearance of A equals the rate of appearance
    of B

15
PRACTICE EXERCISE For the reaction pictured in
Figure 14.3, calculate the average rate of
appearance of B over the time interval from 0 to
40 s. (The necessary data are given in the figure
caption.)
Answer 1.8 ? 10 2 M/s
16
  • Change of Rate with Time
  • Most useful units for rates are to look at
    molarity. Since volume is constant, molarity and
    moles are directly proportional.
  • Consider
  • C4H9Cl(aq) H2O(l) ? C4H9OH(aq) HCl(aq)

17
Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
  • In this reaction, the concentration of butyl
    chloride or chlorobutane, C4H9Cl, was measured at
    various times.

18
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

19
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.

20
Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
  • A plot of concentration 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.

21
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.

22
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.

23
  • C4H9Cl(aq) H2O(l) ? C4H9OH(aq) HCl(aq)
  • We can calculate the average rate in terms of the
    disappearance of C4H9Cl.
  • The units for average rate are mol L-1 s-1 or
    M/s.
  • The average rate decreases with time.
  • We plot C4H9Cl versus time.
  • The rate at any instant in time (instantaneous
    rate) is the slope of the tangent to the curve.
  • Instantaneous rate is different from average
    rate.
  • We usually call the instantaneous rate the rate.

24
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25
Reaction Rates and Stoichiometry
  • What if the ratio is not 11?

2 HI(g) ??? H2(g) I2(g)
  • Therefore,

26
Reaction Rates and Stoichiometry
  • To generalize, then, for the reaction

27
  • Reaction Rate
  • Change in concentration per unit time
  • For this reaction
  • N2 3H2 2NH3

28
  • As the reaction progresses the concentration H2
    goes down

Concentration
H2
Time
29
  • As the reaction progresses the concentration N2
    goes down 1/3 as fast

Concentration
N2
H2
Time
30
  • As the reaction progresses the concentration NH3
    goes up.

Concentration
N2
H2
NH3
Time
31
  • Example
  • N2 3H2 2NH3
  • At a certain temperature, the rate of this
    reaction is -0.13 mol N2 L-1 s-1. What is the
    rate in mol H2 L-1 s-1? What is the rate in mol
    NH3 L-1 s-1?

32
0.39 mol H2 L-1 s-1 0.26 mol NH3 L-1 s-1
33
(b) If the rate at which O2 appears, ?O2? /?t,
is 6.0 ? 105 M/s at a particular instant, at
what rate is O3 disappearing at this same time,
?O3? /?t?
34
Answers (a) 8.4 ? 10 7 M/s, (b) 2.1 ? 10 7 M/s
35
Concentration and Rate
  • One can gain information about the rate of a
    reaction by seeing how the rate changes with
    changes in concentration.

36
Concentration and Rate
  • Comparing Experiments 1 and 2, when NH4
    doubles, the initial rate doubles.

37
Concentration and Rate
  • Likewise, comparing Experiments 5 and 6, when
    NO2- doubles, the initial rate doubles.

38
  • This method requires that a reaction be run
    several times.
  • The initial concentrations of the reactants are
    varied.
  • The goal is to vary only one concentration at a
    time
  • The reaction rate is measured just after the
    reactants are mixed.

39
Concentration and Rate
  • This means
  • Rate ? NH4
  • Rate ? NO2-
  • Rate ? NH NO2-
  • or
  • Rate k NH4 NO2-
  • This equation is called the rate law, and k is
    the rate constant.

40
  • For the reaction
  • NH4(aq) NO2-(aq) ? N2(g) 2H2O(l)
  • we note
  • as NH4 doubles with NO2- constant the rate
    doubles,
  • as NO2- doubles with NH4 constant, the rate
    doubles,
  • We conclude rate ? NH4NO2-.
  • Rate law
  • The constant k is the rate constant.

41
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.
  • This reaction is
  • First-order in NH4
  • First-order in NO2-

42
Rate Laws
  • The overall reaction order can be found by adding
    the exponents on the reactants in the rate law.
  • This reaction is second-order overall.

43
  • The exponents in a rate law indicate how the rate
    is affected by the concentration of each
    reactant.
  • For most reactions the orders are 0, 1 or 2,
    occasionally reactions can have a reaction order
    fractional or even negative.

44
  • 2 NO2 2 NO O2
  • You will find that the rate will only depend on
    the concentrations of the reactants.
  • Rate kNO2n
  • This is called a rate law expression.
  • k is called the rate constant.
  • n is the order of the reactant -usually a
    positive integer.

45
  • Exponents in the Rate Law
  • For a general reaction with rate law
  • we say the reaction is mth order in reactant 1
    and nth order in reactant 2.
  • The overall order of reaction is m n .
  • A reaction can be zeroth order if m, n, are
    zero.
  • This means that the rate does not depend on the
    concentration of (a) reactant(s)
  • Note the values of the exponents (orders) have
    to be determined experimentally. They are not
    simply related to stoichiometry.

46
  • Using Initial Rates to Determine Rate Laws
  • A reaction is zero order in a reactant if the
    change in concentration of that reactant produces
    no effect.
  • A reaction is first order if doubling the
    concentration causes the rate to double.
  • A reacting is nth order if doubling the
    concentration causes an 2n increase in rate.
  • Note that the rate constant does not depend on
    concentration.

47
Example
  • For the reaction
  • BrO3- 5 Br- 6H 3Br2 3 H2O
  • The general form of the Rate Law is
  • Rate kBrO3-mBr-nHp
  • We use experimental data to determine the values
    of m, n,and p

48
Initial concentrations (M)
BrO3-
Br-
H
Rate (M/s)
0.10 0.10 0.10 8.0 x 10-4
0.20 0.10 0.10 1.6 x 10-3
0.20 0.20 0.10 3.2 x 10-3
0.10 0.10 0.20 3.2 x 10-3
  • Now we have to see how the rate changes with
    concentration

49
  • Determine the Rate Law.
  • Calculate the value of k, and determine its
    units.
  • Calculate the initial rate of the reaction if the
    initial concentrations are
  • BrO3- 0.20 M Br- 0.30 M
  • H 0.10 M

50
NH4 NO2- N2 2 H2O
  • Exp NH4 NO2- Initial Rate(M/s)
  • 1 0.100 M 0.0050 M 1.35 x 10-7
  • 2 0.100 M 0.010 M 2.70 x 10-7
  • 3 0.200 M 0.010 M 5.40 x 10-7
  • Determine the rate law.
  • Calculate k, and determine its units.

51
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52
Check Each box contains 10 spheres. The rate law
indicates that in this case B has a greater
influence on rate than A because B has a higher
reaction order. Hence, the mixture with the
highest concentration of B (most purple spheres)
should react fastest. This analysis confirms the
order 2 lt 1 lt 3.
PRACTICE EXERCISE Assuming that rate kAB,
rank the mixtures represented in this Sample
Exercise in order of increasing rate.
Answer 2 3 lt 1
53
  • TYPES OF RATE LAW
  • CONCENTRATION AND TIME
  • FIRST ORDER REACTION
  • INTEGRATED RATE LAW
  • GRAPH METHOD
  • UNITS FOR K
  • GET READY FOR A SURPRISE QUIZ! Reaction rates and
    reaction
  • Hw finish 33 to 43

54
Types of Rate Laws
  • Differential Rate law describes how rate depends
    on reactant concentration (Referred to as simply
    rate law).
  • Rate -D A/D tk A
  • Integrated Rate Law Describes how
    concentration varies with time.

55
For each type of differential rate law there is
an integrated rate law and vice versa. Rate laws
can help us better understand reaction mechanisms
56
  • First Order Reactions
  • Goal convert rate law into a convenient equation
    to give concentrations as a function of time.
  • For a first order reaction, the rate doubles as
    the concentration of a reactant doubles.

Rate Law Integrated Rate Law
57
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. At is
the concentration of A at some time, t, during
the course of the reaction.
58
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
59
Unit for k for a 1st order reaction
  • k t-1

60
Answer 51 torr
61
Solution replace the given values into the
integrated first order rate law equation ln At
-kt ln A0 k 1.45 yr1, t 1.00 yr
insecticide0 5.0 ? 107 g/cm3 and
solve for 1ninsecticide (b) We have k
1.45yr1, insecticide0 5.0 ? 107 g/cm3, and
insecticidet 3.0 ? 107 g/cm3, and so we can
solve the equation for t.
62
First order
  • A certain first order decomposition reaction
    reaches 65 completion in 18.9 seconds. Find the
    rate constant for this reaction.
  • .055 s-1

63
  • Integrated rate law review first order
  • Second order processes
  • Half life

64
HW
  • 33 TO 43 ODD
  • Go to www.sciencegeek.net
  • Follow the links to chapter 12
  • Take quiz

65
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.

66
Graphical method to determine rate order
  • How do we find out if the reaction is first rate
    from the experimental data?
  • Plot the ln reactant vs time.
  • If the plot is a straight line then the reaction
    is first order.

67
First-Order Processes
  • Consider the process in which methyl isonitrile
    is converted to acetonitrile.

68
First-Order Processes
  • This data was collected for this reaction at
    198.9C.

69
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 slope 5.1 ? 10-5 s-1.

70
Second-Order Processes Rate kA2
  • Integrating the rate law for a process that is
    second-order in reactant A, we get

also in the form
y mx b
71
INTEGRATED RATE LAW FOR A 2nd ORDER REACTION
  • If a process is second-order in A, a plot of
    1/A vs. t will yield a straight line, and the
    slope of that line is k.
  • Unit for k M-1 t-1

72
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
73
Second-Order Processes
  • Graphing ln NO2 vs. t yields
  • 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
74
Second-Order Processes
  • Graphing ln 1/NO2 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
75
January 18
  • HALF LIFE
  • SUMMARY OF RATE LAWS
  • TEMPERATURE AND RATE
  • COLLISION MODEL /ACTIVATION ENERGY/ ACTIVATED
    COMPLEX
  • ARRHENIUS EQUATION FOR ACTIVATION ENERGY
  • HW 45 TO 55

76
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.

77
  • Half-Life
  • Half-life is the time taken for the concentration
    of a reactant to drop to half its original value.
  • For a first order process, half life, t½ is the
    time taken for A0 to reach ½A0.

78
Half-Life
  • For a first-order process, this becomes

ln 0.5 -kt1/2
-0.693 -kt1/2
NOTE For a first-order process, the half-life
does not depend on A0.
79
Half life
  • The half life of a reaction is useful to describe
    how fast it occurs.
  • For a first order reaction (like nuclear decay!)
    it does not depend on the initial concentration
    of the reactants.
  • HALF LIFE IS CONSTANT FOR A FIRST ORDER REACTION

80
Half-Life
  • For a second-order process

81
Second Order Reactions
  • The half life depends on the initial
    concentration.
  • As the reactant concentration decreases the half
    life increases.

82
Zero Order Rate Law
  • Rate kA0 k
  • Rate does not change with concentration.
  • Integrated A -kt A0
  • When A A0 /2 t t1/2
  • t1/2 A0 /2k

83
Zero Order Rate Law
  • Most often when reaction happens on a surface
    because the surface area stays constant.
  • Also applies to enzyme chemistry.

84
Summary of Rate Laws
Zero Order 1st Order 2nd Order
Diff. Rate Law K kA KA2
Int. Rate Law A -kt Ao ln A -kt ln Ao 1/A kt 1/Ao
Straight Line Plot A vs. t m -k ln A vs. t m -k 1/A vs. t m k
Half-Life Ao/2k (ln 2)/k 1/kAo
85
Temperature and Rate
  • Generally, as temperature increases, so does the
    reaction rate.
  • This is because k is temperature dependent.

86
  • Most reactions speed up as temperature increases.
    (E.g. food spoils when not refrigerated.)
  • When two light sticks are placed in water one at
    room temperature and one in ice, the one at room
    temperature is brighter than the one in ice.
  • The chemical reaction responsible for
    chemiluminescence is dependent on temperature
    the higher the temperature, the faster the
    reaction and the brighter the light.

87
  • The Collision Model
  • As temperature increases, the rate increases
    (exponentially).

88
  • Since the rate law has no temperature term in it,
    the rate constant must depend on temperature.
  • Consider the first order reaction
  • CH3NC ? CH3CN.
  • As temperature increases from 190 ?C to 250 ?C
    the rate constant increases from 2.52 ? 10-5 s-1
  • to 3.16 ? 10-3 s-1.
  • The temperature effect is quite dramatic. Why?
  • Observations rates of reactions are affected by
    concentration and temperature.

89
  • Goal develop a model that explains why rates of
    reactions increase as concentration and
    temperature increases.
  • The collision model in order for molecules to
    react they must collide.
  • The greater the number of collisions the faster
    the rate.
  • The more molecules present, the greater the
    probability of collision and the faster the rate.

90
  • The higher the temperature, the more energy
    available to the molecules and the faster the
    rate.
  • Complication not all collisions lead to
    products. In fact, only a small fraction of
    collisions lead to product.
  • Effective Collision A collision between
    molecules that leads to products (reaction
    occurs)

91
The Collision Model
  • Furthermore, molecules must collide with the
    correct orientation and with enough energy to
    cause bond breakage and formation.

92
  • Consider
  • 2 NOBr ? 2 NO Br2
  • There are several possible ways that NOBr
    molecules can collide one is effective and the
    others are not.

93
No Reaction
94
The Orientation Factor Cl ClNO Cl2 NO
95
  • Activation Energy
  • Arrhenius molecules must posses a minimum amount
    of energy to react. Why?
  • In order to form products, bonds must be broken
    in the reactants.
  • Bond breakage requires energy.
  • Activation energy, Ea, is the minimum energy
    required to initiate a chemical reaction.

96
Activation Energy
  • 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.

97
Activation Energy
  • The lower the Ae the faster the reaction.
  • Is the energy needed to form the activated
    complex.
  • Activated Complex or Transition State - The
    arrangement of atoms at the top of the energy
    barrier. Is a very unstable arrangement!

98
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.

99
Reaction Coordinate Diagrams
  • It 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.

100
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.

101
MaxwellBoltzmann Distributions
  • As the temperature increases, the curve flattens
    and broadens.
  • Thus at higher temperatures, a larger population
    of molecules has higher energy.

102
MaxwellBoltzmann Distributions
  • If the dotted line represents the activation
    energy, as the temperature increases, so does the
    fraction of molecules that can overcome the
    activation energy barrier.
  • As a result, the reaction rate increases.

103
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-Ea/RT
104
  • The Arrhenius Equation
  • Arrhenius discovered most reaction-rate data
    obeyed the Arrhenius equation
  • k is the rate constant, Ea is the activation
    energy, R is the gas constant (8.314 J/K-mol) and
    T is the temperature in K.
  • A is called the frequency factor.
  • A is a measure of the probability of a favorable
    collision.
  • Both A and Ea are specific to a given reaction.

105
Arrhenius Equation
  • Taking the natural logarithm of both sides, the
    equation becomes
  • ln k -Ea ( ) ln A

y mx b
Therefore, if k is determined experimentally at
several temperatures, Ea can be calculated from
the slope of a plot of ln k vs. 1/T.
106
  • Determining the Activation Energy
  • If we have a lot of data, we can determine Ea and
    A graphically by rearranging the Arrhenius
    equation
  • From the above equation, a plot of ln k versus
    1/T will have slope of Ea/R and intercept of ln
    A.

107
Temperature and Rate
108
  • REACTION MECHANISMS
  • HW 59 TO 67

109
  • If we do not have a lot of data, then we
    recognize

110
Arrhenius Equation Problem
The rate constant for the formation of hydrogen
iodide from the elements H2 (g) I2 (g)? 2HI
(g) is 2.7 x 10-4 L/(mol.s) at 600 K and 3.5 x
10-3 L/(mol.s) at 650 K. Find the activation
energy Ea and solve for k2 when T is 700K
111
SAMPLE EXERCISE 14.10 Relating Energy Profiles to
Activation Energies and Speeds of Reaction
Consider a series of reactions having the
following energy profiles
Assuming that all three reactions have nearly the
same frequency factors, rank the reactions from
slowest to fastest.
Solution The lower the activation energy, the
faster the reaction. The value of ?? does not
affect the rate. Hence the order is (2) lt (3) lt
(1).
PRACTICE EXERCISE Imagine that these reactions
are reversed. Rank these reverse reactions from
slowest to fastest.
Answer  (2) lt (1) lt (3) because Ea values are
40, 25, and 15 kJ/mol, respectively
112
Reaction Mechanisms
  • The balanced chemical equation provides
    information about the beginning and end of
    reaction.
  • The reaction mechanism gives the path of the
    reaction.
  • Mechanisms provide a very detailed picture of
    which bonds are broken and formed during the
    course of a reaction.
  • Elementary Steps
  • Elementary step any process that occurs in a
    single step.

113
Reaction Mechanisms
  • The molecularity of a process tells how many
    molecules are involved in the process.
  • Molecularity is the number of species that must
    collide to produce the reaction indicated by that
    step

114
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.

115
Rate Laws for Elementary Steps
116
  • If a reaction proceeds via several elementary
    steps, then the elementary steps must add to give
    the balanced chemical equation.
  • Intermediate a species which appears in an
    elementary step which is not a reactant or
    product. It is produced in one step and consumed
    in a later step.
  • The mechanism must agree with the experimentally
    determined law.

117
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.

118
RATE DETERMINING STEP
  • The rate is always determined by the slow rate
    step.

119
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.

120
  • NO2(g) NO2(g) ? NO3(g) NO(g)
  • NO3(g) CO(g) ? NO2(g) CO2(g)
  • Notice that if we add the above steps, we get the
    overall reaction
  • NO2(g) CO(g) ? NO(g) CO2(g)

121
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.

122
Fast Initial Step
  • A proposed mechanism is

Step 2 NOBr2 NO ??? 2 NOBr (slow)
Step 1 includes the forward and reverse reactions.
123
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?

124
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

125
Fast Initial Step
  • Because Ratef Rater ,
  • k1 NO Br2 k-1 NOBr2
  • Solving for NOBr2 gives us

126
Fast Initial Step
  • Substituting this expression for NOBr2 in the
    rate law for the rate-determining step gives

k NO2 Br2
127
  • Mechanisms of reaction practice
  • Catalysts
  • Chapter review

128
  • Example
  • NO Cl2 NOCl2 (fast)
  • NOCl2 NO 2 NOCl (slow)
  • 1. Write the overall equation.
  • 2. Determine the rate law.

129
Example
  • At low temperatures, the rate law for the
    reaction
  • CO (g) NO2 (g) ? CO2 (g) NO (g)
  • Rate kNO22. Show why or why not each of the
    following mechanisms is consistent with that rate
    law for this reaction.

130
  • 1. CO NO2 ? CO2 NO
  • 1. 2 NO2 N2O4 (fast)
  • 2. N2O4 2 CO 2 CO2 2 NO (slow)
  • 1. 2 NO2 ? NO3 NO (slow)
  • 2. NO3 CO ? NO2 CO2 (fast)
  • 1. 2 NO2 ? 2 NO O2 (slow)
  • 2. 2 CO O2 ? 2 CO2 (fast)

131
Catalysts
  • Catalysts increase the rate of a reaction by
    decreasing the activation energy of the reaction.
  • Catalyst are never consumed in the chemical
    reaction.
  • Catalysts change the mechanism by which the
    process occurs.

132
Example
  • H2O2 I- ? H2O IO- (slow)
  • H2O2 IO- ? H2O O2 I- (fast)
  • 1. What is the overall equation?
  • 2. What is the catalyst?
  • 3. What is the intermediate?
  • 4. What is the rate law?

133
  • There are two types of catalyst
  • Homogeneous present in the same phase as the
    reacting molecule.
  • Heterogeneous exist in different phase, usually
    a solid.
  • Homogeneous Catalysis
  • The catalyst and reaction is in one phase.
  • Chlorine atoms are catalysts for the destruction
    of ozone.

134
  • Hydrogen peroxide decomposes very slowly
  • 2H2O2(aq) ? 2H2O(l) O2(g)
  • In the presence of the bromide ion, the
    decomposition occurs rapidly
  • 2Br-(aq) H2O2(aq) 2H(aq) ? Br2(aq)
    2H2O(l).
  • Br2(aq) is brown.
  • Br2(aq) H2O2(aq) ? 2Br-(aq) 2H(aq) O2(g).
  • Br- is a catalyst because it can be recovered at
    the end of the reaction.
  • Generally, catalysts operate by lowering the
    activation energy for a reaction.

135
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136
  • Catalysts can operate by increasing the number of
    effective collisions.
  • That is, from the Arrhenius equation catalysts
    increase k be increasing A or decreasing Ea.
  • A catalyst may add intermediates to the reaction.
  • Example In the presence of Br-, Br2(aq) is
    generated as an intermediate in the decomposition
    of H2O2.

137
  • When a catalyst adds an intermediate, the
    activation energies for both steps must be lower
    than the activation energy for the uncatalyzed
    reaction. The catalyst is in a different phase
    than the reactants and products.
  • Heterogeneous Catalysis
  • Typical example solid catalyst, gaseous
    reactants and products (catalytic converters in
    cars).
  • Most industrial catalysts are heterogeneous.

138
  • First step is adsorption (the binding of reactant
    molecules to the catalyst surface).
  • Adsorbed species (atoms or ions) are very
    reactive.
  • Molecules are adsorbed onto active sites on the
    catalyst surface.

139
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140
  • Consider the hydrogenation of ethylene
  • C2H4(g) H2(g) ? C2H6(g), ?H -136 kJ/mol.
  • The reaction is slow in the absence of a
    catalyst.
  • In the presence of a metal catalyst (Ni, Pt or
    Pd) the reaction occurs quickly at room
    temperature.
  • First the ethylene and hydrogen molecules are
    adsorbed onto active sites on the metal surface.
  • The H-H bond breaks and the H atoms migrate about
    the metal surface.

141
  • When an H atom collides with an ethylene molecule
    on the surface, the C-C ? bond breaks and a C-H ?
    bond forms.
  • When C2H6 forms it desorbs from the surface.
  • When ethylene and hydrogen are adsorbed onto a
    surface, less energy is required to break the
    bonds and the activation energy for the reaction
    is lowered.
  • Enzymes
  • Enzymes are biological catalysts.
  • Most enzymes are protein molecules with large
    molecular masses (10,000 to 106 amu).

142
  • Enzymes have very specific shapes.
  • Most enzymes catalyze very specific reactions.
  • Substrates undergo reaction at the active site of
    an enzyme.
  • A substrate locks into an enzyme and a fast
    reaction occurs.
  • The products then move away from the enzyme.

143
  • Only substrates that fit into the enzyme lock
    can be involved in the reaction.
  • If a molecule binds tightly to an enzyme so that
    another substrate cannot displace it, then the
    active site is blocked and the catalyst is
    inhibited (enzyme inhibitors).
  • The number of events (turnover number) catalyzed
    is large for enzymes (103 - 107 per second).

144
Enzymes
145
SAMPLE EXERCISE 14.11 Determining the Energy of
Activation
The following table shows the rate constants for
the rearrangement of methyl isonitrile at various
temperatures
(a) From these data, calculate the activation
energy for the reaction. (b) What is the value of
the rate constant at 430.0 K?
Solution Analyze We are given rate constants, k,
measured at several temperatures and asked to
determine the activation energy, Ea, and the rate
constant, k, at a particular temperature. Plan
We can obtain Ea from the slope of a graph of ln
k versus 1/T. Once we know Ea, we can use ln k
-Ea (1/RT) ln A together with the given rate
data to calculate the rate constant at 430.0 K.
146
Solve (a) We must first convert the temperatures
from degrees Celsius to kelvins. We then take the
inverse of each temperature, 1/T, and the natural
log of each rate constant, ln k. This gives us
the table shown at the right
A graph of ln k versus 1/T results in a straight
line, as shown in Figure 14.17.
147
We report the activation energy to only two
significant figures because we are limited by the
precision with which we can read the graph
148
SAMPLE EXERCISE 14.11 continued
Thus,
Note that the units of k1 are the same as those
of k2.
PRACTICE EXERCISE Using the data in Sample
Exercise 14.11, calculate the rate constant for
the rearrangement of methyl isonitrile at 280C.
Answer 2.2 ? 102s1
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