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Chemical Kinetics Chapter 13

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Title: Chemical Kinetics Chapter 13


1
Chemical Kinetics Chapter 13
Topics 6 and 16
1.
2
What do we mean by kinetics?
  • Kinetics refers to
  • the rate at which chemical reactions occur.
  • The reaction mechanism or pathway through which a
    reaction proceeds.

2.
3
Topics for study in kinetics
Reaction Rates How we measure rates.
Rate Laws How the reaction rate depends on amounts of reactants. Molecularity and order.
Integrated Rate Laws How to calculate amount left or time to reach a given amount of reactant or product.
Half-life How long it takes for 50 of reactants to react
Arrhenius Equation How rate constant changes with temperature.
Mechanisms How the reaction rate depends on the sequence of molecular scale processes.
3.
4
Factors That Affect Reaction Rates
  • The Nature of the Reactants
  • Chemical compounds vary considerably in their
    chemical reactivities.
  • Concentration of Reactants
  • As the concentration of reactants increases, so
    does the likelihood that reactant molecules will
    collide.
  • Temperature
  • At higher temperatures, reactant molecules have
    more kinetic energy, move faster, and collide
    more often and with greater energy.
  • Catalysts
  • Change the rate of a reaction by changing the
    mechanism.

4.
5
Reaction Rates
  • The rate of a chemical reaction can be
    determined by monitoring the change in
    concentration of either reactants or the products
    as a function of time.
  • ?A vs ?t

5.
6
Example 1 Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
C4H9Cl M
  • In this reaction, the concentration of butyl
    chloride, C4H9Cl, was measured at various times,
    t.

?C4H9Cl ?t
Rate
6.
7
Reaction Rates Calculation
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl (aq)
Average Rate, M/s
  • The average rate of the reaction over each
    interval is the change in concentration divided
    by the change in time

7.
8
Reaction Rate Determination
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 the reacting
    molecules.

8.
9
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.

9.
10
Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
  • The reaction slows down with time because the
    concentration of the reactants decreases.

10.
11
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.

11.
12
Reaction Rates Stoichiometry
  • Suppose that the mole ratio is not 11?

Example H2(g) I2(g) ??? 2 HI(g)
2 moles of HI are produced for each mole of H2
used.
The rate at which H2 disappears is only half of
the rate at which HI is generated
12.
13
Concentration and Rate
  • Each reaction has its own equation that gives its
    rate as a function of reactant concentrations.
  • This is called its Rate Law
  • The general form of the rate law is
  • Rate kAxBy
  • Where k is the rate constant, A and B are the
    concentrations of the reactants. X and y are
    exponents known as rate orders that must be
    determined experimentally
  • To determine the rate law we measure the rate at
    different starting concentrations.

13.
14
Concentration and Rate
NH4 (aq) NO2- (aq) ? N2 (g) 2H2O (l)
  • Compare Experiments 1 and 2when NH4 doubles,
    the initial rate doubles.

14.
15
Concentration and Rate
NH4 (aq) NO2- (aq) ? N2 (g) 2H2O (l)
  • Likewise, compare Experiments 5 and 6 when
    NO2- doubles, the initial rate doubles.

15.
16
Concentration and Rate
NH4 (aq) NO2- (aq) ? N2 (g) 2H2O (l)
This equation is called the rate law, and k is
the rate constant.
16.
17
The Rate Law
  • A rate law shows the relationship between the
    reaction rate and the concentrations of
    reactants.
  • For gas-phase reactants use PA instead of A.
  • k is a constant that has a specific value for
    each reaction.
  • The value of k is determined experimentally.
  • Rate K NH4 NO2-
  • Constant is relative here-
  • the rate constant k is unique for each reaction
  • and the value of k changes with temperature

17.
18
The Rate Law
  • Exponents tell the order of the reaction with
    respect to each reactant.
  • This reaction is
  • First-order in NH4
  • First-order in 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.
  • Rate K NH4 1NO2- 1

18.
19
Determining the Rate constant and Order
  • The following data was collected for the
    reaction of substances A and B to produce
    products C and D.
  • Deduce the order of this reaction with
    respect to A and to B. Write an expression for
    the rate law in this reaction and calculate the
    value of the rate constant.

19.
20
The First Order Rate Equation
Consider a simple 1st order reaction A ? B Rate
kA

How much reactant A is left after time t? The
rate equation as a function of time can be
written as
Where At the concentration of
reactant A at time t Ao the concentration of
reactant A at time t 0
K the rate constant
At Ao e-kt Ln At - LnAo - kt
20.
21
First-Order Processes
  • Consider the process in which methyl isonitrile
    is converted to acetonitrile.

How do we know this is a first order reaction?
21.
22
First-Order Processes
  • This data was collected for this reaction at
    198.9C.

Does ratekCH3NC for all time intervals?
22.
23
First-Order Processes
  • When ln P is plotted as a function of time, a
    straight line results.
  • The process is first-order.
  • k is the negative slope 5.1 ? 10-5 s-1.

23.
24
Half-Life of a Reaction
  • 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.

24.
25
Half-Life of a First Order Reaction
  • For a first-order process, set At0.5 A0 in
    integrated rate equation

NOTE For a first-order process, the half-life
does not depend on the initial concentration,
A0.
25.
26
First Order Rate Calculation
Example 1 The decomposition of compound A is
first order. If the initial A0 0.80 mol
dm-3. and the rate constant is 0.010 s-1, what
is the concentration of A after 90 seconds?

26.
27
First Order Rate Calculation
Example 1 The decomposition of compound A is
first order. If the initial A0 0.80 mol
dm-3. and the rate constant is 0.010 s-1, what
is the concentration of A after 90 seconds?
LnAt lnAo -kt
LnAt Ln0.80 - (0.010 s-1 )(90 s) LnAt
- (0.010 s-1 )(90 s) ln0.80 LnAt
-0.90 - 0.2231 lnAt -1.1231 At
0.325 mol dm-3
27.
28
First Order Rate Calculations

Example 2 A certain first order chemical
reaction required 120 seconds for the
concentration of the reactant to drop from 2.00 M
to 1.00 M. Find the rate constant and the
concentration of reactant A after 80 seconds.



28.
29
First Order Rate Calculations

Example 2 A certain first order chemical
reaction required 120 seconds for the
concentration of the reactant to drop from 2.00 M
to 1.00 M. Find the rate constant and the
concentration of reactant A after 80 seconds.


Solution k 0.693/t1/2 0.693/120s 0.005775
s-1 LnA Ln(2.00) -0.005775 s-1 (80 s)
-0.462 ln A - 0.462 0.693 0.231 A
1.26 mol dm-3
29.
30
First Order Rate Calculations

Example 3 Radioactive decay is also a first
order process. Strontium 90 is a radioactive
isotope with a half-life of 28.8 years. If some
strontium 90 were accidentally released, how long
would it take for its concentration to fall to 1
of its original concentration?



30.
31
First Order Rate Calculations

Example 3 Radioactive decay is also a first
order process. Strontium 90 is a radioactive
isotope with a half-life of 28.8 years. If some
strontium 90 were accidentally released, how long
would it take for its concentration to fall to 1
of its original concentration?


Solution k 0.693/t1/2 0.693/28.8 yr 0.02406
yr-1 Ln1 ln(100) - (0.02406 yr-1)t -
4.60 t - 4.60 . - 0.0241
yr-1 t 191 years
31.
32
Second-Order Processes
  • Similarly, integrating the rate law for a process
    that is second-order in reactant A
  • Rate kA2

1 Ao
1 At
kt
Where At the concentration of reactant A at
time t Ao the concentration of reactant A at
time t 0 K the rate constant
32.
33
Second-Order Rate Equation
  • So if a process is second-order in A, a graph of
    1/A vs. t will yield a straight line with a
    slope of k.

33.
34
Determining Reaction OrderDistinguishing Between
1st and 2nd Order
The decomposition of NO2 at 300C is described by
the equation
A experiment with this reaction yields this data
Time (s) NO2, M
0.0 0.01000
50.0 0.00787
100.0 0.00649
200.0 0.00481
300.0 0.00380
34.
35
Determining Reaction OrderDistinguishing Between
1st and 2nd Order
Graphing ln NO2 vs. t yields
  • The graph is not a straight line, so this process
    cannot be 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
Does not fit the first order equation
35.
36
Second-Order Reaction Kinetics
A graph of 1/NO2 vs. t gives this plot.
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
  • This is a straight line. Therefore, the process
    is second-order in NO2.
  • The slope of the line is the rate constant, k.

36.
37
Half-Life for 2nd Order Reactions
  • For a second-order process, set
  • At0.5 A0 in 2nd order equation.

In this case the half-life depends on the initial
concentration of the reactant A.
37.
38
Sample Problem 1 Second Order
  • Acetaldehyde, CH3CHO, decomposes by second-order
    kinetics with a rate constant of 0.334
    mol-1dm3s-1 at 500oC. Calculate the amount of
    time it would take for 80 of the acetaldehyde
    to decompose in a sample that has an initial
    concentration of 0.00750 M.

The final concentration will be 20 of the
original 0.00750 M or 0.00150
1 . .00150
1 . .00750
0.334 mol-1dm3s-1 t
666.7 0.334 t 133.33 0.334 t 533.4
t 1600 seconds
38.
39
Sample Problem 2 Second Order
  • Acetaldehyde, CH3CHO, decomposes by second-order
    kinetics with a rate constant of 0.334
    mol-1dm3s-1 at 500oC. If the initial
    concentration of acetaldehyde is 0.00200 M. Find
    the concentration after 20 minutes (1200 seconds)

Solution
1 . At
1 . 0.00200 mol dm-3
0.334 mol-1dm3s-1 (1200s)
1 . At
0.334 mol-1dm3 s-1 (1200s) 500 mol-1dm3
900.8 mol-1dm3
1 _____. 900.8 mol-1dm3
At
0.00111 mol dm-3
39.
40
Summary of Kinetics Equations
First order Second order Second order
Rate Laws
Integrated Rate Laws complicated
Half-life complicated
40.
41
Temperature and Rate
  • Generally speaking, the reaction rate increases
    as the temperature increases.
  • This is because k is temperature dependent.
  • As a rule of thumb a reaction rate increases
    about 10 fold for each 10oC rise in temperature

41.
42
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.
  • These collisions must occur with sufficient
    energy and at the appropriate orientation.

42.
43
The Collision Model
  • Furthermore, molecules must collide with the
    correct orientation and with enough energy to
    cause bonds to break and new bonds to form

43.
44
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.

44.
45
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.

45.
46
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.

46.
47
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.

47.
48
MaxwellBoltzmann Distributions
  • As the temperature increases, the curve flattens
    and broadens.
  • Thus at higher temperatures, a larger population
    of molecules has higher energy.

48.
49
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.

49.
50
MaxwellBoltzmann Distributions
  • This fraction of molecules can be found through
    the expression
  • where R is the gas constant and T is the
    temperature in Kelvin .

50.
51
Arrhenius Equation
  • Svante Arrhenius developed a mathematical
    relationship between k and Ea
  • where A is the frequency factor, a number that
    represents the likelihood that collisions would
    occur with the proper orientation for reaction.
    Ea is the activation energy. T is the Kelvin
    temperature and R is the universal thermodynamics
    (gas) constant.
  • R 8.314 J mol-1 K-1 or 8.314 x 10-3 kJ
    mol-1 K-1

51.
52
Arrhenius Equation
  • Taking the natural logarithm of both sides, the
    equation becomes

y mx b
When k is determined experimentally at several
temperatures, Ea can be calculated from the slope
of a plot of ln k vs. 1/T.
52.
53
Arrhenius Equation for 2 Temperatures
When measurements are taken for two different
temperatures the Arrhenius equation can be
symplified as follows

Write the above equation twice, once for each of
the two Temperatures and then subtract the lower
temperature conditions from the higher
temperature. The equation then becomes
53.
54
Arrhenius Equation Sample Problem 1
The rate constant for the decomposition of
hydrogen iodide was determined at two different
temperatures 2HI ? H2 I2. At 650 K, k1
2.15 x 10-8 dm3 mol-1s-1 At 700 K, k2
2.39 x 10-7 dm3 mol-1s-1 Find the activation
energy for this reaction.



2.39 x 10-7 Ea Ln ---------------- -
------------------------ x 2.15 x 10-8
(8.314 J mol-1 K-1) Ea 180,000
J mol-1 180 kJ mol-1
1 1 ------ -- ------ 700K 650K
54.
55
Overview of Kinetics Equations
First order Second order Second order
Rate Laws
Integrated Rate Laws complicated
Half-life complicated
Rate and Temp (T)
55.
56
Reaction Mechanisms
  • The sequence of events that describes the actual
    process by which reactants become products is
    called the reaction mechanism.

56.
57
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.

57.
58
Reaction Mechanisms
  • The molecularity of a process tells how many
    molecules are involved in the process.
  • The rate law for an elementary step is written
    directly from that step.

58.
59
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.

59.
60
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.

60.
61
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.

61.
62
Fast Initial Step
  • The rate law for this reaction is found
    (experimentally) to be
  • Because termolecular ( trimolecular) processes
    are rare, this rate law suggests a two-step
    mechanism.

62.
63
Fast Initial Step
  • A proposed mechanism is

Step 1 is an equilibrium- it includes the
forward and reverse reactions.
63.
64
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
  • But how can we find NOBr2?

64.
65
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

65.
66
Fast Initial Step
  • Because Ratef Rater ,
  • k1 NO Br2 k-1 NOBr2
  • Solving for NOBr2 gives us

66.
67
Fast Initial Step
  • Substituting this expression for NOBr2 in the
    rate law for the rate-determining step gives

67.
68
Catalysts
  • Catalysts increase the rate of a reaction by
    decreasing the activation energy of the reaction.
  • Catalysts change the mechanism by which the
    process occurs.
  • Some catalysts also make atoms line up in the
    correct orientation so as to enhance the reaction
    rate

68.
69
Catalysts
  • Catalysts may be either homogeneous or
    heterogeneous
  • A homogeneous catalyst is in the same phase as
    the substances reacting.
  • A heterogeneous catalyst is in a different phase

69.
70
Catalysts
  • One way a catalyst can speed up a reaction is by
    holding the reactants together and helping bonds
    to break.
  • Heterogeneous catalysts often act in this way

70.
71
Catalysts
  • Some catalysts help to lower the energy for
    formation for the activated complex or provide a
    new activated complex with a lower activation
    energy

AlCl3 Cl2 ? Cl AlCl4- Cl C6H6 ? C6H5Cl
H H AlCl4- ? HCl AlCl3 Overall
reaction C6H6 Cl2 ? C6H5Cl HCl
71.
72
Catalysts Stratospheric Ozone
In the stratosphere, oxygen molecules absorb
ultraviolet light and break into individual
oxygen atoms known as free radicals The
oxygen radicals can then combine with ordinary
oxygen molecules to make ozone. Ozone can also
be split up again into ordinary oxygen and an
oxygen radical by absorbing ultraviolet light.


72.
73
Catalysts Stratospheric Ozone
The presence of chlorofluorcarbons in the
stratosphere can catalyze the destruction of
ozone. UV light causes a Chlorine free radical
to be released


The chlorine free radical attacks ozone and
converts it Back to oxygen. It is then
regenerated to repeat the Process. The result is
that each chlorine free radical can Repeat this
process many many times. The result is
that Ozone is destroyed faster than it is formed,
causing its level to drop
73.
74
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.

74.
75
75.
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