Chapter 13 Chemical Kinetics - PowerPoint PPT Presentation

Loading...

PPT – Chapter 13 Chemical Kinetics PowerPoint presentation | free to download - id: 58fe81-MzgyO



Loading


The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
About This Presentation
Title:

Chapter 13 Chemical Kinetics

Description:

Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro Chapter 13 Chemical Kinetics Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA – PowerPoint PPT presentation

Number of Views:197
Avg rating:3.0/5.0
Slides: 104
Provided by: admi523
Learn more at: http://www.mtsac.edu
Category:

less

Write a Comment
User Comments (0)
Transcript and Presenter's Notes

Title: Chapter 13 Chemical Kinetics


1
Chapter 13Chemical Kinetics
Chemistry A Molecular Approach, 1st Ed.Nivaldo
Tro
Roy Kennedy Massachusetts Bay Community
College Wellesley Hills, MA
2008, Prentice Hall
2
Kinetics
  • kinetics is the study of the factors that affect
    the speed of a reaction and the mechanism by
    which a reaction proceeds.
  • experimentally it is shown that there are 4
    factors that influence the speed of a reaction
  • nature of the reactants,
  • temperature,
  • catalysts,
  • concentration

3
Defining Rate
  • rate is how much a quantity changes in a given
    period of time
  • the speed you drive your car is a rate the
    distance your car travels (miles) in a given
    period of time (1 hour)
  • so the rate of your car has units of mi/hr

4
Defining Reaction Rate
  • the rate of a chemical reaction is generally
    measured in terms of how much the concentration
    of a reactant decreases in a given period of time
  • or product concentration increases
  • for reactants, a negative sign is placed in front
    of the definition

5
Reaction Rate Changes Over Time
  • as time goes on, the rate of a reaction generally
    slows down
  • because the concentration of the reactants
    decreases.
  • at some time the reaction stops, either because
    the reactants run out or because the system has
    reached equilibrium.

6
at t 0 A 8 B 8 C 0
at t 0 X 8 Y 8 Z 0
at t 16 A 4 B 4 C 4
at t 16 X 7 Y 7 Z 1
7
at t 16 A 4 B 4 C 4
at t 16 X 7 Y 7 Z 1
at t 32 A 2 B 2 C 6
at t 32 X 6 Y 6 Z 2
8
at t 32 A 2 B 2 C 6
at t 32 X 6 Y 6 Z 2
at t 48 A 0 B 0 C 8
at t 48 X 5 Y 5 Z 3
9
Hypothetical ReactionRed ? Blue
Time (sec) Number Red Number Blue
0 100 0
5 84 16
10 71 29
15 59 41
20 50 50
25 42 58
30 35 65
35 30 70
40 25 75
45 21 79
50 18 82
in this reaction, one molecule of Red turns into
one molecule of Blue
the number of molecules will always total 100
the rate of the reaction can be measured as the
speed of loss of Red molecules over time, or the
speed of gain of Blue molecules over time
10
Hypothetical ReactionRed ? Blue
11
Hypothetical ReactionRed ? Blue
12
Reaction Rate and Stoichiometry
  • in most reactions, the coefficients of the
    balanced equation are not all the same
  • H2 (g) I2 (g) ? 2 HI(g)
  • for these reactions, the change in the number of
    molecules of one substance is a multiple of the
    change in the number of molecules of another
  • for the above reaction, for every 1 mole of H2
    used, 1 mole of I2 will also be used and 2 moles
    of HI made
  • therefore the rate of change will be different
  • in order to be consistent, the change in the
    concentration of each substance is multiplied by
    1/coefficient

13
Average Rate
  • the average rate is the change in measured
    concentrations in any particular time period
  • linear approximation of a curve
  • the larger the time interval, the more the
    average rate deviates from the instantaneous rate

14
Hypothetical Reaction Red ? Blue
Avg. Rate Avg. Rate Avg. Rate
Time (sec) Number Red Number Blue (5 sec intervals) (10 sec intervals) (25 sec intervals)
0 100 0      
5 84 16 3.2    
10 71 29 2.6 2.9  
15 59 41 2.4    
20 50 50 1.8 2.1  
25 42 58 1.6   2.3
30 35 65 1.4 1.5  
35 30 70 1    
40 25 75 1 1  
45 21 79 0.8    
50 18 82 0.6 0.7 1
15
H2
I2
HI
Avg. Rate, M/s Avg. Rate, M/s
Time (s) H2, M HI, M -DH2/Dt 1/2 DHI/Dt
0.000 1.000
10.000 0.819
20.000 0.670
30.000 0.549
40.000 0.449
50.000 0.368
60.000 0.301
70.000 0.247
80.000 0.202
90.000 0.165
100.000 0.135
Avg. Rate, M/s Avg. Rate, M/s
Time (s) H2, M HI, M -DH2/Dt 1/2 DHI/Dt
0.000 1.000 0.000
10.000 0.819 0.362
20.000 0.670 0.660
30.000 0.549 0.902
40.000 0.449 1.102
50.000 0.368 1.264
60.000 0.301 1.398
70.000 0.247 1.506
80.000 0.202 1.596
90.000 0.165 1.670
100.000 0.135 1.730
Avg. Rate, M/s
Time (s) H2, M HI, M -DH2/Dt
0.000 1.000 0.000
10.000 0.819 0.362 0.0181
20.000 0.670 0.660 0.0149
30.000 0.549 0.902 0.0121
40.000 0.449 1.102 0.0100
50.000 0.368 1.264 0.0081
60.000 0.301 1.398 0.0067
70.000 0.247 1.506 0.0054
80.000 0.202 1.596 0.0045
90.000 0.165 1.670 0.0037
100.000 0.135 1.730 0.0030
Avg. Rate, M/s Avg. Rate, M/s
Time (s) H2, M HI, M -DH2/Dt 1/2 DHI/Dt
0.000 1.000 0.000
10.000 0.819 0.362 0.0181 0.0181
20.000 0.670 0.660 0.0149 0.0149
30.000 0.549 0.902 0.0121 0.0121
40.000 0.449 1.102 0.0100 0.0100
50.000 0.368 1.264 0.0081 0.0081
60.000 0.301 1.398 0.0067 0.0067
70.000 0.247 1.506 0.0054 0.0054
80.000 0.202 1.596 0.0045 0.0045
90.000 0.165 1.670 0.0037 0.0037
100.000 0.135 1.730 0.0030 0.0030
Stoichiometry tells us that for every 1 mole/L of
H2 used, 2 moles/L of HI are made.
The average rate is the change in the
concentration in a given time period.
Assuming a 1 L container, at 10 s, we used 0.181
moles of H2. Therefore the amount of HI made is
2(0.181 moles) 0.362 moles
In the first 10 s, the DH2 is -0.181 M, so the
rate is
At 60 s, we used 0.699 moles of H2. Therefore
the amount of HI made is 2(0.699 moles) 1.398
moles
16
average rate in a given time period ? slope of
the line connecting the H2 points and ½ slope
of the line for HI
the average rate for the first 10 s is 0.0181 M/s
the average rate for the first 40 s is 0.0150 M/s
the average rate for the first 80 s is 0.0108 M/s
17
Instantaneous Rate
  • the instantaneous rate is the change in
    concentration at any one particular time
  • slope at one point of a curve
  • determined by taking the slope of a line tangent
    to the curve at that particular point
  • first derivative of the function
  • for you calculus fans

18
H2 (g) I2 (g) ? 2 HI (g)
Using H2, the instantaneous rate at 50 s is
Using HI, the instantaneous rate at 50 s is
19
Ex 13.1 - For the reaction given, the I?
changes from 1.000 M to 0.868 M in the first 10
s. Calculate the average rate in the first 10 s
and the DH.H2O2 (aq) 3 I?(aq) 2 H(aq) ?
I3?(aq) 2 H2O(l)
Solve the equation for the Rate (in terms of the change in concentration of the Given quantity)
Solve the equation of the Rate (in terms of the change in the concentration for the quantity to Find) for the unknown value
20
Measuring Reaction Rate
  • in order to measure the reaction rate you need to
    be able to measure the concentration of at least
    one component in the mixture at many points in
    time
  • there are two ways of approaching this problem
    (1) for reactions that are complete in less than
    1 hour, it is best to use continuous monitoring
    of the concentration, or (2) for reactions that
    happen over a very long time, sampling of the
    mixture at various times can be used
  • when sampling is used, often the reaction in the
    sample is stopped by a quenching technique

21
Continuous Monitoring
  • polarimetry measuring the change in the degree
    of rotation of plane-polarized light caused by
    one of the components over time
  • spectrophotometry measuring the amount of light
    of a particular wavelength absorbed by one
    component over time
  • the component absorbs its complimentary color
  • total pressure the total pressure of a gas
    mixture is stoichiometrically related to partial
    pressures of the gases in the reaction

22
Sampling
  • gas chromatography can measure the concentrations
    of various components in a mixture
  • for samples that have volatile components
  • separates mixture by adherence to a surface
  • drawing off periodic aliquots from the mixture
    and doing quantitative analysis
  • titration for one of the components
  • gravimetric analysis

23
Factors Affecting Reaction RateNature of the
Reactants
  • nature of the reactants means what kind of
    reactant molecules and what physical condition
    they are in.
  • small molecules tend to react faster than large
    molecules
  • gases tend to react faster than liquids which
    react faster than solids
  • powdered solids are more reactive than blocks
  • more surface area for contact with other
    reactants
  • certain types of chemicals are more reactive than
    others
  • e.g., the activity series of metals
  • ions react faster than molecules
  • no bonds need to be broken

24
Factors Affecting Reaction RateTemperature
  • increasing temperature increases reaction rate
  • chemists rule of thumb - for each 10C rise in
    temperature, the speed of the reaction doubles
  • for many reactions
  • there is a mathematical relationship between the
    absolute temperature and the speed of a reaction
    discovered by Svante Arrhenius which will be
    examined later

25
Factors Affecting Reaction RateCatalysts
  • catalysts are substances which affect the speed
    of a reaction without being consumed.
  • most catalysts are used to speed up a reaction,
    these are called positive catalysts
  • catalysts used to slow a reaction are called
    negative catalysts.
  • homogeneous present in same phase
  • heterogeneous present in different phase
  • how catalysts work will be examined later

26
Factors Affecting Reaction RateReactant
Concentration
  • generally, the larger the concentration of
    reactant molecules, the faster the reaction
  • increases the frequency of reactant molecule
    contact
  • concentration of gases depends on the partial
    pressure of the gas
  • higher pressure higher concentration
  • concentration of solutions depends on the solute
    amount to solution ratio (molarity)

27
The Rate Law (A differential equation)
  • the Rate Law of a reaction is the mathematical
    relationship between the rate of the reaction and
    the concentrations of the reactants
  • and homogeneous catalysts as well
  • the rate of a reaction is directly proportional
    to the concentration of each reactant raised to a
    power
  • for the reaction aA bB ? products the rate law
    would have the form given below
  • n and m are called the orders for each reactant
  • k is called the rate constant

28
Reaction Order
  • the exponent on each reactant in the rate law is
    called the order with respect to that reactant
  • the sum of the exponents on the reactants is
    called the order of the reaction
  • The rate law for the reaction

2 NO(g) O2(g) 2 NO2(g) is Rate kNO2O2
The reaction is second order with respect to
NO, first order with respect to O2, and
third order overall
29
Sample Rate Laws
The reaction is autocatalytic, because a product
affects the rate. Hg2 is a negative catalyst,
increasing its concentration slows the reaction.
30
Reactant Concentration vs. TimeA ? Products
31
Half-Life
  • the half-life, t1/2, of a reaction is the length
    of time it takes for the concentration of the
    reactants to fall to ½ its initial value
  • the half-life of the reaction depends on the
    order of the reaction

32
Zero Order Reactions
  • Rate kA0 k
  • constant rate reactions
  • A -kt A0
  • graph of A vs. time is straight line with slope
    -k and y-intercept A0
  • t ½ A0/2k
  • when Rate M/sec, k M/sec

33
First Order Reactions
  • Rate kA
  • lnA -kt lnA0
  • graph lnA vs. time gives straight line with
    slope -k and y-intercept lnA0
  • used to determine the rate constant
  • t½ 0.693/k
  • the half-life of a first order reaction is
    constant
  • the when Rate M/sec, k sec-1

34
(No Transcript)
35
Half-Life of a First-Order ReactionIs Constant
36
Rate Data for C4H9Cl H2O C4H9OH 2HCl
Time (sec) C4H9Cl, M
0.0 0.1000
50.0 0.0905
100.0 0.0820
150.0 0.0741
200.0 0.0671
300.0 0.0549
400.0 0.0448
500.0 0.0368
800.0 0.0200
10000.0 0.0000
37
C4H9Cl H2O ? C4H9OH 2 HCl
38
C4H9Cl H2O ? C4H9OH 2 HCl
39
C4H9Cl H2O ? C4H9OH 2 HCl
slope -2.01 x 10-3 k 2.01 x 10-3 s-1
40
Second Order Reactions
  • Rate kA2
  • 1/A kt 1/A0
  • graph 1/A vs. time gives straight line with
    slope k and y-intercept 1/A0
  • used to determine the rate constant
  • t½ 1/(kA0)
  • when Rate M/sec, k M-1sec-1

41
slope k
1/A
l/A0
time
42
Rate Data For2 NO2 2 NO O2
Time (hrs.) Partial Pressure NO2, mmHg ln(PNO2) 1/(PNO2)
0 100.0 4.605 0.01000
30 62.5 4.135 0.01600
60 45.5 3.817 0.02200
90 35.7 3.576 0.02800
120 29.4 3.381 0.03400
150 25.0 3.219 0.04000
180 21.7 3.079 0.04600
210 19.2 2.957 0.05200
240 17.2 2.847 0.05800
43
Rate Data Graphs For2 NO2 2 NO O2
44
Rate Data Graphs For2 NO2 2 NO O2
45
Rate Data Graphs For2 NO2 2 NO O2
46
Determining the Rate Law
  • can only be determined experimentally
  • graphically
  • rate slope of curve A vs. time
  • if graph A vs time is straight line, then
    exponent on A in rate law is 0, rate constant
    -slope
  • if graph lnA vs time is straight line, then
    exponent on A in rate law is 1, rate constant
    -slope
  • if graph 1/A vs time is straight line, exponent
    on A in rate law is 2, rate constant slope
  • initial rates
  • by comparing effect on the rate of changing the
    initial concentration of reactants one at a time

47
(No Transcript)
48
Practice - Complete the Table and Determine the
Rate Equation for the Reaction A 2 Prod
49
Practice - Complete the Table and Determine the
Rate Equation for the Reaction A 2 Prod
50
(No Transcript)
51
(No Transcript)
52
(No Transcript)
53
Practice - Complete the Table and Determine the
Rate Equation for the Reaction A 2 Prod
the reaction is second order,
54
Ex. 13.4 The reaction SO2Cl2(g) ? SO2(g)
Cl2(g) is first order with a rate constant of
2.90 x 10-4 s-1 at a given set of conditions.
Find the SO2Cl2 at 865 s when SO2Cl20
0.0225 M
SO2Cl20 0.0225 M, t 865, k 2.90 x 10-4
s-1 SO2Cl2
Given Find
Concept Plan Relationships
Solution
Check
the new concentration is less than the original,
as expected
55
Initial Rate Method
  • another method for determining the order of a
    reactant is to see the effect on the initial rate
    of the reaction when the initial concentration of
    that reactant is changed
  • for multiple reactants, keep initial
    concentration of all reactants constant except
    one
  • zero order changing the concentration has no
    effect on the rate
  • first order the rate changes by the same factor
    as the concentration
  • doubling the initial concentration will double
    the rate
  • second order the rate changes by the square of
    the factor the concentration changes
  • doubling the initial concentration will quadruple
    the rate

56
Ex 13.2 Determine the rate law and rate
constant for the reaction NO2(g) CO(g) ? NO(g)
CO2(g) given the data below.
Write a general rate law including all reactants
Examine the data and find two experiments in which the concentration of one reactant changes, but the other concentrations are the same
Expt. Number Initial NO2, (M) Initial CO, (M) Initial Rate (M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
Expt. Number Initial NO2, (M) Initial CO, (M) Initial Rate (M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
Comparing Expt 1 and Expt 2, the NO2 changes
but the CO does not
57
Ex 13.2 Determine the rate law and rate
constant for the reaction NO2(g) CO(g) ? NO(g)
CO2(g) given the data below.
Expt. Number Initial NO2, (M) Initial CO, (M) Initial Rate (M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
Determine by what factor the concentrations and rates change in these two experiments.
58
Ex 13.2 Determine the rate law and rate
constant for the reaction NO2(g) CO(g) ? NO(g)
CO2(g) given the data below.
Expt. Number Initial NO2, (M) Initial CO, (M) Initial Rate (M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
Determine to what power the concentration factor must be raised to equal the rate factor.
59
Ex 13.2 Determine the rate law and rate
constant for the reaction NO2(g) CO(g) ? NO(g)
CO2(g) given the data below.
Expt. Number Initial NO2, (M) Initial CO, (M) Initial Rate (M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
Expt. Number Initial NO2, (M) Initial CO, (M) Initial Rate (M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
Repeat for the other reactants
60
Ex 13.2 Determine the rate law and rate
constant for the reaction NO2(g) CO(g) ? NO(g)
CO2(g) given the data below.
Expt. Number Initial NO2, (M) Initial CO, (M) Initial Rate (M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
Substitute the exponents into the general rate law to get the rate law for the reaction
n 2, m 0
61
Ex 13.2 Determine the rate law and rate
constant for the reaction NO2(g) CO(g) ? NO(g)
CO2(g) given the data below.
Expt. Number Initial NO2, (M) Initial CO, (M) Initial Rate (M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
Substitute the concentrations and rate for any experiment into the rate law and solve for k
62
Practice - Determine the rate law and rate
constant for the reaction NH41 NO2-1 N2 2
H2Ogiven the data below.
Expt. No. Initial NH4, M Initial NO2-, M Initial Rate, (x 10-7), M/s
1 0.0200 0.200 10.8
2 0.0600 0.200 32.3
3 0.200 0.0202 10.8
4 0.200 0.0404 21.6
63
Practice - Determine the rate law and rate
constant for the reaction NH41 NO2-1 N2 2
H2Ogiven the data below.
Rate kNH4nNO2?m
Expt. No. Initial NH4, M Initial NO2-, M Initial Rate, (x 10-7), M/s
1 0.0200 0.200 10.8
2 0.0600 0.200 32.3
3 0.200 0.0202 10.8
4 0.200 0.0404 21.6
64
The Effect of Temperature on Rate
  • changing the temperature changes the rate
    constant of the rate law
  • Svante Arrhenius investigated this relationship
    and showed that

where T is the temperature in kelvins
R is the gas constant in energy units, 8.314
J/(molK)
A is a factor called the frequency factor
Ea is the activation energy, the extra energy
needed to start the molecules reacting
65
(No Transcript)
66
Activation Energy and theActivated Complex
  • energy barrier to the reaction
  • amount of energy needed to convert reactants into
    the activated complex
  • aka transition state
  • the activated complex is a chemical species with
    partially broken and partially formed bonds
  • always very high in energy because partial bonds

67
Isomerization of Methyl Isonitrile
methyl isonitrile rearranges to acetonitrile
68
Energy Profile for the Isomerization of Methyl
Isonitrile
the collision frequency is the number of
molecules that approach the peak in a given
period of time
the activation energy is the difference in energy
between the reactants and the activated complex
the activated complex is a chemical species with
partial bonds
69
The Arrhenius EquationThe Exponential Factor
  • the exponential factor in the Arrhenius equation
    is a number between 0 and 1
  • it represents the fraction of reactant molecules
    with sufficient energy so they can make it over
    the energy barrier
  • the higher the energy barrier (larger activation
    energy), the fewer molecules that have sufficient
    energy to overcome it
  • that extra energy comes from converting the
    kinetic energy of motion to potential energy in
    the molecule when the molecules collide
  • increasing the temperature increases the average
    kinetic energy of the molecules
  • therefore, increasing the temperature will
    increase the number of molecules with sufficient
    energy to overcome the energy barrier
  • therefore increasing the temperature will
    increase the reaction rate

70
(No Transcript)
71
Arrhenius Plots
  • the Arrhenius Equation can be algebraically
    solved to give the following form

this equation is in the form y mx b where y
ln(k) and x (1/T)
a graph of ln(k) vs. (1/T) is a straight line
(-8.314 J/molK)(slope of the line) Ea, (in
Joules)
ey-intercept A, (unit is the same as k)
72
Ex. 13.7 Determine the activation energy and
frequency factor for the reaction O3(g) ? O2(g)
O(g) given the following data
Temp, K k, M-1s-1 Temp, K k, M-1s-1
600 3.37 x 103 1300 7.83 x 107
700 4.83 x 104 1400 1.45 x 108
800 3.58 x 105 1500 2.46 x 108
900 1.70 x 106 1600 3.93 x 108
1000 5.90 x 106 1700 5.93 x 108
1100 1.63 x 107 1800 8.55 x 108
1200 3.81 x 107 1900 1.19 x 109
73
Ex. 13.7 Determine the activation energy and
frequency factor for the reaction O3(g) ? O2(g)
O(g) given the following data
use a spreadsheet to graph ln(k) vs. (1/T)
74
Ex. 13.7 Determine the activation energy and
frequency factor for the reaction O3(g) ? O2(g)
O(g) given the following data
Ea m(-R) solve for Ea
A ey-intercept solve for A
75
Arrhenius EquationTwo-Point Form
  • if you only have two (T,k) data points, the
    following forms of the Arrhenius Equation can be
    used

76
Ex. 13.8 The reaction NO2(g) CO(g) ? CO2(g)
NO(g) has a rate constant of 2.57 M-1s-1 at 701
K and 567 M-1s-1 at 895 K. Find the activation
energy in kJ/mol
T1 701 K, k1 2.57 M-1s-1, T2 895 K, k2
567 M-1s-1 Ea, kJ/mol
Given Find
Concept Plan Relationships
Solution
Check
most activation energies are tens to hundreds of
kJ/mol so the answer is reasonable
77
Collision Theory of Kinetics
  • for most reactions, in order for a reaction to
    take place, the reacting molecules must collide
    into each other.
  • once molecules collide they may react together or
    they may not, depending on two factors -
  • whether the collision has enough energy to "break
    the bonds holding reactant molecules together"
  • whether the reacting molecules collide in the
    proper orientation for new bonds to form.

78
Effective Collisions
  • collisions in which these two conditions are met
    (and therefore lead to reaction) are called
    effective collisions
  • the higher the frequency of effective collisions,
    the faster the reaction rate
  • when two molecules have an effective collision, a
    temporary, high energy (unstable) chemical
    species is formed - called an activated complex
    or transition state

79
Effective CollisionsKinetic Energy Factor
for a collision to lead to overcoming the energy
barrier, the reacting molecules must have
sufficient kinetic energy so that when they
collide it can form the activated complex
80
Effective CollisionsOrientation Effect
81
Collision Theory andthe Arrhenius Equation
  • A is the factor called the frequency factor and
    is the number of molecules that can approach
    overcoming the energy barrier
  • there are two factors that make up the frequency
    factor the orientation factor (p) and the
    collision frequency factor (z)

82
Orientation Factor
  • the proper orientation results when the atoms are
    aligned in such a way that the old bonds can
    break and the new bonds can form
  • the more complex the reactant molecules, the less
    frequently they will collide with the proper
    orientation
  • reactions between atoms generally have p 1
  • reactions where symmetry results in multiple
    orientations leading to reaction have p slightly
    less than 1
  • for most reactions, the orientation factor is
    less than 1
  • for many, p ltlt 1
  • there are some reactions that have p gt 1 in which
    an electron is transferred without direct
    collision

83
Reaction Mechanisms
  • we generally describe chemical reactions with an
    equation listing all the reactant molecules and
    product molecules
  • but the probability of more than 3 molecules
    colliding at the same instant with the proper
    orientation and sufficient energy to overcome the
    energy barrier is negligible
  • most reactions occur in a series of small
    reactions involving 1, 2, or at most 3 molecules
  • describing the series of steps that occur to
    produce the overall observed reaction is called a
    reaction mechanism
  • knowing the rate law of the reaction helps us
    understand the sequence of steps in the mechanism

84
An Example of a Reaction Mechanism
  • Overall reaction
  • H2(g) 2 ICl(g) ? 2 HCl(g) I2(g)
  • Mechanism
  • H2(g) ICl(g) ? HCl(g) HI(g)
  • HI(g) ICl(g) ? HCl(g) I2(g)
  • the steps in this mechanism are elementary steps,
    meaning that they cannot be broken down into
    simpler steps and that the molecules actually
    interact directly in this manner without any
    other steps

85
Elements of a MechanismIntermediates
  • H2(g) 2 ICl(g) ? 2 HCl(g) I2(g)
  • H2(g) ICl(g) ? HCl(g) HI(g)
  • HI(g) ICl(g) ? HCl(g) I2(g)
  • notice that the HI is a product in Step 1, but
    then a reactant in Step 2
  • since HI is made but then consumed, HI does not
    show up in the overall reaction
  • materials that are products in an early step, but
    then a reactant in a later step are called
    intermediates

86
Molecularity
  • the number of reactant particles in an elementary
    step is called its molecularity
  • a unimolecular step involves 1 reactant particle
  • a bimolecular step involves 2 reactant particles
  • though they may be the same kind of particle
  • a termolecular step involves 3 reactant particles
  • though these are exceedingly rare in elementary
    steps

87
Rate Laws for Elementary Steps
  • each step in the mechanism is like its own little
    reaction with its own activation energy and own
    rate law
  • the rate law for an overall reaction must be
    determined experimentally
  • but the rate law of an elementary step can be
    deduced from the equation of the step
  • H2(g) 2 ICl(g) ? 2 HCl(g) I2(g)
  • H2(g) ICl(g) ? HCl(g) HI(g) Rate
    k1H2ICl
  • HI(g) ICl(g) ? HCl(g) I2(g) Rate
    k2HIICl

88
Rate Laws of Elementary Steps
89
Rate Determining Step
  • in most mechanisms, one step occurs slower than
    the other steps
  • the result is that product production cannot
    occur any faster than the slowest step the step
    determines the rate of the overall reaction
  • we call the slowest step in the mechanism the
    rate determining step
  • the slowest step has the largest activation
    energy
  • the rate law of the rate determining step
    determines the rate law of the overall reaction

90
Another Reaction Mechanism
  • NO2(g) CO(g) ? NO(g) CO2(g) Rateobs
    kNO22
  • NO2(g) NO2(g) ? NO3(g) NO(g) Rate
    k1NO22 slow
  • NO3(g) CO(g) ? NO2(g) CO2(g) Rate
    k2NO3CO fast

The first step is slower than the second step
because its activation energy is larger.
The first step in this mechanism is the rate
determining step.
The rate law of the first step is the same as the
rate law of the overall reaction.
91
Validating a Mechanism
  • in order to validate (not prove) a mechanism, two
    conditions must be met
  • the elementary steps must sum to the overall
    reaction
  • the rate law predicted by the mechanism must be
    consistent with the experimentally observed rate
    law

92
Mechanisms with a Fast Initial Step
  • when a mechanism contains a fast initial step,
    the rate limiting step may contain intermediates
  • when a previous step is rapid and reaches
    equilibrium, the forward and reverse reaction
    rates are equal so the concentrations of
    reactants and products of the step are related
  • and the product is an intermediate
  • substituting into the rate law of the RDS will
    produce a rate law in terms of just reactants

93
An Example
2 H2(g) 2 NO(g) ? 2 H2O(g) N2(g) Rateobs
k H2NO2
94
Ex 13.9 Show that the proposed mechanism for the
reaction 2 O3(g) ? 3 O2(g) matches the observed
rate lawRate kO32O2-1
95
Catalysts
  • catalysts are substances that affect the rate of
    a reaction without being consumed
  • catalysts work by providing an alternative
    mechanism for the reaction
  • with a lower activation energy
  • catalysts are consumed in an early mechanism
    step, then made in a later step

mechanism without catalyst O3(g) O(g) ? 2
O2(g) V. Slow
mechanism with catalyst Cl(g) O3(g) ? O2(g)
ClO(g) Fast ClO(g) O(g) ? O2(g) Cl(g)
Slow
96
Ozone Depletion over the Antarctic
97
Energy Profile of Catalyzed Reaction
98
Catalysts
  • homogeneous catalysts are in the same phase as
    the reactant particles
  • Cl(g) in the destruction of O3(g)
  • heterogeneous catalysts are in a different phase
    than the reactant particles
  • solid catalytic converter in a cars exhaust
    system

99
Types of Catalysts
100
Catalytic HydrogenationH2CCH2 H2 ? CH3CH3
101
Enzymes
  • because many of the molecules are large and
    complex, most biological reactions require a
    catalyst to proceed at a reasonable rate
  • protein molecules that catalyze biological
    reactions are called enzymes
  • enzymes work by adsorbing the substrate reactant
    onto an active site that orients it for reaction

102
Enzyme-Substrate BindingLock and Key Mechanism
103
Enzymatic Hydrolysis of Sucrose
About PowerShow.com