1. Ventura Community College
Chemical Kinetics
seconds
minutes
months
Dr. Nick Blake
Ventura Community College

2. Kinetics
“The study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds”
4 factors influence the speed of a reaction:
concentration
how likely is it that reactants will come together to react?
temperature
how quickly can reactants come together and how much excess energy do they have to break bonds and convert to products?
catalysts
substances that normally speed up the reaction without being used up
nature of the reactants
Gases react faster than liquids which react faster than solids, surface area for liquid/solid or gas/solid reactions, big heavy molecules react more slowly than light molecules
In Chemical Kinetics we aim to quantitatively understand how and why each of these factors affect any chemical reaction

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)

4. Defining Reaction Rate
the rate of a chemical reaction is measured as the change of concentration of a reactant (or product) in a given period of time interval
To make the rate > 0, 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. (t1,t2) (s)
A+BC (molecules/s)
X+YZ
(molecules/s)
(0,16)
0.250
0.0625
(16,32)
0.125
(32,48)

10. 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

11. 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

12. H2 I2 HI HI Avg. Rate, M/s t (s) [H2]t, M [HI]t, M -Δ[H2]/Δt
0.000
1.000
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
0.135
1.730
0.0030
Avg. Rate, M/s
t (s)
[H2]t, M
[HI]t, M
-Δ[H2]/Δt
½ Δ[HI]/Δt
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
0.135
Avg. Rate, M/s
t (s)
[H2]t, M
[HI]t, M
-Δ[H2]/Δt
½ Δ[HI]/Δt
0.000
1.000
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
0.135
1.730
Avg. Rate, M/s
t (s)
[H2]t, M
[HI]t, M
-Δ[H2]/Δt
0.000
1.000
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
0.135
1.730
0.0030

13. the average rate for the first 80 s is 0.0108 M/s
average rate = - slope of the line connecting the [H2] points; = ½ slope of the line for [HI]
the average rate for the first 80 s is M/s
the average rate for the first 40 s is M/s
the average rate for the first 10 s is M/s
Tro, Chemistry: A Molecular Approach

14. Instantaneous Rate
the instantaneous rate at a particular t is the change in concentration at that t
Slope of the concentration vs time graph at t
determined by taking the slope of a line tangent to the curve at that particular point
first derivative of the function

15. 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:

16. Example Problem For the reaction given, the [I-] changes from 1
Example Problem For the reaction given, the [I-] changes from M to M in the first 10 s. Calculate the average rate in the first 10 s and the Δ[H+]. 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 reactant here (I-))
Rewrite the Rate in terms of the change in the concentration for the product to find) (H+) and solve for the unknown

17. 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
(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

18. 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
total pressure – the total pressure of a gas mixture is stoichiometrically related to partial pressures of the gases in the reaction

19. 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

20. Factors Affecting Reaction Rate Nature 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

21. Factors Affecting Reaction Rate Temperature
Increasing temperature increases reaction rate
rule of thumb - for each 10°C rise in temperature, the speed of the reaction doubles
there is a mathematical relationship between the absolute temperature and the speed of a reaction discovered by Svante Arrhenius which will be examined later

22. Factors Affecting Reaction Rate Catalysts
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

23. Factors Affecting Reaction Rate Reactant Concentration
generally, the larger the concentration of reactant molecules, the faster the reaction
increases the likelihood of reactant molecules bumping into each other
concentration of gases depends on the partial pressure of the gas
higher pressure = higher concentration
concentration of solutions depends on the solute to solution ratio (molarity)

24. The Rate Law
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

25. Reaction Order 2 NO(g) + O2(g)  2 NO2(g) Rate = k[NO]x[O2]y
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)
Rate = k[NO]x[O2]y
When it is measured experimentally it is found to be
Rate = k[NO]2[O2]1=k[NO]2[O2]
The reaction is:
second order with respect to [NO] (x=2),
first order with respect to [O2] (y=1),
and third order overall = (x + y = = 3)

26. The reaction is autocatalytic, because a product affects the rate.
Sample Rate Laws
The reaction is autocatalytic, because a product affects the rate.
Hg2+ is a negative catalyst, increasing its concentration slows the reaction.

27. Reactant Concentration vs. Time A  Products

28. 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
Tro, Chemistry: A Molecular Approach

29. Zero Order Reactions [A] slope = - k time Rate = k[A]0 = k
constant rate reactions
[A] = -kt + [A]0
graph of [A] vs. time is straight line with
slope = -k and y-intercept = [A]0
t ½ = [A0]/2k
when Rate = M/sec, k = M/sec
[A]0
[A]
time
slope = - k

30. First Order Reactions Rate = k[A] ln[A] = -kt + ln[A]0
graph ln[A] vs. time gives straight line with slope = -k and y-intercept = ln[A]0
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

31. ln[A]0
ln[A]
time
slope = −k

32. Half-Life of a First-Order Reaction Is Constant

33. Rate Data for C4H9Cl + H2O  C4H9OH + HCl
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
0.0000

34. C4H9Cl + H2O  C4H9OH + 2 HCl

35. C4H9Cl + H2O  C4H9OH + 2 HCl

36. C4H9Cl + H2O  C4H9OH + 2 HCl
slope =
-2.01 x 10-3
k =
2.01 x 10-3 s-1

37. Problem: Show the data is consistent with a first order reaction
Problem: Show the data is consistent with a first order reaction. What is the rate constant and what is the half life?
time (s)
NaN3 (M)
0.5 1
0.473 5
0.378 10
0.286 15
0.216 20
0.163
t1/2=0.693/k = 0.693/0.056 s
= s
k = 0.056S-1
time (s)
Ln[NaN3] 1 5 10 15 20

38. Second Order Reactions
Rate = k[A]2
1/[A] = kt + 1/[A]0
graph 1/[A] vs. time gives straight line with slope = k and y-intercept = 1/[A]0
used to determine the rate constant
t½ = 1/(k[A0])
when Rate = M/sec, k = M-1∙sec-1
Tro, Chemistry: A Molecular Approach

39. slope = k
1/[A]
l/[A]0
time
Tro, Chemistry: A Molecular Approach

40. Partial Pressure NO2, mmHg
Rate Data For 2 NO2  2 NO + O2
Time (hrs.)
Partial Pressure NO2, mmHg
ln(PNO2)
1/(PNO2)
100.0
4.605 30
62.5
4.135 60
45.5
3.817 90
35.7
3.576
120
29.4
3.381
150
25.0
3.219
180
21.7
3.079
210
19.2
2.957
240
17.2
2.847

41. Rate Data Graphs For 2 NO2  2 NO + O2

42. Rate Data Graphs For 2 NO2  2 NO + O2

43. Rate Data Graphs For 2 NO2  2 NO + O2

44. 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 ln[A] 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

45. Tro, Chemistry: A Molecular Approach

46. Practice - Complete the Table and Determine the Rate Equation for the Reaction A  2 Prod

47. Practice - Complete the Table and Determine the Rate Equation for the Reaction A  2 Prod
Tro, Chemistry: A Molecular Approach

48. Tro, Chemistry: A Molecular Approach

49. Tro, Chemistry: A Molecular Approach

50. Tro, Chemistry: A Molecular Approach

51. Practice - Complete the Table and Determine the Rate Equation for the Reaction A ® 2 Prod
-D[A] Dt
= 0.1 [A]2
the reaction is second order,
Tro, Chemistry: A Molecular Approach

52. 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 [SO2Cl2]0 = M
Given:
Find:
[SO2Cl2]0 = M, t = 865, k = 2.90 x 10-4 s-1
[SO2Cl2]
Concept Plan:
Relationships:
[SO2Cl2]
[SO2Cl2]0, t, k
Solution:
Check:
the new concentration is less than the original, as expected

53. 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

54. Determine the rate law and rate constant for the reaction NO2(g) + CO(g)  NO(g) + CO2(g) given the data below.
Write the general rate law
Take the ratios of 2 experiments where one reactant is changing but other reactants are fixed
Exp.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s) 1.
0.10
0.0021 2.
0.20
0.0082 3.
0.0083 4.
0.40
0.033
Expt.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s) 1.
0.10
0.0021 2.
0.20
0.0082 3.
0.0083 4.
0.40
0.033
Comparing Exp #1 and Exp #2, the [NO2] changes but the [CO] does not

55. Determine the rate law and rate constant for the reaction NO2(g) + CO(g)  NO(g) + CO2(g) given the data below.
Repeat for the other reactants
Exp.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s) 1.
0.10
0.0021 2.
0.20
0.0082 3.
0.0083 4.
0.40
0.033
Expt.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s) 1.
0.10
0.0021 2.
0.20
0.0082 3.
0.0083 4.
0.40
0.033

56. Determine the rate law and rate constant for the reaction NO2(g) + CO(g)  NO(g) + CO2(g) given the data below.
Substitute the exponents into the general rate law to get the rate law for the reaction
Expt.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s) 1.
0.10
0.0021 2.
0.20
0.0082 3.
0.0083 4.
0.40
0.033
n = 2, m = 0

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.
Exp.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s) 1.
0.10
0.0021 2.
0.20
0.0082 3.
0.0083 4.
0.40
0.033
Substitute the concentrations and rate for any experiment into the rate law and solve for k

58. Practice - Determine the rate law and rate constant for the reaction NH4+1 + NO2-1  N2 + 2 H2O given 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
32.3 3
0.0202 4
0.0404
21.6

59. Practice - Determine the rate law and rate constant for the reaction NH4+1 + NO2-1  N2 + 2 H2O given the data below.
Rate = k[NH4+]n[NO2-]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
32.3 3
0.0202 4
0.0404
21.6

60. Collision Theory H2 + I2  2HI
Explains how reactions occur and why reaction rates differ for different reactions
Reactions occur when reactant atoms/molecules/ions collide (elementary steps)
Some reactions occur in several elementary steps (reaction mechanism)
Some collisions are effective when the geometry of the reactants is favorable, and some are not
H2 + I2  2HI

61. Effective Collisions Orientation Effect

62. Collision Theory
Only reactive collisions can bring about chemical change
Reactive collisions must have exceed the activation energy Ea which goes toward breaking pre-existing bonds

63. Fraction of collisions with enough energy to react
Collision Theory
The rate for an elementary step aA +bBproducts
Probability of aA and bB occupying the same point in space
Aka Exponential factor
# favorable collisions / s
Aka Frequency factor
Fraction of collisions with enough energy to react

64. The Effect of Temperature on Rate
changing the temperature changes the rate constant of the rate law
Arrhenius investigated this relationship and showed that:
where T is the temperature in Kelvin
R is the gas constant in energy units, J/(mol∙K)
A is a factor called the frequency factor (has a weak T dependence)
Ea is the activation energy, the extra energy needed to start the molecules reacting

66. Activation Energy and the Activated 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
in order for the reaction to occur, the H3C-N bond must break; and a new H3C-C bond form

68. Energy Profile for the Isomerization of Methyl Isonitrile

69. The Arrhenius Equation: The 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. Tro, Chemistry: A Molecular Approach

71. Arrhenius Plots
the Arrhenius Equation can be algebraically solved to give the following form:
( J/mol∙K) x (slope of the line) = Ea, (in J/mol)
ey-intercept = A, (unit is the same as k)

72. Determine the activation energy and frequency factor for the reaction O3(g)  O2(g) + O(g) given the following data:
Temp, K
k, M-1∙s-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 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. 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 Equation: Two-Point Form
if you only have two (T,k) data points, the following forms of the Arrhenius Equation can be used:

76. The reaction NO2(g) + CO(g)  CO2(g) + NO(g) has a rate constant of 2
The reaction NO2(g) + CO(g)  CO2(g) + NO(g) has a rate constant of 2.57 M-1∙s-1 at 701 K and 567 M-1∙s-1 at 895 K. Find the activation energy in kJ/mol
Given:
Find:
T1 = 701 K, k1 = 2.57 M-1∙s-1, T2 = 895 K, k2 = 567 M-1∙s-1
Ea, kJ/mol
Concept Plan:
Relationships: Ea
T1, k1, T2, k2
Solution:
Check:
most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable

77. 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 << 1
there are some reactions that have p > 1 in which an electron is transferred without direct collision
Tro, Chemistry: A Molecular Approach

78. 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

79. 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

80. Elements of a Mechanism Intermediates
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

81. 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

82. 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) overall
H2(g) + ICl(g)  HCl(g) + HI(g) Rate = k1[H2][ICl]
HI(g) + ICl(g)  HCl(g) + I2(g) Rate = k2[HI][ICl]

83. Rate Laws of Elementary Steps
Tro, Chemistry: A Molecular Approach

84. 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

85. Another Reaction Mechanism
NO2(g) + CO(g)  NO(g) + CO2(g) Rateobs = k[NO2]2
NO2(g) + NO2(g)  NO3(g) + NO(g) Rate = k1[NO2]2 slow
NO3(g) + CO(g)  NO2(g) + CO2(g) Rate = k2[NO3][CO] 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

86. 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

87. 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

88. An Example 2 NO(g) N2O2(g) Fast
H2(g) + N2O2(g)  H2O(g) + N2O(g) Slow Rate = k2[H2][N2O2]
H2(g) + N2O(g)  H2O(g) + N2(g) Fast k1
k-1
2 H2(g) + 2 NO(g)  2 H2O(g) + N2(g) Rateobs = k [H2][NO]2
for Step 1 Rateforward = Ratereverse

89. O3(g) + O(g)  2 O2(g) Slow Rate = k2[O3][O]
Show that the proposed mechanism for the reaction 2 O3(g)  3 O2(g) matches the observed rate law Rate = k[O3]2[O2]-1
O3(g) O2(g) + O(g) Fast
O3(g) + O(g)  2 O2(g) Slow Rate = k2[O3][O] k1
k-1
for Step 1 Rateforward = Ratereverse

90. Catalysts O3(g) + O(g)  2 O2(g) V. Slow
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
used
regenerated

91. Ozone Depletion over the Antarctic
The Dobson unit (DU) is a unit of measurement of the columnar density of a trace gas in the Earth's atmosphere. One Dobson unit refers to a layer of gas that would be 10 µm thick under standard temperature and pressure.

92. Energy Profile of a Catalyzed Reaction
How does the Cl affect the energy profile of the reaction?
Reaction occurs in 2 elementary steps hence 2 humps in the profile
Significantly reduces Ea increasing the rate

93. 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 car’s exhaust system

94. Types of Catalysts

95. Catalytic Hydrogenation: H2C=CH2 + H2 → CH3CH3
the surface lowers Ea and increases A

96. Enzymes
because many of the molecules are large and complex, most biological reactions require a catalyst to proceed at a reasonable rate, otherwise A will be very small
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 with effectively increases A, and often lowers Ea, making k larger
It is common for the shape of the enzyme to change once it binds with the reactant, and causes a deformation of the molecule that facilitates change

97. Enzyme-Substrate Binding: Lock and Key Mechanism
deformation

98. Enzymatic Hydrolysis of Sucrose
disaccharide + water  2 x monosaccharides
Sucrose is attracted to bind with the enzyme sucrase and when it attaches the bond connecting the two monosaccharides becomes strained and weakens, exposing the bond to attack from H2O lowering Ea and increasing A