1. Chapter 14
Chemical Kinetics

2. Kinetics
The study of reaction rates – the speed at which reactants are converted to products.
Explosions happen in seconds- very fast.
Coal (carbon) will turn into diamond  eventually.
Spontaneous reactions are reactions that will happen  but we can’t tell how fast.
Kinetics is very important because it allows us to manipulate and control reactions.

3. Reactions occur when . . . Molecules find each other
Effective collisions happen between molecules
Frequency of collisions is high

4. Factors that affect Reaction Rate:
1. Temperature
Hot = fast Cold = slow
2. Concentration of Reactants
more concentrated = faster
3. Physical conditions of reactants.
more surface area = faster
4. Presence of a Catalyst –
increases rate without being used up

5. Reaction Rate
The change in concentration of a reactant or a product per unit of time.
Rates decrease with time
Typically rates are expressed as positive values
Instantaneous rate can be determined by finding the slop of a line tangent to a point representing a particular time

6. Reaction Rate - “speed of the reaction”
Rate = Conc. of A at t2  Conc. of A at t t2  t1
Rate = D[A] Dt
Units: Change in concentration per time
For this reaction
N2 + 3H NH3

7. As the reaction progresses the concentration of H2 goes down N2 + 3H2 2NH3
Time

8. As the reaction progresses the concentration N2 goes down 1/3 as fast
As the reaction progresses the concentration N2 goes down 1/3 as fast. Why? N2 + 3H NH3
Concentration
[N2]
[H2]
Time

9. As the reaction progresses the concentration NH3 goes up. N2 + 3H2 2NH3
Time

10. Calculating Rates 1. Average rates are taken over long intervals
2. Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time (Finding the Derivative.)

11. Average slope method
Concentration
D[H2] Dt
Time

12. Instantaneous slope method.
Concentration
D[H2]
D t
Time

13. Find the Instantaneous rate of disappearance:
In order to find the instantaneous rate of disappearance at time 0, a tangent line was drawn and the following information was obtained: the concentration of the unknown changed from M to M from 0 seconds to 200 seconds. What is the instantaneous rate of disappearance at time 0?
2.0 x 10-4 M/s

14. Reaction Rates and Stoichiometry
Remember, the rate of disappearance is a “” slope and the rate of appearance is a “+” slope
For general equation:
aA bB cC + dD
General rate equation:
Rate = -1D[A] = -1D[B] = 1D[C] = 1∆[D] a Dt b Dt c Dt d∆t

15. Defining Rate
We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product.
In our example
N2 + 3H NH3
-D[N2] = -D[H2] = D[NH3] Dt Dt Dt

16. Practice
How is the rate of disappearance of ozone related to the rate of appearance of oxygen in the following equation: 2O3  3O2?
If the rate of appearance of O2, is 6.0x10-5 M/s at a particular instant, what is the value of the rate of disappearance of O3 at this same time?
4.0x10-5 M/s

17. The decomposition of N2O5 proceeds according to the following equation:
2N2O5  4NO2 + O2
If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2x10-7 M/s, what is the rate of appearance of NO2? O2?
NO2 = 8.4x10-7 M/s
O2 = 2.1 x 10-7 M/s

18. Rate Laws
To study the rate of a reaction and determine the rate law, a series of experiments must be done varying the initial concentrations of reactants.
This is called the Initial rate method.
The products do not appear in the rate law because we are only concerned with the very start of the experiment where conc. of reactants are carefully chosen.

19. The Rate Law
The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.
aA + bB cC + dD
Rate = k [A]x[B]y } The rate law expression
k is the rate constant (temperature dependent)
Units of k vary
x and y are integers, usually 0, 1 or 2
reaction is xth order in A, reaction is yth order in B
reaction is (x +y)th order overall

20. F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2]x [ClO2]y Initial [ ]
(All readings
recorded at
same temp.)
Double [F2] with [ClO2] constant: rate doubles
Quadruple [ClO2] with [F2] constant: rate quadruples
Therefore, rate law is: rate = k [F2]1 [ClO2]1
We say, “first order with respect to F2, first order with
respect to ClO2, and 2nd order overall”

21. 1. Rate laws are always determined experimentally.
Rate Laws - Key Points
1. Rate laws are always determined experimentally.
2. Reaction order is always defined in terms of reactant (not product) concentrations.
3. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
F2 (g) + 2ClO2 (g) FClO2 (g)
rate = k [F2]1 [ClO2]1
4. Once you obtain the Rate Law and the
value of k, you can calculate the rate of
reaction for any set of concentrations.

22. Try this. A + B ---> C + D
Run # Initial [A] Initial [B] Initial Rate (v0) ([A]0) ([B]0)
M M x 10-2 M/s
M M x 10-2 M/s
M M x 10-2 M/s
What is the order with respect to A?
What is the order with respect to B?
What is the overall order of the reaction?
Write the rate law expression 1 1
Rate = k [A]0 [B]1

23. Try this. 2NO (g) + Cl2 (g) ---> 2NOCl
[NO(g)] (M) [Cl2(g)] (M) Initial Rate(Ms-1) 
0.250     x 10-6
0.250     x 10-6
0.500     x 10-5
What is the order with respect to Cl2?
What is the order with respect to NO?
What is the overall order of the reaction?
Write the rate law expression
Calculate the value of k 1 2 3
Rate = k [NO]2 [Cl2]1
9.15 x /M2•s

24. Units of k 1st Order overall units: 1/s 2nd Order overall units: 1/M•s
What the Order means:
•Zero order means doubling or tripling the [reactant]o has
NO effect on the rate.
•First order means doubling the [reactant]o doubles the rate,
tripling the [reactant]o triples the rate, etc…
•Second order means doubling the [reactant]o quadruples
the rate (raises to a power of 2), etc. . .
Units of k
1st Order overall units: 1/s
2nd Order overall units: 1/M•s
3rd Order overall units: 1/ M2•s

25. Try this: BrO3- + 5 Br- + 6H+ ----> 3Br2 + 3 H2O
Initial concentrations (M)
Rate (M/s)
[BrO3-]
[Br-]
[H+]
x 10-4
x 10-3
x 10-3
x 10-3
Determine the rate law expression
Calculate the value of k
Rate = k [BrO3]1 [Br-]1 [H+]2
fourth order overall
8.0 1/M3•s

26. NH4+ + NO2-  N2 + 2H2O Determine the Rate Law Expression
Initial Rate (M/s)
0.100 M M
1.35 x 10-7
0.010 M
2.70 x 10-7
0.200 M
5.40 x 10-7
Determine the Rate Law Expression
Determine the value for k

27. Types of Rate Laws
Differential Rate Law – describes how rate depends on concentration
Rate = k [A]x[B]y
Integrated Rate Law – describes how concentration depends on time
For each type of differential rate law, there is an integrated rate law and vise versa.
We can use these laws to help us better understand reaction mechanisms

28. Integrated Rate Law Equations allow us to:
1. Determine the concentration of reactant remaining at any time after the reaction has started.
2. Determine the time required for a fraction of a sample to react.
3. Determine the time for a reactant concentration to fall to a certain level.
4. Verify whether a reaction is first order and determine the rate constant.

29. Integrated Rate Law
Expresses the reaction concentration as a function of time.
Form of the equation depends on the order of the rate law (differential).
A Changing Rate = D[A]n Dt
We will only work with n = 0, 1, and 2

30. First Order For the reaction 2N2O5 4NO2 + O2 Rate = k[N2O5]1
If concentration doubles the rate doubles.
If we integrate this equation with respect to time we get the Integrated Rate Law
ln[N2O5] = - kt + ln[N2O5]0
ln is the natural log
[N2O5]0 is the initial concentration.

31. First Order General form-- Rate = D[A] / Dt = k[A]
ln[A] = - kt + ln[A]0
In the form y = mx + b
y = ln[A] ; m = -k ; x = t ;b = ln[A]0
A graph of ln[A] vs time is a straight line.

32. First Order A graph of ln[A] vs time is a straight line.
By getting the straight line you can prove it is first order

33. First Order For the reaction 2N2O5 4NO2 + O2 Rate = k[N2O5]1
ln[N2O5] = - kt + ln[N2O5]0

34. Try this
The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?
ln[A] = - kt + ln[A]0
ln[A] - ln[A]0 = - kt
ln[0.14 M] - ln[0.88M]0 = - (2.8 x 10-2 s-1) t
 = - (2.8 x 10-2 s-1) t
= - (2.8 x 10-2 s-1) t
66 s = t

35. Second Order (Overall)
Rate = -D[A] / Dt = k[A]2
integrated rate law: 1/[A] = kt + 1/[A]0
y= 1/[A] m = k
x= t b = 1/[A]0
A straight line if 1/[A] vs time is graphed
Knowing k and [A]0 you can calculate [A] at any time t

36. Second Order
Looking at the data alone, you cannot tell if 1st or 2nd order.
To tell the difference you must graph the data, or look at the units of k.

37. Zero Order Rate Law Rate = k[A]0 = k
Rate does not change with concentration.
Integrated [A] = -kt + [A]0
In the form y = mx + b
Most often when reaction happens on a surface because the surface area stays constant.

38. Zero Order
Concentration
Time

39. Zero Order
Concentration
k =
D[A]/Dt
D[A] Dt
Time

40. Zero Order

41. Half Life
The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2

42. Half Life The time required to reach half the original concentration.
If the reaction is first order
t½ = t when [A] = [A]0/2
If the reaction is 2nd order:
If the reaction is zero order: ln
[A]0
[A]0/2 k = t½
ln 2 k =
.693 k =
t½ = 1
k[A]0
t½ =
[A]0 2k

43. Try This
What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?
Which order is it? How do you know? t½
Ln 2 k =
0.693
5.7 x 10-4 s-1 =
= 1200 s = 20 minutes

44. Summary of Rate Laws Order Rate Law Integrated Rate Law Equation
Half-Life

45. Reaction Rates So, WHY do some reactions go fast and others are slow?
Reaction Mechanisms Collision Theory
and
Activation Energy

46. Reaction Mechanisms
The series of steps that actually occur in a chemical reaction.
Each step is called an elementary step.
Kinetics can tell us something about the mechanism
A balanced equation does not tell us how the reactants become products.

47. Reaction Mechanisms -- Example
2NO2 + F NO2F
Rate = k[NO2] [F2] (determined experimentally)
The proposed mechanism is
NO2 + F NO2F + F (slow)
F + NO NO2F (fast)
F is called an intermediate It is formed then consumed in the reaction
The slow step is the rate determining step!

48. Reaction Mechanisms
Each of the two reactions is called an elementary step .
The rate for a reaction can be written from its molecularity
Molecularity is the number of pieces that must collide to produce the reaction indicated by the step.
Molecularity is the number of molecules which must collide effectively for reaction to occur.

49. Unimolecular step involves one molecule - Rate is first order.
Bimolecular step - requires two molecules - Rate is second order
Termolecular step- requires three molecules - Rate is third order
Termolecular steps are almost never heard of because the chances of three molecules colliding effectively at the same time are miniscule.

50. Rate Laws for Elementary Steps
(these ARE based on the stoichiometry)
A products Rate = k[A]
A+A products Rate= k[A]2
2A products Rate= k[A]2
A+B products Rate= k[A][B]
A+A+B Products Rate= k[A]2[B]
2A+B Products Rate= k[A]2[B]
A+B+C Products; Rate= k[A][B][C]

51. Going back to our example:
2NO2 + F NO2F (overall reaction)
Rate = k[NO2] [F2] (determined experimentally)
The proposed mechanism consists of
2 elementary steps:
NO2 + F NO2F + F (slow)
F + NO NO2F (fast)
F is called an intermediate - It is formed then consumed in the reaction.
The slow bimolecular elementary step determines the rate of the reaction and sets the rate law as:
2nd order overall.A + B P rate = k[A] [B]

52. Multistep Example
It has been proposed that the conversion of ozone into O2 proceeds via two elementary steps:
O3  O2 + O
O3 + O  2O2
Describe the Molecularity of each step
Write the equation for the overall reaction.
Identify the intermediate(s)

53. For the Equation: Mo(CO)6 + P(CH3)3  Mo(CO)5P(CH3)3 + CO
The proposed mechanism is:
Mo(CO)6  Mo(CO)5 + CO
Mo(CO)5 + P(CH3)3  Mo(CO)5P(CH3)3
Is the proposed mechanism consistent with the overall reaction?
Identify the intermediate(s)

54. Is a single-step mechanism likely for this reaction? Explain.
Consider the following reaction:
2NO+ Br2  2NOBr
Write the rate law for the reaction assuming it involves a single step.
Is a single-step mechanism likely for this reaction? Explain.

55. Add more mechanism practice…
If intermediate is in the slow step, the rate of the reactants that produced it get plugged into its place for the rate law. (need to show example or two)

56. In summary
Reaction mechanisms describe what really happens during a reaction: how the molecules collide and how many are involved.
There is a bit of guessing as to what the mechanism might be.
Chemists try to match a proposed mechanism to an actual experimentally determined rate law.
The rate of the reaction is the slowest elementary step of the mechanism. The overall rate order is set by this step.

57. What about the role of temperature?
Temperature, Rate, and
Effective Collisions

58. Collision Theory Molecules must collide to react.
Concentration affects rates because collisions are more likely.
Must collide hard enough.
Temperature and rate are related.
Only a small number of collisions produce reactions.
In order to react, colliding molecules must have a total KE equal or greater than some minimum value.

59. O O O O O O O O O O No Reaction N N N N Br Br Br Br Br N N Br Br N Br
2 NOBr NO + Br2 O O O O N N N N
No reaction Br Br Br Br Br N O O N Br Br N O Br O N
an effective collision O N Br Br N O
No Reaction

60. Energy Diagram
Potential
Energy
Reactants
Products
Reaction Coordinate

61. Activation Energy Ea Reaction Coordinate Energy Diagram Potential
Reactants
Products
Reaction Coordinate

62. Reaction Coordinate Energy Diagram Activated complex Potential Energy
Reactants
Products
Reaction Coordinate

63. } DE Reaction Coordinate Energy Diagram Potential Energy Reactants
Products
Reaction Coordinate

64. Reaction Coordinate Br---NO Potential Br---NO Transition State Energy
2NO + Br 2
Reaction Coordinate

65. Terms
Activation energy - the minimum kinetic energy needed to make a reaction happen.
Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier.

66. Exothermic Endothermic A + B C + D
The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction.

67. Exothermic and the Reverse Endothermic
Exothermic Endothermic
(a)Activation energy (Ea) for the forward reaction
(a)Activation energy (Ea) for the reverse reaction
(c) ∆H
50 kJ/mol
300 kJ/mol
150 kJ/mol
100 kJ/mol
-100 kJ/mol
+200 kJ/mol

68. Arrhenius
Said that molecules must possess a certain minimum energy in order to react.
Said the reaction rate should increase with temperature.
At high temperature more molecules have the energy required to get over the barrier.
The number of collisions with the necessary energy increases exponentially.

69. k = Ae-Ea/RT Arrhenius Equation
Number of collisions with the required energy,
worked into an equation:
k = Ae-Ea/RT
k = rate constant
A = frequency factor (total collisions)
e is Euler’s number (opposite of ln)
Ea = activation energy
R = ideal gas law constant
T is temperature in Kelvin

70. To find the activation energy:
Rearrange Arrhenius Equation by taking the natural log of each side
ln k = Ea ln A RT

71. Activation Energy and Rates
The final saga

72. Mechanisms and rates
There is an activation energy (hill to get over) for each elementary step.
Activation energy determines k.
k = Ae- (Ea/RT)
k determines rate
Slowest step (rate determining) must have the highest activation energy.

73. This reaction takes place in three steps
Reaction Progress

74. 1 Ea
First step is fast
Low activation energy

75. High activation energy2 Ea
Second step is slow
High activation energy

76. 3 Ea
Third step is fast
Low activation energy

77. Second step is rate determining

78. Intermediates are present

79. Activated Complexes or Transition States

80. Catalysts - How do they work?
Speed up a reaction without being used up in the reaction.
Enzymes are biological catalysts.
Homogenous Catalysts are in the same phase as the reactants.
Heterogeneous Catalysts are in a different phase as the reactants.

81. How Catalysts Work
Catalysts allow reactions to proceed by a different mechanism - a new pathway.
New pathway has a lower activation energy.
More molecules will have this activation energy.
Do not change E (∆H) of the reaction.

82. No catalyst With a catalyst
A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.
No catalyst With a catalyst
uncatalyzed
catalyzed
Ea , k
ratecatalyzed > rateuncatalyzed

83. Heterogeneous Catalysts
Hydrogen bonds to surface of metal.
Break H-H bonds H H H H H
Pt surface
H2 + C2H > C2H6

84. Heterogeneous Catalysts
Pt surface
H2 + C2H > C2H6

85. Heterogeneous Catalysts
The double bond breaks and bonds to the catalyst. H H H C C H H H H H
Pt surface
H2 + C2H > C2H6

86. Heterogeneous Catalysts
The hydrogen atoms bond with the carbon H H H C C H H H H H
Pt surface
H2 + C2H > C2H6

87. Heterogeneous Catalysts
Pt surface
H2 + C2H > C2H6

88. Catalysts and rate
Catalysts will speed up a reaction but only to a certain point.
Past a certain point adding more reactants won’t change the rate.
Zero Order

89. Catalysts and Rate Rate
Rate increases until the active sites of catalyst are filled.
Then rate is independent of concentration
Concentration of reactants