1. Chemical Kinetics
AP Chem Unit 12

2. Chemical Kinetics Reaction Rates Rate Laws: An Introduction
Determining the Form of the Rate Law
The Integrated Rate Law
Reaction Mechanisms
A Model for Chemical Kinetics
Catalysis

3. Characteristics of a Reaction
Identities of products and reactants
Stoichiometric quantities
Spontaneity
Refers to the inherent tendency for the process to occur. Does not imply anything about speed. Spontaneous does not mean fast. A reaction can be considered spontaneous but take years to occur.

4. Characteristics of a Reaction
The area of chemistry that concerns rates is called chemical kinetics.
One of the main goals of chemical kinetics is to understand the steps by which a reaction takes place.
This series of steps is called the reaction mechanism.
Understanding the mechanism allows us to find ways to change or improve the rate of a reaction.

5. Reaction Rates

6. Example
We start with a flask of NO2 gas at 300°C. But NO2 dioxide decomposes to nitric oxide (a source of air pollution) and O2.
2NO2(g)  2NO(g) + O2(g)
If we were to measure the concentrations of the three gases over time we would see a change in the amount of reactants and products over time.

7. Example
The reactant NO2 decreases with time and the concentrations of the products (NO and O2) increase with time.
2NO2(g)  2NO(g) + O2(g)

8. Reaction Rate
The speed, or rate, of a process is defined as the change in a given quantity (concentration in Molarity) over a specific period of time.
Reaction rate = [concentration of A at time (t2) – concentration of A at time (t1)]/ (t2 – t1)
(Final – initial)

9. Reaction Rate The square brackets indicate concentration in mol/l
In Kinetics, rate is always defined as positive. Since the concentration of reactants decreases over time, a negative sign is added to the equation.

10. Example Problem1
Looking at the NO2 table, calculate the average rate at which NO2 changes over the first 50 seconds of the reaction.
4.2 x 10-5 mol/ls

11. Reaction Rates
Reaction rates are often not constant through the course of a reaction. For example, average rates for NO2 are not constant but decreases with time.

12. Instantaneous Rate
The value of the rate at a particular time can be obtained by computing the slope of a line tangent to the curve at that point in time.

13. Reaction Rates
When considering rates of a reaction you must also take into account the coefficients in the balanced equation for the reaction.
The balanced reaction determines the relative rates of consumption of reactants and generation of products.

14. Reaction Rates 2NO2(g)  2NO(g) + O2(g)
In this example, both the reactant NO2 and the product NO have a coefficient of 2, so NO is produced at the same rate NO2 is consumed.

15. Reaction Rates 2NO2(g)  2NO(g) + O2(g)
The product O2 has a coefficient of 1, which means it is produced half as fast as NO.

16. Introduction to Rate Laws

17. Nature of Reactions
Chemical reactions are reversible. Often times as products are formed, they accumulate and react to form what was the reactant(s).
The previous example:
2NO2(g) <-> 2NO(g) + O2(g)
As NO and O2 accumulate, they can react to re-form NO2.

18. Nature of Reactions
Chemical reactions are reversible. Often times as products are formed, they accumulate and react to form what was the reactant(s).
Now the Δ[NO2] depends on the difference in the rates of the forward and reverse reactions.
In this unit we will not take into account the reverse reactions.

19. Rate Law
If the reverse reaction can be neglected, the reaction rate will depend only on the concentrations of the reactants.
A rate law shows how concentrations relate to the rate of a reaction.
A is a reactant. k is a proportionality constant and n is called the order of the reactant. Both are usually determined by experiment.

20. Rate Law
Most simple reactions, the rate orders are often positive integers, but they can be 0 or a fraction.
The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate.

21. Rate Law
Most simple reactions, the orders are often positive integers, but they can be 0 or a fraction.
The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation.
The rate law constant is dependent upon species in a reaction.

22. Types of Rate Laws There are two types of rate laws:
The differential rate law (often called simply the rate law) shows how the rate of a reaction depends on concentration.
The integrated rate law shows how the concentrations of species in the reaction depend on time.

23. Types of Rate Laws
The differential and integrated rate laws for a given reaction are related and knowing the rate law for a reaction is important because we can usually infer the individual steps involved in a reaction from the specific form of the rate law.

24. Determining the Form of the Rate Law

25. Determining Rate Law Form
The first step in understanding how a given chemical reaction occurs is to determine the form of the rate law.
Reaction rate form is described as orders.
Example: first order, second order, zero order
A first order reaction: concentration of the reactants are reduced by half, the overall rate of the reaction will also be half.
First Order: A direct relationship exists between concentration and rate.

26. Rate Form and Initial Rates
One common method for experimentally determining the form of the rate law for a reaction is the method of initial rates.
The initial rate of a reaction is the instantaneous rate determined just after the reaction begins. Before the initial concentrations of reactants have changed significantly.

27. Method of Initial Rates
NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l)
Rate 1 = 1.35 x10-7 mol/ls = k(.100M)n(.0050M)m

28. Method of Initial Rates
NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l)
n and m can be determined by dividing known rates.

29. Method of Initial Rates Example Problem 2
NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l)
Example: Find the form of the rate law for each reactant and the overall reaction order:
n and m are both 1 (unrelated), overall reaction order is 2 (n+m=2)

30. Method of Initial Rates
NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l)
Example: Find the rate constant for this reaction
k = 2.7 x 10-4 L/mols

31. Practice Problem 1
The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation:
BrO3-(aq) + 5Br-(aq) + 6H+(aq)  3Br2(l) + 3H2O(l)
Using the experimental data, determine the orders for all three reactants, the overall reaction order, and the value of the rate constant.

32. Practice Problem 1 BrO3-(aq) + 5Br-(aq) + 6H+(aq)  3Br2(l) + 3H2O(l)
n=1, m=1, p=2, overall = 4, k=8.0 L3/mol3s

33. The Integrated Rate Law

34. Integrated Rate Law
The rate laws we have considered so far express the rate as a function of the reactant concentrations. The integrated rate law expresses the reactant concentrations as a function of time.

35. First Order Reactions First order rate law:
The above rate law can be put into a different form using calculus (integration)
Integrated rate law for first order:
this equation is of the form y = mx + b
first order slope = -k, ln[A] vs t is a straight line.

36. First Order Reactions First order rate law:
Integrated rate law for first order can also be written:

37. Practice Problem 2
The decomposition of N2O5 in the gas phase was studied at constant temperature: 2N2O5(g)  4NO2(g) + O2(g) The following results were collected: Verify that this is first order for N2O5. Calculate k.
[N2O5] (mol/l)
Time (s)
0.1000
0.0707 50
0.0500
100
0.0250
200
0.0125
300
400

38. Practice Problem 2 2N2O5(g)  4NO2(g) + O2(g)
k= -slope= 6.93 x 10-3s-1
[N2O5] (mol/l)
ln[N2O5]
Time (s)
0.1000
0.0707 50
0.0500
100
0.0250
200
0.0125
300
400

39. Practice Problem 3 2N2O5(g)  4NO2(g) + O2(g)
[N2O5] (mol/l)
Time (s)
0.1000
0.0707 50
0.0500
100
0.0250
200
0.0125
300
400
Practice Problem 3
2N2O5(g)  4NO2(g) + O2(g)
Using the data from practice
problem 2, calculate [N2O5] at
150 s after the start of the reaction.
k= 6.93 x 10-3s-1
.0353 mol/L

40. Half-Life of a First Order Rxn
The time required for a reactant to reach half its original concentration is called the half life of a reactant.
t1/2

41. Half-Life of a First Order Rxn
Using the data from the previous example we can observe half-life.
[N2O5] (mol/l)
Time (s)
0.1000
0.0707 50
0.0500
100
0.0250
200
0.0125
300
400

42. Half-Life of a First Order Rxn

43. Half-Life of a First Order Rxn
The general equation for the half-life of a first-order reaction is
The half-life does not depend on concentration

44. Practice Problem 4
A certain first-order reaction has a half-life of 20.0 minutes.
Calculate the rate constant for this reaction and determine how much time is required for this reaction to be 75% complete?
k=3.47 x 10-2min-1 or 5.78 x 10-4s-1
t= 40 min.

45. Second-Order Rate Laws
integrated second order rate law:
A plot of 1/[A] vs t is a straight line, slope =k
half-life of a second order reaction:

46. Practice Problem 5
Butadiene reacts to form its dimer according to the equation: 2C4H6(g)C8H12(g) The following data was collected: Is this reaction first or second order? What is the rate constant? What is the half-life?
[C4H6] (mol/l)
Time (+/-1s)
1000
1800
2800
3600
4400
5200
6200

47. Practice Problem 5 2C4H6(g)C8H12(g) [C4H6] (mol/l) 1/[C4H6] ln [C4H6]
Time (+/-1s)
1000
1800
2800
3600
4400
5200
6200
2C4H6(g)C8H12(g)

48. Practice Problem 5 2C4H6(g)C8H12(g) second order, k =6.14x10-2L/mols
[C4H6] (mol/l)
1/[C4H6] ln
[C4H6]
Time (+/-1s)
1000
1800
2800
3600
4400
5200
6200
2C4H6(g)C8H12(g)
second order, k =6.14x10-2L/mols
half life= 1630s

49. Half life
It is important to recognize the difference between the half-life for a first-order reaction and the half-life for a second-order reaction.
First order half life depends only on k
half life remains constant throughout the rxn.
Second order half life depends on k and the initial concentration
each successive half-life doubles the preceding one.

50. Half life First order Second order [N2O5] (mol/l) Time (s) 0.1000
0.0707 50
0.0500
100
0.0250
200
0.0125
300
400
[C4H6] (mol/l)
Time (+/-1s)
1000
1800
2800
3600
4400
5200
6200

51. Zero-Order Rate Laws zero order rate law:
integrated zero order rate law:
A plot of [A] vs t is a straight line, slope = -k
half-life of a zero order reaction:

52. Zero-Order Rate Law
Zero-order reactions are most often encountered when a substance such as a metal surface or enzyme (catalyst) is required for the reaction to occur.

53. Integrated Rate Laws for more than one reactant
Most reactions with more than one reactant are simplified by varying the concentrations of the reactants.
the reactant in which the form of the rate law is being determined will have a low concentration.
the other reactants are given a much higher concentration so that the use of the other reactants are not limiting the first.

54. Example The reaction from the first example problem:
BrO3-(aq) + 5Br-(aq) + 6H+(aq)  3Br2(l) + 3H2O(l)
If the rate form for BrO3- is being determined, the concentration of BrO3- is set at 0.001M
The other two reactants have concentrations of 1.0M so that their initial and final concentrations are very similar and can be disregarded.

55. Rate Law Summary
Review p561

56. Reaction Mechanisms

57. Reaction Mechanisms
Most chemical reactions occur by a series of steps called the reaction mechanism.
to really understand a reaction, the mechanism is studied.
Kinetics includes the study of possible steps in a reaction

58. Example NO2(g) + CO(g)  NO(g) + CO2(g)
The rate law for this reaction= k[NO2]2
The reaction is actually more complicated than it appears
The balanced equation only gives us products, reactants and the stoichiometry, but does not give direct information about the mechanism.

59. Example NO2(g) + CO(g)  NO(g) + CO2(g)
For this reaction, the mechanism involves the following steps:
NO2(g) + NO2(g)  NO3(g) + NO(g) , k1
NO3(g) + CO(g)  NO2(g) + CO2(g) , k2
k1 and k2 are the rate constants for the individual reactions

60. Example NO2(g) + CO(g)  NO(g) + CO2(g)
For this reaction, the mechanism involves the following steps:
NO2(g) + NO2(g)  NO3(g) + NO(g) , k1
NO3(g) + CO(g)  NO2(g) + CO2(g) , k2
NO3(g) is an intermediate, a species that is neither a reactant nor a product but is formed and consumed during the reaction sequence.

61. Example NO2(g) + CO(g)  NO(g) + CO2(g)
For this reaction, the mechanism involves the following steps:
NO2(g) + NO2(g)  NO3(g) + NO(g) , k1
NO3(g) + CO(g)  NO2(g) + CO2(g) , k2
Each of these two reactions is called an elementary step.
Elementary steps have a rate law that is written from its molecularity.

62. Molecularity
Molecularity is defined as the number of species that must collide to produce the reaction.
A unimolecular step involves one molecule
A bimolecular reaction involves the collision of two species
A termolecular reaction involves the collision of three species
These steps are rare due to the probability of three molecules colliding simultaneously.

63. Examples of Elementary Steps

64. Reaction Mechanisms
A reaction mechanism is a series of elementary steps that must satisfy two requirements:
The sum of the elementary steps must give the overall balanced equation for the reaction.
The mechanism must agree with the experimentally determined rate law

65. Reaction Mechanism: Rule 1
Example: NO2(g) + CO(g)  NO(g) + CO2(g)
For this reaction, the mechanism involves the following steps:
NO2(g) + NO2(g)  NO3(g) + NO(g) , k1
NO3(g) + CO(g)  NO2(g) + CO2(g) , k2

66. Reaction Mechanism: Rule 2
In order for a mechanism to meet rule 2, the rate-determining step is considered.
Multistep reactions often have one step that is much slower than all the others.
The overall reaction cannot be faster than the slowest, or rate-determining step.

67. Reaction Mechanism: Rule 2
Example: NO2(g) + CO(g)  NO(g) + CO2(g)
If we assume the first step is rate-determining:
NO2(g) + NO2(g)  NO3(g) + NO(g) , k1
NO3(g) + CO(g)  NO2(g) + CO2(g) , k2
The overall rate is equal to the rate of production of NO3.

68. Reaction Mechanism: Rule 2
Example: NO2(g) + CO(g)  NO(g) + CO2(g)
If we assume the first step is rate-determining:
NO2(g) + NO2(g)  NO3(g) + NO(g) , k1
NO3(g) + CO(g)  NO2(g) + CO2(g) , k2
The rate law for the first step is written from its molecularity:
Overall Rate = k1[NO2]2

69. Practice Problem 6
The balanced equation for the reaction of the gases nitrogen dioxide and fluorine is: 2NO2(g) + F2(g)  2NO2F(g) The experimentally determined rate law is: Rate = k[NO2][F2] A suggested mechanism for this reaction is: NO2 + F2 NO2F + F (slow, k1) F + NO2 NO2F (fast, k2) Is this an acceptable mechanism?

70. Practice Problem 6 2NO2(g) + F2(g)  2NO2F(g)
The experimentally determined rate law is:
Rate = k[NO2][F2]
A suggested mechanism for this reaction is:
NO2 + F2 NO2F + F (slow, k1)
F + NO2 NO2F (fast, k2)
The mechanisms satisfy both requirements.

71. A Model for Chemical Kinetics

72. Chemical Kinetics Model
Kinetics depend on concentration, time and reaction mechanisms. There are other factors that affect reaction rates:
Temperature: chemical reactions speed up when the temperature is increased

73. Chemical Kinetics Model
Rate constants show an exponential increase with absolute temperature.

74. Chemical Kinetics Model
This observed increase in reaction rate with temperature is explained with the Collision model.
Molecules must collide to react.
higher concentration means more molecules to collide with each other.
KMT states that an increase in temperature raises molecular velocities and increases collision frequency.

75. Collision Model Only a small amount of collisions produce a reaction
Activation energy is the minimum amount of energy required for a collision to produce a reaction.
The kinetic energy of a molecule is changed into potential energy as the molecules are distorted during a collision
The arrangement of this distortion is called the activated complex or transition state.

76. Activation Energy

77. Activation Energy & Temperature

78. Collision Model
In addition to the activation energy required for a reaction to take place, another variable must be considered.
Molecular orientations must be correct for collisions to lead to reactions.
A correction factor is included in our activation energy formula to account for nonproductive molecular collisions

79. Collision Orientation

80. Collision Model Summary
Two requirements must be satisfied for reactants to collide successfully and rearrange to form products:
The collision must involve enough energy to produce the reaction; the collision energy must equal or exceed activation energy.
The relative orientation of the reactants must allow formation of new bonds.

81. Arrhenius Equation for the Collision Model
The first equation can be rewritten in a linear equation (y=mx+b). Slope = -Ea/R, x=1/T, y=ln(k), A is not concentration but the frequency factor
R = J/Kmol

82. Arrhenius Equation for the Collision Model
The most common procedure for finding Ea is to plot ln(k) vs 1/T and find the slope
It can also be found using only two temperatures:

83. Practice Problem 7
The reaction 2N2O5(g)  4NO2(g) + O2(g) was studied at several temperatures, and the following values of k were obtained:
Calculate the value of Ea for this reaction.
1.0x105 J/mol
k(s-1)
T (°C)
2.0x10-5 20
7.3x10-5 30
2.7x10-4 40
9.1x10-4 50
2.9x10-3 60

84. Catalysis

85. Catalyst
A catalyst is a substance that speeds up a reaction without being consumed in the reaction.
There are two different types of catalysts:
homogeneous – a catalyst that is present in the same phase as the reacting molecules
heterogeneous – exists in a different phase as the reacting molecules (usually a solid).

86. Catalyzed Pathway

87. Effect of Catalysis

88. Heterogeneous Catalysis
Heterogeneous catalysis most often involves gaseous reactants being adsorbed on the surface of a solid catalyst. This process is called adsorption.
Adsorption is the collection of one substance on the surface of another substance.

89. Homogeneous Catalysts
Biologically important reactions in our body are usually assisted by a catalyst/enzyme.
specific proteins are needed by the human body, the proteins in food must be broken into their constituent amino acids that are then used to construct new proteins in the body’s cells.
Without enzymes in human cells these reactions would be much to slow to be useful.

90. The End