1. Chapter 12 Chemical Kinetics
How often does Kinetics appear on the exam?
Multiple-choice 4-8% (2-5 Questions)
Free-response: Almost every year
Kinetics: The rate at which a chemical reaction occurs.
Rate of a reaction describes:
disappearance of a reactant
appearance of a product.
Doppler shift – if the object is moving toward you will observe a blue shift in color, if the object is moving away from you, you will observe a red shift (disappearance = red shift)

2. Chemical Kinetics
Reaction: defined by its reactants and products whose identity must be learned by experiment.
Spontaneity: Inherent tendency for a process to occur, does not imply speed.
Spontaneous reactions are reactions that will happen - but we can’t tell how fast.
Ex. - Diamond will spontaneously turn to graphite – eventually.
Reaction Mechanism- the steps by which a reaction takes place.

3. Factors Affecting Reaction Rates
Physical State of Reaction
Molecules must collide with each other
More homogeneous the faster reaction
Concentration of Reactants
Higher concentration means more molecules to collide
Temperature
Presence of Catalyst
Speed up reactions by changing mechanism of the reaction
Are not consumed during the reaction

4. Reaction Rates
Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

5. Reaction Rate Note for both conc. and time the equation is
Final Value – Initial Value
A is a reactant or product in a balanced equation.
Conc. = Is concentration measured in mol/L = Molarity
 = change in a given quantity. Change can be + or -.
Rate is always defined as a positive value

6. Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times.
The concentration of the products, butyl alcohol and hydrochloric acid are not considered.

7. Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
The average rate of the reaction over each interval is the change in concentration divided by the change in time:
Average rate =
[C4H9Cl] t

8. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
Note that the average rate decreases as the reaction proceeds.
This is because as the reaction goes forward, there are fewer collisions between reactant molecules.

9. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
A plot of [C4H9Cl] vs. time for this reaction yields a curve like this.
The slope of a line tangent to the curve at any point is the instantaneous rate at that time.
The best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction called Initial Reaction Rate

10. Reaction Rates and Stoichiometry
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1 (Hint: coefficients of balanced equation)
Therefore, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.
Rate =
-[C4H9Cl] t =
[C4H9OH]

11. Reaction Rates and Stoichiometry
What if the ratio is not 1:1?
2 HI(g)  H2(g) + I2(g)
In such a case,
Rate = − 1 2
[HI] t =
[I2]
Use the coefficients to build a stoichiometric relationship

12. Reaction Rate for a Generic Reaction
[A] of reactant or product is concentration in mol/L = M
t = time
Negative sign in front of [A] and [B] means concentrations are decreasing, reactants are disappearing
Initial rates will just be [reactants], no product yet

13. Reaction Rate For the reaction: 2NO2(g)  2NO(g) + O2(g)
Start the reaction and measure the concentration of the reactant and products several times
Collect in a data table.
You must have experimental data to do rate problems.

14. Plot the data on a graph: Molarity on y axis, time on x axis
2NO2(g)  2NO(g) + O2(g)
Conc. Of Reactants and Products as Function of Time
Concentration (mol/L)
Time (s)
NO2 NO O2
0.01 50
0.0079
0.0021
0.0011
100
0.0065
0.0035
0.0018
150
0.0055
0.0045
0.0023
200
0.0048
0.0052
0.0026
250
0.0043
0.0057
0.0029
300
0.0038
0.0062
0.0031
350
0.0034
0.0066
0.0033
400
0.0069
Plot the data on a graph: Molarity on y axis, time on x axis

15. Concentrations of Reactant & Products as a Function of Time
2NO2(g)  2NO(g) + O2(g) @300oC
Concentration mol/L 
Time (s)

16. Calculate Average Rate for 1st 50 s
In Kinetics it is customary to report rates as positive number so the expression is usually written with a negative.
How do you get to Carnegie Hall?
Calculate Average Rates for 50→100, 100→150, 150→200, and 200→250

17. Average Rates Data Table
∆[NO2]/∆t
Time Period (s)
4.2*10-5
0-50
2.8*10-5
50-100
2.0*10-5
1.4*10-5
1.0*10-5

18. Calculating Rates Average Rates are taken over long intervals
Instantaneous Rates are determined by finding the slope of a line tangent to the curve at any given point and use 2 points on that line

19. Average Slope Method Dt D[H2] Time Concentration
Like: How long does it take to drive to Atlanta in the middle of rush hour traffic? 2 hours (105 miles)
Now how long does it take to drive to Dalton in the middle of rush hour traffic? 1 hour (only 30 miles) Dt
Time

20. Instantaneous Slope Method.
Concentration
How fast am I going at this particular second in time?
Time

21. Summarizing Rate Information
We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product.
For example N2 + 3H2  2NH3

22. 2 NO2  2 NO + O2 Initial rate only depends on conc. of reactants.
Called RATE LAW EXPRESSION or DIFFERENTIAL RATE LAW or just RATE LAW
k is called the rate constant
Each k is for specific a temperature
n is the order of the reactant, usually a positive integer, NOT Coefficients
Both k and n must be determined experimentally

23. Types of Rate Laws
Differential Rate law - describes how rate depends on concentration = “Rate Law”
Integrated Rate Law - Describes how concentration depends on time.
For each type of Differential Rate law there is an Integrated Rate Law and vice versa.
Rate laws can help us better understand reaction mechanisms.
Rate Laws usually involve reactant concentrations

24. Determining the Order of the Rate Law
Key points
The exponent is the Order of the reaction with respect to each reactant
Order must be determined experimentally, can’t be obtained from the equation
Order can be integer (including 0) or fraction
Need to determine rate for several experiments then determine order

25. Determining the Order of the Rate Law
The first step is to determine the order of the rate law i.e. determine the exponent for each conc. in the rate law.
Must be determined from experimental data.
For this reaction:
2 N2O5 (aq)  4NO2 (aq) + O2(g)
The reverse reaction won’t play a role because O2 escapes from the reaction.
Data for reaction is recorded and we can use that data to calculate the instantaneous rate.

26. Table Conc./Time Data for Reaction 2N2O5  4NO2 + O2
[N2O5] (mol/L)
Time (s) 1
0.88
200
0.78
400
0.69
600
0.61
800
0.54
1000
0.48
1200
0.43
1400
0.38
1600
0.34
1800
0.3
2000

28. Evaluation of the Reaction
Find the Instantaneous Rate for 2 points on the curve: 0.90M and 0.45M by taking the slope of the tangents to the curve at these points.
Record in a data table and analyze
[N2O5]
Instantaneous Rate (mol/L*s)
0.90M
5.4 X 10-4
0.45M
2.7 X 10-4

29. Instantaneous Rate (mol/L*s)
[N2O5]
Instantaneous Rate (mol/L*s)
0.90M
5.4 X 10-4
0.45M
2.7 X 10-4
Notice: when the concentration is halved, the rate is halved. This means the rate is dependent on the concentration.
The differential rate law for this equation is:
What is the value of some number raised to the 1st power? Answer: The number times the constant which is the rate.
This is a 1st order reaction, hence power of 1 in rate law
Order is independent of the coefficient

30. The Method of Initial Rates
This method requires that a reaction be run several times.
The initial concentrations of the reactants are varied.
The reaction rate is measured as soon as the reactants are mixed (Initially)
Eliminates the effect of the reverse reaction.
Compare the rates among experiments to see how the rate depends on the concentration

31. The Method of Initial Rates
For the reaction
NH4+(aq) + NO2-(aq)  N2(g) + 2 H2O(l)
The general form of the Rate Law is:
We use experimental data to determine the values of n and m
We determine the values of n and m by observing how the initial rate depends on initial concentration

32. Experimental Data
Experiment
Initial [NH4+]
Intial [NO2-]
Initial Rate 1
0.100M
0.0050M
1.35*10-7 2
0.010M
2.70*10-7 3
0.200M
5.40*10-7

33. The Method of Initial Rates
Use the rates on the previous slides to determine m and n
The value for m and n are both = 1
OVERALL REACTION ORDER is SUM of n + m = 2
Use an initial rate & concentration values for m and n
to FIND K
k = 2.7 x 10-4 L/mol*s
Carnegie Hall Time: #26, 28, & 29

34. This is Sample Exercise 12.1 on p.537 Study It!
Determining a Rate Law
For the reaction
The general form of the rate law is
This is Sample Exercise 12.1 on p.537 Study It!

35. Integrated Rate Law
Expresses the reaction concentration as a function of time.
Form of the equation depends on the order of the differential rate law.
This is Differential Rate Law generic form.
We will only work order numbers when n=0, 1, and 2 but be aware there are fractional orders.

36. First Order Rate Law For the reaction 2N2O5  4NO2 + O2
The Differential Rate Law is k[N2O5]1
If concentration doubles then rate doubles.
If we integrate this equation with respect to time we get the Integrated Rate Law
ln is the natural log, 0 indicates the initial concentration.
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37. Integrated Rate Law
Generic form of the Integrated First Order Rate Law
In the form y = mx + b
y = ln[A] m = -k (slope)
x = t b = ln[A]0 (y intercept)
If you plot a graph of ln[A] vs time AND it is a straight line, you know the reaction is 1st order.

38. Example 2N2O5(g)4NO2(g) + O2(g) (temp constant)
time (s)
ln[N2O5]
0.1000
-2.303
0.0707 50
-2.649
0.0500
100
-2.996
0.0250
200
-3.689
0.0125
300
-4.382
400
-5.075
Using this data, verify that the rate law is 1st order & calc. k

39. To verify, graph ln[N2O5] vs time
Time (s)
Get straight line proves it is 1st order
To find k, the slope of the line = -k, use 2 points that are ON the line and calc. slope which is the value for k

40. Find k

41. Half Life of First Order
Half Life is defined as the time required to reach half the original concentration.
Symbol is t1/2
Use the Integrated Rate Law
Simplifies to:
Steps are illustrated on p.542

42. Half Life
t1/2 = 0.693/k
The time to reach half the original concentration does not depend on the starting concentration.
An easy way to find k
A certain first-order reaction has a half-life of 20 minutes.
Calculate the rate constant for this reaction.

43. Second Order Differential Rate Law Integrated Rate Law y= 1/[A] m = k
x= t b = 1/[A]0
A plot of 1/[A] vs. time is a straight line
Knowing k and [A]0 you can calculate [A] at any time t

44. The expression for half-life of a 2nd Order Reaction
Second Order Half Life
Rate Law for 2nd order
When t = t1/2 Your Integrated Second Order Rate Law becomes
Intermediate steps are illustrated on p.543
The expression for half-life of a 2nd Order Reaction

45. Zero Order Rate Law Differential Rate Law = k[A]0 = k, [A]0 = 1
Rate does not change with concentration.
Integrated Rate Law [A] = -kt + [A]0
A plot of [A] versus time gives a straight line of slope -k
Half Life defined as [A] = [A]0 /2 when t = t1/2
t1/2 = [A]0 /2k

46. Reactions with Multiple Reactants p.546
BrO Br- + 6H+  3Br2 + 3 H2O
From experimental evidence, found the rate law to be:
Rate = k[BrO3-][Br-][H+]2
Set up experiment so two reactants are in excess.
[BrO3-]0= 1.0 x 10-3 M
[Br-]0 = 1.0 M (excess)
[H+]0 = 1.0 M (excess)

47. Rate = k[BrO3-][Br-][H+]2
As the reaction proceeds
[BrO3-] changes
[Br-] and [H+] does not change
Therefore [Br-] = [Br-]0 and [H+] = [H+]0
Rate law can be rewritten & rearranged
k’ =k[Br-]0[H+]02
Rate Law = k’[BrO3-]

48. PSEUDO FIRST ORDER RATE LAW
Rate Law = k’[BrO3-]
Because this rate law was made by simplifying a more complicated one it is called:
PSEUDO FIRST ORDER RATE LAW
Solve for k
Know [Br-] and [H+] so calc. k

49. Summary of Rate Laws Rate = k Rate = k[A]1 Rate = k[A]2
Summary of the Kinetics for Reactions of the Type aA → Products that are Zero, First, or Second Order in [A]
Order
Zero
First
Second
Diff. Rate law:
Rate = k
Rate = k[A]1
Rate = k[A]2
Integrated rate law:
[A] = –kt + [A]0
ln[A] = –kt + ln[A]0
1/[A]=kt + 1/[A]0
Plot needed to give a straight line:
[A] versus t
ln[A] versus t
1/[A] versus t
Relationship of rate constant to the slope of straight line:
Slope = –k
Slope = k
Half-life:
t½ =[A]0 /2k
t½ =ln2/k
t½ =1/k[A]0

50. Reaction Mechanisms
Defined as a series of steps that actually occur in a chemical reaction.
Kinetics can tell us something about the mechanism
A balanced equation does not tell us how the reactants become products.

51. Slowest reaction in mechanism is RATE DETERMINING STEP
Reaction Mechanisms p.551
2NO2(g) + F2(g)  2NO2F(g)
Rate Law = k[NO2][F2]
The suggested mechanism is
NO2 + F2  NO2F + F (slow)
F + NO2  NO2F (fast)
2NO2 + F2  2NO2F
Requirement: sum of steps = balanced equation
Requirement: mechanism must agree with rate law
Slowest reaction in mechanism is RATE DETERMINING STEP

52. Reaction Mechanisms
Each of the two reactions is called an elementary step .
2nd step is called an intermediate It is formed then consumed in the reaction
The rate for a reaction can be written from its molecularity = number of species that must collide in order to form a product

53. Rate is FIRST ORDER. Rate is SECOND ORDER Rate is THIRD ORDER
UNIMOLECULAR step involves one molecule
Rate is FIRST ORDER.
BIMOLECULAR requires two molecules
Rate is SECOND ORDER
TERMOLECULAR requires three molecules
Rate is THIRD ORDER
Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.

54. How do you get to Carneige Hall? p.571 # 51 &52
Elementary Steps
Molecularity
Rate Law
AProducts
Unimolecular
Rate=k[A]
A+AProducts
Bimolecular
Rate=k[A]2
A+BProducts
Rate=k[A][B]
A+A+BProducts
Termolecular
Rate=k[A]2[B]
A+B+CProducts
Rate=k[A][B][C]
How do you get to Carneige Hall? p.571 # 51 &52

55. Collision theory Molecules must collide to react.
Concentration affects rates because collisions are more likely.
Must collide hard enough.
Must collide with the correct orientation.
Temperature and rate are related.
Only a small number of collisions produce reactions.

56. Potential Energy 
Reactants
Products
Reaction Progress 

57. Activation Energy Ea Reaction Progress  Potential Energy  Reactants
Products
Reaction Progress 

58. Reaction Progress  Activated complex Potential Energy  Reactants
Products
Reaction Progress 

59. Potential Energy 
Reactants DE
Products
Reaction Progress 

60. Reaction Progress  Br---NO Transition State Potential Energy  2BrNO
2NO + Br 2
Reaction Progress 

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

62. Arrhenius 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.

63. Arrhenius Number of collisions with the required energy = ze-Ea/RT
z = total collisions
e is Euler’s number (opposite of ln)
Ea = activation energy
R = ideal gas constant ( J x K-1 x mol-1)
T is temperature in Kelvin

64. Arrhenius Equation z = number of collisions
p = fraction of effective collisions
A = frequency factor = zp
Take natural log of each side

65. Mechanisms and rates
There is an activation energy 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.

66. This reaction takes place in three steps

67. Ea
First step is fast Low activation energy

68. High activation energyEa
Second step is slow
High activation energy

69. Ea
Third step is fast
Low activation energy

70. Second step is rate determining

71. Intermediates are present

72. Activated Complexes or Transition States

73. Catalysts 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.

74. 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

75. Heterogenous Catalysts
Hydrogen bonds to surface of metal.
Break H-H bonds H H H H H
Pt surface

76. Heterogenous Catalysts
Pt surface

77. Heterogenous Catalysts
The double bond breaks and bonds to the catalyst. H H H C C H H H H H
Pt surface

78. Heterogenous Catalysts
The hydrogen atoms bond with the carbon H H H C C H H H H H
Pt surface

79. Heterogenous Catalysts
Pt surface

80. Homogenous Catalysts
Chlorofluorocarbons catalyze the decomposition of ozone.
Enzymes regulating the body processes. (Protein catalysts)

81. 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

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

84. Concentration
Time

85. Concentration
k =
D[A]/Dt
D[A] Dt
Time

86. How to get rid of intermediates
Use the reactions that form them
If the reactions are fast and irreversible - the concentration of the intermediate is based on stoichiometry.
If it is formed by a reversible reaction set the rates equal to each other.

87. Formed in reversible reactions
2 NO + O2  NO2
Mechanism
2 NO  N2O2 (fast)
N2O2 + O2  2 NO2 (slow)
rate = k2[N2O2][O2]
k1[NO]2 = k-1[N2O2]
rate = k2 (k1/ k-1)[NO]2[O2]=k[NO]2[O2]

88. Formed in fast reactions
2 IBr  I2+ Br2
Mechanism
IBr  I + Br (fast)
IBr + Br  I + Br2 (slow)
I + I  I2 (fast)
Rate = k[IBr][Br] but [Br]= [IBr]
Rate = k[IBr][IBr] = k[IBr]2

89. Problems
Observed rate is less than the number of collisions that have the minimum energy.
Due to Molecular orientation
written into equation as p the steric factor.

90. 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

91. 2 NO2  NO + O2
If you define rate in terms of NO2 consumption your rate law is
IF you chose O2 to define rate the rate law is:
Because there are 2 NO2 for every 1 O2
Rate [NO2] = 2 x Rate'
So k[NO2]n = 2 x k'[NO2]n
So k = 2 x k'
Value of k depends on how the rate is defined
‘ notice mark = prime