1. Chapter 14 Chemical Kinetics
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 14 Chemical Kinetics
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

2. Kinetics
Is the branch of chemistry concerned with the rate of chemical reactions and the mechanism by which occur
Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).

3. Collision Theory
In order for two particles to react chemically they must collide. Not only they must collide, but it must be an effective collision. That is, they must have the correct amount of energy and collide with the proper orientation in space.
Any factor which increases the likehood that they will collide will increase the rate of the chemical reaction.

4. Factors That Affect Reaction Rates (bonds must break)
Surface area/contact area (opportunity for collisions).
Concentration (increase frequency)
Temperature (increase frequency)
Catalyst ( Effective collisions)
Nature of reactants (effective collisions)

5. Factors That Affect Reaction Rates
Physical State of the Reactants
In order to react, molecules must come in contact with each other.
The more homogeneous the mixture of reactants, the faster the molecules can react.
For solid reactants increasing the surface area increases the rate of reaction.

6. Factors That Affect Reaction Rates
Temperature
At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.

7. Factors That Affect Reaction Rates
Concentration of Reactants
As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.

8. Factors That Affect Reaction Rates
Presence of a Catalyst
Catalysts speed up reactions by changing the mechanism of the reaction.
Catalysts are not consumed during the course of the reaction.

9. January 10 Reaction rates
Determining the instantaneous rate from a graph.
Stoichiometry and reaction rates
Homework: p 617 q 1to3
and up to what we get to for 11 to 15 , 17 to 31 odd

10. Reaction Rates
Speed of a reaction is measured by the change in concentration with time.
For a reaction A  B
Suppose A reacts to form B. Let us begin with 1.00 mol A.

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

12. At t = 0 (time zero) there is 1
At t = 0 (time zero) there is 1.00 mol A (100 red spheres) and no B present.
At t = 20 min, there is 0.54 mol A and 0.46 mol B.
At t = 40 min, there is 0.30 mol A and 0.70 mol B.
Calculating,

13. For the reaction A  B there are two ways of measuring rate:
the speed at which the products appear (i.e. change in moles of B per unit time), or
the speed at which the reactants disappear (i.e. the change in moles of A per unit time).

14. By convention rates are always positive quantities
By convention rates are always positive quantities. Because A is decreasing with time, we use a negative sign to convert the negative change into a positive rate.
Because one molecule of A is consumed for every molecule of B that forms the average rate of disappearance of A equals the rate of appearance of B

15. SAMPLE EXERCISE 14.1 Calculating an Average Rate of Reaction
From the data given in the caption of Figure 14.3, calculate the average rate at which A disappears over the
time interval from 20 s to 40 s.
Solution
Analyze: We are given the concentration of A at 20 s (0.54 M) and at 40 s (0.30 M) and asked to calculate the average rate of reaction over this time interval.
Plan: The average rate is given by the change in concentration, [A], divided by the corresponding change in time, t.Because A is a reactant, a minus sign is used in the calculation to make the rate a positive quantity.
Solve:
PRACTICE EXERCISE
For the reaction pictured in Figure 14.3, calculate the average rate of appearance of B over the time interval from 0 to 40 s. (The necessary data are given in the figure caption.)
Answer: 1.8  10 –2 M/s

16. Change of Rate with Time
Most useful units for rates are to look at molarity. Since volume is constant, molarity and moles are directly proportional.
Consider:
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)

17. Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
In this reaction, the concentration of butyl chloride or chlorobutane, C4H9Cl, was measured at various times.

18. 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

19. 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.

20. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
A plot of concentration 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.

21. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
All reactions slow down over time.
Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning.

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

23. C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
We can calculate the average rate in terms of the disappearance of C4H9Cl.
The units for average rate are mol L-1 s-1 or M/s.
The average rate decreases with time.
We plot [C4H9Cl] versus time.
The rate at any instant in time (instantaneous rate) is the slope of the tangent to the curve.
Instantaneous rate is different from average rate.
We usually call the instantaneous rate the rate.

24. Using the figure, calculate the instantaneous rate of disappearance of C4H9Cl at t = 0 (the initial rate).
The straight line falls from [C4H9Cl] = M to M in the time change from 0 s to 200 s, as indicated by the tan triangle shown in the figure. Thus, the initial rate is

25. Reaction Rates and Stoichiometry
What if the ratio is not 1:1?
2 HI(g)  H2(g) + I2(g)
Therefore,
Rate = − 1 2
[HI] t =
[I2]

26. Reaction Rates and Stoichiometry
To generalize, then, for the reaction
aA + bB
cC + dD
Rate = − 1 a
[A] t
= − b
[B] = c
[C] d
[D]

27. Reaction Rate
Change in concentration per unit time
For this reaction
N2 + 3H NH3

28. As the reaction progresses the concentration H2 goes down
Time

29. As the reaction progresses the concentration N2 goes down 1/3 as fast
Time

30. As the reaction progresses the concentration NH3 goes up.
Time

31. Example
N2 + 3H NH3
At a certain temperature, the rate of this reaction is mol N2 L-1 s-1. What is the rate in mol H2 L-1 s-1? What is the rate in mol NH3 L-1 s-1?

32. 0.39 mol H2 L-1 s-1
0.26 mol NH3 L-1 s-1

33. SAMPLE EXERCISE 14. 3 Relating Rates at Which Products Appear and
SAMPLE EXERCISE 14.3 Relating Rates at Which Products Appear and Reactants Disappear
(a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction
(b) If the rate at which O2 appears, [O2] /t, is 6.0  10–5 M/s at a particular instant, at what rate is O3
disappearing at this same time, –[O3] /t?
Solution
Analyze: We are given a balanced chemical equation and asked to relate the rate of appearance of the product to the rate of disappearance of the reactant. .
Solve: (a) Using the coefficients in the balanced equation
(b) Solving the equation from part (a) for the rate at which O3 disappears, –[O3] /t we have:
Check: We can directly apply a stoichiometric factor to convert the O2 formation rate to the rate at which
the O3 disappears:

34. SAMPLE EXERCISE 14.3 continued
PRACTICE EXERCISE
The decomposition of N2O5 proceeds according to the following equation:
If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2  10–7 M/s, what is the rate of appearance of (a) NO2, (b) O2?
Answers: (a) 8.4  10 –7 M/s, (b) 2.1  10 –7 M/s

35. Concentration and Rate
One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration.

36. Concentration and Rate
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
Comparing Experiments 1 and 2, when [NH4+] doubles, the initial rate doubles.

37. Concentration and Rate
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
Likewise, comparing Experiments 5 and 6, when [NO2−] doubles, the initial rate doubles.

38. This method requires that a reaction be run several times.
The initial concentrations of the reactants are varied.
The goal is to vary only one concentration at a time
The reaction rate is measured just after the reactants are mixed.

39. Concentration and Rate
This means
Rate  [NH4+]
Rate  [NO2−]
Rate  [NH+] [NO2−] or
Rate = k [NH4+] [NO2−]
This equation is called the rate law, and k is the rate constant.

40. NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
For the reaction
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
we note
as [NH4+] doubles with [NO2-] constant the rate doubles,
as [NO2-] doubles with [NH4+] constant, the rate doubles,
We conclude rate  [NH4+][NO2-].
Rate law:
The constant k is the rate constant.

41. Rate Laws
A rate law shows the relationship between the reaction rate and the concentrations of reactants.
The exponents tell the order of the reaction with respect to each reactant.
This reaction is
First-order in [NH4+]
First-order in [NO2−]

42. Rate Laws
The overall reaction order can be found by adding the exponents on the reactants in the rate law.
This reaction is second-order overall.

43. The exponents in a rate law indicate how the rate is affected by the concentration of each reactant.
For most reactions the orders are 0, 1 or 2, occasionally reactions can have a reaction order fractional or even negative.

44. JANUARY 11 DETERMINING RATE LAWS WITH THE INITIAL CONCENTRATION METHOD
CONCENTRATION AND TIME
FIRST ORDER REACTION
INTEGRATED RATE LAW
GRAPH METHOD
UNITS FOR K

45. 2 NO NO + O2
You will find that the rate will only depend on the concentrations of the reactants.
Rate = k[NO2]n
This is called a rate law expression.
k is called the rate constant.
n is the order of the reactant -usually a positive integer.

46. Exponents in the Rate Law
For a general reaction with rate law
we say the reaction is mth order in reactant 1 and nth order in reactant 2.
The overall order of reaction is m + n + ….
A reaction can be zeroth order if m, n, … are zero.
This means that the rate does not depend on the concentration of (a) reactant(s)
Note: the values of the exponents (orders) have to be determined experimentally. They are not simply related to stoichiometry.

47. Using Initial Rates to Determine Rate Laws
A reaction is zero order in a reactant if the change in concentration of that reactant produces no effect.
A reaction is first order if doubling the concentration causes the rate to double.
A reacting is nth order if doubling the concentration causes an 2n increase in rate.
Note that the rate constant does not depend on concentration.

48. Example For the reaction BrO3- + 5 Br- + 6H+ 3Br2 + 3 H2O
The general form of the Rate Law is
Rate = k[BrO3-]m[Br-]n[H+]p
We use experimental data to determine the values of m, n,and p

49. Now we have to see how the rate changes with concentration
Initial concentrations (M)
[BrO3-]
[Br-]
[H+]
Rate (M/s)
x 10-4
x 10-3
x 10-3
x 10-3
Now we have to see how the rate changes with concentration

50. Determine the Rate Law.
Calculate the value of k, and determine its units.
Calculate the initial rate of the reaction if the initial concentrations are:
[BrO3-] = 0.20 M [Br-] = 0.30 M
[H+] = 0.10 M

51. NH4+ + NO2- N2 + 2 H2O Exp [NH4+] [NO2-] Initial Rate(M/s)
M M x 10-7
M M x 10-7
M M x 10-7
Determine the rate law.
Calculate k, and determine its units.

52. SAMPLE EXERCISE 14.4 Relating a Rate Law to the Effect of Concentration on Rate
Consider a reaction for which rate = k[A][B]2. Each of the following boxes represents a
reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of
increasing rate of reaction.
Solution
Analyze: We are given three boxes containing different numbers of spheres representing mixtures containing different reactant concentrations. We are asked to use the given rate law and the compositions of the boxes to rank the mixtures in order of increasing reaction rates.
Plan: Because all three boxes have the same volume, we can put the number of spheres of each kind into the rate law and calculate the rate for each box.
Solve: Box 1 contains 5 red spheres and 5 purple spheres, giving the following rate:
Box 2 contains 7 red spheres and 3 purple spheres:

53. SAMPLE EXERCISE 14.4 continued
Box 3 contains 3 red spheres and 7 purple spheres:
The slowest rate is 63k (box 2), and the highest is 147k (box 3). Thus, the rates vary in the order 2 < 1 < 3.
Check: Each box contains 10 spheres. The rate law indicates that in this case [B] has a greater influence on rate than [A] because B has a higher reaction order. Hence, the mixture with the highest concentration of B (most purple spheres) should react fastest. This analysis confirms the order 2 < 1 < 3.
PRACTICE EXERCISE
Assuming that rate = k[A][B], rank the mixtures represented in this Sample Exercise in order of increasing rate.
Answer: 2 = 3 < 1

54. Types of Rate Laws
Differential Rate law - describes how rate depends on reactant concentration (Referred to as simply “rate law”).
Rate= -D [A]/D t=k [A]

55. Integrated Rate Law - Describes how concentration varies with 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

56. Integrated Rate Laws
Using calculus to integrate the rate law for a first-order process gives us ln
[A]t
[A]0
= −kt
Where
[A]0 is the initial concentration of A.
[A]t is the concentration of A at some time, t, during the course of the reaction.

57. Integrated Rate Laws Manipulating this equation produces… ln [A]t [A]0
= −kt
ln [A]t − ln [A]0 = − kt
ln [A]t = − kt + ln [A]0
…which is in the form
y = mx + b

58. Unit for k for a 1st order reaction
k = t-1

59. First Order Reactions
Goal: convert rate law into a convenient equation to give concentrations as a function of time.
For a first order reaction, the rate doubles as the concentration of a reactant doubles.
Rate Law
Integrated Rate Law

60. PRACTICE EXERCISE PRACTICE EXERCISE
The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of
6.8  10–4s–1:
If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s?
Note that pressure of a gas is proportional to its concentration so it can be used as the molarity
PRACTICE EXERCISE
The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of
6.8  10–4s–1:
If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s?
Note that pressure of a gas is proportional to its concentration so it can be used as the molarity
Answer: 51 torr

61. SAMPLE EXERCISE 14.7 Using the Integrated First-Order Rate Law
The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr–1
at 12°C. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0  10–7 g/cm3. Assume that the average temperature of the lake is 12°C. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the concentration of the insecticide to drop to
3.0  10–7 g/cm3?
Solution : replace the given values into the integrated first order rate law equation
ln [A]t = -kt + ln [A]0
k = 1.45 yr–1, t = 1.00 yr [insecticide]0 = 5.0  10–7 g/cm3
and solve for 1n[insecticide]
(b) We have k = 1.45yr–1, [insecticide]0 = 5.0  10–7 g/cm3, and [insecticide]t
= 3.0  10–7 g/cm3, and so we can solve the equation for t.
Solve: (a) Substituting the known quantities into the equation , we have
We use the ln function on a calculator to evaluate the second term on the right, giving
To obtain [insecticide]t = 1 yr, we use the inverse natural logarithm, or ex, function on the calculator:
Note that the concentration units for [A]t and [A]0 must be the same.

62. First order
A certain first order decomposition reaction reaches 65% completion in 18.9 seconds. Find the rate constant for this reaction.
.055 s-1

63. First-Order Processes
ln [A]t = -kt + ln [A]0
Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k.

64. Graphical method to determine rate order
How do we find out if the reaction is first rate from the experimental data?
Plot the ln [reactant] vs time.
If the plot is a straight line then the reaction is first order.

65. First-Order Processes
Consider the process in which methyl isonitrile is converted to acetonitrile.
CH3NC
CH3CN

66. First-Order Processes
CH3NC
CH3CN
This data was collected for this reaction at 198.9°C.

67. First-Order Processes
When ln P is plotted as a function of time, a straight line results.
Therefore,
The process is first-order.
k is the negative slope: 5.1  10-5 s−1.

68. Second-Order Processes Rate = k[A]2
Integrating the rate law for a process that is second-order in reactant A, we get 1
[A]t
= kt +
[A]0
also in the form
y = mx + b

69. INTEGRATED RATE LAW FOR A 2nd ORDER REACTION1
[A]t
= kt +
[A]0
If a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k.
Unit for k = M-1 t-1

70. Second-Order Processes
The decomposition of NO2 at 300°C is described by the equation
NO2 (g)
NO (g) + 1/2 O2 (g)
and yields data comparable to this:
Time (s)
[NO2], M
0.0
50.0
100.0
200.0
300.0

71. Second-Order Processes
Graphing ln [NO2] vs. t yields:
The plot is not a straight line, so the process is not first-order in [A].
Time (s)
[NO2], M
ln [NO2]
0.0
−4.610
50.0
−4.845
100.0
−5.038
200.0
−5.337
300.0
−5.573

72. Second-Order Processes
Graphing ln 1/[NO2] vs. t, however, gives this plot.
Because this is a straight line, the process is second-order in [A].
Time (s)
[NO2], M
1/[NO2]
0.0
100
50.0
127
100.0
154
200.0
208
300.0
263

73. Half-Life
Half-life is defined as the time required for one-half of a reactant to react.
Because [A] at t1/2 is one-half of the original [A],
[A]t = 0.5 [A]0.

74. Half-Life
Half-life is the time taken for the concentration of a reactant to drop to half its original value.
For a first order process, half life, t½ is the time taken for [A]0 to reach ½[A]0.

75. Half-Life For a first-order process, this becomes 0.5 [A]0 [A]0 ln
= −kt1/2
ln 0.5 = −kt1/2
−0.693 = −kt1/2
= t1/2
0.693 k
NOTE: For a first-order process, the half-life does not depend on [A]0.

76. Half life
The half life of a reaction is useful to describe how fast it occurs.
For a first order reaction (like nuclear decay!) it does not depend on the initial concentration of the reactants.
HALF LIFE IS CONSTANT FOR A FIRST ORDER REACTION

77. Half-Life For a second-order process 1 0.5 [A]0 = kt1/2 + [A]0 2 [A]0
2 − 1
[A]0
= kt1/2 1 =
= t1/2 1
k[A]0

78. Second Order Reactions
The half life depends on the initial concentration.
As the reactant concentration decreases the half life increases.

79. Zero Order Rate Law Rate = k[A]0 = k
Rate does not change with concentration.
Integrated [A] = -kt + [A]0
When [A] = [A]0 /2 t = t1/2
t1/2 = [A]0 /2k

80. Zero Order Rate Law
Most often when reaction happens on a surface because the surface area stays constant.
Also applies to enzyme chemistry.

81. Summary of Rate Laws Zero Order 1st Order 2nd Order Diff. Rate Law K
k[A]
K[A]2
Int. Rate Law
[A] = -kt + [A]o
ln [A] = -kt + ln [A]o
1/[A] = kt + 1/[A]o
Straight Line Plot
[A] vs. t
m = -k
ln [A] vs. t
1/[A] vs. t
m = k
Half-Life
[A]o/2k
(ln 2)/k
1/k[A]o

82. HW 33 TO 43 ODD Go to www.sciencegeek.net
Follow the links to chapter 12
Take quiz

83. Temperature and Rate
Generally, as temperature increases, so does the reaction rate.
This is because k is temperature dependent.

84. Most reactions speed up as temperature increases. (E. g
Most reactions speed up as temperature increases. (E.g. food spoils when not refrigerated.)
When two light sticks are placed in water: one at room temperature and one in ice, the one at room temperature is brighter than the one in ice.
The chemical reaction responsible for chemiluminescence is dependent on temperature: the higher the temperature, the faster the reaction and the brighter the light.

85. The Collision Model
As temperature increases, the rate increases (exponentially).

86. Consider the first order reaction CH3NC  CH3CN.
Since the rate law has no temperature term in it, the rate constant must depend on temperature.
Consider the first order reaction
CH3NC  CH3CN.
As temperature increases from 190 C to 250 C the rate constant increases from 2.52  10-5 s-1
to 3.16  10-3 s-1.
The temperature effect is quite dramatic. Why?
Observations: rates of reactions are affected by concentration and temperature.

87. The greater the number of collisions the faster the rate.
Goal: develop a model that explains why rates of reactions increase as concentration and temperature increases.
The collision model: in order for molecules to react they must collide.
The greater the number of collisions the faster the rate.
The more molecules present, the greater the probability of collision and the faster the rate.

88. The higher the temperature, the more energy available to the molecules and the faster the rate.
Complication: not all collisions lead to products. In fact, only a small fraction of collisions lead to product.
Effective Collision – A collision between molecules that leads to products (reaction occurs)

89. The Collision Model
Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

90. Consider:
2 NOBr  2 NO + Br2
There are several possible ways that NOBr molecules can collide; one is effective and the others are not.

91. 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

92. The Orientation Factor
Cl + ClNO Cl2 + NO

93. Activation Energy
Arrhenius: molecules must posses a minimum amount of energy to react. Why?
In order to form products, bonds must be broken in the reactants.
Bond breakage requires energy.
Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.

94. Activation Energy
There is a minimum amount of energy required for reaction: the activation energy, Ea.
Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

95. Activation Energy
The lower the Ae the faster the reaction.
Is the energy needed to form the activated complex.
Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier. Is a very unstable arrangement!

96. Reaction Coordinate Diagrams
It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile.

97. Reaction Coordinate Diagrams
It shows the energy of the reactants and products (and, therefore, E).
The high point on the diagram is the transition state.
The species present at the transition state is called the activated complex.
The energy gap between the reactants and the activated complex is the activation energy barrier.

98. Maxwell–Boltzmann Distributions
Temperature is defined as a measure of the average kinetic energy of the molecules in a sample.
At any temperature there is a wide distribution of kinetic energies.

99. Maxwell–Boltzmann Distributions
As the temperature increases, the curve flattens and broadens.
Thus at higher temperatures, a larger population of molecules has higher energy.

100. Maxwell–Boltzmann Distributions
If the dotted line represents the activation energy, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier.
As a result, the reaction rate increases.

101. Maxwell–Boltzmann Distributions
This fraction of molecules can be found through the expression
where R is the gas constant and T is the Kelvin temperature.
f = e−Ea/RT

102. The Arrhenius Equation
Arrhenius discovered most reaction-rate data obeyed the Arrhenius equation:
k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K.
A is called the frequency factor.
A is a measure of the probability of a favorable collision.
Both A and Ea are specific to a given reaction.

103. Arrhenius Equation
Taking the natural logarithm of both sides, the equation becomes
ln k = -Ea ( ) + ln A 1 RT
y = mx + b
Therefore, if k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. 1/T.

104. Determining the Activation Energy
If we have a lot of data, we can determine Ea and A graphically by rearranging the Arrhenius equation:
From the above equation, a plot of ln k versus 1/T will have slope of –Ea/R and intercept of ln A.

105. Temperature and Rate

106. If we do not have a lot of data, then we recognize

107. Arrhenius Equation Problem
The rate constant for the formation of hydrogen iodide from the elements
H2 (g) + I2 (g) 2HI (g)
is 2.7 x 10-4 L/(mol.s) at 600 K and 3.5 x 10-3 L/(mol.s) at 650 K. Find the activation energy Ea obtained in a, and solve for k2 when T is 700K

108. SAMPLE EXERCISE 14.10 Relating Energy Profiles to Activation Energies and Speeds of Reaction
Consider a series of reactions having the following energy profiles:
Assuming that all three reactions have nearly the same frequency factors, rank the reactions from slowest to fastest.
Solution The lower the activation energy, the faster the reaction. The value of  does not affect the rate.
Hence the order is (2) < (3) < (1).
PRACTICE EXERCISE
Imagine that these reactions are reversed. Rank these reverse reactions from slowest to fastest.
Answer:  (2) < (1) < (3) because Ea values are 40, 25, and 15 kJ/mol, respectively

109. Reaction Mechanisms
The balanced chemical equation provides information about the beginning and end of reaction.
The reaction mechanism gives the path of the reaction.
Mechanisms provide a very detailed picture of which bonds are broken and formed during the course of a reaction.
Elementary Steps
Elementary step: any process that occurs in a single step.

110. Reaction Mechanisms
The molecularity of a process tells how many molecules are involved in the process.
Molecularity is the number of species that must collide to produce the reaction indicated by that step

111. Multistep Mechanisms
In a multistep process, one of the steps will be slower than all others.
The overall reaction cannot occur faster than this slowest, rate-determining step.

112. Rate Laws for Elementary Steps

113. If a reaction proceeds via several elementary steps, then the elementary steps must add to give the balanced chemical equation.
Intermediate: a species which appears in an elementary step which is not a reactant or product. It is produced in one step and consumed in a later step.
The mechanism must agree with the experimentally determined law.

114. Slow Initial Step NO2 (g) + CO (g)  NO (g) + CO2 (g)
The rate law for this reaction is found experimentally to be
Rate = k [NO2]2
CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration.
This suggests the reaction occurs in two steps.

115. RATE DETERMINING STEP
The rate is always determined by the slow rate step.

116. Slow Initial Step A proposed mechanism for this reaction is
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)
The NO3 intermediate is consumed in the second step.
As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

117. NO2(g) + NO2(g)  NO3(g) + NO(g) NO3(g) + CO(g)  NO2(g) + CO2(g)
Notice that if we add the above steps, we get the overall reaction:
NO2(g) + CO(g)  NO(g) + CO2(g)

118. Fast Initial Step The rate law for this reaction is found to be
2 NO (g) + Br2 (g)  2 NOBr (g)
The rate law for this reaction is found to be
Rate = k [NO]2 [Br2]
Because termolecular processes are rare, this rate law suggests a two-step mechanism.

119. Fast Initial Step A proposed mechanism is Step 1: NO + Br2
NOBr2 (fast)
Step 2: NOBr2 + NO  2 NOBr (slow)
Step 1 includes the forward and reverse reactions.

120. But how can we find [NOBr2]?
Fast Initial Step
The rate of the overall reaction depends upon the rate of the slow step.
The rate law for that step would be
Rate = k2 [NOBr2] [NO]
But how can we find [NOBr2]?

121. Fast Initial Step NOBr2 can react two ways:
With NO to form NOBr
By decomposition to reform NO and Br2
The reactants and products of the first step are in equilibrium with each other.
Therefore,
Ratef = Rater

122. Fast Initial Step Because Ratef = Rater , k1 [NO] [Br2] = k−1 [NOBr2]
Solving for [NOBr2] gives us k1
k−1
[NO] [Br2] = [NOBr2]

123. Fast Initial Step
Substituting this expression for [NOBr2] in the rate law for the rate-determining step gives
k2k1
k−1
Rate =
[NO] [Br2] [NO]
= k [NO]2 [Br2]

124. Example
NO + Cl2 NOCl2 (fast)
NOCl2 + NO NOCl (slow)
1. Write the overall equation.
2. Determine the rate law.

125. Example At low temperatures, the rate law for the reaction
CO (g) + NO2 (g)  CO2 (g) + NO (g)
Rate = k[NO2]2. Show why or why not each of the following mechanisms is consistent with that rate law for this reaction.

126. 1. CO + NO2  CO2 + NO
1. 2 NO2 N2O4 (fast)
2. N2O4 + 2 CO 2 CO2 + 2 NO (slow)
1. 2 NO2  NO3 + NO (slow)
2. NO3 + CO  NO2 + CO2 (fast)
1. 2 NO2  2 NO + O2 (slow)
2. 2 CO + O2  2 CO2 (fast)

127. Catalysts
Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction.
Catalyst are never consumed in the chemical reaction.
Catalysts change the mechanism by which the process occurs.

128. Homogeneous Catalysis
There are two types of catalyst:
Homogeneous – present in the same phase as the reacting molecule.
Heterogeneous –exist in different phase, usually a solid.
Homogeneous Catalysis
The catalyst and reaction is in one phase.
Chlorine atoms are catalysts for the destruction of ozone.

129. Hydrogen peroxide decomposes very slowly: 2H2O2(aq)  2H2O(l) + O2(g)
In the presence of the bromide ion, the decomposition occurs rapidly:
2Br-(aq) + H2O2(aq) + 2H+(aq)  Br2(aq) + 2H2O(l).
Br2(aq) is brown.
Br2(aq) + H2O2(aq)  2Br-(aq) + 2H+(aq) + O2(g).
Br- is a catalyst because it can be recovered at the end of the reaction.
Generally, catalysts operate by lowering the activation energy for a reaction.

131. Catalysts can operate by increasing the number of effective collisions.
That is, from the Arrhenius equation: catalysts increase k be increasing A or decreasing Ea.
A catalyst may add intermediates to the reaction.
Example: In the presence of Br-, Br2(aq) is generated as an intermediate in the decomposition of H2O2.

132. First step is adsorption (the binding of reactant molecules to the catalyst surface).
Adsorbed species (atoms or ions) are very reactive.
Molecules are adsorbed onto active sites on the catalyst surface.

133. Heterogeneous Catalysis
When a catalyst adds an intermediate, the activation energies for both steps must be lower than the activation energy for the uncatalyzed reaction. The catalyst is in a different phase than the reactants and products.
Heterogeneous Catalysis
Typical example: solid catalyst, gaseous reactants and products (catalytic converters in cars).
Most industrial catalysts are heterogeneous.

134. Example H2O2 + I-  H2O + IO- (slow) H2O2 + IO-  H2O + O2 + I- (fast)
1. What is the overall equation?
2. What is the catalyst?
3. What is the intermediate?
4. What is the rate law?

136. C2H4(g) + H2(g)  C2H6(g), H = -136 kJ/mol.
Consider the hydrogenation of ethylene:
C2H4(g) + H2(g)  C2H6(g), H = -136 kJ/mol.
The reaction is slow in the absence of a catalyst.
In the presence of a metal catalyst (Ni, Pt or Pd) the reaction occurs quickly at room temperature.
First the ethylene and hydrogen molecules are adsorbed onto active sites on the metal surface.
The H-H bond breaks and the H atoms migrate about the metal surface.

137. Enzymes are biological catalysts.
When an H atom collides with an ethylene molecule on the surface, the C-C  bond breaks and a C-H  bond forms.
When C2H6 forms it desorbs from the surface.
When ethylene and hydrogen are adsorbed onto a surface, less energy is required to break the bonds and the activation energy for the reaction is lowered.
Enzymes
Enzymes are biological catalysts.
Most enzymes are protein molecules with large molecular masses (10,000 to 106 amu).

138. Enzymes have very specific shapes.
Most enzymes catalyze very specific reactions.
Substrates undergo reaction at the active site of an enzyme.
A substrate locks into an enzyme and a fast reaction occurs.
The products then move away from the enzyme.

139. Only substrates that fit into the enzyme “lock” can be involved in the reaction.
If a molecule binds tightly to an enzyme so that another substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors).
The number of events (turnover number) catalyzed is large for enzymes ( per second).

140. Enzymes

141. SAMPLE EXERCISE 14.11 Determining the Energy of Activation
The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures
(a) From these data, calculate the activation energy for the reaction. (b) What is the value of the rate constant at K?
Solution
Analyze: We are given rate constants, k, measured at several temperatures and asked to determine the activation energy, Ea, and the rate constant, k, at a particular temperature.
Plan: We can obtain Ea from the slope of a graph of ln k versus 1/T. Once we know Ea, we can use
ln k = -Ea (1/RT) + ln A
together with the given rate data to calculate the rate constant at K.

142. Solve: (a) We must first convert the temperatures from degrees Celsius to kelvins. We then take the inverse of each temperature, 1/T, and the natural log of each rate constant, ln k. This gives us the table shown at the right:
A graph of ln k versus 1/T results in a straight line, as shown in Figure
Figure 14.17  Graphical determination of activation energy. The natural logarithm of the rate constant for the rearrangement of methyl isonitrile is plotted as a function of 1/T. The linear relationship is predicted by the Arrhenius equation giving a slope equal to – Ea/R.

143. The slope of the line is obtained by choosing two well-separated points, as shown, and using the coordinates of each:
Because logarithms have no units, the numerator in this equation is dimensionless. The denominator has the units of 1/T, namely, K–1. Thus, the overall units for the slope are K. The slope equals –Ea/R. We use the value for the molar gas constant R in units of J/mol-K
We report the activation energy to only two significant figures because we are limited by the precision with which we can read the graph
(b) To determine the rate constant, k1, at T = K, we can use with Ea = 160 kJ/ mol, and one of the rate constants and temperatures from the given data, such as k2 = 2.52  10–5s–1 and T2 = 

144. SAMPLE EXERCISE 14.11 continued
Thus,
Note that the units of k1 are the same as those of k2.
PRACTICE EXERCISE
Using the data in Sample Exercise 14.11, calculate the rate constant for the rearrangement of methyl isonitrile
at 280°C.
Answer: 2.2  10–2s–1