1. Rates & Mechanisms of Chemical Reactions
McMurray & Fay ch. 12

2. Rates of Reaction
Different reactions occur at different speeds

3. How do we measure the rate of a reaction?
Two ways:
Measure disappearance of a reactant
Measure appearance of a product

4. Rates are always measured as a function of time
dollars per hour dollars per year miles per hour (miles/hr or miles hr-1) feet per second (feet/second or ft s-1)

5. Reaction rates measure changes in amounts of chemicals over time
In solutions: molarity per time (M s-1, M hr-1, etc.)
In other instances:
moles per time
grams per time

6. Relative rates
Concentrations of reactants may change at different rates, for example: 2 H2 + O2  2 H2O For every O2 used, 2 H2 are used For every mole of O2, 2 moles of H2 are used For every 1 M the concentration of O2 changes, the concentration of H2 changes by 2 M

7. 2 H2 + O2  2 H2O Water is appearing at twice the rate that
oxygen is disappearing
Hydrogen is disappearing
at twice the rate
so half of that rate is equal
to that of oxygen
Oxygen is disappearing
at a certain rate

8. General rate law: aA + bB  cC + dD

9. 2 H2O (l) + CaC2 (s)  C2H2 (g) + Ca(OH)2 (aq)
If H2O is used up at 3.8 mol s-1...
at what rate is CaC2 used up?
at what rate is C2H2 produced?

10. But usually, reaction rate is expressed independently of concentration of any reactant or product
C6H12O6  2 C2H5OH + 2 CO2
[C2H5OH] increases at M hr-1
Reaction rate = M hr -1

11. Instantaneous reaction rates
Reaction rates may change over time
Rate dependent on concentrations
Reactant concentrations decrease over time
Can only measure rates at specific times

12. Instantaneous reaction rate

13. Initial rate of reaction
Find concentration change at t=0

14. Definition: Rate Law
Expression of reaction rate that includes concentrations of all reactants

15. The exponents in the rate law must be determined experimentally.
Generic Rate Law
For the reaction:
A + B  products
The reaction rate can be written as:
rate = k [A]m[B]n
The exponents in the rate law must be determined experimentally.

16. Example Rate Law
Reaction: BrO Br- + 6 H+  3 Br2 + 3 H2O Rate = k [BrO3-] [Br -] [H+]2

17. Reaction Orders
rate = k [A]m[B]n Each rate law includes the concentration of each reactant raised to a power: [A]1 = first order [A]2 = second order [A]0 = zeroth order

18. Reaction order affects how reaction rates change with concentration
rate = k [A]m

19. First order rate = k [A]1 When [A] = 2x: rate = k (2x)1= 2{kx}
First order rule:
Increase conc by factor of n
= Increase rate by factor of n
When [A] = 2x:
rate = k (2x)1= 2{kx}
When [A] = 3x:
rate = k (3x)1 = 3{kx}

20. Second order rate = k [A]2
When [A] = x: rate = k (x)2= {kx2}
Second order rule:
Increase conc by factor of n
= Increase rate by factor of n2
When [A] = 2x:
rate = k (2x)2[B]n= 4{kx2}
When [A] = 3x:
rate = k (3x)2= 9{kx2}

21. Zeroth order rate = k [A]0
When [A] = x: rate = k (x)0= {k}
Zeroth order rule:
Rate does not change if you change concentration
When [A] = 2x:
rate = k (2x)0= {k}
When [A] = 3x:
rate = k (3x)0= {k}

22. Overall Reaction Order
The rate law rate = k [A]m[B]n Is said to be reaction order m + n It is mth order with respect to component A and nth order with respect to component B

23. So, for BrO3- + 5 Br- + 6 H+  3 Br2 + 3 H2O Rate = k [BrO3-] [Br-] [H+]2
This reaction is: First order with respect to [BrO3-] and [Br-] Second order with respect to [H+]

24. Experimental determination of reaction order
Vary initial concentrations
Measure initial rate of formation of products
Compare how concentrations change with how rate changes

25. Determining Rate Laws from Experimental Data
rate = k [A]m[B]n
1. Find two data sets where concentration of first reagent changes but other concentrations remain constant. Use the rules on the right to determine rate order for first reagent
First order rule:
Increase conc by factor of n
= Increase rate by factor of n
Second order rule:
Increase conc by factor of n
= Increase rate by factor of n2
2. Repeat step 1 for subsequent reagents as necessary
Zeroth order rule:
Rate does not change if you change concentration
3. Plug experimental data and rate order exponents determined in steps 1&2 into rate law. Solve for k.

26. 2 NO (g) + O2 (g)  2NO2 (g) Experiment # [NO] (M) [O2] (M)
Initial Rate (M s-1) 1
0.015
0.048 2
0.030
0.192 3
0.096 4
0.384
1. Find two data sets where concentration of first reagent changes but other concentrations remain constant. Use the rules on the right to determine rate order for first reagent
First order rule:
Increase conc by factor of n
= Increase rate by factor of n
Second order rule:
Increase conc by factor of n
=Increase rate by factor of n2
Zeroth order rule:
Rate does not change if you change concentration

27. 2. Repeat step 1 for subsequent reagents as necessary
2 NO (g) + O2 (g)  2NO2 (g)
Experiment #
[NO] (M)
[O2] (M)
Initial Rate (M s-1) 1
0.015
0.048 2
0.030
0.192 3
0.096 4
0.384
2. Repeat step 1 for subsequent reagents as necessary
First order rule:
Increase conc by factor of n
= Increase rate by factor of n
Second order rule:
Increase conc by factor of n
=Increase rate by factor of n2
Zeroth order rule:
Rate does not change if you change concentration

28. 2 NO (g) + O2 (g)  2NO2 (g) Experiment # [NO] (M) [O2] (M)
Initial Rate (M s-1) 1
0.015
0.048 2
0.030
0.192 3
0.096 4
0.384
3. Plug experimental data and rate order exponents
determined in steps 1&2 into rate law. Solve for k.

29. H2O2 (aq) + 3 I- (aq)  I3- (aq) + 2H2O (l)
Experiment #
[H2O2] (M)
[I-] (M)
Initial Rate (M s-1) 1
0.100
1.15 x 10-4 2
0.200
2.30 x 10-4 3 4
4.60 x 10-4
1. Find two data sets where concentration of first reagent changes but other concentrations remain constant. Use the rules on the right to determine rate order for first reagent
First order rule:
Increase conc by factor of n
= Increase rate by factor of n
Second order rule:
Increase conc by factor of n
=Increase rate by factor of n2
Zeroth order rule:
Rate does not change if you change concentration

30. H2O2 (aq) + 3 I- (aq)  I3- (aq) + 2H2O (l)
Experiment #
[H2O2] (M)
[I-] (M)
Initial Rate (M s-1) 1
0.100
1.15 x 10-4 2
0.200
2.30 x 10-4 3 4
4.60 x 10-4
First order rule:
Increase conc by factor of n
= Increase rate by factor of n
2. Repeat step 1 for subsequent reagents as necessary
Second order rule:
Increase conc by factor of n
=Increase rate by factor of n2
Zeroth order rule:
Rate does not change if you change concentration

31. H2O2 (aq) + 3 I- (aq)  I3- (aq) + 2H2O (l)
Experiment #
[H2O2] (M)
[I-] (M)
Initial Rate (M s-1) 1
0.100
1.15 x 10-4 2
0.200
2.30 x 10-4 3 4
4.60 x 10-4
3. Plug experimental data and rate order exponents
determined in steps 1&2 into rate law. Solve for k.

32. NH4+ (aq) + NO2- (aq)  N2 (g) + 2H2O (l)
Experiment #
[NH4+] (M)
[NO2-] (M)
Initial Rate (M s-1) 1
0.24
0.10
7.2 x 10-6 2
0.12
3.6 x 10-6 3
0.15
5.4 x 10-6
1. Find two data sets where concentration of first reagent changes but other concentrations remain constant. Use the rules on the right to determine rate order for first reagent
First order rule:
Increase conc by factor of n
= Increase rate by factor of n
2. Repeat step 1 for subsequent reagents as necessary
Second order rule:
Increase conc by factor of n
=Increase rate by factor of n2
3. Plug experimental data and rate order exponents determined in steps 1&2 into rate law. Solve for k.
Zeroth order rule:
Rate does not change if you change concentration

33. Integrated Rate Laws

34. First-order reactions
rate = k[A]
- d[A] dt
= k[A]
d[A]
[A]
= - k dt

35. First-order reactions
ln [A]t – ln [A]0 = -kt
ln [A]t = -k t + ln [A]0
y = m x + b

36. First-order reactions graph ln [A] vs. t
k = -slope =

37. Second order reaction
-d[A] dt
= k [A]2
d[A]
[A]2
= - k

38. Second-order reactions1
[A] 1
[A]0
= kt 1
[A] 1
[A]0
= k t +
y = m x + b

39. Second order reactions graph 1/[A] vs. T
k = -slope =

40. Zeroth-order reactions
Rate = k [A]0 Rate = k
-d[A] dt
= k
d[A] = -k dt

41. Zeroth-order reactions
[A] = -k t + [A]0
y = m x + b
Graph [A] vs. time
slope = -k

42. Finding reaction order experimentally
Zeroth order: [A] vs. t

43. Finding reaction order experimentally
Is this zeroth, first, or second order?

44. Finding reaction order experimentally

45. You can use integrated rate laws to solve for:
Time at which a given concentration is present
Concentration at a given time
Rate constant

46. Using integrated rate laws to determine elapsed time or rate constant
First order:
ln [A]t – ln [A]0 = -kt
ln = -kt
[A]t
[A]0

47. H2O2 in a lake decays with a rate constant of 0. 10 hr-1
H2O2 in a lake decays with a rate constant of 0.10 hr-1. How long will it take H2O2 to decay to 5% of its present value?
ln = -kt
[A]t
[A]0
1. Start with the integrated rate law
2. Plug the ratio [A]t/[A]0 into the rate law.
We want [A]t to be 5% of the starting value [A]0,
so [A]t/[A0] =0.05
3. Plug this value into the rate law.
4. Solve for t.

48. The concentration of 32P in a cell decreases from 9. 53 at t0 to 7
The concentration of 32P in a cell decreases from 9.53 at t0 to 7.85 at t = 4 days. Find the rate constant for the disappearance of 32P.
ln = -kt
[A]t
[A]0
1. Start with the integrated rate law
2. Plug the concentrations into the rate law.
3. Add the value for time elapsed.
4. Solve for k.

49. A reaction obeys second-order kinetics with rate constant 4. 5 M day-1
A reaction obeys second-order kinetics with rate constant 4.5 M day-1. How long does it take for the concentration of reactant A to drop from 2.1 M to 1.3 M? 1
[A]t
= kt
[A]0
1. Start with the integrated rate law
2. Plug [A]t and [A]0 into the rate law.
3. Solve for t.

50. 251Cf decays via first-order kinetics with a rate constant of 7
251Cf decays via first-order kinetics with a rate constant of 7.7 x 10-4 yr. A metal sample has a 251 Cf concentration of 1.23 %. What will the concentration be in 100 years?
ln = -kt
[A]t
[A]0

51. An enzyme breaks down by zeroth order kinetics
An enzyme breaks down by zeroth order kinetics. At first, the concentration of the enzyme is M. Ten minutes later, the concentration of the enzyme is M. What is the rate constant for the decay?
[A] - [A]0 = -k t

52. Definition: Half-life
The amount of time required for half of the reactant to disappear To find: use [A]t = ½ [A]0

53. Half life for first order reaction
ln [A]t = -k t + ln [A]0
ln (1/2[A]0) = -k t1/2 + ln [A]0
ln [A]0- ln (1/2 [A]0) = k t1/2
[A]0
(1/2 [A]0)
ln = k t1/2

54. Half-life for first order reaction
ln 2 = k t1/2
ln 2 k
t1/2 =

55. ln 2 k
t1/2 =

56. Half-life for zeroth order reactions
[A] = -k t + [A]0
1/2[A]0 = -k t1/2 + [A]0
t1/2 = (1/2k) [A]0
Half-life depends on initial concentration
so is NOT CONSTANT

57. t1/2 = (1/2k) [A]0

58. Half-life for second order reactions1
1/2[A]0 1
[A]0
= kt1/2
Half-life depends on
initial concentration
so is NOT CONSTANT 1
[A]0
= kt1/2 1
k[A]0
t1/2 =

59. First-Order Reactions
For practical reasons,
we usually only talk about half-lives for
First-Order Reactions

60. For first-order reactions:
ln 2 k
0.693 k
t1/2 = =

61. For first-order reactions you can also rearrange to get:
0.693
t1/2
ln 2
t1/2
k = =

62. A compound decays by first-order kinetics and has a half-life of 1
A compound decays by first-order kinetics and has a half-life of 1.6 hours. What is the rate constant, k, of the reaction?

63. All radioactive isotopes decay via first-order kinetics.
Radioactive decay
All radioactive isotopes decay via first-order kinetics.

64. Radioactive decay
Not always easy to measure concentration of radioisotopes.
Substitute one of following into the first-order rate law for concentration:
activity (number of decays per second, N)
isotope ratios

65. 131I is used as a metabolic tracer in hospitals
131I is used as a metabolic tracer in hospitals. It has a half-life, t½ = 8.07 days. How long will it take before the activity falls to 1% of the initial value?
ln = -kt
[A]t
[A]0

66. Vocabulary: CARBON DATING
Method of using 14C/12C ratios to determine the age of an object

67. Carbon Dating: How it works
Radioactive 14C (t1/2 = 5730 years) is constantly produced in the atmosphere
12C is not radioactive
14C:12C ratio in atmosphere constant at 1.2 x 10-12
Living organisms constantly exchange carbon through respiration so 14C:12C ratio similar to atmosphere
After death, this exchange stops; amount of 14C decays by first-order kinetics

68. A fossil is found to have about 0. 36 of the normal 14C:12C ratio
A fossil is found to have about 0.36 of the normal 14C:12C ratio. How old is the fossil?

69. Collision Theory: What affects the rate of a reaction?

70. Reaction: 2 HI  H2 + I2 H I H I
Before new bonds can be formed, the old ones must be broken.
Old bonds don’t just break for no reason—it takes an input of energy to break them

71. H I H I

72. I H H I

73. For a reaction to occur…
Reactants must COLLIDE… …with SUFFICIENT ENERGY …at the right ANGLE ...to break the old bonds

74. Factors that affect the ability to react also affect the speed at which a reaction occurs.
Chemical nature of the reactants
Ability of the reactants to move around
Concentrations of the reactants
Temperature
Catalysts

75. Concentrations affect collision rates
The higher the concentration of the reactants, the higher the likelihood of a good collision

76. Temperature The higher the temperature,
Reactants must collide with enough energy to break bonds
Temperature is a measure of the average kinetic energy
The higher the temperature,
the higher the likelihood of a good collision

77. The more reactants can move, the more likely they are to collide
For a reaction to occur, reactants must come in contact with each other (Most reactions occur in liquid or gas state)
The more reactants can move,
the more likely they are to collide

78. The stronger the bonds holding a molecule together, the more energy a collision must have to break those bonds
The number of collisions with sufficient energy to cause a reaction depends on the chemical properties of the reactants

79. Vocabulary: CATALYST
A substrate that allows reactants to interact more efficiently
Frequently hold reactant molecules so collisions occur from the correct angle

80. (An effective collision is one that actually gives product molecules.)
Collision Theory
The rate of a reaction is proportional to the number of effective collisions per second amongst reactant molecules.
(An effective collision is one that actually gives product molecules.)

81. Vocabulary: ACTIVATION ENERGY
The minimum amount of kinetic energy needed in a collision to cause a reaction to happen

82. Vocabulary: TRANSITION STATE
The moment in a reaction when the old bonds are partially broken and the new bonds are partially formed

83. Reaction: 2 HI  H2 + I2 H I H I

84. Reaction: 2 HI  H2 + I2 H I H I H I

85. Arrhenius equation Temperature (in Kelvin) Rate constant Gas constant
(frequency factor or
pre-exponential factor)
Gas constant
(8.314 J/mol K)
Activation
Energy

86. Arrhenius equation: conversion to logarithmic form

87. Can plot ln k vs. 1/T for experimental data
ln k = ln A – Ea/RT
y = b + m x
Can plot ln k vs. 1/T for experimental data
Slope of line = -Ea/R

88. Experimental Determination of Ea
Run reaction at different temperatures
Determine k at each temperature
Plot ln k vs. 1/T
Multiply slope of best-fit line by R

89. Determination of Ea (Actual Experimental Data)

90. Determination of Ea (Actual Experimental Data)

91. Arrhenius Plot
Ea=42801 J/mol

92. Alternate application of Arrhenius law: Two-point equation

93. For the reaction shown previously (Ea = 42
For the reaction shown previously (Ea = 42.8 kJ/mol), the rate constant k is s-1 at 10.5 oC. What will the rate constant be at 35 oC?

94. Rate constants for the decomposition of gaseous dinitrogen pentoxide are 3.7 x 10-5 s-1 at 25oC and 1.7 x 10-3 s-1 at 55oC. What is the activation energy for this reaction in kJ/mol?

95. Reaction Mechanisms

96. O2 + 4 H+ + 4 e-  2 H2O
O2 + e-  O2-• O2-• + H+  HO2• HO2• + e-  HO2- HO2- + H+  H2O2 H2O2 + e-  H2O2-• H2O2-• + H+  H2O + HO• HO• + e-  HO- HO- + H+  H2O

97. Reaction Mechanisms
Balanced chemical equations represent a net process
Reactions actually occur in steps called elementary processes which only involves one or two reactants
A reaction mechanism shows each of these steps in order

98. Kinetics and reaction mechanisms
Each step in a reaction mechanism has its own rate coefficient
How can we use these to determine the overall reaction rate?

99. Which of the following is an elementary process
Which of the following is an elementary process? (There is more than one correct answer)
C3H8 + 5 O2  3CO2 + 4 H2O
2 NO + H2  N2O + H2O
NO + O3  NO2 + O2
3 H2 + N2  2 NH3
N2O2 + H2 N2O + H2O

100. How fast can you drive on a highway?

101. rate-determining step
The rate at which a reaction can occur is limited by the slowest step
Slowest step (elementary process) is the
rate-determining step

102. Rate Laws for Elementary Processes
Elementary processes will be unimolecular or bimolecular Unimolecular processes are proportional to the concentration of that molecule Bimolecular processes are proportional to the concentration of each molecule

103. Example: O3  O2 + O rate = k[O3]
This elementary process is unimolecular, so
rate = k[O3]

104. Example: CH3Br + OH-  Br- + CH3OH
This elementary process is bimolecular, so
rate = k[CH3Br][OH-]

105. Reactions where the first step is slow
NO2 (g) +NO2 (g) NO(g) + NO3 (g) NO3 (g) + CO (g) N2O (g) + CO2 (g)
Slow
Faster

106. Reactions where the first step is fast
Reversible
Slow
Fast
2 NO (g) N2O2 (g) N2O2 (g) + H2 (g) N2O (g) + H2O (g) N2O (g) + H2 (g) N2 (g) + H2O (g)

107. Catalysts

108. Catalyst
Definition: a substance that changes the rate of a chemical reactant without being used up Divided into two types: Homogeneous catalysts (same phase as the reactants) Heterogeneous catalysts (different phase from the reactants)

109. Homogeneous catalysis
Catalysts are in the same phase as reactants, e.g. all aqueous or all gaseous 2 H2O2 (aq)  2 H2O (l) + O2 (g)

110. Enzymes as catalysts