Kinetics

1
Kinetics

• Ch 14

2
Kinetics

• How fast a reaction occurs
• Factors
• Concentration
• Temperature
• Surface area
• Catalyst

3
Measuring the Rate of a Reaction

• Disappearance of a reactant
• C4H9Cl(aq) H2O(l) ? C4H9OH(aq) HCl(aq)

4

• Draw a graph of the following data (xtime)

5
Rate is the Slope

• y mx b
• C4H9Cl (Rate)(t)
• Rate D C4H9Cl
• D t
• Rate dC4H9Cl
• dt
• (Note how rate decreased with decreasing
  concentration)

6
Ex 1

• Using the graph, calculate the rate from 50.0 s
  to 150.0 s
• ANS 1.64 X 10-4 M/s

7
Ex 2

• Using the graph, estimate the rate at 0 seconds
  and 300 s.
• ANS -2.0 X 10-4 M/s, -1.1 X 10-4 M/s

8
Rates and Stoichiometry

• C4H9Cl(aq) H2O(l) ? C4H9OH(aq) HCl(aq)
• Rate -DC4H9Cl DC4H9OH
• Dt Dt
• Rate of disappearance Rate of appearance
• (only for 11 stoichiometry)

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10

• 2HI(g) ? H2(g) I2(g)
• Rate -1 DHI DH2 DI2
• 2 Dt Dt Dt
• aA bB ? cC dD
• Rate -1 DA -1 DB 1 DC 1 DD
• a Dt b Dt c Dt d Dt

11
Stoichiometry Ex 1

• How is the rate of the disappearance of ozone
  related to the appearance of oxygen according to
• 2O3(g) ? 3O2(g)
• Rate -1 DO3 DO2
• 2 Dt 3 Dt
• -DO3 2 DO2
• Dt 3 Dt

12
Stoichiometry Ex 2

• If the rate of appearance of oxygen at some
  instant is 6.0 X 10-5 M/s, calculate the rate of
  disappearance of ozone.
• -DO3 2 DO2
• Dt 3 Dt
• -DO3 2 (6.0 X 10-5 M/s)
• Dt 3
• -DO3 4.0 X 10-5 M/s
• Dt

13
Stoichiometry Ex 3

• The rate of decomposition of N2O5 at a particular
  instant is 4.2 X 10-7 M/s. What is the rate of
  appearance of NO2 and O2?
• 2N2O5(g) ? 4NO2(g) O2(g)
• Rate -1 DN2O5 1 DNO2 DO2
• 2 Dt 4 Dt Dt
• -DN2O5 1 DNO2
• 2 Dt 4 Dt

14

• DNO2 4 DN2O5
• Dt 2 Dt
• DNO2 2 DN2O5 (-2)(4.2 X 10-7 M/s)
• Dt Dt
• DNO2 8.4 X 10-7 M/s
• Dt
• -DN2O5 DO2
• 2 Dt Dt
• -DO2 (½)(4.2 X 10-7 M/s) 2.1 X 10-7 M/s
• Dt

15
The Rate Law

• aA bB ? cC
• Rate kAmBn
• k Rate constant
• A and B Initial concentrations
• m and n exponents
• k, m, and n must be determined experimentally

16
Rate Law Ex 1

• What is the rate law for the following reaction,
  given the following rate data
• NH4(aq) NO2-(aq) ? N2(g) 2H2O(l)

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18

• Rate kNH4mNO2-n
• Divide Experiment 1 and 2
• Rate2 kNH4mNO2-n
• Rate1 kNH4mNO2-n
• 10.8 X 10-7 k0.0200m0.200n
• 5.4 X 10-7 k0.0100m0.200n
• 2 0.0200m m 1
• 1 0.0100m

19

• Rate kNH4mNO2-n
• Divide Experiment 5 and 6
• Rate6 kNH4mNO2-n
• Rate5 kNH4mNO2-n
• 21.6 X 10-7 k0.20010.0404n
• 10.8 X 10-7 k0.20010.0202n
• 2 0.0404n n 1
• 1 0.0202n

20

• Calculating k
• Rate kNH41NO2-1
• Consider Experiment 1 (can pick anyone)
• Rate1 kNH41NO2-1
• 5.4 X 10-7 k0.010010.2001
• k 2.7 X 10-4 M-1s-1

21
Graphical Curve Fitting

• Consider only experiments where only one
  concentration changes (1-4)
• Rate kNH4mNO2-n
• k and NO2- are constant
• Rate constantNH4m
• Graph Rate(y) vs NH4(x)
• Curve fit (linear or polynomial, consider R2
  value)
• Repeat for NO2- (exp 5-8)

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24
Rate Law Ex 2

• Calculate the rate law for the following general
  reaction A B ? C

25

• ANS Rate (4.0 X 10-3 M-1s-1)A2
• What is the rate of the reaction when A 0.050
  M and B 0.100 M?
• ANS 1.0 X 10-5 M/s

26
Rate Law Ex 3

• A particular reaction varies with H as
  follows. Calculate the Rate Law

27

• Rate 3.2 X 10-8 M2s-1H-1

28
Rate Law Ex 4

• Calculate the rate if the H 0.400 M
• Rate 3.2 X 10-8 M2s-1 H-1
• Rate (3.2 X 10-8 M2s-1 )(0.400)-1
• Rate 8 X 10-8 M/s

29
Reaction Order

• m and n are called reaction orders
• m n . overall reaction order
• Units of k
• k must always have units that allow rate to have
  a unit of M/s

30
Reaction Order Ex 1

• What is the overall reaction order for the
  following reaction. What unit will the rate
  constant (k) have?
• CHCl3 Cl2 ? CCl4 HCl
• RatekCHCl3Cl21/2
• overall reaction order 1 ½ 1.5
• RatekCHCl3Cl21/2
• M/s (k)(M)(M1/2) k M-1/2s-1

31
Concentration and Time First Order

• First Order Reaction reaction whose rate varies
  with the concentration of a single reactant to
  the first power
• Often the decay(chemical or nuclear) or
  decomposition of one substance
• Can solve the equation to calculate concentration
  at any time

32

• Rate - DA kA
• Dt
• - DA kA
• Dt
• dA -kA
• dt
• dA -kdt
• A

33

• ?dA ?-kdt
• A
• ?dA -k?dt (Integrate left from A0 to At
• A and time from 0 to t)
• lnAt lnA0 -kt
• lnAt -kt lnA0
• y mx b
• First order eqns give a linear lnConc vs time
  graph

34
Example of a First Order Reaction

• First order in CH3NC
• lnAt -kt lnA0
• lnCH3NCt -kt lnCH3NC0
• Can use this equation to calculate the
  concentration at any time.

35

• This graph is not linear This graph is

36
An Alternate Form

• lnAt -kt lnA0
• lnAt lnA0 -kt
• lnAt -kt (Now take inverse log)
• A0
• At e-kt
• A0
• At A0e-kt

37

• This graph is not linear This graph is
• At A0e-kt lnAt -kt lnA0

38
First Order Ex 1

• The first order rate constant for the
  decomposition of an insecticide is 1.45 yr-1. If
  the starting concentration of the compound in a
  lake is 5.0 X 10-7 g/cm3, what will be the
  concentration the following year? In two years?

39

• lnAt -kt lnA0
• lnAt -(1.45 yr-1 )(1 yr) ln(5.0 X 10-7
  g/cm3)
• lnAt -15.96
• At e-15.96
• At 1.2 X 10-7 g/cm3
• or
• At A0e-kt
• At (5.0 X 10-7 g/cm3)e-(1.45 yr-1 )(1yr)
• At 1.2 X 10-7 g/cm3
• (A2yr 2.8 X 10-8 g/cm3)

40
First Order Ex 2

• How long will it take the concentration to drop
  to 3.0 X 10-7 g/cm3?
• lnAt -kt lnA0
• ln3.0 X 10-7 g/cm3t -(1.45yr-1)(t) ln5.0
  X 10-7 g/cm3
• Solve for t
• t 0.35 yr

41
First Order Ex 3

• The decomposition of dimethyl ether is a first
  order process with k6.8 X10-4s-1. If the
  initial pressure is 135 torr, what is the partial
  pressure after 1420s?
• (CH3)2O(g) ? CH4(g) H2(g) CO(g)
• (ANS 51 torr)

42
First Order Half-life

• Half-life time required for the concentration
  of a reactant to drop to one half the initial
  value
• Medicine in the body
• Nuclear decay
• At ½A0

43

• ln ½ A0 -kt½
• A0
• ln ½ -kt½
• t½ -ln 1/2
• k
• t½ 0.693
• k

44
Half-Life Ex 1

• From the figure below, calculate the half-life
  for the reaction of C4H9Cl with water.

45
Half-Life Ex 2

• What is the half-life for a reaction with a k
  6.8 X 10-4s-1?
• t½ 0.693
• k
• t½ 0.693
• 6.8 X 10-4s-1
• t½ 1.02 X103s

46
Half-life Ex 3

• Sodium-24 (used in some medical tests) has a
  half-life of 14.8 hours. What is the rate
  constant for its decay?
• (ANS 0.0468 hr-1)

47
Half-Life

• Half-life - The time during which one-half of a
  radioactive sample decays
• Ranges from fraction of a second to billions of
  years.
• You cant hurry half-life.

48
Carbon-14 dating

• By measuring the amount of 14C in traces of
  once-living organisms, one can determine how long
  ago it died.
• E.g., a 5730 years after death, only half of the
  14C remains.
• Reasonable to up to 50,000 years.
• There is a 15 margin of error
• Used in mummies, the Dead Sea Scrolls, Shroud of
  Turin

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Half-Life
The polonium-214 will decay much sooner than the
uranium. The uranium will be radioactive pretty
much until the earth is destroyed when our sun
goes out in 10 billion years.
51
Half-life Example 1

• Carbon-14 has a half-life of 5730 years and is
  used to date artifacts. How much of a 26 g
  sample will exist after 3 half-lives? How long
  is that?

52
Half-life Example 1
53
Half-life Example 2

• Tritium undergoes beta decay and has a half life
  of 12.33 years. How much of a 3.0 g sample of
  tritium remains after 2 half-lives?

54
Solution to Problem
55
Half-life Example 3

• Radon-226 has a half-life of 1600 years? How
  much of a 30 gram sample remains after 6400
  years?

56
Solution to Problem
57
Half-life Example 4

• Cesium-137 has a half-life of 30 years. If you
  start with a 200 gram sample, and you now have 25
  grams left, how much time has passed?

58
Solution to Problem
59
Rate Law Ex 1

• Uranium-238 has a half-life of 4.5 X 109 yr. If
  1.000 mg of a 1.257 mg sample of uranium-238
  remains, how old is the sample?
• k 0.693
• t½
• k 0.693 1.5 x10-10 yr
• 4.5 X 109 yr

60
Rate Law Ex 2

• A wooden object is found to have a carbon-14
  activity of 11.6 disintegrations per second.
  Fresh wood has 15.2 disintegrations per second.
  If the half-life of 14C is 5715 yr, how old is
  the object?

61
Rate Law Ex 2

• A wooden object is found to have a carbon-14
  activity of 11.6 disintegrations per second.
  Fresh wood has 15.2 disintegrations per second.
  If the half-life of 14C is 5715 yr, how old is
  the object?
• ANS 2230 yr

62
Rate Law Ex 3

• After 2.00 yr, 0.953 g of a 1.000 g sample of
  strontium-90 remains. How much remains after 5.00
  years?
• lnNt -kt
• N0
• ln0.953 -k(2.00 yr)
• 1.000
• k 0.0241 yr-1

63
Ex 4

• A sample for medical imaging contains 18 F (1/2
  life 110 minutes). What percentage of the
  original sample remains after 300 minutes?
• ANS 15.1

64
Concentration and Time Second Order

• Second Order Reaction reaction whose rate
  varies with the concentration of a single
  reactant to the second power

65

• Rate - DA kA2
• Dt
• - DA kA2
• Dt
• dA -kA2
• dt
• dA -kdt
• A2

66

• ?dA ?-kdt
• A2
• ?dA -k?dt (Integrate left from A0 to At
• A2 and time from 0 to t)
• 1/At 1/A0 kt
• 1 kt 1
• At A0
• y mx b
• 2nd order eqns give a linear 1/Conc vs time
  graph

67
Second Order Half-life

• 1 kt 1 (At ½A0)
• At A0
• 1 kt½ 1
• ½A0 A0 t½ 1
• 2 kt½ 1 kA0
• A0 A0
• 1 kt½
• A0

68
Second Order Ex 1

• The following data is for the decomposition of
  nitrogen dioxide NO2(g) ? NO(g) ½ O2(g)
• Is the reaction first or second order?

69

• Plot both lnNO2 vs time and 1/NO2 vs time
• Which is linear?

70

• Now calculate k

71

• Pick a convenient point
• 1 kt 1
• At A0
• 150 k(100s) 100
• k 0.5 M-1s-1

72
Second Order Ex 2

• What is the half life for the above reaction?
• t½ 1
• kA0
• t½ 1
• (0.5 M-1s-1)(0.0100 M)
• t½ 200 s

73
Temperature and Rate

• Rate of most chemical reactions increases with
  temperature
• Bread rising
• Plants growing
• Light sticks

74

• Rate kAmBn
• k increases with temperature
• Orders (m, n, etc..) do not change

75
Collision Model

• For an effective collision to occur, molecules
  must collide
• Often (increases with temperature)
• Proper orientation
• Enough Activation Energy (increases with
  temperature)

76
Proper vs. Improper Orientation
77
Insufficient vs. Sufficient Activation Energy
78
Activation Energy

• Energy needed to form activated complex or
  transition state
• Energy needed to start a reaction

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The Arrhenius Equation

• k Ae-Ea/RT
• k Rate constant
• A frequency factor
• Ea Activation Energy
• R 8.314 J/mol-K
• T Temperature in K

81

• k Ae-Ea/RT (Take ln of both side)
• lnk Ea ln A
• RT
• y mx b
• Plot lnk vs. 1/T.
• Slope (m) -Ea/R

82

• Non-graphical method (if you only have two
  points)
• lnk1 Ea ln A lnk2 Ea ln A
• RT1 RT2
• lnk1 - lnk2 Ea lnA - Ea ln A
• RT1 RT2
• lnk1 Ea 1 - 1
• k2 R T2 T1

83
Activation Energy Ex 1

• Given the following rate constants for the
  rearrangement of methyl isonitrile, calculate the
  activation energy.

84

• We wish to graph
• lnk Ea ln A
• RT

85

• slope -Ea/R
• Ea -(slope)(R)
• Ea -(-1.9 X 104 K)(8.314 J/mol-K)
• Ea 160 kJ/mol

86
Activation Energy Ex 2

• If a certain reaction has a k of 2.6 X 10-5 at
  200 oC, calculate the k at 300oC. Assume Ea
  200 kJ/mol.
• lnk1 Ea 1 - 1
• k2 R T2 T1

87

• lnk1 -3.18
• 2.52X10-5
• k1 e-3.18
• 2.52X10-5
• k1 1.0 X 10-6 s-1

88
Activation Energy Ex 3

• Now calculate the rate constant at 280 oC.
• (ANS 2.2 X 10-2 s-1)

89
Reaction Mechanisms

• Elementary Processes only one step involved
• Unimolecular
• Bimolecular
• Termolecular(rare)

90

• Multistep Reactions
• NO2(g) CO(g) ? NO(g) CO2(g)
• NO2(g) NO2(g) ? NO3(g) NO(g)
• NO3(g) CO(g) ? NO2(g) CO2(g)
• NO2(g) CO(g) ? NO(g) CO2(g)
• NO3(g) is an intermediate

91
Reaction Mechanisms Ex 1

• Ozone decomposes in the following steps.
  Describe the molecularity of each step and write
  the overall equation
• O3(g) ? O2(g) O(g)
• O3(g) O2(g) ? 2O2(g)

92

• O3(g) ? O2(g) O(g) unimolecular
• O3(g) O(g) ? 2O2(g) bimolecular
• 2O3(g) ? 3O2(g)
• O(g) is an intermediate

93
Reaction Mechanisms Ex 2

• The following steps
• Mo(CO)6 ? Mo(CO)5 CO
• Mo(CO)5 P(CH3)3 ? Mo(CO)5P(CH3)3
• Are proposed for the reaction
• Mo(CO)6 P(CH3)3 ? Mo(CO)5P(CH3)3 CO
• Is the proposed mechanism consistent with the
  reaction?

94
Rate Determining Step

• Consider two toll plazas
• Plaza A limits the Plaza B limits the
• flow of traffic flow of traffic

95

• St 1 NO2 NO2 ? NO3 NO (slow, k1)
• St 2 NO3 CO ? NO2 CO2 (fast, k2)
• NO2 CO ? NO CO2
• Since Step 1 is much slower, it controls the rate
  (rate determining step)
• Rate k1NO22
• (agrees with experiment)

96
Mechanisms with an Initial Fast Step

• Experimental rate law for 2NO Br2 ? 2NOBr
• is Rate kNO2Br2
• Proposed mechanism
• St 1 NO Br2 ?? NOBr2 (fast)
• St 2 NOBr2 NO ? 2NOBr (slow)
• If Rate kNOBr2NO (involves intermediate)

97

• NOBr2 is intermediate
• Quickly breaks up into NO and Br2
• Rate kNOBr2NO
• Rate kNOBr2NO
• Rate kNO2Br2

98
Reaction Mechanisms Ex 3

• Show that the following mechanism is consistent
  with the rate law Rate kNO2Br2
• St 1 NO NO ?? N2O2 (fast)
• St 2 N2O2 Br2 ? 2NOBr (slow)

99
Catalyst

• Catalyst substance that increases the speed of
  a reaction without being used up in the process
• Can increase the rate of reaction by thousands of
  times
• Lowers the activation energy
• Ex
• 2KClO3(s) ? 2KCl(s) 3O2(g)
• KClO3 is stable
• Adding MnO2 allows the reaction to go fast

100

• Red line shows pressure during a decomposition
  without a catalyst (HCOOH ? CO2 H2)
• Blue line shows with a catalyst

101

• Ex 2H2O2(aq) ?2H2O(l) O2(g)
• Br-(aq) acts as a catalyst
• Brown Br2 formed in between
• Br- regenerated at the end

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103
Catalytic Convertor

• heterogeneous (different phase) catalyst
• Use Platinum and Rhodium ()
• Speed up
• CO O2 ? CO2
• NO, NO2 ? N2

104
Enzymes

• Biochemical catalysts
• Large protein molecules
• Catalase in liver to decompose H2O2

105

• Active Site lock and key model
• Turnover numbers of 1000 to 10 million per second

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107

• a) -DH2O/2Dt DH2/2Dt DO2/Dt
• b) -DSO2/2Dt -DO2/Dt DSO3/2Dt
• c) -DNO/2Dt -DH2/2Dt DN2/Dt
  DH2O/2Dt
• a) Rate kClO22OH-
• b) k 2.3 X 102 M-2s-1
• c) Rate 0.12 M/s
• 32)a) Rate kNO2O2
• b and c) k 7.11 X 103 M-2s-1
• d) Rate 0.400 M/s
• e) Rate 0.200 M/s

108

• a) 1st order (b) k 0.0101 s-1 (c ) 68.7 s
• 46. a) 1st order (b) k 3.68 X 10-3 min-1