The document discusses kinetics and reaction rates. It defines kinetics as the study of reaction rates, and explains that thermodynamics determines if a reaction will occur but not how fast. It then covers factors that affect reaction rates like temperature, concentration, and catalysts. The document provides examples of calculating rates from concentration data and determining rate laws from experimental results. It also introduces integrated rate laws and how to determine reaction order from different rate law equations.

1. Kinetics
Reaction Rates

2. Kinetics
• The study of reaction rates.
• Thermodynamics will tell us whether or not
the reaction will occur
• Spontaneous reactions (ΔG = neg) are
reactions that will happen - but we
can’t tell how fast.
• Diamond will spontaneously turn to
graphite – eventually.
• Kinetics is about how fast

3. Review- Collision Model of Chemical
Reactions
Particles must collide
with enough energy
to break bonds
EFFECTIVE COLLISIONS

4. Factors that affect the rate of reaction
Increase number of effective collisions
 increase rate of reaction
• Temperature
• Concentration
• Surface area
• Pressure
• Agitation
• Catalyst

5. Reaction Rate
• Rate = Conc. of A at t2 -Conc. of A at t1
t2- t1
• Rate =∆[A]
t
• Change in concentration per unit time
• Consider the reaction
N2(g) + 3H2(g)  2NH3(g)

6. • As the reaction progresses the
concentration H2 goes down
C
o
n
c
e
n
t
r
a
t
i
o
n
Time
[H[H22]]
N2 + 3H2 → 2NH3
Δ[H2]/time = neg

7. • As the reaction progresses the
concentration N2 goes down 1/3 as fast
Δ[H2]/time = neg
C
o
n
c
e
n
t
r
a
t
i
o
n
Time
[H[H22]]
[N[N22]]
N2 + 3H2 → 2NH3
Δ[N2]/time = neg
Δ[H2]/time = neg

8. • As the reaction progresses the
concentration NH3 goes up 2/3 times
C
o
n
c
e
n
t
r
a
t
i
o
n
Time
[H[H22]]
[N[N22]]
[NH[NH33]]
N2 + 3H2 → 2NH3
Δ[NH3]/time = pos

9. Reaction Rates are always positive
• Over time
– the concentration of products increases
– the concentration of reactants decreases
• So Δ[P]/time is positive
• But Δ[R]/time is negative, so the rate
would be expressed as - Δ[R]/time

10. If rate is measured in terms of
concentration of Hydrogen
-Δ[H2] - mol H2
time L time
and N2 + 3H2  2NH3
Then using the stoichiometry of the reaction
-Δmol H2 1 mol N2
L time 3 mol H2
Then
-Δmol H2 2 mol NH3
L time 3 mol H2

11. Calculating Rates
• Average rates are taken over a period of
time interval
• Instantaneous rates are determined by
finding the slope of a line tangent to the
curve at any given point because the rate
can change over time

12. Average Rate – Find the slope of the
line between the two points
Note: slope =
C
o
n
c
e
n
t
r
a
t
i
o
n
Time
[H[H22]]
tt
Note: slope = Δy / Δx
Δconcentration / Δtime
THE RATE

13. Instantaneous Rate – Find the slope of
the line tangent to that point
C
o
n
c
e
n
t
r
a
t
i
o
n
Time
∆∆[H[H22]]
 tt
d[Hd[H22]]
dtdt

14. 12_291
0.0003
70s
O2
0.0025
0.005
0.0075
0.0100
0.0006
70s
0.0026
110 s
NO2
NO
50 100 150 200 250 300 350 400
Concentrations(mol/L)
Time (s)
∆[NO2 ]
∆t
INSTANTANEOUS RATES
2NO2  2NO + O2

17. C4H9Cl + H2O → C4H9OH + HCl

18. Rate Laws
• Reactions are reversible.
• As products accumulate they can begin to
turn back into reactants.
• Early on the rate will depend on only the
amount of reactants present.
• We want to measure the reactants as
soon as they are mixed.
• In addition, as the reaction progresses the
concentration of reactants decrease
• This is called the initial rate method.

19. • Rate laws show the effect of concentration
on the rate of the reaction
• The concentration of the products do not
appear in the rate law because this is an
initial rate.
• The order (exponent)
–must be determined experimentally,
–cannot be obtained from the equation
Rate LawsRate Laws

20. • You will find that the rate will only depend
on the concentration of the reactants.
(Initially)
• 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.
2 NO2 2 NO + O2

21. NO2  NO + ½ O2
The data for
the reaction at 300ºC
Plot the data to show
concentration as a
function of time.
Time (s)
[NO2]
(mol/L)
0.00 0.01000
50.0 0.00787
100.0 0.00649
200.0 0.00481
300.0 0.00380

22. NO2  NO + ½ O2
To determine the rate law for the reaction, we
must find the exponent in the equation.
Rate = k [NO2]x
Experiment [NO2] M Rate, M s-1
1 0.060 0.0360
2 0.030 0.0090
3 0.020 0.0040

23. NO2  NO + ½ O2
To determine the rate law for the reaction, we
must find the exponent in the equation.
Rate = k [NO2]x
Exp [NO2] M Rate, M s-1
1 0.060 0.0360
2 0.030 0.0090
3 0.020 0.0040
2x [NO2] 4x rate

24. NO2  NO + ½ O2
The exponent for the concentration of NO2
must be 2 to show the effect of changing
the concentration of NO2 on the rate.
Rate = k [NO2]2
Exp [NO2] M Rate, M s-1
1 0.060 0.0360
2 0.030 0.0090
3 0.020 0.0040
3x [NO2] 9x rate

25. NO2  NO + ½ O2
Rate = k [NO2]x
Pick two experiments to plug into
the equation.
Experiment [NO2] M Rate, M s-1
1 0.060 0.0360
2 0.030 0.0090
3 0.020 0.0040

26. NO2  NO + ½ O2
Use the rate and one of the experiments
to calculate the rate constant.
Experiment [NO2] M Rate, M s-1
1 0.060 0.0360
2 0.030 0.0090
3 0.020 0.0040

27. 2ClO2 + 2OH-
→ ClO3
-
+ ClO2
-
+ H2O
• Determine the rate law for the reaction.
• Calculate the rate constant.
• What is the overall reaction order?
Experiment [ClO2
-
] M [OH-
] M Rate, M s-1
1 0.060 0.030 0.0248
2 0.020 0.030 0.00276
3 0.020 0.090 0.00828

28. Types of Rate Laws
• Differential Rate Law - describes how rate depends on
concentration. Method to determine- change initial
concentration and measure effect on rate
• Integrated Rate Law - describes how concentration
depends on time. Method to determine- measure the
concentration of reactants as function of time
• For each type of differential rate law there is an
integrated rate law and vice versa.

29. Determining Rate Laws
• The first step is to determine the form of
the rate law (especially its order).
• 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 the gas leaves

30. [N[N22OO55] (mol/L)] (mol/L) Time (s)Time (s)
1.001.00 00
0.880.88 200200
0.780.78 400400
0.690.69 600600
0.610.61 800800
0.540.54 10001000
0.480.48 12001200
0.430.43 14001400
0.380.38 16001600
0.340.34 18001800
0.300.30 20002000
Now graph the data

31. 0
0.2
0.4
0.6
0.8
1
1.2 0
200
400
600
800
1000
1200
1400
1600
1800
2000
• To find rate we have to find the slope at two points
• We will use the tangent method.
[N[N22OO55]]
(mol/L)(mol/L)
Time (s)

32. 0
0.2
0.4
0.6
0.8
1
1.2
0
200
400
600
800
1000
1200
1400
1600
1800
2000
At .80 M the rate is (.88 - .68) = 0.20 =- 5.0x 10 -4
(200 - 600) -400

33. 0
0.2
0.4
0.6
0.8
1
1.2
0
200
400
600
800
1000
1200
1400
1600
1800
2000
At .40 M the rate is (.52 - .32) = 0.20 =- 2.5 x 10 -4
(1000-1800) -800

34. • At 0.8 M rate is 5.0 x 10-4
• At 0.4 M rate is 2.5 x 10-4
• Since the rate is twice as fast when the
concentration is twice as big the rate law
must be.. First power
• Rate = -[N2O5] = k[N2O5]1
= k[N2O5]
t
• We say this reaction is first order in N2O5
• The only way to determine order is to run
the experiment.

35. 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 just after
the reactants are mixed.
• Eliminates the effect of the reverse
reaction.

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

37. Initial concentrations (M)
Rate (M/s)
BrOBrO33
--
BrBr--
HH++
0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10-4-4
0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10-3-3
0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10-3-3
0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10-3-3
Calculate the rate law and k

38. Integrated Rate Law
• Expresses the reaction concentration as a
function of time.
• Form of the equation depends on the
order of the rate law (differential).
• Changes Rate = ∆[A]n
∆t
• We will only work with n = 0, 1, and 2

39. First Order
• For the reaction 2N2O5  4NO2 + O2
• We found the Rate = k[N2O5]1
• If concentration doubles rate doubles.
• If we integrate this equation with respect to time
we get the Integrated Rate Law
[N2O5]0 = initial concentration of N2O5
[N2O5]t = concentration of N2O5 after some time (t)
ln = natural log
= - ktln ( )[N2O5]0
[N2O5]t

40. First Order rate = k[N2O5]
• For the reaction 2N2O5  4NO2 + O2
• Rearrange this equation to y = mx + b form
– Since- log of quotient equals difference of logs
• SO. ln[N2O5]t = - k t + ln[N2O5]0
- ln[N2O5]t=ln ( )[N2O5]0
[N2O5]t
= ktln ( )[N2O5]0
[N2O5]t
ln[N2O5]t
y mx b+=

41. First Order
• Linear form integrated rate law
• ln[N2O5]t = - k t + ln[N2O5]0
= ktln ( )[N2O5]0
[N2O5]t
y mx b+=
ln[N2O5]t
t
m = slope = -k

42. • Graphing natural log of concentration of
reactant as a function of time
• If you get a straight line, you may
determine it is first order
• Integrated First Order
First Order
= - ktln ( )[R]0
[R]t

43. Half Life
• The time required to reach half the
original concentration.
• [R]t = [R]0/2 when t = t1/2
• If the reaction is first order
= - ktln ( )[R]0
[R]t
Since [R]t = [R]0/2
[R]t = [R]0 equals 2
ln (2) = kt1/2

44. 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

45. • The highly radioactive plutonium in
nuclear waste undergoes first-order
decay with a half-life of approximately
24,000 years. How many years must
pass before the level of radioactivity due
to the plutonium falls to 1/128th (about
1%) of its original potency?

46. Second Order
• Rate = -∆[R]/∆t = k[R]2
• integrated rate law
• 1/[R]t = kt + 1/[A]0
• y= 1/[A] m = k
• x= t b = 1/[A]0
• A straight line if 1/[A] vs t is graphed
• Knowing k and [A]0 you can calculate [A]
at any time t

47. Second Order Rate = k [R]2
integrated rate law
1/[R]t = k t + 1/[A]0
y = m x + b
t
1/[R] Slope = k

48. Second Order Half Life
• [A] = [A]0 /2 at t = t1/2
1
2
0
2[ ]A
= kt +
1
[A]1
0
2
2
[ [A]
-
1
A]
= kt
0 0
1
t
k[A]1 =
1
0
2
1
[A]
= kt
0
1 2

49. 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

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

51. Time
C
o
n
c
e
n
t
r
a
t
i
o
n

52. Time
C
o
n
c
e
n
t
r
a
t
i
o
n
∆[A]/∆t
∆t
k =
∆[A]

53. Zero order First order Second order
Rate
law
Graph
[A] vs
time
Integrated
rate law

54. Zero order First order Second order
Linear
graph
with
time
Integrated
rate law
Rearrange
to y=mx+b

55. Summary of Rate Laws

56. Reaction Mechanisms
• The 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.

57. • 2NO2 + F2 2NO2F
Rate = k[NO2][F2]
• The overall equation is a summary.
• In order for this reaction to occur as
written, three molecules must collide
simultaneously
• An unlikely event.
• So a reaction mechanism is proposed
showing a series of likely steps
Reaction Mechanisms

58. • 2NO2 + F2 2NO2F
• Rate = k[NO2][F2]
• The proposed mechanism is
• NO2 + F2  NO2F + F (slow)
• F + NO2  NO2F (fast)
• F is called an intermediate It is formed
then consumed in the reaction
Reaction Mechanisms

59. • Each of the two reactions is called an
elementary step .
• The rate for a reaction can be written from
its molecularity. The reaction rate reflects
the stoichimetry.
• Molecularity is the number of pieces that
must come together.
• Elementary steps add up to the balanced
equation
Reaction Mechanisms

60. • Unimolecular step involves one molecule -
Rate is first order.
• Bimolecular step - requires two molecules -
Rate is second order
• Termolecular step- 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.

61. • A  products Rate = k[A]
• A+A  products Rate= k[A]2
• 2A  products Rate= k[A]2
• A+B  products Rate= k[A][B]
• A+A+B  products Rate= k[A]2
[B]
• 2A+B  products Rate= k[A]2
[B]
• A+B+C  products Rate= k[A][B][C]
Molecularity and Rate Laws
For an elementary step –
the rate law reflects the stoichiometry

62. Reaction Mechanisms
• Proposed series of elementary steps.
– Describing each collision
• The rate of the reaction is determined by
the rate limiting step.
• The overall experimentally determined
rate law must be consistent with the rate
law that reflects the rate limiting step.

63. • The proposed mechanism is
NO2 + F2  NO2F + F (slow) rate = k [NO2][F2]
F + NO2  NO2F (fast) rate = k [F][NO2]
Since the first step is the slow step, it is the rate
limiting step and thus the rate of the overall
reaction depends upon the rate of this step.
Note: The rate of this step is also the rate of the
overall reaction.
2NO2 + F2  2NO2F
Rate = k[NO2][F2]

64. The steps must add up to the
overall reaction.
• 2 NO2
Cl → 2 NO2
+ Cl2
• Proposed steps
• NO2
Cl → NO2
+ Cl (slow)
NO2
Cl + Cl → NO2
+ Cl2
(fast)
• Write the overall rate law that would
be consistent with this proposed
reaction mechanism.

65. How to get rid of intermediates
• They can’t appear in the rate law.
• Slow step determines the rate and the rate
law
• 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.

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

67. 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] because
the first step is fast
• Rate = k[IBr][IBr] = k[IBr]2

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

69. P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Reaction Coordinate
Reactants
Products

70. P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Reaction Coordinate
Reactants
Products
Activation
Energy Ea

71. P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Reaction Coordinate
Reactants
Products
Activated
complex

72. P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Reaction Coordinate
Reactants
Products
∆E
}

73. P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Reaction Coordinate
2BrNO
2NO + Br
Br---NO
Br---NO
2
Transition
State

74. 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.

75. Arrhenius
• Said that 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.

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

77. Problem with this
• Observed rate is too small
• Due to molecular orientation- they have to
be facing the right way
• written into equation as p the steric factor.

78. O
N
Br
O
N
Br
O N Br ONBr ONBr
O NBr
O N BrONBr No
Reaction
O NBr
O NBr

79. Arrhenius Equation
•k = zpe-Ea/RT = Ae-Ea/RT
• A is called the frequency factor = zp
• k is the rate constant
• ln k = -(Ea/R)(1/T) + ln A
• Another line !!!!
• ln k vs 1/T is a straight line
• With slope Ea/R so we can find Ea
• And intercept ln A

80. Arrhenius Equation- Another line !!!!
ln k = -(Ea/R)(1/T) + ln A
y = m x + b
ln k
1/T
slope = -(Ea/R)

81. Activation Energy and Rates
The final saga

82. Mechanisms and rates
• There is an activation energy for each
elementary step.
• Activation energy determines k.
• k = Ae- (E
a/RT)
• k determines rate
• Slowest step (rate determining) must have
the highest activation energy.

83. • This reaction takes place in three steps

84. 
Ea
First step is fast
Low activation energy

85. Second step is slow
High activation energy
Ea


86. Ea
Third step is fast
Low activation energy


87. Second step is rate determining

88. Intermediates are present

89. Activated Complexes or
Transition States

90. 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.

91. 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.
• Does not change E
• Show up as a reactant in one step and a
product in a later step

92. Pt surface
HH
HH
HH
HH
• Hydrogen bonds to
surface of metal.
• Break H-H bonds
Heterogenous Catalysts

93. Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH

94. Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
• The double bond breaks and bonds to the
catalyst.

95. Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
• The hydrogen atoms bond with the
carbon

96. Pt surface
H
Heterogenous Catalysts
C HH C
HH
H HH

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

98. 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

99. Catalysts and rate.
Concentration of reactants
R
a
t
e
• Rate increases until the active
sites of catalyst are filled.
• Then rate is independent of
concentration