Chemical Kinetics
1
Chemical Kinetics
2
• “It is not enough to understand the stoichiometry
and thermodynamics of a reaction; we also must
understand the factors that govern the rate of the
reaction.”
Chapter 12
• We can use thermodynamics to tell if a
reaction is product or reactant favored.
• But this gives us no info on HOW FAST reaction
goes from reactants to products.
H2O2 decomposition in
an insect
• KINETICS — the study of REACTION RATES
H2O2 decomposition
catalyzed by MnO2
and their relation to the way the reaction proceeds,
i.e., its MECHANISM.
• The reaction mechanism is our goal!
Reaction Rates
3
Determining a Reaction Rate
Blue dye is oxidized
with bleach.
Its concentration
decreases with time.
The rate — the
change in dye conc
with time — can be
determined from the
plot.
• So, if chemical kinetics is the study of
reaction rates, it might be a good idea to
understand what reactions rates are.
• We define the reaction rate as the change
in concentration of a reactant or product
per unit time.
concentration of A at time t 2 − concnetration of A at time t 1 ∆[A]
=
t 2 − t1
∆t
Dye Conc
Rate =
• Where A is the reactant or product being
considered.
Example
4
Time
Example Concentration Data
5
We have the follow set of data. What method of
visualizing data might be useful to use here?
• Using the following reaction we will
explorer the world of reaction rates.
2NO2(g) → 2NO(g) + O2(g)
Page 1
6
2NO2(g) → 2NO(g) + O2(g)
7
• A graph can help
you visualize
what is going on
a lot easier than
looking at a table
of data.
• You know the definition of a rate is as
follows.
• Therefore to determine the average
rate at which the concentration of NO2
changes over the first 50 seconds you
would?
• Using a
combination of
the graph and the
table we can start
determining
some rates.
Rate =
Instantaneous Rate
Rate =
∆[A]
∆t
∆[A] 0.0079 M − 0.0100 M
=
= −4.2 M/sec
∆t
50 sec − 0 sec
We can see that the rate of the reaction is not
constant, but changes with time.
2NO2(g) → 2NO(g) + O2(g)
9
• Sometimes it is important to know what the
rate of a reaction is at a certain time.
• This is known as an instantaneous rate.
• Instantaneous rates can be determined by the
slope of a tangent line at a certain point on the
curvature.
• For example: if we wanted to know the
instantaneous rate at t = 100 sec. of NO2.
10
1. What is the change in y?
•
- 0.0026 M
2. What is the change in x?
•
110 sec.
3. What is the slope?
•
- 0.0026 M / 110 sec.
•
2.4×10-5 M/sec
4. What happened to the
negative sign?
– Draw a tangent line at the point on the curve.
– Determine the slope of that tangent line
•
change in y
Slope =
change in x
2NO2(g) → 2NO(g) + O2(g)
8
Average Rate
11
It is customary to
work with positive
reaction rates.
Rate Laws: Introduction
What else can you
see?
• Fact: All chemical reactions are reversible.
• So let us examine the decomposition of NO2
2NO2(g) → 2NO(g) + O2(g)
{Forward Reaction}
• If all reactions are reversible at some point in time the
reverse reaction will be come important. (Think of a
closed container.)
If you look at the
balanced chemical
reaction and the graph
you should be able to
notice that the rate at
which NO2 is used up is
equal to the rate that
NO is produced.
{Reverse Reaction}
2NO(g) + O2(g) → 2NO2(g)
• As you can probably already see this makes life a little
more complicated. So, we avoid this by studying the
reaction when the reverse reaction makes only a
negligible contribution.
• What is it called when the forward and reverse reaction
rates are equal? (CH13)
Looking at O2 you can
notice that the reaction
rate is half of that of
NO2 or NO.
Page 2
12
Rate Laws: Introduction
Rate Laws: Introduction
13
• At what point in time is the reverse reaction negligible?
• We defined a rate as a change in concentration per unit
time.
• However, which reactant or product concentration do we
choose in defining the rate?
– At a point soon after the reactants are mixed.
• Therefore the reaction rate will only depend on the
concentration of the reactants.
• Rate = k[NO2]n
– We could use any of the three species.
2NO2(g) → 2NO(g) + O2(g)
∆[ NO2 ]
Rate = −
= k[ NO2 ]n
∆T
∆[O2 ]
Rate ' = −
= k '[ NO2 ]n
∆T
Rate = 2 × Rate '
k = rate constant
n = rate order (can be an integer including zero or a fraction)
Important Points
1. The concentration of the products do no appear in the rate law.
Why?
k[ NO2 ]n = 2k '[ NO2 ]
2. The value of the exponent n must be determined by experiment; it
CANNOT be written from the balanced equation.
Rate Laws: Types
•
•
•
•
{Forward Reaction}
Because 2 NO2
molecules are
consumed for every O2
molecule produced.
k = 2× k'
15
Determining the Form of the Rate Law
(What is n?)
There are two types of rate laws.
1. The differential rate law (often called simply the rate law) shows
how the rate of a reaction depends on concentrations.
2. The integrated rate law shows how the concentrations of species in
the reaction depend on time.
•
14
16
• We need to be able to determine more than just the rate
equation.
• We need to be able to determine the value of n and also the
value of k.
• The method we use is called the Method of Initial Rates
• Let us consider the following reaction.
Because we typically consider reactions only under
conditions where the reverse reaction is unimportant, out
rate laws will involve only concentrations of reactants.
Because the differential and integrated rate laws for a
given reaction are related in a well-defined way, the
experimental determination of either of the rates laws is
sufficient.
Experimental convenience usually dictates which type of
rate law is determined experimentally.
Knowing the rate law for a reaction is important mainly
because we can usually infer the individual steps involved
in the reaction from the specific form of the rate law.
NH4+(aq) + NO2-(aq) → N2(g) + 2H2O(l)
17
18
Order
Determination
Rate 1 = 1.35 × 10-7 M ⋅ s = k(0.100 M)n (0.0050 M)m
+
∆[ NH 4 ]
+
−
Rate = −
= k[ NH 4 ]n [ NO2 ]m
∆t
Rate 2 = 2.70 × 10 -7 M ⋅ s = k(0.100 M)n (0.010 M)m
1. Look at the data and determine the differences
• You doubled the concentration of NO2- from experiment
1 to 2, but kept the concentration of NH4+ the same.
• This leads to a doubling of the initial rate.
• You doubled the concentration of NH4+ from experiment
2 to 3, but kept the concentration of NO2- the same.
• This leads to a doubling of the initial rate.
Rate 2 2.70 × 10 -7 M ⋅ s k(0.100 M)n (0.010 M)m
=
= 2.00
=
Rate 1 1.35 × 10 -7 M ⋅ s k(0.100 M)n (0.0050 M)m
∴m = 1
Rate 3 5.40 × 10 -7 M ⋅ s (0.200 M)n
=
=
= 2.00
Rate 2 2.70 × 10 -7 M ⋅ s (0.100 M)n
∴n = 1
Page 3
19
20
Rate Orders and Ratios
How the initial rate
Rate Order
changes when the
concentration is doubled
• Therefore we know the rate law is first order
in terms on NH4+ and NO2-.
Rate = k[NH4+][NO2-]
No change
0
Rate Doubles
1
Rate Quadruples
2
• We can then calculate k from the data.
+
-
Rate = k[NH4 ][NO 2 ]
1.35 × 10-7 M ⋅ sec = k(0.100 M)(0.0050 M)
k = 2.7 × 10-4 M ⋅ sec
21
22
Overall Reaction Order
• Sum of the order of each component in the rate
law.
• rate = k[H2SeO3][H+]2[I−]3
• The overall reaction order is 1 + 2 + 3 = 6.
Determine the rate law and rate constant for the following data
Get it?
Rate = k[BrO3-][Br-][H+]2
OVER-ALLS
8.0 × 10 − 4 M ⋅ sec = k(0.10 M)(0.10 M)(0.10 M) 2
k = 8.0 M − 3 ⋅ sec
Integrated Rate Law
What is the overall reaction order?
First Order Rate Law
23
From the differential format we
can rewrite the equation as follows
• Up until now we have been talking about differential rate
laws.
• It is also useful to be able to express the reactant
concentrations as a function of time.
Rate = -
– The Integrated Rate Law
• To start we will look at reactions involving a single
reactant
−
aA → products
d[A]
= k[A] (first order)
dt
d[A]
= k(dt)
[A]
[A]
−
∆[A]
Rate = −
= k[A]n
∆t
where n = 0, 1, 2 (zero, first, and second order)
t
d[A]
= k(dt)
[A] t ∫=0
[A]0
∫
ln[A] - ln[A]0 = -kt ⇒ ln[A] = −kt + ln[A]0
or
[A] = [A]0 e -kt
Page 4
24
First Order Integrated Rate Law
ln[A] = -kt + ln[A]0
25
Example
2N2O5(g) → 4NO2(g) + O2(g)
• The equation shows how the concentration of A
depends on time.
[N2O5] (M)
0.1000
0.0707
0.0500
0.0250
0.0125
0.00625
– If the initial concentration of A and the rate constant k are known,
the concentration of A at any time can be calculated.
• The equation can also take the form y=mx+b.
–
–
–
–
y=ln[A]
x=t
m=-k
b=ln[A]
1. Verify that the rate
law is first order.
2. Calculate the rate
constant.
• If it is first order we need to be able to plot ln [N2O5] vs.
time
• So, we construct a new table and plot the data.
ln [N2O5]
-2.303
-2.649
-2.996
-3.689
-4.382
-5.075
27
Example
Time (sec)
0
50
100
200
300
400
Time (sec)
0
50
100
200
300
400
• Well, we have no information on the initial rates; so, we
start thinking about the integrated rate laws…
• So, we have discovered a way to test if the
reaction is first order. If we plot ln[A] vs time and
get a straight line we know the reaction is first
order.
ln [N2O5]
-2.303
-2.649
-2.996
-3.689
-4.382
-5.075
26
2N2O5(g) → 4NO2(g) + O2(g)
Time (sec)
0
50
100
200
300
400
Example
28
2N2O5(g) → 4NO2(g) + O2(g)
Since it is a first order reaction,
we know that in the first order
integrated rate law that k is
equal to the -slope of the line.
∆Y ∆(ln[N 2O 5 ])
=
∆X
∆t
- 5.075 - (-2.303) − 2.772
Slope =
=
= −6.93 × 10 − 3 s −1
400 − 0
400
k = −(slope) = 6.93 × 10 − 3 s −1
Slope =
The plot proves it
is a straight line
and so it is a first
order reaction.
Example (1st order Half-Life)
29
Half-Life of a First-Order Reaction
•
We can also determine what is known as a half-life.
•
Half-life is defined as the time required for a reactant
to reach half of its original concentration.
t1/2 =
•
30
A certain first-order reaction has a half-life of
20.0 minutes.
1. Calculate k for the reaction
2. How much time is required for this reaction to be 75 %
complete?
0.693
k
• t1/2 = half-life of the reaction
• k = rate constant
• For a first-order reaction, the half-life does not
depend on concentration.
k=
0.693
t 1/2
k=
0.693
= 3.47 × 10 − 2 min −1
20.0min
If the reaction is 75% complete
that means there is 25% of the
reactant left.
ln[A] = - kt + ln[A]0
If we assume that the orignal
concentration is 1 then the concentration
at 75% completion is 0.25.
ln 0.25 = -kt + ln 1
ln 0.25 - ln 1 = −(3.47 × 10- 2 min -1 )t
- 1.3863 = -(3.47 × 10- 2 min -1 )t
t = 39.95 min ≈ 40 min
See pages 570 – 572 for mathematical proof.
Page 5
Second-Order Rate Law
31
32
Half-Life of a Second-Order Reaction
• We will skip the calculus lesson and just give
you the second-order integrated rate law.
• For aA → products in a second-order reaction,
Rate =
t1/2 =
−∆ A
=k A 2
∆t
o
• t1/2 = half-life of the reaction
• k = rate constant
• Ao = initial concentration of A
• Integrated rate law is
1
1
= kt +
A
Ao
• The half-life is dependent upon the
initial concentration.
Therefore
•
1
kA
A plot of 1/[A] vs. time with produce a straight line
with the slope equal to k
Example
t(s)
0
1000
1800
2800
3600
4400
5200
6200
33
2 C4H6(g) → C8H12(g)
With the following data determine:
1. Is the reaction first of second order?
2. What is the value of the rate constant for the
reaction?
3. What is the half-life for the reaction under the
conditions of this experiment
[C4H6] (M)
0.01000
0.00625
0.00476
0.00370
0.00313
0.00270
0.00241
0.00208
Time (sec)
0
1000
1800
2800
3600
4400
5200
6200
Time (sec)
0
1000
1800
2800
3600
4400
5200
6200
34
Example
From the previous data you
can do some calculations and
obtain the following table.
• From the graphs we will
be able to ascertain what
order the reaction is.
• So, what order is the
reaction?
Example
35
36
• We can do the same thing to calculate the half life or use the secondorder half life equation.
• Since we know that the reaction is second order we can use the
second order integrated rate-law to determine k.
[C4H6] (M)
0.01000
0.00625
0.00476
0.00370
0.00313
0.00270
0.00241
0.00208
ln[C4H4]
-4.605
-5.075
-5.348
-5.599
-5.767
-5.915
-6.028
-6.175
• If we plot this data we will
obtain the following two
graphs.
Since we know that first order
reactions create a straight line
when you plot ln [A] vs. time
and second order reactions
create a straight line when you
plot 1/[A] vs. time. We can
create a new table and graphs.
Example
1/[C4H6]
100
160
210
270
320
370
415
481
1
1
= kt +
[A]0
[A]
1
1
= kt +
[A]
[A]0
1
1
= k(6200) +
0.00208
0.01000
1
1
= (6.14 × 10 −2 M−1 ⋅ sec)(t 1/2 ) +
0.00500
0.01000
481 = k(6200) + 100
100 = (6.14 × 10 −2 M−1 ⋅ sec)(t 1/2 )
381
= k = 6.14 × 10 −2 M−1 ⋅ sec
6200
t 1/2 = 1628.66 sec ≈ 1.63 × 10 3 sec
Page 6
Zero-Order Rate Laws
37
Example
• However, there can be zero-order reactions.
• We can do the same thing to calculate the half life or use the secondorder half life equation.
t 1/2 =
1
k[A]0
t 1/2 =
1
(6.14 × 10 −2 M−1 ⋅ sec)(0.010 00 M)
38
• Most reactions involving a single reactant show either first or
second order kinetics.
• The rate is constant.
• It does not change with concentration.
Rate = k[A]0 = k(1) = k
Example:
[A] = −kt + [A]0
t 1/2 =
t 1/2 = 1628.66 sec ≈ 1.63 × 10 3 sec
The following reaction takes
place on a hot platinum
surface.
[A]0
2k
39
2N2O(g) → 2N2(g) + O2(g)
A Summary
40
• Simplification: Conditions are set such that only
forward reaction is important.
• Two types:
» differential rate law
» integrated rate law
• Which type? Depends on the type of data collected differential and integrated forms can be interconverted.
• Most common: method of initial rates.
• The platinum surface is completely covered with N2O
molecules.
• Concentration v. time: used to determine integrated rate
law, often graphically.
• If we increase the concentration, it has no effect on the rate.
• For several reactants: choose conditions under which
only one reactant varies significantly (pseudo first-order
conditions).
• Since only those N2O molecules on the surface can react.
• An Example of a zero-order reaction.
41
Reaction Mechanisms
42
• The sequence of events at the molecular level that
control the speed and outcome of a reaction.
• A chemical equation does not tell us how reactants
become products - it is a summary of the overall
process.
Br from biomass burning destroys
stratospheric ozone.
(See R.J. Cicerone, Science, volume 263, page 1243, 1994.)
Page 7
Step 1:
Br + O3 ---> BrO + O2
Step 2:
Cl + O3 ---> ClO + O2
Step 3:
BrO + ClO + light ---> Br + Cl + O2
NET:
2 O3 ---> 3 O2
The reaction has many steps in
the reaction mechanism.
43
44
Often Used Terms
• Intermediate: formed in one step and used up in a
subsequent step and so is never seen as a product.
• Molecularity: the number of species that must collide
to produce the reaction indicated by that step.
• Elementary Step: A reaction whose rate law can be
written from its molecularity.
• uni, bi and termolecular
Example
45
46
The sum of the elementary steps must give
the overall balanced equation for the reaction.
NO2(g) + CO(g) → NO(g) + CO2(g)
• Here is a basic reaction.
• Rate = k[NO2]2
• But the mechanism is thought to involve the
following steps.
NO2(g) + NO2(g) → NO3(g) + NO(g)
k1
NO3(g) + CO(g) → NO2(g) + CO2(g)
NO2 (g) + NO2 (g) → NO3 (g) + NO(g)
--------------------------------------------------NO2(g) + CO(g) → NO(g) + CO2(g)
k2
NO3 (g) + CO(g) → NO2 (g) + CO2 (g)
47
48
The mechanism must agree with the
experimentally determined rate law.
Rate-Determining Step
• In a multistep reaction, it is the
slowest step. It therefore
determines the rate of reaction.
NO2(g) + NO2(g) → NO3(g) + NO(g)
slow
NO3(g) + CO(g) → NO2(g) + CO2(g)
fast
Rate of formation of NO3 =
∆[NO3 ]
= k 1[NO2 ]2
∆t
Overall rate = k 1[NO2 ]2
Which agrees with the experimentally
determined rate law.
Page 8
5
6
49
50
Concentration
Collision Model
• What is an easy way to increase the number
of collisions?
• Increase the concentration
• If we have a higher concentration, then we
will have more molecules, which in turn
gives us more collisions per unit time.
• If we have more collisions, then the reaction
rate must increase.
• We will discuss a model that accounts
for the observed characteristics of
reaction rates, the collision model.
• The basic idea of the model is that in
order for molecules to react, they must
collide.
Concentration can affect the
rate of a chemical reaction
7
51
Factors Affecting Rates
• Concentrations
Temperature
•
Rate with 0.3 M HCl
•
Another way to change the
number of collisions is to
change the temperature.
Let us consider temperature of a
gas:
–
Rate with 6.0 M HCl
52
If we increase the temperature:
1. We increase the molecular velocity
2. Which would increase the frequency
of collisions between molecules
3. Which increases the reaction rate
Temperature can affect the
rate of a chemical reaction
8
53
Factors Affecting Rates
Example of Change in Temperature
and Concentration
• Temperature
Bleach at 54 ˚C
54
• As the "temperature" increases, the spheres hit
the edges more frequently.
• When more spheres are added, the interparticle
collisions are more frequent.
Bleach at 22 ˚C
http://chemed.chem.purdue.edu/demos/main_pages/4.4.html
Page 9
55
56
Factors Affecting Rates
Problems with Collision Model
• Physical state of reactants
• It has been shown that the reaction
rate is much smaller than the
calculated collision frequency.
• In other words, only a small fraction of
all collisions produces a reaction.
57
58
Svante Arrhenius
Example of Activation Energy
• Let us look at an example of activation
energy.
• We will use the decomposition of BrNO(g) as
an example:
• In the 1880’s, this problem was first
addressed by Svante Arrhenius.
• He proposed that there is a threshold
energy which must be overcome in
order for the chemical reaction to
take place.
• This threshold energy is called the
activation energy.
Example of Activation Energy
(cont’d)
2BrNO(g) → 2NO(g) + Br2(g)
• In this reaction, two Br-N bonds need to be
broken and one Br-Br bond needs to form.
• The energy required to break a Br-N bond is
considerable (243 kJ/mol).
• This energy must be supplied from
somewhere, but from where?
59
60
Svante Arrhenius
• Based upon the collision model, the energy comes from
the kinetic energy of the molecules.
• Once the molecules collide, the kinetic energy is changed
into potential energy which breaks the bonds and
rearranges the atoms.
Number of collisions with the activation energy =
(total number of collisions) e -Ea/RT
Transition
State
Where do you think
Activation
the
products would
be Energy
if the reaction
was endothermic?
Reactants
• Arrhenius postulated that the number
of collisions having the activation
energy is a fraction of the total number
of collisions.
Because the products
are lower in energy
than the reactants,
this chemical reaction
Products (it
is exothermic
produces heat).
Ea is the activation energy
R is the universal gas constant
T is the temperature in Kelvin
e –Ea/RT is the fraction of collisions with
energy Ea or greater at temperature T
Page 10
Molecular Orientation
61
More Details
• It has to do with molecular orientation.
• When molecules collide, they must be lined up a
certain way for the reaction to proceed.
• Experiments have shown that the
observed reaction rate is considerably
smaller than the rate of collisions with
enough energy to surmount the barrier.
• In other words, there are enough collisions
with enough energy to cause the reaction,
but the reaction does not occur.
• Why (does chemistry have to be so
difficult)?
Line up so that the
two Br atoms can
react with each
other and NO(g)
can be ejected.
2BrNO(g) → 2NO(g) + Br2(g)
Remember this is the
decomposition reaction
Main Points
63
– Concentration (the more molecules, the more collisions).
– Temperature (if we speed up the movement of the molecules, this will
help the reaction to occur and speed it up).
– Molecular orientation (molecules must be properly orientated in order
to react)
– Activation energy (must provide enough energy to get over the hill)
If we consider the Arrhenius Equation, we can easily see
how temperature affects the rate of a chemical reaction.
What other factors could there be?
k = A e -Ea/RT
Can we change the
activation energy?
A is the frequency factor that is made up of the collision
frequency and the steric factor (molecular orientation).
•
–
–
Potential Energy
•
Temperature
can change
Figure 12.18: The exhaust gases from
an automobile engine are passed
through a catalytic converter to
minimize environmental damage.
65
Yes, We Can Change the Activation Energy.
•
64
We have seen and reviewed different factors that affect
the rate of a chemical reaction.
k = A e -Ea/RT
This is done through
the use of a catalyst.
A catalyst is a
substance that speeds
up a reaction without
being consumed.
It does this by
providing a new
reaction pathway that
has a lower activation
energy.
Two different
classifications of
catalysts:
The Br atoms are
not lined up to
allow for the
reaction to
efficiently occur.
Factors So Far that Affect the Rate
of a Chemical Reaction
For a reaction to occur:
• The collision must involve enough energy to
produce the reaction or the collision energy
must be equal or greater than the activation
energy.
• The relative orientation of the reactants must
allow efficient formation of any new bonds
necessary to produce the products.
We can represent these factors as a rate
constant (k) with one equation called the
Arrhenius Equation.
•
62
Reaction Progress
Homogenous
Heterogeneous
Page 11
66
67
Heterogeneous Catalyst
Factors Affecting Rates
68
(hydrogenation of ethylene)
H
Catalysts: catalyzed decomp of H2O2
2 H2O2 --> 2 H2O + O2
H
H
C
H
(g) + H2(g)
C
H
H C
H
H
C H (g)
H
A heterogeneous catalyst is one that exists
in a different phase relative to the
reactants, usually a solid.
Heterogeneous catalytic reactions typically
involves four steps:
1. Adsorption of the reactants
2. Migration of the adsorbed reactants on
the surface
3. Reaction of the adsorbed substances
4. Escape, or desorption, of the products
Homogeneous Catalyst
69
21
The Take HOME Message
• A homogeneous catalyst is one that exists in
the same phase as the reacting molecules.
• An example of this is the breakdown of ozone
(O3).
•
There are five major factors that
influence a chemical reaction.
1.
2.
3.
4.
5.
– Chlorine atoms are released from CCl2F2 (Freon-12)
when it reacts with light.
– These chlorine atoms then go on to break down
ozone.
•
Cl(g) + O3(g) → ClO(g) + O2(g)
O(g) + ClO(g) → Cl(g) + O2(g)
O(g) + O3(g) → 2O2(g)
•
Therefore, the chlorine gas is the homogeneous catalyst.
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Concentration of Reactants
Temperature
Activation Energy
Molecular Orientation
Use of Catalyst
The Arrhenius equation (k = A e –Ea/RT)
describes how some of these factors
affect the reaction rate (k).
A catalyst lowers the activation energy
and is not consumed.
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