Transcript Chemical Kinetics

Chemical Kinetics
The area of chemistry concerned
with the speeds, or rates, at which a
chemical reaction occurs.
1
Reaction Rate
The reaction rate is the change in the
concentration of a reactant or a product
with time, (M/s or M . s-1), where M is
molarity and s represents seconds.
Another way to represent rate is mol . L-1 s-1
2
Factors that Influence Reaction
Rate
Under a given set of conditions, each reaction
has its own characteristic rate, which is
ultimately determined by the chemical nature
of the reactants. (You will remember this
from Chem I - potassium and water have a
different rate of reaction than iron and
oxygen.)
For a given reaction (using the same reactants),
we can control four factors that affect its
rate: the concentration of reactants, their
physical state, the temperature at which the
reaction occurs, and the use of a catalyst.
3
The Collision Theory of Chemical
Kinetics
The kinetic molecular theory of gases postulates that gas
molecules frequently collide with one another.
Therefore, it seems logical to assume – and it is
generally true - that chemical reactions occur as a
result of collisions between reacting molecules.
In terms of the collision theory of chemical kinetics,
then, we expect the rate of a reaction to be directly
proportional to the frequency of the collisions (number
of molecular collisions per second).
Rate =
Number of collisions
s
4
The Collision Theory of Chemical
Kinetics
This simple relationship explains the
dependence of reaction rate on
concentration.
•Increasing the concentration
increases the likelihood that molecules
will collide.
Rate =
Number of collisions
s
5
Concentration
Since molecules must collide in order to
react, the more frequently they collide, the
more often a reaction occurs. Thus,
reaction rate is proportional to the
concentration of reactant
Means “proportional to”
Rate
collision frequency
concentration
Therefore, if we increase the
concentration… we increase the collision
frequency, which…
increases
the rate
6
Physical State
Molecules must mix in order to collide. When
reactants are in the same phase, as in aqueous
solution, occasional stirring keeps them in
contact. When they are in different phases,
more vigorous mixing is needed. The more
finely divided a solid or liquid reactant, the
greater the surface are per unit volume, the
more contact it makes with the other
reactant, and the faster the reaction.
7
Temperature
Molecules must collide in order to react.
Since the speed of a molecule depends
on its temperature, more collisions will
occur if the temperature is increased.
Speed of a molecule
Number of collisions
8
Temperature
Molecules must also collide with enough energy
to react. Increasing the temperature
increases the kinetic energy of the molecules,
which in turn increases the energy of the
collisions.
Therefore, at a higher temperature, more
collisions occur with enough energy to react.
Thus, raising the temperature increases the
reaction rate by increasing the number and
especially the energy of the collisions.
9
Temperature
Two familiar kitchen
appliances employ this
effect: a
refrigerator slows
down chemical
processes that spoil
food, whereas an oven
speeds up other
chemical processes to
cook it.
10
Expressing Reaction Rate
Before we can deal quantitatively with the effects of
concentration and temperature on reaction rate, we
must be able to express the rate mathematically. A
rate is a change in some variable per unit of time.
For example, the rate of motion of a car is the change
of position of the car divided by time. A car that
travels 57 miles in 60. minutes is traveling at…
57 miles/60. minutes = .95 miles/min
In the case of chemical reactions, the
positions of the substances do not change
over time, but their concentrations do.
11
We know that any reaction can be
represented by the general equation
reactants  products
This equation tells us that during the
course of a reaction, reactants are
consumed while products are formed.
As a result, we can follow the progress
of a reaction by monitoring either the
decrease in concentration of the
reactants or the increase in
concentration of the products.
12
The following figure shows the progress
of a simple reaction in which A
molecules are converted to B
molecules:
A  B
B
A
13
The decrease in number of A molecules and the
increase in the number of B molecules with
time are shown below.
14
In general, it is more convenient to express the
reaction rate in terms of the change in
concentration with time. Thus, for the
reaction A  B we can express the rate as:
Rate = -
D[A]
[A]final – [A]initial
Dt
Because the concentration of A decreases
during the time interval, D[A] is a negative
quantity. The rate of a reaction is a positive
quantity, so a minus sign is needed in the rate
expression to make the rate positive.
15
or
Rate =
D[B]
Dt
The rate of product formation does not require
a minus sign because D[B] ([B]final – [B]initial) is a
positive quantity (the concentration increases
with time).
16
Rate = -
D[A]
Dt
or
Rate =
D[B]
Dt
Where D[A] and D[B] are the changes in
concentration (molarity) over a time
period Dt.
These rates are average rates because
they are averaged over a certain time
period (Dt).
17
Reaction Rates and Stoichiometry
We have seen that for stoichiometrically simple
reactions of the type A  B, the rate can
either be expressed in terms of the decrease
in reactant concentration with time, -D[A]/Dt,
or the increase in product concentration with
time, D[B]/Dt.
For more complex reactions, we must be careful
in writing the rate expressions.
Consider the reaction
2A  B
Two moles of A
disappear for
each mole of B
that forms 18
Another way to think of this is to say
that the rate of disappearance of A is
twice as fast as the rate of appearance
of B. We write the rate as either
Rate = -
1 D[A]
2 Dt
or
Rate =
D[B]
Dt
For the reaction
2A  B
19
In general, for the reaction
aA + bB  cC + dD
The rate is given by
Rate = -
1 D[A]
a Dt
= -
1 D[B]
b Dt
=
1 D[C]
c Dt
=
1 D[D]
d Dt
20
Write the expression for the following
reactions in terms of the disappearance
of the reactants and the appearance of
the products:
3O2(g)  2O3(g)
Rate = -
1 D[O2]
3 Dt
=
1 D[O3]
2 Dt
21
If O2 is disappearing at .25 M/s, what is
the rate of formation of O3?
-
1 D[O2]
3 Dt
1 -.25M
3
s
=
=
.50Ms = 3(x)s
1 D[O3]
2 Dt
1
x
2
s
x = .17M
22
By definition, we know that to determine
the rate of a reaction we have to
monitor the concentration of the
reactant (or product) as a function of
time.
• For reactions in solution, the concentration of a
species can often be measured by spectroscopic
means.
• If ions are involved, the change in concentration can
also be detected by an electrical conductance
measurement.
• Reactions involving gases are most conveniently
followed by pressure measurements.
23
Reaction of Molecular Bromine
and Formic Acid
In aqueous solutions, molecular bromine reacts
with formic acid (HCOOH) as follows:
Br2(aq) + HCOOH(aq)  2Br-(aq) + 2H+(aq) + CO2(g)
Reddishbrown
colorless
colorless
colorless
colorless
The rate of Br2 disappearance can be determined by
monitoring the color over time.
As the reaction proceeds, the color of the solution …
goes from brown to colorless
24
Reaction of Molecular Bromine
and Formic Acid
Br2(aq) + HCOOH(aq)
2Br-(aq) + 2H+(aq) + CO2(g)
As the reaction proceeded, the
concentration of Br2 steadily decreased
and the color of the solution faded.
25
Measuring the change (decrease) in bromine
concentration at some initial time ([Br2]0) and
then at some other time, ([Br2]t) allows us to
determine the average rate of the reaction
during that interval:
Average rate =
Average rate =
D[Br2]
Dt
[Br2]t – [Br2]0
tfinal – tinitial
26
Use the data in the following table to
calculate the average rate over the
first 50 second time interval.
Time
(s)
[Br2]
(M)
Rate
(M/s)
0.0
0.0120
4.20 x 10-5
50.0
0.0101
3.52 x 10-5
100.0
0.00846
2.96 x 10-5
150.0
0.00710
2.49 x 10-5
200.0
0.00596
2.09 x 10-5
250.0
0.00500
1.75 x 10-5
300.0
0.00420
1.48 x 10-5
350.0
0.00353
1.23 x 10-5
400.0
0.00296
1.04 x 10-5
27
Average rate =
(0.0101 – 0.0120)M
(50.0 – 0.0)s
= 3.80 x 10-5 M/s
Time
(s)
[Br2]
(M)
Rate
(M/s)
0.0
0.0120
4.20 x 10-5
50.0
0.0101
3.52 x 10-5
100.0
0.00846
2.96 x 10-5
150.0
0.00710
2.49 x 10-5
200.0
0.00596
2.09 x 10-5
250.0
0.00500
1.75 x 10-5
300.0
0.00420
1.48 x 10-5
350.0
0.00353
1.23 x 10-5
400.0
0.00296
1.04 x 10-5
28
Now use the data in the same table to
calculate the average rate over the
first 100 second time interval.
Time
(s)
[Br2]
(M)
Rate
(M/s)
0.0
0.0120
4.20 x 10-5
50.0
0.0101
3.52 x 10-5
100.0
0.00846
2.96 x 10-5
150.0
0.00710
2.49 x 10-5
200.0
0.00596
2.09 x 10-5
250.0
0.00500
1.75 x 10-5
300.0
0.00420
1.48 x 10-5
350.0
0.00353
1.23 x 10-5
400.0
0.00296
1.04 x 10-5
29
Average rate =
(0.00846 – 0.0120)M
= 3.54 x 10-5 M/s
(100.0 – 0.0)s
Time
(s)
[Br2]
(M)
Rate
(M/s)
0.0
0.0120
4.20 x 10-5
50.0
0.0101
3.52 x 10-5
100.0
0.00846
2.96 x 10-5
150.0
0.00710
2.49 x 10-5
200.0
0.00596
2.09 x 10-5
250.0
0.00500
1.75 x 10-5
300.0
0.00420
1.48 x 10-5
350.0
0.00353
1.23 x 10-5
400.0
0.00296
1.04 x 10-5
30
These calculations demonstrate that the
average rate of the reaction depends on
the time interval we choose.
By calculating the average reaction rate
over shorter and shorter intervals, we
can obtain the rate for a specific
instant in time, which gives us the
instantaneous rate of the reaction at
that time.
31
The figure below shows the plot of [Br2] versus time, based on
the data table given previously. Graphically, the instantaneous
rate at 100 seconds after the start of the reaction is the slope
of the line tangent to the curve at that instant.
Unless otherwise stated, we will refer to the instantaneous rate
as simply “the rate”.
The instantaneous rate at any other time can be determined in a
similar manner.
32
At a specific
temperature, a rate
constant (k) is a
constant of
proportionality
between the reaction
rate and the
concentrations of
reactants.
k, the Rate Constant
rate
[Br2]
Means “proportional to”
rate = k[Br2]
k is specific for a given reaction at a
given temperature; it does not change as
the reaction proceeds.
33
Rearrange the equation
rate = k[Br2]
To solve for k
Since reaction
rate
has
the
units
Rate
M/s, and [Br2] is in
k=
[Br2]
M, the unit of k
for this first
order reaction is
1/s or s-1.
34
Calculate the rate constant for the following reaction
Br2(aq) + HCOOH(aq)  2Br-(aq) + 2H+(aq) + CO2(g)
Time
(s)
[Br2]
(M)
Rate
(M/s)
0.0
0.0120
4.20 x 10-5
50.0
0.0101
3.52 x 10-5
100.0
0.00846
2.96 x 10-5
150.0
0.00710
2.49 x 10-5
200.0
0.00596
2.09 x 10-5
250.0
0.00500
1.75 x 10-5
300.0
0.00420
1.48 x 10-5
350.0
0.00353
1.23 x 10-5
400.0
0.00296
1.04 x 10-5
k = rate/[Br2]
(s-1)
35
Because k is a constant (for this reaction at this specific
temperature), it doesn’t matter which row we consider, so
let’s consider the data at time 0.0 seconds…
Time
(s)
[Br2]
(M)
Rate
(M/s)
k = rate/[Br2]
(s-1)
0.0
0.0120
4.20 x 10-5
3.50 x 10-3
50.0
0.0101
3.52 x 10-5
100.0
0.00846
2.96 x 10-5
150.0
0.00710
2.49 x 10-5
200.0
0.00596
2.09 x 10-5
250.0
0.00500
1.75 x 10-5
300.0
0.00420
1.48 x 10-5
350.0
0.00353
1.23 x 10-5
400.0
0.00296
1.04 x 10-5
rate = k[Br2]
k=
4.20 x 10-5
0.0120
= 3.50 x 10-3
36
To prove that k is a constant, calculate k at time 200.0
seconds
Time
(s)
[Br2]
(M)
Rate
(M/s)
0.0
0.0120
4.20 x 10-5
50.0
0.0101
3.52 x 10-5
100.0
0.00846
2.96 x 10-5
150.0
0.00710
2.49 x 10-5
200.0
0.00596
2.09 x 10-5
250.0
0.00500
1.75 x 10-5
300.0
0.00420
1.48 x 10-5
350.0
0.00353
1.23 x 10-5
400.0
0.00296
1.04 x 10-5
k = rate/[Br2]
(s-1)
3.50 x 10-3
3.51 x 10-3
The slight variations in the values of k are due to
experimental deviations in rate measurements.
37
Filling in the rest of the table…
Time
(s)
[Br2]
(M)
Rate
(M/s)
k = rate/[Br2]
(s-1)
0.0
0.0120
4.20 x 10-5
3.50 x 10-3
50.0
0.0101
3.52 x 10-5
3.49 x 10-3
100.0
0.00846
2.96 x 10-5
3.50 x 10-3
150.0
0.00710
2.49 x 10-5
200.0
0.00596
2.09 x 10-5
3.51 x 10-3
3.51 x 10-3
250.0
0.00500
1.75 x 10-5
3.50 x 10-3
300.0
0.00420
1.48 x 10-5
3.52 x 10-3
350.0
0.00353
1.23 x 10-5
3.48 x 10-3
400.0
0.00296
1.04 x 10-5
3.51 x 10-3
38
Units of the Rate Constant k for
Several Overall Reaction Orders
Overall
Reaction
Order
0
Units of k (when t is seconds)
1
1/s (or s-1)
2
L/mol . s (or L mol-1 s-1)
3
L2/mol2 . s (or L2 mol-2 s-1)
mol/L . s (or mol L-1 s-1)
39
It is important to understand that k is NOT
affected by the concentration of Br2.
• The rate is greater at a higher
concentration and smaller at a lower
concentration of Br2, but the ratio of
rate/[Br2] remains the same provided the
temperature doesn’t change.
40
The Rate Law
The rate law expresses the relationship of the rate of a reaction to
the rate constant (k) and the concentrations of the reactants
raised to a power. For the general reaction
aA + bB  cC + dD
The rate law takes the form
Rate = k[A]x[B]y
Where x and y are numbers that must be determined
experimentally.
Note – in general, x and y are NOT equal to the stoichiometric
coefficients a and b from the overall balanced chemical
equation. When we know the values of x, y and k, we can use the
rate equation shown above to calculate the rate of the reaction,
given the concentrations of A and B.
41
Rate = k[A]x[B]y
The reaction orders define how the rate is
affected by the concentration of each
reactant.
This reaction is xth order in A, yth order in B.
42
The following rate law was determined for
the formation of nitrogen (VI) oxide and
molecular oxygen from nitrogen (IV)
oxide and ozone
Rate = k[NO2][O3]
How would the rate of this reaction be
affected if the concentration of NO2
increased from 1.0 M to 2.0 M?
43
Rate = k[NO2][O3]
This reaction is first order with respect to
both NO2 and O3. This means that doubling
the concentration of either reactant would
double the rate of the reaction.
(2)1 = 2
How many times
greater the
concentration is
How many times
greater the rate of the
reaction will be
What the order is for
that reactant
44
The reaction between nitrogen monoxide
and molecular oxygen is described by a
different rate law.
Rate = k [NO]2[O2]
How would the rate of this reaction be
affected if the concentration of NO
increased from 1.0 M to 2.0 M?
45
Rate = k[NO]2[O2]
This reaction is second order with
respect to NO and first order with
respect to O2.
Doubling the concentration NO would
increase the rate by a factor of four
(2)2 = 4 How many times
How many times
greater the
concentration is
greater the rate of the
reaction will be
What the order is for
that reactant
46
Rate = k [NO]2[O2]
How would the rate of this reaction be
affected if the concentration of NO
increased from 1.0 M to 5.0 M?
In this case, when the concentration of
NO is multiplied by 5, the rate increases
by a factor of 25!
(5)2 = 25
47
The exponents x and y specify the relationships
between the concentrations of reactants A
and B and the reaction rate. Added together,
they give us the overall reaction order,
defined as the sum of the powers to which all
reactant concentrations appearing in the rate
law are raised. For the equation
Rate = k[A]x[B]y
The overall reaction order is x + y.
48
For the following reaction
(CH3)3CBr(l) + H2O(l)  (CH3)3COH(l) + HBr(aq)
The rate law has been found to be
rate = k[(CH3)3CBr]
This reaction is first order in 2-bromo-2-methylpropane.
Note that the concentration of H2O does not even appear in the rate
law. Thus, the reaction is zero order with respect to H2O. This
means that the rate does not depend on the concentration of H2O;
we could also write the rate law for this reaction as
rate = k[(CH3)3CBr][H2O]0
What is the overall order of this reaction?
1st order overall (1+0=1)
49
Reaction orders are usually positive
integers or zero, but they can also be
fractional or negative. In the reaction
CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g)
A fractional order appears in the rate law:
rate = k[CHCl3][Cl2]1/2
This order means that the reaction depends on
the square root of the Cl2 concentration. If
the initial Cl2 concentration is increased by a
factor of 4, for example, the rate increases by
V4 (= 2), therefore the rate would double.
50
A negative exponent means that the reaction rate
decreases when the concentration of that
component increases. Negative orders are often
seen for reactions whose rate laws include
products. For example, in the atmospheric
reaction
2O3(g)
3O2(g)
The rate law has been shown to be
Rate = k[O3]2[O2]-1 ; or
[O3]2
rate = k [O ]
2
If the [O2] doubles, the reaction proceeds
half as fast.
51
To see how to determine the rate law of a
reaction, let us consider the reaction
between fluorine and chlorine dioxide:
F2(g) + 2ClO2(g)  2FClO2(g)
52
One way to study the effect of reactant
concentration on reaction rate is to determine
how the initial rate depends on the starting
concentrations. It is preferable to measure
the initial rates because as the reaction
proceeds, the concentrations of the reactants
decrease and it may become difficult to
measure the changes accurately. Also, as the
reaction continues, the product concentrations
increase,
products  reactants
so the reverse reaction becomes increasingly
likely. Both of these complications are virtually
absent during the earliest stages of the
reaction.
53
The following table shows three rate
measurements for the formation of
FClO2.
[F2]0
(M)
0.10
[ClO2]0
(M)
0.010
Initial Rate
(M/s)
1.2 x 10-3
0.10
0.040
4.8 x 10-3
0.20
0.010
2.4 x 10-3
54
Looking at entries 1 and 3, we see that as we
double [F2]0 while holding [ClO2]0 constant,
the reaction rate doubles. Thus the rate is
directly proportional to [F2], and the
reaction is first order with respect to F2.
[F2]0
(M)
0.10
[ClO2]0
(M)
0.010
Initial Rate
(M/s)
1.2 x 10-3
0.10
0.040
4.8 x 10-3
0.20
0.010
2.4 x 10-3
55
Similarly, the data in entries 1 and 2 show that as
we quadruple [ClO2] while holding [F2] constant,
the rate increases by four times, so the rate is
also directly proportional to [ClO2], making the
reaction 1st order with respect to [ClO2]
[F2]
(M)
0.10
[ClO2]
(M)
0.010
Initial Rate
(M/s)
1.2 x 10-3
0.10
0.040
4.8 x 10-3
0.20
0.010
2.4 x 10-3
56
We can summarize our observations by
writing the rate law as
Rate = k[F2][ClO2]
Because both [F2] and [ClO2] are raised to
the first power, the reaction is second
order overall.
57
What is the rate constant for this
reaction at this temperature?
58
From the reactant concentrations and the
initial rate, we can calculate the rate
constant. Using the data from the
previous table we can write
Rate = k[F2][ClO2]
k=
rate
[F2][ClO2]
1.2 x 10-3 M/s
(0.10M)(0.010M)
= 1.2 L/mol.s
59
The reaction of nitric oxide with hydrogen at
1280 oC is
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
From the following data that was collected experimentally
at this temperature, determine the rate law and calculate
the rate constant.
Experiment
[NO]
[H2]
Initial Rate (M/s)
1
5.0 x 10-3
2.0 x 10-3
1.3 x 10-5
2
10.0 x 10-3
2.0 x 10-3
5.0 x 10-5
3
10.0 x 10-3
4.0 x 10-3
10.0 x 10-5
60
We assume that the rate law will take the form
Rate = k[NO]x[H2]y
Experiments 1 and 2 show that when we double the concentration
of NO at constant concentration of H2, the rate quadruples. Thus
the reaction is second order in NO. Experiments 2 and 3 indicate
that doubling [H2] at constant [NO] doubles the rate; the reaction
is first order in H2. The rate law is therefore given by
Rate = k[NO]2[H2]
Which shows that it is a (1 + 2) or third-order reaction overall.
Experiment
[NO]
[H2]
Initial Rate (M/s)
1
5.0 x 10-3
2.0 x 10-3
1.3 x 10-5
2
10.0 x 10-3
2.0 x 10-3
5.0 x 10-5
3
10.0 x 10-3
4.0 x 10-3
10.0 x 10-5
61
The rate constant can be calculated using the data from any one of
the experiments. Since
k=
rate
[NO]2[H2]
k=
5.0 x 10-5 M/s
(10.0 x 10-3 M)2(2.0 x 10 -3M)
k = 2.5 x 102/ M2 . s
Experiment
[NO]
[H2]
Initial Rate (M/s)
1
5.0 x 10-3
2.0 x 10-3
1.3 x 10-5
2
10.0 x 10-3
2.0 x 10-3
5.0 x 10-5
3
10.0 x 10-3
4.0 x 10-3
10.0 x 10-5
62
The following points summarize our
discussion of the rate law:
• Rate laws are ALWAYS determined
experimentally.
• Reaction order should be defined in
terms of reactant (not product)
concentrations.
• The order of a reactant is not related
to the stoichiometric coefficient of the
reactant in the overall balanced
equation.
63
Relationship Between Reactant
Concentration and Time
Rate law expressions enable us to
calculate the rate of a reaction from
the rate constant and reactant
concentrations.
The rate laws can also be used to
determine the concentrations of
reactants any time during the course of
a reaction.
64
First-order reactions
A first order reaction is a reaction whose rate depends
on the reactant concentration raised to the first
power.
rate = k[A]
In a first-order reaction of the type
A  product
Rate can be expressed as
D[A]
rate = - Dt
65
Through the methods of calculus, this expression
is integrated over time to obtain the
integrated rate law for a first-order reaction:
ln
Since
[A]0
[A]t
ln
= kt
[A]0
[A]t
= ln[A]0 – ln[A]t
We can write the integrated rate
law for a first order reaction as
ln[A]0 – ln[A]t = kt
66
Cyclobutane decomposes at 1000 oC to two moles of ethylene
-1.
(C2H4) with a very high rate constant, 87 s-1
If the initial concentration of cyclobutane is 2.00 M, what is
the concentration after 0.010 s?
First order reaction, therefore kt = ln[C4H8]0 – ln[C4H8]t
Rearranging the equation so that
ln [C4H8]t can be solved for
ln [C4H8]t = ln[C4H8]0 - kt
Substituting the data
ln [C4H8]t = ln(2.00 mol/L) - (87 s-1)(0.010 s)
ln [C4H8]t = 0.69 – 0.87 = -0.18
Taking the antilog (inverse log) of both sides
[C4H8]t = 0.010 s = e-0.18 = 0.84 mol/L C4H8
67
What percent of the cyclobutane has
decomposed in this time?
2.00 M - .84 M = 1.16 M
1.16 M/2.00 M x 100 = 58% has
decomposed (42 % remains)
68
Reaction half-life (t1/2)
The t1/2 of a reaction is the time required
to reach half the initial reactant
concentration.
For a first order reaction, the formula
for determining t1/2 is
Note – t1/2 of a
t1/2 =
ln 2
k
= 0.693
k
first order
reaction is
constant, it is
independent of
reactant
concentration! 69
A plot of [N2O5] vs time for three half-lives.
0.060
[N2O5]
0.050
0.040
0.030
0.020
0.010
0.000
0
24
Time (min)
48
72
70
Determining t1/2 for a first-order
reaction
Cyclopropane is the smallest cyclic
hydrocarbon. Because its 60o bond angles
allow only poor orbital overlap, its bonds
are weak. As a result, it is thermally
unstable and rearranges to propene at 100
oC via the following reaction:
CH2
H2C
CH2
 CH3 – CH = CH2
71
The rate constant is 9.2 s-1. How long does
it take for the initial concentration of
cyclopropane to decrease by one-half?
1st order reaction.
The label tells you this is a _____
t1/2 =
ln 2
k
= 0.693 = 0.693-1 = 0.075 s
k
9.2 s
It takes 0.075 s for half the
cyclopropane to form propene.
72
Second-order reactions
For a general second-order rate equation, the
expression including time can become quite
complex, so let’s consider only the simplest case,
one in which the rate law contains only one
reactant
2A  product
The integrated rate law for a second-order
reaction involving one reactant:
1 - 1
= kt
[A]t [A]0
73
At 25 oC, hydrogen iodide breaks down
very slowly to hydrogen and iodine
according to the following:
rate = k[HI]2
and k = 2.4 x 10-21 L/mol . s
If 0.0100 mol HI(g) is placed in a 1.0 L
container, how long will it take for the
concentration of HI to reach 0.00900 mol/L?
74
rate = k[HI]2
and k = 2.4 x 10-21 L/mol . s
1 - 1
= kt
[A]t [A]0
1
1
0.00900
0.0100
= (2.4 x 10-21 )(t)
111 – 100 = (2.4 x 10-21)(t)
11/ 2.4 x 10-21 = t = 4.6 x 1021 seconds
75
In contrast to the half-life of a firstorder reaction, the half-life of a
second-order reaction DOES depend on
reactant concentration:
t1/2 =
1
k[A]0
76
In the simple decomposition reaction
AB(g)  A(g) + B(g)
rate = k[AB]2 and k = 0.20 L/mol . s
How long will it take for [AB] to reach half
of its initial concentration of 1.50 M?
77
t1/2 =
1
k[A]0
t1/2 =
1
(0.20)(1.50)
t1/2 = 3.3 seconds
78
Determining the Reaction Order
from the Integrated Rate Law
Suppose you don’t know the rate law for a
reaction and don’t have the initial rate
data needed to determine the reaction
orders. Another method for finding
reaction orders is a graphical technique
that uses concentration-time data
directly.
79
To find the reaction order from the
concentration-time data, some trialand-error graphical plotting is required:
• If you obtain a straight line when you
plot [reactant] vs. time, the reaction is
zero order with respect to that
reactant.
80
Integrated rate laws
and reaction order.
A plot of ln[A] vs t
gives a straight line
for a reaction that is
first order in A
A plot of 1/[A] vs t
gives a straight
line for a reaction
that is second
order in A
81
Graphical determination of the reaction
order for the decomposition of N2O5.
A plot of 1/[N2O5] vs t
is curved, indicating
that the reaction IS
NOT second order in
N2O5.
Time
[N2O5]
ln[N2O5] 1/[N2O5]
0
0.0165
-4.104
60.6
10
0.0124
-4.390
80.6
20
0.0093
-4.68
1.1 x 102
30
0.0071
-4.95
1.4 x 102
40
0.0053
-5.24
1.9 x 102
50
0.0039
-5.55
2.6 x 102
60
0.0029
-5.84
3.4 x 102
82
Graphical determination of the reaction
order for the decomposition of N2O5.
A plot of ln[N2O5] vs t gives
a straight line, indicating the
reaction is first order in
N2O5
Time
[N2O5]
ln[N2O5] 1/[N2O5]
0
0.0165
-4.104
60.6
10
0.0124
-4.390
80.6
20
0.0093
-4.68
1.1 x 102
30
0.0071
-4.95
1.4 x 102
40
0.0053
-5.24
1.9 x 102
50
0.0039
-5.55
2.6 x 102
60
0.0029
-5.84
3.4 x 102
83
Activation Energy
According to Collision Theory of Chemical Kinetics,
molecules must collide to react. However, not all
collisions lead to reactions. Energetically
speaking, there is some minimum collision energy
below which no reaction occurs. Any molecule in
motion possesses kinetic energy; the faster it
moves, the greater its kinetic energy. When
molecules collide, part of their kinetic energy is
converted to vibrational energy. If the initial
kinetic energies are large enough, the colliding
molecules will vibrate so strongly that some of
the chemical bonds will break. This bond
fracture is the first step toward product
formation. If the initial kinetic energies are too
small, the molecules will merely bounce off each
other intact, and no change results from the
collision.
84
We postulate that in order to react, the
colliding molecules must have a total
kinetic energy equal to or greater than
the activation energy, (Ea), which is
defined as the minimum amount of
energy required to initiate a chemical
reaction.
85
Molecules must also be oriented in a
favorable position – one that allows the
bonds to break and atoms to rearrange.
86
NO(g) + NO3(g)  2NO2(g)
The picture to the right
shows a few of the
possible collision
orientations for this
simple gaseous reaction:
Of the five collisions shown, only one has an orientation in which the N
of NO collides with an O of NO3. Actually, the probability factor (p)
for this reaction is 0.006; only 6 collisions in every thousand have an
orientation that leads to a reaction.
87
Collisions between individual atoms have p values
near 1: almost no matter how they hit, they
react. In such cases, the rate constant
depends only on the frequency and energy of
the collisions.
At the other extreme are biochemical
reactions, in which the reactants are often
two small molecules that can react only when
they collide with a specific tiny region of a
giant molecule-a protein or nucleic acid. The
orientation factor for such reactions is often
less than 10-6: fewer than one in a million
sufficiently energetic collisions leads to
product formation.
88
When reactants collide at the proper angle with energy
equal to the activation energy (Ea), they undergo an
extremely brief interval of bond disruption and bond
formation called a transition state.
Endothermic/Exothermic (choose one)
During this transition
state, the reactants form
a short-lived complex that
is neither reactant nor
product, but has partial
bonding characteristics of
both. This transitional
structure is called an
activated complex.
89
An activated complex is a
highly unstable species with a
high potential energy. (It
was energized by the particle
collision.) Once formed, it
will break up almost
immediately.
Activation energy is the
energy required to
achieve the transition
state and form the
activated complex
The activated complex exists
along the reaction pathway at
the point where the energy is
greatest – at the peak
indicated by the activation
energy.
90
We can think of activation energy, (Ea), as a barrier that
prevents less energetic molecules from reacting…
…because only the molecules who have enough kinetic
energy to exceed the activation energy can take part in
the reaction.
91
Maxwell Boltzmann Diagram
Because the number of reactant molecules in
an ordinary reaction is very large, the speeds,
and hence also the kinetic energies of the
molecules, vary greatly.
92
Normally, only a small fraction of the colliding
molecules -- the fastest-moving ones – have enough
kinetic energy to exceed the activation energy.
At higher temperature, more molecules can surpass
the activation energy, therefore, the rate of
product formation is greater at the higher
temperature.
As temperature
increases, the
probability of finding
molecules at higher
temperature increases.
93
With very few exceptions, reaction rates
increase with increasing temperature.
As a general rule of thumb, you can expect a
10 oC increase in temperature to result in
a doubling of the reaction rate.
94
Arrhenius Equation
The dependence of the rate constant of a reaction on
temperature can be expressed by the following equation,
known as the Arrhenius equation:
k = Ae-Ea/RT
Where Ea is the activation energy (in J/mol), R the gas
constant (8.314 J/K . mol), T the absolute temperature, and
e the base of the natural logarithm scale. The quantity A
represents the collision frequency and is called the
frequency factor.
95
Frequency Factor (A)
The frequency factor is the product of the
collision frequency (Z) and an orientation
factor (p) which is specific for each reaction.
The factor p is related to the structural
complexity of the colliding particles. You can
think of it as the ratio of effectively oriented
collisions to all possible collisions.
In the activation energy problems we are solving,
the actual value of A need not be known
because A can be treated as a constant for a
given reacting system over a fairly wide
temperature range.
96
k = Ae-Ea/RT
• As the activation energy increases, k
decreases,
and as k decreases, rate decreases
•As the temperature increases, k
increases
• And as k increases, rate
increases
97
You can derive the following equation
from the Arrhenius equation:
ln k =
-Ea
R
1
T
+ ln A
98
ln k =
Y=
-Ea
R
mx
m
1
T
+ ln A
+
b
Thus, a plot of ln k versus 1/T gives a
straight line whose
•slope
slope (m) is equal to –Ea/R and whose
•intercept (b) with the y-axis is ln A.
99
Given k and temperature, you can use your
graphing calculator to determine the
activation energy of a reaction.
See the following example…
100
The rate constants for the decomposition
of acetaldehyde
CH3CHO(g)  CH4(g) + CO(g)
were measured at five different
temperatures. The data are shown in the
following table.
101
Plot ln k versus 1/T to determine the activation
energy (-slope) in kJ/mol for the previous reaction.
T (K)
k
700
0.011
730
0.035
760
0.105
790
0.343
810
0.789
To determine the activation energy, we
need to graph 1/T on the x axis and ln k
on the y-axis.
102
ln k
1/T
1.43 x 10-3
-4.51
1.37 x 10-3
-3.35
1.32 x 10-3
-2.254
1.27 x 10-3
-1.070
1.23 x 10-3
0.237
Graph these
values using your
graphing
calculator.
Slope = -Ea/R
Ea = -Slope . R
Ea = -(-2.16 x 104 K)(8.314 J/K . Mol)
Ea = 1.80 x 105 J/mol
Ea = 1.80 x 102 kJ/mol
103
The slope of the line can be
determined using linear
regression.
104
An equation relating the rate constants k1
and k2 at temperature T1 and T2 can be
used to calculate the activation energy
or to find the rate constant at another
temperature if the activation energy is
known.
k2
Ea 1 1
ln
=
k1
R T2 T1
105
The rate constant of a first-order reaction is 3.46 x
10-2 s-1 at 298 K. What is the rate constant at 350 K
if the activation energy for the reaction is 50.2
kJ/mol?
k2 =
k1 = 3.46 x 10-2s-1
T1 = 298
k2
ln k =
1
Ea 1
R T
2
T2 = 350
1
T1
k2
50,200 J/mol
1
1
3.46 x 10-2= 8.314 J/K . mol
350 298
k2
ln
3.46 x 10-2 = 3.01 Taking the antilog of both sides
k2
-1
3.01 = 20.3
k
=
0.702
s
2
=
e
3.46 x 10-2
ln
106
Reaction Mechanisms
As we mentioned earlier, an overall balanced
chemical equation does not tell us much about
how a reaction actually takes place. In many
cases, it merely represents the sum of
several elementary steps, or elementary
reactions, that represent the progress of the
overall reaction at the molecular level. The
term for the sequence of elementary steps
that leads to product formation is reaction
mechanism.
The reaction mechanism is comparable to the
route of travel followed during a trip; the
overall chemical equation specifies only the
origin and the destination.
107
As an example of a reaction mechanism,
let us consider the reaction between
nitrogen monoxide and oxygen:
2NO(g) + O2(g)  2NO2(g)
We know that the products are not
formed directly from the collision
of two NO molecules with an O2
molecule because N2O2 is detected
during the course of the reaction.
108
Let us assume that the reaction actually takes
place via two elementary steps as follows:
Elementary Step 1
NO(g) + NO(g)  N2O2(g)
Elementary Step 2
N2O2(g) + O2(g)  2NO2(g)
Overall reaction
2NO + N2O2 + O2  N2O2 + 2NO2
Each of the elementary steps listed above is called a bimolecular
reaction because each step involves two reactant molecules. A step
that just involves one reactant molecule is a unimolecular reaction.
Very few termolecular reactions, reactions that involve the
participation of three reactant molecules in one elementary step,
are known, because the simultaneous encounter of three molecules
is a far less likely event than a bimolecular collision. (There are no
known examples of reactions involving the simultaneous encounter of
four molecules.)
109
Elementary Step 1
NO(g) + NO(g)  N2O2(g)
Elementary Step 2
N2O2(g) + O2(g)  2NO2(g)
Overall reaction
2NO + N2O2 + O2  N2O2 + 2NO2
Species such as N2O2 are called intermediates because
they appear in the mechanism of the reaction (that is,
in the elementary steps) but not in the overall balanced
equation. Keep in mind that an intermediate is always
formed in an early elementary step and consumed in a
later elementary step. Note – an intermediate ≠
activated complex!
110
You can propose a mechanism for a reaction if
you consider…
• The elementary steps in a multistep reaction
mechanism must always add to give the
balanced chemical equation of the overall
process. (Any intermediates that are formed
in earlier steps must be consumed in later
steps.)
• Unimolecular and bimolecular reactions are
more common than termolecular reactions.
• The rate of the overall reaction is limited by
the rate of the slowest elementary step, (For
that reason, the slowest elementary step is
typically called the rate-determining step.
111
Gaseous nitrogen monoxide reacts with fluorine gas to
produce nitrogen hypofluorite (NOF). The
intermediate product NOF2(g) has been isolated as an
intermediate in this reaction. Propose a two step
mechanism consistent with this intermediate product.
Step 1: NO(g) + F2(g) NOF2(g)
Step 2: NOF2(g) + NO(g)  2NOF(g)
•The elementary steps add to give the overall
balanced chemical equation
•The first and the second step are bimolecular
112
Even when a proposed mechanism is
consistent with the rate law, later
experimentation may show it to be
incorrect or only one of several
alternatives.
113
Knowing the elementary steps of a reaction enables us to
propose a rate law.
Suppose we have the following elementary step:
A  products
Because there is only one reactant molecule present, this is
a/n ___molecular
reaction. It follows that the larger
uni
the number of A molecules present, the faster the rate
of product formation.
Thus the rate of a unimolecular reaction is directly
proportional to the concentration of A, or is first order
in A:
Rate = k[A]
114
For a bimolecular elementary reaction
involving A and B molecules
A + B  product
the rate of product formation depends
on how frequently A and B collide, which
in turn depends on the concentration of
A and B. Thus we can express the rate
as
Rate = k[A][B]
115
Similarly, for a bimolecular elementary
reaction of the type
A + A  products
or
2A  products
the rate becomes
Rate = k[A]2
116
Rate Laws for Elementary Steps
Elementary Step
Molecularity
Rate Law
A  product
unimolecular rate = k[A]
2A  product
bimolecular
rate = k[A]2
A + B  product
bimolecular
rate = k[A][B]
2A + B  product termolecular
rate = k[A]2[B]
117
Remember, when we study a reaction that
has more than one elementary step, the
rate law for the overall process is given
by the rate-determining step, which is
the slowest step in the sequence of
steps leading to product formation.
118
Example: The gas-phase decomposition of
dinitrogen monoxide (N2O) is believed to
occur via two elementary steps:
Step 1: N2O N2 + O
Step 2: N2O + O  N2 + O2
Experimentally the rate law is found to be
Rate = k[N2O]
(a) Write the equation for the overall reaction.
(b) Identify the intermediates.
(c) What can you say about the relative rates
of steps 1 and 2?
119
(a) Adding the equations for steps 1 and 2 gives
the overall reaction
2N2O  2N2 + O2
(b) Since the O atom is produced in the first
elementary step and it does not appear in the
overall balanced equation, it is an
intermediate.
(c) Step 1 must be the rate-determining (slower)
step because that step is also first order
with respect to N2O. (The rate law for the
rate-determining step should match the rate
law that is determined experimentally.)
120
Example 2: Hydrogen Peroxide Decomposition
Does the decomposition of hydrogen
peroxide occur in a single step?
The overall reaction is
I-
2H2O2(aq)  2H2O(l) + O2(g)
By experiment, the rate law is found to be
Rate = k[H2O2][I-]
121
From this alone you can see that H2O2
decomposition does not occur in a single
elementary step corresponding to the
overall balanced equation. If it did, the
rate law would be
Rate = [H2O2]2
or in other words, the reaction would be
second order in H2O2. Remember, the
experimentally determined rate law for
this reaction was shown to be
Rate = k[H2O2][I-]
122
Catalysis
I-
2H2O2(aq)  2H2O(l) + O2(g)
Rate = k[H2O2][I-]
For the decomposition of hydrogen peroxide we see
that the reaction rate depends on the concentration
of iodide ions even though I- does not appear in the
overall equation. I- is a catalyst for this reaction, a
substance that increases the rate of a chemical
reaction without itself being consumed.
123
A catalyst exists before the reaction
occurs and can be recovered and reused
after the reaction is complete. This is
the opposite of intermediates, which
are produced in one step of a mechanism
and consumed in another.
124
In many cases, a catalyst
increases the rate by
providing a set of elementary
steps with more favorable
kinetics than those that exist
in its absence.
In other words, a catalyst
typically lowers the
activation energy for the
reaction by forming an
activated complex that has
less potential energy.
125
Forward reaction
C+D
C+D
A+B
A+B
Reverse reaction
Because the activation energy for the reverse
reaction is also lowered, a catalyst enhances the
rates of the forward and reverse reaction
equally.
126
There are three general types of
catalysis: heterogeneous catalysis,
homogeneous catalysis, and enzyme
catalysis.
127
In heterogeneous catalysis the reactants
and the catalyst are in different
phases. Usually the catalyst is a solid
and the reactants are either gases or
liquids. Heterogeneous catalysis is by
far the most important type of catalysis
in industrial chemistry, especially in the
synthesis of many key chemicals.
128
Ammonia is an extremely valuable inorganic substance
used in the fertilizer industry and many other
applications.
N2(g) + 3H2(g)  2NH3(g)
DH = -92.6 kJ
This reaction is extremely slow at room temperature,
and although raising the temperature accelerates the
above reaction, it also promotes the decomposition of
NH3 molecules into N2 and H2, thus lowering the yield
of NH3.
In 1905, after testing literally hundreds of compounds
at various temperatures and pressures, Fritz Haber
discovered that iron plus a few percent of oxides of
potassium and aluminum catalyze the reaction. This
procedure is known as the Haber process.
129
Haber Process
First the H2 and the
N2 molecules bind to
the surface of the
catalyst. This
interaction weakens
the covalent bonds
within the molecules
and eventually causes
the molecules to
dissociate. The highly
reactive H and N
atoms combine to
form NH3 molecules,
which then leave the
surface.
130
Nitric acid is one of the most important
inorganic acids. It is used in the production
of fertilizers, dyes, drugs, and in many other
products. The major industrial method of
producing nitric acid is the Ostwald process.
The starting materials, ammonia and molecular
oxygen, are heated in the presence of a
platinum-rhodium catalyst to about 800 oC.
131
In homogeneous catalysis, the reactants
are dispersed in a single phase, usually
liquid. Acid and base catalyses are the
most important types of homogeneous
catalysis in liquid solution.
Homogeneous catalysis can also take place
in the gas phase. A well-known example
of catalyzed gas-phase reactions is the
lead chamber process, which for many
years was the primary method of
manufacturing sulfuric acid.
132
Homogeneous catalysis has several advantages
over heterogeneous catalysis. For one thing,
the reactions can often be carried out under
atmospheric conditions, thus reducing
production costs and minimizing the
decomposition of products at high
temperatures. In addition, homogeneous
catalysts can be designed to function
selectively for a particular type of reaction,
and homogeneous catalysts cost less than the
precious metals (for example, platinum and
gold) used in heterogeneous catalysis.
133
Of all the intricate processes that have evolved
in living systems, none is more striking or
more essential than enzyme catalysis.
Enzymes are biological catalysts. Enzymes can
increase the rate of a biochemical reaction by
a factor ranging from 106 to 1012 times!
An enzyme acts only on certain molecules, called
substrates while leaving the rest of the
system unaffected. It has been estimated
that an average living cell may contain some
3000 different enzymes, each of them
catalyzing a specific reaction in which a
substrate is converted into the appropriate
products.
134
An enzyme is typically a large protein molecule
that contains one or more active sites where
interactions with substrates takes place.
These sites are structurally compatible with
specific substrate molecules, in much the
same way as a key fits a particular lock. In
fact, the notion of a rigid enzyme structure
that binds only to molecules whose shape
exactly matches that of the active site was
the basis of an early theory of enzyme
catalysis, the so-called lock-and-key theory.
135
This theory accounts for the specificity of
enzymes, but it contradicts research evidence
that a single enzyme binds to substrates of
different sizes and shapes.
Chemists now know that an enzyme molecule (or at least its
active site) has a fair amount of structural flexibility and
can modify its shape to accommodate more than one type
of substrate.
136
The
End
137