Chemical Kinetics
Chapter 14
Chemistry, The Central Science, 7th& 8theditions
Theodore L. Brown; H. Eugene LeMay, Jr.; and
Bruce E. Bursten
1.
What do we mean by kinetics?
Kinetics refers to
 the rate at which chemical
reactions occur.
 The reaction mechanism or
pathway through which a reaction
proceeds.
2.
Topics for study in kinetics
Reaction Rates
How we measure rates.
Rate Laws
How the reaction rate depends
on amounts of reactants.
Molecularity and order.
Integrated Rate Laws
How to calculate amount left or
time to reach a given amount of
reactant or product.
Half-life
How long it takes for 50% of
reactants to react
Arrhenius Equation
How rate constant changes with
temperature.
Mechanisms
How the reaction rate depends
on the sequence of molecular
scale processes.
3.
Factors That Affect Reaction Rates

The Nature of the Reactants


Concentration of Reactants


As the concentration of reactants increases, so does
the likelihood that reactant molecules will collide.
Temperature


Chemical compounds vary considerably in their
chemical reactivities.
At higher temperatures, reactant molecules have
more kinetic energy, move faster, and collide more
often and with greater energy.
Catalysts

Change the rate of a reaction
by changing the mechanism.
4.
Reaction Rates
The rate of a chemical reaction can be determined
by monitoring the change in concentration of
either reactants or the products as a function of
time.
[A] vs t
5.
Example 1: Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
[C4H9Cl] M
In this reaction, the
concentration of
butyl chloride,
C4H9Cl, was
measured at various
times, t.
Rate =
[C4H9Cl]
t
6.
Reaction Rates Calculation
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl (aq)
Average Rate, M/s
The average rate of the
reaction over each
interval is the change in
concentration divided by
the change in time:
7.
Reaction Rate Determination
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)


Note that the average
rate decreases as the
reaction proceeds.
This is because as the
reaction goes forward,
there are fewer
collisions between the
reacting molecules.
8.
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)


A plot of concentration
vs. time for this
reaction yields a curve
like this.
The slope of a line
tangent to the curve at
any point is the
instantaneous rate at
that time.
9.
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)

The reaction slows
down with time
because the
concentration of the
reactants decreases.
10.
Reaction Rates and Stoichiometry
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)


In this reaction, the
ratio of C4H9Cl to
C4H9OH is 1:1.
Thus, the rate of
disappearance of
C4H9Cl is the same as
the rate of appearance
of C4H9OH.
Rate =
-[C4H9Cl]
=
t
[C4H9OH]
t
11.
Reaction Rates & Stoichiometry
Suppose that the mole ratio is not 1:1?
Example
H2(g) + I2(g)  2 HI(g)
2 moles of HI are produced for each mole of H2 used.
The rate at which H2 disappears is only half of the
rate at which HI is generated
12.
Concentration and Rate
Each reaction has its own equation that gives its rate as
a function of reactant concentrations.
This is called its Rate Law
The general form of the rate law is
Rate = k[A]x[B]y
Where k is the rate constant, [A] and [B] are the
concentrations of the reactants. X and y are exponents
known as rate orders that must be determined
experimentally
To determine the rate law we measure the rate at
different starting concentrations.
13.
Concentration and Rate
NH4+ (aq) + NO2- (aq)  N2 (g) + 2H2O (l)
Compare Experiments 1 and 2:
when [NH4+] doubles, the initial rate doubles.
14.
Concentration and Rate
NH4+ (aq) + NO2- (aq)  N2 (g) + 2H2O (l)
Likewise, compare Experiments 5 and 6:
when [NO2-] doubles, the initial rate doubles.
15.
Concentration and Rate
NH4+ (aq) + NO2- (aq)  N2 (g) + 2H2O (l)
This equation is called the rate law, and k is the
rate constant.
16.
The Rate Law

A rate law shows the relationship between the
reaction rate and the concentrations of reactants.
 For gas-phase reactants use PA instead of [A].

k is a constant that has a specific value for each
reaction.

The value of k is determined experimentally.
Rate = K [NH4+ ][NO2- ]
“Constant” is relative herethe rate constant k is unique for each reaction
and the value of k changes with temperature
17.
The Rate Law




Exponents tell the order of the reaction
with respect to each reactant.
This reaction is
First-order in [NH4+]
First-order in [NO2−]
The overall reaction order can be found by
adding the exponents on the reactants in
the rate law.
This reaction is second-order overall.
Rate = K [NH4+ ]1[NO2- ]1
18.
Determining the Rate constant and
Order
The following data was collected for the reaction of
substances A and B to produce products C and D.
Deduce the order of this reaction with respect to A and to B.
Write an expression for the rate law in this reaction and
calculate the value of the rate constant.
[NO] mol dm-3
0.40
0.40
0.20
[O2] mol dm-3
0.50
0.25
0.25
Rate mol dm-3 s-1
1.6 x 10-3
8.0 x 10-4
2.0 x 10-4
19.
The First Order Rate Equation
Consider a simple 1st order reaction: A  B
Rate = k[A]
How much reactant A is left after time t?
The rate equation as a function of time can
be written as
Where
[A]t =[A]o e-kt
Ln [A]t - Ln[A]o = - kt
[A]t = the concentration of
reactant A at time t
[A]o = the concentration of
reactant A at time
t=0
K = the rate
constant
20.
First-Order Processes
Consider the process in
which methyl isonitrile is
converted to acetonitrile.
CH3NC
CH3CN
How do we know this is a
first order reaction?
21.
First-Order Processes
CH3NC
CH3CN
This data was
collected for this
reaction at
198.9°C.
Does
rate=k[CH3NC]
for all time intervals?
22.
First-Order Processes

When Ln P is plotted as a function of time, a
straight line results.


The process is first-order.
k is the negative slope: 5.1  10-5 s-1.
23.
Half-Life of a Reaction


Half-life is defined
as the time
required for onehalf of a reactant
to react.
Because [A] at
t1/2 is one-half of
the original [A],
[A]t = 0.5 [A]0.
24.
Half-Life of a First Order Reaction
For a first-order process, set [A]t=0.5 [A]0 in
integrated rate equation:
NOTE: For a first-order
process, the half-life
does not depend on the
initial concentration, [A]0.
25.
First Order Rate Calculation
Example 1: The decomposition of compound A is first
order. If the initial [A]0 = 0.80 mol dm-3. and the rate
constant is 0.010 s-1, what is the concentration of [A]
after 90 seconds?
26.
First Order Rate Calculation
Example 1: The decomposition of compound A is first
order. If the initial [A]0 = 0.80 mol dm-3. and the rate
constant is 0.010 s-1, what is the concentration of [A]
after 90 seconds?
Ln[A]t – Ln[A]o = -kt
Ln[A]t – Ln[0.80] = - (0.010 s-1 )(90 s)
Ln[A]t = - (0.010 s-1 )(90 s) + Ln[0.80]
Ln[A]t = -0.90 - 0.2231
Ln[A]t = -1.1231
[A]t =
0.325 mol dm-3
27.
First Order Rate Calculations
Example 2: A certain first order chemical reaction required
120 seconds for the concentration of the reactant to drop from
2.00 M to 1.00 M. Find the rate constant and the concentration
of reactant [A] after 80 seconds.
28.
First Order Rate Calculations
Example 2: A certain first order chemical reaction required
120 seconds for the concentration of the reactant to drop from
2.00 M to 1.00 M. Find the rate constant and the concentration
of reactant [A] after 80 seconds.
Solution
k =0.693/t1/2 =0.693/120s =0.005775 s-1
Ln[A] – Ln(2.00) = -0.005775 s-1 (80 s)= -0.462
Ln A = - 0.462 + 0.693 = 0.231
A = 1.26 mol dm-3
29.
First Order Rate Calculations
Example 3: Radioactive decay is also a first order process.
Strontium 90 is a radioactive isotope with a half-life of 28.8
years. If some strontium 90 were accidentally released, how
long would it take for its concentration to fall to 1% of its original
concentration?
30.
First Order Rate Calculations
Example 3: Radioactive decay is also a first order process.
Strontium 90 is a radioactive isotope with a half-life of 28.8
years. If some strontium 90 were accidentally released, how
long would it take for its concentration to fall to 1% of its original
concentration?
Solution
k =0.693/t1/2 =0.693/28.8 yr =0.02406 yr-1
Ln[1] – Ln(100) = - (0.02406 yr-1)t = - 4.065
t = - 4.062
.
- 0.02406 yr-1
t = 168.8 years
31.
Second-Order Processes
Similarly, integrating the rate law for a
process that is second-order in reactant A:
Rate = k[A]2
1
[A]t
= kt +
1
[A]o
Where
[A]t = the concentration of reactant A at time t
[A]o = the concentration of reactant A at time
t=0
K = the rate constant
32.
Second-Order Rate Equation
So if a process is second-order in A,
a graph of 1/[A] vs. t will yield a
straight line with a slope of k.
33.
Determining Reaction Order
Distinguishing Between 1st and 2nd Order
The decomposition of NO2 at 300°C is described by the
equation:
NO2 (g)
NO (g) + 1/2 O2 (g)
A experiment with this reaction yields this data:
Time (s)
[NO2], M
0.0
0.01000
50.0
0.00787
100.0
0.00649
200.0
0.00481
300.0
0.00380
34.
Determining Reaction Order
Distinguishing Between 1st and 2nd Order
Graphing ln [NO2] vs. t yields:
• The graph is not a straight
line, so this process cannot
be first-order in [A].
Time (s)
[NO2], M
ln [NO2]
0.0
0.01000
-4.610
50.0
0.00787
-4.845
100.0
0.00649
-5.038
200.0
0.00481
-5.337
300.0
0.00380
-5.573
Does not fit the first order equation:
35.
Second-Order Reaction Kinetics
A graph of 1/[NO2] vs. t gives
this plot.
Time (s)
[NO2], M
1/[NO2]
0.0
0.01000
100
50.0
0.00787
127
100.0
0.00649
154
200.0
0.00481
208
300.0
0.00380
263
• This is a straight line.
Therefore, the process
is second-order in [NO2].
• The slope of the line is
the rate constant, k.
36.
Half-Life for 2nd Order Reactions
For a second-order process, set
[A]t=0.5 [A]0 in 2nd order equation.
In this case the half-life
depends on the initial
concentration of the
reactant A.
37.
Sample Problem 1: Second Order
Acetaldehyde, CH3CHO, decomposes by second-order
kinetics with a rate constant of 0.334 mol-1dm3s-1 at 500oC.
Calculate the amount of time it would take for 80 % of the
acetaldehyde to decompose in a sample that has an initial
concentration of 0.00750 M.
The final concentration will be 20% of the original 0.00750 M
or = 0.00150
1
.
.
= 0.334 mol-1dm3s-1 t + 1
.00150
.00750
666.7 = 0.334 t + 133.33
0.334 t = 533.4
t = 1600 seconds
38.
Sample Problem 2: Second Order
Acetaldehyde, CH3CHO, decomposes by second-order
kinetics with a rate constant of 0.334 mol-1dm3s-1 at 500oC. If
the initial concentration of acetaldehyde is 0.00200 M. Find
the concentration after 20 minutes (1200 seconds)
Solution
1
[A]t
1
[A]t
. = 0.334 mol-1dm3s-1 (1200s) +
1
.
0.00200 mol dm-3
. = 0.334 mol-1dm3 s-1 (1200s) + 500 mol-1dm3
= 900.8 mol-1dm3
[A]t
=
1 _____. = 0.00111 mol dm-3
900.8 mol-1dm3
39.
Summary of Kinetics Equations
First order
Second order
Second order
Rate
Laws
Integrat
ed Rate
Laws
complicated
Halflife
complicated
40.
Temperature and Rate



Generally speaking, the
reaction rate increases
as the temperature
increases.
This is because k is
temperature dependent.
As a rule of thumb a
reaction rate increases
about 10 fold for each
10oC rise in temperature
41.
The Collision Model



In a chemical reaction, bonds are
broken and new bonds are formed.
Molecules can only react if they
collide with each other.
These collisions must occur with
sufficient energy and at the
appropriate orientation.
42.
The Collision Model
Furthermore, molecules must collide with
the correct orientation and with enough
energy to cause bonds to break and new
bonds to form
43.
Activation Energy


In other words, there is a minimum amount of energy
required for reaction: the activation energy, Ea.
Just as a ball cannot get over a hill if it does not roll
up the hill with enough energy, a reaction cannot
occur unless the molecules possess sufficient
energy to get over the activation energy barrier.
44.
Reaction Coordinate Diagrams
It is helpful to
visualize
energy changes
throughout a
process on a
reaction
coordinate
diagram like
this one for the
rearrangement
of methyl
isonitrile.
45.
Reaction Coordinate Diagrams


It shows the energy
of the reactants and
products (and,
therefore, E).
The high point on the
diagram is the
transition state.
• The species present at the transition state is
called the activated complex.
• The energy gap between the reactants and
the activated complex is the activation
energy barrier.
46.
Maxwell–Boltzmann Distributions

Temperature is
defined as a
measure of the
average kinetic
energy of the
molecules in a
sample.
• At any temperature there is a wide
distribution of kinetic energies.
47.
Maxwell–Boltzmann Distributions


As the temperature
increases, the
curve flattens and
broadens.
Thus at higher
temperatures, a
larger population of
molecules has
higher energy.
48.
Maxwell–Boltzmann Distributions

If the dotted line represents the activation energy,
as the temperature increases, so does the fraction
of molecules that can overcome the activation
energy barrier.
• As a result, the
reaction rate
increases.
49.
Maxwell–Boltzmann Distributions
This fraction of molecules can be found through the
expression:
where R is the gas constant and T is the temperature in Kelvin
.
50.
Arrhenius Equation
Svante Arrhenius developed a mathematical relationship
between k and Ea:
where A is the frequency factor, a number that
represents the likelihood that collisions would occur
with the proper orientation for reaction. Ea is the
activation energy. T is the Kelvin temperature and R is
the universal thermodynamics (gas) constant.
R = 8.314 J mol-1 K-1 or 8.314 x 10-3 J mol-1 K-1
51.
Arrhenius Equation
Taking the natural
logarithm of both
sides, the equation
becomes
1
RT
y = mx + b
When k is determined experimentally at
several temperatures, Ea can be calculated
from the slope of a plot of ln k vs. 1/T.
52.
Arrhenius Equation for 2 Temperatures
When measurements are taken for two different
temperatures the Arrhenius equation can be
symplified as follows:
Write the above equation twice, once for each of the two
Temperatures and then subtract the lower temperature
conditions from the higher temperature. The equation then
becomes:
53.
Arrhenius Equation Sample Problem 1
The rate constant for the decomposition of
hydrogen iodide was determined at two different
temperatures
2HI  H2 + I2.
At 650 K, k1 = 2.15 x 10-8 dm3 mol-1s-1
At 700 K, k2 = 2.39 x 10-7 dm3 mol-1s-1
Find the activation energy for this reaction.
2.39 x 10-7
Ea
Ln ---------------- = - ------------------------ x
2.15 x 10-8
(8.314 J mol-1 K-1)
[
1
1
------ -- -----700K 650K
]
Ea = 180,000 J mol-1 = 180 kJ mol-1
54.
Overview of Kinetics Equations
First order
Second order
Second order
Rate
Laws
Integrated
Rate Laws
Half-life
complicated
complicated
Rate and
Temp (T)
55.
Reaction Mechanisms
The sequence of events that describes
the actual process by which reactants
become products is called the
reaction mechanism.
56.
Reaction Mechanisms


Reactions may occur all at once or
through several discrete steps.
Each of these processes is known as
an elementary reaction or elementary
process.
57.
Reaction Mechanisms
•
The molecularity of a process tells how many
molecules are involved in the process.
•
The rate law for an elementary step is written
directly from that step.
58.
Multistep Mechanisms


In a multistep process, one of the steps will
be slower than all others.
The overall reaction cannot occur faster
than this slowest, rate-determining step.
59.
Slow Initial Step
NO2 (g) + CO (g)  NO (g) + CO2 (g)

The rate law for this reaction is found experimentally
to be
Rate = k [NO2]2


CO is necessary for this reaction to occur, but the
rate of the reaction does not depend on its
concentration.
This suggests the reaction occurs in two steps.
60.
Slow Initial Step

A proposed mechanism for this reaction is
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)

The NO3 intermediate is consumed in the second step.

As CO is not involved in the slow, rate-determining step, it
does not appear in the rate law.
61.
Fast Initial Step

The rate law for this reaction is
found (experimentally) to be

Because termolecular (= trimolecular)
processes are rare, this rate law
suggests a two-step mechanism.
62.
Fast Initial Step

A proposed mechanism is
Step 1 is an equilibriumit includes the forward and reverse reactions.
63.
Fast Initial Step

The rate of the overall reaction depends
upon the rate of the slow step.
The rate law for that step would be

But how can we find [NOBr2]?

64.
Fast Initial Step



NOBr2 can react two ways:
 With NO to form NOBr
 By decomposition to reform NO and Br2
The reactants and products of the first step
are in equilibrium with each other.
Therefore,
Ratef = Rater
65.
Fast Initial Step

Because Ratef = Rater ,
k1 [NO] [Br2] = k−1 [NOBr2]
Solving for [NOBr2] gives us
k1
[NO]
[Br
]
=
[NOBr
]
2
2
k−1
66.
Fast Initial Step
Substituting this expression for [NOBr2] in the
rate law for the rate-determining step gives
67.
Catalysts



Catalysts increase the rate of a reaction by
decreasing the activation energy of the reaction.
Catalysts change the mechanism by which the
process occurs.
Some catalysts also make atoms line up in the
correct orientation so as to enhance the reaction
rate
68.
Catalysts
Catalysts may be
either homogeneous or
heterogeneous
A homogeneous
catalyst is in the same
phase as the
substances reacting.
A heterogeneous
catalyst is in a different
phase
69.
Catalysts
One way a
catalyst can
speed up a
reaction is by
holding the
reactants
together and
helping bonds
to break.
Heterogeneous
catalysts often
act in this way
70.
Catalysts
Some catalysts
help to lower
the energy for
formation for
the activated
complex or
provide a new
activated
complex with a
lower activation
energy
AlCl3 + Cl2  Cl+ + AlCl4Cl+ + C6H6  C6H5Cl + H+
H+ + AlCl4-  HCl + AlCl3
Overall reaction
C6H6 + Cl2  C6H5Cl + HCl
71.
Catalysts & Stratospheric Ozone
In the stratosphere, oxygen molecules absorb ultraviolet
light and break into individual oxygen atoms known as free
radicals
The oxygen radicals can then combine with ordinary oxygen
molecules to make ozone.
Ozone can also be split up again into ordinary oxygen and
an oxygen radical by absorbing ultraviolet light.
72.
Catalysts & Stratospheric Ozone
The presence of chlorofluorcarbons in the
stratosphere can catalyze the destruction of ozone.
UV light causes a Chlorine free radical to be released
The chlorine free radical attacks ozone and converts it
Back to oxygen. It is then regenerated to repeat the
Process. The result is that each chlorine free radical can
Repeat this process many many times. The result is that
Ozone is destroyed faster than it is formed, causing its
level to drop
73.
Enzymes


Enzymes are
catalysts in
biological
systems.
The substrate fits
into the active
site of the
enzyme much like
a key fits into a
lock.
74.
75.