Presentation Week By Sabanci University Bekir Dizman

SABANCI UNIVERSITY
Faculty of Engineering and Natural Sciences
Chemical Kinetics
CHEM202
Assoc. Prof. Bekir DIZMAN
bekirdizman@sabanciuniv.edu
0535 966 6083 (cell), 0216 300 1312 (office)
1

6
DRG
Research Areas and Projects
Sustainable
materials
ISOPREP
ROCOVERY
PETR
RECOMP
Composites
Thermal latent
curing agents
1001-1
1001-2
IP
Cure kinetics,
cure cycle,
reaction
mechanism
ATLAS
KEAS-2244
Nanoengineered
materials
ATLAS
Energy storage
1001-3
PCB substrate
material
development
PCB
Coatings and
surface-active
agents
HK-2244
POZCOATS
Advanced
materials for
environmental
applications
Solutions to
Water
Security
PFASTER
CO2 / N
Management
and Valorization
Membranes
1001-4
KEFİS
SecureDrink
SONICA
HK-1505
http://myweb.sabanciuniv.edu/bekirdizman

POZ Copolymer-IM TLCs
7
R1
R2
Curing agent

Cure Kinetics and Cure Cycle Determination
8

Cure kinetics and reaction mechanism in wood composites (KEAS-2244)
9

10
https://sucourse.sabanciuniv.edu/plus/course/view.php?id=6151

Syllabus
11
Semester : Spring 2023
Instructor : Assoc. Prof. Dr. Bekir Dızman
Assistants : Asu Ece Ateşpare, Saeed Salamat Gharamaleki, Belkıs
Güneş, Sezen Öztürk
Exams and Grading : 2 Midterms/Take-home exams (20% each)
1st Midterm/Take-home exam: April 17, 2022
2nd Midterm/Take-home exam: May 22, 2022
Final Exam/Take-home exam (25%)
Laboratories (15%)
Assignments (20%)
Course Schedule : Monday 14:40-15:30 FENS L030
Wednesday 10:40-12:30 FENS L067
Laboratory : Tuesday 14:40-17:30 FENS G049
Office Hours : Outside lecture hours (Common hour: Wed. 12:40-13:30)
Textbook : An Introduction to Chemical Kinetics (Claire Vallance)

Syllabus
12
Rules
1. Attendance to all classes is required and will change the grading as follows:
a. Full attendance (≤ 3 misses): Students will be awarded with one grade higher (i.e., A- A)
b. Partial attendance (= 4-7 misses ): Grades will not be affected by attendance (i.e., A- A-)
c. Poor attendance (> 7 misses): Students will receive one grade lower (i.e., A- B+)
2. Laboratory dates will be announced by the TAs of the class.
3. Experiments will be performed in groups (3-5 students in each group). Each student will submit an
individual experimental report. Students are required to complete all experiments. Students not
completing all experiments will get an incomplete grade.
4. Assignments will be returned one week after they are given. Late returns will get a 20% reduction in the
grade.
5. Cellphones are not allowed during the class hours.
6. Attendance to the synchronous Zoom lectures and recitations, etc. and real-time online exams with SU
email account is required.

Introduction to Chemical Kinetics
13
Chapters
1. Elementary reactions
2. Rate Laws: Relating the reaction rate to reactant concentrations
3. Determining the rate law and obtaining mechanistic information from
experimental data
4. Experimental techniques for measuring reaction rates
5. Introduction to complex reactions
6. Using the steady-state approximation to derive rate laws for complex reactions
7. Chain reactions and explosions

Course Content
1
4
Week Week of Subject Midterm / Assignment Lab.
#1 27-Feb
Elementary reactions
#2 6-Mar
Rates of elementary reactions
Activation Energy/Collision Theory
#3 13-Mar
Arrhenius Equation/Catalysts
Rate laws
Assignment 1
#4 20-Mar Determining the rate law from experimental data
#5 27-Mar Experimental techniques for measuring reaction rates Assignment 2
#6 3-Apr Introduction to complex reactions
#7 10-Apr The steady state approximation
#8 17-Apr Semester Break
#9 24-Apr Unimolecular reactions 1st Midterm Exp. 1
#10 1-May Third order reactions Assignment 3
#11 8-May Enzyme reactions Exp. 2
#12 15-May Linear chain reactions Assignment 4
#13 22-May Explosions and branched chain reactions 2nd Midterm Exp. 3
#14 29-May Polymerization kinetics

Learning Outcomes
1
5
1. Students will:
a. learn about elementary reactions, methods to obtain reaction rates, and factors
affecting reaction rates,
b. describe empirical kinetics and simple collision theory,
c. identify the order of a simple reaction,
d. understand the differences between overall and elementary reactions.
2. Students will:
a. learn about rate laws, rate constants, and the mathematical framework for
understanding chemical reaction rates,
b. identify the steady state and rate determining step approximations,
c. calculate the individual and overall orders, rate constants and activation energies of
Arrhenius reactions,
d. use the main features of collision theory to calculate the reaction kinetic parameters
for simple systems,
e. develop an ability to describe and undertake appropriate experiments to determine
the rate laws and activation energies of simple reactions.

Learning Outcomes
1
6
3. Students will:
a. apply the steady state and rate determining step approximations to more complicated
systems,
b. describe the kinetic principles underlying complex reactions (e.g. polymerizations and
photochemical reactions).
4. Students will learn about:
a. various methods for determining the rate law for a reaction from data recorded during
experimental measurements of the reaction rate,
b. how data on the temperature dependence of the rate constant can provide information on
activation barriers along the reaction pathway.
5. Students will:
a. work on experimental methods to follow reactant and product concentrations over a broad
range of timescales,
b. demonstrate proficiency in assembling basic laboratory glassware, perform fundamental
laboratory techniques,
c. make and record relevant experimental observations, interpret the results,
d. work with other students in small groups to complete clearly defined tasks,
e. adopt a systematic approach to problem solving.

Chemical Kinetics - Introduction
1
7
Chemical kinetics is the study of the rates of chemical reactions, the factors that affect these
rates, and the reaction mechanisms by which reactions occur

Chemical Kinetics - Introduction
1
8

Chemical Kinetics
2
7
Timescale of reactions (fs-thousands of years)
 Fossilization (thousands of years)
 Reactions in combustion, atmospheric chemistry, and biology (100-300 fs)

Types of Chemical Reactions
2
8
 Very fast (instantaneous) reaction:
• Reaction is so fast that the determination
of rate of reaction is difficult.
• Involve ionic species.
• Occurs within 10-14 to10-16 seconds.
 Very slow reaction:
• Reaction is slow, extremely slow
• Determination of rate of reaction is a
difficult task.
• Take months to show measurable change.
 Moderate reactions:
• Occurs at moderate and measurable rates
at room temperature.
• Study of kinetics is simple.
• Molecular in nature.
AgNO3 + NaCl → AgCl + NaNO3
BaCl2 + H2SO4 → BaSO4 + 2HCl
HCl + NaOH → NaCl + H2O
Examples
2H2O2 → 2H2O + O2
2N2O5 → 2N2O4 + O2
CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
C12H22O11 + H2O → C6H12O6 + C6H12O6
Cane sugar Glucose Fructose
NO2 + CO → NO + CO2
2FeCl3 + SnCl2 → 2FeCl2 + SnCl4 (aqueous phase)

Chemical Kinetics
29
Reaction rate
• Rate at which the concentrations of the
reactants and products change with time
• Slope of the graph (not constant with
time)
• Reaction rate decreases with time
(concentration dependent)
 Mechanisms by which chemical
reactions proceed (deduce the
mechanism if possible)

Factors Influencing A Simple Bimolecular Reaction
30
Two reactants collide and react to form products
The rate depends on:
1. Collision frequency
 Number of collisions per second
 Governed mostly by the reactant concentrations
2. Reaction probability
 Fraction of those collisions leading to a successful reaction
 Determined primarily by the presence of any energetic barriers
along the reaction pathway and the amounts of energy
available for the reactants to overcome them
A + B C

Factors Affecting The Reaction Rate
31

32
The nature of the participating substances: Reactions that appear similar may have different rates
under the same conditions, depending on the identity of the reactants. For example, when small pieces
of the metals iron and sodium are exposed to air, the sodium reacts completely with air overnight,
whereas the iron is barely affected.
https://www.youtube.com/watch?v=jI__JY7pqOM

33
Concentrations and pressure of reactants: Reaction rates generally increase as the concentrations of
the reactants are increased. Reaction rates increase as the pressure of reactants is increased.
Reason
• An increase in concentration of solutions or pressure of gases
means there are more reactant particles in a given space.
• More particles in a given space will result in more frequent
collisions.
• This will consequently increase the rate of reaction.
Examples
• Acids that have a higher concentration react more vigorously with
other substances.
• Gases stored under high pressure are more explosive when ignited.
• The industrial production of ammonia is performed at high pressure
to increase the rate of reaction and therefore the efficiency of
production.

34
Temperature: Reaction rates generally increase rapidly as the temperature is increased. An
increase in temperature increases the frequency of intermolecular collisions.
Reason
• An increase in temperature means that reactant particles move faster and have more energy.
• An increase in the movement of reactant particles will result in more frequent collisions.
• An increase in the energy of reactant particles means that there will be a greater proportion of
successful collisions – that is, collisions will more likely involve sufficient energy to break bonds.
• These two factors will consequently increase the rate of reaction.
Examples
• Magnesium reacts more vigorously with water when heated over a Bunsen burner.
• The industrial production of nitric acid is performed at high temperature to increase the rate of
reaction and therefore the efficiency of production.
• The cold temperatures of fridges and freezers reduces the rate of food spoilage.
https://www.youtube.com/watch?v=u_1uLP30uxY

35
Surface area: For reactions that occur on a surface rather than in solution, the rate increases
as the surface area is increased
Reason
• An increase in surface area of solids or liquids in
heterogeneous reaction mixtures means that more
reactant particles are exposed.
• Greater exposure of reactant particles will result in
more frequent collisions.
Examples
• A steel bar will not burn when placed in a flame but
steel wool will burn readily.
• Dust in flour mills and coal mines can be a serious
safety issue as it is highly flammable, whereas
grain and coal do not ignite readily.
• Fuel injectors create a fine mist from liquid fuels,
enabling efficient combustion in car engines.

36
Stirring: Stirring heterogeneous reaction mixtures will lead to an increase in reaction rate.
Reason
• Stirring or agitating a heterogeneous reaction
mixture means that more reactant particles are
exposed.
• Greater exposure of reactant particles will result
in more frequent collisions.
• This will consequently increase the rate of
reaction.
Examples
• Many industrial food and chemical processes,
such as fermentation reactions, are carried out
in large batch reactors that use stirrers to
continually agitate the reaction mixture.

37
Catalyst enhances reaction rate by reducing the activation energy
Reason
• Catalysts are substances that lower the energy required to
break bonds (activation energy), by orienting reactant
molecules in a way that makes bonds easier to break.
• Lower activation energy means a greater proportion of
successful collisions between reactant particles.
• Some reactions do not proceed at all without a catalyst.
• Catalysts increase the rate of a chemical reaction, but are
not themselves changed during the reaction.
Examples
• Hydrogen peroxide breaks down slowly if left on the shelf,
but in the presence of manganese dioxide catalyst, it breaks
down rapidly.
• A platinum catalyst enables the rapid reaction between
hydrogen and oxygen in fuel cells.
• Most chemical reactions that take place in living things
require specialised biological catalysts called enzymes,
otherwise they would not proceed.

38
Chapters
experimental data

39
Chapters
experimental data

Elementary Reactions
40
Elementary reactions
The rates of elementary reactions: energetic considerations
The rates of elementary reactions: simple collision theory and
the Arrhenius equation
• Encounter rate or collision rate
• Energy requirement
• Steric requirement
• Putting everything together
• Shortcomings of simple collision theory
• Arrhenius equation
• Catalysis

Elementary reaction
• The simplest kind
• Occurs in a single step
• Classification is done in
terms of molecularity
(the number of reactants
involved)
 Unimolecular
 Bimolecular
 Termolecular
41
Complex reaction
• Involves more than one
elementary step
• Most chemical reactions are
complex reactions
• Overall reaction equation is the
net result of the elementary
steps
• There is often no single
chemical process that
corresponds to the reaction
equation

 Unimolecular (a single reactant)
 Bimolecular (two reactants, the most common type)
 Termolecular (three reactants)
42
The molecularity of an elementary
reaction refers to the number of
free atoms, ions, or molecules that
enter into the reaction
https://www.youtube.com/watch?v=oC3klPMRnwo

Elementary Reactions (Truly Single Step)
43
Unimolecular reactions
Bimolecular reactions
Termolecular reactions

Complex Reactions
44
A complex reaction is a multi-step reaction composed of elementary reactions
Haber-Bosch process for
producing ammonia

Complex Reactions
45
producing ammonia

The Rates of Elementary Reactions: Energetic Considerations
46
Activation barrier (activation energy, potential
barrier)
 Determines the rate at which the reaction will
occur at a particular temperature
 Low barrier  reaction is fast
 High barrier  reaction is slow
 Increase temperature  increase reaction rate

Activation Energy
47
It is defined as the least possible amount of energy
(minimum) which is required to start a reaction or the
amount of energy available in a chemical system for a
reaction to take place.
Svante Arrhenius

51
Activation Energy (Ea)
Exothermic reaction Endothermic reaction

54
Example
Q: Sketch a potential energy diagram for a general reaction A + B C + D
Given that ΔHreverse = -10 kJ and Ea forward = +40 kJ

55
Example
Answer the following questions based on the
potential energy diagram shown here:
1. Does the graph represent an endothermic or
exothermic reaction?
2. Label the position of the reactants, products,
and activated complex.
3. Determine the heat of reaction, ΔH, (enthalpy
change) for this reaction.
4. Determine the activation energy, Ea for this
reaction.
5. How much energy is released or absorbed
during the reaction?
6. How much energy is required for this reaction
to occur?

56
Example
1. The graph represents an endothermic reaction
2. The reactants, products, and activated complex are
shown
3. ΔH = +50 kJ
Since this is an endothermic reaction, ΔH will have a
positive value. ΔH is the difference in energy between
the energy levels of the initial reactants (50 kJ) and the
final products (100 kJ), and does not depend on the
actual pathway
4. Ea = +200 kJ
Activation energy is the amount of energy required to
go from the energy level of the reactants (50 kJ) to the
highest energy point on the graph, the activated
complex (250 kJ)
5. 50 kJ of energy is absorbed during this
endothermic reaction (this is the value of ΔH)
6. 200 kJ of energy is required for this reaction to
occur (Ea).

Simple Collision Theory
58
What is necessary for a reaction to occur on the molecular level?
For molecules to react, they must collide in the correct orientation and with
more kinetic energy than the activation energy of the reaction

59
 Before atoms, molecules, or ions can react, they
must first come together, or collide
 An effective collision between two molecules
puts enough energy into key bonds to break them
 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
 As the temperature increases, the curve flattens
and broadens
 Thus at higher temperatures, a larger population
of molecules has higher energy
 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
 This fraction of molecules can be
found through the expression:
where R is the gas constant and T
is the temperature in Kelvin

61
 Simplest model to understand and predict the reaction rate for an elementary
bimolecular reaction
A + B  P
 Provides insight into the temperature dependence of chemical reaction rates
 The factors that a reaction rate depends upon:
1. Encounter rate (collision rate)
2. Energy requirement (energy needs to be present in a particular form)
3. Steric requirement (collision needs to be in a particular orientation)

Encounter Rate (Collision Rate)
62
The rate of collisions (the number of collisions per second) depends on:
1. Number density (the number of reactant molecules per unit volume, gas-phase equivalent of
concentration, nA and nB).
The more particles per unit volume, the more often collisions will occur.
2. The sizes of the two collision partners.
The larger the particles in a sample of gas, the more often
they will collide with each other.
Target area that one collision partner presents to the second
Collision cross section (target area):
Collision diameter:
3. The relative velocity of the collision partners
The faster the particles are moving around in a gas, the more often they collide.
For gas molecules at thermal equilibrium, the distribution of the relative velocities
is given by the Maxwell-Boltzmann distribution

Encounter Rate (Collision Rate)
63
Maxwell-Boltzmann distribution
Reduced mass of collision partners
: Boltzman constant
: Mean relative velocity
: Collision frequency A + B
: Collision frequency A + A
: Distribution of relative velocities
To determine the collision frequency, we need to determine
the number of particles within the volume corresponding to
a time of t = 1 s.
https://www.youtube.com/watch?v=KhujN3TUYko
l
l

65
 Energy requirement
• The activation energy (Ea) is the minimum energy that must be supplied by collisions for a reaction
to occur
• The fraction of collisions for which the collision energy is high enough to overcome any Ea for the
reaction is given by the factor e-Ea/kBT
 Steric requirement
• Experimentally measured rates are smaller than those calculated from simple collision theory
• The spatial orientations of the colliding species affect the reaction rate
• Introduction of a steric factor (P) into the reaction rate expression
• Reaction cross section ( )
Energy and Steric Requirements

Putting Everything Together
66
 Simple collision theory expression for the reaction rate is:
 Rate constant (k or k(T)):
• is temperature dependent
• relates the rate to the concentrations
• determines how fast the reaction will proceed relative to the other reactions
• is most dominantly affected by the energetic factor (determined by Ea)

Shortcomings of Simple Collision Theory
67
 It does not predict absolute rate constants in quantitative agreement with experiment.
 Temperature dependence of the rate constant is described well with the Arrhenius
equation
: Pre-exponential factor
 It does not account for the fact that, unless the collision is head on, not all of the
kinetic energy of the two reactants is available for reaction. Conservation of linear
and angular momentum means that only the kinetic energy corresponding to the
velocity component along the relative velocity vector of the reactants actually
contributes to the collision energy.
 The energy stored in the internal degrees of freedom in the reactants (vibrations,
rotations, etc.) has been ignored. For reactions involving large molecules, this often
leads to a sizeable discrepancy between simple collision theory and experiment,
though this is partly corrected for by the inclusion of the steric factor, P.

Arrhenius Equation
68
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.
The expression e–Ea/RT represents the fraction
of molecular collisions sufficiently energetic to
produce a reaction
Taking the natural logarithm of both sides, the
equation becomes:
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.
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
A is the frequency factor

Example
70
15.88
17.22
18.18
18.95
19.58
y = -2767.2x + 25.11
R² = 1
15.00
16.00
17.00
18.00
19.00
20.00
21.00
22.00
23.00
24.00
25.00
0 0.001 0.002 0.003 0.004
lnk
1/T
lnk vs 1/T
T 1/T k lnk
300 0,003333333 7,90E+06 15,88
350 0,002857143 3,00E+07 17,22
400 0,0025 7,90E+07 18,18
450 0,002222222 1,70E+08 18,95
500 0,002 3,20E+08 19,58
slope -2767,2
intercept 25,09 7,88E+10

ln
k2
k1
=
Ea
R
-
1
T2
1
T1
-
Effect of Temperature on the Reaction Rate
71
k  Ae
Ea
RT ln k = ln A - Ea/RT
Example:
The activation energy of a first order reaction is 50.2 kJ/mol at 25 oC. At what
temperature will the rate constant double?
ln
k2
k1
=
Ea
R
-
1
T2
1
T1
-
ln
k1
=
50.2
R
-
1
T2
1
298
-
k2 = 2k1 T1 = 25 + 273 = 298 K
T2 = 308 K = 308-273 = 35 oC

Example
72
What is the activation energy of a reaction that has a rate
constant of 2.50 x 102 kJ/mol at 325K and a rate constant of
10.00 x 102 kJ/mol at 375K?

73

Catalysts
74
 Catalyst
• is a substance that is added to a reaction mixture in order to speed up the reaction
• provides an alternative reaction mechanism with a lower activation barrier
• is regenerated as part of the reaction mechanism, allowing it to catalyze a reaction many times over

Catalysts
75
A catalyst increases the rate of a chemical reaction without itself being consumed.
Ea
k
uncatalyzed catalyzed
ratecatalyzed > rateuncatalyzed
k  Ae
Ea
RT

Catalysts
76
How do catalysts increase reaction rates?
 In general, catalysts operate by lowering the
overall activation energy, Ea, for a reaction. (It
lowers the “hill”.)
 However, catalysts can operate by increasing
the number of effective collisions.
 A catalyst usually provides a completely
different mechanism for the reaction.
 In the preceding peroxide decomposition
example, in the absence of a catalyst, H2O2
decomposes directly to water and oxygen.
 In the presence of Br–, Br2(aq) is generated as
an intermediate.
 When a catalyst adds an intermediate, the
activation energies for both steps must be
lower than the activation energy for the
uncatalyzed reaction.

Catalysts
77
There are two types of catalyst:
1. Heterogeneous--one that is present in a different phase as the reacting
molecules.
2. Homogeneous-- one that is present in the same phase as the reacting
molecules.
Example:
Hydrogen peroxide decomposes very slowly in the absence of a catalyst:
2H2O2(aq)  2H2O(l) + O2(g)
In the presence of bromide ions, the decomposition occurs rapidly in an acidic
environment:
2Br–
(aq) + H2O2(aq) + 2H+
(aq)  Br2(aq) + 2H2O(l)
Br2(aq) + H2O2(aq)  2Br–
(aq) + 2H+
(aq) + O2(g)
Br– is a homogeneous catalyst.
The net reaction is still…2H2O2(aq)  2H2O(l) + O2(g)

Homogeneous Catalysts
79
Reaction profile for the uncatalyzed and
catalyzed decomposition of ozone

Heterogeneous Catalysts
82
Many reactions are
catalyzed by the
surfaces of
appropriate solids

83
 Often we encounter a situation involving a solid catalyst in contact with
gaseous reactants and gaseous products…
 Example: catalytic converters in cars.
• Many industrial catalysts are heterogeneous.
How do they do their job?
 The first step is adsorption (the binding of reactant molecules to the
catalyst surface).
 Adsorption occurs due to the high reactivity of atoms or ions on the surface
of the solid.
 Molecules are adsorbed onto active sites on the catalyst surface.
 The number of active sites on a given amount of catalyst depends on
several factors such as:
• The nature of the catalyst.
• How the catalyst was prepared.
• How the catalyst was treated prior to use.

84
Example: C2H4(g) + H2(g)  C2H6(g)
 In the presence of a metal catalyst (Ni, Pt or Pd) the reaction occurs
quickly at room temperature.
Here are the steps…
• First, the ethylene and hydrogen molecules are adsorbed onto active sites on the metal
surface.
• Second, the H–H bond breaks and the H atoms migrate about the metal surface and runs
into a C2H4 molecule on the surface.
• Third, when an H atom collides with a C2H4 molecule on the surface, the C−C π-bond
breaks and a C–H σ-bond forms.
• Lastly, When C2H6 forms it desorbs from the surface.
• When ethylene and hydrogen are adsorbed onto a surface, less energy is
required to break the bonds.
• The Ea for the reaction is lowered, thus the reaction rate increases.

Enzyme Catalysis
87
Enzymes are high-molecular-mass proteins that usually catalyze one
specific reaction—or a set of quite similar reactions—but no others

Catalysts
88
 Addition of the catalyst increases the reaction rate by a factor of

90
Chapters
experimental data

Rate Laws: Relating Reaction Rate to Reactant Concentrations
91
Rate of reaction
Rate laws
The units of the rate constant
Integrated rate laws
Half lives

Rate of Reaction
92
 Rate of a chemical reaction is defined as the rate at which reactants are
used up, or equivalently, the rate at which products are formed
 Reaction rate is the change in the concentration of a reactant or a product
with time (M/s)
A B
rate = -
D[A]
Dt
rate =
D[B]
Dt
D[A] = change in concentration
of A over time period Dt
D[B] = change in concentration
of B over time period Dt
Because [A] decreases with time,
D[A] is negative.
time

Rate- Units and Dependence on Stoichiometric Coefficients
93
 Rate units
• Rate has units of concentration per unit time
mol/dm3s = mol/Ls = M/s
• For gas phase reactions, alternative units of concentration are used, usually units of pressure, such
as torr, mbar, or Pa.
Rate has units of torr/s, mbar/s, or Pa/s
• A rate in pressure units can be converted to a rate in concentration units simply by dividing by RT
pV = nRT n/V = p/RT
 Dependence of rate on stoichiometric coefficients
 Changes in concentrations of the reactants and/or products are inversely proportional to their
stoichiometric proportions
Haber-Bosch process
 In general, for a reaction that looks like this…

Example
94
Dinitrogen pentoxide decomposes to nitrogendioxide and oxygen. If the rate
of appearance of NO2 is 8Ms-1 after 2hours, calculate:
1. The rate of appearance of O2 at the same time
2. The rate of disappearance of N2O5 at the same time
Hint – Balance the equation and write the rate equation before solving the
question.

A Conceptual Example
96
2 H2O2  2 H2O + O2
t
O
t
O
H
Rate
D
D

D
D


]
[
]
[
2
1 2
2
2
General reaction rate:
calculated by dividing rate
expressions by stoichiometric
coefficients

Average Reaction Rate
97
 Rates of chemical reaction tend to slow down as time
goes on in the reaction
 The average rate of the reaction is calculated by dividing
the change in concentration over the time interval of the
reaction
 At the beginning of the reaction, the rate is faster than the
average and near the end of the reaction, the rate is
slower than the average

Measuring Reaction Rates
98
 In general, the greater the
concentration of a reactant, the faster
the reaction goes

Measuring Reaction Rates
99
 The average rate of
reaction during an
experiment is the
negative of the slope of
the reaction rate
 The instantaneous rate at
the beginning of a
reaction is called the
initial rate of reaction

Rate Laws
10
0
Rate = k[A]m[B]n
k:rate constant
m: the reaction order with respect to A
n: the reaction order with respect to B
m + n = the overall order
Example:
m = 1 1st order with respect to A (the reaction is directly proportional
to a change in [A])
n = 0 0th order with respect to B (the rate of the reaction does not
change with [B])
m + n = 1 (overall order)
Rate = k[A]1[B]0 = k[A]
A + B → C + D

Example
10
1
2nd order with respect to A
1st order with respect to B
Overall order: 3
1st order with respect to A
1/2 order with respect to B
Overall order: 1.5

Reaction Order and Concentration
10
2
The effects of doubling one initial concentration:
For zero-order reactions, no effect on rate (20 = 1)
For first-order reactions, the rate doubles (21 = 2)
For second-order reactions, the rate quadruples (22 = 4)
For third-order reactions, the rate increases eightfold (23 = 8)

Example: Concentration and Rate
10
3
 Compare Experiments 1 and 2: when [NH4
+] doubles, the initial rate doubles.
 Likewise, compare Experiments 5 and 6: when [NO2
-] doubles, the initial rate doubles.

Rate Laws- Elementary and Complex Reactions
10
5
 Elementary reactions always follow simple rate laws, with the overall order reflecting
molecularity of the process
 Rate law for an elementary reaction can be written directly from the reaction equation, since
the rate is simply proportional to the concentrations of each reactant
Examples:
 Complex reactions
• Simple rate laws
• Complex rate laws (multi-step reaction mechanism)
Overall order = 1
Overall order = 2
Overall order = 2
Overall order = 2
Overall order = 2
k
.
.
. .
. .
. .

The Units of the Rate Constant
10
6
 The units of a rate constant depend on the form of the particular rate law in which it appears
 The first order rate law
 The second order rate law
2
 The 3/2 order rate law

Integrated Rate Laws
10
7
 A rate law is a differential equation that describes the rate of change of a reactant or product
concentration with time.
 If we integrate the rate law, then we obtain an expression for the concentration as a function of time.
This is generally the type of data obtained in an experiment, allowing a direct comparison between the
rate law and experimental data.
 In many simple cases, the rate law may be integrated analytically.
 Otherwise, numerical (computer-based) techniques may be used.
 Four of the simplest rate laws are given in table below in both their differential and integrated form.

Integration Rules
10
9
Common Functions Function Integral
Constant ∫a dx ax + C
Variable ∫x dx x2/2 + C
Square ∫x2 dx x3/3 + C
Reciprocal ∫(1/x) dx ln|x| + C
Exponential ∫ex dx ex + C
∫ax dx ax/ln(a) + C
∫ln(x) dx x ln(x) − x + C
Trigonometry (x in radians) ∫cos(x) dx sin(x) + C
∫sin(x) dx -cos(x) + C
∫sec2(x) dx tan(x) + C
Rules
Function Integral
Multiplication by constant ∫cf(x) dx c∫f(x) dx
Power Rule (n≠-1) ∫xn dx xn+1/n+1 + C
Sum Rule ∫(f + g) dx ∫f dx + ∫g dx
Difference Rule ∫(f - g) dx ∫f dx - ∫g dx
Integration by Parts See Integration by Parts
Substitution Rule See Integration by Substitution

Zero-Order Integrated Rate Law and Half Life
11
0
Zero Order Reaction-Half-life
Substitute the values t = t1/2 and [A] = [A]0/2
[A]0/2 - [A]0 = -kt1/2
-[A]0/2 = -kt1/2
t1/2 = [A]0/2k

First Order Integrated Rate Law and Half Life
11
1
First Order Reaction - Half-life

Second Order Integrated Rate Law
11
2
( )
Second Order Reaction - Half-life
A → P Second

Integrated Rate Laws
117
A → P Second

Zeroth Order Integrated Rate Law
118

Zeroth Order Integrated Rate Law
119

First Order Integrated Rate Law
120

First Order Integrated Rate Law
121
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
Methyl isonitrile is converted to acetonitrile

122

123

Example
124
The decomposition of NO2 at 300 °C is described by the equation
NO2 (g) NO (g) + 1/2 O2 (g) and yields these 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
0
0.002
0.004
0.006
0.008
0.01
0.012
0 50 100 150 200 250 300 350
[NO
2
],
M
Time, s
The plot is not a straight line, so the
process is not zero-order in [NO2].

Example
125
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
The plot is not a straight line, so the
process is not first-order in [NO2].

Example
126
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].

127
Chapters
experimental data

Determining Rate Law and Obtaining Mechanistic
Information From Experimental Data
12
8
Isolation method
Differential methods
Integral methods
Determining orders and rate constants from a half life analysis
Examples
Exploring the temperature dependence of the rate constant

Introduction
12
9
A kinetics experiment consists of measuring the concentrations of one or more
reactants or products at a number of different times after initiation of the
reaction. This information is important in two contexts:
1. Determining the absolute rate of the reaction and/or the individual
elementary steps in the mechanism.
2. Investigating the mechanism of the reaction by comparing the experimental
data with rate laws predicted for different reaction mechanisms.
Kinetic data analysis to determine:
• The reaction rate,
• Orders with respect to each reactant,
• The rate constant for the reaction
Measurements at different temperatures to determine:
• Activation energy
• Pre-exponential factor
Rate law

Isolation Method
13
0
• An approach to performing experiments that simplifies the rate law in order to
determine its dependence on the concentration of a single reactant at a time.
• The dependence of the reaction rate on the chosen reactant concentration is
isolated by having all other reactants present in a large excess, so that their
concentration remains essentially constant throughout the course of the reaction.
A + B → P
[B] = 1000x[A]
If a is 1 (pseudo-first order)
If a is 2 2 (pseudo-second order)
keff: effective rate constant

Differential Methods: Determining reaction orders from the differential form
of the rate law
13
1
There are two ways to obtain data to plot in this way:
1. Measure the concentration of the reactant, [A], as a function of time and use
this data to calculate the rate, −d[A] dt, as a function of [A]. A plot of log(rate)
against log[A] then yields the reaction order with respect to A. This method
has the advantage that all of the data may be acquired in a single
experiment; however, it can become difficult to interpret the rate data if
secondary reactions occur.
2. Make a series of measurements of the initial rate of the reaction when
different initial reactant concentrations [A]0 are employed. These may then
be plotted as above to determine the order, a. This is a commonly used
technique known as the initial rates method. This approach has the
advantage that, because the rate measurement is made at the start of the
reaction, secondary reactions will not influence the measurement. However,
it does require several experiments to be carried out with different initial
concentrations of each reactant, while the other initial reactant
concentrations are held constant.

Differential Methods
13
2

13
3
• A plot of log(rate) against log[A] gives a straight line graph whose intercept is
the value for log k and the gradient is equal to the order of the reaction.
• This treatment is valid for any order values. The gradient (slope) of the graph
line is always equal to the order.

13
6

Determination of Rate Laws
13
8

13
9

14
0

14
1

14
2

14
3

Example 1
144
The measurement involved
tracking the volume of CO2
evolved from 0.01 dm3 of a
0.01 mol dm−3 solution of the
acid, RCOOH, as a function of
time. The measured data are
shown below, and were
obtained at atmospheric
pressure and a temperature
of 298 K.

Example 1
145
nCO2, 1500= (1 atm x 0.64 (10-1 dm)3) / (0.082 L.atm/mol.K) x 298K
nCO2, 1500= 2.62 x 10-5 mol
M = n/V = n / 10-2 dm3

Example 1- Integrated Rate Law Method
146
 If the reaction is first order
 If the reaction is second order
A plot of ln [RCOOH] against t should be linear, with a slope of -k
A plot of 1 / [RCOOH] against t should be linear, with a slope of -k
0

147

148
 If the reaction is zeroth order  [RCOOH] = [RCOOH]0 – k.t
y = -9E-07x + 0.0089
R² = 0.941
0
0.002
0.004
0.006
0.008
0.01
0.012
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
[RCOOH]
time, s
[RCOOH] vs time Plot

Example 1- Differential Method
149

Example 1- Half Life Analysis
150
t1/2 (1st) = 3466 s
t1/2 (2nd) = 6932 s -3466 s =3466 s
1st Order reaction

Example 2
151
Determining the rate law for the reaction CO + Cl2 ⟶ COCl + Cl using the method of
initial rates

Determining Orders and Rate Constants From
A Half Life Analysis
153
To perform a half-life analysis, we determine a series of successive half lives from a plot of
reactant concentration versus time. The initiation of the reaction, t = 0, is used as the start time
from which to measure the first half life, t1/2. Then, t = t1/2 is used as the start time from which to
measure the second half life, t1/2, and so on. The behavior of the half lives for zeroth, first, and
second order reactions are summarized below:
/ k

Example-Half Life
154
What is the half-life of N2O5 if it decomposes with a rate constant of
5.7 x 10-4 s-1?
t½
ln 2
k
=
0.693
5.7 x 10-4 s-1
= = 1200 s = 20 minutes
How do you know decomposition is first order?
units of k (s-1)

Exploring Temperature Dependence of Rate Constant
157

Overall Activation Energies For Complex Reactions
158

Summary of the Kinetics of 0th-, 1st-, and 2nd-Order Reactions
159

Concentration vs Time Plots for 0th , 1st , and 2nd Order Reactions
161

Concentration-Time and Concentration-Rate Relationship
162

16
5

166
Chapters
experimental data

Experimental Techniques For Measuring Reaction Rates
167
Techniques for mixing the reactants and initiating reaction
• Slow reactions
• Fast reactions
 Flow techniques
 Flash photolysis and laser pump probe techniques
 Relaxation methods
 Shock tubes
 Lifetime methods
Techniques for monitoring concentrations as a function of time
• Slow reactions
• Fast reactions
 Absorption spectroscopy and the Beer-Lambert law
 Resonance fluorescence
 Laser-induced fluorescence
Temperature control and measurement

168
 Monitoring chemical reaction rates
• Slow reactions (easy to monitor)
• Fast reactions (specialized techniques)
 Kinetics experiment consists of
• Mixing reactants and initiating the reaction on a timescale that is negligible
relative to that of the reaction
• Monitoring one or more of [Reactants] or [Products] as a function of time
 Monitoring and controlling reaction temperature is important

169
 Batch techniques
• Reaction is initiated at a single chosen point in time
• Concentrations are then followed as a function of time
 Continuous techniques
• Reaction is continuously initiated
• Time dependence of the reaction mixture composition is inferred from,
for example, the concentrations in different regions of the reaction
vessel

Techniques for mixing the reactants and initiating reaction
170
Slow reactions
The reaction is initiated by mixing reactants by hand, with a
magnetic stirrer or other mechanical device
Fast reactions
A wide range of techniques have been developed:
1. Flow techniques
2. Flash photolysis and laser pump probe techniques
3. Relaxation methods
4. Shock tubes
5. Lifetime methods

Techniques for mixing the reactants and initiating reaction
171
1. Flow techniques: Timescale (ms-s)
Disadvantages
 Large quantities of reactants
are needed
 Very high flow velocities are
required
Reactions of atomic or radical
species may be studied using the
discharge flow method, in which the
reactive species are generated by a
microwave discharge immediately
prior to injection into the flow tube.

Flow Techniques
172
Stopped-flow method
 Reactants are rapidly flowed
into a fixed-volume reaction
chamber and mixed by the
action of a syringe fitted with
an end stop
 Composition of the reaction
mixture is then monitored
spectroscopically as a
function of time after mixing
at a fixed position in the
reaction chamber
 Very small sample volumes are
needed
 Biochemical reactions are studied
with this method (enzyme-catalyzed
reactions)
 All flow techniques have the common problem of heterogeneous reactions at the walls of the flow tube

2. Flash Photolysis and Laser Pump Probe Techniques
174
Flash photolysis technique
 A pulse of light (flash) initiates the reaction by dissociating a suitable
precursor molecule in the reaction mixture to a reactive species
 [Reactive species] is monitored with t, usually via a spectroscopic technique
 The shortest timescale of reactions is determined by the duration of the flash
 The flash is provided by a discharge lamp (μs-ms) or a laser pulse (fs-ns)
Pulse radiolysis is a variation on flash photolysis in which a short pulse of high
energy electrons, from ns to μs in duration, is passed through the sample in
order to initiate a reaction
 Advantages:
 There is no mixing time to reduce the time resolution of the technique
 There are no wall reactions
 Laser pump probe technique: pulsed lasers are employed both to initiate the
reaction (the ‘pump’) and to detect the products via a pulsed spectroscopic
technique (the ‘probe’). The time separation between the two pulses can be
varied either electronically or with an optical delay line, down to a resolution of
around 10 fs

Flash Photolysis and Laser Pump Probe Techniques
175

3. Relaxation Methods
176
:
 Relaxation, the return to equilibrium.
 Temperature jump, a procedure in which a sudden
temperature rise is imposed and the system returns to
equilibrium.
 Pressure-jump techniques, as for temperature jump,
but with a sudden change in pressure.
 Relaxation methods overcome the mixing problems

Shock Tubes
179
Drawbacks:
 Side reactions due to high temperatures
 Signal-to-noise levels are low because there is no signal averaging

Range of Half-Life for Different Methods
181

182
Techniques for mixing the reactants and initiating reaction
• Flow techniques
• Flash photolysis and laser pump probe techniques
• Relaxation methods
• Shock tubes
• Lifetime methods
Techniques for monitoring concentrations as a function of time
• Absorption spectroscopy and the Beer-Lambert law
• Resonance fluorescence
• Laser-induced fluorescence
Temperature control and measurement

Techniques for Monitoring Concentrations as a Function of Time
183
Slow reactions
Fast reactions
1. Absorption spectroscopy and the Beer-Lambert law
2. Resonance fluorescence
3. Laser-induced fluorescence

184
Slow reactions
1. Real-time analysis: performed while the reaction is in progress by
 Withdrawing a small sample
 Monitoring the bulk
2. Quenching method: reaction is stopped after a certain time from initiation.
The key requirement: reaction must be slow enough- or quenching fast
enough- for little reaction to occur during the quenching process itself
Different ways for quenching:
 sudden cooling
 adding a large amount of solvent
 rapid neutralization of an acid reagent
 removal of a catalyst
 addition of a quencher
3. Combination of quenching and real-time analysis
 Withdrawing and quenching small samples of the reaction mixture at a series of times
during the reaction.

185
Ways to follow composition of reaction mixture:
 For reactions in which at least one reactant or product is a gas, the progress
of the reaction may be followed by monitoring the pressure in a sealed
system, the volume of gas evolved, or the change in mass of the reaction
mixture;
 For reactions involving ions, conductivity or pH measurements may often be
 employed;
 If the reaction is slow enough, the reaction mixture may be titrated;
 If one of the components is colored, then colorimetry may be appropriate;
 Absorption or emission spectroscopy are common.
 For reactions involving chiral compounds, polarimetry (measurement of
optical activity) may be useful;
 Other techniques include mass spectrometry, gas chromatography, NMR/ESR,
and many more.

186
Fast reactions: require a fast measurement technique, are monitored
spectroscopically
1. Absorption spectroscopy and the Beer-Lambert law
2. Resonance fluorescence
3. Laser-induced fluorescence

187
Absorption spectroscopy is widely used to track reactions in which the
reactants and products have different absorption spectra.
A monochromatic light source, often a laser beam, is passed through the reaction mixture,
and the ratio of the transmitted to incident light intensity, I / I0, is measured as a function of
time. The quantity T = I / I0 is known as the transmittance.
1. Absorption Spectroscopy and the Beer-Lambert Law

188
Transmittance may be related to the changing concentration of the absorbing
species using the Beer–Lambert law, which has both ‘decadic’ and ‘exponential’
forms, as follows:
Absorption Spectroscopy and the Beer-Lambert Law
}
or
( )

189
Resonance fluorescence is a widely used spectroscopic method for detecting
atomic species such as H, N, O, Br, Cl, F.
The light source is a discharge lamp filled with a mixture of helium and a molecular precursor
for the atom of interest. A microwave discharge inside the lamp dissociates the precursor.
2. Resonance Fluorescence

190
 In laser-induced fluorescence, a laser is used to excite a chosen species in a reaction mixture to an
electronically excited state.
 The excited states then emit photons to return to the ground state, and the intensity of this fluorescent
emission is measured.
 Because the number of excited states produced by the laser pulse is proportional to the number of
ground state molecules present in the reaction mixture, the fluorescence intensity provides a measure
of the concentration of the chosen species.
 This technique is exquisitely sensitive, but does rely on the molecule to be detected having a suitable
fluorescent excited state.
 The method is widely used to study reactions involving OH, NO, and other small radical species.
3. Laser-induced Fluorescence

Temperature Control and Measurement
191
 The temperature of a reaction mixture is most commonly monitored using a
thermocouple, due to its wide range of operations and the potential for
automation; however, standard thermometers are also commonly used.
 There are numerous ways in which the temperature of a reaction mixture
may be controlled.
• Reactions in the liquid phase may be carried out in a temperature-controlled
thermostat.
• Reactions in the gas phase are usually carried out inside a stainless steel
vacuum chamber, in which thermal equilibrium at the temperature of the
chamber is maintained through collisions of the gas molecules with the
chamber walls.
• High temperatures up to 1300 K may be obtained using conventional heaters.
• Low temperatures may be achieved by flowing cooled liquid through the walls
of the reaction vessel, and very low temperatures may be reached by using
cryogenic liquids, such as liquid nitrogen (∼77 K) or liquid helium (∼4 K).

192
Chapters
experimental data

Introduction to Complex Reactions
193
 Introduction
 Reaction mechanisms
 Equilibrium
 Consecutive reactions
• Case 1: k1 >> k2
• Case 2: k2 >> k1
 Pre-equilibria
 Moving on to more complicated mechanisms

194
 A complex reaction is:
A reaction whose mechanism comprises more than one
elementary step
 Types of complex reactions
• Consecutive reactions (Ch.5)
• Pre-equilibria (Ch.5)
• Unimolecular reactions (Ch.6)
• Third order reactions (Ch.6)
• Enzyme reactions (Ch.6)
• Chain reactions and explosions (Ch.7)
• Polymerizations

195
 Compare an experimentally determined rate law with the
predicted rate law for a given mechanism in order to determine
whether the proposed mechanism is indeed the correct one
 If they disagree Mechanism is not correct
 If they agree Mechanism may be correct
Any proposed mechanism must also be able to account for all
other properties of the reaction, which may include quantities
such as the product distribution, product stereochemistry, kinetic
isotope effects, temperature dependence, and so on.

Complex Reactions
196
producing ammonia

Reaction Mechanisms
197
 Reactions may occur all at once or through several discrete
steps.
 The sequence of events that describes the actual process by
which reactants become products is called the reaction
mechanism.
 Each of these processes is known as an elementary reaction or
elementary process.
 Intermediates are species that appear in a reaction mechanism
but not in the overall balanced equation.

Rate Determining Step
199
 Rate Determining Step: For a complex reaction there may be many steps
involved with different rates. Then, rate of the overall reaction is controlled
by the slowest step, which is called the rate-determining step.
Case a: slow step is rate determining
Case b: slow step is rate determining
Case c: slow step is not rate determining;
because an alternative route also exists

Example 1
200
Rate determining step 12

Example 2
201
The experimental rate law for the reaction between NO2 and CO to produce NO
and CO2 is as follows: rate = k[NO2]2. The reaction is believed to occur via two
steps:
What is the equation for the overall reaction?
What is the intermediate?
What can you say about the relative rates of steps 1 and 2?

Consecutive Reactions
206
 The simplest complex reaction consists of two consecutive,
irreversible elementary steps.
 Uncommon in chemistry, but common in processes such as
radioactive decay
 Fairly straightforward to solve the rate equations

207
 Identify a number of initial conditions for the reactant and product
concentrations
• At time t = 0 the concentrations of A, B, and C are [A] = [A]0, [B] = [C] = 0
• Since the reaction transforms A into B and C in a stepwise manner and in
a 1:1 ratio, we also have the condition that at all times [A] + [B] + [C] = [A]0
 Rate equations are:
Substituting this into the second
rate equation gives another first-
order differential equation.

208
These equations allow us to predict the way in which the concentrations of
reactant A, intermediate B, and product C will change with time for any values of
the rate constants k1 and k2. Considering the extreme cases, when either k1 ≫ k2
or k2 ≫ k1, proves to be rather instructive in allowing us to introduce a number
of additional features of complex reactions.

Case 1: k1 >> k2
212
When the rate of the first step in the reaction sequence is much faster than that
of the second, all of the A initially present is rapidly converted into B, which is
then slowly used up to form C. We can treat k2 as negligible in comparison with
k1 in the denominator of equation, and the expression simplifies to
We see that the rate of production of C, and therefore the overall rate of the
two-step reaction, becomes independent of k1 and only depends on the rate
constant k2. We say that the second, ‘slow’ step is the rate determining step in
the mechanism.

Case 2: k2 >> k1
214
When we reverse the situation so that the second step in the mechanism
becomes much faster than the first, the intermediate B is consumed as soon
as it is produced. We can now treat k1 as negligible in relation to k2 in the
denominator of equation, and the expression now simplifies to
The overall rate of production of C now depends only on the rate constant
k1, and the first step is now rate determining.

Pre-equilibria
216
In our second example, we will make the mechanism only slightly more
complicated than the sequential reaction mechanism in our first example by
making the first step reversible.
The rate equations are now:
These equations cannot be
solved analytically, and in
general, would have to be
integrated numerically to
obtain an accurate solution.

Pre-equilibria
217
 However, the situation simplifies considerably for the special case in
which k−1 ≫ k2. In this case, an equilibrium is reached between the
reactants A and B and the intermediate C, and the equilibrium is only
perturbed very slightly by C ‘leaking away’ very slowly to form the
product D.
 If we assume that we can neglect this small perturbation of the equilibrium,
then, once equilibrium is reached, the rates of the forward and reverse
reactions must be equal.
Equilibrium constant

Pre-equilibria
218
 The rate of the overall reaction is simply the rate of formation of the product
D, so, using the above result, we can write
 The reaction is therefore first order in each of the reactant concentrations [A]
and [B], and second order overall, with an effective rate constant keff = k2K.
It is worth noting that because this rate law has been derived based on the
assumption of a preequilibrium between A, B, and C, it will not be accurate in
the very early stages of the reaction, before the equilibrium has been
established.

Example
219
 Show that the pre-equilibrium mechanism in which 2A I (K)
followed by I + B  P (kb) results in an overall third-order reaction.

Moving on to More Complicated Mechanisms
220
 At this point, we have considered two simple reaction mechanisms for which
the rate equations can be solved exactly—or nearly exactly in the case of the
pre-equilibrium.
 The rate equations for all complex reaction mechanisms comprise a
complicated system of coupled differential equations that cannot be solved
analytically.
 In state-of-the art kinetic modelling studies, sophisticated software is used to
obtain numerical solutions to the rate equations in order to determine the time-
varying concentrations of all species involved in a reaction sequence.
 By making a few simple assumptions about the nature of reactive
intermediates within a reaction sequence, very good approximate solutions to
the rate equations may be derived.

221
Chapters
experimental data

Using the Steady-State Approximation to Derive Rate Laws
for Complex Reactions
222
 The steady-state approximation: an example
 The steady-state approximation: a general approach
 ‘Unimolecular’ reactions: the Lindemann–Hinshelwood mechanism
• High pressure
• Low pressure
• Testing the rate law against experimental data
 Third-order reactions
 Enzyme reactions and the Michaelis–Menten mechanism

The Steady-State Approximation
223
Steady-state approximation (SSA)
 Allows us to derive approximate rate laws for reactions with very complex
mechanisms
 Relies on assumptions relating to reactive intermediates involved in the
reaction of interest.
A reactive intermediate: a species that is formed during the reaction but is
completely transformed into other species during one or more elementary
steps, and so does not appear in the overall reaction equation.
• is used up virtually as soon as it is formed
• its concentration remains very low and essentially constant throughout the
course of the reaction- true at all times except
 at the very start of the reaction, when the concentration [R] of the
reactive intermediate must necessarily build up from zero to some
small non-zero value
 at the very end of the reaction (at least in the case of a reaction that
goes to completion), when [R] must return to zero

The Steady-State Approximation
224
 During the period of time when [R] is essentially constant, we find that
because d[R] / dt is so much smaller than the rates of change of the reactant
and product concentrations, it is a good approximation to set d[R] / dt = 0.
 Applying SSA
• Converts a mathematically intractable set of coupled differential equations
into a system of simultaneous algebraic equations, one equation for each
species involved in the reaction.
• The algebraic equations may be solved to find the concentrations of the
reactive intermediates, and these may then be substituted back into the
equation for the rate of change of the product in order to obtain an
expression for the overall rate law.

The Steady-State Approximation: An Example
225
 Investigate the case in which k2 ≫ k1, so that C is now a reactive intermediate
 The overall reaction rate is the rate of formation of the product, D.

Limiting Case 1: k-1 << k2 (neglecting k-1)
226
 The rate of the overall reaction is the same as the rate of the first
elementary step in the reaction mechanism. This is not all that
surprising.
 If k2 is much larger than k1 and k−1, then as soon as the A + B ⟶ C
step has occurred, C is immediately converted into products, and
there is virtually no chance for the reverse reaction, C ⟶ A + B, to
occur.
 The initial elementary step is rate determining, and therefore
dominates the kinetics.

Limiting Case 2: k-1 >> k2 (neglecting k2)
227
 This is the same result as we obtained in our previous treatment by
assuming a pre-equilibrium, with K = k1 / k−1.

The Steady-State Approximation: A General Approach
228
1. Write down a steady-state equation for each reactive intermediate, Ri, by
setting d[Ri] / dt = 0.
2. Solve the resulting set of simultaneous equations to obtain expressions for
the concentrations of each intermediate in terms of the reactant and product
concentrations. It is worth looking carefully at the equations in order to
identify the simplest route to solving them. For example:
a. If one of the equations contains only one reactive intermediate, it may
simply be rearranged to give the concentration of that intermediate in
terms of reactant and product concentrations. The resulting expression
can often be substituted into other equations to obtain the corresponding
expressions for other reactive intermediates.
b. If the equations depend on more than one reactive intermediate, and share
terms, look for sums or differences of the equations that will simplify
matters. Often a steady-state problem that initially appears extremely
complicated becomes trivial when you simply add together two of the
steady state equations.

The Steady-State Approximation: A General Approach
229
3. Write down an expression for the overall rate (usually the rate of change
of one of the products). This will generally involve the concentrations of
one or more reactive intermediates.
4. Substitute your expressions for the reactive intermediate concentrations
into the overall rate equation in order to eliminate reactive intermediates
from the equation. The result should be an overall rate equation that
depends only on the reactant and product concentrations.
Concentrations of reactive intermediates must not appear in the final
rate law. If they do, then you have not finished solving the steady-state
equations.

Example 1. Decomposition of N2O5
230

Example 1. Decomposition of N2O5
233

Example 2
235
Use the steady-state approximation to deduce the rate law for
the consumption of A2.

Example 3
237
Find the dependence of the rate of decomposition of R2 on the
concentration of R2.

Unimolecular Reactions: the Lindemann-Hinshelwood Mechanism
239
 A number of gas-phase reactions follow first order kinetics and apparently only
involve one chemical species.
 Examples:
• Structural isomerisation of cyclopropane to propene,
• Decomposition of azomethane (CH2N2CH3⟶ C2H6 + N2, with experimentally
determined rate law
d[CH2N2CH3] / dt = k[CH3N2CH3]
 The mechanism by which the reactant molecules acquire enough energy to
react remained a puzzle for some time, particularly since the rate law seemed to
rule out a bimolecular step. The puzzle was solved by Lindemann in 1922, when
he proposed the following mechanism for ‘thermal’ unimolecular reactions.

240
The reactant, A, acquires enough energy to
react by colliding with another molecule, M
(note that in many cases M will actually be
another A molecule). The excited reactant A*
then undergoes unimolecular reaction to
form the products, P. To determine the
overall rate law arising from this
mechanism, we can apply the steady state
approximation to the excited state (reactive
intermediate) A*.

241

Rate Law at High Pressure
242
At high pressure there are many collisions, and collisional de-excitation of A* is
therefore much more likely than a unimolecular reaction of A* to form products,
i.e. k−1[A*][M] ≫ k2[A*]. In this limit, we can neglect the k2 term in the denominator,
and the rate law simplifies to
This is a first order rate law, with a first order (‘unimolecular’) rate constant
kuni = k1k2 / k−1. The proposed mechanism therefore explains the observed first-
order kinetics when the reaction is carried out at sufficiently high pressures,
under conditions for which the unimolecular step in the mechanism becomes rate
determining.

Rate Law at Low Pressure
243
At low pressures, there are few collisions, and A* will generally undergo the
unimolecular reaction before it undergoes collisional de-excitation, i.e.,
i.e. k2[A*] ≫ k−1[A*][M] . In this case, we can neglect the k−1[M] term in the
denominator, and the rate law simplifies to
Under low-pressure conditions, the kinetics are therefore second order.
Formation of the excited species A*, a bimolecular process, is now the rate
determining step.

Testing The Rate Law Against Experimental Data
244
with
The rate constant keff is a 1st-order rate constant, but it depends on the
concentration of the collision partner M, and only truly becomes constant at
sufficiently high pressure, when it reduces to the value kuni = k1k2 / k−1, as shown
previously.
If experimental measurements of the rate constant keff are made as a function of
pressure (equivalent to [M]), then the Lindemann–Hinshelwood mechanism may
be tested.

245
Take the reciprocal of
A plot of 1/keff against 1/[M]
should therefore be linear,
with an intercept of k-1 / (k1k2)
and a slope of 1/k1.
y = ax + b

246
 Usually there is a reasonable fit between theory and experiment at low
pressure, but a pronounced deviation at high pressure, with experimental
values of keff being larger than the values predicted by the Lindemann–
Hinshelwood mechanism.
 The proposed mechanism therefore correctly predicts some, but not all,
features of the reaction.
 It turns out that while the general idea of a collisional activation process is
correct, the true mechanism of ‘unimolecular’ reactions is slightly more
involved.
 The principal failing of the Lindemann–Hinshelwood mechanism is that it
assumes that any excited reactant A* will undergo a unimolecular reaction to
produce products. In practice, however, excitation is generally required in a
degree of freedom that is coupled to the reaction coordinate in some way. For
example, vibrational excitation may be required in a bond that breaks during
the reaction.

Example
247
Subtracting from each other
-4 s-1

Third-Order Reactions
249
A number of reactions are found experimentally to have third-order kinetics.
An example is the oxidation of NO, for which the overall reaction equation and
experimentally-determined rate law are given below.
 One possibility for the reaction mechanism is a single elementary step
involving a three-body collision, i.e. a true termolecular reaction. However,
such collisions are exceedingly rare, and certainly too unlikely to explain the
observed rate at which this reaction proceeds.
 An added complication is that the reaction rate is found to decrease with
increasing temperature, a certain indication of a complex mechanism.

250
An alternative mechanism that leads to the same rate law consists of two
elementary steps involving a pre-equilibrium.

251
A very common situation in which third-order kinetics are observed involves
reactions in which two reactants combine to form a single product. Such
reactions require a so-called ‘third body’ to take away some of the excess
energy from the reaction product. An example is the formation of ozone.
If the mechanism consisted of a single elementary step, as written, this
reaction would barely occur. To understand the reason for this, we need to turn
to some basic classical mechanics, namely the fact that in any collision, energy
and momentum must both be conserved. To demonstrate the problem,
consider the somewhat contrived situation in which the O atom and O2
molecule initially have equal and opposite momenta, and collide head-on to
react. The following arguments apply equally well to any other situation, but are
clearest to see in this simple case.

252
 Since total momentum must be conserved, and initially the total momentum is zero
(because the momenta of the O and O2 exactly cancel each other out), the final
momentum of the O3 product must also be zero, i.e. the product molecule must be
stationary.
 Now consider the conservation of energy. Since we are forming a bond, the
reaction is exothermic, so by the conservation of energy, the total kinetic energy
possessed by the O3 product must be the sum of the reaction exothermicity and
the kinetic energies of the reactants. We have already determined that conservation
of momentum requires the newly formed O3 molecule to be stationary, so all of this
kinetic energy must be channelled into the vibrational and/or rotational motion of
the molecule, and not into translational motion. Highly vibrationally excited
molecules are extremely unstable, and under the circumstances described the O3
will very quickly dissociate back into reactants.
 The only way for the O3 to survive is for it to transfer some of its vibrational energy
to another molecule, M, in a collision. The molecule M is known as a third body. The
energy may end up as internal excitation (rotation or vibration) of M, or simply as
kinetic energy as the two molecules fly away from each other after the collision.

253
The actual mechanism is therefore
 A third body is only required for reactions in which a single product is
formed from two or more reactants, since this is the only time that the
conservation of momentum forces a large amount of energy to be released
into the internal excitation of the product.
 If two products are formed, they can both carry away almost arbitrary
amounts of energy as translational kinetic energy, while still conserving
the total momentum.

Enzyme Reactions and the Michaelis–Menten Mechanism
254
An enzyme is a protein that catalyzes a
chemical reaction by lowering the
activation energy.
Each enzyme has an active site that is carefully designed by
nature to bind a particular reactant molecule, known as the
substrate.
The activation energy of the reaction for the enzyme-bound
substrate is lower than for the free substrate molecule, due to the
fact that the interactions involved in binding shift the substrate
geometry closer to that of the transition state for the reaction.
Once the reaction has occurred, the product molecules are
released from the enzyme.

An Enzyme-Substrate Complex
255

Enzyme Reactions
257
Enzyme-catalyzed reactions occur millions of times faster than
the corresponding uncatalyzed reactions.
Virtually every chemical reaction in biology requires an enzyme in
order to occur at a significant rate, and each enzyme is specific to
a particular reaction (enzyme-specificity).
Many drugs work by binding to a carefully targeted enzyme in
place of the normal substrate molecule, thereby inhibiting enzyme
activity and slowing the reaction rate.
Enzyme kinetics is an extremely important and complex field, but
the basic kinetics of a simple enzyme catalysis process may be
modelled quite simply.

Enzyme Reactions
258
https://europepmc.org/backend/ptpmcrender.fcgi?accid=PMC4692135&blobtype=pdf

Turnover Number
259
Turnover rate = Turnover number = kcat = vmax/[E]o
The number of substrate molecules that can be converted to product by a
single enzyme molecule per unit time (usually per minute or per second).

Initial Velocity of an Enzyme-Catalyzed Reaction
260

Dependence of Rate on Enzyme Concentration
261

Dependence of Rate on Substrate Concentration
262

Dependence of Enzyme-Catalyzed Reactions on pH and Temperature
263

Enzyme Reactions
264
Assumptions:
1. Solutions are behaving ideally
2. Our constants are indeed constant
a) Enzyme concentration is constant (not affected by protein synthesis and
degradation)
b) Rate constant is constant (not affected by environmental factors such as T)
3. S  P (is negligible without enzyme)

Enzyme Reactions
265
 In an enzyme-catalyzed reaction, a substrate S is converted to products P in a
reaction that is catalyzed by an enzyme E. For many such reactions, the rate is
found experimentally to follow the Michaelis–Menten equation
 The constant KM is called the Michaelis constant, and νmax is the maximum
reaction rate, which is found to be linearly proportional to the total enzyme
concentration. The constant of proportionality, kcat is known as the turnover
number, and represents the maximum number of molecules of substrate that
each enzyme molecule can convert into products (or ‘turn over’) per second.
This maximum rate occurs when the substrate is present in large excess.

Kinetic Model
266
 The rate depends on the enzyme concentration [E], even though there is no net
change in its concentration over the course of the reaction. The simplest trial
mechanism involves the formation of a bound enzyme–substrate complex ES,
followed by conversion of the complex into the products and free enzyme
(which may then go on to catalyze further reactions).
 In the previous cases, reactive intermediates had concentrations much lower
than those of the reactants. In the case of an enzyme-catalyzed reaction, the
concentration of the reactive intermediate ES is not much less than the free
enzyme concentration, [E]. However, because the enzyme is regenerated in the
second step of the mechanism, both [E] and [ES] change much more slowly
than [S] and [P], and so the steady-state approximation is valid.

Kinetic Model
267
 Applying the steady-state approximation to [ES], we have
 If the total enzyme concentration is [E]0, and the enzyme is present either as
free enzyme, E, or enzyme–substrate complex, ES, then the amount of free
enzyme must be [E] = [E]0 - [ES]. Substituting this into the above equation gives

Kinetic Model
268
 The overall rate of reaction is found from the rate of formation of product, P.
 While the definitions of k and KM above may seem fairly arbitrary, we have
chosen these particular combinations of rate constants in order to finish with a
rate equation in the same form as the experimentally-derived Michaelis–Menten
equation. The two equations agree if we define k2 = kcat.
(KM: Michaelis constant)

Kinetic Model
269
The rate of enzyme-catalyzed reaction depends linearly on the enzyme
concentration, [E], but in a more complicated way on the substrate
concentration, [S]. The dependence on [S] simplifies under certain conditions.
1. When [S] ≫KM, then the rate becomes
and the overall rate is independent of the substrate concentration. There is
so much substrate present that only a tiny fraction is bound up in the
enzyme–substrate complex and used up in the reaction, and the
concentration of the free substrate remains essentially constant as the
reaction proceeds. The enzyme is saturated with substrate, and the reaction
rate reaches a maximum.

Kinetic Model
270
2. When [S] ≪ KM, the reaction rate becomes
and the rate is first order in both [E]0 and [S].

Kinetic Model
271
 To determine k2 and KM from experimental rate
data, we invert the expression for k to obtain a
new form of the equation that is expected to
generate a straight line plot known as a
Lineweaver–Burk plot.
 A plot of 1/k against 1/[S] has a slope of KM/k2, a
‘y’ intercept of 1/k2, and an ‘x’ intercept of 1/KM.
We can therefore find KM and k2 directly from the
x and y intercepts, or from one intercept and the
slope. Usually, the initial rates method is used to
measure k, in order to preclude any
complications that may arise from secondary
reactions of the products.

Reaction Rate vs. [S]
272
kb = k2
ka = k1
ka’ = k-1

Maximum Rate
273
kb = k2
ka = k1
ka’ = k-1

Catalytic Efficiency
274
kb = k2
ka = k1
ka’ = k-1
a

Example 2
278
[E ]0 from this equation is substituted in equation below

281
Chapters
experimental data

Chain Reactions and Explosions
282
 Chain reactions
 Linear chain reactions
• The hydrogen-bromine reaction
• The hydrogen-chlorine reaction
• The hydrogen-iodine reaction
 Comparison between the hydrogen-halogen reactions
 Explosions and branched chain reactions

Chain Reactions
283
 Chain reactions are complex reactions that involve chain carriers,
the reactive intermediates that react to produce further reactive
intermediates.
 Examples of chain reactions
• combustion of a fuel gas
• development of rancidity in fats
• the polymerase chain reaction used to amplify DNA samples for analysis
• many polymerization reactions, e.g. polymerization of ethylene to
polyethylene
• nuclear fission caused by neutron bombardment
 The elementary steps in a chain reaction
• Initiation
• Propagation
• Termination

Elementary Steps in A Chain Reaction
284
1. Initiation step
The reaction is initiated either thermally or photochemically. The first reactive intermediates / chain
carriers (in this case a Cl radical) are produced.
2. Propagation step
The reaction of a radical leads to the formation of another radical, i.e. another reactive intermediate. In
the first propagation step above, Cl reacts to form ClO; in the second step ClO reacts to form Cl.
3. Termination step
Chain carriers are deactivated. Often this occurs through radical–radical recombination, reactions with
walls, or reactions with another molecule to create an inactive product. Note that termination products
may go on to be involved in other reactions, but are not involved further in the chain reaction of interest.
Some chain reactions also involve inhibition steps, in which product molecules are destroyed.
Inhibition steps are sometimes also referred to as retardation or de-propagation steps.

Chain Reactions
285
 The chain length in a chain reaction is defined as the number of propagation
steps per initiation step, or alternatively as the rate of propagation divided by
the rate of initiation. Chain lengths can be very long: in the above example a
single Cl radical can destroy around 106 molecules of ozone
 The reaction of Cl atoms with ozone is an example of a cyclic chain reaction.
Atomic chlorine acts as a catalyst and is continuously regenerated until it is
removed by a termination step. It is also possible to have non-cyclic chain
reactions, involving many reactive species and elementary steps. Non-cyclic
chain reactions can have extremely complicated kinetic mechanisms.

Linear and Branched Chain Reactions
286
Chain reactions in which each propagation step produces only one
reactive intermediate are called linear chain reactions.
Branched chain reactions are also possible, in which a chain carrier
reacts to form more than one chain carrier in a single elementary
step.
The hydrogen-bromine reaction
The hydrogen-chlorine reaction
The hydrogen-iodine reaction

Linear and Branched Chain Reactions
287
Linear chain reactions
Chain reactions without branching steps
Examples: H2 + Br2 reaction,,
alkane pyrolysis and polimerization
reactions
Branched chain reactions
Chain reactions that include branching
reaction steps
Examples: H2+O2 reaction,
hydrocarbonair explosions and flames

The Hydrogen-Bromine Reaction
288
 Overall reaction equation
 The measured order of 1/2 with respect to Br2 indicated that the reaction
proceeds via a complex reaction mechanism, rather than a simple bimolecular
collision.
 Further investigation showed that this rate law in fact only holds for the early
stages of the reaction, and that the true rate law takes the form

The Hydrogen-Bromine Reaction
289
 Any proposed mechanism for the reaction must agree with both of these
observations.
 The first step in any chain reaction is the initiation step. The reaction between
H2 and Br2 can be initiated by either thermally-induced or photon-induced
dissociation of Br2
or
 We will concentrate on the thermal mechanism for the purposes of deriving a
rate law for the overall reaction, but the steps following the initiation step are
the same for both cases.

The Hydrogen-Bromine Reaction- Mechanism
290
 In the second step, because the H–H bond is stronger than the H–Br bond, the
reverse (inhibition) reaction becomes possible once an appreciable amount of
HBr has built up.
 The reaction chain contains two radical chain carriers, H and Br.
 In order to arrive at an overall rate law for the reaction, we apply the steady-
state approximation to the two chain carriers.

291

292
Concentrationtime profiles of the H2Br2 reaction
(stoichiometric mixture, T= 600 K, p= 1 atm)

293
rates of R1 and R5 << rates of R2
and R3
rate of R1 = rate of R5
In the case of small [HBr] :
rate of R2 = rate of R3
production rates
d[H2]/dt -100.1
d[Br2]/dt -100.1
d[HBr]/dt +200.2
d[H]/dt +0.0014
d[Br]/dt +0.0026
rates of reaction steps
R1 Br2+M2 Br+M 1.0
R2 Br+H2HBr+H 100.2
R3 H+Br2HBr+Br 100.1
R4 H+HBrH2+Br 0.1
R5 2 Br+M  Br2+M 1.0
Relative rates at t = 1 second (all rates are normalized )with respect to R1
 
4
3
2
d
H
d
v
v
v
t



 
5
4
3
2
1 2
2
d
Br
d
v
v
v
v
v
t





0.0014 = +100.2 –100.1 –0.1
0.0026 = 2.0 – 100.2 + 100.1 + 0.1 – 2.0
 
4
3
2
d
HBr
d
v
v
v
t



200.2 = +100.2 +100.1 –0.1 M
Br
2
M
Br
1 2 


H
HBr
H
Br
2 2 


Br
HBr
Br
H
3 2 


Br
H
HBr
H
4 2 


M
Br
M
Br
2
5 2 



Chain Length
294
Mean number of propagation steps which occur before termination =
1
.
50
2
2
.
100
v
2
v
5
2



consumption rate of the chain carrier in the propagation step

consumption rate of the chain carrier in the termination step
The chain length at t =1 s
in the H2Br2 reaction
at the defined conditions

295
 Solve these two equations to obtain expressions for the concentrations of H
and Br in terms of the reactant and product concentrations and the various
rate constants.
 Simplify the required algebra considerably through careful inspection.
 The two equations each depend on both chain carrier concentrations, and
also share terms.
 We can simplify the process of solving the equations by adding them together
to give:

296

297

298
In the early stages of the reaction, the concentration of the HBr product is much
lower than that of the reactant Br2, and the second term in the denominator
becomes negligible. The rate law then reduces to

Presentation Week By Sabanci University Bekir Dizman

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