What have we learned about kinetics so far?
(1) Reaction kinetics are described by rate laws
(2) Reaction rates are expressed quantitatively by rate
constants and rate orders, both of which must be
determined experimentally.
(3) First order reactions show an exponential decrease of
the reactant concentration with time; the half-life is
independent of the initial reactant concentration
(4) Second-order reactions show a curved decrease of the
reactant concentration with time; the half-life is inversely
related to the initial reactant concentration
(5) Zero-order reactions show a linear decrease of the
reactant concentration with time; the half-life is directly
related to the initial reactant concentration
What have we learned about kinetics?
(1) Rate: change in concentration per unit of time
(2) Rate expression: for aA + bB  cC + dD
rate = - d[A] = - d[B] = d[C] = d[D]
a*dt
b*dt
c*dt
d*dt
(3) Differential rate law: rate=k[A]n[B]m , where k, m, and n have to be
experimentally determined
(4) Order of reaction w.r.t. A:
1st: rate = k[A]; 2nd: rate = k[A]2; zero: rate = k
(5) Integrated rate law: 1st: plot of ln[A] vs t is linear 
ln[A] = -kt + ln[A]o
2nd: plot of 1/[A] vs t is linear 
1/[A] = kt + 1/[A]o
zero: plot of [A] vs t is linear 
[A] = -kt + [A]o
(6) Integrated rate law for reactions with more than 1 reactant:
Pick one reactant to change and hold the others constant (use large excess).
Combine rate constant and constant concentrations into k’, the pseudo rate
constant, simplifying the rate law with respect to the chosen reactant.
Chapter #15 – Chemical Kinetics
15.1) Reaction Rates
15.2) Rate Laws: Introduction
15.3) Determining the Form of the Rate Law
15.4) Integrated Rate Law
15.5) Rate Laws: Summary
15.6) Reaction Mechanisms
15.7) The Steady-State Approximation
15.8) A Model for Chemical Kinetics
15.9) Catalysis
Why do we study reaction kinetics?
To fully understand or apply any chemical reaction, we must
know more than just the identities of the reactants and the
products.
We must know:
• if the reaction will occur (thermodynamically favorable?)
• how long it will take to occur (kinetically feasible?)
For a complete understanding, we should also know how it
occurs…reaction mechanisms and a model of chemical
kinetics (collisions, effect of temperature…).
When chemists asks: "how does this reaction occur ?"
they are asking for a reaction mechanism.
A reaction mechanism is the series of bondbreaking and bond-forming steps that occur
during the conversion of reactants to products.
Chemical reactions are often represented by a single
balanced equation, but that does not mean that the reaction
will take place in only one step. Most reactions occur in more
than one step.
Consider for example the reaction:
NO2(g) + CO(g)  NO(g) + CO2(g)
How does this reaction occur? Kinetics measurements are
key experiments used to elucidate reaction mechanisms.
NO2(g) + CO(g)  NO(g) + CO2(g)
If this reaction proceeds in one step as written, the
rate law should be
Rate = k[NO2][CO]
Experimental results show: Rate = k[NO2]2
The rate of this reaction is actually independent of
[CO] and 2nd order w.r.t. [NO2].
The reaction, therefore, does NOT proceed in one
elementary step as written above.
An elementary step is a reaction for which the
rate law can be written from the molecularity...
MOLECULARITY of an elementary step:
the number of species (i.e. reactants) that must
collide to produce the reaction indicated by that step
Examples (Table 15.7):
A (1 molecule)  products
Unimolecular Rate = k[A]
A+A (2 molecules)  products
(2A)
Bimolecular
Rate = k[A]2
A+B (2 molecules)  products
Bimolecular
Rate = k[A][B]
2A+B (3 molecules)  products
Termolecular Rate = k[A]2[B]
A+B+C (3 molecules)  products
Termolecular Rate = k[A][B][C]
All elementary steps in a reaction mechanism will
always add up to give the balanced chemical equation
or the net equation that describes the overall reaction.
Consider the following reaction:
3 ClO- (aq)  ClO3- (aq) + 2 Cl- (aq)
Species that are formed in one step and consumed in
another are called intermediates – they do not appear in the
overall reaction equation or the rate law.
A note about elementary steps:
Unlike the rate law for an overall reaction, which is
experimentally determined, the rate law for an
elementary step can be obtained from the
stoichiometry of the species reacting in that step.
The coefficients in the balanced chemical equation
of an elementary step will appear in the rate law of
the elementary step.
What have we learned about how this reaction occurs?
k
NO2(g) + CO(g)  NO(g) + CO2(g)
Rate = k[NO2]2
• The overall reaction must proceed in multiple elementary
steps.
• The slowest step must involve the collision of two NO2
molecules (because we know the reaction is second order
w.r.t. NO2):
k
NO2 + NO2  intermediate(s)
The slowest elementary step determines the rate of
the overall reaction. This step is, therefore, called the
Rate Determining Step.
We can use these results to develop a detailed
reaction mechanism…
Necessary criteria for a valid proposed mechanism:
1. The sum of the elementary steps must equal the overall
balanced equation for the reaction.
2. The mechanism must agree with the experimentally
determined rate law.
Sometimes multiple mechanisms exist that meet
both of these two criteria.
These criteria are, therefore, necessary but not
sufficient to prove a mechanism.
The proposed mechanism:
k1
NO2 + NO2  NO3(g) + NO(g)
k2
NO3(g) + CO(g)  NO2(g) + CO2(g)
Overall:
k
NO2(g) + CO(g)  NO(g) + CO2(g)
k1 is SMALL (slow step)
k2 is LARGE (fast step)
Rate = k[NO2]2
The predicted Rate = k[NO2]2 of the overall reaction
is now consistent with experiment, but it does not
prove the mechanism is correct.
Note: The species NO3(g) is not a reactant or a
product. It is "an intermediate."
Rate Laws and Intermediates
Reaction intermediates cannot be part of the rate
law for a reaction.
This means that you need to use algebra and
substitutions to replace any intermediates that are
present as reactants in the rate determining step.
Consider the following reaction:
H2 (g) + CO (g) H2CO (g)
H2 (g) + CO (g) H2CO (g)
Rate Laws and Mechanisms
H2 (g) + CO (g) H2CO (g)
Rate = k [H2]1/2[CO]
• Rate law agrees with experimentally observed law
• No intermediates appear in the rate law
• Sum of the elementary steps = the overall reaction
The mechanism is plausible!!
Developing a Mechanism
1. Experimentally determine the rate law
2. Use chemical intuition to propose plausible
mechanisms
3. Perform more experiments to eliminate least likely
mechanisms
The Steady-State Approximation
(Zumdahl Section 15.7)
Helpful with complex, multi-step reactions, where a
specific rate-determining step cannot be chosen.
Assume that the concentrations of any intermediates
remain constant as the reaction proceeds.
Example: 2 NO(g) + H2(g)  N2O(g) + H2O(g)
Proposed Mechanism:
Bimolecular
2NO
k1
N2O2
k-1
Unimolecular
N2O2 + H2
k2
N2O + H2O
The double arrows in the first step indicate that both the
forward and reverse of this step are important on the time
scale of the overall reaction.
Let's make the approximation: [N2O2] is constant
Example (cont)
If [N2O2] is constant, then d[N2O2]/dt = 0
2 NO
N 2 O 2 + H2
k1
k-1
k2
Rate of production
of N2O2
=
d[N2O2]/dt = k1[NO]2
=
N 2O 2
N 2O + H 2O
Total rate of consumption
of N2O2
-d[N2O2]/dt = k-1[N2O2] +
k2[N2O2][H2]
Steady-State Condition:
k1[NO]2 = k-1[N2O2] + k2[N2O2][H2]
Solve for [N2O2]:
k1[NO]2 = [N2O2] (k-1 + k2[H2])
[N2O2] =
k1[NO]2
k-1+ k2[H2]
Substitute into rate law…
Example (cont):
We now use this information to determine the rate
law for the overall reaction:
2NO(g) + H2(g)  N2O(g) + H2O(g)
Rate of reaction = -d[H2]/dt = k2[H2][N2O2]
(from 2nd step of mechanism)
Substituting the steady-state equation for [N2O2]
into this rate equation gives:
k1[NO]2
Rate = -d[H2]/dt = k2[H2]
k-1 + k2[H2]
Example (cont):
2NO(g) + H2(g)  N2O(g) + H2O(g)
k1[NO]2
Rate = -d[H2]/dt = k2[H2]
k-1 + k2[H2]
To test this mechanism, we can test the rate law by
changing the concentrations of H2 and NO
For large [H2], k2[H2] >> k-1
Rate ≈ k1[NO]2
For large [NO], k2[H2] << k-1
Rate ≈ k [H2][NO]2
The simplified rate laws obtained from the
steady-state approximation provide
predictions that we can test.
What have we learned about mechanisms?
(1) Reaction mechanism is a series of elementary steps with
simple reaction orders
(2) Reaction mechanisms must satisfy the following:
a) the sum of the elementary steps must give the overall
balanced equation for the reaction
b) the mechanism must agree with the experimentally
determined rate law
(3) Intermediates: are formed and consumed within the overall
reaction, so cannot appear in the overall reaction equation nor
in the rate law for the reaction
(4) Steady state approximation: allows us to simplify the reaction
kinetics of complex, multi-step reactions by assuming that the
intermediates are at equilibrium
Chapter #15 – Chemical Kinetics
15.1) Reaction Rates
15.2) Rate Laws: Introduction
15.3) Determining the Form of the Rate Law
15.4) Integrated Rate Law
15.5) Rate Laws: Summary
15.6) Reaction Mechanisms
15.7) The Steady-State Approximation
15.8) A Model for Chemical Kinetics
15.9) Catalysis
A Model for Chemical Kinetics
(Zumdahl Section 15.8)
The rates of reactions are influenced by many factors:
• Concentrations of reactants affect rates
• Temperature
• Catalysts
Obviously, a rate constant is not a universal constant, but is
constant only for a fixed set of experimental conditions.
How do we explain the dependence of reaction rates on
temperatures and catalysts?
i.e. Where do rate constants come from?
Collision Theory of Reactions
Consider a gas phase reaction: 2A(g)
products
This reaction requires collision of 2 reactants.
• We can calculate the collision frequency from the ideal
gas law. (Zumdahl, Section 5.9)
• We can measure the reaction rate experimentally.
For such reactions, the reaction rate is typically several
orders of magnitude slower than that predicted from the
collisional frequency alone.
We can conclude that not every collision of
reactants results in a successful reaction.
What factors might keep the two reactants
from reacting once they have collided?
One obvious candidate is molecular orientation: only the
correct orientation of reactants will lead to successful product
formation.
Figure15.13
This could account for perhaps 1 order of magnitude, but not
for the several orders of magnitude reduction in rates observed
experimentally…
the answer must be more complex.
Arrhenius believed that for molecules to react upon collision,
they must become "activated," and the parameter Ea has
become known as the Activation Energy.
Figure 15.11a
Only some collisions occur with enough energy to
overcome the activation barrier (Ea).
Transition state and DE vs Ea
• DE tells you if the rxn is exothermic or endothermic, but is not
related to rate and doesn’t tell you anything about the
rate…leave that to Ea.
• Ea for the reverse reaction can be calculated using Ea for the
forward reaction and DE.
The fraction of effective collisions increases
exponentially with T.
Figure 15.12
Number of Collisions
Recall properties of gases from Ch 5 – distribution of velocities
The Pop Bottle Demo…
H2 (g) + ½ O2 (g)  H2O (l)
DH = DHf(H2O) = - 286 kJ/mol
H2 flammble, O2 strong oxidizer
• Water is very stable relative to its elements.
• This reaction is exothermic and spontaneous, so why are
the reactants just sitting there?
• Initial energy (heat) released by the reaction expands the
gases greatly, creating an explosive force within the
confined space of the thick-walled glass bottle.
Reaction rates have a strong dependence on temperature.
This gives us a clue that molecular collisions overcoming Ea
might be involved…
Note that this temperature
dependence is not linear.
In the late 1800s, Svante
Arrhenius suggested that k
varies exponentially with 1/T
k = Ae-Ea/RT
Ea is the activation energy (a constant with units of energy),
R is the gas constant (in Joules), and A (frequency factor) is
a constant with the same units as k.
Intuitive understanding of Arrhenius plots…
Even if we don’t know about Maxwell-Boltzmann distributions,
we intuitively expect reactions to proceed faster at higher
temperatures.
k = Ae-Ea/RT
Taking the natural logarithm of
this equation gives:
ln k = ln A - Ea/RT
ln k = - Ea/RT + ln A
y = mx + b
so that a plot of ln k vs. 1/T
should be a straight line with
slope of -Ea/R.
Many rate constants do follow
this behavior.
Finding the activation energy from 2 data points
Instead of plotting ln k vs. 1/T to get Ea from the slope,
you can calculate Ea from two sets of data at different T:
ln k2 = ln A - Ea/RT2
ln k1 = ln A - Ea/RT1
Taking the difference of these two equations:
ln k2 - ln k1 = (-Ea/RT2) - (-Ea/RT1)
ln (k2/k1) = -(Ea/R)(1/T2 - 1/T1)
or
ln (k2/k1) = (Ea/R)(1/T1 - 1/T2)
Example:
2HI(g)
H2(g) + I2(g)
Rate = k[HI]2
T1 = 500 K
T2 = 600 K
k1 = 9.51  10-9 M-1 s-1
k2 = 1.10  10-5 M-1 s-1 (big change !!!)
𝑘2
𝐸𝑎 1
1
ln = −
−
𝑘1
𝑅 𝑇2
𝑇1
1.1 𝑥 10−5
𝐸𝑎
1
1
ln
=−
−
−9
9.51 𝑥 10
8.3145 𝐽/𝑚𝑜𝑙 ∙ 𝐾 600 𝐾
500 𝐾
Rearrange and solve:
Ea = 176 kJ/mol
Demo…
Potassium Permanganate and Glycerol
14 KMnO4 (s) + 4 C3H5(OH)3 (s) 
7 K2CO3 (s) + 7 Mn2O3 (s) + 5 CO2 (g) + 16 H2O (g)
• Very exothermic redox reaction
• Glycerol easily oxidized
• Large KMnO4 crystals…lots of surface area
• Takes a little bit to get going
 Once there’s a little heat, glycerol boils and KMnO4 ignites
Why do we care about Ea?
• It gives us an idea of the size of an energy barrier a
reaction has to overcome.
– Large barriers require a great deal of energy to get the
reaction started
– Low barriers require less energy
• Can chemists do anything to influence activation
energies?
YES!
Catalysis
• 16 of the top 20 synthetic chemicals in the US are
produced via catalysis.
• Most of these catalysts are inorganic transition
metal compounds.
Why are these chemicals important?
• Sulfuric acid – car batteries, ore processing,
oil refining, chemical synthesis, fertilizers
• Ethene (ethylene) – feedstock for the polymer
polyethylene, oxidized to make detergents, as
a starting material for nearly every polymer
• Propene (propylene) – feedstock for
polypropylene (packaging, clothing)
• Polyethylene – soda bottles, most widely
used plastic in the world
Why are these chemicals important?
• Cumene – used to produce phenol and acetone
– Phenol – used in the production of polycarbonates,
resins and eventually nylon
– Acetone – primarily used as a solvent and a precursor
to PMMA (Plexiglas)
• Acetic acid – produced catalytically (Monsanto
Process and Cativa Process)
– Primary use – vinyl acetate monomer (paints,
adhesives)
– Other uses – acetic anhydride (aspirin, cellulose film,
heroin) and esters (inks, paints, varnish, lacquers)
What is a catalyst?
• A catalyst speeds up a reaction by
– changing the reaction pathway
– lowering the activation energy
• A catalyst is not consumed in the reaction,
but is regenerated.
Catalytic pathways
• Catalysts create new
reaction pathways without
affecting the overall
thermodynamics of a
reaction.
• Catalysts cannot make a
thermodynamically
unfavorable reaction more
favorable.
Catalytic Cycles
Catalysts are not consumed in the reaction – they are
regenerated through the catalytic cycle.
Catalysis definitions
• Homogeneous catalysis:
– The catalyst and the reagents are all in the same
phase (usually liquid).
– Advantage: Usually more selective, easier to
dissipate heat from a solution
• Heterogeneous catalysis:
– The catalyst and the reagents are in different phases
(the catalyst is usually solid, while the reagents are
liquid or gas)
– Advantage: Easier to separate the catalyst from the
reaction mix, more robust
Catalysts
• Catalysts are used in a huge variety of ways because they
can enhance reaction rates by many orders of magnitude!
• In general, they work by lowering the activation barrier to a
reaction.
Note that the energies of the
reactants and products do not
change, only the energy
barrier changes.
The transition state is
stabilized by the catalyst!
Catalysts and T.S. Stabilization
Catalysts
Fig 15.16. Effect of Catalysts on Number of Effective Collisions
How do catalysts increase the number of effective collisions?
Often by holding molecules in the correct orientation for a
reaction to occur.
Example of Heterogeneous Catalysts
Figure 15.16: Exhaust gases
2N2O(g)
2N2(g) + O2(g)
Figure 15.17: Hydrogenation of ethylene
C2H4(g) + H2(g)
C2H6(g)
Experts at UW – see the “Research”
page on the Chem Dept website (here).
Two demos…
Cobalt chloride catalysis: combining tartrate and hydrogen
peroxide in the presence and absence of the CoCl2 catalyst;
reusing catalyst
CoCl2
NaK[O2CCH(OH)]2 + 5 H2O2  4 CO2 + 6 H2O + NaOH + KOH
(aq)
(aq)
(g)
(g)
(aq)
(aq)
Catalytic decomposition of H2O2 with MnO2: hydrogen
peroxide will decompose on it’s own to water and oxygen, but
it will take a while
MnO2
H2O2  ½ O2 + H2O
(l)
(g)
(l)
Key Catalytic Processes
• Alkene metathesis
– Redistribution of carbon-carbon double bonds
• Acetic acid synthesis
– Highly selective process using Rh or Ir
• Fischer-Tropsch Synthesis
– Conversion of hydrogen and carbon monoxide to
hydrocarbons and water
• Polymerization
– The majority of plastics we use every day are the
result of catalytic polymerization reactions
Alkene Metathesis
• Alkene metathesis
is important
because it
redistributes double
bonds.
• Nobel Prize in
Chemistry – 2005
to Robert Grubbs,
Yves Chauvin,
Richard Schrock
Acetic Acid Synthesis
A rhodium catalyst is used in the Monsanto process for the
production of acetic acid from methanol.
Fischer-Tropsch Synthesis
• Syngas (H2 and CO) is converted to
hydrocarbons and water:
CO + 2H2  —CH2 — + H2O
(2n+1) H2 + n CO  CnH(2n+2) + H2O
• Provides a non-petroleum source of fuels
(especially low-sulfur diesel) – used by Germany
in WWII.
• Where does the CO and H2 come from?
– Water-gas shift reaction (H2O+COH2+CO2)
– Steam reforming (H2O+CH4CO+3H2)
Steam reforming
• Also known as ―gasification‖
H2O + CH4  CO + 3H2
• Coal was a common source of the methane used.
Polymerization
• The linking together of monomers to create chains of CC bonds has drastically altered life since the 1950s.
• Polymers such as
polyethylene,
polystyrene, and
polypropylene make
up common items.
• Properties at the
molecular level
determine polymer
properties.
Enzymes are Catalysts!
Telomeres…
• Special DNA sequences
at ends of eukarytoic
chromosomes
(TTAGGG in humans)
• Have roles in structure,
function, and protection of
chromosomes
• DNA replication problem:
telomeres shorten with
each cell replication cycle
Photos courtesy of:
1)NIGMS, NIH, US Dept. Health & Human Services
2)Telomere picture - docinthemachine.com
Human chromosomes
(blue) capped by
telomeres (white)
Enzymes are Catalysts!
Shortening of Telomeres…
Photo courtesy of:
Pharminox (www.pharminox.com)
Enzymes are Catalysts!
Telomerase…