Chemistry 232
Kinetics of Complex
Reactions
The Pre-Equilibrium
Approximation

Examine the following process
A
k1
B
k 1
v1  k1  A 
v 1  k 1 B 
k2
B C
v 2  k 2 B 
d C 
 k 2 B 
dt
Pre-Equilibrium (II)

B is obviously an intermediate in the
above mechanism.
• Could use SSA.

What if the initial equilibrium is fast?
• Step 2 is the rds!
k1
  A   K
B    A  
k 1
Pre-Equilibrium (III)

We now have a simple expression for
the [B]; hence
d C 

k1 
 k 2 B    k 2 
  A   k '  A 
dt
k 1 

Lindemann-Hinshelwood
Mechanism

An early attempt to explain the kinetics
of complex reactions.
Mechanism
k1
Rate Laws
A  AA  A

k 1
A  A 2 A

k2
A P

v 1  k 1 A 
2
 
v 1  k 1 A  A
 
v 2  k 2 A


The ‘Activated’ Intermediate


Formation of the product depends
directly on the [A*].
Apply the SSA to the net rate of
formation of the intermediate [A*]
d A
dt

  k A 
2
1
 
 
 k 1 A A   k 2 A  0


Is That Your ‘Final Answer’?

Substituting and rearranging
d P 
k 2 k 1 A 

dt
k 2  k 1 A 
2
The ‘Apparent Rate Constant’
Depends on Pressure

The rate laws for the LindemannHinshelwood Mechanism are pressure
dependent.
High Pressure Case
Low Pressure Case
d P  k 2 k 1 A 

dt
k 1
d P 
2
 k 1 A 
dt
2
/
 k A 
 k / A 
The Pressure
Dependence of k’

In the Lindemann-Hinshelwoood
Mechanism, the rate constant is
pressure dependent.
1
1
k 1


/
k 1 A  k 1 k 2
k
Catalysts

So far, we have considered one way of
speeding up a reaction (i.e. increasing T
usually increases k). Another way is by
the use of a catalyst.

A catalyst - a substance that speeds up
the rate of the reaction without being
consumed in the overall reaction.

look at the following two reactions
A+B  C

rate constant k
A+B  C rate constant with catalyst is
kcat
NOTE: RATE WITH CATALYST >
RATE WITHOUT CATALYST
Types of Catalyst

We will briefly discuss three types of
catalysts. The type of catalyst depends
on the phase of the catalyst and the
reacting species.
• Homogeneous
• Heterogeneous
• Enzyme
Homogeneous Catalysis


The catalyst and the reactants are in the
same phase
e.g. Oxidation of SO2 (g) to SO3 (g)
2 SO2(g) + O2(g)  2 SO3 (g)

SLOW
Presence of NO (g), the following occurs.
NO (g) + O2 (g)  NO2 (g)
NO2 (g) + SO2 (g)  SO3 (g) + NO (g)
FAST



SO3 (g) is a potent acid rain gas
H2O (l) + SO3 (g)  H2SO4 (aq)
Note the rate of NO2(g) oxidizing SO2(g)
to SO3(g) is faster than the direct
oxidation.
NOx(g) are produced from burning fossil
fuels such as gasoline, coal, oil!!
Heterogeneous Catalysis

The catalyst and the reactants are in
different phases
• adsorption the binding of molecules on a
surface.

Adsorption on the surface occurs on
active sites
• Places where reacting molecules are
adsorbed and physically bond to the metal
surface.

The hydrogenation of ethene (C2H4 (g))
to ethane
C2H4 (g) + H2(g)  C2H6 (g)


Reaction is energetically favourable
• rxnH
= -136.98 kJ/mole of ethane.
With a finely divided metal such as Ni
(s), Pt (s), or Pd(s), the reaction goes
very quickly .

There are four main steps in the process
• the molecules approach the surface;
• H2 (g) and C2H4 (g) adsorb on the surface;
• H2 dissociates to form H(g) on the surface;
•
the adsorbed H atoms migrate to the
adsorbed C2H4 and react to form the product
(C2H6) on the surface
the product desorbs from the surface and
diffuses back to the gas phase
Simplified Model for Enzyme
Catalysis


E  enzyme; S  substrate; P  product
E + S  ES
ES P + E
rate = k [ES]
The reaction rate depends directly on the
concentration of the substrate.
Enzyme Catalysis



Enzymes - proteins (M > 10000 g/mol)
High degree of specificity (i.e., they will
react with one substance and one
substance primarily
Living cell > 3000 different enzymes
The Lock and Key Hypothesis


Enzymes are large, usually floppy
molecules. Being proteins, they are
folded into fixed configuration.
According to Fischer, active site is rigid,
the substrate’s molecular structure
exactly fits the “lock” (hence, the “key”).
The Lock and Key (II)
The Michaelis-Menten
Mechanism

Enzyme kinetics – use the SSA to examine
the kinetics of this mechanism.
E S
k1
k 1
k2
ES  P  E
ES – the enzyme-substrate complex.
Applying the SSA to the
Mechanism


Note that the formation of the product
depends directly on the [ES]
What is the net rate of formation of [ES]?
d ES 
 k 1 E S o  k 1 ES   k 2 ES 
dt
ES – The Intermediate

Apply the SSA to the equation for
d[ES]/dt = 0
d ES 
0
dt
k 1 E S o  k 1 ES   k 2 ES 
k 1 E S o
ES  
k 1  k 2
Working Out the Details

Let [E]o = [E] + [ES]
Initial enzyme
concentration
Free enzyme
concentration
Note that [E] = [E]o - [ES]
Complex
concentration
k 1 E o S o
ES  
k 1  k 2  k 1 S o
The Final Equation

Substituting into the rate law vp.
v p  k 2 ES 
vp

k 1 E o S o
 k 2 
 k 1  k 2  k 1 S o
k 2 E o S o

S o  K M



The Michaelis Constant and the
Turnover Number

The Michaelis Constant is defined as
KM


k 2  k 1

k1
The rate constant for product
formation, k2, is the turnover number for
the catalyst.
Ratio of k2 / KM – indication of catalytic
efficiency.
The Maximum Velocity

As [S]o gets very large.
lim v p
S o 
 k 2 E o  v max
Note – Vmax is the maximum velocity for
the reaction. The limiting value of the
reaction rate high initial substrate
concentrations.
Lineweaver-Burk Equation

Plot the inverse of the reaction rate vs.
the inverse of the initial substrate
concentration.
1
1
KM 1


v o v max v max S o
Chain Reactions

Classifying steps in a chain reaction.
• Initiation
• C2H6 (g) 2 CH3•
• Propagation Steps
• C2H6 + •CH3 
•C2H5 + CH4
• Branching Steps
• H2O + •O•  2 •OH
Chain Reactions (Cont’d)



Retardation Step
• HBr + H•
 H2 + Br•
Terminations Steps
• 2 CH3CH2•
 CH3CH2CH2CH3
Inhibition Steps
• R• + CH3•
 RCH3
The H2 + Br2 Reaction

The overall rate for the reaction was
established in 1906 by Bodenstein and Lind
d HBr 
k H 2 Br2  2

/
Br2   k HBr 
dt
3
The Mechanism

The mechanism was proposed independently
by Christiansen and Herzfeld and by Michael
Polyani.
Mechanism
Rate Laws
Br2  2Br 
v1  k1Br2 
Br  H2  HBr  H  v 2  k2 Br H2 
H  Br2  HBr  Br  v 2  k2 Br2 H 

H  HBr  H2  Br  v 3  k3 H HBr
2
v 4  k 4 Br 
Br  Br   Br2
Using the SSA

Using the SSA on the rates of formation
of Br• and H•
3
k


2k 2  1 H 2 Br2  2
k4 
d HBr 


k
dt
Br2   3 k / HBr 
2
Hydrogenation of Ethane

The Rice-Herzfeld Mechanism
Mechanism
C2H6  2CH3 
C2H6  CH3   CH3CH2  CH4
CH3CH2   CH2CH2  H 
CH3CH3  H   CH3CH2  H2
CH3CH2  H   C2H6
Rate Laws for the RiceHerzfeld Mechanism

The rate laws for the elementary
reactions are as follows.
v1  k1C2H6 
v 2  k 2 C2H6 CH3 
v 2  k 2 CH3CH2 
v 2 /  k 2 / H CH3CH3 
v 3  k3 H CH3CH2 
Explosions

Thermal explosions
• Rapid increase in the reactions rate with
temperature.

Chain branching explosions
•
chain branching steps in the mechanism lead
to a rapid (exponential) increase in the
number of chain carriers in the system.
Photochemical Reactions


Many reactions are initiated by the
absorption of light.
Stark-Einstein Law – one photon is absorbed
by each molecule responsible for the primary
photochemical process.
vI  I
I = Intensity of the absorbed radiation
Primary Quantum Yield

Define the primary quantum yield, 
# of primary products

# of photons absorbed
 Define the overall quantum yield, 
# of reactant molecules that react

# of photons absorbed
Photosensitization

Transfer of excitation energy from one
molecule (the photosensitizer) to
another nonabsorbing species during a
collision..
254 nm
Hg  Hg 
Hg  H 2  Hg  2H 
Hg   H 2  HgH  H 
Polymerization Kinetics

Chain polymerization
• Activated monomer attacks another
monomer, chemically bonds to the monomer,
and then the whole unit proceeds to attack
another monomer.

Stepwise polymerization
• A reaction in which a small molecule (e.g.,
H2O) is eliminated in each step.
Chain Polymerization


The overall polymerization rate is first order
in monomer and ½ order in initiator.
The kinetic chain length, kcl
• Measure of the efficiency of the chain
propagation reaction.
 kcl
vp
# of monomer units consumed


vi
# of active centres produced
Mechanism

Initiation
I  2 R•
Or
M + R•  M1 •

Rate Laws
v i  ki I 
Propagation
M + M1•  M2 •
M + M2•  M3 •
M + M3•  M4 •
Etc.
v p  k p M M n 1 
Mechanism (Cont’d)

Termination
M + M 3 •  M4 •
v t  kt M 
2
Note – Not all the initiator molecules produce chains
Define  = fraction of initiator molecules that produce chains
d M 

 2k i I 
dt
Return to Kinetic Chain Length

We can express the kinetic chain length
in terms of kt and kp
 kcl

k p M  M  

2
2k t M  
k p M  I 
2  k i k t

1
1
2
2
Stepwise Polymerization

A classic example of a stepwise
polymerization – nylon production.
NH2-(CH2)6-NH2 + HOOC-(CH2)4COOH 
NH2-(CH2)6-NHOC-(CH2)4COOH + H2O

After many steps
H-(NH-(CH2)6-NHOC-(CH2)4CO)n-OH
The Reaction Rate Law

Consider the condensation of a generic
hydroxyacid
OH-M-COOH

Expect the following rate law
v poly  k poly  OH  COOH 
The Reaction Rate Law (Cont’d)



Let [A] = [-COOH]
A can be taken as any generic end
group for the polymer undergoing
condensation.
Note 1 –OH for each –COOH
v poly  k poly  OH A
 k poly A
2
The Reaction Rate Law (Cont’d)

If the rate constant is independent of the
molar mass of the polymer

 COOH o
 COOH t 
1  k poly t  COOH o

Ao

1  k poly t Ao
The Fraction of
Polymerization

Denote p = the fraction of end groups
that have polymerized

Ao  At
p
Ao
p
k poly t Ao
1  k poly t Ao
Statistics of Polymerization

Define Pn = total probability that a
polymer is composed of n-monomers
Pn  p 
n 1
1  p 
The Degree of
Polymerization

Define <n> as the average number of
monomers in the chain
 1  Ao
 
n  
 1  p  At
Degree of Polymerization
(cont’d)

The average polymer length in a
stepwise polymerization increases as
time increases.
k poly t Ao
 1 
  1 
n  
1  k poly t Ao
 1 p 
 1  k poly t Ao
Molar Masses of Polymers


The average molar mass of the polymer
also increases with time.
Two types of molar mass distributions.
• <M>n = the number averaged molar mass of
•
the polymer.
<M>w = the mass averaged molar mass of the
polymer.
Definitions of <M>n

Two definitions!
M

n
1

Mo
1  p 
1
n
M

J
J
J
n
Mo = molar mass of monomer
n = number of polymers of mass Mn
MJ = molar mass of polymer of length nJ
Definitions of <M>w

<M>w is defined as follows
M
 1  p  Mo  j x n  p
2
w
2
nJ M J

J nJ MJ
Note - xn the number of monomer
units in a polymer molecule
x n 1
The Dispersity of a Polymer
Mixture


Polymers consists of many molecules of
varying sizes.
Define the dispersity index () of the
mass distribution.

M
M
w
n
Note – monodisperse sample
ideally has <M>w=<M>n
The Dispersity Index in a Stepwise
Polymerization

The dispersity index varies as follows in
a condensation polymerization
1  
M
M
w
n
Note – as the polymerization
proceeds, the ratio of <M>w/<M>n
approaches 2!!!
Mass Distributions in
Polymer Samples

For a random polymer sample
Monodisperse Sample
Polydisperse Sample
Pn
09
11
13
15
17
19
21
23
25
27
29
31
33
35
37
Molar mass / (10000 g/mole)
39
41