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ME529 Combustion and Air Pollution
Topic 05a. Chemical Kinetics
5.1 Introduction
Chemical thermodynamics tells us if a reaction will proceed spontaneously at a given T
and P. Thermodynamics also tells us the composition of the reaction products that would
be reached eventually if the system were maintained at that T and P for a long enough
time to reach equilibrium. But to learn how fast the reaction will occur, and/or how long
it will take to reach equilibrium, chemical kinetics is required.
The rates of chemical reactions and the mechanism by which one chemical species is
converted into another form the subject matter of chemical kinetics.
The rate is the mass (or moles) of a product produced, or reactant consumed, per unit
time.
A mechanism is a sequence of individual chemical events whose overall result produces
the observed reaction.
For example, if the observed overall combustion reaction is
H2 + 1/2 O2 ==> H2O
the mechanism can include the following elementary reactions:
H2 <==> 2H
O2 <==> 2O
H + O <==> OH
HO + H <==> H2O
The rate of a chemical reaction depends on the temperature, pressure and concentrations
of species involved. Of course, a catalyst or inhibitor can alter the rate by orders of
magnitude.
Consider, for example, the reaction
CO(g) + NO(g) ===> CO2(g) + 1/2 N2(g)
The Gibbs free energy change for this reaction at 25C and 1 atm is -82.2 kcal, indicating
that the reaction is spontaneous under these conditions. Indeed, Kc at 25C is so large
(~1060) that these two toxic gases, even at the low concentrations found in automobile
exhausts, should combine almost completely to form carbon dioxide and molecular
nitrogen. Unfortunately, this reaction occurs so slowly under ordinary conditions that it
does not offer a practical method of removing carbon monoxide and nitrogen oxide from
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polluted air. However, when that polluted air can be passed over a suitable catalyst at a
higher temperature, the reaction can efficiently remove CO and NO. This reaction is the
basis for 2 of the 3 pollutants removed by the catalytic converters on gasoline-powered
automobiles (the third pollutant removed is HC by another reaction - more on 3-way
catalytic converters later when we discuss spark ignition engines).
Hence, there is no direct correlation between the rate of a reaction and the thermodynamic
driving force as expressed by the free energy change of the equilibrium constant. A
different set of principles, which encompass chemical kinetics, must be used to predict
how rapidly a reaction occurs.
5.2 Rate Constant
Consider the overall reaction between CO and nitrogen oxide:
CO(g) + NO(g) ===> CO2(g) + 1/2 N2(g)
The rate of this reaction is the change in concentration (rate of formation) of one of the
products, or the negative change in concentration (rate of destruction) of one of the
reactants, i.e.,
rate 
d CO2 
d CO

dt
dt
Because of the 1:1 stoichiometry of the reaction, these two rate expressions are
equivalent. For every mole of CO2 formed, one mole of CO disappears.
The rate is proportional to the concentrations of the reactants.
rate  kCONO
The proportionality constant k is the rate constant for the reaction. k is a function of
temperature only and for overall reactions it is independent of the concentrations of
reactants.
For a general reaction
aA(g) + bB(g) ===> products
the rate expression takes the form
rate = k [A]m [B]n
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The powers m and n describe the order of the reaction. The reaction is mth order with
respect to A and nth order with respect to B. The overall order of the reaction is the sum
of the exponents m and n.
The order of overall reactions must be determined experimentally and cannot be
deduced from the coefficients of the balanced reaction equation.
5.3 Dependence of the reaction rate on concentration - reaction order and integrated rate
equation
The order of a reaction indicates the dependence of the rate on concentration.
a) zeroth order reaction - the rate is independent of concentration; either
 the rate is intrinsically independent of concentration; or
 the species is in such abundant supply that its concentration is nearly constant
during reaction
d A
 k o A0  k o
dt
 d A  k o dt
 A
t
 d A  k o dt
 Ao
0
Ao  A  k o t
A  Ao  k o t
R


The half-life of a reaction is the time required for 1/2 of the reactant to be consumed.
1
Ao  Ao  ko t1/ 2
2
b) first order reaction
R
 A


 Ao
d A
 k1 A
dt
d A
 k dt
A 1
t

0
 A 
  k1t
ln 
 Ao 
A  Ao exp  k1t 
3
 t1 / 2 
Ao
2k o
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The half life:
1
Ao  Ao exp  k1t1 / 2   t1 / 2  1 ln 2
2
k1
An example: consider the decomposition of dinitrogen pentoxide
ME529 - Combustion and Air Pollution - Kinetics - 1st order rxn example - decomposition of dinitrogen
pentoxide
2 N 2O 5
==> 4  NO 2 O 2
0
1
1
0.765
2
t
3
(min)
N2O5
0.497
mol
0.349
lit
4
0.246
5
0.173
A plot of the log of concentration vs. time is a straight line (which means the reaction is
1st order):
0
lgN2O5
i
y
i
0.5
1
0
2
4
6
t t
i i
Plot of logX vs time is linear ==> 1st order rxn
6
1
Hence, by inspection:
1
N2O5
i
N2O5
i
0.5
4
ln A  ln Ao  k1t
2
or:
0
0
0
2
4
t
i
6
ln
 
Ao
0A
4
2
 k1t
4
t
i
6
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or:
ln
Ao
A
 k1t
c.i) second order reaction for [A] + [A] ==> products
R
 A


d A
 k 2 A2
dt
d A
 Ao A
2
t

 k 2 dt
0
1
1

k t
A Ao 2
The half life:
1
1

 k 2 t1 / 2
1 2 Ao Ao
1

Ao
 k 2 t1/ 2
 t1/ 2 
1
k 2 Ao
c.ii) second order reaction for a[A] + b[B] ==> products
Let [X] be the amount of A that has reacted (expressed as concentration). Then,
[A] = [A] o - [X]
[B] = [B] o - b/a [X]
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R
R
R
d Ao  X 
b


 k 2 AB   k 2 Ao  X B o  X 
dt
a


d X 
b
b


 k 2 Ao B o  B o X   Ao X   X 2 
dt
a
a




d X 
b
b


 k 2 Ao B o   B o  Ao X   X 2 
dt
a
a






 Ao  X  
 A 
 A  bAo  aB o
ln 
 ln   
k 2 t  ln  o 

a
 B  
 B o 
 B o  b X  
a


The half life (in terms of [A]/[A] o = 1/2):
t1 / 2 

aB o
1
ln 
k 2 bAo  aB o   2aBo  bAo



For example, consider the following reaction (decomposition of acetaldehyde)
CH3CHO(g) ===> CH4(g) + CO(g)
The following data is available from experiments:
[CH3CHO]
(mol/lit)
Rate
(mol/lit-sec)
0.1
0.2
0.3
0.4
0.02
0.081
0.182
0.318
The reaction rate is: rate = - d[CH3CHO]/dt = k [CH3CHO] m
a) find the order of the reaction
We have plenty of data to do this calculation. The reaction order m is constant. So just
use two sets of data:
rate2 = k [CH3CHO] 2m
rate1 = k [CH3CHO] 1m
Divide them and solve for m to see that this reaction is 2nd order:
rate2  conc2 


rate1  conc1 
m

0.081  0.2 

0.02  0.1 
6
m
 4  2m
 m2
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b) find the reaction rate constant k
rate = k [CH3CHO] 2
k
rate
conc
2

0.182
0.3
2
 2.0
lit
mol sec
The log of the rate plotted vs. the log of the concentration is a straight line with slope
equal to the order m and intercept equal to log(k).
0.4
0
log rate
i
2
1
2
1
1.3
0.5
log conc
i
0.3
m
slope( logc logr)
m  1.999
b
interceptlogc
(
 logr)
b  0.302
Note that 10 0.303 = 2.
c) Find the rate of reaction at [CH3CHO] = 0.15 mol/lit
rate = k [CH3CHO] m = 2.0 lit/mol-sec [0.15 mol/lit] 2 = 0.045 mol/lit-sec
SUMMARY: Characteristics of zero, 1st and 2nd order reactions
Order Rate
Units
Conc-time
t 1/2
0
k
mol/(lit-sec)
Xo - X = k t
Xo / 2 k
1
kX
1/sec
ln(Xo/X) = k t
0.693 / k
2
2
kX
lit/(mol-sec)
1/X - 1/Xo = k t
1 / k Xo
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Linear plot
X vs. t
ln X vs. t
1/X vs. t
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Third order reactions are rare. There is one, however, that is involved in smog formation:
2 NO(g) + O2 (g) ==> 2 NO2 (g)
rate = k [NO] 2[O2 ]
5.4 Rates of more complex reactions
Let's see how to determine the rate of more complex reactions:
a) opposing reactions (or reversible reaction)
b) concurrent reactions
c) consecutive reactions
a) simple opposing 1st order reactions
k1
A B
k1'
d A
 k1 A  k '1 B 
dt
note : B   B o  Ao  A
R
R
d A
 k1  k '1 A  k '1 Ao  B o 
dt
For steady state reactions at equilibrium, d[A]/dt = 0. Hence,
d A
eq  0  k1  Aeq  k '1 B eq
dt
 G o 
k
 1  K c T   exp 

k '1
 RT 
Req  
B eq
Aeq
Observe that we now have a relationship between the forward and reverse reaction rate
constants and the equilibrium constant.
k 
k
d A 
  k1  1 A  1 Ao  B o 
dt
Kc 
Kc

 K  1 

d A
1
A 
Ao  B o 
R
 k1  c
dt
Kc
 K c 

R
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Note that:
Kc 
Beq
Aeq
Aeq K c
Bo  Ao  Aeq
Aeq
d A
 1  
 A  B

o
dt
o
So we can write:
R
R

K 1
d A
 K  1 
A  c
Aeq 
 k1  c
dt
Kc
 K c 

 K 1
d A
 A  Aeq
 k1  c
dt
 Kc 


For a particular reaction, we could find [A] eq using STANJAN or another equilibrium
code. Then we can make the following substitutions and solve to obtain a simple
expression:
A'  A  Aeq
 K 1

k R  k1  c
 Kc 

 A  Aeq 

ln 
  k R t

 Ao  Aeq 

What if we have a small displacement y from equilibrium in concentration?
For example, given equilibrium concentrations [A] eq and [B] eq. Then,



dy
 k1 Aeq  y  k '1 Beq  y
dt

Eventually, dy/dt ==> 0 as the system shifts back to equilibrium. Hence,




dy
 k1 Aeq  k '1 Beq  k1  k1' y  0  k1  k1' y
dt
dy
  k1  k1' y
dt


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



 y 
ln     k1  k1' t
 yo 
y  yo exp  k1  k1' t
The relaxation time is defined as the time required for the change to 'relax' to 1/e of the
initial value.


yo
 y o exp  k1  k1' 
e
b) concurrent reactions or parallel reactions
k1
A B
k2
A C
d A
 k1 A  k 2  A  k1  k 2  A
dt
d B 
 k1  A
dt
d C 
 k 2 A
dt

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A  Ao exp  k1  k 2 t
d B 
 k1 Ao exp  k1  k 2 t
dt
B  k1 Ao 1  exp  k1  k 2 t
k1  k 2
C  
k2
Ao 1  exp  k1  k 2 t
k1  k 2
B  k1
C  k 2
c) consecutive reactions – see problem set
k1
A B
k2
B C
The steady state hypothesis: The concentrations of reactive intermediates are
approximately constant and their rate of change of concentration is approximately zero.
We can arrive at the steady state solution from the exact solution or use the steady state
assumption from the beginning. The steady state assumption applies to compounds
formed slowly but consumed quickly. The steady state assumption simplifies
mathematics, but you still have to write out the reaction mechanism.
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5.4 Chain Reactions
Chain reactions consist of three elementary steps:
I. Initiation
II. Propagation (carrying)
III. Termination (breaking)
For example, consider the following reaction: H2 + Br2 ==> 2HBr
Reaction mechanism for H2 + Br2 ==> 2HBr
Initiation
Br2 + M ==> 2 Br + M Endothermic
Propagation
Br + H2 ==> HBr + H
Endothermic
H + Br2 ==> HBr + Br Exothermic
H + HBr ==> H2 + Br
Exothermic
Termination
2Br + M==> Br2 + M
Exothermic
+45.2 kcal
+16.4 kcal
- 40.5 kcal
- 16.4 kcal
- 45.2 kcal
When chain branching occurs, more chain carriers are produced than consumed. For
example, during the combustion of hydrogen:
Initiation
Propagation
Branching
Termination
H2 + M ==> 2 H + M
OH + H2 ==> H2O + H
H + O2 ==> OH + O
2H + M ==> H2 + M
two radicals are produced from one radical
As branching occurs, entropy increases until the maximum entropy is reached - thermal
equilibrium.
5.5 The partial equilibrium hypothesis: Combustion is a complex of fast and slow
chemical reactions. Fast reactions tend to be chain branching or chain propagating. If
these reactions are treated as if they are in equilibrium (as opposed to slow recombination
reactions), the concentrations of reactive radical species can be expressed in terms of
major stable species.
Contrast the steady state hypothesis (net production of a species is zero) with the partial
equilibrium hypothesis (a reaction is in equilibrium).
5.6 Dependence of the reaction rate on temperature and pressure (not to be confused
with Le Châtlier’s Principle that explains in general how the thermodynamic equilibrium
state of a chemical reaction responds to a change in pressure or temperature).
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Pressure
a) 1st order chemical reaction involving ideal gases:
k1
A B
rate  
d A
 k1 A
dt
A  mass of
A
volume
rate  k1

nA
p
x p
 A  A
V
RT
RT
xA p
RT
From this analysis, you can see that as the pressure increases, the rate of the reaction also
increases.
b) 2nd order chemical reaction involving ideal gases
d A
 x p
rate  
 k 2 A2  k 2  A 
dt
 RT 
2
So as pressure increases for a 2nd order reaction, the reaction rate increases in proportion
to the order. This can be generalized to an mth order reaction. One method of finding the
order or a reaction is to pressurize the reaction vessel, measure the rate, and plot ln(rate)
vs. ln(p). The slope will be the reaction order.
For complex reactions, like chain reactions, increasing pressure decreases the reaction
rate.
Temperature
An increase in temperature greatly increases the fraction of molecules having high kinetic
energy. These are the molecules that are the most likely to react when they collide.
Hence, an increase in temperature increases the rate of most chemical reactions.
For example, for the reaction
CO(g) + NO2(g) ===> CO2(g) + NO(g)
the following data are obtained:
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Temperature (K)
600
650
700
750
800
k (lit/mol sec)
0.028
0.22
1.3
6.0
23
A plot of ln k vs. 1/T is a straight line.
5
ln k
i
0
5
0.0012 0.0014
0.0016
0.0018
1
T
i
Arrhenius plot
Hence,
ln k  A 
B
T
where A and B are constants for a particular reaction, and are determined from the slope
and intercept of the plot. We have Svandte Arrhenius, a Swedish physical chemist, to
thank for this relationship - which he demonstrated in 1887.
The constant B is related to the activation energy, Ea, for the reaction. B = Ea / R, where
R is the universal ideal gas constant. Hence, the slope of the line plotted in the above
figure is
slope = -Ea / R
The activation energy for a reaction is physically interpreted as the amount of energy that
must be absorbed to weaken the bonds holding the reactant molecules together. Ea
represents the energy required to bring the reactants to the point where they can rearrange
to form products. Very fast reactions are characterized by small activation energies. If Ea
is small, A will predominate and k will be large.
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Concept of activation energy
Stable molecules, before being converted to products, must pass through an unstable,
high energy intermediate state. The molecule in the state is referred to as an activated
complex. Ea is the difference in energy between the activated complex and the reactants.
The energy between the reactants and the products we already know as the enthalpy of
reaction. The Ea for the reverse reaction, Ea’, is the sum of the enthalpy of reaction and
Ea.
For an exothermic reaction, Ea’ must be greater than Ea; vice versa for an endothermic
reaction. And this analysis assumes that the activation energy is independent of
temperature - otherwise, a straight line would not fit the data.
If we exponentiate the rate equation we get:
k  Aexp

E
RT
The constant A is called the pre-exponential factor. For some reactions, experimental
data shows the temperature dependence of the pre-exponential factor and you will see the
reaction rate constant reported as (see Table 4.1 on p. 117 in the text):
k  AT exp
b

E
RT
References
Flagan, R. C., and Seinfeld, J. H., Fundamentals of Air Pollution Engineering, Prentice
Hall, 1988.
15