LECTURE 7
AOSC 434
AIR POLLUTION
RUSSELL R. DICKERSON
RECAP. THERMODYNAMICS
G  H  TS
G   RT ln( K eq )
K eq
  G 
 e

 RT 
 H 0  G 0 

GT  H  T 
298


0
III. KINETICS
a) Rates, Order, Lifetime
Thermodynamics tells what is forbidden and provides equilibrium
concentrations, but tells us nothing about what can happen, how fast it
happens, or even if it happens in a finite amount of time. Kinetics
provides the information needed to find out how fast reactions occur.
Much of air pollution is the result of slow kinetics causing reactions to fail
to go to completion. If thermodynamics alone ruled the universe, we
would all be little more than puddles of warm Perrier.
Consider the indoor air pollutant radon 222 (²²²Rn). The EPA has
accused this noble gas of causing a great deal of lung cancer through its
radioactive decay.
The red line
describes first
order loss with
a rate constant
of 1 min-1
The blue line
describes the
rate of
formation of the
product.
minutes
© R. R. Dickerson 2011
4
222
Rn  218Po  
An inert gas cannot stick in your lungs, but particles can. Not only is
radon itself radioactive, but its daughters, starting with Polonium 218, are
also alpha emitters. These daughters have a very low vapor pressure, and
stick readily to aerosol particles. The particles then can stick to your
lungs, as we shall see in a later lecture.
If an unvented room contained a known amount of radon 222, how
long would it take for the radon to decay? The concentration changes as
shown in the following image. It looks as if the decay of Rn is
exponential, and we can test for this by plotting the logarithm of the
concentration as a function of time
We see that the following relationship holds.
ln([ Rn ]t )  k  t  ln([ Rn ]0 )
Where k is the slope and ln([Rn]₀) is the Y-intercept. The rate of decay
of radon seems to depend only on the concentration of radon. What is the
origin of this relationship?
d[Rn]/dt = -k [Rn]
We can solve for the concentration of radon as a function of time by
first separating the variables:
1
d [ Rn ]  k dt
[ Rn ]
Integrating both sides from time zero to t yields:
t
t
1
0 [ Rn ] d [ Rn ]  0  k dt
ln[ Rn ]t  ln[ Rn ]0   kt
 [ Rn ]t 
   kt
ln 
 [ Rn ]0 
The most useful form of this expression is:
[ Rn ]t
 e (  kt )
[ Rn ]0
This is reaction is said to be first order because its rate depends on
only one reactant, in this case radon. The order of a reaction is the
number of reactants actually involved in the chemistry. In general a first
order reaction takes the form:
A → products
The slope mentioned earlier, k, is the rate constant for the reaction.
The units of any reaction are concentration per unit time, thus the units of
the rate constant for a first order reaction must be time⁻¹. For example,
the rate of loss of a species could be expressed in: M/s, ppm/min, (μg
m⁻³)/hr, or (molecules/cm ³)/s. The most common units for gas-phase
kinetics are molecular number density or molecules per cubic cm.
Remember that air has 2.7x1019 cm⁻³, at STP and 2.5x1019 cm⁻³ at RTP.
The half-life of a species is the time it takes for the original
concentration to be reduced to one-half. Thus after n half lives:
[ Rn ]t
1
 n
[ Rn ]0
2
For Rn the half-life is 3.8 days, but for first order reactions, a more
natural expression of the lifetime is the time equivalent to the inverse of
the first order rate constant. If t = 1/k, then:
[ Rn ]
 e ( tk )  e 1  0.37
[ Rn ]0
The lifetime (1/k) is given the symbol τ. Tau is sometimes referred to
as the “e-folding” lifetime. The half-life and lifetime are related by a
constant.
[ A]t
 1 / 2  e (  k t1 / 2 )
[ A]0
ln( 1 / 2)   k  t1/ 2
t1/ 2 
ln( 2)
 ln( 2) 
k
The lifetime of radon is five and a half days;  Rn  5.5 d
Nitric acid anhydride, N₂O₅, decomposition at sea-level pressure
provides another example of a process with first order kinetics.
N 2O5  NO2  NO3
d [ N 2O5 ] / dt  d [ NO2 ] / dt  [ NO3 ] / dt  k [ N 2O5 ]
[ N 2O5 ]t
 e(  k t )
[ N 2O5 ]0
A general expression for the product is:
[ NO 2 ]t 2
 e (  k t )
[ NO2 ]t1
Many reactions leading to or removing air pollution from the
atmosphere require more than one chemical. Those with two
reactants are second order. In general
A + B → Products
The rate equation is
d[A]/dt = d[B]/dt = – k[A][B]
If A and B are the same, we can integrate the rate equation to find the
concentration of the reactant as a function of time.
A  A  prod .
d [ A] / dt  2k [ A]2
t
t
1
0 [ A]2 d [ A]  0  2kdt
1
1

 2kt
[ A]0 [ A]t
This can be rewritten with the initial concentration as a constant.
1 /[ A]t  2kt  1 /[ A]0
A few reactions of combustion or atmospheric interest follow such
kinetics, for example at constant pressure:
O + O → O₂
N + N → N₂
Two different reactants are more common, in fact the majority of reaction in
air pollution have the kinetic form of the following examples.
NO + O₃ → NO₂ + O₂
CO + OH → CO₂ + H
H₂S + OH → H₂O + HS
The units of the rate of a reaction are still concentration per unit time, thus
second order rate constants must have units of (conc⁻¹ time⁻¹ ), such as
(M ⁻¹ s ⁻¹ ), or (ppm⁻¹ min⁻¹), or ( (molecules/cm³) ⁻¹ s⁻¹ ). These last units
are the most common in gas-phase kinetics, but they are usually written with
the “molecules” understood, i.e. cm³ s⁻¹ . Notice that the rate for all
irreversible reactions is independent of the concentration of the products.
Because the rate of a second order reaction depends on two
concentration, we cannot define a lifetime in the same sense as a
first order lifetime. If, however, one of the reactants is in great
excess over the other, we can define a pseudo first order rate
constant. If [B] is approximately constant, then
k '  k  [B ]
1
1
A 

k  [B ] k '
If A is constant or in great excess over B then we can calculate
the lifetime of B the same way.
Some reactions proceed only when three species come into contact
simultaneously. These reactions are third order.
A + B + C → Products
The rate equation is:
d[A]/dt = d[B]/dt = d[C]/dt = – k[A][B][C]
Prominent examples of atmospherically significant third order reactions
are:
2 NO  O  2 NO
2
2
O  O2  M  O3  M 
d [O ] / dt  d [O2 ] / dt  d [O3 ] / dt  k [O ][O2 ][ M ]
The “M” in the formation of ozone appears on both sides of the
equation, so in not necessary for the stoichiometry of the reaction, but it
plays a critical role in the kinetics of the reaction. The formation of ozone
proceeds with the release of considerable energy, thus third body is
required to remove the excess energy; otherwise the reaction will not
proceed. Most addition reactions require a third body. Sulfur dioxide
from coal burning can start down the road to sulfuric acid with the
following reaction.
SO₂ + OH + M → HOSO₂ + M
The units of the rate of the reaction are still concentration per unit time,
but the units of a third order rate constant must be conc⁻² time ⁻¹. The
most common units are cm⁶ s ⁻¹. As a general rule, for a reaction of order
n, the units of k are:
conc  ( n 1)time1.
Again “lifetime” has no meaning for a third order reaction, but if two of
the three reactants are constant, we can define a pseudo first order rate
constant k’. Let A and B be in great excess over C, then
k '  k [ A ][ B ]
This situation is not as unlikely as it might seem. For example pseudo
first order conditions nearly always hold true for O atoms in air wherever
the density and mixing ratio of oxygen are constant.

1
1

(k  [O2 ]  [ M ]) k '
Problem left for students: Where [M] is the sum of [N₂] and [O₂].
How does the lifetime of O atoms vary with pressure?
A + B + C  PRODUCTS
d[A]/dt = -k[A][B][C]
Examples
2NO + O2 = 2NO2
O + O2 + M = O3 + M†
M is any third body (usually N2) needed to dissipate excess energy. From the ideal
gas law and Avogo's number:
P T0
P T0
M  M0x
 2.69 x1019 x
P0 T
P0 T
Where Mo is the molecular number density at STP in molecules cm-3.
Third order rate constants have units of conc-2 time-1. These are usually (cm-3)-2 s-1.
kNO-O2 = 2.0 x 10-38 cm6 s-1
kO-O2 = 4.8 x 10-33 cm6 s-1
© R. R. Dickerson 2014
17
Useful idea:
For the following reversible reaction:
A+B↔C+D
d[C]/dt = kf [A][B] - kr [C][D]
At steady state d[C]/dt = 0, by definition.
Thus:
[C ]  [ D]

 K EQ
kr [ A]  [ B]
kf
© R. R. Dickerson 2011
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For second and third order reactions we can sometime approximate first order
conditions – or use pseudo first order kinetics.
A + B → Prod.
If [B] >> [A]
Then k[B] is approximately constant. We call this pseudo first order rate
constant k’.
k'  k[B]
[A] t
 e -k[B]t 
[A ]0
1
A 
k[B]
© R. R. Dickerson 2011
19
For second and third order reactions we can sometime approximate first order
conditions – or use pseudo first order kinetics.
For example:
NO + O3  NO2 + O2
k = 1.9 x 10-14 cm3 s-1
Assume: [O3] >> [NO] and d[O3]/dt ~ 0.0
Let: mean [O3] = 50 ppb (a reasonable value for air near the surface).
 NO
1
1


 42s
14
9
19
k[O3 ] 1.9 x10  50 x10  2.5x10
CONCLUSION: any NO injected into such an atmosphere (by a car for
example) will quickly turn into NO2 , if there are no other reactions that play
a role. We will call k[O3] the pseudo first order rate constant.
© R. R. Dickerson 2011
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For third order reactions we must assume that two components are constant.
  k[ A][ B]
1
For example:
O + O2 + M  O 3 + M
k = 4.8x10-33 cm6 s-1
ASSUME:
d[O2]/dt = d[M]/dt = 0.0
We know that [O2] = 0.21 and that [M] ~ [O2] + [N2] ~ 1.00.
At RTP P02 = 0.21 atm and PM ~ 1.0 atm. Therefore the lifetime of O atoms is
1/k[O2][M]M02 where M0 is the conversion to molecules per cm3.
 O  4.8 x10
33

19 2 1
x0.21x1.0 x(2.5 x10 )
= 1.6x10-6 s, short indeed!
© R. R. Dickerson 2011
21
Example
Same reaction at stratospheric temperature and pressure.
P30km ~ P0 exp(-30/7) = 0.014 atm
 O  4.8 x10
33

19 2 1
x0.21x1.0 x(0.014 x 2.5 x10 )
= 2.1x10-3 s
This is still short, but it is a thousand times longer than in the troposphere!
The pressure dependence has a major impact on the formation and destruction
of tropospheric and stratospheric ozone.
Problem for the student: Compare the rate of loss of O atoms to reaction with O 2 vs. NO2 in the
troposphere where [NO2] ~ 10-8 and in the stratosphere where [NO2] ~ 10-6 .
© R. R. Dickerson 2014
22
If two of the reactants in a third order reaction are the same, we can
derive a useful expression for the rate of loss of the reactant.
A + A + B  PROD
For a great excess of B:
d[A]/dt = -(2k[B])[A]2
[A]-2 d[A] = -(2k[B])dt
t
t
0
0
-2
A
 dA   - (2kB)dt
-[A]t-1 + [A]0-1 = -(2k[B])t
[A]t-1 = 2k[B]t + [A]0-1
Now we can calculate the concentration at any time t in terms of the
initial concentration and the rate constant k.
© R. R. Dickerson 2011
23
The method works for the self reaction of nitric oxide:
NO + NO + O2 → 2NO2
and shows that this reaction can ruin NO in N2 calibration
standards if any air gets into the cylinder.
d[O2]/dt = -k[NO]2[O2]
d[NO]/dt = -2k[NO]2[O2] = 2d[NO2]/dt
k = 3.3x10-39 exp (530/T) cm6 molecules-2 s-1
In the atmosphere it is only important for highly concentrated
exhaust gases.
© R. R. Dickerson 2011
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Lecture Summary
• Changes in enthalpy and entropy, H and S, are nearly independent
of temperature.
• Gibbs free energy provides the criterion of feasibility.
• The residence time (lifetime), t, is the inverse of the first order rate
constant, k.
• If second or third order reactions can be approximated as first order
then lifetimes can be estimated.
• For reversible reactions, kf/kr = Keq
© R. R. Dickerson 2011
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