Introduction to Kinetics
AOSC 620
Fall 2015
R. Dickerson
See Finlayson-Pitts Chapter 5
Seinfeld and Pandis Chapters 4 & 9
Copyright © 2015 R.R. Dickerson
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Kinetics
A. Rates, rate constants, and reaction order.
Thermodynamics tells us if a reaction can proceed and gives equilibrium
concentration. Kinetics tells us how fast reactions proceed. If
thermodynamics alone controlled the atmosphere it would be dissolved
in the oceans as nitrates - we would be warm puddles of carbonated
water.
1. First Order Reactions
A  PRODUCTS
EXAMPLES
-> 218Po + 
N2O5  NO2 + NO3 (forward reaction only)
222Rn
Copyright © 2013 R.R. Dickerson
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The red line
describes first order
loss with a rate
constant of 1 min-1
The blue line is the
rate of formation of
the product.
minutes
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Radon is important source of indoor air pollution, and N2O5
is nitric acid anhydride, important in air pollution nighttime
chemistry. The rate equations take the form:
d[Prod.]/dt = k[A] = -d[React.]/dt
For example:
d[Po]/dt = kRn [Rn] = -d[Rn]/dt
Where k is the first order rate constant and k has units of
time-1 such as s-1, min-1, yr-1. We usually express
concentration, [Rn], in molecules cm-3 and k in s-1.
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Another example, nitric acid anhydride.
d[NO2] /dt = k [N2O5]
and
d[N2O5]/dt = -k [N2O5]
Integrating
ò
N 2O5
dt =
N 2O5
ò -k dt
ln ([ N2O5 ]t / [ N2O5 ]o) = -k t
If we define the starting time as zero:
[N2O5]t / [N2O5]o = exp(-kt)
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The rate constants for these first order reactions are:
kRn = 0.182 days-1
kN2O5 = 0.26 s-1 (at room temperature)
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2. Second Order Reactions
A + B  PRODUCTS
EXAMPLES
NO + O3  NO2 + O2
HCl + OH  H2O + Cl
Examples of the rate equations are as follows:
d[NO]/dt = -k[NO][O3]
d[Cl]/dt = k[OH][HCl]
Units of k are {conc-1 time-1}.
1/(molecules/cm3) (s-1) = cm3 s-1
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For our second order kinetics examples
Units of k are {conc-1 time-1}.
1/(molecules/cm3) (s-1) = cm3 s-1
Rate constants have the following values:
kNO-O3 = 1.8x10-14 cm3 s-1
kHCl-OH = 8.0x10-13 cm3 s-1
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3. Third Order Reactions
A + B + C  PRODUCTS
d[A]/dt = -k[A][B][C]
Examples
2NO + O2  2NO2
O + O2 + M  O3 + M
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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 Avg's number:
æ P ö æ T0 ö
19 æ P ö æ 273 ö
M Z = M o * ç ÷ * ç ÷ = 2.69x10 * ç ÷ * ç
÷ ~ M 0 PZ
è1ø è T ø
è P0 ø è T ø
Where Mo is the molecular number density at STP.
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
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USEFUL IDEA:
For the following reaction:
A+BC+D
d[C]/dt = kf [A][B] - kr [C][D]
At steady state d[C]/dt = 0, by definition. Thus:
{ kf }/{ kr } = {[C][D]}/{[A][B]} = Keq
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Halflife and Lifetime
Definition: Halflife, t1/2 the time such that:
[A]t1/2 / [A]0 = 1/2
Definition of lifetime or residence time,  , comes from kinetics, where k is
the first order rate constant with units of time-1.
We know that:
[A]t/[A]0 = exp(-k t)
The lifetime, , is when t = 1/k so
  1/k
We can link half-life and lifetime:
t1/2 = ln(2)/k  0.69/k
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For radon 222 (222Rn) the lifetime is 5.5 days, but the half-life is only 3.8 days.
For second order reactions we need pseudo first order conditions
t = ( k éë Aùû)
-1
For example:
NO + O3  NO2 + O2
k = 1.8 x 10-14 cm3 s-1
ASSUME: [O3] >> [NO] and d[O3]/dt ~ 0.0
Let:
NO
éëO3 ùû = 50 ppb (a reasonable value for air near the surface).
-1
= k éëO3 ùû = 1/{1.8 x 10-14 x 50x10-9 x 2.5x1019} = 44 s
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.
Copyright © 2013 R.R. Dickerson
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For third order reactions we must assume that two components
are constant.
-1
t = k éë AùûéëBùû
(
)
For example:
O + O2 + M  O3 + M
k = 4.8x10-33 cm6s-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
 = [4.8x 10-33 x 0.21 x 1.0 x (2.5x1019) 2]-1
= 1.6x10-6 s VERY SHORT!
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Example 2
Same reaction at stratospheric temperature and pressure.
P30km ~ P0 exp(-30/7) = 0.014 atm
 = [4.8x10-33 x 0.21 x 1.0 x (0.014 x 2.5x1019) 2]-1 = 2.1x10-3 s
The lifetime of an O atom is 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.
Copyright © 2013 R.R. Dickerson
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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
Integrating from 0 to time t
 [A]
-2
d[A] =
-(2k[B])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.
Copyright © 2013 R.R. Dickerson
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Transport and Evolution of a Pollution
Plume from Northern China:
A Satellite-based Case Study
Can Li1, Nick Krotkov2, 3, Russ Dickerson1, Zhanqing Li1, 4, and Mian Chin2
1AOSC, UMD 2GSFC, NASA 3GEST, UMBC 4ESSIC, UMD
Why China? A lot of emissions… Local, regional, and global effects
Why transport and evolution? Key factors determining the largescale impact of air pollution (e.g., conversion from SO2 to sulfate
aerosols, adding CCN to the system)
Why satellites? Transport episodes associated with synoptic weather
system, and are of regional scale – new satellite sensors provide
great spatial coverage, daily observation, and sensors of different
strengths can be combined
Copyright © 2013 R.R. Dickerson
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On April 5,2005 on the aircraft:
Lots of SO2
Lots of dust
~1 hr later and from space:
Lots of SO2
Lots of dust
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Satellite Snapshots
Trajectory
Projection
Trajectory model projects
movement of the plume,
satellite sensors take
snapshots every day
AMF Correction to
operational product,
combination of satellites
and models, and MODIS
AOD data
SO2 lifetime: 1-4 day
SO2 to sulfate: ~0.1-0.2 increase in
AOD near plume core in one day
Uncertainties? Details? Suggestion?
Copyright © 2013
Please come to see our poster.
R.R. Dickerson
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The comparison demonstrates that operational OMI algorithm can distinguish
between heavy pollution ( April 5 ahead of cold front ):
SO2
AQUA- MODIS RGB
OMI SO2
…and background SO2 conditions (on April 7 behind cold front)
SO2
MODIS RGB
OMI SO2
SO2 lifetime estimate
By following the path of the air mass, and correcting for the altitude
effect, the amount of SO2 lost to dry deposition and chemical
reactions was approximated.
Ln (C/C0) = -kt
0
-0.1
y = -0.26x - 0.01
ln(C/C 0 )
-0.2
-0.3
-0.4
-0.5
-0.6
0
0.5
1
1.5
2
2.5
Days after 5 April 2005
The mass decreases from the 5th to the 7th looks to be first order with a
lifetime of ~4 d.
The mass on April 8 shows no change from April 7 because the air mass was
very disperse (large box) and mixed with other plumes.
SO2 + OH  prod (H2SO4)
If attack by OH is the only sink for SO2, and if the
lifetime is ~4 d =~ 3x105 s.
Lifetime =  = ([OH]*M*k)-1
k = 2x10-12 cm-3 s-1
Molecular number density of OH = 1.4x106 cm-3