# The Atmosphere

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```					Introduction to Photochemical
Smog Chemistry
Basic Reactions that form O3
Distinguish between O3 formation in the
troposphere and stratosphere
How hydrocarbons       and aldehydes
participate in the formation of smog ozone
Nitrogen loss mechanisms
Secondary aerosol formation
Running simple simulation models
Ozone
ozone is a form of oxygen; it has three
atoms of oxygen per molecule
It is formed in the lower troposphere
(the atmosphere we live up to 6 km)
from the photolysis of NO2
NO2 + light --> NO + O.
O. + O2 -----> O3 (ozone)
its concentration near the earth’s
surface ranges from 0.01 to 0.5 ppm
Ozone
background ranges from 0.02 to 0.06
ppm
What is a ppm??
A ppm in the gas phase is one
molecule per 106 molecules air or
1x10-6 m3 O3 per 1 m3 air or
1x10-6 atmospheres per 1 atmosphere
of air
A ppm in water is 1x10-3grams /L
water
Ozone

let’s convert 1 ppm ozone to grams/m3
we need to convert the volume 1x10-6
m3 of O3 to grams
let’s 1st convert gas volume to moles
and from the molecular weight convert
to grams
at 25oC or 298K one mole of a gas=
24.45liters or 24.45x10-3 m3
Ozone

 we have 1x10-6 m3 of ozone in one
ppm
 so:      1x10-6 m3
--------------------- = #moles O3
24.45x10-3 m3/mol
O3 has a MW of 48 g/mole
so # g O3 in 1ppm =
#moles Ox 48g/mole per m3
= 4.1x10-5 g/m3
Ozone Health Effects
Ozone causes dryness in the throat,
irritates the eyes, and can predispose the
lungs to bacterial infection.
It has been shown to reduce the volume or
the capacity of air that enters the lungs
School athletes perform worse under high
ambient O3 concentrations, and asthmatics
have difficulty breathing
The current US standard has been just
reduced from 0.12 ppm for one hour to 0.08
ppm for one hour
Lung function after exposure
to O.32 ppm O3

8

6
Liters/sec

4

2

0
0   1         2           3   4
Liters after
before
Athletic performance
decreased performance %   80
70
60
50
1962-1964
40
30
1959-1961
20
10
0
0.1 0.15 0.2 0.25 0.3 0.35 0.4
O3 in ppm
How do we measure Ozone

40 years ago chemists borrowed
techniques that were developed for water
sampling and applied them to air sampling
for oxidants, of which O3 is the highest
portion, a technique called “neutral
buffered KI was used.
a neutral buffered solution of potassium
iodide was placed in a bubbler
How do we measure Ozone

a neutral buffered solution of potassium
iodide is placed in a bubbler

KI + O3 --> I2

measure I2
How do we measure Ozone

Air goes in through the top of the bubbler
and oxidants are trapped in the KI liquid
and form I2
Air goes in

KI solution +
I2
How do we measure Ozone

The absorbance of the I2 in the KI solution
is then measured with a spectrophotometer

KI solution +
I2
How do we measure Ozone

The absorbance of the I2 in the KI solution
is then measured with a spectrophotometer

KI solution +
I2
How do we measure Ozone

The absorbance of the I2 in the KI solution
is then measured with a spectrophotometer

Spectrophotometer

KI solution +
I2
A calibration curve

A standard curve is constructed from
known serial dilutions of I2 in KI solution

to do this I2 is weighed out on a 4 place
balance and diluted with KI solution to a
known volume
A calibration curve

A standard curve is constructed from
known serial dilutions of I2 in KI solution

to do this I2 is weighed out on a 4 place
balance and diluted with KI solution to a
known volume

I2
Serial dilutions from stock solution

I2

5     3   2   1
mg/Liter
absorbances are measured for each of
the serially diluted standards
Spectrophotometer

absorbance
Standard Curve

I2 absorbances are plotted vs. concentration
absorbance

1   2     3     4     5
concentration (mg/liter)
How do we measure Ozone

The absorbance of the I2 in the KI solution
is then measured with a spectrophotometer

Spectrophotometer

KI solution +
I2
We then compare our sample
absorbance to the standard curve
I2 absorbances are plotted vs. concentration
absorbance

air sample

1   2     3     4     5
concentration (mg/liter)
Problems

anything that will oxidize KI to I2 will give a
false positive response
NO2, PAN, CH3-(C=O)-OO-NO2, give
positive responses

SO2 gives a negative response
Instrumental techniques of
measuring Ozone
Chemilumenescene became popular in the
early 1970s
For ozone, it is reacted with ethylene
ethylene forms a high energy state of
formaldehyde, [H2C=O]*
[H2C=O]*--> light + H2C=O
A photomultiplyer tube measures the light
The amount of light is proportional O3
Chemilumenescence measurement
of Ozone
pump

O3      sample
PM                        air with
tube   {H2C=O}*
waste                                  O3
ethylene
ethylene
catalytic
converter       CO2 + H2O
Using UV photometry to measure
Ozone
This is the most modern technique for
measuring ozone
sample air with O3 enters a long cell and a
254 nm UV beam is directed down the cell.
at the end of the cell is a UV photometer
which is looking at 254 nm light

we know that:
light Intensityout= light intensityin e- a LC
Photochemical Reactions
 Oxygen (O2) by itself does not react very
fast in the atmosphere.
 Oxygen can be converted photochemically
to small amounts of ozone (O3). O3 is a very
reactive gas and can initiate other
processes.
In the stratosphere O3 is good, because it
filters uv light. At the earth's surface,
because it is so reactive, it is harmful to
living things
In the stratosphere O3 mainly forms from
the photolysis of molecular oxygen (O2)
O2 + uv light -> O.
O. + O2 +M --> O3 + M
In the troposphere nitrogen dioxide from
combustion sources photolyzes
 NO2 + uv or visible light -> NO + O.
O. + O2 +M --> O3 (M removes excess
energy and stabilizes the reaction)
O3 can also react with nitric oxide (NO)

O3 + NO -> NO2 + O2
both oxygen and O3 photolyzes to give O.

O2 + hn -> O. +O.       (stratosphere)

O3 + hn ->   O. + O2

O. can react with H2O to form OH. radicals
O. + H2O -> 2OH.
OH. (hydroxyl radicals) react very quickly
with organics and help “clean” the
atmosphere; for example:

 OH. + H2C=CH2  products ;very very fast

If we know the average OH. radical
concentration, we can calculate the half-life
or life time of many organics [org] in the
atmosphere.
from simple kinetics we can show that:

d[org]/dt = -krate [org] [OH]

If [OH.] is constant

ln [org]t = ln [org]t=o -krate[OH.]x time1/2

Let’s say we want to know the time it takes
for the organic to go to 1/2 its original
[conc].
ln [org]t = ln [org]t=o -krate[OH.]x time1/2

rearranging
{
ln [org]t / [org]t=o }= -krate[OH.]x t1/2

The time that it takes for the conc to go to
half means [org]t will be 1/2 of its starting
conc. [org]t=o .
This means [org]t / [org]t=o = 1/2

and ln (1/2) = -0.693= -krate[OH.]x t1/2
if we use CO as an example, it has a known
rate constant for reaction with OH.

CO + OH. -> CO2         krate= 230 ppm-1 min-1

If the average OH. conc. is 3 x10-8 ppm
for t1/2 we have:

ln(1/2) = -krate[OH.] x t1/2

 -0.693= -230 ppm-1 min-1 x 3 x10-8ppm x t1/2
 t1/2 = 100456 min or 69.7 days
What this means is that if we emit CO from
a car, 69.7 days later its conc. will be 1/2 of
the starting amount. In another 69.7 days it
will be reduced by 1/2 again.

For the same average OH. conc. that we
used above, what would be the t1/2 in years
for methane and ethylene, if their rate
constants with OH. radicals are 12.4 and
3840 ppm-1 min-1 respectively?

CH4      H2C=CH2
Why is the reaction of OH. with ethylene so
much faster than with methane?

H                H
1.      H-C-H....OH . -> H-C. + .H OH .
H                H

2. H2C=CH2 attack by OH.is at the double
bond, which is rich in electrons
What happens in urban air??
In urban air, we have the same
reactions as we discussed before

NO2 + uv light -> NO + O.
O. + O2 +M --> O3 + M
 O3 + NO -> NO2 + O2
This is a do nothing cycle (Harvey
Jeffries)
 What is the key reaction that
generates ozone at the surface of
the earth?

What is the main reaction that
generates it in the stratosphere?

How would you control O3
formation?
In the urban setting there are a lot of
ground base combustion sources

Exhaust
hydrocarbons
NO & NO2
CO
If organics are present they can photolyze
H2C=O + hn -> .HC=O + H.
H. + O2 -> .HO2
if we go back to the cycle
NO2 + uv light -> NO + O.
O. + O2 +M --> O3 + M
O3 + NO -> NO2 + O2

 .HO2 can quickly oxidize NO to NO2
 NO + .HO2 -> NO2 + OH.
(This is a key reaction in the cycling of NO to
NO2,Why??)
OH. + can now attack hydrocarbons such
which makes formaldehyde and other

for ethylene
CH2=CH2 + OH. -> OHCH2CH2.
OHCH2CH2. + O2 -> OHCH2CH2O2.
OHCH2CH2O2. + NO ->NO2 + OHCH2CH2O.

OHCH2CH2O. + O2 -> H2C=O + .CH2OH

 O2 + .CH2OH -> H2C=O + .HO2
These reactions produce a host of radicals
which “fuel” the smog reaction process
First OH radicals attack the electron rich
double bond of an alkene

oxidize NO to NO2 ,just like HO2 can
Further reaction takes place resulting in
carbonyls and HO2 which now undergo
further reaction; the process then
proceeds…
There is similar chemistry for
alkanes

OH. + H 3-C-CH3 --> products

and for aromatics
OH. + aromatics --> products
O=CH
CH3
OH                  CH3

Aromatic                     + H2O
+ HO2                      O
benzaldehyde
Reactions                                                 o-cresol
NO                 NO2
+O2                       CH3
CH2.                                                    CH3
OH
*
OH                             OH
H
.    H

toluene                              +
O2
CH3                                    CH3                              CH3
.O                 OH    NO2                         OH                                  O
.
H           +O2                    H
H                                 H                                  O
NO
oxygen bridge
rearrangement
OH
H

O.               +O2
H            + HO2
H
O +
ring cleavage
?
H
OH                                            H                        CH3
H                                butenedial
methylglyoxal
O
H
+ HO2
Nitrogen Storage (warm vs. cool)
OH      H3C-C=O + H2O
H3C-C=O                   .
H

NO2
PAN

warm
cool
Nitrogen Loss (HNO3 formation)

• NO2 + O3  NO3.+ O2
• NO3.+ NO2  N2O5
• N2O5 + H2O  2HNO3 (surface)

• NO2 + OH.  HNO3 (gas phase)
Nitrogen Loss (alkylnitrates)

-C-C-C-C- + OH. --> -C-C-C-C- + H 2O
.
butane
O2
NO
-C-C-C-C-

NO2
2-butylnitrate

NO2
-C-C-C-C- + H.                 -C-C-C-C-

2-butanal
How can we easily estimate O3
if we know NO and NO2?
The rate of of formation of O3 is governed
by the reaction:
 NO2 + uv light -> NO + O. and its rate
const k1 because:

O. + O2 +M --> O3 + M is very fast
so the rate of formation O3 is:

rateform = +k 1 [NO2]
The rate of removal of O3 is governed by
the reaction:
 O3 + NO -> NO2 + O2 and its rate const k3

so the rate of removal of O3 is:

rateremov = -k 3 [NO] [O3]

the overall ratetot =rateform +rateremov
ratetot = -k3 [NO] [O3] +k1 [NO2]
if ratetot at steady state = 0, then

k1 [NO2]= k3 [NO][O3]

and
 [O3] = k1 [NO2] / {k3 [NO] }

This means if we know NO, NO2, k1 and k3
we can estimate O3
Calculate the steady state O3 from the
following:

NO2 = 0.28 ppm
NO = 0.05 ppm
k1 = 0.4 min-1
k3 = 26 ppm-1min-1
 What is the key reaction that
generates ozone at the surface of
the earth?
What reactions remove nitrogen?

What is the main reaction that
generates it in the stratosphere?

How would you control O3
formation?
 Can we use computers to predict
the amount of ozone formed if we
know what is going into the
atmosphere?
yes
but we need to create experimental
systems to see of our models are
working correctly.
In 1972 we built the first large outdoor
smog chamber, which had an interior
volume of 300 m3.

We wanted to predict oxidant
formation in in the atmosphere.

The idea was to add different
hydrocarbon mixtures and NO + NO2,
to the chambers early in the morning.
Samples would be taken through out
the day. We would then compare our
data to the predictions from chemical
mechanisms.

If we could get a chemical mechanism
to work for many different conditions,
we would then test it under real out
door- urban conditions.

Or Darkness       300 m3          NO
chamber         &NO2

Teflon Film
walls                               propylene

Formaldehyde
Example experiment with the
following chamber concentrations:

•   NO = 0.47
•   NO2 = 0.11 ppm
•   Propylene = 0.99 ppmV
•   temp = 15 to 21oC
TSR

1.2

1

0.8
cal cm-2 min-1

0.6

0.4

0.2

0

-0.2
6   8   10          12   14   16
time in hours
Example Mechanism
NO2+ hn -> NO + O.          k1 keyed to sunlight
O. + O2 --> O3              k2
O3 +NO2 --> NO + O3         k3
H2C=O + hn --> .HC=O + H.   k4 keyed to sunlight
 H. +O2 --> HO2.            k5
HO2. + NO --> NO2+OH.       k6 (fast)
OH.+ C=C ---> H2C=O + HO2
+ H2COO.      k7

• dNO2/dt = -k1[NO2];   DNO2=-k1 [NO2] Dt
Photochemical System
NOx-O3: model vs. data
0.6
O3
0.5                         NO2
0.4            NO
ppm

0.3

0.2
PAN
0.1
NO2
0
10          11         12           13       14          15
time in hours
NO-data     O3-mod      NO2-data
O3-data     NO-mod      NO2-mod
Photochemical System
Propylene: data vs. model
1.2

1

0.8
ppmV

0.6

0.4

0.2

0
10      11    12         13    14   15
Time in hours
 The fact that - dT/dz = g/ cp = 9.8 oK/kilometer is constant is
consistent with observations

 And this is called the dry adiabatic lapse rate so that - dT/dz = d
 When - dT/dz > d the atmosphere will be unstable and air will
move (convection) to re-establish a stability
The quantity d is called the dry

Air that contains water is not as heavy and has
a smaller lapse rate  and this will vary with the
amount of water

If the air is saturated with water the lapse rate
is often called s

 Near the surface sis ~ 4 oK/km and at 6 km
and –5oC it is ~6-7 oK/km
At midday, there is generally a reasonably well-
mixed layer lying above the surface layer into
which the direct emissions are injected.

As the sun goes down, radiative cooling results in
the formation of a stable nocturnal boundary layer,

Sun-down         more cooling at
midday        earth cools      surface at night

           

}   Inversion layer

temp          temp             temp
What happens to the material above the inversion layer??

more cooling at
surface at night

}  residual layer

} Inversion layer
temp

These materials are in a residual layer that contains the
species that were well-mixed in the boundary layer during
the daytime. These species are trapped above and do not
mix rapidly during the night with either the inversion
boundary layer below or the free troposphere above.
When the sun comes up the next day it heats the earth an
the air close to the earth.

more cooling at              Heating at surface
surface at night             during the nest day

} Inversion layer
}   Inversion layer

temp                       temp

During the next day heating of the earth's surface results
in mixing of the contents of the nocturnal boundary layer
and the residual layer above it
How do we get mixing height
in the morning?
that they take at the airport each morning.

• In the morning the temperature usually
increases with height for a few hundred
meters and then starts to decrease with
height (see the green curve) according to the
temperature sensor on the balloon

• The the break in the curve is usually defines
the inversion height in the early morning
Mixing height in the morning

Balloon temperature

}Inversion height
Temp in oC
Mixing height in the morning
• There are another set of lines called the dry adiabatic lines, which
are thermodynamically calculated, and represent the ideal
decrease in temperature with height for dry air starting from the
ground.

• In the morning, the mixing height is estimated by taking the
lowest temperature just before sunrise and adding 5oC to it, and
then moving up the dry adiabatic line at that temperature until it
intersects the balloon temperature line or the green curve.

• Let’s say the lowest temperature just before sunrise was 20oC. We
would add 5oC to it and get 25oC. We then move up the 25oC dry
adiabatic line. We then go straight across to the right, to the
height in kilometers and get a morning mixing height of ~350
meters (0.35 km). This is illustrated in the next slide. It is
animated so you can see it more easily
Mixing height in the morning

Balloon temperature                           1.5

1.1

lines
0.4
0.3
0.2
0.1
20   25   30   35      0.0

Temp in oC
Mixing height in the afternoon

• To get the mixing height in the afternoon, you just
take the highest temperature between 12:00pm and
15:00 pm
• Do not add anything to it, but as before run up the dry
adiabatic curve and intersect the morning balloon
temperature curve.
• Let say the highest afternoon temperature is 35oC, we
would estimate an afternoon a mixing height of ~1.67
km
Afternoon Mixing height

Balloon temperature                           1.5

1.1

lines
0.4
0.3
0.2
0.1
20   25   30   35      0.0

Temp in oC
let’s see how this kinetics model works
1st we will look at a mechanism
2nd we will look at the model inputs
3rd we will run the model with reduced
hydrocarbons (formaldehyde) to see the
effect of reducing HC
run the model with reduced NOx
Before this, however, let’s see how you get
light into the model
How do we get light into the
mechanism??
A molecule photolyzes or breaks apart when
it absorbs photons that have energy that is
greater than the bond strength
Let’s look at the energy in a mole of photons
which have a wavelength 288 nm
The energy E, in this light is
E= 6.02x1023x hc/l
c= 3x108m/s; h=6.63x10-34Js, l=288x10-9m
E= 416kJ/mole
If all this light was absorbed it would break C-
H bond
Light and rate constants
The question is, is all the light absorbed??
Actually not, but this brings up the concept of
quantum yields, f, and light absorption s

f= # molecules reacted/# photons absorbed
What about the light flux, j at a given l?
This is the # of photons of light cm-2 sec-1
The rate constant for photolyis can be written as
kratel= Jl x f l x absorption coefl
Light and rate constants
kratel= Jl x f l x absorption coefl
the absortion coef. s has units of
cm2/molecule
and comes from Beer’s law I=Io e-sl[C]
kratel= Jl x f l x s l
This is at one wavelength l; what do we do
when we have two wavelengths l and l1?
kratel1 = the rate const. at a different
wavelength l1 and kratel1 = J1 x f l1 x s l1
krateltotal = Jl   x   f l x s l + J1 x f l1x s l1
Light and rate constants
krateltotal = Jl   x   f l x s l + Jl1 x f l1x s l1
so across all wavelengths
 so   krateltotal = S Jl        x   flxsl
What this says is that if we know the
light flux or “intensity” at each
wavelength, Jl , the absorption coef.,
s l at each wavelength and the
quantum yield s l , we can calculate
krateltotal for the real atmosphere
Light and rate constants
Lets calculate kratel for NO2 at the
wave length of 400-405 nm and a
zenith angle of 20 degrees
J400-405nm= photons cm-2 sec-1 =
1.69x1015
f400nm = quantum yield = ~0.65
s400-405nm = ~6x10-19 cm2 molecule-1
 kratel= Jl x f l x s l= 0.00067sec-1
Light and rate constants

so in the reaction
NO2 + light at 400-405nm -> NO +
O.
 kratel= Jl x f l x s l= 0.00067sec-1

dNO2/dt = krate l [NO2]
Light and rate constants
There are tables that give J at each
wave length as a function of the
angle of the sun
The angle of the sun is called the
zenith angle. When the sun is
directly over head the zenith angle
is zero degrees
when it has just gone down it is
90o
Light and rate constants

Sun

q
Light and rate constants
This means for a given latitude and time of
year we can know when the sun comes up
and how high in the sky it will go at noon
in the winter time it will not go as high in the
sky as in summer.
from these tables if we know f and s for a
compound we can calculate the photolysis
rate constants for any compound over the
course of the day as the zenith angle changes
NO2, H2C=O, O3, acetaldehyde
Extending this kinetics approach to
simulate secondary Aerosols
formation by linking gas and particle
phase chemistry

An exploratory model for aerosol
formation from biogenic hydrocarbons using a
gas-particle partitioning/thermodynamic model-
Kamens Research Group, ES&T, 1999 and 2001
Global Emissions of hydrocarbons

1150 x1012 grams of biogenic hydrocarbons emitted
each year
of the biogenics ~ 10 -15% can produce particles in
the atmosphere (terpenes)
man made emissions of volatile non methane
hydrocarbons ~ same as terpenes… don’t produce
particles
Reasons to study biogenic secondary
aerosol formation
Global model calculations are sensitive to
fine particles in the atmosphere
Biogenic particles serve as sites for the
condensation of other reacted urban organics
This leads to haze and visibility reductions
There is a great need to develop predictive
models for secondary aerosol formation from
naturally emitted hydrocarbons
Objective
• To describe a new predictive technique for the
formation of aerosols from biogenic hydrocarbons
based on fundamental principals.

• Have the ability to embrace a range of different
atmospheric chemical and physical conditions

Chemical System

+ NOx+ sunlight ----> aerosols
a-pinene
a-pinene was selected because it is generally the most
prevalently emitted terpene from trees and other plants
Overview
• The reactions of biogenic hydrocarbons
produce low vapor pressure reaction
products that distribute between gas and
particle phases.

Gas Particle Partitioning
O
O gas phase products
OH
pinonic acid

atmospheric particle
• Equilibrium partitioning can be represented as an
between the rate of oxidized terpene product up-
take and rate of terpene product loss from the
aerosol system.

•    Kinetically this is represented as forward and
backward reactions Kp = kon/koff

• Gas and particle phase reactions were linked in one
mechanism and a chemical kinetics solver provided
by Professor Jeffries, was used to simulate the
reaction over time

• This was compared with aerosol concentrations
obtained by reacting a-pinene with either O3 or NOx
in sunlight in an outdoor chamber.
OH attack on a-pinene
a-pinene
OH                    O2
OO OH
OH

HOO           CHO          HOO O
+ OH
O    CHO
pinonaldehyde

+ CO, HOOH
2,
O   CHO
norpinonaldehyde
O3 attack on
a-pinene                                                  COOH
O       O                     O
O3                      O                         norpinonic
Criegee1                          acid
O    COOH
OO
O                          pinonic acid
a-pinene
O CH CHO
3
O
COOH
+ other
products
Criegee2        COOH
pinic acid
Reactions of product pinonaldehyde with OH and
light                                               O
O
+ hn                                     (c)
pinonaldehyde
+
O 2 OH                                             (b)
(a)                                                                                               +CO+HO           2
O2                                                           =o
=o        O                                                                                        O                          OO   .
H 2 O+                  OO   .     +NO     2
=o                              =o
CO     +                       pin-OO
O                                          acetone                      2
(d)                       OONO 2                           OO    .     O O
pinonald-oo                 +HO      2                                                                                       NO             NO 2
pinonald-PAN                      pin-O       2        + methyll
(f)                 (e)           =o          O                                                    glyoxa                    =o
NO                   NO 2                                     +HO     2
NO                                                  OH
NO 2
pinonic acid                                           =o                                               =o
=o
+HO     2                        +h n
O                                                                                  =o
O  .                                                norpinonaldehyde                                                  =o
+CO+HO   2
OH
O2                                       OO .
C 8 -oo.
=o                                                        (g)               =o
+HO   2                                                    NO                 NO   2
CO 2 +                                                                 OH                                      NO
OO   .   NO                       =o                  O2
OO .
C 8 -oo.                        =o
pin-O     2                                                                                       NO 2
=o                    +CO
NO      2                                               +H 2 O
norpinonaldehyde                                                                O
O=C        8=O
Particle formation-self nucleation
 Criegee biradicals can react with aldehydes and
carboxylic groups to form secondary ozonides and
anhydrides.                   CH 3
CH 3               C=O
C=O
.
C   +
O=C
CH   3
CH3                    C=O
C=O
C
O C

Creigee + pinaldehyde --> seed1

The equilibrium between the gas and particle
phases is:

• Kp = kon/koff
• The equilibrium constant Kp can be
calculated (Pankow, Atmos. Environ, 1994)

7.501 RT fom
Kp = o
pL g Mw 10    9

• poL is the liquid vapor pressure and g the activity
coefficient of the partitioning organic in the liquid
portion of the particle, fom is the raction of organic
mass in the particle and Mw is the average
molecular weight of the organic mass
• Rates that Gases enter and leave the
particle can be estimated from
• Kp = kon/koff

•   where koff = {kbT/h} e -Ea/RT

• Ea can be estimated and with Boltzman’s
(kb) and Planck’s constants (h) and
temperature,T. koff can be calculated and
with Kp, kon can also be evaluated
Overall Mechanism linked gas and particle phase rate
expressions
Representitive a-pinene gas phase reactions + rate constants (#) min-1 or ppm-1 min-1
1] OH + a-pinene --> 0.95 ap-oo + 0.05 acetone + 0.04357 vol-oxy # 17873
2] ap-oo + NO --> 0.8 NO2 + 0.6 pinald + 0.8 HO2 + 0.2 HCHO
+ 0.13 vol-oxy + 0.015 oxypinacid + 0.2 OH-apNO3 +0.1 acetone # 3988 exp(-360/T)
3] ap-oo + ap-oo --> 0.4 pinald + 0.3 HCHO
+1.57 vol-oxy +0.3 HO2                                #1226
4] a-pinene + NO3 --> apNO3-oo                                    # 544 exp (818/T)
6] a-pinene + O3 --> 0.4 crieg1 + 0.6 crieg2                      # 1.492 exp (-732/T)
7] Criegee1 --> 0.35 pinacid + + 0.3pinald + 0.15 stabcrieg1      # 1e6,
+ 0.05 oxypinald + 0.14 vol-oxy + 0.5 HO2+ 0.8 OH + 0.03 O + 0.4CO

{Representitive pinonaldehyde gas phase chemistry}
12] pinald --> 0.65 pinO2 {+ 1.35 CO} + 1.35 HO2 + 0.35 C8O2 # HVpinald
13] pinO2 + NO--> 0.72 pinald + 0.8 HO2 + 0.2 MGLY
+0.15 vol-oxy + NO2                           # 3988 exp (360/T),
14] C8O2 + NO  NO2 + 0.8vol-oxy +HO2                     # 3988 exp (360/T),
15] C802 + C8O2 1.5 vol-oxy + HO2 + 0.05 seed1           # 2.4 exp (1961/T)
16] pinald + OH --> 0.9 pinald-oo + 0.05 pinO
+ 0.043 C2O3 + 0.05 CO2 +0.032vol-oxy        #132000,
19] pinald-oo + NO2 --> pinald-PAN                       # 0.000118 exp (5500/T),
20] pinald-PAN --> 0.9 pinald-oo+ 0.05 oxypin-oo
+0.05pred-oo+ NO2                            # 1.0.6x1011 exp (-864/T),
23] pinald-oo + HO2 --> pinacid                         # 211 exp (1380/T),

{Representative Partitioning reactions}
25] stabcrieg1 + pinald --> seed1                    # 29.5,
28] stabcrieg2 + oxypinacid --> seed1                # 29.5,
29] diacidgas + seed --> seed + diaacidpart          # 70,
Particle formation from a-pinene + NOx in the
presence of Sunlight; symbols are data and lines are
model predictions
a-pinene                    A      Particle phase                        B
model TSP
NO                      O3model

O3data                                    Sum products
(data)
NOy                                   filter data
NO2

Gas phase                             C          Particle phase                    D
O O
pinonaldehyde
pinaldmodel
pinonaldehyde              pinaldmodel

norpinonaldehyde

Particle phase        pinic aciddata    E      Particle phase                        F

diacidmodel                                         pinonic aciddata
pinacidmodel
Soxypinald
norpinonic acid

Time in hours (EDT)                             Time in hours (EDT)
Reaction of a-pinene with O3 at different concentrations in the
dark; top experiment #1, chamber temperature 23oC; middle
experiment #2, 12oC; bottom experiment #3, 27oC; symbols are
data, and lines are model predictions

Initial reactants                                                            Particle formation
0.4                                                                             2
1a          a
1b Reacted a -pinene
0.3                                 -pinene                                   1.5

mg/m3
ppmV

0.2                                                                             1
O3                                                                                    EAA         model
0.1                                                                           0.5
Filter mass data
0                                                                          0
19           19.5                  20           20.5                         19           20          21            22
time in hours (pm)                                                         time in hours (pm)
0.7                                                                            3.5
0.6               2a                                                                 3        2b           Reacted a -pinene
0.5
O3                                                  2.5

mg/m3
filter mass data
ppmV

0.4
2
0.3
1.5
0.2
1                        model
0.1                                a -pinene                                   0.5
0
19.5    19.7      19.9        20.1         20.3   20.5                     0
time in hours (pm)                                           19    19.5   20   20.5    21    21.5   22   22.5
time in hours (pm)

3a                                                                               3b
a-pinene
0.15                                                                                             filter mass data
ppmV

model
O3

time in hours (pm)                                                               time in hours (pm)
Summary
Models vs. experimental aerosol yields illustrate that reasonable predictions of
secondary aerosol formation are possible from both dark ozone and light-NOx/a-
pinene systems over a variety of different outdoor conditions. On average,
measured gas and particle phase products accounted for ~40% to 60% of the
reacted a-pinene carbon. Model predictions suggest that organic nitrates accounts
for another 25-35% of the reacted carbon, and most of this is in the gas phase.
Measured particle phase products accounted for 60 to 100% of the particle filter
mass. Measurements show that pinic acid is one of the primary aerosol phase
products. In the gas phase, pinonaldehyde and pinonic acid are major products.
Model simulations of these products and others show generally good fits to the
experimental data from the perspective of timing and concentrations.
These results are very encouraging for a compound such as pinonaldehyde, since it
is being formed from OH attack on a-pinene, and is also simultaneously, photolyzed
and reacted with OH. Additional work is need to determine the quantum yields of
product aldehydes, the measurement of nitrates on particles, and possible particle
phase reactions

Acknowldegements
This work was supported by a Grant from National Science Foundation, the USEPA
STAR Research Gramt Program, Fulbright fellowship support for R. Kamens in
Thailand, a gift of a GC-FTIR-MS system from the Hewlett Packard Corporation and
from the Varian Corp of a Saturn GC-ITMS. We appreciate the help of that ESE
students Sangdon Lee, Sirakarn Leungsakul, and Bharad Chandramouli provided
with the outdoor chamber experiments.

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