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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
  Formation of free radicals
  Nitrogen loss mechanisms
  Secondary aerosol formation
  Running simple simulation models
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
background ranges from 0.02 to 0.06
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

let’s convert 1 ppm ozone to grams/m3
start with: 1x10-6 m3 per 1 m3 air
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

 we have 1x10-6 m3 of ozone in one
 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





                 0   1         2           3   4
                            Liters after
                          Athletic performance
decreased performance %   80
                           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 +
      How do we measure Ozone

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

  KI solution +
      How do we measure Ozone

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

  KI solution +
      How do we measure Ozone

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


  KI solution +
            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

Serial dilutions from stock solution


               5     3   2   1
absorbances are measured for each of
    the serially diluted standards

                 Standard Curve

I2 absorbances are plotted vs. concentration

               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


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

                                air sample

                1   2     3     4     5
                concentration (mg/liter)

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

                               O3      sample
             PM                        air with
             tube   {H2C=O}*
waste                                  O3
     converter       CO2 + H2O
Using UV photometry to measure
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
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
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
ln [org]t = ln [org]t=o -krate[OH.]x time1/2

      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 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
 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
In the urban setting there are a lot of
  ground base combustion sources

                      NO & NO2
If organics are present they can photolyze
 or generate radicals
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
OH. + can now attack hydrocarbons such
 which makes formaldehyde and other
 radical products

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

   Oxygen then add on the hydroxy radical
   forming a peroxy-hydroxy radical

    the peroxy-hydroxy radical radical can
    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
There is similar chemistry for

OH. + H 3-C-CH3 --> products

and for aromatics
OH. + aromatics --> products
                                                                       OH                  CH3

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

                                            toluene                              +
                        CH3                                    CH3                              CH3
                 .O                 OH    NO2                         OH                                  O
                                    H           +O2                    H
                                     H                                 H                                  O
                                                          oxygen bridge

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


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


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

 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]

 [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

    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
 Can we use computers to predict
 the amount of ozone formed if we
 know what is going into the
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

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

              Or Darkness       300 m3          NO
                                chamber         &NO2

Teflon Film
walls                               propylene

Example experiment with the
following chamber concentrations:

     •   NO = 0.47
     •   NO2 = 0.11 ppm
     •   Propylene = 0.99 ppmV
     •   temp = 15 to 21oC
          Solar Radiation Profile



cal cm-2 min-1





                        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
0.5                         NO2
0.4            NO


          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






         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
 the dry adiabatic lapse rate

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,
corresponding to a radiation inversion.

              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

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?
• We start with the balloon temperature curve
  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

• 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


                              Dry adiabatic
                       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
         Afternoon Mixing height

Balloon temperature                           1.5


                              Dry adiabatic
                       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
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
 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 =
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 +
 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
Light and rate constants


        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
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
• 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
  which bring about aerosol formation.

                 Chemical System

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

         Gas Particle Partitioning
                         O gas phase products
          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
                   OH                    O2
                                               OO OH

     HOO           CHO          HOO O
                                + OH
            O    CHO

                                          + CO, HOOH
                                     O   CHO
     O3 attack on
     a-pinene                                                  COOH
                         O       O                     O
         O3                      O                         norpinonic
                         Criegee1                          acid
                                         O    COOH
              O                          pinonic acid
                         O CH CHO
                                                       + other
                             Criegee2        COOH
                                          pinic acid
Reactions of product pinonaldehyde with OH and
light                                               O
                                                             + hn                                     (c)
                                  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
                                                                                                           +HO     2                        +h n
                   O                                                                                  =o
                  O  .                                                norpinonaldehyde                                                  =o
                                                                                                                       +CO+HO   2
                                                                                                 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   +
                                                          CH   3
                                        CH3                    C=O
                                                        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
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
                                 NOy                                   filter data

    Gas phase                             C          Particle phase                    D
                                              O O
                              pinonaldehyde              pinaldmodel


      Particle phase        pinic aciddata    E      Particle phase                        F

                        diacidmodel                                         pinonic aciddata
                                                                         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


               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
                                         O3                                                  2.5

                                                                                                                               filter mass data

                                                                                                   1                        model
              0.1                                a -pinene                                   0.5
                        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
                0.15                                                                                             filter mass data


                                        time in hours (pm)                                                               time in hours (pm)
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

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.