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 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 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

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

Athletic performance

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 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 sample air with O3 O3 PM tube {H2C=O}* waste 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] 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 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 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 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 NO2,Why??)

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 proceeds…

There is similar chemistry for alkanes OH. + H 3-C-CH3 --> products and for aromatics OH. + aromatics --> products

Aromatic Reactions * toluene ring cleavage benzaldehyde o-cresol CH 3 OH 2 . O=CH NO +O H * O + HO + H + toluene o-cresol benzaldehyde rearrangement methylglyoxal butenedial oxygen bridge radical ? ring cleavage Aromatic Reactions

Nitrogen Storage (warm vs. cool) OH H C-C=O + H2O H C-C=O 3 . 3 H 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) butane O 2 -C-C-C-C- 2-butylnitrate NO -C-C-C-C- -C-C-C-C- + H. 2 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 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 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.

The Chamber had two sides Or Darkness NO &NO2 300 m3 chamber 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

Solar Radiation Profile

Example Mechanism dNO2/dt = -k1[NO2]; DNO2=-k1 [NO2] Dt 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

Photochemical System

The fact that - dT/dz = g/ cp = 9 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. altitude temp  Sun-down earth cools Inversion layer altitude temp more cooling at surface at night } altitude temp  midday

} } What happens to the material above the inversion layer?? more cooling at surface at night } residual layer altitude } 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. Inversion layer altitude temp more cooling at surface at night } Inversion layer altitude temp Heating at surface during the nest day } 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 height in kilometers 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 Temp in oC 20 25 30 35 Dry adiabatic lines height in kilometers 0.0 0.1 0.2 0.3 0.4 1.1 1.5

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 height in kilometers 0.0 0.1 0.2 0.3 0.4 1.1 1.5 Balloon temperature Temp in oC 20 25 30 35 Dry adiabatic lines

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 f l x s l 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 Chemical System 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 a-pinene was selected because it is generally the most prevalently emitted terpene from trees and other plants

Gas Particle Partitioning Overview The reactions of biogenic hydrocarbons produce low vapor pressure reaction products that distribute between gas and particle phases. Gas Particle Partitioning OH 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 -pinene with either O3 or NOx in sunlight in an outdoor chamber.

OH attack on a-pinene a-pinene O3 attack on a-pinene + OH O CHO O OH HOO + OH O2 OO a-pinene pinonaldehyde a CHO O CH 3 Criegee2 Criegee1 -pinene COOH pinic acid + other products pinonic acid + CO, HO 2, OH norpinonaldehyde norpinonic acid O3 attack on a-pinene

Reactions of product pinonaldehyde with OH and light 2 + (a) (b) (c) (d) (e) pinonaldehyde acetone OO . NO + hn + +CO+HO O 2 2 =o =o O OO . +NO CO H O+ 2 =o =o 2 O pin-OO 2 OONO 2 . pinonald-oo OO O O +HO NO NO 2 pinonald-PAN pin-O + 2 2 methyl (f) =o glyoxal O =o NO NO 2 +HO OH 2 =o pinonic acid =o +HO + h n =o 2 =o . norpinonaldehyde =o +CO+HO 2 OH O 2 OO . C -oo. =o 8 =o (g) +HO NO NO CO + 2 2 NO 2 . OH OO NO =o OO . O 2 C -oo. =o pin-O 8 2 =o +CO NO 2 NO 2 +H norpinonaldehyde 2 O O O=C =O 8

Particle formation-self nucleation Criegee biradicals can react with aldehydes and carboxylic groups to form secondary ozonides and anhydrides. O=C C=O CH 3 + C CH3 O . oo Creigee + pinaldehyde --> seed1 The equilibrium between the gas and particle phases is: Kp = kon/koff

The equilibrium constant Kp can be calculated (Pankow, Atmos The equilibrium constant Kp can be calculated (Pankow, Atmos. Environ, 1994) 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, 31] pinacid + pinacidpart --> pinaidpart + pinacidpart # 25, 33] oxypinaldgas + pinaldpart --> oxypinaldpart + pinaldpart # 20 36] pinald-PANgas + oxypinaicdpart --> pinald-PANpart + oxypinaicdpart # 25, 37] OH-apNO3gas + pinald-PANpart --> OH-apNO3part + pinald-PANpart # 25, 38] diacidpart --> diacidgas # 3.73e14 exp (-10350/T), 39] pinacidpart --> pinacidgas # 3.73e14 exp (-9650/T), 45] OH-apNO3part --> OH-apNO3gas # 3.73e14 exp (-9200/T),

Particle formation from a-pinene + NOx in the presence of Sunlight; symbols are data and lines are model predictions O3data NOy NO2 NO a-pinene pinonaldehyde Particle phase pinonic aciddata norpinonic acid ppm mg/m3 model TSP filter data Time in hours (EDT) O3model Sum products (data) A norpinonaldehyde pinaldmodel Gas phase C E F D pinacidmodel ppmV a-pinene B diacidmodel Soxypinald pinic aciddata O

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 0.1 0.2 0.3 0.4 0.5 0.6 0.7 ppmV 19.5 19.7 19.9 20.1 20.3 20.5 time in hours (pm) O 3 a - pinene 19 20 1 1.5 2 2.5 3.5 mg/m3 21 21.5 22 22.5 Reacted model filter mass data EAA Filter mass data O3 a-pinene 0.15 1a 1b 2a 2b 3b 3a Initial reactants Particle formation

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.