1. Introduction to Photochemical Smog Chemistry Basic Reactions that form O 3 Distinguish between O 3 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

2. Ozone uozone is a form of oxygen; it has three atoms of oxygen per molecule uIt is formed in the lower troposphere (the atmosphere we live up to 6 km) from the photolysis of NO2 uNO 2 + light --> NO + O. uO. + O 2 -----> O 3 (ozone) uits concentration near the earth’s surface ranges from 0.01 to 0.5 ppm

3. Ozone  background ranges from 0.02 to 0.06 ppm  What is a ppm??  A ppm in the gas phase is one molecule per 10 6 molecules air or  1x10 -6 m 3 O 3 per 1 m 3 air or  1x10 -6 atmospheres per 1 atmosphere of air  A ppm in water is 1x10 -3 grams /L water

4. Ozone  let’s convert 1 ppm ozone to grams/m 3  start with: 1x10 -6 m 3 per 1 m 3 air  we need to convert the volume 1x10 -6 m 3 of O 3 to grams  let’s 1 st 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 m 3

5. Ozone  we have 1x10 -6 m 3 of ozone in one ppm  so: 1x10 -6 m 3 --------------------- = #moles O 3 24.45x10 -3 m 3 /mol  O 3 has a MW of 48 g/mole  so # g O3 in 1ppm = #moles Ox 48g/mole per m 3  = 4.1x10 -5 g/m 3

6. 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 O 3 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

7. Lung function after exposure to O.32 ppm O 3

8. Athletic performance

9. 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 O 3 is the highest portion, a technique called “neutral buffered KI was used.  a neutral buffered solution of potassium iodide was placed in a bubbler

10. How do we measure Ozone  a neutral buffered solution of potassium iodide is placed in a bubbler  KI + O 3 --> I 2  measure I 2

11. How do we measure Ozone  Air goes in through the top of the bubbler and oxidants are trapped in the KI liquid and form I 2 Air goes in KI solution + I 2

12. How do we measure Ozone  The absorbance of the I 2 in the KI solution is then measured with a spectrophotometer KI solution + I 2

13. How do we measure Ozone  The absorbance of the I 2 in the KI solution is then measured with a spectrophotometer KI solution + I 2

14. How do we measure Ozone  The absorbance of the I 2 in the KI solution is then measured with a spectrophotometer KI solution + I 2 Spectrophotometer

15. A calibration curve  A standard curve is constructed from known serial dilutions of I 2 in KI solution  to do this I 2 is weighed out on a 4 place balance and diluted with KI solution to a known volume

16. A calibration curve  A standard curve is constructed from known serial dilutions of I 2 in KI solution  to do this I 2 is weighed out on a 4 place balance and diluted with KI solution to a known volume I2I2

17. Serial dilutions from stock solution I2I2 5321 mg/Liter

18. absorbances are measured for each of the serially diluted standards Spectrophotometer absorbance

19. Standard Curve I 2 absorbances are plotted vs. concentration 12345 concentration (mg/liter) absorbance

20. How do we measure Ozone  The absorbance of the I 2 in the KI solution is then measured with a spectrophotometer KI solution + I 2 Spectrophotometer

21. We then compare our sample absorbance to the standard curve I 2 absorbances are plotted vs. concentration 12345 concentration (mg/liter) absorbance air sample

22. 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

23. 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 O 3

24. Chemilumenescence measurement of Ozone PM tube pump waste ethylene catalytic converter ethylene CO 2 + H 2 O O3O3 sample air with O 3 {H 2 C=O}*

25. Using UV photometry to measure Ozone  This is the most modern technique for measuring ozone  sample air with O 3 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 Intensity out = light intensity in e -  LC

26. Photochemical Reactions  Oxygen (O 2 ) by itself does not react very fast in the atmosphere.  Oxygen can be converted photochemically to small amounts of ozone (O 3 ). O 3 is a very reactive gas and can initiate other processes.  In the stratosphere O 3 is good, because it filters uv light. At the earth's surface, because it is so reactive, it is harmful to living things

27.  In the stratosphere O 3 mainly forms from the photolysis of molecular oxygen (O 2 )  O 2 + uv light -> O.  O. + O 2 +M --> O 3 + M  In the troposphere nitrogen dioxide from combustion sources photolyzes  NO 2 + uv or visible light -> NO + O.  O. + O 2 +M --> O 3 (M removes excess energy and stabilizes the reaction)

28.  O 3 can also react with nitric oxide (NO)  O 3 + NO  NO 2 + O 2  both oxygen and O 3 photolyzes to give O. O 2 + h  O. +O. (stratosphere) O 3 + h  O. + O 2  O. can react with H 2 O to form OH. radicals O. + H 2 O  2OH.

29.  OH. (hydroxyl radicals) react very quickly with organics and help “clean” the atmosphere; for example:  OH. + H 2 C=CH 2   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.

30.  from simple kinetics we can show that: d[org]/dt = -k rate [org] [OH] If [OH.] is constant  ln [org] t = ln [org] t=o -k rate [OH.]x time 1/2  Let’s say we want to know the time it takes for the organic to go to 1/2 its original [conc].

31.  ln [org] t = ln [org] t=o -k rate [OH.]x time 1/2  rearranging ln { [org] t / [org] t=o } = -k rate [OH.]x t 1/2  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= -k rate [OH.]x t 1/2

32.  if we use CO as an example, it has a known rate constant for reaction with OH.  CO + OH.  CO 2 k rate = 230 ppm -1 min -1  If the average OH. conc. is 3 x10 -8 ppm  for t 1/2 we have: ln(1/2) = -k rate [OH.] x t 1/2  -0.693= -230 ppm -1 min -1 x 3 x10 -8 ppm x t 1/2  t 1/2 = 100456 min or 69.7 days

33.  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 t 1/2 in years for methane and ethylene, if their rate constants with OH. radicals are 12.4 and 3840 ppm -1 min -1 respectively? CH 4 H 2 C=CH 2

34.  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. H 2 C=CH 2 attack by OH.is at the double bond, which is rich in electrons

35.  In urban air, we have the same reactions as we discussed before  NO 2 + uv light  NO + O.  O. + O 2 +M  O 3 + M  O 3 + NO  NO 2 + O 2  This is a do nothing cycle (Harvey Jeffries) What happens in urban air??

36.  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 O 3 formation?

37. In the urban setting there are a lot of ground base combustion sources Exhaust hydrocarbons NO & NO2 CO

38.  If organics are present they can photolyze or generate radicals  H 2 C=O+ h  ->. HC=O + H.  H. + O 2 . HO 2  if we go back to the cycle NO 2 + uv light -> NO + O. O. + O 2 +M  O 3 + M O 3 + NO  NO 2 + O 2 . HO 2 can quickly oxidize NO to NO 2  NO +. HO 2  NO 2 + OH. (This is a key reaction in the cycling of NO to NO 2,Why??)

39.  OH. + can now attack hydrocarbons such which makes formaldehyde and other radical products  for ethylene CH 2 =CH 2 + OH.  OHCH 2 CH 2. OHCH 2 CH 2. + O 2  OHCH 2 CH 2 O 2.  OHCH 2 CH 2 O 2. + NO  NO 2 + OHCH 2 CH 2 O.  OHCH 2 CH 2 O. + O 2  H 2 C=O +. CH 2 OH  O 2 +. CH 2 OH  H 2 C=O +. HO 2

40. 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 NO 2,just like HO 2 can

41. Further reaction takes place resulting in carbonyls and HO 2 which now undergo further reaction; the process then proceeds…

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

43. Aromatic Reactions

44. Nitrogen Storage (warm vs. cool) H 3 C-C=O OH H 3 C-C=O + H 2 O. H PAN warm cool

45. Nitrogen Loss (HNO 3 formation) NO 2 + O 3  NO 3. + O 2 NO 3. + NO 2  N 2 O 5 N 2 O 5 + H 2 O  2HNO 3 (surface) NO 2 + OH.  HNO 3 (gas phase)

46. Nitrogen Loss (alkylnitrates) 2-butanal butane -C-C-C-C- O 2 + H. NO 2 -C-C-C-C- 2-butylnitrate

47.  The rate of of formation of O 3 is governed by the reaction:  NO 2 + uv light -> NO + O. and its rate const k 1 because:  O. + O 2 +M  O 3 + M is very fast  so the rate of formation O 3 is:  rate form = +k 1 [NO 2 ] How can we easily estimate O 3 if we know NO and NO 2 ?

48.  The rate of removal of O 3 is governed by the reaction:  O 3 + NO  NO 2 + O 2 and its rate const k 3  so the rate of removal of O 3 is:  rate remov = -k 3 [NO] [O 3 ]  the overall rate tot =rate form +rate remov

49.  rate tot = -k 3 [NO] [O 3 ] +k 1 [NO 2 ]  if rate tot at steady state = 0, then  k 1 [NO2]= k 3 [NO][O 3 ] and  [O 3 ] = k 1 [NO 2 ] / {k 3 [NO] }  This means if we know NO, NO 2, k 1 and k 3 we can estimate O 3

50.  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

51.  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 O 3 formation?

52.  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.

53.  In 1972 we built the first large outdoor smog chamber, which had an interior volume of 300 m 3.  We wanted to predict oxidant formation in in the atmosphere.  The idea was to add different hydrocarbon mixtures and NO + NO 2, to the chambers early in the morning.

54.  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.

55. The Chamber had two sides Or Darkness Formaldehyde propylene 300 m 3 chamber Teflon Film walls NO &NO 2

56. Example experiment with the following chamber concentrations: NO = 0.47 NO 2 = 0.11 ppm Propylene = 0.99 ppmV temp = 15 to 21 o C

57. Solar Radiation Profile

58. Example Mechanism  NO 2 + h  NO + O. k 1 keyed to sunlight  O. + O 2 --> O 3 k 2  O 3 +NO 2 --> NO + O 3 k 3  H 2 C=O + h -->.HC=O + H. k 4 keyed to sunlight  H. +O 2 --> HO 2. k 5  HO 2. + NO --> NO 2 +OH. k 6 (fast)  OH.+ C=C ---> H 2 C=O + HO 2 + H 2 COO. k 7 dNO 2 /dt = -k 1 [NO 2 ];  NO2=-k 1 [NO 2 ]  t

59. Photochemical System

61.  The fact that - dT/dz = g/ c p = 9.8 o K/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

62.  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  s is ~ 4 o K/km and at 6 km and –5 o C it is ~6-7 o K/km The quantity  d is called the dry the dry adiabatic lapse rate

63. 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  midday altitude temp  Sun-down earth cools Inversion layer altitude temp more cooling at surface at night }

64. What happens to the material above the 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. Inversion layer altitude temp more cooling at surface at night residual layer } }

65. 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 } 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 Inversion layer altitude temp Heating at surface during the nest day }

67. 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

68. Mixing height in the morning Balloon temperature height in kilometers Temp in o C Inversion height }

69. 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 5 o C 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 20 o C. We would add 5 o C to it and get 25 o C. We then move up the 25 o C 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

70. Mixing height in the morning Balloon temperature Temp in o C 20253035 Dry adiabatic lines height in kilometers 0.0 0.1 0.2 0.3 0.4 1.1 1.5

71. 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 35 o C, we would estimate an afternoon a mixing height of ~1.67 km

72. Afternoon Mixing height Balloon temperature Temp in o C 20253035 Dry adiabatic lines height in kilometers 0.0 0.1 0.2 0.3 0.4 1.1 1.5

73.  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

74. How do we get light into the mechanism?? uA molecule photolyzes or breaks apart when it absorbs photons that have energy that is greater than the bond strength uLet’s look at the energy in a mole of photons which have a wavelength 288 nm uThe energy E, in this light is  E= 6.02x10 23 x hc/ c= 3x10 8 m/s; h=6.63x10 -34 Js,  =288x10 -9 m uE= 416kJ/mole uIf all this light was absorbed it would break C- H bond

75. Light and rate constants uThe question is, is all the light absorbed??  Actually not, but this brings up the concept of quantum yields, , and light absorption    = # molecules reacted / # photons absorbed  What about the light flux, j at a given uThis is the # of photons of light cm -2 sec -1 uThe rate constant for photolyis can be written as  k rate = J x  x absorption coef

76. Light and rate constants  k rate = J x  x absorption coef  the absortion coef.  has units of cm 2 /molecule and comes from Beer’s law I=I o e -  l[C]  k rate = J x  x   This is at one wavelength  what do we do when we have two wavelengths and 1?  k rate  = the rate const. at a different wavelength   and k rate  = J 1 x   x    k rate total = J x  x  + J 1 x   x  

77. Light and rate constants  k rate total = J x  x  + J 1 x   x   uso across all wavelengths  so k rate total =  J x  x   What this says is that if we know the light flux or “intensity” at each wavelength, J, the absorption coef.,  at each wavelength and the quantum yield , we can calculate k rate total for the real atmosphere

78. Light and rate constants  Lets calculate k rate  for NO 2 at the wave length of 400-405 nm and a zenith angle of 20 degrees uJ 400-405nm = photons cm -2 sec -1 = 1.69x10 15   400nm = quantum yield = ~0.65   400-405nm = ~6x10 -19 cm 2 molecule -1  k rate = J x  x   0.00067sec -1

79. Light and rate constants uso in the reaction uNO2 + light at 400-405nm -> NO + O.  k rate = J x  x   0.00067sec -1  dNO 2 /dt = k rate [NO2]

80. Light and rate constants uThere are tables that give J at each wave length as a function of the angle of the sun uThe angle of the sun is called the zenith angle. When the sun is directly over head the zenith angle is zero degrees uwhen it has just gone down it is 90 o

81. Light and rate constants Sun 

82. Light and rate constants uThis 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 uin the winter time it will not go as high in the sky as in summer.  from these tables if we know  and  for a compound we can calculate the photolysis rate constants for any compound over the course of the day as the zenith angle changes uNO 2, H 2 C=O, O 3, acetaldehyde

83. 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

84. Global Emissions of hydrocarbons  1150 x10 12 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

85. 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

86. 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 which bring about aerosol formation. Chemical System + NOx + sunlight ----> aerosols  -pinene  -pinene was selected because it is generally the most prevalently emitted terpene from trees and other plants

87. Overview The reactions of biogenic hydrocarbons produce low vapor pressure reaction products that distribute between gas and particle phases. Gas Particle Partitioning atmospheric particle gas phase products pinonic acid OH O O

88. 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 K p = k on /k off 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 O 3 or NOx in sunlight in an outdoor chamber.

89. OH attack on  -pinene

90. . pinonald-oo OH pinonic acid O pin-O 2 OO. NO 2 norpinonaldehyde +HO 2 +NO 2 O pinonald-PAN =o OO. OH O 2 =o O O=C 8 =O C 8 -oo. =o OONO 2 +HO 2 O 2 NO 2 norpinonaldehyde =o +HO 2 CO =o OO. O O O + h + methy l glyoxa l + OH + CO+HO 2 =o OO. O 2 NO 2 =o +HO 2 + h NO 2 =o OO. C 8 -oo. +CO+HO 2 NO 2 (f) (g) CO 2 + pin-O 2 pin-OO 2 +CO H 2 O+ +H 2 O Reactions of product pinonaldehyde with OH and light

91. Particle formation-self nucleation uCriegee biradicals can react with aldehydes and carboxylic groups to form secondary ozonides and anhydrides. O=C C=O CH 3 + C C=O CH 3 C C=O CH 3 O C C=O. CH 3 oo Creigee + pinaldehyde --> seed1 The equilibrium between the gas and particle phases is: K p = k on /k off

92. The equilibrium constant K p can be calculated (Pankow, Atmos. Environ, 1994) p o L is the liquid vapor pressure and  the activity coefficient of the partitioning organic in the liquid portion of the particle, f om is the raction of organic mass in the particle and Mw is the average molecular weight of the organic mass

93. Rates that Gases enter and leave the particle can be estimated from K p = k on /k off where k off = {k b T/h} e -Ea/RT E a can be estimated and with Boltzman’s (k b ) and Planck’s constants (h) and temperature,T. k off can be calculated and with K p, k on can also be evaluated

94. Overall Mechanism linked gas and particle phase rate expressions Representitive  -pinene gas phase reactions + rate constants (#) min -1 or ppm-1 min-1 1 ] OH +  -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]  -pinene + NO3 --> apNO3-oo # 544 exp (818/T) 6]  -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 # 1.0.6x10 11 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] diacid gas + seed --> seed + diaacid part # 70, 31] pinacid + pinacid part --> pinaid part + pinacid part # 25, 33] oxypinaldgas + pinald part --> oxypinald part + pinald part # 20 36] pinald-PAN gas + oxypinaicd part --> pinald-PAN part + oxypinaicd part # 25, 37] OH-apNO3 gas + pinald-PAN part --> OH-apNO3 part + pinald-PAN part # 25, 38] diacid part --> diacid gas # 3.73e14 exp (-10350/T), 39] pinacid part --> pinacid gas # 3.73e14 exp (-9650/T), 45] OH-apNO3 part --> OH-apNO3 gas # 3.73e14 exp (-9200/T),

95. O 3 data NO y NO 2 NO  -pinene pinonaldehyde Particle phase pinonaldehyde pinonic acid data norpinonic acid Particle phase ppm mg/m 3 Particle phase model TSP filter data mg/m 3 Time in hours (EDT) mg/m 3 O 3 model Sum products (data) A norpinonaldehyde pinald model Gas phase mg/m 3 C E F D pinald model pinacid model ppmV  - pinene B diacid model  oxypinald pinic acid data Particle phase Time in hours (EDT) Particle formation from  -pinene + NOx in the presence of Sunlight; symbols are data and lines are model predictions O O

96. Reaction of  -pinene with O 3 at different concentrations in the dark ; top experiment #1, chamber temperature 23 o C ; middle experiment #2, 12 o C ; bottom experiment #3, 27 o C; symbols are data, and lines are model predictions

97. Summary Models vs. experimental aerosol yields illustrate that reasonable predictions of secondary aerosol formation are possible from both dark ozone and light-NOx/  - 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  -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  -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.