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Objective

Envr 890 Organic Aerosols: Lecture 2. What questions do we have to start asking if we are to build a kinetics model to predict aerosol formation from isoprene reactions? kamens@unc.edu http://www.unc.edu/~kamens. Objective.

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Objective

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  1. Envr 890 Organic Aerosols:Lecture 2.What questions do we have to start asking if we are to build a kinetics model to predict aerosol formation from isoprene reactions?kamens@unc.eduhttp://www.unc.edu/~kamens

  2. Objective • To describe a predictive techniquefor the formation of aerosols from biogenic hydrocarbons based on fundamental principals and apply it to isoprene • Have the ability to embrace a range of different atmospheric chemicaland physical conditions which bring about aerosol formation.

  3. Chemical System + NOx+ sunlight + ozone----> aerosols a-pinene

  4. Overview • This reaction system produces low vapor pressurereaction products that distribute between gas and particle phases.

  5. COOH O cis-pinonic acid Gas Particle Partitioning gasphase products particle

  6. Equilibrium partitioning between the gas and particle phases is assumed for all products. • Kinetically this is represented as forwardand backward reactions. • Kp = kon/koff

  7. [gas] + [surface] [particle] The equilibrium constant Kp can be calculated from

  8. Overview • Gas and particle phase reactions were linked in one mechanism and a chemical kinetics solver was used to simulate the reaction over time • Model simulations were compared with aerosol concentrations obtained by reacting -pinene with either O3 or NOx in sunlight in an outdoor chamber.

  9. -pinene-O3 experiments in the UNC 190m3outdoor chamber 1. added O3 O3 -pine O3 2. added -pinene O3 O3 -pine O3 O3 -pine

  10. -pinene-O3 experiments in the UNC 190m3outdoor chamber 1. added O3 O3 -pine O3 2. added -pinene O3 O3 -pine 3. immediate burst of particles O3 O3 -pine

  11. O COOH • pinald - • pinacid - • diacid - • oxy-pinald - • oxy-acid- Pinonic acid O CHO COOH pinonaldehyde HOOC nor-pinonaldehyde Pinic acid O CHO COO HOOC C8 diacid O COOH OH O O CHO COOH Norpinonic acid OH Products

  12. + CO, HO OH 2, CHO O norpinonaldehyde COOH O O O O norpinonic O acid Criegee1 3 COOH O O O pinonic acid O a -pinene CH CHO O 3 + other O COOH products Criegee2 COOH pinic acid Mechanism

  13. Now you need to know about the reactions of alkenes with O3 • C=C +O3 • See pages Pitts and Pitts, 2000, pages 179-206

  14. How do we chemically represent his reaction scheme? 1st the reaction of O3 with an alkene generates an aldehyde +a high energy bi-radical called a Criegee bi-radical 2nd a certain fraction of these “hot” Criegee’s stabilize and the rest decomposes. (see page 198 of Pitts and Pitts, Table 6.10) 3rd “Hot” Criegee’s R-(H)C.-00. decompose to OH, CO and an R. group

  15. For the propylene +O3 reaction (page 198 Pitts and Pitts) propylene +O3  two Criegee’s C-C=C +O3  H2C.OO.* + CH3C.HOO.* + what two aldehydes? CH3C.HOO.* 0.15 Stabilized Criegee  0.54 (CH3. + CO +OH.)  CH3. + CO2+ H.  HCO +CH3O 0.17  0.14 (CH4 + CO2)

  16. How do we chemically represent this for the a-pinene system? apine + O3 0.4*crieg1 + 0.6*crieg2min-1 or ppm-1min-1 1.492 e-732/T More highly substituted Criegee’s are more stable crieg1 = 0.47*pinacid + 0.13*stabcrieg1 1x106 + 0.8*OH + 0.3*HO2 + 0.37*pinald + 0.01*oxypinald + 0.03*O. + 0.19*CO

  17. What would we do for isoprene (see Kamens et al, IJKS, 1982?

  18. a-pinene O 2 OH OO OH OH HOO HOO CHO O + OH O CHO OH attack on a-pinene pinonaldehyde

  19. Each Reaction can be represented as a differential equation • a-pinene + ozone --> pinald • d[a-pinene]/dt = -krate1[a-pinene][O3] • d[pinald]/dt = -krate2[pinal][OH]

  20. Particle formation-self nucleation

  21. CH 3 C=O CH 3 oo. C=O . + C O=C CH 3 C=O CH 3 C=O oo C C O Particle formation-self nucleation • Criegee’s can react with aldehydes and carboxylic groups to form secondary ozonides and anhydrides.

  22. Estimating rates that gas phase products enter and leave the particle • The equilibrium between the gas and particle phases is: • Kp = kon/koff

  23. The equilibrium constant Kp can be calculated from • [gas] + [surface] [particle] Kp = kon/koff

  24. A general Fluxexpression for evaporation from a liquid surface is • is the energy barrier for single molecule to evaporate from the surface (Joules) • koff = k evap in koff =  e -Ea/RT is: kbT/h • From koff and Kp the rate coefficient kon can now be calculated

  25. Overall Mechanism • linked gas and particle phase rate expressions

  26. Diacid is generated in the gas phase. How do we get it on and off the particle phase?? diacidgas + pinacidpart --> pinacidpart + diacidpart 68 diacidpartdiacidgas 3.73x1014 e-10285/T Where do the rate coefficients come from??

  27. Kp= kon/koff koff =  e -Ea/RT We need vapor pressure (poL ) to calculate Kp for koff , we need kbT/h for , and an activation energy, Ea To calculate vapor pressures see Chapter 4 of (Schwarzenbach et al, Environmental Organic Chemistry, 2003)

  28. if we substitute DSvap Tb= 88J mol-1 K-1 and R =8.31 Jmol-1 K-1 This means we need Tbor boiling point

  29. Boiling points can be estimated based on chemical structure (Joback, 1984) Tb= 198 + S DTb DT (oK) -CH3 = 23.58 K -Cl = 38.13 -NH2 = 73.23 C=O = 76.75 CbenzH- = 26.73 C-OH = 93 Joback obs (K) (K) acetonitrile 347 355 acetone 322 329 benzene 358 353 amino benzene 435 457 benzoic acid 532 522 toluene 386 384 pentane 314 309 methyl amine 295 267 trichlorethylene 361 360 phenanthrene 598 613

  30. Stein, S.E., Brown, R.L. Estimating Normal boiling Points from Group Contributions, J.Chem. Inf Comput. Sci, 34, 581-587, 1994

  31. Chamber Data • First look at experiments with different temperatures and gas phase concentrations of -pinene and ozone

  32. A high concentration warm experiment • 0.82 ppmV -pinene + 0.60 ppm O3 • Temp = ~295K • %RH = ~61%

  33. Gas Phase -pinene and O3data and model A -pinene MODEL 

  34. Particle concentration. Warm high concentration experiment Reacted -pinene A mg/m3 Observed model

  35. Other Conditons 0.88 ppmv a-pinene + 0.47 ppm O3(269K) 0.82 ppmv a-pinene + 0.60 ppm O3(295K) 0.60 ppmv a-pinene + 0.65 ppm O3 (284K) 0.35 ppmv a-pinene + 0.25 ppm O3(295K)

  36. Lower concentration experiments

  37. 0.20 ppm a-pinene + 0.11 ppm O3 model data

  38. How do we start to build an Isoprene + O3 mechanism??

  39. The initail attack of O3 on an olefin (alkene) generates two possible Criegee high energy bi-radicals and two possible aldehydes.

  40. A general rule is that the more highly substituted carbon will form a higher fraction of Criegee’s

  41. H H H + O3 MACR + H2C.OO. H H CH3  CI1.OO. + HCHO C=C-C=C  MVK + H2C.OO.  CI2.OO. + HCHO

  42. What would we do for isoprene (see Kamens et al, IJKS, 1982?

  43. There are 4 general product channels by which high energy Criegee’s (Criegee intermediates) decompose: • Stabilized Criegee biradicals • an Ester Channel • O-atom elimination • Hydroperoxide channel

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