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Introduction to Photochemical Smog Chemistry
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Introduction to Photochemical Smog Chemistry

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  1. 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

  2. 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

  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 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

  4. 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

  5. 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

  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 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

  7. Lung function after exposure to O.32 ppm O3

  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 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

  10. How do we measure Ozone • a neutral buffered solution of potassium iodide is placed in a bubbler • KI + O3 --> I2 • measure I2

  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 I2 Air goes in KI solution + I2

  12. How do we measure Ozone • The absorbance of the I2 in the KI solution is then measured with a spectrophotometer KI solution + I2

  13. How do we measure Ozone • The absorbance of the I2 in the KI solution is then measured with a spectrophotometer KI solution + I2

  14. How do we measure Ozone • The absorbance of the I2 in the KI solution is then measured with a spectrophotometer Spectrophotometer KI solution + I2

  15. 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

  16. 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

  17. Serial dilutions from stock solution I2 5 3 2 1 mg/Liter

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

  19. Standard Curve I2 absorbances are plotted vs. concentration absorbance 1 2 3 4 5 concentration (mg/liter)

  20. How do we measure Ozone • The absorbance of the I2 in the KI solution is then measured with a spectrophotometer Spectrophotometer KI solution + I2

  21. 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)

  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 O3

  24. Chemilumenescence measurement of Ozone pump sample air with O3 O3 PM tube {H2C=O}* waste ethylene ethylene catalytic converter CO2 + H2O

  25. 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- aLC

  26. 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

  27. 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)

  28. 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. radicalsO. + H2O -> 2OH.

  29. 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.

  30. 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].

  31. 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

  32. 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 or69.7days

  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 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

  34. Why is the reaction of OH. with ethylene so much faster than with methane? H H1. H-C-H....OH . -> H-C. + .HOH . H H 2. H2C=CH2attack by OH.is at the double bond, which is rich in electrons

  35. 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)

  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 O3 formation?

  37. In the urban setting there are a lot of ground base combustion sources Exhaust hydrocarbonsNO & NO2 CO

  38. 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 cycleNO2 + 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??)

  39. 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

  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 NO2,just like HO2 can

  41. Further reaction takes place resulting in carbonyls and HO2 which now undergo further reaction; the process then proceeds…

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

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

  44. PAN warm cool Nitrogen Storage (warm vs. cool) OH H C-C=O + H2O H C-C=O . 3 3 H

  45. Nitrogen Loss (HNO3 formation) • NO2 + O3  NO3.+ O2 • NO3.+ NO2  N2O5 • N2O5 + H2O  2HNO3 (surface) • NO2 + OH. HNO3 (gas phase)

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

  47. How can we easily estimate O3if 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]

  48. 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

  49. 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

  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