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

Why are we interested in the Smog Chemistry??? Let’s start with the formation of ozone Then continue on to fine particles that form in a smog atmosphere

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 > O 3 (ozone) uits 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 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

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

Ozone  we have 1x10 -6 m 3 of ozone in one ppm  so: 1x10 -6 m = #moles O x10 -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

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

Lung function before exposure to O.32 ppm O 3

Lung function after exposure to O.32 ppm O 3

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 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

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

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 KI solution

How do we measure Ozone  A top is added to the bubbler so that air can enter the KI solution  KI + O 3 --> I 2  measure I 2 KI solution

How do we measure Ozone  a pump is attached to the bubbler KI solution pump

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

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

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

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

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

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

Serial dilutions from stock solution I2I2

I2I2 5 mg/Liter

Serial dilutions from stock solution I2I2 53 mg/Liter

Serial dilutions from stock solution I2I2 532 mg/Liter

Serial dilutions from stock solution I2I mg/Liter

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

Making a plot I 2 adsorbances are plotted vs. concentration

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

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

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

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

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

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 O 3

Chemilumenescence measurement of Ozone PM tube

Chemilumenescence measurement of Ozone PM tube pump

Chemilumenescence measurement of Ozone PM tube ethylene pump waste ethylene

Chemilumenescence measurement of Ozone PM tube ethylene O3O3 sample air with O 3 PM tube pump waste ethylene

Chemilumenescence measurement of Ozone PM tube pump ethylene O3O3 sample air with O 3 {H 2 C=O}* PM tube picks up light from {H 2 CH=O} * waste ethylene

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}*

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

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

 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)

 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.

 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. O. can react with H 2 O to form OH. Radicals O. + H 2 O  2OH.

 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  = -230 ppm -1 min -1 x 3 x10 -8 ppm x t 1/2  t 1/2 = 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 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

 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

 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??

 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?

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  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??)

 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

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

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

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

Aromatic Reactions

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

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)

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

 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 ?

 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

 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

 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 O 3 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 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.

 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.

New UNC Aerosol Smog Chamber

The Chamber had two sides Or Darkness 300 m 3 chamber Teflon Film walls

The Chamber had two sides Or Darkness Formaldehyde propylene 300 m 3 chamber Teflon Film walls

The Chamber had two sides Or Darkness Formaldehyde propylene 300 m 3 chamber Teflon Film walls

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

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

Solar Radiation Profile

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

Photochemical System

 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

What we have described is a box model or a batch reactor….materials are added one time at the beginning and they react… Is this how the atmosphere works? Is there a change in the mixing volume of the atmosphere with time or does the air volume over a city stay constant?

We need to get mixing height into the model Balloon temperature Temp in o C Dry adiabatic lines height in kilometers

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 a mole of light at 288 nm is  E= 6.02x10 23 x hc/ c= 3x10 8 m/s; h=6.63x Js,  =288x10 -9 m uE= 416kJ/mole uIf all this light was absorbed it would break C- H bond

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 a mole of light at 288 nm is  E= 6.02x10 23 x hc/ c= 3x10 8 m/s; h=6.63x Js,  =288x10 -9 m uE= 416kJ/mole uIf all this light was absorbed it would break C- H bond

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 uThe light flux is the # of photons of light cm -2 sec -1  I have called this I but often it is called J or F at a given  The rate constant for photolyis can be written as k rate = J x  x absorption coef

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 2 ?  k rate total = J  x   x   + J 2 x   x  

Light and rate constants uso across all wavelengths  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

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

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

Solar flux striking the earth

Light and rate constants 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

Light and rate constants (zenith angle) Sun 

Light and rate constants uWhen the sun is at lower angles in the sky, less energy strikes the surface of the earth; sun up and sun down the zenith angle is zero uAlgorithms have been written to relate zenith angle to light flux at each wavelength

Light and rate constants uThe zenith angle for given latitude and time of year can known, as well as the time that 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.

Light and rate constants  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