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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

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introduction to photochemical smog chemistry
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
  • 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
ozone3
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
ozone4
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
ozone5
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 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
how do we measure ozone
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 ozone10
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 ozone11
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 ozone12
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 ozone13
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 ozone14
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 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 curve16
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

standard curve
Standard Curve

I2 absorbances are plotted vs. concentration

absorbance

1 2 3 4 5 concentration (mg/liter)

how do we measure ozone20
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
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
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
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
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
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
slide26
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
slide27
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)
slide28
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.
slide29
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.
slide30
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].
slide31
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
slide32
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
slide33
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
slide34
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
slide35

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)
slide36
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
In the urban setting there are a lot of ground base combustion sources

Exhaust

hydrocarbonsNO & NO2

CO

slide38
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??)
slide39
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
slide40

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

slide41

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

slide42
There is similar chemistry foralkanes
  • OH. + H 3-C-CH3 --> products
  • and for aromatics
  • OH. + aromatics --> products
slide43

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

nitrogen storage warm vs cool

PAN

warm

cool

Nitrogen Storage (warm vs. cool)

OH

H

C-C=O

+ H2O

H

C-C=O

.

3

3

H

nitrogen loss hno 3 formation
Nitrogen Loss (HNO3 formation)
  • NO2 + O3  NO3.+ O2
  • NO3.+ NO2  N2O5
  • N2O5 + H2O  2HNO3 (surface)
  • NO2 + OH. HNO3 (gas phase)
nitrogen loss alkylnitrates

-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

slide47

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]
slide48
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
slide49
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
slide50
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
slide51
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?
slide52
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.
slide53
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.
slide54
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
The Chamber had two sides

Or Darkness

NO &NO2

300 m3 chamber

Teflon Film walls

propylene

Formaldehyde

example experiment with the following chamber concentrations
Example experiment with the following chamber concentrations:
  • NO = 0.47
  • NO2 = 0.11 ppm
  • Propylene = 0.99 ppmV
  • temp = 15 to 21oC
example mechanism
Example Mechanism
  • 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
slide61
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
slide62

The quantity dis 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
slide63

Sun-down earth cools

more cooling at surface at night

midday

altitude

altitude

altitude

}

Inversion layer

temp

temp

temp

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.

slide64

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.

slide65

more cooling at surface at night

Heating at surface during the nest day

altitude

altitude

}

Inversion layer

}

Inversion layer

temp

temp

When the sun comes up the next day it heats the earth an the air close to the earth.

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
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

}

Inversion height

Mixing height in the morning

height in kilometers

Temp in oC

mixing height in the morning69
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 morning70

Balloon temperature

1.5

1.1

height in kilometers

Dry adiabatic

lines

0.4

0.3

0.2

0.1

0.0

20

25

30

35

Temp in oC

Mixing height in the morning
mixing height in the afternoon
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

Balloon temperature

1.5

1.1

height in kilometers

Dry adiabatic

lines

0.4

0.3

0.2

0.1

0.0

20

25

30

35

Temp in oC

Afternoon Mixing height
slide73
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
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
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, jat a given l?
  • This is the # of photons of light cm-2 sec-1
  • The rate constant for photolyis can be written as
  • kratel= Jlxflx absorption coefl
light and rate constants76
Light and rate constants
  • kratel= Jlx fl x absorption coefl
  • the absortion coef. s has units of cm2/moleculeand comes from Beer’s law I=Io e-sl[C]
  • kratel= Jlxflxsl
  • 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 l1and kratel1 = J1 xfl1 xsl1
  • krateltotal = Jlxflxsl + J1 xfl1xsl1
light and rate constants77
Light and rate constants
  • krateltotal = Jlxflxsl + Jl1 xfl1xsl1
  • so across all wavelengths
  • sokrateltotal= SJlxflxsl
  • What this says is that if we know the light flux or “intensity” at each wavelength, Jl , the absorption coef., sl at each wavelength and the quantumyield sl , we can calculate krateltotalfor the real atmosphere
light and rate constants78
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= Jlxflxsl= 0.00067sec-1
light and rate constants79
Light and rate constants
  • so in the reaction
  • NO2 + light at 400-405nm -> NO + O.
  • kratel= Jlxflxsl= 0.00067sec-1
  • dNO2/dt = krate l [NO2]
light and rate constants80
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 constants82
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 sfor 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
slide83

Extending this kinetics approachto simulate secondary Aerosols formation by linking gas and particle phase chemistryAn exploratory modelfor aerosol formation from biogenic hydrocarbons using a gas-particle partitioning/thermodynamic model-Kamens Research Group, ES&T, 1999 and 2001

slide84

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
slide85

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
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

a-pinene

a-pinenewas selected because it is generally the most prevalently emitted terpene from trees and other plants

overview

O

O

OH

Overview
  • The reactions of biogenic hydrocarbons produce low vapor pressurereaction products that distribute between gas and particle phases.

Gas Particle Partitioning

gasphase products

pinonic acid

atmospheric particle

slide88
Equilibriumpartitioning 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 forwardand 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

O2

OH

OO

OH

OH

CHO

HOO

HOO

O

+ OH

O

CHO

pinonaldehyde

OH

+ CO, HO

2,

CHO

O

O3 attack on

a-pinene

norpinonaldehyde

COOH

O

O

O

O

O

norpinonic

3

acid

Criegee1

COOH

O

O

O

pinonic acid

O

a

-pinene

CH

CHO

O

3

+ other

O

COOH

products

Criegee2

COOH

pinic acid

OH attack on a-pinene
slide90

O

O

(c)

pinonaldehyde

OH

O

(b)

2

(a)

+

O

.

OO

acetone

(d)

(e)

NO

NO

2

O

O

Reactions of product pinonaldehyde with OH and light

+ hn

+

+CO+HO

O

2

2

=o

=o

.

O

OO

CO

+NO

2

=o

=o

H

O+

pin-OO

O

2

2

.

OONO

O

2

OO

O

pinonald-oo

NO

+HO

NO

2

2

+

pinonald-PAN

pin-O

methyl

2

=o

glyoxal

(f)

=o

O

NO

NO

+HO

2

2

OH

=o

pinonic acid

=o

+HO

n

+ h

2

=o

=o

.

=o

norpinonaldehyde

+CO+HO

2

OH

O

OO

.

2

C

-oo.

8

=o

(g)

=o

NO

+HO

NO

2

2

CO

+

NO

OH

2

.

=o

OO

.

OO

NO

O

=o

C

-oo.

2

8

pin-O

2

NO

=o

+CO

2

NO

2

+H

O

2

norpinonaldehyde

O

O=C

=O

8

particle formation self nucleation

CH

3

CH

C=O

3

oo

C=O

.

+

C

O=C

CH

3

CH3

C=O

C=O

oo

C

C

O

Particle formation-self nucleation
  • Criegee biradicals can react with aldehydes and carboxylic groups to form secondary ozonides and anhydrides.

Creigee + pinaldehyde --> seed1

The equilibrium between the gas and particle phases is:

  • Kp= kon/koff
slide92
The equilibrium constant Kp can be calculated (Pankow, Atmos. Environ, 1994)
  • poLis the liquid vapor pressure and g the activity coefficient of the partitioning organic in the liquid portion of the particle, fomis the raction of organic mass in the particle and Mw is the average molecular weight of the organic mass
slide93
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. koffcan be calculated and with Kp, kon can also be evaluated
overall mechanism linked gas and particle phase rate expressions
Overall Mechanism linked gas and particle phase rate expressions

Representitive a-pinene gas phase reactions + rate constants (#) min-1or 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),

slide95

A

B

a-pinene

Particle phase

model TSP

NO

O3model

ppm

mg/m3

ppmV a-pinene

Sum products

(data)

O3data

NOy

filter data

NO2

O

O

C

Particle phase

D

Gas phase

pinonaldehyde

mg/m3

pinaldmodel

mg/m3

pinonaldehyde

pinaldmodel

norpinonaldehyde

E

Particle phase

Particle phase

F

pinic aciddata

mg/m3

mg/m3

pinonic aciddata

diacidmodel

pinacidmodel

Soxypinald

norpinonic acid

Time in hours (EDT)

Time in hours (EDT)

Particle formation from a-pinene + NOx in the presence of Sunlight; symbols are data and lines are model predictions

slide96

0.4

a

-

pinene

0.3

Initial reactants

Particle formation

ppmV

0.2

1a

1b

O

3

0.1

0

19

19.5

20

20.5

time in hours (pm)

3.5

3

a

Reacted

-

pinene

2.5

filter mass data

2

mg/m3

2a

0.7

2b

1.5

0.6

model

1

O

0.5

3

0.5

0.4

ppmV

0

0.3

19

19.5

20

20.5

21

21.5

22

22.5

time in hours (pm)

0.2

a

-

pinene

0.1

2

0

a

Reacted

-

pinene

19.5

19.7

19.9

20.1

20.3

20.5

1.5

time in hours (pm)

mg/m3

1

3a

3b

EAA

model

0.5

a-pinene

Filter mass data

filter mass data

0.15

0

mg/m3

19

20

21

22

ppmV

time in hours (pm)

model

O3

time in hours (pm)

time in hours (pm)

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

slide97
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 nitratesaccounts 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 fromthe 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.