smog print

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 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

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

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

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 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

Lung function after exposure to O.32 ppm O3

0

2

4

6

8

Lite

rs/s

ec

0 1 2 3 4 Litersbefore after

Athletic performance

0

10

20

30

40

50

60

70

80 de

crea

sed

perf

orm

ance

%

0.1 0.15 0.2 0.25 0.3 0.35 0.4 O3 in ppm

1962-1964

1959-1961

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


a neutral buffered solution of potassium iodide is placed in a bubbler

KI + O3 --> I2

measure I2


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


The absorbance of the I2 in the KI solution is then measured with a spectrophotometer

KI solution + I2



KI solution + I2

Spectrophotometer

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 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

Serial dilutions from stock solution

I2

5 3 2 1mg/Liter

absorbances are measured for each of the serially diluted standards

Spectrophotometer

absorbance

Standard Curve

I2 absorbances are plotted vs. concentration

1 2 3 4 5 concentration (mg/liter)

abso

rban

ce



KI solution + I2

Spectrophotometer

We then compare our sample absorbance to the standard curve

I2 absorbances are plotted vs. concentration

1 2 3 4 5 concentration (mg/liter)

abso

rban

ce

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 O3

Chemilumenescence measurement of Ozone

PMtube

pump

waste ethylene

catalytic converter

ethylene

CO2 + H2O

O3 sample air with O3

{H2C=O}*

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- LC

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

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)

O3 can also react with nitric oxide (NO)

O3 + NO NO2 + O2

both oxygen and O3 photolyzes to give O.

O2 + h O. +O. (stratosphere)

O3 + h O. + O2

O. can react with H2O to form OH. radicalsO. + H2O 2OH.

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.

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].

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

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 ppmfor 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 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 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

Why is the reaction of OH. with ethylene so much faster than with methane?

H H1. H-C-H....OH . -> H-C. + .H OH .

H H

2. H2C=CH2 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

NO2 + uv lightNO + O.

O. + O2 +MO3 + M

O3 + NO NO2 + O2

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 O3 formation?

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

Exhaust

hydrocarbonsNO & NO2

CO

If organics are present they can photolyze or generate radicals H2C=O + h-> .HC=O + H.

H. + O2.HO2

if we go back to the cycle NO2 + uv light -> NO + O. O. + O2 +MO3 + 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??)

OH. + can now attack hydrocarbons such which makes formaldehyde and other radical products

for ethylene CH2=CH2 + OH. OHCH2CH2

.

OHCH2CH2. + O2OHCH2CH2O2

.

OHCH2CH2O2. + NONO2 + OHCH2CH2O.

OHCH2CH2O. + O2 H2C=O + .CH2OH

O2 + .CH2OH H2C=O + .HO2

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

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

There is similar chemistry foralkanes

OH. + H 3-C-CH3 --> products

and for aromaticsOH. + aromatics --> products

CH3

OH OH

CH2.

O=CH

NO NO2+O2

CH3

H

OH

H .

CH3OH

*

O2

+ HO2+ H2O

CH3

H

OH

H

CH3

O

O

+

.

CH3

O

CH3

H

OH

H

.O

NO

NO2+O2

toluene

o-cresolbenzaldehyde

rearrangementOH

HO .

H

HO

H

OH

HO

H

+ HO2

+O2

CH3H

+

methylglyoxalbutenedial

oxygen bridgeradical

+ HO2

?ring cleavageradical

Aromatic Reactions

Nitrogen Storage (warm vs. cool)

H3C-C=OOH H3C-C=O+ H2O.

H O2

H3C-CNO2

H3C-CPAN

H3C-C + NO2

warmcool

Nitrogen Loss (HNO3 formation)

• NO2 + O3 NO3.+ O2

• NO3.+ NO2 N2O5

• N2O5 + H2O 2HNO3 (surface)

• NO2 + OH. HNO3 (gas phase)

Nitrogen Loss (alkylnitrates)

2-butanal

butane

-C-C-C-C- + OH. --> -C-C-C-C- + H2O .

-C-C-C-C-

O2NO

NO2

-C-C-C-C-

-C-C-C-C- + H.NO2 -C-C-C-C-

2-butylnitrate

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 +MO3 + M is very fastso the rate of formation O3 is:

rateform = +k 1 [NO2]

How can we easily estimate O3 if we know NO and NO2?

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

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

Calculate the steady state O3 from the following:

NO2 = 0.28 ppmNO = 0.05 ppmk1 = 0.4 min-1k3 = 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 O3 formation?

Can we use computers to predict the amount of ozone formed if we know what is going into the atmosphere?

yesbut 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 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.

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

Or Darkness

Formaldehyde

propylene

300 m3 chamber

Teflon Film walls

NO &NO2

Example experiment with the following chamber concentrations:

• NO = 0.47 • NO2 = 0.11 ppm• Propylene = 0.99 ppmV• temp = 15 to 21oC

Solar Radiation Profile

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

cal c

m-2

min

-1

6 8 10 12 14 16 time in hours

TSR

Example Mechanism NO2+ hNO + O. k1 keyed to sunlight

O. + O2 --> O3 k2

O3 +NO2 --> NO + O3 k3

H2C=O + h --> .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]; NO2=-k1 [NO2] t

Photochemical System

0

0.1

0.2

0.3

0.4

0.5

0.6 pp

m

10 11 12 13 14 15 time in hours

NO-data O3-mod NO2-data

O3-data NO-mod NO2-mod

NOx-O3: model vs. data

PAN

NO2O3

NO

NO2

Photochemical System

0

0.2

0.4

0.6

0.8

1

1.2

ppm

V

10 11 12 13 14 15 Time in hours

Propylene: data vs. model

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

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

The quantity d is called the dry the dry adiabatic lapse rate

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.

altit

ude

temp

midday

altit

ude

temp

Sun-down earth cools

Inversion layeraltit

ude

temp

more cooling at surface at night

}

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.


ude

temp


residual layer

}}

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


ude

temp


}

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

altit

ude

temp

Heating at surface during the nest day

}

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

heig

ht i n

ki

lom

eter

s

Temp in oC

Inversion height}

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 morning

Balloon temperature

Temp in oC

20 25 30 35

Dry adiabaticlines

heig

ht i n

ki

lom

eter

s

0.00.10.20.30.4

1.1

1.5

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

Temp in oC

20 25 30 35

Dry adiabaticlines

heig

ht i n

ki

lom

eter

s

0.00.10.20.30.4

1.1

1.5

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??

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/

c= 3x108m/s; h=6.63x10-34Js,=288x10-9m E= 416kJ/mole If all this light was absorbed it would break C-H bond

Light and rate constants The 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

This is the # of photons of light cm-2 sec-1

The rate constant for photolyis can be written as krate= J x x absorption coef

Light and rate constants krate= J x x absorption coef

the absortion coef. has units of cm2/moleculeand comes from Beer’s law I=Io e-l[C]

krate= J x x This is at one wavelength what do we do

when we have two wavelengths and 1? krate= the rate const. at a different wavelength

and krate = J1 x x

kratetotal = J x x + J1 x x

Light and rate constants kratetotal = J x x + J1 x x so across all wavelengths

so kratetotal = 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 kratetotal for the real atmosphere

Light and rate constants

Lets calculate kratefor 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

400nm = quantum yield = ~0.65 400-405nm = ~6x10-19 cm2 molecule-1

krate= J x x 0.00067sec-1


so in the reaction NO2 + light at 400-405nm -> NO + O. krate= J x x 0.00067sec-1

dNO2/dt = krate [NO2]


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


Sun


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 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

NO2, H2C=O, O3, acetaldehyde

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

Global Emissions of hydrocarbons

1150 x1012 grams of biogenic hydrocarbons emitted each yearof 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

Reasons to study biogenic secondary aerosol formationGlobal model calculations are sensitive to fine particles in the atmosphereBiogenic particles serve as sites for the condensation of other reacted urban organicsThis leads to haze and visibility reductionsThere is a great need to develop predictive models for secondary aerosol formation from naturally emitted hydrocarbons

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

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

OO

• 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 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.

CHOOO CH3

OOO

Criegee2

Criegee1OOO

-pinene

O3

COOHCOOH

pinic acid

+ otherproducts

Opinonic acid

CHOO

COOH

+ CO, HO2, OH

COOHO

norpinonaldehyde

norpinonic acid

O3 attack on-pinene

CHOO

OH

HOO OHOO CHO

OH

+ OH

O2

OO OH

-pinene

pinonaldehyde

OH attack on -pinene

OH

OO

O2

+

(a)(b)

(c)

(d)

(e)

pinonaldehyde

acetoneO

OO .

NO2NO

OO .

pinonald-oo

OH

pinonic acid

O

pin-O 2

OO .

NO2

NO

norpinonaldehyde

+HO 2

+NO 2 O

pinonald-PAN

=o

=o=o

=o

=o

=o

=o

OO .OHO2 =o

OO=C 8 =O

C8 -oo.

=oOONO 2

+HO 2

O2

NO2NO

norpinonaldehyde=o

=o+HO 2

CO=oOO .

O

O O

+ h+

methylglyoxal+

OH

+CO+HO 2=o

OO .

O2

NO2NO

=o=o +HO 2

+ h

NO2NO

=o

OO .C8 -oo.

+CO+HO 2

NO2

NO

(f)

(g)CO 2 +

pin-O 2

pin-OO 2

+CO

H2 O+

+H2 O

Reactions of product pinonaldehyde with OH and light

Particle formation-self nucleation Criegee biradicals can react with aldehydes and

carboxylic groups to form secondary ozonides and anhydrides.

O=C

C=OCH 3

+

CC=O

CH3

C

C=OCH 3

O

CC=O .

CH 3

oo

oo

Creigee + pinaldehyde --> seed1

The equilibrium between the gas and particle phases is:

• Kp = kon/koff

• The equilibrium constant Kp can be calculated (Pankow, Atmos. Environ, 1994)

• poL is the liquid vapor pressure and the activity

coefficient of the partitioning organic in the liquid portion of the particle, fom is the raction of organic mass in the particle and Mw is the average molecular weight of the organic mass

K R Tp M wpLo

7 5 0 110 9

. fom

• 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. koff can be calculated and with Kp, kon can also be evaluated

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 # 1.0.6x1011 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 # 2036] 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),

O3data

NOyNO2

NO

-pinene

pinonaldehyde

Particle phasepinonaldehyde

pinonic aciddata

norpinonic acid

Particle phase

ppm

mg/

m3

Particle phase model TSP

filter data

mg/

m3

mg/

m3

Time in hours (EDT)

mg/

m3

O3model

Sum products (data)

A

norpinonaldehyde

pinaldmodel

Gas phase

mg/

m3

C

E F

D

pinaldmodel

pinacidmodel

ppm

V

-pi

nene

B

diacidmodel

oxypinald

pinic aciddataParticle phase

Time in hours (EDT)

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

OH

pinonic acid

O=o

OH

O

pinic acid

OOH

OO

Reaction of -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

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

ppm

V

19.5 19.7 19.9 20.1 20.3 20.5 time in hours (pm)

O3

-pinene

0

0.1

0.2

0.3

0.4

ppm

V

19 19.5 20 20.5 time in hours (pm)

-pinene

O3

0 0.5

1 1.5

2 2.5

3 3.5

mg/

m3

19 19.5 20 20.5 21 21.5 22 22.5 time in hours (pm)

Reacted -pinene

model

filter mass data

0

0.5

1

1.5

2

mg/

m3

19 20 21 22time in hours (pm)

Reacted -pinene

EAA model

Filter mass data

filter mass data

model

time in hours (pm)

g/m

3

O3

-pinene

ppm

V 0.15

time in hours (pm)

1a 1b

2a 2b

3b3a

Initial reactants Particle formation

SummaryModels 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

AcknowldegementsThis 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.

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