LECTURE 1/2
AOSC 637 Spring 2011
Atmospheric Chemistry
Russell R. Dickerson
Copyright © R. R. Dickerson 2011
1
Topics to be covered this semester
Organic and biochemistry for physicists and meteorologists.
Laboratory techniques for detection and properties of aerosols.
Weak acids and bases.
Basic chemical thermodynamics
Experimental design.
Spectroscopy of polyatomic molecules and photochemistry.
Kinetics theory and lab techniques.
Biogeochemical cycles of Ox, NOx, SOx, HOx, CH4 and halogens.
Measurements of cloud properties.
Cloud microphysics
Dry Deposition and micrometeorology
Unanswered questions on the formation and properties of aerosols:
SOA
SO2 oxidation
Absorption and mixing
Copyright © R. R. Dickerson 2011
2
Not covered (awaiting results of
diagnostic exam.)
Basic thermodynamics of dry and wet air.
Parcel theory and stability.
General circulation and synoptic circulations.
Stefan-Boltzmann and Wien.
Basic ozone photochemistry.
Biogeochemical cycle of C.
Convection and chemistry
Unit conversions
Copyright © R. R. Dickerson 2011
3
TODAY’S OUTLINE
Ia. Chemistry (Concentration Units):
1. Gas-phase
2. Aqueous-phase
Ib. Atmospheric Physics
1. Pressure
2. Atmospheric structure and circulation
A. pressure and temp. profiles
B. thermo diagram and stability
C. circulation (winds)
Copyright © R. R. Dickerson 2011
4
Ia. 1. GAS-PHASE
Atoms
He Ar
monatomic
N₂ O₂
Molecules
CO₂ O₃
Radicals
H₂CO CCl₂F₂ OH HO₂
diatomic
triatomic
polyatomic
UNITS OF CONCENTRATION
Mole Fraction – for ideal gas this is the same as volume fraction. Also
called mixing ratio, or volume mixing ratio.
fraction
[O₂] = 1/5
percent
[Ar] = 1%
[H₂O] = up to 4%
parts per million (10⁶)
[CH₄] = 1.7 ppm
parts per billion (10⁹)
[O₃] = 30 ppb
parts per trillion (10¹²)
[CCl₂F₂] = 100 ppt
Copyright © R. R. Dickerson 2011
5
ATMOSPHERIC CO2 INCREASE OVER PAST 1000 YEARS
Jacob: Intergovernmental Panel on Climate Change (IPCC) document, 2001
Concentration units: parts per million (ppm)
number of CO2 molecules per 106 molecules of air
CO2 CONCENTRATION IS MEASURED HERE AS MIXING RATIO
Copyright © R. R. Dickerson 2011
6
For an ideal gas these concentrations are constant regardless temperature
and pressure.
Ideal Gas Law: PV = nRT
For example if T₂ = 2T₁ and if dP = 0 then V₂ = 2V₁.
Meteorologists favor the ideal gas law for a kg of air:
pα = R’T
Where R’ has units of J kg-1K-1 and α is the specific volume (volume
occupied by 1 kg of air; Mwt 29 g/mole).
If air in New York is brought to Denver (P = 83% atm) there will be no
change in the concentration of pollutants as long as the concentration is
expressed as a volume (moler) mixing ratio.
MASS PER UNIT VOLUME
Best for particles (solid or liquid)
Weigh a filter – suck 1.00 m⁻³ air through it – reweigh it
Change in weight is conc “dust” in mass/unit volume or μgm⁻³
Copyright © R. R. Dickerson 2011
7
EXAMPLE
If you find 10 μg/m³ “dust” of which 2 μg/m³ are nitrate (NO₃‾), how
much gas phase HNO₃, expressed as a mixing ratio, was there in the air
assuming that all the nitrate was in the form of nitric acid? We must
convert 2.0 μg/m³ HNO₃ to ppb:
2.0 10 6 /63
1000/22.4
gm 3 / g / mole
L / m 3 / L / mole
Remember one mole of an ideal gas is 22.4 liters at STP = 0 C & 1.0 atm.
2.0 μg/m³ HNO₃ = 7.1 x 10⁻¹º = 0.71 ppb
In general: 1.0 μg/m³ HNO₃ = 0.35 ppb
Notice that the concentration in μg/m³ changes with P and T of the air.
Copyright © R. R. Dickerson 2011
8
Mixing ratios area also good for writing reactions:
NO + O₃ → NO₂ + O₂
1 ppm + 1 ppm = 1 ppm + 1 ppm
Note: the [O₂] in air is not 1 ppm, rather it is 0.2 x 10⁶ + 1.0 ppm.
Above is an example of an irreversible reaction. There are also
reversible reactions.
EXAMPLE
Equilibrium of ammonium nitrate
NH₃ + HNO₃ ↔ NH4 NO3(S)
[NH 4 NO3 ]
Keq 
[NH3 ][HNO 3 ]
Ammonium nitrate is a solid, and thus has a concentration defined
as unity.
Copyright © R. R. Dickerson 2011
9
Number density nX [molecules
-3
cm ]
Proper measure for
• calculation of reaction rates
• optical properties of atmosphere
# molecules of X
nX 
unit volume of air

Column concentration =  n( z )dz
0
Proper measure for absorption of radiation by
atmosphere
Column concentrations are measured in molecules cm-2 , atm*cm, and
Dobson Units, DU.
1 atm*cm = 1000 DU = 2.69x1019 cm-2.
Copyright © R. R. Dickerson 2011
10
STRATOSPHERIC OZONE LAYER (Jacob’s book)
1 “Dobson Unit (DU)” = 0.01 mm ozone at STP = 2.69x1016 molecules cm-2
THICKNESS OF OZONE LAYER IS MEASURED AS A COLUMN CONCENTRATION
Copyright © R. R. Dickerson 2011
11
AQUEOUS-PHASE CHEMISTRY
HENRYS LAW
The mass of a gas that dissolves in a given amount of liquid as a given
temperature is directly proportional to the partial pressure of the gas above
the liquid. This law does not apply to gases that react with the liquid or
ionized in the liquid. See Finlayson p.151 or Chameides, J. Geophys. Res.,
4739, 1984. Check out also
www.mpch-mainz.mpg.de/~sander/res/henry.html
Copyright © R. R. Dickerson 2011
12
HENRY’S LAW CONSTANT
(M /atm at 298 K)
GAS
OXYGEN
O₂
OZONE
O₃
NITROGEN DIOXIDE
NO₂
CARBON DIOXIDE
CO₂
SULFUR DIOXIDE
SO₂
NITRIC ACID (effective)
HNO₃
HYDROGEN PEROXIDE
H₂O₂
HYDROPEROXY RADICAL
HO₂
ALKYL NITRATES
(RONO₂)
1.3 x 10⁻²
9.4 x 10⁻³
1.0 x 10⁻²
3.1 x 10⁻²
1.3
2.1 x 10⁺⁵
9.7 x 10⁺⁴
9.0 x 10³
1.3
Copyright © R. R. Dickerson 2011
13
HENRY’S LAW EXAMPLE
What was the pH of fresh water in the preindustrial atmosphere? What
would be the pH of pure rain water in Washington, D.C. today? Assume
that the atmosphere contains only N₂, O₂, and CO₂ and that rain in
equilibrium with CO₂.
Remember:
H₂O = H⁺ + OH⁻
[H⁺][OH⁻] = 1 x 10⁻¹⁴
pH = -log[H⁺]
In pure H₂O pH = 7.00
We can measure:
[CO₂] = 280 in a preindustrial world
~ 390 ppm today
Copyright © R. R. Dickerson 2011
14
Today’s barometric pressure is 993 hPa = 993/1013 atm = 0.98 atm. Thus
the partial pressure of CO₂ is
PCO 2  280 106 (0.98)  2.74 104 atm
[CO 2 ]aq  H  P(CO 2 )  3.4 10 2  2.74 10 4
 9.33 10 6 M (1.3 x10 5 M currently)
In water CO₂ reacts slightly, but [H₂CO₃] remains constant as long as the
partial pressure of CO₂ remains constant.
CO 2  H 2 O  H 2 CO3
 H 2 CO3  H   HCO 3
[H  ][HCO 3 ]
 Keq  4.3 10 7
[H 2 CO3 ]
Copyright © R. R. Dickerson 2011
15
We know that:
and
THUS
[H 2CO3 ]  9.33 10 6 M
[H  ]  [HCO 3 ]
[H  ]  K a * H 2CO3
H+ = 2.09x10-6 → pH = -log(2.09x10-6) = 5.68
H+ = 2.5x10-6 → pH = -log(2.5x10-6) = 5.60 today.
The pH of the ocean today is ~8.1 so [H+] = 7.9x10-9.
[H+] * [HCO3-]/[H2CO3] = 7.9x10-9 * [HCO3-]/1.28x10-5 = 4.7 x10-7
[HCO3-] = 7.6x10-4 M
Most of the C in the oceans is tied up as bicarbonate.
Copyright © R. R. Dickerson 2011
16
EXAMPLE 2
If fog water contains enough nitric acid (HNO₃) to have a pH of 4.7, can any
appreciable amount nitric acid vapor return to the atmosphere? Another way to ask this
question is to ask what partial pressure of HNO₃ is in equilibrium with typical “acid
rain” i.e. water at pH 4.7? We will have to assume that HNO₃ is 50% ionized.
pH  log[H  ]
[H  ]  10  4.7  2  10 5
PHNO3  [HNO 3 ]aq /H
  2  10 5 /2.1  105
  9.0  10 11 atm
This is equivalent to 90 ppt, a small amount for a polluted environment, but the actual
[HNO₃] would be even lower because nitric acid ionized in solution. In other words,
once nitric acid is in solution, it wont come back out again unless the droplet
evaporates; conversely any vapor-phase nitric acid will be quickly absorbed into the
aqueous-phase in the presence of cloud or fog water.
Which pollutants can be rained out?
Copyright © R. R. Dickerson 2011
17
We want to calculate the ratio of the aqueous phase to the gas phase concentration
of a pollutant in a cloud. The units can be anything , but they must be the same.
We will assume that the gas and aqueous phases are in equilibrium. We need
the following:
Henry’s Law Coefficient: H (M/atm)
Cloud liquid water content: LWC (gm⁻³)
Total pressure: PT (atm)
Ambient temperature: T(K)
LET:
X aq be the concentration of X in the aqueous phase in moles/m³
X gas be the concentration of X in the gas phase in moles/m³
[X] aq  HPX
Where [X] aq is the aqueous concentration in M, and Px is the partial pressure
expressed in atm. We can find the partial pressure from the mixing ratio and
total pressure.
Copyright © R. R. Dickerson 2011
18
PX  [X] gas PT
For the aqueous-phase concentration:
Xaq  [X]aq LWC 103
units:
moles/m³ = moles/L(water) x g(water)/m³(air) x L/g
Xaq  H[X] gas PT LWC 103
For the gaseous content:
X gas 
units:
moles/m³ =
[X] gas

T 1
 22.4 10 3

273 PT 

L(X)/L(air )
L/mole  m 3 /L

Copyright © R. R. Dickerson 2011

19
X aq
X gas

1
3 T
/[X] gas
 H[X] gas PT LWC 10  22.4 10
273
P
T 

3
X aq
X gas
 H  LWC 
T
 22.4  10 6
273
Notice that the ratio is independent of pressure and concentration. For a
species with a Henry’s law coefficient of 400, only about 1% will go into a
cloud with a LWC of 1 g/m³. This points out the need to consider aqueous
reactions.
Copyright © R. R. Dickerson 2011
20
What is the possible pH of water in a high cloud (alt. ≃ 5km) that absorbed
sulfur while in equilibrium with 100 ppb of SO₂?
SO 2  H 2 O  SO 2  H 2 O
[SO 2 ]  100ppb
PSO 2  [SO 2 ]PTotal  [SO 2 ]P5km
In the next lecture we will show how to derive the pressure as a function of
height. At 5km the ambient pressure is 0.54 atm.
PSO 2  100 10 9 0.54  5.4 10 8 atm
[SO 2 ]aq  HPSO 2
  7 10 8 M
This SO₂ will not stay as SO₂•H₂O, but participate in a aqueous phase
reaction, that is it will dissociate.
SO2  H 2O  H   HOSO2
Copyright © R. R. Dickerson 2011
21
The concentration of SO₂•H₂O, however, remains constant because more SO₂
is entrained as SO₂•H₂O dissociates. The extent of dissociation depends
on [H⁺] and thus pH, but the concentration of SO₂•H₂O will stay constant
as long as the gaseous SO₂ concentration stays constant. What’s the pH for
our mixture?


[H ][HOSO 2 ]
Ka 
[H 2O  SO 2 ]
If most of the [H⁺] comes from SO₂•H₂O dissociation, then
[H  ]  [HOSO 2 ]
[H  ]  K a [H 2 O  SO 2 ]  3 10 5
Note that there about 400 times as much S in the form of HOSO₂⁻ as in the
form H₂O•SO₂. HOSO₂⁻ is a very weak acid, ant the reaction stops here.
The pH of cloudwater in contact with 100 ppb of SO₂ will be 4.5
Copyright © R. R. Dickerson 2011
22
Because SO₂ participates in aqueous-phase reactions, Eq. (I) above will
give the correct [H₂O•SO₂], but will underestimate the total sulfur in
solution. Taken together all the forms of S in this oxidation state are called
sulfur four, or S(IV).
If all the S(IV) in the cloud water turns to S(VI) (sulfate) then the hydrogen
ion concentration will approximately double because both protons come off
H₂O•SO₄, in other words HSO₄⁻ is a strong acid.
This is fairly acidic, but we started with a very high concentration of SO₂,
one that is characteristic of urban air. In more rural areas of the eastern US
an SO₂ mixing ratio of a 1-5 ppb is more common. As SO₂•H₂O is
oxidized to H₂O•SO₄, more SO₂ is drawn into the cloud water, and the
acidity continue to rise. Hydrogen peroxide is the most common oxidant
for forming sulfuric acid in solution; we will discuss H₂O₂ later.
Copyright © R. R. Dickerson 2011
23
Second example - alkylnitrates
Can alkyl nitrates, R-ONO2, be removed from the atmosphere by rain (wet
deposition)? Consider the relative amount of an alkyl nitrate in the gas phase
vs. the aqueous phase in a cloud. If most of the alkyl nitrate is in the aqueous
phase, than precipitation must be important. We need the following
information:
1. Henry's Law Coef. (KH) for R-ONO2  2 M/atm at 298 K (Luke et al., 1989).
2. A thick cloud has 1.0 g liquid water per cubic meter.
3. The typical temperature of a cloud is near 0 C.
4. The typical altitude of a cloud is about 5 km thus the pressure is about 0.5 atm.
5. The most alkyl nitrate one might find in the atmosphere over a continent is
about 1.0 ppb.
First we apply Henry's law to find out what the aqueous concentration of R-ONO2
would be is the cloud is in equilibrium with the vapor phase.
[R-ONO2 ]aq = KH x [R-ONO2 ]gas x Ptotal
Where [R-ONO2 ]gas must be expressed in partial pressure, atm.
= 2.0 x 10-9 x Ptotal
= 2.0 x10-9 x 0.5
[R-ONO2 ]aq = 10-9 M
Copyright © R. R. Dickerson 2011
24
How do we compare this to the gas phase concentration? Change both values into
moles/m3.
[R-ONO2 ]aq = 10-9 x 10-3 = 10-12
UNITS: moles/L(water) x L(water)/m3 (air) = moles /m3
[R-ONO2 ]gas = 10-9 x 103 x 0.5/22.4 = 2.2 x 10-8
UNITS: L (R-ONO2 )/ L (air) x L/m3 x atm/(L atm/mole) = moles/m3
We see that the vapor phase concentration is 22,000 higher than the aqueous phase
concentration. Rainout will still matter, however, if R-ONO2 reacts in solution
and thus is removed. This is the case for SO2 in water containing H2O2 where
H2SO4 is produced, but aqueous reactions of R-ONO2 with species commonly
found in rainwater are as yet unknown. This implies that alkyl nitrates may have
a residence time long enough to be important in regional or global atmospheric
chemistry.
For species X, a general solution to the "rain out" question is given by an
expression for the ratio of moles of gas-phase X to moles of aqueous-phase X in
a given volume of air.
Xaq /Xgas = KH x LWC x 2.24 x 10-5
Where KH is the Henry's Law coefficient in M/atm, and LWC is the liquid water
content in g/m3. This equation is valid at 273 K; to correct for temperature
multiply the right side by (T/273). The equation above shows that the ratio is
independent of pressure and concentration. For alkyl nitrates this ratio is about
4.4 x10-5. For a species with a Henry's law coefficient of 4 x102, about 1% will
go into a cloud with a LWC of 1g/m3.
Copyright © R. R. Dickerson 2011
25