copyright © r. r. dickerson 20111 lecture 1/2 aosc 637 spring 2011 atmospheric chemistry russell r....

Copyright © R. R. Dickerson 2011 1

LECTURE 1/2

AOSC 637 Spring 2011

Atmospheric Chemistry

Russell R. Dickerson


Topics to be covered this semesterOrganic and biochemistry for physicists and meteorologists.Laboratory techniques for detection and properties of aerosols.Weak acids and bases.Basic chemical thermodynamicsExperimental 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 microphysicsDry Deposition and micrometeorologyUnanswered questions on the formation and properties of aerosols:

SOASO2 oxidationAbsorption and mixing


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


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)


Ia. 1. GAS-PHASE

Atoms Molecules RadicalsHe Ar N₂ O₂ CO₂ O₃ H₂CO CCl₂F₂ OH HO₂monatomic diatomic triatomic polyatomic

UNITS OF CONCENTRATIONMole Fraction – for ideal gas this is the same as volume fraction. Also called mixing ratio, or volume mixing ratio.

fraction [O₂] = 1/5percent [Ar] = 1%

[H₂O] = up to 4%parts per million (10⁶) [CH₄] = 1.7 ppmparts per billion (10 )⁹ [O₃] = 30 ppbparts per trillion (10¹²) [CCl₂F₂] = 100 ppt

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 RATIO6Copyright © R. R. Dickerson 2011


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


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:

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.

1000/22.4

/63102.0 6moleLmL

moleggm

///

//3

3


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 ↔

Ammonium nitrate is a solid, and thus has a concentration defined as unity.

(S)34NONH

]][HNO[NH

]NO[NHKeq

33

34

Number density nX [molecules cm-3]

# molecules of X

unit volume of airXn Proper measure for• calculation of reaction rates• optical properties of atmosphere

0

Column concentration = ( )n z dz

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.

10Copyright © R. R. Dickerson 2011

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 CONCENTRATION11Copyright © R. R. Dickerson 2011


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


GAS HENRY’S LAW CONSTANT

(M /atm at 298 K)OXYGEN

O₂ 1.3 x 10⁻²

OZONE

O₃ 9.4 x 10⁻³

NITROGEN DIOXIDE

NO₂ 1.0 x 10⁻²

CARBON DIOXIDE

CO₂ 3.1 x 10⁻²

SULFUR DIOXIDE

SO₂ 1.3

NITRIC ACID (effective)

HNO₃ 2.1 x 10 ⁵⁺HYDROGEN PEROXIDE

H₂O₂ 9.7 x 10 ⁴⁺HYDROPEROXY RADICAL

HO₂ 9.0 x 10³

ALKYL NITRATES

(RONO₂) 1.3


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


Today’s barometric pressure is 993 hPa = 993/1013 atm = 0.98 atm. Thus the partial pressure of CO₂ is

In water CO₂ reacts slightly, but [H₂CO₃] remains constant as long as the partial pressure of CO₂ remains constant.

atm46CO 1074.2)98.0(10280P

2

currently) M x10(1.3 M 109.33

102.74103.4)P(COH][CO56

422aq2

7eq

32

3

332

3222

104.3K]CO[H

]][HCO[H

HCOHCOH

COHOHCO


We know that:

and

THUS

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.

32a

3

632

COH*K ][H

][HCO][H

M109.33 ]CO[H


EXAMPLE 2If 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.

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?

atm109.0

10/2.1102

/H][HNOP

10210][H

]log[HpH

11

55

aq3HNO

54.7

3


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: (atm)

Ambient temperature: T(K)

LET:

be the concentration of X in the aqueous phase in moles/m³

be the concentration of X in the gas phase in moles/m³

Where is the aqueous concentration in M, and is the partial pressure expressed in atm. We can find the partial pressure from the mixing ratio and total pressure.

TP

aqX

gasX

aq[X] xPXaq HP[X]


For the aqueous-phase concentration:

units: moles/m³ = moles/L(water) x g(water)/m³(air) x L/g

For the gaseous content:

units: moles/m³ =

TgasX P[X]P

3aqaq 10LWC[X]X

3Tgasaq 10LWCPH[X]X

T

3

gasgas

P1

273T

1022.4

[X]X

/LmL/mole

)L(X)/L(air3


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.

gasT

33Tgas

gas

aq /[X]P

1

273

T1022.410LWCPH[X]

X

X

6

gas

aq 1022.4273

TLWCH

X

X


What is the possible pH of water in a high cloud (alt. 5km) that absorbed ≃sulfur while in equilibrium with 100 ppb of SO₂?

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.

This SO₂ will not stay as SO₂•H₂O, but participate in a aqueous phase reaction, that is it will dissociate.

5km2Total2SO

2

2222

]P[SO]P[SOP

100ppb][SO

OHSOOHSO

2

M107

HP][SO

atm105.40.5410100P

8

SOaq2

89SO

2

2

222 HOSOHOHSO


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?

If most of the [H ] comes from SO₂⁺ •H₂O dissociation, then

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

]SOO[H

]][HOSO[HK

22

2a

522a

2

103]SOO[HK][H

][HOSO][H


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.


Second example - alkylnitratesCan 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


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 20111 lecture 1/2 aosc 637 spring 2011 atmospheric chemistry russell r....

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