1. Copyright © R. R. Dickerson 20111 LECTURE 1/2 AOSC 637 Spring 2011 Atmospheric Chemistry Russell R. Dickerson

2. Copyright © R. R. Dickerson 20112 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, CH 4 and halogens. Measurements of cloud properties. Cloud microphysics Dry Deposition and micrometeorology Unanswered questions on the formation and properties of aerosols: SOA SO 2 oxidation Absorption and mixing

3. Copyright © R. R. Dickerson 20113 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

4. Copyright © R. R. Dickerson 20114 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)

5. Copyright © R. R. Dickerson 20115 Ia. 1. GAS-PHASE AtomsMolecules Radicals He Ar N ₂ O ₂ CO ₂ O ₃ H ₂ CO CCl ₂ F ₂ OH HO ₂ monatomic 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[O ₂ ] = 1/5 percent[Ar] = 1% [H ₂ O] = up to 4% parts per million (10 ⁶ )[CH ₄ ] = 1.7 ppm parts per billion (10 ⁹ )[O ₃ ] = 30 ppb parts per trillion (10¹²)[CCl ₂ F ₂ ] = 100 ppt

6. ATMOSPHERIC CO 2 INCREASE OVER PAST 1000 YEARS Jacob: Intergovernmental Panel on Climate Change (IPCC) document, 2001 Concentration units: parts per million (ppm) number of CO 2 molecules per 10 6 molecules of air CO 2 CONCENTRATION IS MEASURED HERE AS MIXING RATIO 6Copyright © R. R. Dickerson 2011

7. 7 For an ideal gas these concentrations are constant regardless temperature and pressure. Ideal Gas Law: PV = nRT For example if T ₂ = 2T ₁ and if dP = 0 then V ₂ = 2V ₁. Meteorologists favor the ideal gas law for a kg of air: pα = R’T Where R’ has units of J kg -1 K -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

8. Copyright © R. R. Dickerson 20118 EXAMPLE If you find 10 μg/m³ “dust” of which 2 μg/m³ are nitrate (NO ₃ ‾), how much gas phase HNO ₃, 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 HNO ₃ to ppb: Remember one mole of an ideal gas is 22.4 liters at STP = 0 C & 1.0 atm. 2.0 μg/m³ HNO ₃ = 7.1 x 10 ⁻ ¹º = 0.71 ppb In general: 1.0 μg/m³ HNO ₃ = 0.35 ppb Notice that the concentration in μg/m³ changes with P and T of the air.

9. Copyright © R. R. Dickerson 20119 Mixing ratios area also good for writing reactions: NO + O ₃ → NO ₂ + O ₂ 1 ppm + 1 ppm = 1 ppm + 1 ppm Note: the [O ₂ ] in air is not 1 ppm, rather it is 0.2 x 10 ⁶ + 1.0 ppm. Above is an example of an irreversible reaction. There are also reversible reactions. EXAMPLE Equilibrium of ammonium nitrate NH ₃ + HNO ₃ ↔ Ammonium nitrate is a solid, and thus has a concentration defined as unity.

10. Number density n X [molecules cm -3 ] Proper measure for calculation of reaction rates optical properties of atmosphere 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 = 2.69x10 19 cm -2. 10Copyright © R. R. Dickerson 2011

11. STRATOSPHERIC OZONE LAYER (Jacob’s book) 1 “Dobson Unit (DU)” = 0.01 mm ozone at STP = 2.69x10 16 molecules cm -2 THICKNESS OF OZONE LAYER IS MEASURED AS A COLUMN CONCENTRATION 11Copyright © R. R. Dickerson 2011

12. 12 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

13. Copyright © R. R. Dickerson 201113 GASHENRY’S LAW CONSTANT (M /atm at 298 K) OXYGEN O ₂ 1.3 x 10 ⁻ ² OZONE O ₃ 9.4 x 10 ⁻ ³ NITROGEN DIOXIDE NO ₂ 1.0 x 10 ⁻ ² CARBON DIOXIDE CO ₂ 3.1 x 10 ⁻ ² SULFUR DIOXIDE SO ₂ 1.3 NITRIC ACID (effective) HNO ₃ 2.1 x 10 ⁺⁵ HYDROGEN PEROXIDE H ₂ O ₂ 9.7 x 10 ⁺⁴ HYDROPEROXY RADICAL HO ₂ 9.0 x 10³ ALKYL NITRATES (RONO ₂ ) 1.3

14. Copyright © R. R. Dickerson 201114 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 N ₂, O ₂, and CO ₂ and that rain in equilibrium with CO ₂. Remember: H ₂ O = H ⁺ + OH ⁻ [H ⁺ ][OH ⁻ ] = 1 x 10 ⁻ ¹ ⁴ pH = -log[H ⁺ ] In pure H ₂ O pH = 7.00 We can measure: [CO ₂ ] = 280 in a preindustrial world ~ 390 ppm today

15. Copyright © R. R. Dickerson 201115 Today’s barometric pressure is 993 hPa = 993/1013 atm = 0.98 atm. Thus the partial pressure of CO ₂ is In water CO ₂ reacts slightly, but [H ₂ CO ₃ ] remains constant as long as the partial pressure of CO ₂ remains constant.

16. Copyright © R. R. Dickerson 201116 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 + ] * [HCO 3 - ]/[H 2 CO 3 ] = 7.9x10 -9 * [HCO 3 - ]/1.28x10 -5 = 4.7 x10 -7 [HCO 3 - ] = 7.6x10 -4 M Most of the C in the oceans is tied up as bicarbonate.

17. Copyright © R. R. Dickerson 201117 EXAMPLE 2 If fog water contains enough nitric acid (HNO ₃ ) 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 HNO ₃ is in equilibrium with typical “acid rain” i.e. water at pH 4.7? We will have to assume that HNO ₃ is 50% ionized. This is equivalent to 90 ppt, a small amount for a polluted environment, but the actual [HNO ₃ ] 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?

18. Copyright © R. R. Dickerson 201118 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.

19. Copyright © R. R. Dickerson 201119 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 =

20. Copyright © R. R. Dickerson 201120 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.

21. Copyright © R. R. Dickerson 201121 What is the possible pH of water in a high cloud (alt. ≃ 5km) that absorbed sulfur while in equilibrium with 100 ppb of SO ₂ ? 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 SO ₂ will not stay as SO ₂ H ₂ O, but participate in a aqueous phase reaction, that is it will dissociate.

22. Copyright © R. R. Dickerson 201122 The concentration of SO ₂ H ₂ O, however, remains constant because more SO ₂ is entrained as SO ₂ H ₂ O dissociates. The extent of dissociation depends on [H ⁺ ] and thus pH, but the concentration of SO ₂ H ₂ O will stay constant as long as the gaseous SO ₂ concentration stays constant. What’s the pH for our mixture? If most of the [H ⁺ ] comes from SO ₂ H ₂ O dissociation, then Note that there about 400 times as much S in the form of HOSO ₂⁻ as in the form H ₂ OSO ₂. HOSO ₂⁻ is a very weak acid, ant the reaction stops here. The pH of cloudwater in contact with 100 ppb of SO ₂ will be 4.5

23. Copyright © R. R. Dickerson 201123 Because SO ₂ participates in aqueous-phase reactions, Eq. (I) above will give the correct [H ₂ OSO ₂ ], 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 H ₂ OSO ₄, in other words HSO ₄⁻ is a strong acid. This is fairly acidic, but we started with a very high concentration of SO ₂, one that is characteristic of urban air. In more rural areas of the eastern US an SO ₂ mixing ratio of a 1-5 ppb is more common. As SO ₂ H ₂ O is oxidized to H ₂ OSO ₄, more SO ₂ 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 H ₂ O ₂ later.

24. Copyright © R. R. Dickerson 201124 Second example - alkylnitrates Can alkyl nitrates, R-ONO 2, 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. (K H ) for R-ONO 2  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-ONO 2 would be is the cloud is in equilibrium with the vapor phase. [R-ONO 2 ] aq = K H x [R-ONO 2 ] gas x P total Where [R-ONO 2 ] gas must be expressed in partial pressure, atm. = 2.0 x 10 -9 x P total = 2.0 x10 -9 x 0.5 [R-ONO 2 ] aq = 10 -9 M

25. Copyright © R. R. Dickerson 201125 How do we compare this to the gas phase concentration? Change both values into moles/m 3. [R-ONO 2 ] aq = 10 -9 x 10 -3 = 10 -12 UNITS: moles/L(water) x L(water)/m 3 (air) = moles /m 3 [R-ONO 2 ] gas = 10 -9 x 10 3 x 0.5/22.4 = 2.2 x 10 -8 UNITS: L (R-ONO 2 )/ L (air) x L/m 3 x atm/(L atm/mole) = moles/m 3 We see that the vapor phase concentration is 22,000 higher than the aqueous phase concentration. Rainout will still matter, however, if R-ONO 2 reacts in solution and thus is removed. This is the case for SO 2 in water containing H 2 O 2 where H 2 SO 4 is produced, but aqueous reactions of R-ONO 2 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. X aq /X gas = K H x LWC x 2.24 x 10 -5 Where K H is the Henry's Law coefficient in M/atm, and LWC is the liquid water content in g/m 3. 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 x10 2, about 1% will go into a cloud with a LWC of 1g/m 3.