1. IPCC [2007] AEROSOL CLIMATE FORCING

2. SCATTERING OF RADIATION BY AEROSOLS: “DIRECT EFFECT” By scattering solar radiation, aerosols increase the Earth’s albedo Scattering efficiency is maximum when particle diameter =  particles in 0.1-1  m size range are efficient scatterers of solar radiation

3. Mt. Pinatubo eruption 1991 1992 1993 1994 -0.6 -0.4 -0.2 0 +0.2 Temperature Change ( o C) Observations NASA/GISS general circulation model Temperature decrease following large volcanic eruptions EVIDENCE OF AEROSOL EFFECTS ON CLIMATE:

4. SCATTERING vs. ABSORBING AEROSOLS Scattering sulfate and organic aerosol over Massachusetts Partly absorbing dust aerosol downwind of Sahara Absorbing aerosols (black carbon, dust) warm the climate by absorbing solar radiation

5. AEROSOL “INDIRECT EFFECT” FROM CLOUD CHANGES Clouds form by condensation on pre-existing aerosol particles (“cloud condensation nuclei”) when RH>100% clean cloud (few particles): large cloud droplets low albedo efficient precipitation polluted cloud (many particles): small cloud droplets high albedo (1 st indirect) suppressed precipitation (2 nd indirect)

6.  Particles emitted by ships increase concentration of cloud condensation nuclei (CCN)  Increased CCN increase concentration of cloud droplets and reduce their avg. size  Increased concentration and smaller particles reduce production of drizzle  Liquid water content increases because loss of drizzle particles is suppressed  Clouds are optically thicker and brighter along ship track N~ 100 cm -3 W~ 0.75 g m -3 r e ~ 10.5 µm N~ 40 cm -3 W~ 0.30 g m -3 r e ~ 11.2 µm from D. Rosenfeld EVIDENCE OF INDIRECT EFFECT: SHIP TRACKS

7. AVHRR, 27. Sept. 1987, 22:45 GMT US-west coast NASA, 2002 Atlantic, France, Spain SATELLITE IMAGES OF SHIP TRACKS

8. Aircraft condensation trails (contrails) over France, photographed from the Space Shuttle (©NASA). OTHER EVIDENCE OF CLOUD FORCING: CONTRAILS AND “AIRCRAFT CIRRUS”

9. AQUEOUS PHASE CHEMISTRY

10. DEFINITIONS AND ISSUES Heterogeneous chemistry: chemistry involving more than one phase Aqueous-phase chemistry: heterogeneous chemistry occurring in or on particles (aerosols, fog droplets, cloud droplets, etc) Bulk solutions Cloud/fog droplets Aerosol particles Not too different Aerosols may have high ionic strengths Considerable uncertainty applying equilibrium and rate constants obtained from dilute solutions in the lab to atmospheric particles Can also exchange material b/w phases (large reservoir in gas phase) Can be very different!

11. AQUEOUS PHASE REACTION MECHANISM STEP 1: Diffusion to the surface STEP 4: Chemical Reaction STEP 2: Dissolution STEP 3: Diffusion in aqueous phase X XX X X+Y  ? STEP 2’: Ionization (for some species), VERY fast  A + + B -

12. SOLUBILITY AND HENRY’S LAW Henry’s Law: Distribution of species between aqueous and gas phases (for dilute solutions) H A = Henry’s Law Constant Units here are mol/L/atm OR M/atm Some Henry’s Law Constants of Atmospheric Relevance: Chemical Species Henry’s Law Constant @ 25°C (mol/L/atm) HNO 3 2.1x10 5 H2O2H2O2 7.5x10 4 HCHO3.5x10 3 NH 3 57.5 SO 2 1.2 CO9.6x10 -4 Note: H A ↑ as T↓ STEP 2 Can use ideal gas law to obtain the “dimensionless” Henry’s Law Constant:

13. THE ROLE OF LIQUID WATER Diameter (  m) L (cm 3 /m 3 )L (m 3 /m 3 )pH haze0.05-0.510 -5 – 10 -4 10 -11 – 10 -10 1-8 clouds100.1-110 -7 – 10 -6 3-6 fog100.05-0.55x10 -8 – 5x10 -7 2-6 rain500-50000.1-110 -7 – 10 -6 4-5 The liquid water amount affects the partitioning of species between the gas and the aqueous phase (esp for very soluble species) L = liquid water content of the atmosphere (cm 3 of water / m 3 of air) Consider, the distribution factor of a species: =1, there are equal amounts of A in each phase <<1, A is predominantly in the gas phase >> 1, A is predominantly in the aqueous phase Generally, L~10 -6, then f A =1 for H A ~4x10 4 M/atm. If H A << than this, most of A in gas phase All of gas in solution: STEP 2

14. NON-IDEAL SOLUTIONS Rain/Clouds = diluteHaze/plume = concentrated Henry’s Law (approximate activities using concentrations) Calculate activities (a): Undissociated species A: Species BX which dissociates: m A = molality [moles A/kg solvent]  = molal activity coefficient = f(ionic strength of solution, I) z i = charge on each ion (i) Challenge: calculate activity coefficients in the multi-component, high ionic strength solutions characteristic of atmospheric aerosols For example, use Debye-Hückel limiting law: STEP 2

15. IONIZATION REACTIONS The most fundamental ionization reaction: H 2 O(l) ↔ H + (aq) + OH - (aq) K a = acid dissociation constant (the larger the value, the stronger the acid, and thus the more acid is dissociated) pKa = -log[K a ] If pH > pKa a molecule is more likely to donate a proton (deprotonate) Electroneutrality (charge balance): in pure water [H + ]=[OH - ] pH = -log[H + ]  the activity of H + < 7 = acidic > 7 = basic 7 = neutral Some species (eg. O 3 ) simply dissolve in water and do not undergo reactions. Others do, and in some cases, reaction with liquid water does not change the essential proportionality of the liquid phase [X] to the gas phase P x. For example, when formaldehyde dissolves in water it forms a gem-diol: CH 2 O (aq) + H 2 O (l) ↔ CH 2 (OH) 2 (aq) Here [CH 2 (OH) 2 ] ~ P CH2O But not always so straight-forward for acidic or basic gases… STEP 2’

16. ACIDIC/BASIC IONIZATION REACTIONS, EXAMPLE: SO 2 Illustrate with SO 2 dissolved in a cloud drop: [SO 2 (aq)]=H SO2 P SO2  from Henry’s Law However, SO 2 is an acid in aqueous solution: SO 2 (aq)+H 2 O(l) ↔H + (aq)+HSO 3 - (aq) HSO 3 - (aq) ↔H + (aq)+SO 3 2- (aq) Acid dissociation constants (K a1, K a2 ): Solve for equilibrium concentrations of bisulphite and sulphite: With fast equilibria often group: [S(IV)]=[SO 2 (aq)]+[HSO 3 - ]+[SO 3 2- ]  all have same oxidation state H* =“effective” of “modified” Henry’s Law constant Solubility of S(IV) increases as pH increases. Effect of ionization in solution is to increase the effective solubility of the gas. STEP 2’

17. SOLVING THE SULFUR DIOXIDE / WATER EQUILIBRIUM From equilibrium we had: Add the electroneutrality equation: [H + ]=[OH - ]+[HSO 3 - ]+2[SO 3 2- ] If S(IV) is the only species in solution we can solve this for [H + ], with one more piece of information (for example P SO2 =1ppb, T=298K, pH=5.4, could then calc [S(IV)]) If other species are present need to modify electroneutrality equation, for example with sulfate: STEP 2’