QUESTIONS 1.What molar fraction of HNO 3 do you expect to partition into fog droplets at room temperature? How does this compare to the fraction that would partition to a typical cloud/rain droplet (with L=10 -6 as discussed in class)? 2.a) Notice that fog is generally a bit more acidic than clouds. Why? b)If a fog droplet and a cloud droplet were exposed to the same atmospheric concentration of SO 2 in which droplet would [S(IV)] be higher?

REMINDER: 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 -

AQUEOUS PHASE CHEMICAL KINETICS Units for aqueous phase rate (R a ): M/s (moles/L solution /s), if multiplied by L (vol/vol) then the rate is converted to moles/L air /s  equivalent to a gas-phase conversion rate - occasionally see fractional rates (% per hour) Overall kinetics are described by combining Henry’s Law dissolution and ionization with aqueous phase chemical reactions in liquid atmospheric particles. Simplest case: if drop is in equilibrium with the gas phase (as we will assume) STEP 4 Dissolved gases and ions in particles can undergo chemical rxn  oxidizing Important aqueous phase reaction: conversion of S(IV) to S(VI) - possible reagents: O 2 (aq), H 298 =1.3x10 -3 O 3 (aq), H 298 =9.4x10 -3 H 2 O 2 (aq),H 298 =7.1x10 4 Important reaction in atmosphere: O 3 and H 2 O 2

AQUEOUS PHASE REACTIONS: S(IV) TO S(VI), OXIDATION BY O 3 STEP 4 SO 3 2- (aq) + O 3 (aq)  SO 4 2- (aq)+O 2 (aq) Rate of sulfate production: Liquid phase concentration of O 3 : Using previous expression for sulfite: Original production expression did not depend on pH, but see in final form the strong pH dependence (due to equilibrium b/w gas phase SO 2 and liquid phase SO 3 2- ). As SO 3 2- is oxidized, it is replaced by SO 2 dissolving from the gas phase. Since this is an acidic gas, both the pH and the rate of oxidation decrease as the reaction proceeds  reaction is self-limiting All 3 S(IV) species react with ozone, the empirical rate equation is: k 2 > k 1 > k 0

AQUEOUS PHASE REACTIONS: S(IV) TO S(VI), OXIDATION BY H 2 O 2 STEP 4 HSO 3 - (aq) + H 2 O 2 (aq) ↔ SO 2 OOH - (aq) SO 2 OOH - (aq) + H + (aq)  H 2 SO 4 (aq) Hoffman and Calvert (1985) suggest the following rate expression: First reaction accelerated by partitioning of more S(IV) to bisulfite at higher pH. Second reaction becomes faster as pH declines. Thus, little net effect of pH in atm. Organic peroxides have also been proposed as potential aqueous S(IV) oxidants, however concentrations and solubilities are low enough that they are predicted to be of minor importance. Comparison of S(IV) oxidation paths (Seinfeld and Pandis, 2006) Rates of sulfate production are per unit volume, and thus must be multiplied by total liquid water content (therefore more important in clouds)

AQUEOUS SULFUR OXIDATION SO 2 SO 2. HO 2 HSO 3 - SO 3 2- H 2 O 2 (g) S(IV) S(VI) O 3 (g) H 2 SO 4 HSO 4 - SO 4 2-

MASS TRANSFER LIMITATIONS XX+Y  ? We have assumed that drops are in equilibrium with the gas phase, or that this:is fast compared to this: This is often true because particles are small. BUT gas phase diffusion tends to limit the rates of gas-particle reactions for very large particles, such as cloud droplets. Mass transfer limitations tend to be most important for relatively insoluble gases (for SO 2 oxidation in droplets, O 3 is the species most subject to mass transfer limitations) X Diffusion in liquids is relatively slow. Dissolved gases are unlikely to reach the center of the drop for fast reactions. Thus, the average liquid phase concentration is less than expected on the basis of Henry’s Law and overall rate is reduced. EXTREME example: N 2 O 5 hydrolysis happens so quickly that the reaction appears to happen as soon as molecule enters solution, thus rate ~A drop (as if occurring at the surface) X+Y  ? X STEP 1/3

EXAMPLE FROM SEINFELD AND PANDIS Pure water droplet (pH=7), L=10 -6 m 3 water /m 3 air is exposed to environment : Chemical Species Initial (gas phase) After equilibrium (gas phase) HNO 3 1 ppb10 -8 ppb H2O2H2O2 1 ppb0.465 ppb O3O3 5 ppb NH 3 5 ppb1.87 ppb SO 2 5 ppb3.03 ppb After equilibrium conditions solved with equations for Henry’s Law, dissociation equilibrium, electroneutrality and mass balance, such as: [H + ]+[NH 4 + ]=[OH - ]+[HO 2 - ]+[HSO 3 - ]+2[SO 3 2- ]+[NO 3 - ] New gas phase concentrations are effective initial conditions. pH=6.17 As reaction proceeds, will convert S(IV)  S(VI), will change pH and thus effective Henry’s Law constant (for example for S(IV) dissolution). So at each time step must re- calculate electroneutrality, and now include sulfate formation as well: [H + ]+[NH 4 + ]=[OH - ]+[HO 2 - ]+[HSO 3 - ]+2[SO 3 2- ]+[NO 3 - ]+[HSO 4 - ]+2[SO 4 2- ] Integrate chemical rate equations in time, assuming instantaneous equilibrium and either depleting (closed) or holding constant (open) the gas phase concentrations.

CLOSED VS. OPEN SYSTEM [Seinfeld & Pandis] In an open system, partial pressures remain constant (gases are replenished) In a closed system, as gas is transferred to the liquid phase it is depleted