QUESTIONS

1. What molar fraction of HNO3 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 SO2 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

XX X

XX+Y ?

STEP 2’: Ionization (for some species), VERY fast

A+ + B-

AQUEOUS PHASE CHEMICAL KINETICS

Units for aqueous phase rate (Ra): M/s (moles/Lsolution/s), if multiplied by L (vol/vol) then the rate is converted to moles/Lair/s equivalent to a gas-phase conversion rate

- occasionally see fractional rates (% per hour)

aa

1a

A

R L R LRTR ,

[A(g)] P[s ]

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: O2 (aq), H298=1.3x10-3

O3(aq), H298=9.4x10-3

H2O2(aq),H298=7.1x104

Important reaction in atmosphere: O3 and H2O2

AQUEOUS PHASE REACTIONS: S(IV) TO S(VI), OXIDATION BY O3

STEP 4

SO32-(aq) + O3(aq) SO4

2-(aq)+O2(aq)

3

3 3

2 2

3 3 3

224

O 3 3

3 O O

2a1 a2 SO SO4

O O O 2

d[SO ]k [O (aq)][SO ]

dt[O (aq)] H P

K K H Pd[SO ]k H P

dt [H ]

Rate of sulfate production:

Liquid phase concentration of O3:

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 SO2 and liquid phase SO3

2-).

As SO32- is oxidized, it is replaced by SO2 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:

2

240 2 1 3 2 3 3

d[SO ]k [SO (aq)] k [HSO ] k [SO ] [O (aq)]

dt

k2 > k1 > k0

AQUEOUS PHASE REACTIONS: S(IV) TO S(VI), OXIDATION BY H2O2

STEP 4

HSO3-(aq) + H2O2(aq) ↔ SO2OOH-(aq)

SO2OOH-(aq) + H+(aq) H2SO4(aq)

22 2 34 k[H ][H O ][HSO ]d[SO ]

dt 1 K[H ]

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

SO2

SO2.HO2 HSO3

- SO32-

H2O2(g)

S(IV)

S(VI)

O3(g)

H2SO4 HSO4- SO4

2-

MASS TRANSFER LIMITATIONS

X X+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 SO2 oxidation in droplets, O3 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: N2O5 hydrolysis happens so quickly that the reaction appears to happen as soon as molecule enters solution, thus rate ~Adrop (as if occurring at the surface)

X+Y ?

X

STEP 1/3

EXAMPLE FROM SEINFELD AND PANDISPure water droplet (pH=7), L=10-6 m3

water/m3air is exposed to environment :

Chemical Species

Initial (gas phase)

After equilibrium (gas phase)

HNO3 1 ppb 10-8 ppb

H2O2 1 ppb 0.465 ppb

O3 5 ppb 5 ppb

NH3 5 ppb 1.87 ppb

SO2 5 ppb 3.03 ppb

After equilibrium conditions solved with equations for Henry’s Law, dissociation equilibrium, electroneutrality and mass balance, such as:

[H+]+[NH4+]=[OH-]+[HO2

-]+[HSO3-]+2[SO3

2-]+[NO3-]

New gas phase concentrations are effective initial conditions. pH=6.17

3

3

3 3O

O 3 T 3

[O (aq)] [O (aq)]H

P [O ] [O (aq)]

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+]+[NH4

+]=[OH-]+[HO2-]+[HSO3

-]+2[SO32-]+[NO3

-]+[HSO4-]+2[SO4

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