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AQUEOUS PHASE REACTION MECHANISM

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AQUEOUS PHASE REACTION MECHANISM Powered By Docstoc
					                            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 2: Dissolution         STEP 2’:
                                                                 Ionization
                                                                 (for some
X                                                                species),
                                            X X  A+ + B-        VERY fast




STEP 3: Diffusion in aqueous phase   STEP 4: Chemical Reaction


                                                  X+Y ?
           X
                                                                                  STEP 4
               AQUEOUS PHASE CHEMICAL KINETICS
   Dissolved gases and ions in particles can undergo chemical rxn  oxidizing

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)
                RaL    R LRT
        Ra            a    , [s 1 ]
               [A(g)]    PA


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)

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
                                                                                          STEP 4
AQUEOUS PHASE REACTIONS: S(IV) TO S(VI), OXIDATION BY O3

                      SO32-(aq) + O3(aq)  SO42-(aq)+O2(aq)

                                         d[SO4 2 ]
             Rate of sulfate production:             k O3 [O3 (aq)][SO32 ]
                                            dt
      Liquid phase concentration of O3: [O3 (aq)]  H O3 PO3
                                         d[SO4 2 ]                 K a1K a 2 HSO2 PSO2
  Using previous expression for sulfite:             k O3 H O3 PO3
                                            dt                           [H  ]2

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 SO32-).

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:
               d[SO4 2 ]
                  dt
                                                                    
                           k 0 [SO2 (aq)]  k1[HSO3 ]  k 2 [SO32 ] [O3 (aq)]   k2 > k1 > k0
                                                                                   STEP 4
AQUEOUS PHASE REACTIONS: S(IV) TO S(VI), OXIDATION BY H2O2
                     HSO3-(aq) + H2O2(aq) ↔ SO2OOH-(aq)
                     SO2OOH-(aq) + H+(aq)  H2SO4(aq)
    Hoffman and Calvert (1985) suggest the following rate expression:

                        d[SO 4 2 ] k[H  ][H 2 O 2 ][HSO3  ]
                                   
                           dt              1  K[H  ]


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
                                                                        Comparison of S(IV)
solubilities are low enough that they are                                    oxidation paths
predicted to be of minor importance.                             (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



                                        S(IV)

              SO2.HO2   HSO3-   SO32-


H2O2(g)                                 O3(g)


                                        S(VI)
              H2SO4     HSO4-   SO42-
                                                                                  STEP 1/3
                      MASS TRANSFER LIMITATIONS
    We have assumed that drops are in equilibrium with the gas phase, or that

    this:                                 is fast compared to this:
                 X                                                       X+Y ?

            X


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)

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              X+Y ?
basis of Henry’s Law and overall rate is reduced.
EXTREME example: N2O5 hydrolysis happens so quickly that the            X
reaction appears to happen as soon as molecule enters solution,
thus rate ~Adrop (as if occurring at the surface)
               EXAMPLE FROM SEINFELD AND PANDIS
Pure water droplet (pH=7), L=10-6 m3water/m3air is exposed to environment :
   Chemical   Initial       After         After equilibrium conditions solved with equations
   Species    (gas phase)   equilibrium
                            (gas phase)   for Henry’s Law, dissociation equilibrium,
   HNO3       1 ppb         10-8 ppb      electroneutrality and mass balance, such as:
                                                [O (aq)]        [O3 (aq)]
   H2O2       1 ppb         0.465 ppb
                                          HO3  3         
   O3         5 ppb         5 ppb                 PO3       [O3 ]T  [O3 (aq)]
   NH3        5 ppb         1.87 ppb
                                          [H+]+[NH4+]=[OH-]+[HO2-]+[HSO3-]+2[SO32-]+[NO3-]
   SO2        5 ppb         3.03 ppb

                                          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+]+[NH4+]=[OH-]+[HO2-]+[HSO3-]+2[SO32-]+[NO3-]+[HSO4-]+2[SO42-]

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
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




                                                              [Seinfeld & Pandis]

				
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posted:3/21/2012
language:English
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