sulfur dioxide reactions

[Hamed]

Dear all,

I am considering the dissolution of sulfur dioxide in my simulation. Considering the llnl.dat database, I observe the precipitation of Anhydrite in my model, which is not expected since there is no oxygen present in the system. In the literature, it has been indicated that SO2 dissolution in brine cannot lead to sulfate mineral precipitation, and it needs oxygen. I wonder how it is controlled in PHREEQC that I have such precipitation. I have included the relevant reactions that are in llnl.dat database.

Thank you

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SOLUTION_MASTER_SPECIES
S(+4)  SO3-2            0.0   S  32
S(+2) S2O3-2 0.0 S
S(+3) S2O4-2 0.0 S
S(+5) S2O5-2 0.0 S
S(+7) S2O8-2 0.0 S
SOLUTION_SPECIES

2.0000 H+ + 1.0000 SO3-2  =  SO2 +1.0000 H2O   
      -log_k           +9.0656
-delta_H 26.7316 kJ/mol

 1.0000 SO4-2  =  SO3-2 + 0.5000 O2     
       -log_k           -46.6244
-delta_H 267.985 kJ/mol

2.0000 SO3-2  + 1.500 O2  + 2.0000 H+  = S2O8-2  + H2O 
        -log_k           70.7489
  -delta_H 0    # Not possible to calculate enthalpy of reaction S2O8-2
        -analytic 0.18394E+03    0.60414E-01    0.13864E+05   -0.71804E+02    0.21628E+03

 2.0000 H+  + 2.0000 SO3-2  = S2O4-2  + .500 O2  + H2O
       -log_k           -25.2076
-delta_H 0 # Not possible to calculate enthalpy of reaction S2O4-2

2.0000 H+  + 2.0000 SO3-2  = S2O3-2  + O2  + H2O   
        -log_k           -40.2906

2.0000 H+  + 2.0000 SO3-2  = S2O5-2  + H2O   
        -log_k           9.5934

[dlparkhurst]

Basically, thermodynamic equilibrium produces the following reaction:

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SO2 + H2O = 3/4H2S + 1/4SO4-2 + 1/2H+

If you don't want SO4-2 to be produced, you must change the system so that thermodyamic equilibrium does not occur. To define an redox-unreactive SO2, you will need to define a new "element" with SOLUTION_MASTER_SPECIES that is in the S(4) state, and appropriate SOLUTION_SPECIES for the S(4) state. See the definition of Sg in phreeqc.dat for an unreactive H2S definition.

[Hamed]

Does this mean that sulfate can be produced in the system even in the absence of oxygen (underground condition)?

[dlparkhurst]

My mistake on the equation. Should have been

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SO2 + H2O = 1/4H2S + 3/4SO4-2 + 3/2H+

But, yes, thermodynamic equilibrium says that when you dissolve SO2, S(-2) and S(6) are the stable redox states. Some of the S gets reduced to H2S and some gets oxidized to SO4-2. No oxygen is needed; sulfur can be described as disproportionating to two oxidation states.

Now a system may not reach thermodynamic equilibrium. The atmosphrere should make nitric acid, but N2 and O2 remain metastable. So, S(4) species, or other combinations of sulfur redox states could form and remain metastable.

[Hamed]

Thank you very much. My last question is whether this reaction requires microbial activity, such as that of sulfate-reducing bacteria, to transfer electrons and make the reaction happen.

[dlparkhurst]

I do not know what would actually happen when SO2 dissolves. PHREEQC calculates thermodynamic equilibrium, which indicates you would form S(-2) and S(6). I do not know the rate at which this transformation would occur. I don't know whether it is slow or fast; you would need to look for some empirical data. I do know that in most natural environments, concentrations of redox states other than S(-2) and S(6) are quite small.