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Calculate Zn at eq with sphalerite with RXN and REACT


Silvain Rafini
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Hi Brian/all

I'm trying to predict the concentration of Zn at equilibrium with sphalerite at neutral pH and Eh = -50 mV (groundwater), taking into account only the components that are present into the dissolution reaction :

ZnS   +   2 O2   <-->   Zn++   +  SO4--

 

My objectives, after that, are to complexify the system and investigate the influence of some major ligands that are present in natural groundwater (PO4, SO4, Cl), on Zn solubility in the presence of sphalerite. In order to predict natural Zn concentrations in various types of groundwater.

 

Below are the input parameters in REACT, water is defined only by components Zn-SO4-pH-Eh and Na for charge balance, reacting with 10 g of sphalerite :

image.png.d828c568e1960c9525a6dc430d35bfeb.png

    The command panel is :

> # React script, saved Fri Feb 28 2020 by Utilisateur
     > data = thermo.tdat verify
     > conductivity = conductivity-USGS.dat
     > time start = 0 day, end = 1 yr
     > temperature = 25 C
     > H2O          = 1 free kg
     > pH           = 7
     > Zn++         = .1 ug/l
     > SO4--        = 1 mg/l
     > swap e- for O2(aq)
     > Eh           = -50 mV
     > balance on Na+
     > react 10 g of Sphalerite

 

This model predicts as low value as Zn = 1.01 E-9 mol/L, as shown below :

image.thumb.png.de26a9eb3423c2bc3a6a242c2e6d80c0.png

Trying to corroborates this result with reaction balancing using RXN :     

 Sphalerite  + 2 O2(aq)  = Zn++  + SO4--

       Log K's:
                   0 ? 140.0535        150 ? 81.8395
                  25 ? 126.8808        200 ? 69.6068
                  60 ? 111.2955        250 ? 59.0203
                 100 ? 96.7075        300 ? 49.4066

       Polynomial fit:
         log K = 140 - .5635 T + .001593 T^2 - 3.229e-6 T^3 + 2.744e-9 T^4

       Log K at 25 C =  126.8808

       Assumptions implicit in equilibrium equation:
         temperature            = 25 C
         activity of SO4--      = 10^-3     

       Equilibrium equation:
         129.9 = log a[Zn++] - 2 log a[O2(aq)]

 

This seems to me a unrealistic result...

This would give Zn concentrations as high as log aZn++ = 315.46, if we make the conversion from Eh = -50 mV (pE = -0.84) to log aO2 = -92.78  (see below RXN window with H2O dissociation reaction)

 

image.png.275717006778adf4e420ec3c3afd8b6b.png

 

At this point, I guess I need help... It seems to me that I should obtain the same results with these two approaches, but I obtain unrealistic values.

 

Moreover, if I specify e- activity ae-, or Pe = -0.84 in the above RXN model, the reaction changes (see below) and the resulting Zn value also, quite drastically...

Sphalerite  + 4 H2O  = Zn++  + SO4--  + 8 H+  + 8 e-

       Log K's:
                   0 ? -47.3593        150 ? -40.0723
                  25 ? -45.1228        200 ? -39.5912
                  60 ? -42.9247        250 ? -39.7529
                 100 ? -41.2689        300 ? -40.7818

       Polynomial fit:
         log K = -47.35 + .1002 T - .0005183 T^2 + 1.452e-6 T^3 - 1.981e-9 T^4

       Log K at 25 C =  -45.1228

       Assumptions implicit in equilibrium equation:
         temperature            = 25 C
         activity of SO4--      = 10^-3     
         activity of H+         = 10^-7     
         activity of e-         = 10^.8451  
         activity of H2O        = 10^0      

       Equilibrium equation:
         7.116 = log a[Zn++]

 

 

Could you please help me by explaining what is wrong in these approaches, and what would be the more appropriate one.

 

Thank you very much in advance.

 

Silvain

 

 

 

 

 

 

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