PhreeqcUsers Discussion Forum

Processes => Oxidation and reduction equilibria => Topic started by: Rensburg on 14/04/26 13:42

Title: The causes of the sudden change and methods to maintain stability
Post by: Rensburg on 14/04/26 13:42

Dear dlparkhurst ,
  I am currently refining my model and have one final issue that I am struggling with. I would like to seek your help again.

  My SOLUTION is defined based on measured data. In the measured data, S(-2) and Fe(2) are present in relatively stable forms, and pe is consistently around 4.5. However, I don’t understand why, after initial equilibration of the solution, pe suddenly jumps to above 13, and both S(-2) and Fe(2) are oxidized ,although I tried removing the O(0) definition (although I think this is unreasonable because dissolved oxygen is indeed present in the measured data). That's why?

Code: [Select]

SOLUTION_MASTER_SPECIES
    Doc       Doc          0     CH2O         30

SOLUTION_SPECIES
Doc = Doc
    log_k     0
    delta_h   0 kJ
2 NO3- + 12 H+ + 10 e- = N2 + 6 H2O
#-log_k 207.08
       -log_k 100
# H2O + 0.01e- = H2O-0.01
#        -log_k -9 # aids convergence

SOLUTION 1
    temp      8.4
    pH        8.38
    pe        4.52
    redox     pe
    units     mg/l
    density   1
    # O(0)      4.9
    Amm       0.04128
    N(5)      7.0857
    N(3)      0
    P         0.07
    Doc       10
    Fe(2)     0.04
    S(-2)     0.02
    Ca        41.416 charge
    Mg        7.629
    Na        0.426
    K         0.354
    Cl        4.6866
    S(6)      23.6671
    F         0.2
    Alkalinity  50   as HCO3
    -water    1 # kg
END

USE SOLUTION 1
REACTION 1
NaCl  1
0
END  I would like to find a way to keep S(-2) and Fe(2) stable (with a slight decrease) and maintain pe around 4.5 (with a slight fluctuation).How should I improve my defination?

Title: Re: The causes of the sudden change and methods to maintain stability
Post by: dlparkhurst on 14/04/26 16:54

Redox disequilibrium is allowed in the SOLUTION definition; however, when the solution reacts in any way (KINETICS, EQUILIBRIUM_PHASES, REACTION, etc), redox equilibrium will obtain. S(-2) and N(5), for example, will not coexist at redox equilibrium, which explains the results that you observe. Natural systems do not tend to have both O(0) or N(5) and S(-2) or Fe(2). Samples with both oxidized and reduced species are likely mixtures of two or more environments and are transitory.

If you really want to preserve redox disequilibrium, you must separate all of the redox states and react them kinetically as we have discussed before.

You should step back and consider what are the main reactions that you want to consider. Are S(-2) and Fe(2) important to your system? or are you really interested in denitrification?

Title: Re: The causes of the sudden change and methods to maintain stability
Post by: Rensburg on 14/04/26 17:36

  I understand. Thank you very much. As I mentioned before, I want to understand how changes in oxygen under nitrification and denitrification affect the concentrations of other redox species. I will carefully consider my next steps and try to make corrections. Thank you again.