Kinetic of Fe(OH)3(a) precipitation and Kinetic of ferrous iron oxidation

[woonghee89]

Hi all,

I tried to incorporate Fe(OH)3(a) kinetic process on the Example 9 (of Kinetic of ferrous iron oxidation) to simulate both kinetic processes simultaneously. However, I am not convincing that this code was well formulated or not. I would be grateful if you could review my PHREEQC code, specifically below points.
 
1.   I incorporated the NO3- concentration into the kinetic of ferrous iron oxidation and changed O2 pressure to dissolved O2 concentration.

2.   I made a kinetic expression of Fe(OH)3(a) precipitation by modification of a kinetic expression of Quartz dissolution (of an example in Geochemistry, groundwater and pollution, C.A.J. Appelo and D. Postma).

The PHREEQC code is as follows.
-----------------------------------------------------------
SOLUTION_MASTER_SPECIES
Fe_di      Fe_di+2    0.0       Fe_di       55.847
Fe_tri      Fe_tri+3    0.0    Fe_tri      55.847

SOLUTION_SPECIES
Fe_di+2 = Fe_di+2
   log_k   0.0

Fe_di+2 + H2O = Fe_diOH+ + H+
   log_k   -9.5
   delta_h 13.20   kcal

Fe_tri+3 = Fe_tri+3
   log_k   0.0

Fe_tri+3 + H2O = Fe_triOH+2 + H+
   log_k   -2.19
   delta_h 10.4   kcal

Fe_tri+3 + 2 H2O = Fe_tri(OH)2+ + 2 H+
   log_k   -5.67
   delta_h 17.1   kcal

Fe_tri+3 + 3 H2O = Fe_tri(OH)3 + 3 H+
   log_k   -12.56
   delta_h 24.8   kcal

PHASES
Fe_tri(OH)3(a)
   Fe_tri(OH)3 + 3 H+ = Fe_tri+3 + 3 H2O
   log_k   4.891

END

SOLUTION 1
   units mg/L
   pH     7.0

   O(0)   8.63
   N(5)   6.498
   Fe_di  25.27
    Fe_tri 0
   Na     373.5
   Cl     647 charge
END

RATES
Fe_di_ox
-start
10 Fe_di = TOT("Fe_di")
20 if (Fe_di <= 0) then goto 200
30 mo2 = mol("O2")
40 mNO3 = mol("NO3-")
50 moles = 1.9e+15 * (ACT("OH-"))^2 * (mNO3 + mo2) * Fe_di * TIME
200 SAVE moles
-end

Fe_tri(OH)3(a)
-start
10 A0 = parm(1)
20 V = parm(2)
30 rate = 10^-14 * (1 - SR("Fe_tri(OH)3(a)")) * A0/V * (m/m0)^0.67
40 moles = rate * time
50 SAVE moles
-end

END

USE SOLUTION 1
EQUILIBRIUM_PHASES
   O2(g)           -0.67

KINETICS 1
Fe_di_ox
   -formula Fe_di -1.0  Fe_tri 1.0

Fe_tri(OH)3(a)
    -formula Fe_tri(OH)3(a)
    -m0 0.1       # initial Fe_tri(OH)3(a), mol/L
    -parms 22.6 0.162  # A0 in m2, V in dm3 (=L)
    -step 14688000 in 171

SELECTED_OUTPUT
   -file Iron_ox_kinetic_HFO_kinetic_ex9.sel
   -reset false
USER_PUNCH
-headings Days Fe(2) Fe(3) pH  SI_HFO
10 PUNCH SIM_TIME/3600/24, TOT("Fe_di")*1000*55.847, TOT("Fe_tri")*1000*55.847, -LA("H+"), SI("Fe_tri(OH)3(a)")

USER_GRAPH
  -axis_titles Days "Fe [mg/L]"
  -start
  10 graph_x total_time/24/3600
  20 graph_y TOT("Fe_di")*1000*55.847
  30 graph_y TOT("Fe_tri")*1000*55.847
  -end
END
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Thank you in advance for your time and help.

Woonghee

[dlparkhurst]

Maybe someone else will review it, but I try to avoid reviewing or checking code. You need to convince yourself that it is doing what you intended by looking carefully at the output and doing whatever is necessary to check your calculations. You have a much better understanding of what you want to calculate than I.

[woonghee89]

Many thanks for your comment.