Kinetics for ferrihydrite Fe(OH)3(a) in mining canal

[M.Gundlach]

Dear All,

I´m using PHREEQC for modelling geochemical processes in connection with a construction of a mining canal. In the pit water, there are high concentrations of especially iron. The purpose of the modelling is how much of the ferrihydrite precipitates while flowing through the canal. To have reasonable results, I have to integrate the kinetics in the model, but for now I have problems with doing that. In other threads on this platform I found some hints for the kinetic modelling for ferrihydrite, but I don´t know to what extent I can use them for my problem.


With best regard,

M.Gundlach

[dlparkhurst]

Just some preliminary thoughts, so don't take them as the final answer.

You don't give too many specifics. My guess is that precipitation of ferrihydrite is fast once iron is oxidized,  so perhaps the rates are dependent on the rate of oxidation of Fe(2), which in turn would depend on temperature, oxygen concentration, sunlight, possibly oxygen transfer into the canal water, biological activity, and other factors. The rate is not very well constrained, and in the end you probably need some empirical data.

[John Mahoney]

I would start with a simple rate law, and Grundl and Delwiche comes to mind although looking it up now I realize it is 25 years old.

Grundle, T., Delwiche, J. 1993. Kinetics of ferric oxyhydroxide precipitation. Journal of Contaminant Hydrology 14(1, 71-97).

The good thing about this paper is it provides a precipitation rate. 

As I recall (and it has been about 20 years since I dealt with this type of process) it uses a simple rate constant that converts the dissolved Fe(OH)3 complex into ferrihydrite.  It is a simple law that could easily fit into a RATES in PHREEQC.  You could set this up as a TRANSPORT problem quite easily.  As you indicated, there are other sources of data in the newer literature.  if you have data you could fit a better rate constant using PhreePlot (See below for another issue and Application) 

 NOW there are going to be issues about the stability of the Ferrihydrite you form.  In a flowing reacting system the precipitation reactions can be more complicated because you are constantly adding fresh solution, which may not have enough time to fully react.   This could be reflected in your adjusted rate constant or by the effective stability of the Fe(OH)3 that forms.  If you have enough and the appropriate data you could even use PhreePlot to adjust the Ksp as a second fitting parameter.   

I just saw David Parkhurst's comment he raises some good points about Fe+2 oxidation.    And so I have another question as to how is this solution going to be neutralized, will that be rate limiting? 
 

[M.Gundlach]

First, thank you for the comments.

After some research, I made a first attempt for my modelling of kinetic iron-precipitation in my closed canal with some pre-assembled terms from "example 9" from the manual. The model "works", but I have some issues with combining the kinetics with the precipitation of the Fe(OH)3(a) with equilibrium_phases. In my selected_output, the whole Fe(OH)3(a) precipitates at the beginning, although it should precipitate slowly and not completly after one step.
I don´t know if it´s necessary, but I show my Input-file if it helps finding a solution. I guess it would help if i can simply use the equilibrium_phases for every step of the modelling.

TITLE KINETIC PRECIPITATION COSED MINING SEWER

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 6.48
temp 13.7
pe 4.8
density 0.9933

O(0) 2.1 as O2

Cl 184
C(4) 311.5 as CO2
P 0.11 as PO4-3
S(6) 1760   
Pb 0.0022
B 0.11
Cd 0.011
Ca 310
Fe 163
K 11
Mg 136
Mn 12
Na 252
Sr 1.0
Zn 0.73

EQUILIBRIUM_PHASES

O2(g)              -0.68
Fe(OH)3(a)       0.0 0.0

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 = moles * (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

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)
    -m  0                    # Fe(OH)3(a) at the beginning of the modelling
    -m0 0                                # primary Fe_tri(OH)3(a), mol/L 
    -parms 22.6 0.162        # A0 in m2, V in dm3 (=L)
    -steps 6480 in 60        # 6480s in 60s steps = discharge duration through the canal (1,8 Stunden)
   

SELECTED_OUTPUT

-reset false
-file Kinetik.txt
-totals Fe O
-saturation_indices Fe(OH)3(a)
-equilibrium_phases Fe(OH)3(a)
-kinetic_reactants Fe(OH)3(a)
-step


END

Thank you in advance,

M. Gundlach

[dlparkhurst]

(1) Because you are using Fe_di and Fe_tri, you should not use Fe or Fe(OH)3(a). All of your iron reactions should only involve Fe_di and Fe_tri.

(2) Signs are tricky with KINETICS define M=0. If the amount SAVEd is positive, then M is calculated to decrease, but not below 0. Thus, if you want the amount of kinetic reactant to increase, the amount SAVEd must be positive. You must adjust the signs of the formula accordingly. If the product of the SAVEd quantity and the coefficient in -formula is positive, the amount in solution increases. If the product is negative, then the amount in solution decreases.

(3) 50 moles = moles * (ACT("OH-"))^2 * (mNO3 + mo2) * Fe_di * TIME 

A couple of issues here. moles is zero until set, so line 50 will always result in zero for moles. Second, the activity of OH- is about 1e-7 or smaller, so the product (excluding moles) is on the order of 1e-20, so withoud a very large TIME (seconds) you will get little reaction. Normally, there is also a rate constant multiplied times the product of concentrations. So, reconsider your rate equation formulation.

(4) Keep things simple to start. Do the calculation with 1 step until you get things working. Also, you can debug your rate equation by adding PRINT statements in the RATES definition.

[M.Gundlach]

Thanks to your help and after changing my input, I now get some hopefully "reasonable" results.

Now I´m left with one problem. The highest amount of the Fe_tri_ox precipitates after one kinetic step.

For example, for one "scenario", the total flow period is 720 seconds divided in 60 steps a 12 seconds.
After the first of the 60 steps,  1,8075e-05 mol/l precipitates. After the last of the 60 steps, 1,9071e-05 is the amount of what precipitates in whole.
I don´t know if the modelling is still correct, or if there is another fault in the input, or if this is because of the circumstances of the solution in general.

If needed, I post the "new" input file.

TITLE Grubenwasser Dickenberger Stollenmundloch 03.07.2017 #Westfeld

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

SOLUTION 1

units mg/l

pH 6.48
pe 4.8
temp 13.7
density 0.9933

O(0) 2.1 as O2

Cl 184
C(4) 311.5 as CO2
P 0.11 as PO4-3
S(6) 1760   
Pb 0.0022
B 0.11
Cd 0.011
Ca 310         
Fe_di 162            
Fe_tri 1
K 11
Mg 136
Mn 12
Na 252
Sr 1.0
Zn 0.73

EQUILIBRIUM_PHASES

Fe_tri(OH)3(a) 0.0 0.0
O2(g)      -0.68

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 = moles * (ACT("OH-"))^2 *(mNO3 + mo2)*Fe_di * TIME   
50 moles = 10^13 * (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

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)
    -m  1            
    -m0 1                
    -parms 22.6 0.162     
    -steps 720 in 60        


SELECTED_OUTPUT

-file Westfeld1AusfällungenATMKinetik0,5.txt
-reset false
-totals Fe O
-saturation_indices Fe_tri(OH)3(a)
-equilibrium_phases Fe_tri(OH)3(a)
-kinetic_reactants Fe_tri(OH)3(a)
-step

END

Thanks in advance,

M. Gundlach

 

[dlparkhurst]

I would experiment with the rate constants. First, I would add one in RATES for the Fe(II) oxidation. Then, you should precipitate more slowly if you decrease the rate constant for Fe(II) oxidation or Fe(OH)3 precipitation.

[pia said]

Hi I`m looking for information about the Fe(OH)3 or FeOOH precipitation rate. Where did you find the rate of your code? Or if someone knows where find information can you please tell me

[pia said]

Hi, i'm confused i don't know what value to use in A0 (initial surface area) in the case of precipitation, because initially there is not solid in my solution, so i tend to think that A0=0. If someone can help me I would be very grateful.

[pia said]

Hi, i'm confused i don't know what value to use in A0 (initial surface area) in the case of precipitation, because initially there is not solid in my solution, so i tend to think that A0=0. If someone can help me I would be very grateful.

[dlparkhurst]

True, if you use zero as a multiplier of the rate, you will never precipitate anything. But, I don't have any easy answer for you. You will have to decide how you want to handle the problem. You could start with a small amount, which would make your calculation arbitrary. The literature usually indicates a certain supersaturation is needed to begin precipitation. It also depends on whether there is another substrate on which to nucleate. The literature may have some expressions that are used. Assuming you have some empirical data, I would probably  assume the rate is proportional to (1 - SR("Ferrihydrite") and try to pick a rate constant that approximates the empirical data.

[pia said]

Thanks for your answer I have experimental data that I can use.

But now I'm confuse again with the value of m0 in  the case of precipitation, because m0 is the initial moles of solid so I tend to think that is cero but if it cero the formula is indeterminate, so I don’t know what to do, can you help me please?

[dlparkhurst]

The factor (M/M0)^n is used to adjust the surface area as a mineral dissolves, sometimes called the shrinking core model. You are right that it would not apply in precipitation, so either eliminate it from the equation or choose an arbitrary value and a value of zero for n.

[pia said]

Thanks for your answer !!!