Goethite - Precipitation Rates

[lourdsa]

Dear all,

I am not a chemist but have a basic understanding of inorganic chemistry. I am currently trying to understand rate of mineral precipitation in our system.

We currently have water running through a system at the following pressure, temperature and chemical signature as defined below:

REACTION_PRESSURE 10
    11

SOLUTION 10
    temp      10
    pH        8.74
    pe        -3.8
    redox     pe
    units     mg/l
    density   1
    Al        12 ug/L
    Alkalinity 75
    Ba        20.2 ug/L
    Ca        15
    Cl        1.5
    F         0.17
    Fe        33.5 ug/L
    K         2.1
    Mg        6.1
    Mn        56.6 ug/L
    O(0)      0.03
    S(6)      1.9
    Si        13.5
    Zn        1.22 ug/L
    -water    1 # kg

This solution showed the water was theoretically supersaturated with respect to the iron and manganese oxides/hydroxides. 

USE solution 10
EQUILIBRIUM_PHASES 1
    Fe(OH)3(a) 0 0 precipitate_only
    Goethite  0 0 precipitate_only
    Hausmannite 0 0 precipitate_only
    Hematite  0 0 precipitate_only
    K-mica    0 0 precipitate_only
    Manganite 0 0 precipitate_only
    Pyrolusite 0 0 precipitate_only
    Talc      0 0 precipitate_only

We have then used EQUILIBRIUM PHASES data block to understand Hematite, K-mica, Pyrolusite and Talc can potentially reach equilibrium and precipitate by using a SI and initial moles of 0. However, we want to understand whether kinetics will allow this to occur. I understand the mathematical expressions for the rates of the kinetics reactions are defined in the

RATES data block:
RATES
Mineral #rates name
10 rate = k * (1 - SR("Mineral"))
20 moles = rate * TIME
30 SAVE moles
END

We have found a dissolution rate in the PHREEQC kinetics database for our example mineral, Goethite, and assumed the following:
•   The dissolution rate specified in the PHREEQC kinetics database is the same as the precipitation rate;
•   The dissolution rate, used as the precipitation rate, is assumed to be the empirical constant (k);
•   The dissolution rate, used as the precipitation rate, was calculated empirically at a temperature ranging from 20-25 degrees and a pH of 0.4 The rate is assumed to be the same for our in-situ conditions (T= 13.6, pH = 8.7)

We have then multiplied this empirical constant (k) with (1-SR) where (IAP/Kk = SR). Note we have not modeled total rate (Rk) as we do not know A, V, m or m0. Results shows the saturation ratio becomes 1, indicating it reaches equilibrium after 160 seconds.

My questions are:
1) Is our assumption that empirical constant (k) in the rates equation the same as the specific rate (taken from the PHREEQC kinetics database) correct?
2) have we defined our KINETICS data block correctly?
3) Water is being pumped at a rate of 40 l/s, how can we corporate this into our script to understand amount of precipitation occurring in the system?

Any help/advice on the script and how to progress this further will be greatly appreciated.

Thank you,
Kind regards,
LA

[dlparkhurst]

I don't see any problems with your analysis. In another post, I think I mentioned that assuming the Fe is Fe(2), the rate determining step may be the oxidation of Fe(2) to Fe(3). But, your rate uses an indeterminant "k", which simply lumps all of the rates together anyway.
The rate is could be faster or slower than your calculations. Only some additional empirical data in your system would allow you to improve the estimate of k.

It is within an order of magnitude, but there is a sufficient amount of oxygen to oxidize all of the iron, so at equilibrium with goethite, the Fe concentration is calculated to be vanishingly small. The extreme case is that all of the iron precipitates quickly, so the rate of Fe accumulation is equal to the rate of Fe inflow (flow rate times concentration). Ultimately, the answer to your questions is simply what fraction of the iron accumulates comparing input to output.

Note that the pressure should have little effect on the calculations.

[John Mahoney]

There is a lot more going on that you need to sort out. First Normally the iron hydroxide [Fe(OH)3 phase] will precipitate first and then over time age into hematite.  But in some cases I might allow for hematite to form directly.  If you can find it look up Grundl and Delwiche

Kinetics of ferric oxyhydroxide precipitation
Tim Grundl, Jim Delwiche Published 1993
DOI:10.1016/0169-7722(93)90042-Q

Journal of Contaminant Hydrology
Volume 14, Issue 1, August 1993, Pages 71-87

They presented a kinetic model for ferrihydrite precipitation it is probably dated now but it is a start.  I think I wrote about this some other time in this forum. 

See this kinetics and rate controlling factors /re: Kinetics for ferrihydrite Fe(OH)3(a) in mining canal for some more discussion. 


So maybe there are some better rates out there.

[lourdsa]

Thank you both for your replies, very much appreciated.

[lourdsa]

Hi both,

Can you please tell me what's wrong with my script for Calcite:

I have used the dissolution rate from the Kinetics Database and assumed this is the same for precipitation rate.

My code is as follows:

SOLUTION 1 TS-6
    temp      13.61
    pressure  10.7
    pH        8.74
    pe        -3.8
    redox     pe
    units     mg/l
    density   1
    Al        12 ug/L
    Alkalinity 75
    Ba        20.2 ug/L
    Ca        15
    Cl        1.5
    F         0.17
    Fe        33.5 ug/L
    K         2.1
    Mg        6.1
    Mn        56.6 ug/L
    O(0)      0.03
    S(6)      1.9
    Si        13.5
    Zn        1.22 ug/L
    -water    1 # kg

REACTION TEMPERATURE 1-3
   13.61 10 7.5

RATES
  Calcite
  -start
  10 k1 = ((2.13*(10e-2))*(exp(-16000/(8.31*TK)))*(ACT("H+")^0.33))
  20 k2 = ((3.47*(10e-3))*(exp(-46000/(8.31*TK)))*(ACT("H2CO3")^0.33))
  30 k3 = ((1.17*(10e-2))*(exp(-46000/(8.31*TK)))*(ACT("H2O")^0.33))
  40 k = k1 + k2 + k3
  50 moles= k * time
  60 SAVE moles
  -end
Use SOLUTION 1
KINETICS
  Calcite
   -cvode
      -tol    1e-8
   -steps 86400 in 10
INCREMENTAL_REACTIONS true
SELECTED_OUTPUT
   -file Calcite_Rate.txt
   -reset false
USER_PUNCH
-headings Seconds  Ca SR_Ca
10 PUNCH SIM_TIME TOT("Ca"),SR("Calcite")

The output text file shows that the Ca concentration is increasing rather than decreasing and the saturation ratio is getting progressively further away from 0 rather than getting closer to it.

This is also the case with Hematite.

SOLUTION 1 TS-6
    temp      13.61
    pressure  10.7
    pH        8.74
    pe        -3.8
    redox     pe
    units     mg/l
    density   1
    Al        12 ug/L
    Alkalinity 75
    Ba        20.2 ug/L
    Ca        15
    Cl        1.5
    F         0.17
    Fe        33.5 ug/L
    K         2.1
    Mg        6.1
    Mn        56.6 ug/L
    O(0)      0.03
    S(6)      1.9
    Si        13.5
    Zn        1.22 ug/L
    -water    1 # kg

REACTION TEMPERATURE 1-3
   13.61 10 7.5
RATES
  Hematite
  -start
  10 k = 10^-17
  20 moles= k * time
  30 SAVE moles
  -end
Use SOLUTION 1
KINETICS
  Hematite
   -cvode
      -tol    1e-8
   -steps 10000 in 10
INCREMENTAL_REACTIONS true
SELECTED_OUTPUT
   -file Hematite_Rate.txt
   -reset false
USER_PUNCH
-headings Seconds  Fe(3) SR_Hematite
10 PUNCH SIM_TIME TOT("Fe(3)"),SR("Hematite")
END

Thank you,
Kind regards,
LA

[dlparkhurst]

It looks to me like your rates are always positive, which adds calcite to the solution, in other words always dissolution.

The rate in the database is below. It uses SI("Calcite") in a way that the rate is positive when undersaturated and negative when supersaturated [(1 - 10^(2/3*si_cc))].

Code: [Select]

Calcite
   -start
1   REM   PARM(1) = specific surface area of calcite, cm^2/mol calcite
2   REM   PARM(2) = exponent for M/M0

10  si_cc = SI("Calcite")
20  IF (M <= 0  and si_cc < 0) THEN GOTO 200
30  k1 = 10^(0.198 - 444.0 / TK )
40  k2 = 10^(2.84 - 2177.0 /TK )
50  IF TC <= 25 THEN k3 = 10^(-5.86 - 317.0 / TK)
60  IF TC > 25 THEN k3 = 10^(-1.1 - 1737.0 / TK )
80  IF M0 > 0 THEN area = PARM(1)*M0*(M/M0)^PARM(2) ELSE area = PARM(1)*M
110 rate = area * (k1 * ACT("H+") + k2 * ACT("CO2") + k3 * ACT("H2O"))
120 rate = rate * (1 - 10^(2/3*si_cc))
130 moles = rate * 0.001 * TIME # convert from mmol to mol
200 SAVE moles
   -end