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Author Topic: Kinetics and solid solution  (Read 2446 times)

Flopi

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  • Posts: 26
Kinetics and solid solution
« on: 04/06/20 16:50 »
Dear Forum,

I was wondering if it is possible to use kinetics with the solid solution. If for example I define a solid solution

Code: [Select]
Carbonate
-comp Calcite 0
-comp Magnesite 0

and I want to follow its precipitation out of a supersaturated solution, should I simply define 2 rates of Magnesite and Calcite precipitation separately and then 'pretend' they belong to one single solid solution or is there a way to get the precipitation of the solid_solution directly? (I am not sure I am clear, as it is not really clear in my head either...). Ideally, it would be interesting to compare the precipitation kinetics of the solid solution with the pure minerals.

Thanks for the help,
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dlparkhurst

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  • Posts: 4000
Re: Kinetics and solid solution
« Reply #1 on: 05/06/20 04:04 »
The equilibrium condition for a component of an ideal solid solution is

K = IAP/x

where x is the mole fraction of the component in the solid solution.

Here is a simulation that kinetically approaches equilibrium for a calcite-magnesite solid solution. However, the rate constants (-parm in KINETICS) are completely arbitrary. Also, calcite and magnesite do not form an ideal solid solution. Finally, I have assumed the initial solution is supersaturated with the solid solution; there would probably be problems if the solution were undersaturated.

This script provides a starting point, and you can consider how you want to proceed.

Code: [Select]
PHASES
Magnesite           
        MgCO3 = Mg+2 + CO3-2
        log_k           -8.029
        delta_h -6.169 kcal
END
RATES
Calcite
-start
10 k = PARM(1)
20 cal = KIN("Calcite")
30 if cal <= 0 then cal = 1e-6
40 mag = KIN("Magnesite")
50 if mag <= 0 then mag = 1e-6
60 x = cal/(cal + mag)
70 rate = k * (1 - SR("Calcite")/x)
80 SAVE rate * time
-end
Magnesite
-start
10 k = PARM(1)
20 cal = KIN("Calcite")
30 if cal <= 0 then cal = 1e-6
40 mag = KIN("Magnesite")
50 if mag <= 0 then mag = 1e-6
60 x = mag/(cal + mag)
70 rate = k * (1 - SR("Magnesite")/x)
80 SAVE rate * time
-end

SOLUTION
pH  7 charge
Ca 10
Mg 10
C  40
KINETICS
Calcite
-m  1e-6
-parm 1e-8
Magnesite
-m  1e-6
-step  864000 in 10
-parm 1e-8
USER_PRINT
10 cal = KIN("Calcite")
20 mag = KIN("Magnesite")
30 xcal = cal/(cal + mag)
40 xmag = mag/(cal + mag)
50 PRINT "SRcal/xcal: ", SR("Calcite") / xcal
60 PRINT "SRmag/xmag: ", SR("Magnesite") / xmag
USER_GRAPH 1
    -headings               days SIcal SImag Cal Mag
    -axis_titles            "Days" "log10(SatRatio/x)" "Moles"
    -initial_solutions      false
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
10 cal = KIN("Calcite")
20 mag = KIN("Magnesite")
30 xcal = cal/(cal + mag)
40 xmag = mag/(cal + mag)
50 GRAPH_X total_time / 86400
60 GRAPH_Y LOG10(SR("Calcite") / xcal)
70 GRAPH_Y LOG10(SR("Magnesite") / xmag)
80 GRAPH_SY cal, mag
  -end
    -active                 true
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Flopi

  • Top Contributor
  • Posts: 26
Re: Kinetics and solid solution
« Reply #2 on: 05/06/20 08:40 »
Dear David,
Once more, thanks a lot for you quick answers.

Just one question: when you say that "the rate constants are arbitrary", do you mean that you put arbitrary values in the code, or that nobody knows what are the actual values when it comes to solid solutions?
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dlparkhurst

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  • Posts: 4000
Re: Kinetics and solid solution
« Reply #3 on: 05/06/20 15:47 »
I arbitrarily picked the rate constants to reach equilibrium in about 10 days.

Seems like if Mg fraction is small (I think natural mag-calcites are a few percent Mg), then the precipitation rate might be similar to calcite precipitation. Even with that assumption, precipitation rates are tough to estimate. Also, if mag calcite is your interest, I think you would need to consider a non-ideal solid solution (and a miscibility gap?). Your puzzler may get sore.
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