mineral phase transitions

[Sen]

Dear all,

I'm encountering some issues during the calculation process of mineral phase transitions. This time, I need to complete the phase diagram drawing for the phase transitions among three types of clay minerals, which are: Kaolinite, Illite, and Montmor-Ca. I hope to identify the horizontal and vertical coordinates of the mineral phase transitions through thermodynamic calculations. Although I have achieved consistency in the vertical coordinates, the horizontal coordinates are different. I also need to use phreeqc to complete this process. Could you please help me?

The calculation process as followed:


Phase transition calculation of clay minerals

1. Thermodynamic data of clay minerals

Kaolinite
Al2Si2O5(OH)4 + 6H+  =  2Al3+ + 2SiO2(aq) + 5H2O(l)
        log_k           6.8101
       delta_H           -151.779    kJ/mol

Illite
K0.6Mg0.25Al1.8Al0.5Si3.5O10(OH)2 + 8H+ = 0.25Mg2+ + 0.6K+ + 2.3Al3+ + 3.5SiO2(aq) + 5H2O(l)
        log_k           9.0260
       delta_H          -171.764   kJ/mol
   
Montmor-Ca
Ca0.165Mg0.33Al1.67Si4O10(OH)2 + 6H+ = 0.1650Ca2+ + 0.33Mg2+ + 1.67Al3+ + 4H2O(l) + 4SiO2(aq)
        log_k           2.4952
       delta_H          -100.154   kJ/mol   



2.Calculation process
2.1 The reaction between calcium montmorillonite and kaolinite is:

The phase transition reaction is as follows:
2Ca0.165Mg0.33Al1.67Si4O10(OH)2(Montmor-Ca) + 1.98H+ + 0.35H2O(l) = 1.67Al2Si2O5(OH)4(Kaolinite) + 0.33Ca2+ + 0.66Mg2+ + 4.66SiO2(aq)
log10 K (Montmor-Ca→Kaolinite) = 1.67 ? log10K(Kaolinite) -2 ?log10K(Montmor-Ca)
=0.165log10([Ca2+]/[Na+]2) + 0.165log10([Ca2+][Na+]2[Mg2+]4) + 4.66log10[SiO2] + 1.98pH
= 0.165 Y + X

setting：
X = 0.165log10([Ca2+][Na+]2[Mg2+]4) + 4.66log10[SiO2] + 1.98pH
Y = log10([Ca2+]/[Na+]2)

By substituting the thermodynamic data, the following can be easily obtained for Montmor-Ca→Kaolinite:
Y = -6.0606 X - 38.6818 (25℃)
Y = -6.0606 X - 43.1436 (10℃)

Use the same method
2.2 The reaction between calcium montmorillonite and Illite is:

2.3Ca0.165Mg0.33Al1.67Si4O10(OH)2(Montmor-Ca) + 1.002K+ + 0.44H+ = 1.67K0.6Mg0.25Al1.8Al0.5Si3.5O10(OH)2(Illite) + 0.3795Ca2+ + 0.3415Mg2+ + 3.355SiO2(aq) + 0.85H2O(l)

We can get:
 log10 K (Montmor-Ca→Illite) = 0.18975 Y + X

Setting:
X = 0.18975log10 ([Ca2+] [Na+]2) + 0.3415log10 [Mg2+] + 3.355log10 [SiO2] + 0.44pH - 1.002log10[K+]
Y = log10([Ca2+]/[Na+]2)


Substituting thermodynamic data can be easily obtained Montmor-Ca→Illite:
Y = - 5.2701 X - 49.1937 (25 ℃)
Y = - 5.2701 X - 52.8353 (10 ℃)


Best regards,
Sen

[dlparkhurst]

I'm not a phase diagram guy, and PHREEQC may not be the best tool for what you want to do. You might consider PHREEPLOT from David Kinniburgh, which is based on PHREEQC but designed to make predominance and phase diagrams.

I think the following is a translation of what you posted. Note that you must write Ca+2 and Al+3, not Ca2+ and Al3+. Also, SiO2(aq) is the master species for llnl.dat. You will have to change the equations to H4SiO4(aq) if you want to use phreeqc.dat and most other databases.

Code: [Select]

PHASES
Kaolinite
Al2Si2O5(OH)4 + 6H+  =  2Al+3 + 2SiO2 + 5H2O
        log_k           6.8101
       delta_H           -151.779    kJ/mol

Illite
K0.6Mg0.25Al1.8Al0.5Si3.5O10(OH)2 + 8H+ = 0.25Mg+2 + 0.6K+ + 2.3Al+3 + 3.5SiO2 + 5H2O
        log_k           9.0260
       delta_H          -171.764   kJ/mol
   
Montmor-Ca
Ca0.165Mg0.33Al1.67Si4O10(OH)2 + 6H+ = 0.1650Ca+2 + 0.33Mg+2 + 1.67Al+3 + 4H2O + 4SiO2
        log_k           2.4952
       delta_H          -100.154   kJ/mol   
END
SOLUTION
END
USE solution 1
EQUILIBRIUM_PHASES
Kaolinite 0 0
Montmor-Ca 0 0
REACTION
Montmor-Ca  1
4e-5 in 10 steps
USER_GRAPH
10 GRAPH_X  0.165*LOG10[ACT("Ca+2")*ACT("Na+")^2*ACT("Mg+2")^4] + 4.66*LA("SiO2") - 1.98*LA("H+")
20 GRAPH_Y   LOG10(ACT("Ca+2")/ACT("Na+")^2)
END