Calcite Disolution in open and close system

[pia said]

Hi i'm modeling calcite disolution in 2 systems, similar example 5.6 book Appelo y Postma 2da edition but including the kinetics. My [Ca] and pH results aren't consistents with literature. In a close system the moles of Ca in solution are higher than in a open system and the pH is lower. According to my knowledge it should be the opposite. Can you help me please ? Beforehand thank you very much.

Open system

Code: [Select]

RATES
    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
120 rate = area * (k1 * ACT("H+") + k2 * ACT("CO2") + k3 * ACT("H2O")) * (1 - 10^(2/3*si_cc))
130 moles = rate * 0.001 * TIME
200 SAVE moles
-end
END

SOLUTION 1
    temp      25
    pH        2
    pe        6.33 O2(g) -0.68
    redox     pe
    units     ppm
    density   1
    As(5)     4
    B         30
    Fe(2)     50
    S(6)      1000 charge
    -water    0.24 # kg #volumen solución según datos de Henry 2015

EQUILIBRIUM_PHASES 1
    CO2(g) -1.54
    Goethite 0 0   
    Fe(OH)3(a) 0 0
    O2(g) -0.68

KINETICS 1
Calcite
    -formula  Calcite  1
    -m        0.015
    -m0       0
    -parms    380000 0.6 #area BET 0.38 m2/gr, parm(2) según manual
    -tol      1e-08
-steps        3600 in 10 steps # seconds
-step_divide 1
-runge_kutta 3
-bad_step_max 500
INCREMENTAL_REACTIONS True

EQUILIBRIUM_PHASES 1
    CO2(g) -1.54 10
    Goethite 0 0   
    Fe(OH)3(a) 0 0
    O2(g) -0.68 

USER_GRAPH 1
    -headings               Tiempo Ca pH
    -chart_title            "Modelo PWP sistema abierto, Datos Henry 2015"
    -axis_titles            "Tiempo, min" "Ca en solución, mol" ""
    -axis_scale y_axis      0 0.0035
    -axis_scale sy_axis     0 8
    -initial_solutions      true
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
10 GRAPH_X TOTAL_TIME/60
20 GRAPH_y TOT("Ca")*0.24 #se multiplica por el volumen de la solucion para dejarlo en moles
30 pH = -LA("H+")
40 graph_sy pH
  -end
    -active                 true
END
Open then Close system

Code: [Select]

RATES
    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
120 rate = area * (k1 * ACT("H+") + k2 * ACT("CO2") + k3 * ACT("H2O")) * (1 - 10^(2/3*si_cc))
130 moles = rate * 0.001 * TIME
200 SAVE moles
-end
END

SOLUTION 1
    temp      25
    pH        2
    #pe        6.33 O2(g) -0.68
    redox     pe
    units     ppm
    density   1
    As(5)     4
    B         30
    Fe(2)     50
    S(6)      1000 charge
    -water    0.24 # kg #volumen solución según datos de Henry 2015

EQUILIBRIUM_PHASES 1 #la solucion inicial en contacto con el aire
    CO2(g) -1.54 10
    O2(g) -0.68
SAVE solution 2
END

USE solution 2

KINETICS 1
Calcite
    -formula  Calcite  1
    -m        0.015
    -m0       0
    -parms    380000 0.6 #area BET 0.38 m2/gr, parm(2) según manual
    -tol      1e-08
-steps        3600 in 10 steps # seconds
-step_divide 1
-runge_kutta 3
-bad_step_max 500
INCREMENTAL_REACTIONS True

EQUILIBRIUM_PHASES 2
    Goethite 0 0   
    Fe(OH)3(a) 0 0

USER_GRAPH 1
    -headings               Tiempo Ca pH
    -chart_title            "Modelo PWP solución en contacto con la atmosfera y luego sistema cerrado, Datos Henry 2015"
    -axis_titles            "Tiempo, min" "Ca en solución, mol" ""
    -axis_scale y_axis      0 0.0035
    -axis_scale sy_axis     0 8
    -initial_solutions      true
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
10 GRAPH_X TOTAL_TIME/60
20 GRAPH_y TOT("Ca")*0.24 #se multiplica por el volumen de la solucion para dejarlo en moles
30 pH = -LA("H+")
40 graph_sy pH
  -end
    -active                 true

[dlparkhurst]

The general thought that pH will be lower an open system assumes that the open system will allow more CO2(g) to dissolve, resulting in more dissolution of calcite.

You are starting at a pH of 2, and CO2(g) is exsolving from solution in the open system and remaining in solution in the closed system. Loss of CO2 results in a lower partial pressure of CO2 in the open system than in the closed system. In the open system, pH is higher and less calcite dissolves, whereas in the closed system, pH is lower and more calcite dissolves.

[pia said]

Thank you very much.