Oxidation rate of manganese by KMnO4

[Nick]

Dear Dr. Parkhurst

Some time ago I set up a model for the precipitation of manganese from tailings water using potassium permanganate as the oxidant. Now I would like to build kinetics into the model, but am finding this much more difficult and haven’t been able to get this to work as of yet. Could you or someone on your team please provide me with a little guidance regarding how to go about setting this up.

I understand that I must use the RATES and KINETICS data blocks, however, I’m not too sure how to set these up as so far my attempts have not worked. The rate equation I would like to use is:

d[〖Mn2+]/dt= -ak1[〖Mn2+][KMnO4] [OH-]^a-k2[〖Mn2+][MnO2]

How do I go about setting this up? I’m not sure how to call out KMnO4 and OH- concentrations from the solution as the reaction progresses. When KMnO4 is added as a reactant it is broken up and added to the solution in its elemental form. For Mn2+ and MnO2 I assume I can use the TOTMOLE and EQUI statements respectively.

The following code is what I am using to model the oxidation of manganese by KMnO4.  I would like to now model the kinetics of this reaction now that I have determined the dose required to remove manganese by precipitation to a solution concentration below 1mg/L.

Any advice or guidance would be much appreciated.

Thanks in advance,
Nick

SOLUTION 1 Pore water
    temp      25
    pH        7.5
    pe        4
    redox     pe
    units     mg/l
    density   1
    C(4)      20
    Ca        500
    Mg        2120
    Mn(2)     80
    Na        175
    S(6)      9880
    -water    1 # kg

EQUILIBRIUM_PHASES 1 Raise pH with Lime
    Birnessite 0 0
    Bixbyite  0 0
    Brucite   0 0
    Calcite   0 0
    Dolomite(disordered) 0 0
    Dolomite(ordered) 0 0
    Fix_H+    -7.5 portlandite 1
    Gypsum    0 0
    Hausmannite 0 0
    Magnesite 0 0
    Manganite 0 0
    Mg(OH)2(active) 0 0
    Nsutite   0 0
    Pyrochroite 0 0
    Pyrolusite 0 0
    Rhodochrosite 0 0

#PHASES
#Permanganate
#    KMnO4 = K+ + MnO4-
#    log_k     -20.21
#    delta_h   -121.692 kJ

REACTION 1
    KMnO4
    0.0005 moles in 50 steps
   
PHASES
Fix_H+
    H+ = H+
    log_k     0
#SELECTED_OUTPUT
#        -file   Permanganate-Lime treatment.sel
#        -pH
#        -EQUILIBRIUM_PHASES Gypsum Pyrolusite Rhodochrosite Bixbyite Dolomite(ordered) Hausmannite
    # -Molalities Mn+2, Mn+3, Mn+7, Mn+6, Ca, S-2, S+6, Mg, C+4, Na,
    # -Totals Ca Na Mg S Mn
    # -reaction
       # -Solution

[dlparkhurst]

There are examples and documentation in the manual. Tony Appelo's web site also has many examples (http://hydrochemistry.eu/). Here is an example of calcite kinetics. Some relevant Basic functions are TOT, MOL, ACT.



DATABASE ../database/phreeqc.dat
RATES
########
#Calcite
########
# Example of KINETICS data block for calcite rate,
#   in mmol/cm2/s, Plummer et al., 1978, AJS 278, 179; Appelo et al., AG 13, 257.
# KINETICS 1
# Calcite
# -tol   1e-8
# -m0    3.e-3
# -m     3.e-3
# -parms 1.67e5   0.6  # cm^2/mol calcite, exp factor
# -time  1 day

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 )
70  t = 1
80  IF M0 > 0 THEN t = M/M0
90  IF t = 0 THEN t = 1
100 area = PARM(1) * M0 * t^PARM(2)
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

SOLUTION 1
   C   1 CO2(g) -2.0
END
USE solution 1
KINETICS 1
Calcite
   -tol   1e-8
   -m0    3.e-3
   -m     3.e-3
   -parms 1.67e5   0.6  # cm^2/mol calcite, exp factor
   -time  0.1 day
END

[Nick]

Thank you for your quick reply.

I have made a few more attempts to model the oxidation rate and still don't seem to be getting the answer I expect.

for simplicity I am currently trying to model the removal of manganese from solution using first order kinetics [Mn]*exp(-a*Time). When I calculate the manganese concentrations using this equation I get the results I would expect (exponential decay of Mn concentration in solution). However, In PHREEQC I see an initial drop in Mn concentration for the first time step and then it remains constant for the remaining time steps. Once I get this working I will then improve the model to include kinetic equations reported in literature.

Model is as follows:

Any advice would be much appreciated.

SOLUTION 1 Pore water
    temp      25
    pH        7.5
    pe        4
    redox     pe
    units     mg/l
    density   1
    C(4)      20
    Ca        500
    Mg        2120
    Mn(2)     80
    Na        175
    S(6)      9880
    -water    1 # kg

  RATES
        Mn oxidation
     -start
     10 a = parm(1)
    # 20 k1 = parm(2)
    # 30 k2 = param(3)
     20 Mn = TOT("Mn(2)")
     #40 Ox = TOT("MnO4-")
     #40 OH = TOT("OH-")
     #50 MnO2 = EQUI("Pyrolusite")    
     30 mole = Mn*EXP(-a*Time)
     40 save moles
      -end

    KINETICS 1
       Mn oxidation   
        -formula   KMnO4  0.1
        -m        0.1
        -m0       0.1
        -parms    1
        -tol      1e-008
        -Time_step  300 seconds in 20 # seconds
        -step_divide 1
       -runge_kutta 3
       -bad_step_max 500
     INCREMENTAL_REACTIONS True

PHASES
Fix_H+
    H+ = H+
    log_k     0

Permanganate
    KMnO4 = K+ + MnO4-
    log_k     -20.21
    delta_h   -121.692 kJ

EQUILIBRIUM_PHASES 1 Raise pH with Lime
    Birnessite 0 0
    Bixbyite  0 0
    Brucite   0 0
    Calcite   0 0
    #Dolomite(disordered) 0 0
    #Dolomite(ordered) 0 0
    Fix_H+    -7.5 portlandite 1
    Gypsum    0 0
    Hausmannite 0 0
    Magnesite 0 0
    Manganite 0 0
    #Mg(OH)2(active) 0 0
    #Nsutite   0 0
    #Pyrochroite 0 0
    Pyrolusite 0 0
    Rhodochrosite 0 0
    #KMnO4 0 0.1

#REACTION 1
#    KMnO4
#    0.1 0
   
USER_GRAPH 1
 -headings
 -chart_title "Mn in solution vs Permanganate dose"
 -axis_titles "Time" "Mn (mg/L)"
 -axis_scale x_axis  0 180
 -axis_scale y_axis 0   80
 -connect_simulations true
 -initial_solutions true
 -start
 20 graph_X  total_time
 30 graph_y  TOT("Mn")*54.9*1000

END

[dlparkhurst]

First, thermodynamically, MnO7- is unstable relative to O2(aq) and Mn+2 using the wateq4f.dat databasee. You can either define the redox states as separate elements (there is an example in the manual), or I have made O2(aq) less stable, allowing Mn(7) to persist (SOLUTION_SPECIES).

You said you wanted Mn+2 to oxidize, so I have added O2 as an oxidant (assuming O2(aq) is nonreactive), so that as O2 is added, Mn(2) oxidizes to the higher redox states 3, 6, and 7. However, Mn(7) is dominant. The time frame is arbitrarily in the range of 10 days.

SOLUTION_SPECIES
2H2O = O2 + 4H+ + 4e-
#   log_k           -86.08
    log_k     -100
    delta_h   134.79 kcal

SOLUTION 1 Pore water
    temp      25
    pH        7.5
    pe        4   
    redox     pe
    units     mg/l
    density   1
    C(4)      20
    Ca        500
    Mg        2120
    Mn(2)     80
    Na        175
    S(6)      9880
    -water    1 # kg

RATES
    Mn
-start
10 k = parm(1) / (60*60*24)
20 Mn = TOT("Mn(2)")
30 rate = k * Mn
40 save rate * TIME
-end

KINETICS 1
Mn
    -formula  O2  2
    -m        100
    -m0       100
    -parms    0.5
    -tol      1e-08
-steps       864000 in 10 steps # seconds

INCREMENTAL_REACTIONS True

USER_GRAPH 1
    -headings               Days Mn(2) Mn(3) Mn(6) Mn(7)
    -axis_titles            "Time, in days" "Mn (mg/L)" ""
    -chart_title            "Mn in solution"
    -axis_scale x_axis      0 10 auto auto
    -axis_scale y_axis      0 80 auto auto
    -initial_solutions      true
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
20 graph_X  total_time / 86400
30 graph_y  TOT("Mn(2)") * 54.9 * 1000
40 graph_y  TOT("Mn(3)") * 54.9 * 1000
50 graph_y  TOT("Mn(6)") * 54.9 * 1000
60 graph_y  TOT("Mn(7)") * 54.9 * 1000
  -end
    -active                 true

END

[Nick]

Thanks for this, I really appreciate the help and I now have a much better understanding of how the kinetic key word works.

Thanks again.
Nick