Kinetic Oxidation of Cr (VI) to Cr(III) by Fe0

[Rozeran]

Hello,

The removal rate for Cr(VI) (kinetically controlled) is given by the following formula:
dC/dt = -kobs*C
where C is the conc. of Cr(VI)
kobs is taken as 6.5

I checked Example 9 in the manual. There, Fe(2) oxidation to Fe(3) is studied and valence states of iron are decoupled. I tried to do the same with Cr(VI) and Cr(III), with above rate equation. My code is below. The equilibrium reactions for Cr is given in minteq and minteqv4 database. So checked those databases to complete the SOLUTION_SPECIES section.

[TITLE Cr VI Oxidation to Cr III
SOLUTION_MASTER_SPECIES
Cr_six   Cr_sixO4-2      0.0   Cr_six   51.996
Cr_tri   Cr_tri(OH)+2   0.0   Cr_tri   51.996

SOLUTION_SPECIES
Cr_sixO4-2 = Cr_sixO4-2
        log_k   0
        delta_h 0       kcal
        -gamma  4.0  0.0
Cr_sixO4-2 + H+ = HCr_sixO4-
        log_k   6.5089
        delta_h 0.9     kcal
Cr_sixO4-2 + 2H+ = H2Cr_sixO4
        log_k   5.6513
        delta_h 0       kcal
2Cr_sixO4-2 + 2H+ = Cr_six2O7-2 + H2O
        log_k   14.5571
        delta_h -2.995  kcal
#Cr_tri(OH)2+ + 2H+ + e- = Cr_two + 2H2O
        #log_k   2.947
        #delta_h 6.36    kcal
Cr_sixO4-2 + 6H+ + 3e- = Cr_tri(OH)2+ + 2H2O
        log_k   67.376
        delta_h -103    kcal
Cr_tri(OH)2+ + 2H+ = Cr_tri+3 + 2H2O
        log_k   9.62
        delta_h -20.14  kcal
Cr_tri(OH)2+ + H+ = Cr_tri(OH)+2 + H2O
        log_k   5.62
        delta_h 0       kcal
Cr_tri(OH)2+ + H2O = Cr_tri(OH)3 + H+
        log_k   -7.13
        delta_h 0       kcal
Cr_tri(OH)2+ + 2H2O = Cr_tri(OH)4- + 2H+
        log_k   -18.15
        delta_h 0       kcal
Cr_tri(OH)2+ = Cr_triO2- + 2H+
        log_k   -17.7456
        delta_h 0       kcal

PHASES
Cr_tri(OH)3
        Cr_tri(OH)3 + H+ = Cr_tri(OH)2+ + H2O
        log_k   1.7005
        delta_h -7.115  kcal
END
SOLUTION 1
        pH  7.0
        Cr_six  0.1
        Na  10.
        Cl  10.  charge
RATES
Cr_six_ox
  -start
  10  Cr_six = TOT("Cr_six") # Total molality of Cr_six
  20  if (Cr_six <= 0) then goto 200 #the amount of Cr_six is less than zero i.e. not enough
  30  kobs = 6.5
  40  moles = kobs* Cr_six * TIME #rate expression
  200 SAVE moles
  -end

KINETICS 1
Cr_six_ox
        -formula  Cr_six  -1.0  Cr_tri  1.0
        -steps 100 400 3100 10800 21600 5.04e4 8.64e4 1.728e5 1.728e5 1.728e5 1.728e5
        -step_divide 1e-8

INCREMENTAL_REACTIONS true

SELECTED_OUTPUT
        -file CrOX.sel
        -reset false
USER_PUNCH
        -headings Days  Cr(6)  Cr(3)  pH  si_Chromiumhydroxide
  10 PUNCH SIM_TIME, TOT("Cr_six")*1e6, TOT("Cr_tri")*1e6, -LA("H+"), \
           SI("Cr_tri(OH)3")
END]

The code didn't work. The errors include:
ERROR: Equation does not balance for element, Cr_six: right - left =  1.0000 moles
ERROR: Equation does not balance for element, Cr_tri: right - left = -1.0000 moles
ERROR: Equation for species Cr_tri(OH)2+ does not balance.
ERROR: Calculations terminating due to input errors.

I cross-checked the equations and could not find where it is not balancing. Basically the problem should be within SOLUTION_SPECIES. While going through databases, I selected the equations with reactants present in my solution composition. Thus I didn't include reactions taking place between Cr and Br for example.

Do you see where the problem is here? Thanks in advance,

[dlparkhurst]

Minteq.dat and Minteq.v4.dat have Cr(OH)2+ as the master species, while you have Cr(OH)+2. The following script works with the minteq master species and reactions.

Your decay rate is 6.5 1/s, so Cr_six is removed in less than a second.

Code: [Select]

SOLUTION_MASTER_SPECIES
Cr_six   Cr_sixO4-2      0.0   Cr_six   51.996
Cr_tri   Cr_tri(OH)2+    0.0   Cr_tri   51.996

SOLUTION_SPECIES
Cr_sixO4-2 = Cr_sixO4-2
        log_k   0
        delta_h 0       kcal
        -gamma  4.0  0.0
Cr_sixO4-2 + H+ = HCr_sixO4-
        log_k   6.5089
        delta_h 0.9     kcal
Cr_sixO4-2 + 2H+ = H2Cr_sixO4
        log_k   5.6513
        delta_h 0       kcal
2Cr_sixO4-2 + 2H+ = Cr_six2O7-2 + H2O
        log_k   14.5571
        delta_h -2.995  kcal
#Cr_tri(OH)2+ + 2H+ + e- = Cr_two + 2H2O
#       log_k   2.947
#       delta_h 6.36    kcal
#Cr_sixO4-2 + 6H+ + 3e- = Cr_tri(OH)2+ + 2H2O
#        log_k   67.376
#        delta_h -103    kcal
Cr_tri(OH)2+ = Cr_tri(OH)2+   
        log_k 0.0
Cr_tri(OH)2+ + 2H+ = Cr_tri+3 + 2H2O
        log_k   9.62
        delta_h -20.14  kcal
Cr_tri(OH)2+ + H+ = Cr_tri(OH)+2 + H2O
        log_k   5.62
        delta_h 0       kcal
Cr_tri(OH)2+ + H2O = Cr_tri(OH)3 + H+
        log_k   -7.13
        delta_h 0       kcal
Cr_tri(OH)2+ + 2H2O = Cr_tri(OH)4- + 2H+
        log_k   -18.15
        delta_h 0       kcal
Cr_tri(OH)2+ = Cr_triO2- + 2H+
        log_k   -17.7456
        delta_h 0       kcal

PHASES
Cr_tri(OH)3
        Cr_tri(OH)3 + H+ = Cr_tri(OH)2+ + H2O
        log_k   1.7005
        delta_h -7.115  kcal
END
SOLUTION 1
        pH  7.0
        Cr_six  0.1
        Na  10.
        Cl  10.  charge
RATES
Cr_six_ox
  -start
  10  Cr_six = TOT("Cr_six") # Total molality of Cr_six
  20  if (Cr_six <= 0) then goto 200 #the amount of Cr_six is less than zero i.e. not enough
  30  kobs = 6.5
  40  moles = kobs* Cr_six * TIME #rate expression
  200 SAVE moles
  -end
KINETICS 1
Cr_six_ox
        -formula  Cr_six  -1.0  Cr_tri  1.0
        -steps 1
INCREMENTAL_REACTIONS true
END


[Rozeran]

Thank you so much @dlparkhurst!

Master Specie issue must be a typo, I revised it.

And I see that you eliminated (#) the equation:
#Cr_sixO4-2 + 6H+ + 3e- = Cr_tri(OH)2+ + 2H2O
#        log_k   67.376
#        delta_h -103    kcal

When this equation is included, it also worked.

[dlparkhurst]

You do not want to include that reaction. The point of defining Cr_six and Cr_tri is to split the two redox states. You do not want any SOLUTION_SPECIES reactions that connect the two "elements". You want the only way to convert Cr_six to Cr_tri to be with your KINETICS reaction.

[Rozeran]

Okay, now it is clear.

Thank you