Processes > Oxidation and reduction equilibria

Temperature dependent speciation calculation problem

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Leo:
Hi,

doing some tests regarding another problem I have in PHAST, I set up a simple PHREEQC model:


--- Code: ---SOLUTION 1

-pH 7.1281273
-pe 11.775556
-temp 10
-units mol/l
C                              0.005106919
Ca                            0.00260352
Cl                            0.000912093
Fe                            1.89319E-05
K                              0.000173173
Mg                            0.000348927
N                              0.001102245
Na                            0.000754766
S                              0.000625511
--- End code ---

This results in very low N(5) species concentrations. If I change the temperature in above initial solution to 25C, the N(5) concentration is noticeably higher.

Then, I used

--- Code: ---REACTION_TEMPERATURE 1
    10 25 in 7 steps
--- End code ---
to model the temperature effect on N speciation.

However, the N speciation seems to solely depend on the initial SOLUTION 1, not on the changed temperatures due to REACTION_TEMPERATURE. I also tried using pH ... charge, because I got the idea from another thread in this forum, that this could have to to with charge imbalances. Here, the problem persists.

Attached, you will find a file which runs all four tests (10C, 25C; 10C + pH charge; 25C + pH charge) with REACTION_TEMPERATURE (10...25C) and creates a selected_output file. I used wateq4f.dat for the calculations.

Why do N species behave differently, only depending on the initial solution's temperature?

Thanks,
Leo

dlparkhurst:
Let's simplify and consider only N(5) and N(0). The relevant reaction is

        2 NO3- + 12 H+ + 10 e- = N2 + 6 H2O

Ignoring the charge balance calculations, changing temperature of the initial solution changes the log K for this reaction, but the pH and pe are fixed. Consequently, the N2/NO3- ratio must adjust, depending on the temperature dependence of log K.

Now, when you use REACTION_TEMPERATURE, you start with a ratio of N2/NO3-, depending on the temperature you use in the initial solution calculation. pe and pH are no longer fixed; they adjust to achieve a charge balance (or imbalance) and redox balance. With no reductant or oxidant added, the amounts of the two redox states do not change. Hence, the same N2/NO3- ratio, while pH and pe vary.


--- Code: ---USER_GRAPH 1
    -headings               TC N(5) N(0) pH pe
    -axis_titles            "Temperature" "Log molality" "pH and pe"
    -initial_solutions      true
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
10 GRAPH_X TC
20 GRAPH_Y LOG10(TOT("N(5)"))
40 GRAPH_Y LOG10(TOT("N(0)"))
90 GRAPH_SY -LA("H+")
100 GRAPH_SY -LA("e-")
SOLUTION 1
-pH 7.1281273
-pe 11.775556
-temp 10
-units mol/l
Cl     .001 charge
N                              0.001102245
END
SOLUTION 2
-pH 7.1281273
-pe 11.775556
-temp 15
-units mol/l
Na     .001 charge
N                              0.001102245
END
SOLUTION 2
-pH 7.1281273
-pe 11.775556
-temp 20
-units mol/l
Na     .001 charge
N                              0.001102245
END
SOLUTION 2
-pH 7.1281273
-pe 11.775556
-temp 25
-units mol/l
Na     .001 charge
N                              0.001102245
END
SOLUTION 2
-pH 7.1281273
-pe 11.775556
-temp 30
-units mol/l
Na     .001 charge
N                              0.001102245
END
USER_GRAPH 1
-detach
USER_GRAPH 2
    -headings               TC N(5) N(0) pH pe
    -axis_titles            "Temperature" "Log molality" "pH and pe"
    -initial_solutions      true
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
 10 GRAPH_X TC
 20 GRAPH_Y LOG10(TOT("N(5)"))
 40 GRAPH_Y LOG10(TOT("N(0)"))
 90 GRAPH_SY -LA("H+")
100 GRAPH_SY -LA("e-")
  -end
    -active                 true
END
USE solution 1
REACTION_TEMPERATURE
10 30 in 5
END


--- End code ---

Leo:
Thank you, David. The graphs and your explanation were quite insightful. However, I have two follow-up questions:

1) in the second case (REACTION_TEMPERATURE): the K should still be temperature-dependent, or am I wrong here? Why doesn't the equilibrium NO3- <> N2 shift accordingly, or, why dies it only shift for pH and pe?

2) what would happen in "reality"? If I had a solution containing Nitrate and, let's say, Sodium at certain concentrations at a specific temperature (10C) in a closed system, if I were heating the solution to 25C, what changes in water composition should I be expecting? I assume that both pH/pe and N species distribution change slightly since all K are temperature dependent and the reactions are interlinked (H+).

dlparkhurst:
If NO3- is reduced to N2(aq), then something must be oxidized. In the pure nitrogen system, there is only H2O that would be oxidized to O2. That only happens to a limited extent under these conditions. The following script expands the temperature range, and you can see a little increase in O2, but an imperceptible decrease in NO3-.


--- Code: ---USER_GRAPH 2
    -headings               TC N(5) N(0) O(0) pH pe
    -axis_titles            "Temperature" "Log molality" "pH and pe"
    -initial_solutions      false
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
 10 GRAPH_X TC
 20 GRAPH_Y LOG10(TOT("N(5)"))
 30 GRAPH_Y LOG10(TOT("N(0)"))
 60 GRAPH_Y LOG10(TOT("O(0)"))
 90 GRAPH_SY -LA("H+")
100 GRAPH_SY -LA("e-")
  -end
    -active                 true
END
SOLUTION 1
-pH 7.1281273
-pe 11.775556
-temp 10
-units mol/l
Cl    0.001 charge
N     0.001102245
REACTION_TEMPERATURE
10 300 in 5
END
--- End code ---

Reality is something else. Thermodynamically, the atmosphere should be nitric acid. In natural system, NO3- is stable under oxygenated conditions. If there are reductants available (organic matter) O2 will be reduced to H2O and NO3- will be reduced to N2 as carbon is oxidized. N2 is pretty inert once it forms, so natural systems show NO3- reduction to N2, but little or no reduction of N2 to NH3.

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