Dissolution of NO2

[Hamed]

Dear all,

Hope this message finds you well. I have a question regarding the dissolution of NO2 in the reservoir brine. In the reaction defined in the database, O2 is required for NO2 dissolution. My question is when we inject CO2, which has NO2 impurity, and there is no O2 in the reservoir condition, does NO2 dissolution happen, and where does the oxygen come from? I have made a model, and I don't include oxygen in it; however, I see that regardless of the presence of oxygen in the model, the NO2 dissolves completely in the brine.

Thank you in advance for your time and consideration,

Sincerely

Code: [Select]

NO2(g)
        NO2 +0.5000 H2O + 0.2500 O2  =  1.0000 H+ + 1.0000 NO3-
        -log_k           8.3673
  -delta_H -94.0124 kJ/mol #Calculated enthalpy of reaction NO2(g)
        -analytic 9.4389e+001 -2.7511e-001 -1.6783e+004 2.1127e+001 -2.6191e+002
        -T_c  431
  -P_c  99.68
  -Omega 0.8340

[dlparkhurst]

I don't think there is an NO2(aq) species in any of the databases; that is no N(4) aqueous state. In that case, dissolution of NO2 must disproportionate into a more oxidized (N(5)) and more reduced state (perhaps N(3), NO2-) when it dissolves into solution.

You can either accept that these redox reactions will occur on dissolution or perhaps what you want to do is to define a new "element" that is represented by NO2(aq). It needs to be a new element so that the redox reactions with other N species do not occur. Something like the following:

Code: [Select]

SOLUTION_MASTER_SPECIES
    [NO2]         [NO2]            0     NO2             46
SOLUTION_SPECIES
[NO2] = [NO2]
PHASES
[NO2](g)
        #[NO2] +0.5000 H2O + 0.2500 O2  =  1.0000 H+ + 1.0000 NO3-
        #-log_k           8.3673
  #-delta_H -94.0124 kJ/mol #Calculated enthalpy of reaction NO2(g)
        #-analytic 9.4389e+001 -2.7511e-001 -1.6783e+004 2.1127e+001 -2.6191e+002
[NO2] = [NO2]
        -log_k -2 # Pick a Henry's law constant
        -T_c  431
  -P_c  99.68
  -Omega 0.8340
END
SOLUTION
EQUILIBRIUM_PHASES
[NO2](g) -3 10
END

If you want the redox reactions to occur kinetically, then you will need to define RATES and KINETICS to transfer [NO2] to NO2.

[Hamed]

Thank you for the response. If we consider that the redox reaction is valid and NO2 can dissociate to NO3-, should oxygen be available in the system so that NO2 can dissolve into the solution? At underground conditions where O2 may not be available, is the dissolution of NO2 into brine likely to occur? Where is O2 coming from?

Thank you

[dlparkhurst]

It will dissolve whether oxygen is preset or not. It will dissolve more if oxygen is prent and redox reactions are allowed.

[Hamed]

Thank you. In the reactants of the reaction, oxygen is present. So, my question is, when oxygen does not exist, how is this reaction still producing NO3-, and where does the oxygen come from? From the reduction of water?

[dlparkhurst]

Code: [Select]

2NO2 + H2O = NO2- + NO3- + 2H+


[dlparkhurst]

These definitions are no different than N in Amm.dat if done correctly and will produce equilibrium reactions.

If you want NO2 I to dissolve and not react, use the definitions I gave you before.

If you want something else, you will have to come up with a concept for the reactions you want and do not want to occur.