Crude Oil as equilibrium phase

[swilke]

Hello everyone,

I have a problem and was hoping somebody might have an idea for a solution (haha). I want to model how the presence of crude oil (or gasoline or something comparable) affects the solubility of other phases, e.g. by drastically altering the redox-situation. To do so I would like to just create a SOLUTION and bring it into contact with two EQUILIBRIUM PHASES: the crude oil and the mineral to be actually solved.

The problem ist that I was so far unable to find an equilibrium phase that could act as a proxy for petroleum. I would be thankfull for suggestions.

So far I've been employing the REACTION keyword with C8H16 but that way I am forcing a more or less arbitrary amount of C-2 into the solution and end up having no idea how severe the presence of oil is actually gonna kick in.

[dlparkhurst]

First, if you add C8H16 in REACTION, the C will end up as C(4) species or C(-4) (methane), depending on the presence of electron acceptors.

Addition of oil has many aspects. It does react to some unknown extent as an electron source, causing reduction. Adding CH2O or C8H16 will accomplish reduction to the extent you add these reactants. If you want organic species not to react to C(4) and C(-4) you need to define them as separate "elements" that do not include the element C. See ligands in minteq.dat and llnl.dat.

Dissolution of oil components that do not react, will affect the activity of water (a bit) by dissolution of polar and nonpolar species. The polar species will interact with ions in the solution as ligands. minteq.dat, minteq.v4.dat, and llnl.dat have collections of ligands (not necessarily complete) with some data for formation of ion pairs and complexes.  The small change in the activity of water will probably have little effect on your mineral unless organic solubility is very high, but the ion pairing may have more of an affect on the activity of the dissolved species of the mineral, and thus the solubility.

[swilke]

What happens to the C, once it is dissolved, technically doesn't really matter. I am more interested in the redox condition created by the oxidation of the oil and the extend to which that will reduce sulfate to sufide and affect the dissolution of sulfate-minerals. I have been browsing the databases you mention, but none of the organic components listed there seem to be of the kind that makes up the bulk of a crude oil.

But then again, me not being much of an organic chemist, I might have missed something. I guess some alkanes between pentane and hexadecane would do, but this is really unknown territory for me.

[swilke]

So, I've been contemplating this all day and this is what I came up with:

PHASES
Pseudo-Octane
    C8H16 + 16H2O + 16e- = 8CH4 + 16OH-
    log_k     -195

SOLUTION 1
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mmol/kgw
    density   1
    -water    1 # kg

EQUILIBRIUM_PHASES 1
    Pseudo-Octane    0 10
    Barite    0 10
END

What I did was basically just manually vary the log_k until i was able to reproduce the solubility of octane in water at standard-conditions of 0.007 mg/l. This seems to work. I am not convinced it is wise, though.

[dlparkhurst]

I think this is closer to what you have in mind. I set the log K to produce the solubility you stated. However, the solubility of octane done in this way will have almost no effect on the solubility of barite because it is so sparingly soluble.

You will have to consider how to deal with the oxidation of octane. The most straightforward approach is to run REACTION or KINETICS with the formula

REACTION
Octane -1 C8H16 1

As Octane reacts (REACTION or KINETICS) it eventually will reduce sulfate, which will affect the amount of barite that dissolves.

Code: [Select]

SOLUTION_MASTER_SPECIES
    Octane        Octane           0     C8H16           112.22
SOLUTION_SPECIES
Octane = Octane
    log_k     0
PHASES
Octane
    Octane = Octane
    log_k     -7.205
END
SOLUTION 1
    -units mg/L
END
USER_PRINT
-start
10 PRINT "Octane, mg/kgw: ", TOT("Octane")*GFW("C8H16")*1000
-end
USE solution 1
EQUILIBRIUM_PHASES 1
    Octane    0 10
END


[swilke]

Thank you for your help, but as you said: this way sulfate remains the dominant sulfur species in the solution and barite will remain mostly insoluble. This is very unlike the effect I see from adding C8H16 via REACTION.

I would like to place this EQUILIBRIUM PHASE somewhere in a PHAST model, therefore REACTION or KINETICS would be really unhandy.

[dlparkhurst]

As long as you realize your phase has nothing to do with octane; your mineral could just as easily be CH2:

2CH2 +  2H2O = CO2 + CH4 + 4e- + 4H+

So what you are really doing is setting the following relation:

Log K = Log ([CO2]) + Log([CH4]) - 4pe - 4pH - 2Log([H2O])

That seems pretty arbitrary to me, but suit yourself.