Conceptual Models > Design of conceptual models

Selected moles for EQUILIBRIUM_PHASES

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Tom:
A topic posted in the FAQs board on water to rock ratio as defined in PHREEQC, raises a number of questions worth considering when specifying the number of moles of EQUILIBRIUM_PHASES. We encourage users to comment with their interpretations.

See the original post in the FAQ section: http://phreeqcusers.org/forum/index.php?topic=30.0

The following questions may be considered:
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Q. I have calculated the mass of rock (pure limestone) in contact with 1 litre of water. Assuming a porosity of 30% and dry density of 2.7 kg/L (calcite, CaCO3), I have a rock mass of 6.3 kg (as in the above example). What is the effect of using the following EQUILIBRIUM_PHASES data blocks?

SOLUTION 1
-water 1 kg

EQUILIBRIUM_PHASES 1
Calcite 0 10


SOLUTION 2
-water 1 kg

EQUILIBRIUM_PHASES 2
Calcite 0 100

Comments:
In the first case, 10 moles of calcite equates to approximately 1 kg of calcite (molecular mass 100.09 g/mol). In this case, the mass of calcite available is less than the rock mass (only about 15 % of the mass).

In the second case, 100 moles equates to 10 kg, around 4 kg more than the calculated rock mass.

How do these extremes effect the conceptual model? It does not appear consistent to specify more or less moles of a phase than is possible in accordance with 1 litre of water and the calculated rock mass.

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Q. Considering the above, how should I go about calculating the proportions of minerals in a given rock mass?

For example, I receive bulk-rock XRD analysis from the lab which states my sample has the following composition:

50 % calcite
10 % pyrite
20 % illite
20 % quartz

The lab also states the average dry density of the sample is 2.9 kg/L. The porosity is 30 %. I calculate the saturated rock mass using the equations in the above post at 6.8 kg.

To accurately model the system, should I calculate the moles of each phase for EQUILIBRIUM_PHASES as a percentage of this rock mass?

I.e., (rough mole calculations):

Calcite = 50 % of 6.8 kg = 3.4 kg. Moles = 3400/100.09 = 34 moles
Pyrite = 10 % of 6.8 kg = 0.68 kg. Moles = 680/119.98 = 5.7 moles
Illite = 20 % of 6.8 kg = 1.4 kg. Moles = 1400/389.34 = 3.5 moles
Quartz = 20 % of 6.8 kg = 1.4 kg. Moles = 1400/60.08 = 23 moles

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Any comments and thoughts on the above appreciated.

dlparkhurst:
Assuming your solution is 1 kg water and about 1 L in volume, the outline calculation from wt percent to moles is reasonable. However, solubility of these minerals is on the order of millimoles, so it takes many, many pore volumes of water to make a significant change in the moles of minerals. For most calculations, a mole of mineral is likely to be more than enough to maintain equilibrium through most calculations. Evaporite minerals are a different case because their solubilities may be moles per liter.

Most solutions are about 1 L in volume when defined with SOLUTION; however, it is possible to have different volumes, SOLUTION; -water 0.5 would define a solution of about 0.5 L. A different volume would require adjustment of the moles of minerals to maintain a similar mineral mole to water volume ratio.

JoniDH:
How was the rock mass calculated here? (the link to the original post is broken)
Assuming a porosity of 30% and a dry density of 2.7 kg/L I would expect a rock mass of 1/0.3*2.9=9.6 kg and not 6.8 as stated.

Similarly I don't arrive to the same mineral concentrations in the example 14 "Example 14—Advective Transport, Cation Exchange, Surface Complexation,
and Mineral Equilibria" pg 366 in the phreeqc version 3 manual.
Here, a porosity of 0.22 and a rock density of 2.7 kg/l would yield a rock mass of 2.7*1/0.22=12.27 kg

For calcite (gfw 100 g/mol): 0.1% gewichtsprocent would then be 0.001*12.27*1000/100=0.12mol/L (could be correct)
Voor dolomiet (gfw 184.41 g/mol): 3 %; 0.03*12.27*1000/1854.41= 1.9 mol/L (vs 1.6 in de manual)

What am I overlooking here?

dlparkhurst:
> Assuming a porosity of 30% and a dry density of 2.7 kg/L I would expect a rock mass of 1/0.3*2.9=9.6 kg and not 6.8 as stated.

0.7*2.7/0.3 = 6.3 
(L rock) * (kg rock / L rock) / (L water) = (kg rock) / (L water)

> For dolomite
0.78*2.7 * 1000 * 0.03 / 184.41 / 0.22   = 1.56
(L rock) * (kg rock/L rock) * (g rock/kg rock) * (g dolomite/g rock) / (g dolomite/mol dolomite) / (L water) = (mol dolomite) / (L water)

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