Beginners > PHREEQC manual examples
Gas solubility in a solution
dat:
Hi,
I am new to PHREEQC and I want to check the solubility of gasses (CO2,H2) in solutions. For the first time I used the H20 solution.
In the example 22, there is a CO2 solubility graph, but in the output file I couldn't find those values. For example, the co2 solubility at 100 atm is 1.4mol/kgw in the graph and in the output file that value is not there. But the correct value is graphical value according to the literature.
Could you please assist me with this? Really appreciate your help.
I hereby attach the code in the manual.
--- Code: ---[code]
GAS_PHASE 1
-fixed_volume
CO2(g) 0
SOLUTION 1
REACTION
CO2 1; 26*1
INCREMENTAL_REACTIONS true
USER_GRAPH
-axis_titles "CO2 Pressure / atm" "CO2 / (mol/kgw)"
-chart_title "CO2 solubility at high CO2 pressure"
-initial_solutions true
10 DATA 10, 25, 35, 50, 50, 75, 100, 100, 135, 150, 175, 200, 203, 250, 300, 400
12 DIM dx(16) : RESTORE 10
20 FOR i = 1 to 16 : READ dx(i) : NEXT i
30 DATA 0.3, 0.707, 0.936, 1.21, 1.22, 1.39, 1.42, 1.43, 1.47,\
1.49, 1.53, 1.52, 1.56, 1.58, 1.62, 1.72
32 DIM dy(16) : RESTORE 30
40 FOR i = 1 to 16 : READ dy(i) : NEXT i
50 PLOT_XY PR_P("CO2(g)"), tot("C(4)"), color = Red, symbol = None
60 IF STEP_NO < 16 THEN PLOT_XY dx(STEP_NO + 1), dy(STEP_NO + 1), color = Red, symbol = Square , line_width = 0
END
--- End code ---
dlparkhurst:
The script has experimental data defined in the DATA statements and then plots calculated pressures and solubility by adding increments of CO2 to the system.
The closest point in the output is at 95 atm, with solubility 1.4 mol/kgw, using the phreeqc.dat database. If you add CO2 in REACTION in smaller increments, you will get pressures closer to 100 atm.
--- Code: ---Total pressure: 95.42 atmospheres (Peng-Robinson calculation)
Gas volume: 1.00e+00 liters
Molar volume: 5.69e-02 liters/mole
P * Vm / RT: 0.22207 (Compressibility Factor Z)
Moles in gas
----------------------------------
Component log P P phi Initial Final Delta
CO2(g) 1.98 9.542e+01 0.481 1.658e+01 1.756e+01 9.791e-01
-----------------------------Solution composition------------------------------
Elements Molality Moles
C 1.438e+00 1.438e+00
--- End code ---
The following uses a fixed-pressure GAS_PHASE to determine the solubility of CO2 at 100 atm.
--- Code: ---SOLUTION 1
-temp 25
-pressure 100
GAS_PHASE 1
-fixed_pressure
-pressure 100
CO2(g) 100
END
--- End code ---
dat:
Hi,
Thank you for the reply and that helps. I just have some more questions, if you don't mind.
1. When finding the solubility do we have to get the C value in the solution composition or molality of CO2 in the distribution of species?
--- Code: --- Moles in gas
----------------------------------
Component log P P phi Initial Final Delta
CO2(g) 1.98 9.542e+01 0.481 1.658e+01 1.756e+01 9.791e-01
-----------------------------Solution composition------------------------------
Elements Molality Moles
C 1.438e+00 1.438e+00
--- End code ---
--- Code: ---
----------------------------Distribution of species----------------------------
Log Log Log mole V
Species Molality Activity Molality Activity Gamma cm³/mol
H+ 8.392e-04 8.134e-04 -3.076 -3.090 -0.014 0.00
OH- 1.365e-11 1.321e-11 -10.865 -10.879 -0.014 -3.77
H2O 5.551e+01 9.761e-01 1.744 -0.010 0.000 17.99
C(-4) 9.281e-18
CH4 9.281e-18 9.283e-18 -17.032 -17.032 0.000 35.58
C(4) 1.438e+00
CO2 1.368e+00 1.369e+00 0.136 0.136 0.000 34.39
(CO2)2 3.437e-02 3.438e-02 -1.464 -1.464 0.000 68.78
HCO3- 8.392e-04 8.126e-04 -3.076 -3.090 -0.014 24.77
CO3-2 5.965e-11 5.244e-11 -10.224 -10.280 -0.056 -4.43
--- End code ---
2. When I tried to find the solubility of H2 it doesn't give me a value in solution composition. Why is that ?
--- Code: ---GAS_PHASE 1
-fixed_volume
-pressure 1
-volume 1
-temperature 25
H2(g) 0
SOLUTION 1
REACTION 1
H2(g) 1;26*1
INCREMENTAL_REACTIONS true
--- End code ---
Output
--- Code: --- Moles in gas
----------------------------------
Component log P P phi Initial Final Delta
H2(g) 2.00 9.935e+01 1.030 2.945e+00 3.928e+00 9.825e-01
-----------------------------Solution composition------------------------------
Elements Molality Moles
Pure water
--- End code ---
3. And in the gas phase why we are inputting the gas partial pressure as 0. Because it has some value no ?
Really appreciate your help
dlparkhurst:
1. When there is no carbon in the solution initially, it seems clear to me that the solubility is equal to the amount of carbon that dissolves into the solution, which will be equal to the total dissolved carbon and, also, the sum of all the carbon species [not just CO2(aq)].
2. "Pure water" is printed in the location where the total concentrations of elements is given. Hydrogen is not included because virtually all of the hydrogen in solution is in liquid H2O; total hydrogen is printed under "Description of Solution", but usually is not very useful. The solubility of H2 will be given under H(0) in the Distribution of Species section. I suspect you want to consider H2 as an inert gas and aqueous species, in which case, you may want to use the "element" Hdg, which is defined in phreeqc.dat and pitzer.dat. Whereas H2 will react as a reductant with other redox active elements, Hdg will not. As a bonus, if you use Hdg, it will show up in the section "Solution composition".
3. In concept, you are adding a head space of 1 L that is initially devoid of gas. As you add Hdg (or H2), it partitions between the gas phase and the liquid phase to obtain equilibrium. So, yes an initial gas phase partial pressure of zero for Hdg(g) makes sense. For your particular calculation, you could have a nonzero partial pressure of Hdg(g) in the gas phase, and as Hdg is added it would still partition between the gas and the liquid to obtain equilibrium; you would just get a different sequence of pressure and concentration that accounts for the initial amoung of Hdg in the gas phase.
dat:
Thank you for your explanation.
Actually I want H2 as a reactant. I am going to observe the H2 solubility. In that case I think I should add H2 right.
If that so I have to get the molality of H(0). Is that correct?
Thank you.
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