Hydrogen solubility in distilled water vs pressure

[MMMM]

Hi everyone
I'm trying to simulate hydrogen solubility in distilled water versus temp and pressure. I've written below code but I think it doesnt work correctly. I assume 1 mol hydrogen is contacted with 1 kg water so brine will fully saturated with hydrogen. According to literature, solubility of H2 increases with pressure but the curve generated with this script shows that solubility of H2 decreases with pressure.
I would appreciate it if someone could help me.

Code: [Select]

[TITLE     Solubility of Hydrogen in DistilledWater
SOLUTION  1     

EQUILIBRIUM_PHASES  1
H2(g)   0  1

###################################################################
USE SOLUTION 1
USE EQUILIBRIUM_PHASES 1

REACTION_PRESSURE 11
0.987  987 in 6 steps
REACTION_TEMPERATURE 2
30

USER_GRAPH 3
-headings    temp 30C
-axis_titles "Pressure atm"  "H2-dissolution, in mole/kgW"
-axis_scale x_axis auto
-axis_scale y_axis auto   
 -start
 10 graph_x Pressure 
 20 graph_y TOT("H(0)")
 #30 graph_sy TOTMOLE("H(0)")
 -end
END

USE SOLUTION 1
USE EQUILIBRIUM_PHASES 1
REACTION_PRESSURE 11
0.987  987 in 6 steps
REACTION_TEMPERATURE 3
60
USER_GRAPH 3
-headings    temp 60C
END

USE SOLUTION 1
USE EQUILIBRIUM_PHASES 1
REACTION_PRESSURE 11
0.987  987 in 6 steps
REACTION_TEMPERATURE 4
90
USER_GRAPH 3
-headings    temp 90C
END

USE SOLUTION 1
USE EQUILIBRIUM_PHASES 1
REACTION_PRESSURE 11
0.987  987 in 6 steps
REACTION_TEMPERATURE 5
120
USER_GRAPH 3
-headings    temp 120C
END

USE SOLUTION 1
USE EQUILIBRIUM_PHASES 1
REACTION_PRESSURE 11
0.987  987 in 6 steps
REACTION_TEMPERATURE 6
150
USER_GRAPH 3
-headings    temp  150C
END

USE SOLUTION 1
USE EQUILIBRIUM_PHASES 1
REACTION_PRESSURE 11
0.987  987 in 6 steps
REACTION_TEMPERATURE 7
180
USER_GRAPH 3
-headings    temp 180C
END

USE SOLUTION 1
USE EQUILIBRIUM_PHASES 1
REACTION_PRESSURE 11
0.987  987 in 6 steps
REACTION_TEMPERATURE 8
210
USER_GRAPH 3
-headings     temp 210C
END

USE SOLUTION 1
REACTION_PRESSURE 11
0.987  987 in 6 steps
REACTION_TEMPERATURE 8
210
USER_GRAPH 3
-headings     temp 210C
END/code]

[dlparkhurst]

You have specified a partial pressure of 1 atm of H2(g) for all of your calculations, regardless of total pressure. I think you intended the H2(g) partial pressure to be equal to the total pressure.

You could specify a partial pressure equal to the total pressure for each of your data points, but that would be tiresome.

Here is a calculation using GAS_PHASE that lets a gas bubble equilibrate with the solution art the specified temperatures and pressures. I have used Hdg(g), which is defined in phreeqc.dat and pitzer.dat, and represents H2(g) and H2(aq), but avoids any reactions between H2 and any other dissolved or gaseous constituents. I have taken the extra step to consider that the gas phase would contain some H2O(g) in addition to H2(g). However, you must be above the steam curve for water to retain a liquid phase, so as the temperature increases, the minimum pressure in REACTION_PRESSURE increases. If you don't want to consider H2O(g), you can remove it from the GAS_PHASE definition.

Code: [Select]

TITLE     Solubility of Hydrogen in DistilledWater
SOLUTION  1     
END
GAS_PHASE 1
-fixed_pressure
Hdg(g) 1e2
H2O(g) 0
END
REACTION_PRESSURE 1
#0.987  987 in 6 steps
.987 2 5 10 20 50 100 200 500 1000
END
###################################################################
USE solution  1
USE gas_phase 1
USE reaction_pressure 1
REACTION_TEMPERATURE 1
30

USER_GRAPH 3
-headings    temp 30C
-axis_titles "Pressure atm"  "H2-dissolution, in mole/kgW"
-axis_scale x_axis auto auto auto auto #log
-axis_scale y_axis auto auto auto auto #log
 -start
 10 graph_x Pressure
 20 graph_y MOL("Hdg")
 -end
END
USE solution  1
USE gas_phase 1
REACTION_PRESSURE 1
.987 2 5 10 20 50 100 200 500 1000
REACTION_TEMPERATURE 1
60
USER_GRAPH 3
-headings    temp 60C
END

USE solution  1
USE gas_phase 1
REACTION_PRESSURE 1
.987 2 5 10 20 50 100 200 500 1000
REACTION_TEMPERATURE 1
90
USER_GRAPH 3
-headings    temp 90C
END
KNOBS
-step 10
-pe 2
-iter 400
USE solution  1
USE gas_phase 1
REACTION_PRESSURE 1
2 5 10 20 50 100 200 500 1000
REACTION_TEMPERATURE 1
120
USER_GRAPH 3
-headings    temp 120C
END
USE solution  1
USE gas_phase 1
REACTION_PRESSURE 1
5 10 20 50 100 200 500 1000
REACTION_TEMPERATURE 1
150
USER_GRAPH 3
-headings    temp 150C
END
USE solution  1
USE gas_phase 1
REACTION_PRESSURE 1
10 20 50 100 200 500 1000
REACTION_TEMPERATURE 1
180
USER_GRAPH 3
-headings    temp 180C
END
USE solution  1
USE gas_phase 1
REACTION_PRESSURE 1
20 50 100 200 500 1000
REACTION_TEMPERATURE 1
210
USER_GRAPH 3
-headings    temp 210C
END

[MMMM]

I very much appreciate your help.

[MMMM]

I have another question.
I want to add calcite mineral and calculate the hydrogen gas loss due to reaction and/or dissolution. I added 100 moles of H2 gas, but only 60 moles of gas entered. When I plot the number of moles of H2 gas, there are only 60 moles at the initial time. I cannot understand where the other 40 moles are.
For now, I want to do the calculations assuming equilibrium without considering the reaction rates.
Please help me understand this.
Many thanks.

Code: [Select]

[SOLUTION 1
end
EQUILIBRIUM_PHASES  1
Calcite     0 10     
Save Solution 1
Save Equilibrium_Phases 1
END
Use Solution 1
Use Equilibrium_Phases 1
GAS_PHASE 1
     -fixed_pressure
     -pressure 1
     -temperature 20
CO2(g) 0
CH4(g) 0
H2(g) 0
H2S(g) 0
H2O(g) 0
REACTION 1
H2 1
100 in 50 steps


USER_GRAPH 1
    -headings     rxn Hydrogen Gas mols
    -chart_title "H2 gas"
    -axis_titles "H2 added, mol", "Present H2(g), mol"
    -axis_scale x_axis auto auto auto aut
    -axis_scale y_axis auto auto auto aut
    -start
   
10 GRAPH_X RXN
20 GRAPH_Y gas("H2(g)")
/code]

[dlparkhurst]

Look at the output file. I did warn you in a previous post.

Code: [Select]

4H2 + CaCO3 = Ca+2 + 2OH- + CH4 + H2O

You have 10 moles of calcite in your system, so you end up with 10 moles of CH4 and consume 40 moles of H2.

[MMMM]

Thank you so much for your help.