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Question about dissolution kinetics
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Topic: Question about dissolution kinetics (Read 1587 times)
Sidol
Frequent Contributor
Posts: 11
Question about dissolution kinetics
«
on:
March 20, 2020, 03:59:20 PM »
Hello!
I am studying kinetics in PHREEQC at this moment, and I have a question about the RATES block of an albite dissolution kinetics example file (I found it in the Environmental Applications of Geochemical Modeling, Zhu and Anderson 2002, on page 240.).
Here is the code:
TITLE Albite dissolution kinetics
SOLUTION 1
temp 25
pH 6
pe 4
redox pe
units mol/kgw
density 1
C(4) 0.02
Ca 0.05
Cl 0.3 charge
Na 0.3
-water 1 # kg
EQUILIBRIUM_PHASES 1
Quartz 0 0
Kaolinite 0 0
K-mica 0 0
KINETICS 1 Define Albite parameters
Albite
-formula NaAlSi3O8 1
-m 0.03814
-m0 0.03814
-parms 18370 117200
-tol 1e-008
-steps 63113850000 in 100 steps # seconds
-step_divide 1
-runge_kutta 3
-bad_step_max 500
INCREMENTAL_REACTIONS True
RATES
Albite
-start
10 rem M = current no. moles of Ab
20 rem at satart M (= m) = M0 = 0.03814
30 rem Initial area (A0) is 1000 [cm^2/g] for 10 g (=1e4 cm2)
40 rem so A06V [cm^-1] is 10
50 rem pre-exp const A [mol/cm^2.s] is 18370 = PARM(1)
60 rem act energy Ea [J/mol] is 117200 = PARM(2)
70 rem this gives the rate const of 5.50e-17 mol/[cm^2.s] at 25C
80 A0 = 1e4
90 V = 1000
100 sr_ab = SR("Albite")
110 if (M <= 0) then goto 200
120 R = 8.31451
130 T = 298.15
140 area = (M/M0)^(2/3)*A0
150 rate_const = PARM(1)*EXP(-PARM(2)/(R*T))
160 rate = (area/V)*rate_const*(1-sr_ab)
170 rem multiply by 1000 to change mol/(cm^3.s) to mol/(L.s)
180 rate = rate*1000
190 moles = rate*TIME
200 PUT(area,1)
210 PUT(rate,2)
220 SAVE moles
-end
SELECTED_OUTPUT 1
-file D:\Beton\FRIKSZI\Albite_kinetics_01.sel
-reset false
USER_PUNCH 1
-headings Ab_left area rate Time(years) a_Na+ SI(Albite)
-start
10 PUNCH M GET(1) GET(2) TOTAL_TIME/31556925 10*(LA("Na+")) SI("Albite")
-end
END
I understand each rows at the RATES block but on the row 170 it draw attention to conversion between mol/(cm3*s) to mol/(L*s). Why do we need this conversion?
Is it neccessary because we have 1L (1 kg) solution (SOLUTION 1)? Is it the same value (V = 1000) as observable at row 90?
Thank You for Your help in advance.
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dlparkhurst
Top Contributor
Posts: 3585
Re: Question about dissolution kinetics
«
Reply #1 on:
March 20, 2020, 06:34:23 PM »
Formally, I think the units of V are cm^3, and I am guessing the units of 1000 are cm^3/L, with an implicit assumption of 1 L of solution. This may be numerically correct because the two 1000s cancel, but I think the rate may depend only on surface area and not the volume of solution.
In any case, the SAVEd quantity in RATES must be moles, not moles per liter. As my high school chemistry teacher taught me, "follow the units".
With version 3, we made some changes to allow rates to scale properly with PHAST and PhreeqcRM. With these programs, there is some scaling to determine the amount of kinetic reactant in the representative volume of a cell. When KINETICS is scaled, the moles of reactant (M) are adjusted, but the parameters (-parms) are not. Thus, it is convenient to make the parameters have units of per mole, m^2/mol for area for example, rather than absolute area (m^2). Then the appropriate area can be calculated in the RATES definition from the parameter and the number of moles of kinetic reactant in a cell. See the Albite definition in phreeqc.dat (version 3) for an example.
For surface controlled reactions, the literature rates may logically be mol/m^2 of surface. For aqueous reactions, say degradation of an organic compound, the literature rate may be mol/L. If the rate is actually mol/L, it is necessary to multiply mol/L by the solution volume (SOLN_VOL) to get the number of moles of reaction to be SAVEd.
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