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Re: question about PHAST/PHREEQC-- with attachments

Try the file below. Here are the major changes:

(1) You should define Benzene as a separate "element" if you want it to
react kinetically. If you define it as part of "C", it will react to
equilibrium, which in this case means that all of the C6H6 would probably
disappear instantly to come to equilibrium with the rest of the C system.

(2) Signs for kinetic reactions are tricky. The product of the "moles"
saved and the reaction coefficients in -formula determine whether an
element(s) enter or leave solution; negative indicates decreasing
concentration in solution, positive increasing. Here, the number of "moles"
is negative, Benzene 1 is positive and so benzene concentration decreases;
C6H6 -1 is negative so the product of "moles"*-1 is positive and C6H6
increases in solution.

The kicker is that positive "moles" decreases the number of moles of a
kinetic reactant (-M in KINETICS definition and Basic variable M in a rate
expression). The program stops the reaction when 0 moles of reactant are
left. By using a negative "moles", the amount of reactant increases, so the
reaction will continue indefinitely, which is what you want in this case I
think. The number of moles of the kinetic reactant (Basic variable M) will
be the total number of moles of benzene reacted in a given cell.

(3) You can't specify both a concentration of O2 and a pe with a specified
O2 concentration. In this case, the pe (in equilibrium with a specified
pO2) will determine the O2 concentration.


#C(6)     C6H6           0.0     C6H6            78.108
Benzene   Benzene       0     C6H6  78.108
#C6H6 = C6H6
Benzene = Benzene
 log_k           0.000
        -gamma    5.4000    0.0000

#6 CO3-2 + 12 H+ = C6H6 + 3 H2O + 7.5 O2
# log_k           0.000
#        -gamma    5.4000    0.0000
 units mol/kgw
 pH  7.0 charge
 temp 25
# C(6)    0.001
Benzene     .001
 units mol/kgw
 pH  7.0 charge
 pe 12.0 O2(g) -0.67
 temp 25
# O(0)    0.0001

# -formula C (H) 6.0 12.0
 -formula Benzene 1 C6H6 -1
 -m 0
 -m0 0
10 Ks = 1.0d-4
20 Ka = 3.125d-6
30 qm = 1.407e-3/3600
#40 f1 = MOL("C(6)")/(Ks + MOL("C(6)"))
40 f1 = MOL("Benzene")/(Ks + MOL("Benzene"))
50 f2 = MOL("O2")/(Ka + MOL("O2"))
60 rate = -qm * f1 * f2
70 moles = rate * TIME
90 SAVE moles


David Parkhurst (dlpark@xxxxxxxx)
U.S. Geological Survey
Box 25046, MS 413
Denver Federal Center
Denver, CO 80225

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