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. David SOLUTION_MASTER_SPECIES #C(6) C6H6 0.0 C6H6 78.108 Benzene Benzene 0 C6H6 78.108 SOLUTION_SPECIES #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 SOLUTION 1 units mol/kgw pH 7.0 charge temp 25 # C(6) 0.001 Benzene .001 END SOLUTION 2 units mol/kgw pH 7.0 charge pe 12.0 O2(g) -0.67 temp 25 # O(0) 0.0001 END KINETICS 1 C6H6_degradation # -formula C (H) 6.0 12.0 -formula Benzene 1 C6H6 -1 -m 0 -m0 0 RATES C6H6_degradation -start 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 -end END David Parkhurst (dlpark@xxxxxxxx) U.S. Geological Survey Box 25046, MS 413 Denver Federal Center Denver, CO 80225 Project web page: https://wwwbrr.cr.usgs.gov/projects/GWC_coupled
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