Hello Dr. Parkhurst, My name is Barb Petrunic and I am a student at the University of New Brunswick working under the supervision of Kerry MacQuarrie and Tom Al. I am trying to work through example problem 15 that is provided in the Phreeqc program however, I am having trouble understanding the rates data block for the Biomass in the problem. I was wondering if you could answer a couple questions with regards to this example problem: 1) why does the "moles" of biomass have to be negative? The signs get a little confusing, but positive numbers indicate addition to solution and negative numbers mean removal from solution. Still "positive number" is a product of three factors, the "moles" saved in the basic program (-3 in the following example), the coefficient in the -formula of KINETICS (positive or negative, -1 in the following example), and possibly a stoichiometric coefficient in the reactant part of -formula (always positive, 1 for C and O and 2 for H in the following example). So a "saved" moles of -3, of the -formula CH2O -1, would result in 3 mol C, 6 mol H, and 3 mol O being added to solution. If saved moles is 3 and -formula CH2O -1, then the result would remove C, H, and O from solution. Similarly, if saved moles is -3 and -formula CH2O 1, the result would remove moles from solution. Finally, a saved moles of 3, and -formula CH2O 1, would add the elements to solution. Now the formula for Biomass is 0.0 H, so nothing is added or removed from solution. However, a cumulative moles of kinetic reactant is calculated as given by the "M" Basic function, and initialized with -M in KINETICS, is calculated. The quantity of biomass is needed to calculate the rate reactions because the rates of substrate usage depend on the amount of biomass present. By convention, if the saved reaction moles is negative, M is incremented; if the saved reaction moles is positive, M is decremented. Note, that this is independent of the stoichiometry given in -formula. The intention was that if something comes out of solution, then the moles of the kinetic reactant increased, and if it goes into solution, the moles of the kinetic reactant is decreased. In this case, nothing really comes out of solution, but the convention is used to cause the biomass to increase, and to increase the biomass (M), the saved moles must be negative. One other point is that M can not be negative. If M is decremented to zero, then a kinetic reaction that consumes the kinetic reactant will stop because it has run out of reactant. 2) why is the "if" statement required in line 60 in the rates data block? It is probably not needed, but it does point out that you can not have less than 0 biomass. I think the program would automatically limit the removal of reactant so that M can not be less than 0, but I'm not positive. Line 60 may help with the numerical integration of the rate equation. David
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