RATES output unit: mol/kgw?

[asainz]

I am working with kinetic dissolution of calcite with the rate of Plummer et al (1978) in a problem where the amount of water varies and I have doubts about the units.

If I understood correctly, the output of RATES is mol/dm3 water (more precisely: mol/kgw assuming a water density of 1kg/dm3). The output of RATES is multiplied then (in the main routine of PHREEQC) by the mass of water to calculate the moles dissolved/precipitated in the time step. Is that right?

Are the output of RATES always multiplied by the amount of water?

Thanks in advance for the help,
Alvaro

--------
I add the thread from PHREEQC Q+A where I found this sentence:

“The volume defaults to 1 kg = 1 dm3, but can be adjusted with the identifier -water in keyword SOLUTION”

(http://wwwbrr.cr.usgs.gov/projects/GWC_coupled/phreeqc/mail/msg00195.html)

[dlparkhurst]

########
#Calcite
########
We have been a little careless in the past by assuming a liter of kilogram of water. We are planning to revise the RATES definitions to be a bit more careful. Some rates are proportional to the surface area, like the Plummer and others calcite example. Other rates are for homogeneous solution reactions that are proportional to the volume of solution. We will try to be more clear in the future.

In any event, the value SAVE'd in a RATES definition is moles, not moles per liter. The formula defined in KINETICS is gives a stoichiometry that is multiplied by the SAVE'd moles and added (or removed if the product of moles and stoichiometric coefficient is negative) to the solution, regardless of the solution volume.

Hopefully, this definition of kinetics for calcite is clear.


# Example of KINETICS data block for calcite rate,
#   in mmol/cm2/s, Plummer et al., 1978, AJS 278, 179; Appelo et al., AG 13, 257.
# KINETICS 1
# Calcite
# -tol   1e-8
# -m0    3.e-3
# -m     3.e-3
# -parms 5e5   0.6  # cm^2, exp factor
# -time  1 day

Calcite
   -start
    1 REM   parm(1) = surface area of calcite, cm^2, per kg water in SOLUTION
    2 REM   parm(2) = exponent for m/m0

    10  si_cc = SI("Calcite")
    20  if (m <= 0  and si_cc < 0) then GOTO 200
    30  k1 = 10^(0.198 - 444.0 / TK )
    40  k2 = 10^(2.84 - 2177.0 /TK )
    50  if TC <= 25 then k3 = 10^(-5.86 - 317.0 / TK)
    60  if TC > 25 then k3 = 10^(-1.1 - 1737.0 / TK )
    70  t = 1
    80  if m0 > 0 then t = M/M0
    90  if t = 0 then t = 1
    100 area = PARM(1) * (t)^PARM(2)
    110 rate = area * (k1 * ACT("H+") + k2 * ACT("CO2") + k3 * ACT("H2O"))
    120 rate = rate * (1 - 10^(2/3*si_cc))
    130 moles = rate * 0.001 * TIME # convert from mmol to mol
    200 SAVE moles
   -end

[asainz]

Thanks for the explanation Mr Parkhurst, I think finally I understood how RATES is working.