McKay, Donald. "Front matter" Multimedia Environmental Models ...

It is important to note that the reaction rate is controlled by the product V, C,
and k. A large value of any one of these quantities may convey the wrong impression
that the reaction is important.
6.3.5 Level II Using Fugacity and D Values for Reaction
We can now follow the same process as used when treating advection and define
D values for reactions. If the rate is V C k or V Z f k, it is also D Rf, where D R is V
Z k. Note that D R has units of mol/m 3 Pa identical to those of D A or G Z, discussed
earlier. If there are several reactions occurring to the same chemical in the same
phase, then each reaction can be assigned a D value, and these D values can be
added to give a total D value. This is equivalent to adding the rate constants. The
Level II mass balance becomes
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E = SV iC ik i = SV iZ ifk i = fSV iZ ik = fSD R
Thus, f can be deduced, followed by concentrations, amounts, the total amount M,
and the rates of individual reactions as V C k or D f. We can repeat Example 6.2
in fugacity format.
Air V A = 10 4 Z A = 4 ¥ 10 –4 k A = 0.01 D RA = 0.04
Water V W = 100 Z W = 0.1 k W = 0.1 D RW = 1.0
Sediment V S = 1.0 Z S = 1.0 k S = 0.001 D RS = 0.001
Worked Example 6.3
f = E/SD Ri = 10/1.041 = 9.606
C A = 0.0038 rate = D f = 0.384
C W = 0.9606 = 9.606
C S = 9.6060 = 0.010
Total = 1.041
An evaluative environment consists of 10000 m 3 air, 100 m 3 water, and 10 m 3
soil. There is input of 25 mol/h of chemical, which reacts with half-lives of 100
hours in air, 75 hours in water, and 50 hours in soil. Calculate the concentrations
and amounts given the Z values below:
Phase
Volume
V (m 3 ) Z k
VZk
or D
C
(mol/m 3 )
m
(mol)
Rate
(mol/h)
Air 10000 4 ¥ 10 –4 0.00693 0.0277 0.0386 386 2.68
Water 100 0.1 0.00924 0.0924 9.66 966 8.93
Soil 10 1.0 0.0139 0.1386 96.6 966 13.39
Total 2318 25.0