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

eaction situation in which there is no advection, and there is a constant inflow of
chemical in the form of an emission, as depicted in Figure 6.1b. When a steady state
is reached, there must be an equivalent loss in the form of reactions. Starting from
a clean environment, the concentrations would build up until they reach a level such
that the rates of degradation or loss equal the total rate of input. We further assume
that the phases are in equilibrium, i.e., transfer between them is very rapid. As a
result, the concentrations are related through partition coefficients, or a common
fugacity applies. The equations are as follows:
Using partition coefficients,
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E = V 1C 1k 1 + V 2C 2k 2 etc. = SV iC ik i
E = SV iC wK iwk i = C wSV iK iwk i
from which C w can be deduced, followed by other concentrations, amounts, rates
of reaction, and the persistence. In the general expression, K WW, the water-water
partition coefficient is unity.
Worked Example 6.2
The evaluative environment in Example 6.1 is subject to emission of 10 mol/h
of chemical, but no advection. The reaction half-lives are air, 69.3 hours; water, 6.93
hours; and soil, 693 hours. Calculate the concentrations. Recall that K AW = 0.004
and K SW = 10.
The rate constants are 0.693/half-lives or air, 0.01; water, 0.1; soil, 0.001; h –1 .
Therefore,
The rates of reaction then are
air = 0.38
water = 9.61
soil = 0.01
which add to the emission of 10.
E = V AC Ak A + V WC Wk W + V SC Sk S
= C W(V AK AWk A + V Wk W + V SK SWk S)
= C W(0.4 + 10 + 0.01) = C W(10.41) = 10
C W = 0.9606 mol/m 3 , C A = 0.0038, C S = 9.606