Fick's differential equations describing diffusion can serve to determine the instantaneous rate of release of a semiochemical from a capillary test-tube:

release rate = -3.141593 * r * r * D * (C2 - C1) / x

where:

r = radius of the tube opening

D = diffusion coefficient

C2 = liquid concentration

C1 = 0 (assuming convection carries vapor away)

x = distance between micro test-tube opening and meniscus level of liquid

(Villars and Benedek, 1974) (* = BASIC symbol for multiply). More complicated equations (Brooks, 1980) are needed to describe the release over time as the level of the liquid decreases in the tube. In practice, however, it is usually more accurate to measure the release rate over the expected experimental period because one does not know precisely the diffusion coefficient (D) and other contributing factors (e.g., meniscus curvature, surface tension, and temperature effects). Tilden and Bedard (1985) and others (Browne, 1978; Byers and Wood, 1980; byers, 1982; Tilden et al., 1983) have used 52-mm-long x 3.5-mm-ID glass tubes sealed at the bottom to dispense western pine beetle pheromone and aggregation components,

The concentration (C2) is actually the vapor pressure of the semiochemical, and this can be varied according to Raoult's law, which states that the vapor pressure (release rate) of a volatile substance (semiochemical) is proportional to its mole fraction in a solvent. The following equation can then be derived for purposes of diluting semiochemicals with solvent in order to obtain a specific semiochemical release rate:

mls = fws * (gsem / fwsem - fsem * gsem / fwsem) / fsem / gs

where:

mls = milliliters of solvent

fws = formula weight (or molecular weight) of solvent

gs = grams solvent per milliliter (density)

gsem = grams of semiochemical fwsem = formula weight of semiochemical

fsem = mole fraction of semiochemical or the proportion of the release rate when neat (0 < fsem <= 1)

For example, a stock solution of E in ethanol that would yield a release rate 10% that of a neat solution of E and that is to be made using 0.5 g of E would require 0.52 ml of E (0.5 g * ml/0.96 g) and 1.6 ml ethanol

([46 g/mole * (0.5 g/156 g/mole - 0.1 * 0.5 g/156 g/mole)/0.1] * ml/0.828 g).

Stock solutions of E in ethanol that should release about 1%, 0.1% and 0.01%, and so forth, that of a neat solution (pure semiochemical) can then be made simply from the 10% solution by serial 1:10 (1 + 9) dilutions.

Chemical Ecology