## Info

The sum of the second and last terms on the right-hand side of this equation is the integral of d{[A][B]}, so the above equation becomes:

Equality of the chemical potentials of water in the two aqueous phases is expressed by:

gwi[W]i = gwii[W]ii or lngwi + ln[W]i = lngwii + ln[W]n (11.58)

The molar concentration of water, [W], can be expressed in terms of mole fractions by:

[W] = Xw/Vw = (1 - Xa - Xb)/Vw so that ln[W] = -lnvw + ln(1 - xA - xB) @ -lnvw - xA - xB = -lnvw - ([A] + [B])vw Substituting this equation and Eq (11.57) into (11.58) yields:

[A], + [B], + 1 bAA [A]2 + 1 bBB [B]2 + b AB [A], [B], = [A]jj + [B]jj + 2 b AA [A]2i + 2 b BB [B]2 + bAB [A]ii [B]n

Equations (11.55a), (11.55b) and (11.59) provide three equations with four unknowns, [A]i, [B]i, [A]n and [B]ii. The fourth equation is a statement of species conservation.

Suppose that the system consists of a total volume V of water to which nA moles of A and nB moles of B are added. After splitting into two phases of volumes Vi and Vii, the species conservation equations are:

nA = [A]iVi + [A]nVn or [A]M = f [A]i + (1-f)[A]n nB = [B]iVi + [B]nVn or

[B]tot = f [B]i + (1-f)[B]n where f = VI/Vtot is the fraction of the total water volume converted to phase I and [A]tot = nA/Vtot and [B]tot = nB/Vtot. These are specified. Eliminating f between the two species conservation equations yields:

Solution of the phase-equilibrium equations Dimensionless forms of these equations result from the definitions:

X = bAA[A]i; Y = bAA[A]n; U = bee[B]i; V = bee^n ZA = bAA[A]tot;

Equations (11.57a) and (11.57b) become:

With some rearrangement, Eq (11.59) is converted to:

(1 + X) 2 J B + (1 + U) 2 JA + 2XU _ (1 + Y)2 J B + (1 + V) 2 J A + 2YV 