Simple Liquid-Liquid Equilibrium Calculation
One kilogram of liquid 2,6-dimethyl pyridine \left( C _{7} H _{9} N , MW =107.16\right) is mixed with 1 kg of water, and the mixture is heated to 80°C. Determine the compositions and total amounts of the two coexisting liquid phases.
From the equilibrium phase diagram, Fig. 11.2-1, we have
x_{ P }^{ I }=0.0170 \quad \text { and } \quad x_{ P }^{ II }=0.233
as the mole fractions of 2,6-dimethyl pyridine in the two equilibrium phases. The water mole fractions in the phases can be found from
x_{ H _{2} O }^{ J }=1-x_{ P }^{ J }
Also, note that 1 kg of 2,6-dimethyl pyridine is equal to 9.332 mol, and 1 kg of water is equal to 55.51 mol. To compute the amounts of each of the phases, we use the mass balances of Eq. 11.2-1b on a molar basis, which yield
\begin{aligned}M_{ i } &=\omega_{ i }^{ I } M^{ I }+\omega_{ i }^{ II } M^{ II } \quad(\text { mass basis) }\\N_{ i }&=x_{ i }^{ I } N^{ I }+x_{ i }^{ II } N^{ II } \quad(\text { molar basis) }\end{aligned} (11.2-1b)
9.332 mol 2,6-dimethyl pyridine =x_{ P }^{ I } x^{ I }+x_{ P }^{ II } x^{ II }=0.0170 N^{ I }+0.233 N^{ II }
and
55.51 \text { mol water }=\left(1-x_{ P }^{ I }\right) N^{ I }+\left(1-x_{ P }^{ II }\right) N^{ II }=0.983 N^{ I }+0.767 N^{ II }
These equations have the solution
N^{ I }=26.742 mol \quad \text { and } \quad N ^{ II }=38.10 mol
Now 1 mol of a solution with x_{ P }=0.0170 \text { weighs } 0.0170 \times 107.66+0.987 \times 18.015=19.53 g, and 1 mol of a solution with x_{ P }=0.233 \text { weighs } 0.233 \times 107.16+0.767 \times 18.015=38.786 g. Therefore, there are 26.742 mol \times 19.53 g / mol =522 g =0.522 kg kg of phase I and 38.10 mol \times38.786 g / mol =1478 g =1.478 kg of phase II.
