Question

Physical Chemistry Problem: Below is the water-nicotine phase map. Estimate the composition of the two phases. If 0.50 moles of each water and nicotine are mixed at 125^oC (where the tie line is shown) and allowed to equilibrate, what are the compositions of the two phases, how many total moles are in each phase and how many moles of each component are in each phase?

H2O Nicotine uc 210e P=2 T. 61 Ic 0 0.2 0.4 0.6 0.8 Mole fraction of nicotine, xv

Answer 1

The tie line at 125^oC intersects the phase boundaries at X_N = 0.83 and X_N = 0.17. So the composition of the phase 1 is X_N = 0.83 and X_W = 0.17 and that of phase 2 is X_N = 0.17 and X_W = 0.83.

Now according to the lever rule, the ratio of amounts of each phase is equal to the ratio of the distances of intersection point from the mixing mole fraction

that is

n1/n2 = 0.83 - 0.5/ 0.5 - 0.17 = 0.33/ 0.33 = 1

this means that the total moles in each phase are equal and is equal to 0.50 moles

Therefore

no. of moles of nicotine in phase 1 = 0.50 x 0.83 = 0.415

and no. of moles of water in phase 1 = 0.50 x 0.17 = 0.085

and

no. of moles of nicotine in phase 2 = 0.50 x 0.17 = 0.085

and no. of moles of water in phase 2 = 0.50 x 0.83 = 0.415