Answer to Question #206170 in Physical Chemistry for Kareena Max

Question

I-Butanol and chlorobenzene form a minimum-boiling azeotropic system. The mole fraction of I-butanol in the liquid (x) and vapour (y) phases at 1.000 atm is given below for a variety of boiling temperatures (H. Artigas, C.

Lafuente, P. Cea, F.M. Royo, and J.S. Urieta,J. Chem. Eng. Data 42,132 (1997))

T/K 396.57 393.94 391.60 390.15 389.03 388.66 388.57

X 0.1065 0.1700 0.2646 0.3687 0.5017 0.6091 0.7171

Y 0.2859 0.3691 0.4505 0.5138 0.5840 0.6409 0.7070

Pure chlorobenzene boils at 404.86 K.

(a) Construct the chlorobenzene-rich portion of the phase diagram from the data.

(b) Estimate the temperature at which a solution whose mole fraction of I-butanol is 0.300 begins to boil.

(c) State the compositions and relative proportions of the two phases present after a solution initially 0.300 l-butanol is heated to 393.94 K

Answer 1

These are the properties of MgO and NiO phase diagram plot. Liquidus temperature is the temperature above which the system is entirely liquid, and the solidus is the temperature below which the system is completely solid. Between these two points the liquid and solid phases are in equilibrium.

a) at x= .30 the melting point is 2150.

b)The composition of the solid can be found by drawing a vertical line to the base of the diagram. Thus it is seen that the first crystals precipitated from composition X will have the composition.

% solid = [x/(x + y)] x 100

c) the y=.70 is 2650