1

 Equilibrium in vapor–liquid systems is restricted by the phase rule:

F=C–P+2
 Consider the ammonia (A) – water (B), vapor-liquid system
F=2
 4 variables are T, P, xA and yA (xB and yB will be fixed if xA and yA are specified)
 If P is fixed, then only 1 other variable can be set
 For ideal systems, Raoult’s law holds
pA = PA xA

2
 T-x-y diagram

 For the system benzene (A)– 98oC

toluene (B) at a total pressure
of 101.32 kPa
 Saturator vapor line is the
dew-point line
 Saturated liquid line is the
bubble-point line

0,318 0,532
3
 The ideal system benzene-toluene follows Raoult’s law so the BP diagram can be
generated from the pure vapor-pressure data
𝑝𝐴 + 𝑝𝐵 = 𝑃
Raoult’s law: 𝑃𝐴 𝑥𝐴 + 𝑃𝐵 1 − 𝑥𝐴 = 𝑃
𝑝𝐴 = 𝑃𝐴 𝑥𝐴
𝑝𝐴 𝑃𝐴 𝑥𝐴
𝑦𝐴 = =
𝑃 𝑃
Vapor Pressure
Mole-Fraction
Temperature Benzene Toluene Benzene at
101.325 kPa
K °C kPa mmHg kPa mmHg xA yA
353.3 80.1 101.32 760 1.000 1.000
358.2 85 116.9 877 46.0 345 0.780 0.900
363.2 90 135.5 1016 54.0 405 0.581 0.777
368.2 95 155.7 1168 63.3 475 0.411 0.632
373.2 100 179.2 1344 74.3 557 0.258 0.456
378.2 105 204.2 1532 86.0 645 0.130 0.261
4
383.8 110.6 240.0 1800 101.32 760 0 0
5
6
Calculate the vapor and liquid compositions in equilibrium at 95°C (368.2 K) for
benzene–toluene using the vapor pressure from the Table at 101.32 kPa.

7
 Benzene (A) – Toluene (B)
system
 45o line shows that yA is richer in
component A than is xA
 Typical diagram for ideal
systems

8
 The y vs x plot will cross the 45o line at yAz = xAz

9
 Tray type towers (sieve tray, valve tray, bubble cap tray)

10
  Packed towers

Types of packing

Typical random or dumped tower packings:

(a) Raschig ring; (b) Berl saddle; (c) Pall ring;
(d) Intalox metal, IMTP; (e) Jaeger Metal Tri-Pack

11
 two entering phases of known amounts and compositions, L0 and V2, enter the stage;
mixing and equilibration occur; and the two exit streams, L1 and V1, leave in
equilibrium with each other.
 Total mass/mole balance:
 If 2 components (A & B), then the component balance (mass/mole) is:

 A balance for B is not required since xA + xB = 1

 If sensible heat effects are small and the latent heats of both compounds are the same,
then when 1 mol of A condenses, 1 mol of B must vaporize
 Hence, the total moles of vapor V2 entering will equal V1 leaving. Also, moles L0 = L1
 This case is called one of constant molal overflow. An example is the benzene–toluene
system. 12
A vapor at the dew point and 101.32 kPa containing a mole fraction of 0.40 benzene
(A) and 0.60 toluene (B) and 100 kg mol total is brought into contact with 110 kg mol
of a liquid at the boiling point containing a mole fraction of 0.30 benzene and 0.70
toluene. The two streams are contacted in a single stage, and the outlet streams leave
in equilibrium with each other. Assume constant molal overflow. Calculate the
amounts and compositions of the exit streams.

13
 Introduction
 All components are present in both phases.
 The vapor phase is created from the liquid phase by vaporization at the boiling point.
 Basic requirement is that the composition of the vapor be different from the
composition of the liquid with which it is in equilibrium at the boiling point of the
liquid.
 Distillation is concerned with solutions where all components are appreciably volatile,
such as ammonia–water or ethanol–water solutions, where both components will be in
the vapor phase.
 By contrast, in evaporation of a solution of salt and water, for example, the water is
vaporized but the salt is not.
 The process of absorption differs from distillation in that one of the components in
absorption is essentially insoluble in the liquid phase eg. absorption of ammonia from
air by water, where air is insoluble in the water–ammonia solution.
15
 Relative Volatility
 In the equilibrium diagram for a binary mixture of A and B, the greater the distance
between the equilibrium line and the 45° line, the greater the difference between the
vapor composition yA and liquid composition xA, hence, the separation is more easily
made.
 Relative volatility αAB is the ratio of the concentration of A in the vapor to the
concentration of A in the liquid divided by the ratio of the concentration of B in the
vapor to the concentration of B in the liquid:

 If the system obeys Raoult’s law:

16
Using the data from the vapor pressure table, calculate the relative volatility for the
benzene–toluene system at 85°C (358.2 K) and 105°C (378.2 K).

17
 Distillation methods (2 types)
1. involves the production of a vapor by boiling the liquid mixture to
be separated in a single stage and recovering and condensing the
vapors. No liquid is allowed to return to the single-stage still to
contact the rising vapors.
2. involves the return of a portion of the condensate to the still. The
vapors rise through a series of stages or trays, and part of the
condensate flows downward through the series of stages or trays
countercurrent to the vapors. This second method is called
fractional distillation, distillation with reflux, or rectification.

18
 occurs in a single stage where a liquid mixture is partially vaporized.
 The vapor is allowed to come to equilibrium with the liquid, and the
vapor and liquid phases are then separated (can be done batchwise or
continuously).

Since L = F – V:
• Usually F, V and L (mol/hr) are known or
set. So the two unknowns are x and y
• A convenient method to use is to plot
this equation on the x-y equilibrium
diagram. The intersection of the
equation and the equilibrium line is the
desired solution 19
 Liquid is first charged to a heated kettle.
 The liquid charge is boiled slowly and the vapors are withdrawn as rapidly as they form to
a condenser, where the condensed vapor (distillate) is collected.
 The first portion of vapor condensed will be richest in the more volatile component A.
 As vaporization proceeds, the vaporized product becomes leaner in A.

• Originally, a charge of L1 moles of components A

and B with a composition of x1 mole fraction of A is
placed in the still.
• At any given time, there are L moles of liquid left in
the still with composition x, and the composition of
the vapor leaving in equilibrium is y. A differential
amount dL is vaporized.
20
 The composition in the still pot changes with time
 We assume that a small amount of dL is vaporized
 The composition of the liquid changes from x to x–dx and the amount of liquid from L to
L–dL
 A material balance on A can be made, where the original amount = the amount left in
the liquid + the amount of vapor:

 Neglecting the term dx dL and rearranging,

 Integrating, (Rayleigh equation)

 where L1 is the original moles charged, L2 is the moles left in the still, x1 is the original
composition, and x2 is the final composition of liquid
 The average composition of total material distilled, yav can be obtained by material
balance:

21
A mixture of 100 mol containing 50 mol % n-pentane and 50 mol % n-heptane is
distilled under differential conditions at 101.3 kPa until 40 mol are distilled. What is
the average composition of the total vapor distilled and the composition of the liquid
left? The equilibrium data are as follows, where x and y are mole fractions of n-
pentane:

x y x y x y
1.000 1.000 0.398 0.836 0.059 0.271
0.867 0.984 0.254 0.701 0 0
0.594 0.925 0.145 0.521

22