Lecture 5:
VLE, LLE, and SLE
Phase Diagrams
Physical Chemistry 2 for Engineers Lecture
Refresher on asynchronous lecture
•
Raoult’s Law provides a relation between vapor and liquid phase compositions at a
given T and P for ideal solution of volatile liquids.
Four main calculations: Bubble T/P, Dew T/P
𝑥𝑖 𝑃𝑖𝑠𝑎𝑡 = 𝑦𝑖 𝑃
•
Henry’s Law provides relation between vapor and liquid phase compositions at a
given T and P for very dilute mixtures.
Usually, the dilute component in the liquid phase follows the Henry’s Law.
The dominant component in the liquid phase follows the Raoult’s Law (ideally).
𝑥𝑖 ℋ𝑖 = 𝑦𝑖 𝑃
Refresher on asynchronous lecture
•
Vapor pressures are dependent of temperature and are calculated using the
Antoine equation.
Antoine coefficients are different for each chemical species.
Forms of the equation and units of temperature and pressure are usually provided with the
data set.
Review of phase diagrams
•
Shows the regions of pressure and temperature at
which its various phases are thermodynamically
stable.
In fact, any two intensive variables such as temperature
and magnetic field but we will focus on temperature and
pressure
•
A single phase is represented by an area.
•
Lines separating the regions are called phase
boundaries. This is where two phases are at
equilibrium.
•
So far, the phase diagrams we tackled are for
pure substances. What happens when we are
dealing with solutions/mixtures?
Total vapor pressure of binary solution
•
For a binary solution of volatile liquids A
and B for a given temperature,
𝑦𝐴 𝑃 = 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡 ,
𝑦𝐵 𝑃 = 𝑥𝐵 𝑃𝐵𝑠𝑎𝑡
•
If we add these:
𝑦𝐴 + 𝑦𝐵 𝑃 = 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡 + 𝑥𝐵 𝑃𝐵𝑠𝑎𝑡
𝑃 = 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡 + (1 − 𝑥𝐴 )𝑃𝐵𝑠𝑎𝑡
𝑃 = 𝑃𝐵𝑠𝑎𝑡 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴
•
This equation says that the total vapor
pressure (or simply pressure) is equal linear
with respect to liquid phase compositions.
Vapor phase compositions
𝑃 = 𝑃𝐵𝑠𝑎𝑡 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴
•
Substituting this expression into Raoult’s Law:
𝑦𝐴 𝑃 = 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡
𝑦𝐴 𝑃𝐵𝑠𝑎𝑡 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴 = 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡
𝑥𝐴 𝑃𝐴𝑠𝑎𝑡
𝑦𝐴 = 𝑠𝑎𝑡
𝑃𝐵 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴
•
This equation allows us to plot vapor phase
composition 𝑦𝐴 with regards to the liquid
phase composition 𝑥𝐴 .
In turn, we can plot pressure-vapor composition
plots.
Pressure in terms of vapor composition
𝑥𝐴 𝑃𝐴𝑠𝑎𝑡
𝑦𝐴 = 𝑠𝑎𝑡
𝑃𝐵 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴
𝑦𝐴 𝑃𝐵𝑠𝑎𝑡 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴 𝑦𝐴 = 𝑥𝐴 𝑃𝐴𝑠𝑎𝑡
𝑃𝐵𝑠𝑎𝑡 𝑦𝐴
𝑥𝐴 = 𝑠𝑎𝑡
𝑃𝐴 + 𝑃𝐵𝑠𝑎𝑡 − 𝑃𝐴𝑠𝑎𝑡 𝑦𝐴
𝑃=
𝑃𝐵𝑠𝑎𝑡
+
𝑃𝐴𝑠𝑎𝑡
−
𝑃𝐵𝑠𝑎𝑡
𝑥𝐴 =
𝑃𝐵𝑠𝑎𝑡
𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑃𝐵𝑠𝑎𝑡 𝑦𝐴
+ 𝑠𝑎𝑡
𝑃𝐴 + 𝑃𝐵𝑠𝑎𝑡 − 𝑃𝐴𝑠𝑎𝑡 𝑦𝐴
𝑃𝐴𝑠𝑎𝑡 𝑃𝐵𝑠𝑎𝑡
𝑃 = 𝑠𝑎𝑡
𝑃𝐴 + 𝑃𝐵𝑠𝑎𝑡 − 𝑃𝐴𝑠𝑎𝑡 𝑦𝐴
Pressure-composition (Pxy) diagram
•
By having two expressions for pressure in terms
of both vapor and liquid composition of a binary
volatile solution, we can plot two curves for
pressure corresponding to vapor and liquid
compositions.
𝑃 = 𝑃𝐵𝑠𝑎𝑡 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴
𝑃𝐴𝑠𝑎𝑡 𝑃𝐵𝑠𝑎𝑡
𝑃 = 𝑠𝑎𝑡
𝑃𝐴 + 𝑃𝐵𝑠𝑎𝑡 − 𝑃𝐴𝑠𝑎𝑡 𝑦𝐴
Pxy diagram
•
In these diagrams, the vapor curve lies at lower
pressures compared to the liquid curve.
Consistent with the notion that lowering the
pressure will drive the equilibrium towards the side
of the vapor.
•
Horizontal axis can be interpreted as liquid (𝑥),
vapor (𝑦), or overall (𝑧) composition.
For liquid and vapor compositions, the difference is
in which curve these compositions will be projected
into to find the corresponding pressure.
•
Vapor/liquid region is where the system is purely
vapor/liquid.
•
Area in between liquid and vapor curves is where
the system is at vapor-liquid equilibrium.
•
At the vapor/liquid curve, there
infinitesimal amount of liquid/vapor.
is
an
Pxy diagram
VLE benzene
+ toluene
A – benzene
B – toluene
𝑃 = 𝑃𝐵𝑠𝑎𝑡 + 𝑃𝐴𝑠𝑎𝑡 − 𝑃𝐵𝑠𝑎𝑡 𝑥𝐴
𝑃𝐴𝑠𝑎𝑡 𝑃𝐵𝑠𝑎𝑡
𝑃 = 𝑠𝑎𝑡
𝑃𝐴 + 𝑃𝐵𝑠𝑎𝑡 − 𝑃𝐴𝑠𝑎𝑡 𝑦𝐴
Pure toluene
Pure benzene
Using Pxy diagrams
•
Determine
the
pressure
and
liquid
composition at 20 oC when the vapor is 60%
benzene.
P ≈ 38 mmHg
𝒙𝑨 ≈ 𝟎. 𝟐𝟖
Using Pxy diagrams
•
A vapor containing 40% benzene and 60%
toluene are in a sealed container at a pressure
of 20 mmHg. Determine the vapor and liquid
compositions when the pressure is increased
to 35 mmHg.
𝒙𝑨 ≈ 𝟎. 𝟐𝟑 𝒚𝑨 ≈ 𝟎. 𝟓𝟒
Using Pxy diagrams
•
For the previous example, if there are initially
10 moles of vapor, how many moles will turn
to liquid when the pressure is increased to 35
mmHg?
𝒙𝑨 ≈ 𝟎. 𝟐𝟑 𝒚𝑨 ≈ 𝟎. 𝟓𝟒
𝒛𝑨 = 𝟎. 𝟒
Inverse Lever Arm Rule (ILAR)
•
Used in phase diagrams to measure the
ratio of amounts between two phases in
equilibrium.
Applicable in VLE, LLE, SLE, and other phase
diagrams.
𝑉 𝑧𝐴 − 𝑥𝐴
=
𝑀 𝑦𝐴 − 𝑥𝐴
𝐿
𝑉 𝑦𝐴 − 𝑧𝐴
=1− =
𝑀
𝑀 𝑦𝐴 − 𝑥𝐴
𝒙𝑨 ≈ 𝟎. 𝟐𝟑 𝒚𝑨 ≈ 𝟎. 𝟓𝟒
L
Tie line
Isopleth
𝒛𝑨 = 𝟎. 𝟒
M V
Inverse Lever Arm Rule (ILAR)
•
For the previous example, if there is initially
10 moles of vapor, how many moles will
turn to liquid when the pressure is increased
to 35 mmHg?
𝒙𝑨 ≈ 𝟎. 𝟐𝟑 𝒚𝑨 ≈ 𝟎. 𝟓𝟒
𝐿
𝑦𝐴 − 𝑧𝐴
=
𝑀 𝑦𝐴 − 𝑥𝐴
𝐿
0.54 − 0.4
=
= 0.452
10 0.54 − 0.23
L
𝐿 = 4.52 𝑚𝑜𝑙
4.52 moles of the original vapor will turn to
liquid.
𝒛𝑨 = 𝟎. 𝟒
M V
Temperature-composition (Txy)
diagrams
•
A Txy diagram shows the composition of phases that are in equilibrium at various
temperatures at a given pressure.
•
Unlike in Pxy diagrams, the vapor curve lies at higher temperatures compared to
the liquid curve.
•
ILAR is also applicable in Txy diagrams.
Txy diagrams
•
Although Txy diagrams can be generated analytically by expressing vapor pressures
in terms of temperature, Txy diagrams are usually generated using experimental
data and fitting temperature-composition data into a cubic polynomial model.
Generating Txy diagrams
•
The following temperature-composition data were gathered for a hexane-heptane
system.
T (oC)
65
66
70
77
85
100
xhexane
0
0.20
0.40
0.60
0.80
1
yhexane
0
0.02
0.08
0.20
0.48
1
Generating Txy diagrams
•
Fit data into cubic polynomial model and plot:
•
Models: T vs x, and T vs y
Txy diagrams
•
Temperature at xHx = 0 is the boiling point of
pure heptane while temperature at xHx = 1 is
the boiling point of pure hexane.
•
Below liquid curve (blue), the liquid solution is
below its boiling point or saturation
temperature and is therefore purely liquid
(subcooled liquid).
•
Above the vapor curve (purple), the vapor
mixture is above its saturation temperature
and is therefore purely vapor (superheated
vapor).
•
In between the two curves, there is coexistence of vapor and liquid.
Txy diagrams
•
The inverse lever arm rule (ILAR) applies to Txy
diagram as well to determine ratios between vapor
and liquid phases.
𝑉 𝑧𝐴 − 𝑥𝐴
=
𝑀 𝑦𝐴 − 𝑥𝐴
𝐿
𝑉 𝑦𝐴 − 𝑧𝐴
=1− =
𝑀
𝑀 𝑦𝐴 − 𝑥𝐴
V
M
L
Distillation
•
Simple distillation
Vapor is withdrawn and condensed to separate a
volatile liquid from a non-volatile solute or solid.
•
Fractional distillation
Boiling and condensation cycles are
successively to separate two volatile liquids.
repeated
Azeotropes
•
An azeotropic state is characterized by equal
vapor and liquid compositions.
•
Azeotropes lie at a lower or higher
temperature compared to the individual
boiling points of the constituents.
Ethanol-water azeotrope is at 78.2 oC. Ethanol
boiling point is 78.4 oC while water boils at 100
oC. (low–boiling azeotrope)
•
When an azeotrope is reached during
distillation, no more changes in the
composition is expected due to equal
compositions of vapor and liquid.
Maximum concentration of ethanol in ethanolwater mixture that can be achieved through
fractional distillation is 97.2% by volume due to
the azeotrope.
•
Existence of azeotropes depend on
deviation of the solution from ideality.
the
Immiscible liquids
•
For the distillation of two immiscible liquids A and B, the boiling
commences when the sum of the individual pressures reaches the
prevailing pressure.
𝑃 = 𝑃𝐴𝑠𝑎𝑡 + 𝑃𝐵𝑠𝑎𝑡
•
For the system in (b), the two liquids will boil at different
temperatures
•
For the system in (a), the two liquids will boil at the same
temperature but at a lower temperature compared to their individual
boiling points.
Example
•
Determine the boiling point of a mixture of octane (1) and water (2) at 101.325
kPa.
Antoine Equation:
ln 𝑃𝑖𝑠𝑎𝑡 𝑘𝑃𝑎 = 𝐴 −
𝐵
𝑇 °𝐶 + 𝐶
Antoine coefficients:
Water:
𝐴 = 16.3872, 𝐵 = 3885.70, 𝐶 = 230.170
Octane:
𝐴 = 13.9346, 𝐵 = 3123.13, 𝐶 = 209.635
Solution
𝑃 = 𝑃1𝑠𝑎𝑡 + 𝑃2𝑠𝑎𝑡
101.325 = exp 13.9346 −
3123.13
𝑇 + 230.170
+ exp 16.3872 −
𝑻 = 𝟖𝟏. 𝟕𝟑 °𝑪
Octane boiling point at 1 atm: 125.6 oC
Water boiling point at 1 atm: 100 oC
3885.70
𝑇 + 230.170
Liquid-liquid equilibrium
(LLE)
Liquid-liquid equilibrium (LLE)
•
Consider temperature-composition diagrams for
systems that consist of pairs of partially miscible
liquids.
These liquids do not mix in all proportions at all
temperatures.
Same principles in VLE phase diagrams may be applied.
•
Suppose small amounts of liquid B is added to pure
liquid A at temperature T
There will come a time when successive additions will not
dissolve but will form a second phase.
This is called phase separation.
•
A dissolves slightly into the added B and the two
phases co-exist at compositions 𝑎′ and 𝑎′′.
•
If more and more B is added, system crosses the
phase boundary and becomes a single phase system
again.
Example
•
The phase diagram for the system nitrobenzene (N) and hexane (H) at 1 atm is
shown in the figure. A mixture of 0.59 mol H and 0.41 mol N was prepared at 290
K. What are the compositions of the phases, and in what proportions do they
occur?
𝑥𝑁 = 0.41
Solution
𝑙𝛽
𝛼
=
𝛼 + 𝛽 𝑙𝛼 + 𝑙𝛽
𝛽
𝑙𝛼
=
𝛼 + 𝛽 𝑙𝛼 + 𝑙𝛽
𝛼
𝛼 𝛼 + 𝛽 𝑙𝛽 0.83 − 0.41
=
= =
=7
𝛽
𝛽
𝑙𝛼 0.41 − 0.35
𝛼+𝛽
There is 7 times more hexane-rich phase than
nitrobenzene-rich phase.
𝑥𝑁 = 0.35
𝑥𝑁 = 0.83
Critical solution temperature
•
The upper critical solution temperature 𝑇𝑢𝑐 is the
maximum temperature at which phase separation
can occur.
•
Some systems exhibit a minimum temperature
𝑇𝑙𝑐 or lower critical solution temperature where
phase separation can occur.
Ternary systems
•
For ternary liquid mixtures (3 partially miscible liquids), the conventional Cartesian
coordinates cannot represent the system fully.
•
Thus, a triangular diagram will be used.
•
For ternary systems: 𝑥𝐴 + 𝑥𝐵 + 𝑥𝐶 = 1
Ternary systems
Ternary phase diagrams
•
Miscibility of liquids are highly dependent on
temperature.
•
A face of the triangular prism in the figure
represents a binary system.
A ternary system phase diagram is a horizontal
section of this triangular prism at a given constant
temperature.
Ternary phase diagrams of partially
miscible liquids
•
The figure is a ternary phase diagram of a liquid
mixture.
•
A is fully miscible with both C and B, but C and B
are only partially miscible.
•
In this diagram, the two phases formed are C-rich
and B-rich phases.
•
The point P is called the plait point.
This is where the compositions of the two phases
become identical.
•
Tie lines are identified experimentally.
Usually, several tie lines are included in ternary phase
diagrams.
P
Example of a ternary phase diagram
•
Line a corresponds to a H2O-CHCl3 ratio of 3:2.
Adding acetic acid will retain the ratio between water
and trichloromethane but the system will move from the
bottom to the acetic acid corner through line a.
Acetic acid will dissolve in both phases which will also
change the composition water-rich and
trichloromethane-rich phases.
After some amount is added, the system will be fully
miscible.
•
Line b corresponds to a H2O-CHCl3 ratio of 2:3.
Same goes for line b.
Extension of ternary phase diagrams to
solids.
Solid-liquid equilibrium
(SLE)
Eutectics
•
Consider a binary liquid mixture with composition a 1.
•
The cooling of the binary liquid mixture will undergo
several stages:
a2 – B begins to precipitate.
a3 – pure B solid is in equilibrium with a liquid of
composition b3.
a4 – The system freezes into a solid with pure A and pure
B phases.
•
The point 𝑒2 is called the eutectic point. It is the
lowest temperature where the liquid is present in the
system.
•
Relative amounts of liquids and solids can be
elucidated using ILAR.
Cooling curves
•
The solid-liquid boundary is given by the points at
which the rate of cooling changes.
•
Points of interest:
a1 – liquid is being cooled.
a2 – B begins to precipitate.
a3 – eutectic begins solidifying. It takes some time to
solidify the eutectic. Liquid begins freezing to produce
two-phase solid.
a4 – mixture is completely frozen.
That’s it for the coverage of LE 3
•
LE 3 schedule: October 18, 2021 (class hours)
•
Synchronous exam. Only those present in Zoom meeting will be given a grade.
•
LE policies still apply.
•
Multiple choice on BB. (30 minutes)
•
Problem solving will be passed through Gform in a PDF format. (2.5 hours)
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