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VLE, LLE, and SLE Phase Diagrams

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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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