Embed Size (px)
Citation preview
Chapter 11:Other Types of
Phase Equilibria in Fluid Mixtures
(selected topics)
Partitioning a solute among two coexisting liquid phases
• Two partially miscible or completely immiscible liquids
• How the solute distributes between the two phases
• Purification– LL extraction, Partition chromatography
• Drug distribution– Lipids, body fluids
• Pollutant distribution– air.,water,soil
Partition of a solute
Distribution coefficient
Concentration of solute in phase I
Concentration of solute in phase IIK
I I II IIi i i ix x
The LLE equilibrium condition is:
Then: ,
, ,
, ,
IIIIIii
x i II IIi i
T P xxK
x T P x
Case I: solute does not affect solubility of the
solvents• Little amount of solute or totally
immiscible liquids
• I-a) Ni moles of solute completely dissolved and distributed between immiscible solvents
Case I-b: some undissolved solute (solid or gas) in equilibrium with two
immiscible solvents
Case Ic: partially miscible liquids
Case II: solute affects the LLE (partially miscible
solvents)
Liquid-liquid equilibrium (LLE)
Extraction problems involve at least three components: the solute and two solvents. It is usual to represent their phase behavior in triangular diagrams.
Liquid-liquid equilibrium (LLE)
Extraction problems involve at least three components: the solute and two solvents. It is usual to represent their phase behavior in triangular diagrams.
Binodal curve
Tie line
Plait point
Liquid-liquid equilibrium (LLE)
Reading the scale in a triangular diagram
Example 4
It is desired to remove some acetone from a mixture that contains 60 wt% acetone and 40 wt% water by extraction with methyl isobutyl ketone (MIK). If 3 kg of MIK are contacted with 1 kg of this acetone+water solution, what will be the amounts and compositions of the phases in equilibrium?
Solution
Example 4
60 wt% acetone and 40 wt% water (1 kg) Pure MIK (3 kg)
Example 4
Next…
Find the point that represents the global composition of the system.
Based on the information given, the total amounts of acetone, water, and MIK are equal to 0.6 kg, 0.4 kg, and 3 kg.
The corresponding weight fractions are: 0.15, 0.10, and 0.75.
Locate this point in the diagram.
Example 4
Global composition
Example 4
Approximated tie line
Example 4
Approximated tie line
MIK-rich phase
80.5% MIK15.5% Acetone4.0% Water
Water-rich phase
2.0% MIK8.0% Acetone90.0% Water
Example 4
Calculation of the phase amounts (LI and LII)
Global mass balance
One component mass balance (water for example)
4 4I II II IL L L L
0.04 0.90 0.04 0.90 4
0.4 3.721
I II I I
I
L L L L
kg L kg
Liquid-liquid equilibrium (LLE)
Liquid-liquid equilibrium (LLE)
Osmotic equilibrium
Consider two cells at the same temperature, separated by a membrane permeable to some of the species present, but impermeable to others.
For simplicity, assume a binary solute+solvent system and that the membrane is permeable to the solvent only. Cell I contains the pure solvent and cell II contains the mixture.
At equilibrium, the following equation is valid:
,III
solvent solventf T P f
Osmotic equilibrium
, ,I II II IIsolvent solvent solvent solventf T P x f T P
,III
solvent solventf T P f
, exp
I
satsolvent
LPI sat sat solvent
solvent solvent solvent
P
Vf T P P dP
RT
, exp
II
satsolvent
LPII sat sat solvent
solvent solvent solvent
P
Vf T P P dP
RT
Osmotic equilibrium
1 exp
II
I
LPII II solventsolvent solvent
P
Vx dP
RT
Assuming the liquid is incompressible:
1 exp
L II IsolventII II
solvent solvent
V P Px
RT
Osmotic equilibrium
Applying logarithm:
0 ln ln
L II IsolventII II
solvent solvent
V P Px
RT
ln lnII I II IIsolvent solventL
solvent
RTP P x
V
: osmotic pressure
Example 5
Compute the osmotic pressure at 298.15 K between an ideal aqueous solution 98 mol% water and pure water.
For an ideal solution, the activity coefficient is equal to 1 and the molar volume of water is approximate equal to 18x10-6 m3/mol.
Solution
Example 5
Compute the osmotic pressure at 298.15 K between an ideal aqueous solution 98 mol% water and pure water.
For an ideal solution, the activity coefficient is equal to 1 and the molar volume of water is approximate equal to 18x10-6 m3/mol.
Solution
36
8.314 298.15. ln 0.98
18 10
JK
mol Kmmol
Osmotic equilibrium
For ideal solutions:
ln IIsolventL
solvent
RTx
V
For a dilute ideal solution, by using a Taylor series expansion of the logarithm of the solvent mole fraction, the following approximated expression can be derived:
1 II IIsolvent soluteL L
solvent solvent
RT RTx x
V V
Osmotic equilibriumII
II solutesoluteL L II II
solvent solutesolvent solvent
IIsoluteIIsolution
L II IIsolvent solutesolvent
IIsolution
nRT RTx
n nV V
n
VRT
n nVV
Assuming: and II IIsolution solventV V II II II
solvent solute solventn n n
IIsoluteII II IIsolution solute solute
L II IIsolvent solution solutesolventIIsolvent
n
V n CRTRT RT
n V mVV
Osmotic equilibrium
For simplicity, let us drop superscript II, then:
solute
solute
CRTm
where msolute is the solute’s molar mass.
A practical application of this equation is to use it to find the molar mass of polymers and proteins.
Osmotic equilibrium
Schematics of an osmometer
Example 6
Polyvinyl chloride (PVC) is soluble in cyclohexanone. At 25oC, if a solution of PVC batch with 2 g/L of solvent is placed in an osmometer, the height h in the osmometer is 0.85 cm. Knowing that the density of pure cyclohexanone is 0.98 g/cm3, estimate the molar mass of this PVC batch.
Example 6
Solution
At the membrane, in the mixture side:
II atmsolutionP P g h H
At the membrane, in the pure solvent side:
I atmsolventP P gH
H
II IP P gh
Assuming the density of the solution and of the solvent are equal:
Example 6
Solution
H
6 3
3 3 2
1 10 10.98 9.81 0.85 81.72
1000 1 100
g kg cm m mgh cm Pa
cm g m s cm
Example 6
Solution
H
solute
solute
CRTm
solutesolute
Cm RT
3
1000 18.314 298.15 2
. 100060.67
81.72solute
J g L kgK
kgmol K L m gm
Pa mol
Recommendation
Read the sections of chapter 11 covered in these notes and review the corresponding examples.
