soluciones y mezclas
TRANSCRIPT
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Polymer Solutions and Blends
Why do some things mix and others don’t?.
A Phase Separated System
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Δ S total = + Δ S system Δ S surroundings Remove Barrier
The second law
- T Δ S total = - T Δ S Δ H
The Gibbs Free Energy
Would Δ Stotal be > 0 ? Heat
Δ S total > 0
Δ G = - T Δ S total = - T Δ S Δ H
= - Δ Q system T = Δ H system
T Δ S surroundings -
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The Free Energy of Mixing
Mix Δ S total > 0 Δ Gmix < 0
Polymer + Solvent For Mixing
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The Mean Field Assumption
Each molecule is assumed to be acted upon by a potential that is an average taken over all the interactions in the system, rather than one determined by local composition . This allows us to treat the entropy and enthalpy changes upon mixing as separate and additive quantities. We will start by treating the entropy.
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The Entropy of Mixing
First examine the simplest possible problem, mixing spherical molecules of equal size. If we can figure out the total number of ways all the molecules can be arranged, Ω, then we can use Boltzmann’s equation;
S = k ln Ω to calculate the entropy. But where on earth do you even start on a problem like this, when you have the jumbled up, randomly arranged mess that is the liquid state ?
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The Entropy of Mixing
Lattice Model
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The Entropy of Mixing Small Molecules of Equal Size
If we start with nA molecules of type A and nB molecules of type B, then also assume that there are no “holes” on the lattice*, then there are just n0 = nA + nB lattice sites. Now we take all the molecules off the lattice and put them back on randomly one at a time, not caring, at this point, if it’s an “A” or a “B”. How many ways are there to put the first molecule on the lattice? A. nA B. nB C. n0
*This is equivalent to assuming that there is no free volume, or the fluid is incompressible.
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The Entropy of Mixing Small Molecules of Equal Size
How many ways are there of putting two molecules on the lattice?
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The Entropy of Mixing Small Molecules of Equal Size
It must be (n0)(n0 - 1). OK, how many ways can you put 3 molecules back on the lattice? - you’ve got it, (n0)(n0 - 1) )(n0 - 2). So, the number of ways of putting all n0 molecules back on the lattice is, (n0)(n0 - 1) )(n0 - 2) (n0 - 3) - - - - (3)(2)(1). or n0 factorial, n0!
There are n0 - 1 arrangements of
the second molecule with first molecule placed
here
Another n0 - 1 arrangements with first molecule now
placed here
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The Entropy of Mixing Small Molecules of Equal Size
Do we get a new distinguishable
arrangement if we let these two molecules
switch positions?
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The Entropy of Mixing Small Molecules of Equal Size
Ω = ( n A + n B )!
n A ! n B !
- Δ S m = k ( n A ln x A + n B ln x B )
x A = n A
n A + n B x B =
n B n A + n B
ln( n A ! ) = n A ln n A - n A , etc
S = k lnΩ
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- Δ S m = R ( n A ln x A + n B ln x B )
- Δ S m = k ( n A ln x A + n B ln x B )
The Entropy of Mixing Small Molecules of Equal Size
Molecules
Moles
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The Entropy of Mixing Molecules of Different Size
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The Entropy of Mixing a Polymer and a Solvent
Model -the polymer is a chain of connected
segments, each equal in size to a solvent
molecule.
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The Entropy of Mixing a Polymer and a Solvent
A much more complex problem. Flory’s result;
- Δ S m = R ( n A ln φ A + n B ln φ B ) where φA and φB are the volume fractions of the A and B components (say solvent and polymer), respectively. This is often called the combinatorial entropy of mixing.
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- Δ S m = R
( n A ln x A + n B ln x B )
- Δ S m = R
( n A ln φ A + n B ln φ B )
The Entropy of Mixing
Regular Solution Theory
Flory - Huggins Theory
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Mole Fraction and Volume Fraction
The mole fraction of polymer is;
x p = n p
n p + n s = ������ 1
76
φp = = ������ 25 100 np mVs + nsVs
np mVs
The volume fraction is;
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The Entropy of Mixing
- ΔSm = k(25 ln0.25 +75 ln0.75)
- ΔSm = k(1 ln0.25 +75 ln0.75)
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The Entropy of Mixing
- Δ S m R
V r V =
V r V
n A ln φ A + n B ln φ B [ ]
n A V r V
= n A m A V r Vm A
= φ A m A
n B V r V =
n B m B V r Vm B
= φ B m B
- Δ S m R
V r V
= - Δ S ‘ m R
= φ A m A
ln φ A + φ B m B
ln φ B ⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
The small entropy of mixing polymers is more seen if we express it on a per mole of lattice sites basis (i.e. divide by the number of moles of lattice sites = V/Vr)
Using
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The Enthalpy of Mixing
• Dispersion Forces • Dipole/dipole Interactions • Hydrogen Bonding • Coulombic Interactions
Increasing Interaction Strength Non Polar
Highly Polar
Focus on dispersion and weak polar
forces }
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I can feel these guys But not
these
Nearest Neighbor Assumption
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B B
B B
B B B
B
Let the attractive interaction energy between a pair of small A molecules be εAA. If we assume that we can simply add the interactions between all pairs, then we can say that the interaction of a chosen A molecule with all its nearest neighbors is zεAA, where z is the number of nearest neighbors. A
A
A A
A A
A A
Interaction energy
Do the same for the B molecules
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B B What we really want, however, is the change in energy on going from the pure to the mixed state. In order to calculate this we first consider the change in energy when we replace interactions between a pair of A molecules, AA, and pairs of B molecules, BB, with AB pairs. The energy change per AB pair is given by;
A A
B A B A Δ ε AB = ε AB -
1 2 ε AA + ε BB [ ]
Change in Interaction Energy
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Δ H m = n A φ B z Δ ε AB [ ] A
A A
A
B B
B A
Probability that a B is next to a chosen A is φB
The Heat of Mixing
Δ H m = n AB Δ ε AB
Now all we have to do is multiply this by the number of AB pairs
Then
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The Flory Parameter χ
χ = z Δ ε AB kT
Δ H m = kT n A φ B χ [ ]
Flory defined the following interaction parameter, χ, which he made dimensionless by dividing by kT;
The heat of mixing is then simply
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The Flory-Huggins Equation
Δ G m = Δ H m - T Δ S m = kT n A φ B χ + n A ln φ A + n B ln φ B [ ]
Δ G m RT
= n A ln φ A + n B ln φ B + n A φ B χ
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Cohesive Energy Density
A
A A
A A
A A
A
B B
B B
B B B
B
C AA = Δ E vap
V = n A z ε AA
2 V A
C BB = Δ E vap
V = n B z ε BB 2 V B
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Cohesive Energy Density
A A
A A
A A
A B
C AB = ?
C AB = C AA 0 . 5 C BB
0 . 5
Δ E ~ - ( 2 C AB - C AA - C BB ) = ( C AA 0 . 5 - C BB 0 . 5 ) 2
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δ A = C AA 0 . 5
δ B = C BB 0 . 5
Δ E ~ ( δ A - δ B ) 2
Δ H m = ( n A + n B ) V m φ A φ B ( δ A - δ B ) 2
χ ≈ 0 . 34 + V r RT
δ A - δ B [ ] 2
Solubility Parameters
χ ≈ V r RT
δ A - δ B [ ] 2
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The Phase Behaviour of Polymer Solutions and Blends
Example of a phase separated or immiscible
system
Δ S total > 0 Δ Gmix < 0
One Condition for Mixing
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The Phase Behaviour of Polymer Solutions and Blends
Δ G ' m RT
= Δ G m RT
V r V
= φ A m A
ln φ A + φ A m A
ln φ B + φ A φ B χ
χ ≈ V r RT
δ A - δ B [ ] 2 Always Positive
Positive
Δ G ' m RT
= Δ G m RT
V r V
= φ A m A
ln φ A + φ A m A
ln φ B + φ A φ B χ
Small
Polymer Blends
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Phase Behaviour
Single Phase
(Miscible Mixture)
Tem
pera
ture
Composition
Phase Separated
(Immiscible)
χ ≈ V r RT
δ A - δ B [ ] 2
χ = a T
+ b
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xB1 xB2
B1
B2 Q
Q*
xB
ΔG
0 0
Phase Behaviour
ΔG1
ΔGtotal = ΔGQ = ΔG1 + ΔG2
ΔG2
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xB1 xB2
B1
B2 Q
Q*
xB
ΔG
0 0
Phase Behaviour
ΔGQ*
ΔGQ* < ΔGQ
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xB1 xB2
B1
B2 Q
Q*
xB
ΔG
0 0
The Chemical Potential and the Conditions for Phase Separation
ΔG1 ΔG2
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xB1 xB2
B1
B2
xB
ΔG
0 0
The Chemical Potential and the Conditions for Phase Separation
∂ Δ G ∂ n B
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ n B 1
= ∂ Δ G ∂ n B
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ n B 2
∂ Δ G ∂ x B
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ x B 1
= ∂ Δ G ∂ x B
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ x B 2
Δ µ B 1 = Δ µ B
2
or
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The Chemical Potential
Δ µ B 1 = Δ µ B
2 Composition B1
Composition B2
Δ µ A 1 = Δ µ A
2
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The Flory-Huggins Chemical Potential
Composition B1
Composition B2
Δ G m RT
= n s ln φ s + n p ln φ p + n s φ p χ
Δ µ s RT = ln( 1 - φ p ) + 1 -
1 M
⎡ ⎣
⎤ ⎦ ⎥ φ p + φ p
2 χ ⎥
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Phase Diagrams
Δ µ s 1 = Δ µ s
2 B1
B2
xB
ΔG Te
mpe
ratu
re
UCST
Composition
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xB1 xB2
B1
B2
xB
ΔG
S1 S2 C
0 0
More Phase Behaviour
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B1
B2
xB
ΔG
S1 S2
Tem
pera
ture
Composition
0 0
UCST Binodal
Spinodal
The Binodal and Spinodal
∂ 2 Δ G
∂ x B 2 = 0
∂ 3 Δ G
∂ x B 3 = 0
At the UCST we also have
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Tem
pera
ture
Composition
The Conditions for Forming a Miscible Mixture
Single Phase
Phase Separated
We can now summarize our conditions for forming a single phase or miscible mixture; a) The free energy change on mixing should be negative. b) The second derivative of the free energy of mixing should be positive (which means a point of inflection on the free energy curve has not been reached and it is concave upwards across the composition range).
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Tem
pera
ture
Tc
[φp]c Composition
The Critical Value of χ
∂ 2 ( Δ G m ' / RT )
∂ φ A 2 =
1 φ A m A
+ 1
φ B m B - 2 χ = 0
∂ 3 ( Δ G m
' / RT ) ∂ φ A 3
= - 1
φ A 2 m A +
1 φ B 2 m B
= 0
χ c = 1 2
1 + 1
m p 1 / 2
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
2
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The Critical Value of - Blends χ
∂ 2 ( Δ G m ' / RT )
∂ φ A 2 =
1 φ A m A
+ 1
φ B m B - 2 χ = 0
∂ 3 ( Δ G m
' / RT ) ∂ φ A 3
= - 1
φ A 2 m A +
1 φ B 2 m B
= 0
φ A [ ] c = m B
1 / 2
m A 1 / 2 + m B
1 / 2
χ c = 1 2
1 m A
1 / 2 + 1
m B 1 / 2
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
2
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UCST
Upper Critical Solution Temperature
Tem
pera
ture
Composition
The Critical Value of the Solubility Parameter Difference for Polymer Solutions
χ c = 0 . 34 + V r RT
δ p - δ s [ ] c 2
= 0 . 5
δ p - δ s [ ] ≈ ± 1
χ c = 1 2
1 + 1
m p 1 / 2
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
2
χ c ~ 1 / 2
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0.15 0.10 0.05 0.00 -10 -5 0 5
10 15 20
Volume Fraction of Polymer
T (0
C)
Flory - Huggins Theory
Experimental Data points
A Comparison of Theory and Experiment
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Values of for Solutions of ���Polystyrene in Cyclohexane
χ
1.0 0.8 0.6 0.4 0.2 0.0 0.4
0.6
0.8
1.0
1.2
1.4
VOLUME FRACTION OF POLYMER
χ
FLORY - HUGGINS
χ = χ 1 + χ 2 φ p + χ 3 φ p 2 + - - - - -
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The Lower Critical Solution Temperature
Tem
pera
ture
Composition
LCST - Lower Critical Solution Temperature
Phase Separated
Single Phase
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10,300 4800
19,800 10,300
4800 0.1 0.2 0.3 Polymer volume fraction
Tem
pera
ture
(°C) 190
140 100
50
-50 0
The Lower Critical Solution Temperature
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The Lower Critical Solution Temperature
UCST - Upper Critical Solution Temperature
Tem
pera
ture
Composition
LCST - Lower Critical Solution Temperature
An aside is in order here. The fact that the lower critical solution temperature is at a higher temperature than the upper critical solution temperature can be a bit confusing. The word “lower” in LCST is chosen to mean at the bottom (lowest temperature) of a two phase region, while the word “upper” in UCST designates the top of a two phase region.
Phase Separated
Phase Separated
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Some Limitations of the Flory - Huggins Theory
1. Based on a lattice model that uses various approximations in the “counting” process 2. Ignores "free volume” and only accounts for combinatorial entropy 3. Assumes random mixing of chains in calculating the entropy and segments in calculating the enthalpy 4. Only applies to non - polar molecules 5. Does not Apply to Dilute Solutions
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Dilute Solutions
• There is an intermolecular excluded volume effect in a good solvent (meaning χ << 0.5, the critical value). • There is also an intramolecular excluded volume effect, again in a good solvent, such that the chain expands its dimensions relative to the usual end-to-end distance found in concentrated solutions or the melt. • The chain expansion varies with χ, hence temperature, and there is a temperature, the theta temperature, where the chain again becomes ideal, or can be described by Gaussian statistics.
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Tem
pera
ture
The Theta Temperature
α 5 ~ m 0 . 5 1 2 - χ ⎡ ⎣ ⎢
⎤ ⎦ ⎥ +
φ p 3 + - - - - - -
R 2 0 . 5
~ α m 0 . 5 ~ m 0 . 6
At high temperatures
At the θ temperature
1 2
χ ∼
R 2 0 . 5
~ m 0 . 5