phase stability - university of cincinnati
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Phase Stability
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DG = DH – T DS
Supercool/Superheat/Supersaturate
At the equilibrium m elting /freezing point or equilibrium
concentration for precipitation the driving force for phase
separation, DG = DH – T DS = 0. So a phase will only form by
lowering the tem perature below the m elting point or increasing
the concentration above the solubility lim it. This is related to
“critical slowing down”. At the critical point nothing happens, or
anything can happen, there is no therm odynam ic direction for
the process.
Cooling below the m elting point produces a supercooled liquid,
raising the concentration above the saturation lim it produces a
supercritical solution. You can also heat a solid above the
equilibrium m elting point and produce a superheated solid.
The rate of phase separation depends on two factors:
DG and kinetics. Kinetics slows exponentially at lower
tem peratures D ~ exp(K/(T-T0)). The therm odynam ic driving
force, exp(-DG/kT), decays at higher tem peratures, so a
m axim um in grow th rate results below the equilibrium in the
super cooled liquid state.
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Kauzm ann Paradox, a therm odynam ic basis for the glass transition
The entropy of the liquid becom es sm aller than the entropy of the
solid at the Kauzm anntem perature, TK. This could be the infinite cooling glass transition tem perature.
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Superheating and M eltingSuperheating can occur since m elting occurs at surfaces and if the surfaces are stabilized then superheated solids can be produced
Grow th of a liquid phase relies on grow th of a m echanical instabilityA m echanical instability w ill not spontaneously grow if it occurs in a m eta-stable region in T and P:(dG/dx)=0 defines equilibrium or binodal; (d2G/dx2) = 0 defines the m etastable lim it or spinodal(d3G/dx3) = 0 defines the critical point
G = -ST + Vp, dG = -SdT + Vdp(d2G/dp2)T = (dV/dp)T < 0 and (d2G/dT2)p = -(dS/dT)p < 0
First requires that the bulk m odulus be positive,
Second requires positive heat capacity, (dS/dT)p = Cp/T > 0
Shear m odulus goes to 0 at
highest possible supercritical solid
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Therm al (energy) fluctuations are the basis of therm odynam ics
M ap of energy in 3d space at one tim e slice
https://upload.wikim edia.org /wikipedia/com m ons/2/2a/Quantum _Fluctuations.gif
Therm odynam ic Hypothesis:
System s are always probing free energy space (tem perature, com position, pressure, m agnetic field, electrical field, energy, extent of reaction, etc.) through random fluctuations. This
enables the attainm ent of the lowestfree energy at an equilibrium state.
This hypothesis leads to consideration of the partial derivatives of free energy
as defining features in free energy space particularly critical points (d3G/dx3=0; d2G/dx2=0; dG/dx=0), equilibrium (binodal) points (dG/dx=0), and spinodal points (d2G/dx2=0).
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The book considers first a reversible chem ical reaction A <=> BCyclohexane from boat to chair conform ation for instance
As tem perature changes you can observer a different m ix of states, E = kBT ~ 2.5 kJ/m ole at RTBut fluctuations allow for 0.1 % boat conform ation. At 1073K 30% boat. Probability is exp(-E/kT).
The percent in boat can be m easured using NM R spectroscopy.
Chair Chair
Boat
TransitionState
M etastable
TransitionState
The equilibrium point depends on tem perature,
kBT
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Extent of reaction would be the conversion from chair (A) to boat (B), for instance, at a tem perature T. The m inim um , (dG/dx) = 0 is the equilibrium
point. The free energy plot can be concave up (stable equilibrium ) concave down (unstable), flat (critical point/tem perature), and other shapes
A <=> B
Gibbs-Duhem Equation
Affinity
At tem perature T, the reaction is like a spring bouncing between
A and B oscillating about the equilibrium point since any deviation from equilibrium increases the free energy.
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a) Stable dG/dx = 0b) Unstable
c) Spinodal d2G/dx2 = 0d) M etastable (This can depend on your vantage point)
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Affinity as a Taylor series in the reaction coordinate
z is a fluctuation in a reaction coordinate, T, p, m , x , etc.
At equilibrium the affinity is 0 so = 0
If then the equilibrium is stable
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Construction of a Phase Diagram Based on Fluctuations
Consider the Hildebrand M odel
Binodal phase equilibria is defined by
xA = 1-xB
Spinodal is defined by
Critical Point is defined by
d3G m/dxB3 = RT(1/(1-xB)2 - 1/xB2) = 0 or xA = xB = 0.5
Using this com position in d2G m/dxB2 = 0 yields
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Hildebrand Equation
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If W = -A + BT, and the entropy term is sm all (for polym ers for instance) B is non-com binatorial entropyPhase separation occurs on heating: Lower Critical Solution Tem perature (LCST)
Flory-Huggins Equation
fA is the volum e fraction of AN A is the num ber of “c” units in A
nc is the total num ber of c volum e units in the systemC is an average of A and B unitsVolum e fraction has replace m ole fractionc is an average interaction energy per c site per RT
c ~ W /RT
Sym m etric blend N A = N B
Entropy part is sm all since N i ~ 100,000c ~ a - b/T leads to LCST behavior, a is non-com binatorial entropy, b is the enthalpy of specific
interactions that leads to m iscibilty
2 Phase
1 Phase
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Since cN depends on 1/T specifying cN specifies the tem perature. Large cN is low tem pearature.
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Joon Roe
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Spinodal
BB
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Spinodal
BB
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Higgins JS, Cabral JT A Thorny
Problem? Spinodal Decomposition in Polymer Blends M acromolecules 534137−4140 (2020)
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Series expansion of the free energy density yields,
Positive second derivative leads to increase (Stable)Negative to decrease (Phase Separation)
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Diffusion in Nucleation and Grow th and in Spinodal Decom position
Fickian diffusion down a concentration gradient
Spinodal diffusion up a concentration gradient driven by free energy
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Phase grows from fluctuations
Higgins JS, Cabral JT A Thorny Problem? Spinodal Decomposition in Polymer Blends M acromolecules 53 4137−4140 (2020)
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Activation Barrier for Nucleation and Growth
Balance between bulk and surfaceFree energies
(Next chapter)
Different relative bulk to surface free
energies (sm all to large)
Rate of Nucleation
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Critical Scattering
A polym er chain in a m elt is a
random walk like a diffusion
path so M ass ~ size2
By Bragg ’s law size is related to
inverse angle
d = 2/(l sinq)
Reduced angle is q = 4p/l sinqIntensity (S) scales with M ass
So S ~ q-2
Then a plot of 1/S vs q2 should
be a line that reflects the
inverse of contrast at q = 0
At the critical point contrast
goes to ∞ so 1/S goes to 0.
Deuteration provides contrast
for neutrons, l = 10Å.
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