Phase Diagram sGibbs Phase Rule

Eutectic Phase Diagram Ag + CuUnivariant Equilibrium

LiquidusSolidus

Invariant EquilibriumEutectic

Lever RuleTieline (conode)

and

Silver acts like a solvent to

copper and copper acts like a solvent to silver w ith lim ited solubility that is

a function of tem perature with a solubility lim it at the eutectic point (3

phases in equilibrium )

1

L => a + B

Phase Diagram sGibbs Phase Rule

Eutectic Phase Diagram Ag + CuUnivariant Equilibrium

LiquidusSolidus

Invariant EquilibriumEutectic

Lever RuleTieline (conode)

and

2

Phase Diagram s

3

Phase Diagram sSlow Cool M orphologyConsider the m orphology in a slow cool

from T1 (liquid) to T3 (solid). Just below T1 dom ains of Cu/Ag at about 98% Cu form in the liquid m atrix. The fraction Cu/Ag solid increases as tem perature drops, probably growing on the seeds form ed at T1. By T2

the m atrix is solid with dom ains of liquid. The com position of the m atrix is richer in Ag at the outside, close to 95% Cu. At the eutectic dom ains of Cu/Ag with 15% Cu becom e im m ersed in a m atrix of about 95%

Cu/Ag. The entire system is solid so diffusion essentially stops and the structure is locked.

Rapid Quench M orphologyFor a rapid quench to T3, dom ains of 10% Cu form in a m atrix of 97% Cu.

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Com m on Tangent Construction

Does dG/dxi = dG/dni = µ i?NodG/dni = dG/dxi dxi/dnidxi/dni = d(ni/(ni+nj))/dni = 1/(ni+nj) – ni/(ni+nj)2

T3

Solid Solutions

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Si-Ge phase diagramSi-Ge Gibbs Free Energy

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Solve for xBSS xBliq since xA + xB =1

Ideal

Cp(T) is constant

Calculate the Phase Diagram for a Solid Solution

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8

https://webbook.nist.gov/chemistry/name-ser/

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Solid solution is flatter than ideal (Pos. deviation or destabilized) Liquid is deeper than ideal (Neg. Deviation or stabilized)

Deviations are associated with m inim a in phase diagram10

Solid solution is flatter than ideal (Pos. deviation or destabilized) Liquid is deeper than ideal (Neg. Deviation or stabilized)

Deviations are associated with m inim a in phase diagram11

Azeotrope

Heteroazeotrope

Liquid/Vapor Equilibria

Ideal

Solid/Liquid Equilibria

Ideal

Eutectic

Non-IdealCongruent m elting solid solution

Phase diagram is split into two phase

diagram s with a special com position that acts as a pure com ponent

Same crystallographic structure in solid solution phase

Different crystallographic structures in solid solution phases

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L => a

L => a + b

L => a + L

Eutectic Phase Diagram w ith Three Crystallographic Phases, Two Cubic and One Hexagonal

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Regular Solution M odeling for Phase Diagram s

G E/RT = xA ln gA + xB ln gB Generic expression using activity coefficient

G E/RT = W xA xB Hildebrand Regular Solution expression

Two unknowns & Two equations:

We found that a solution of RT ln gA = W xB2 and RT ln gB = W xA

2 worked

Two unknowns, xBSS, xB

liq & Two equations

Solve num erically

Generate Phase diagram since Dµ is a function of T

Positive W is necessary for phase separationDG m = RT(xA ln(xA) + xB ln(xB)) + W xAxB 14

Two Different Crystallographic

Phases at Equilibrium

One Crystallographic

Phase

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g => a + b

Polyvinylm ethyl Ether/Polystyrene (LCST Phase behavior)

DG m = RT(xA ln(xA) + xB ln(xB)) + WxAxB

W m ust have a tem perature

dependence for UCSTW = A + B/T so that it gets sm aller w ith increasing tem perature this is a non-com binatorial entropy i.e. ordering on m ixing

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2-Phase

Single Phase

2-Phase

Single Phase

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Pure A and B melt at 800 and 1000 K with the entropy of fusion of both compounds set to 10 J K –1 mol–1 typical for metals. The interaction coefficients of the two solutions have been varied systematically in order to generate nine different phase diagrams.

Solid & liquid IdealW liq = -15 kJW ss = 0

W liq = 0 W ss = 0

W liq = +15 kJW ss = 0

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Pure A and B melt at 800 and 1000 K with the entropy of fusion of both compounds set to 10 J K –1 mol–1 typical for metals. The interaction coefficients of the two solutions have been varied systematically in order to generate nine different phase diagrams.

Solid & liquid IdealW liq = -15 kJW ss = 0

Liquid stabilized

W liq = 0 W ss = 0

W liq = +15 kJW ss = 0

Liquid destablized

congruently m elting solid solution

congruently m elting solid solution

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Pure A and B melt at 800 and 1000 K with the entropy of fusion of both compounds set to 10 J K –1 mol–1 typical for metals. The interaction coefficients of the two solutions have been varied systematically in order to generate nine different phase diagrams.

Solid destabilized

W liq = -15W ss = 15 kJ

Solid destabilized & liquid stabilized

W liq = 0 kJ W ss = 15 kJ

W liq = +15 kJW ss = +15 kJ

Solid & Liquiddestablized

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Pure A and B melt at 800 and 1000 K with the entropy of fusion of both compounds set to 10 J K –1 mol–1 typical for metals. The interaction coefficients of the two solutions have been varied systematically in order to generate nine different phase diagrams.

Solid stabilized liquid destabilized

W liq = 0W ss = -10 kJ

Solid stabilized

W liq = +10 kJ W ss = -10 kJ

W liq = -10 kJW ss = -10 kJ

Solid & Liquidstablized

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If it is assum ed that

Then you need only one W(The solid state doesn’t m ix)

Regular solution m odel has only a few param eters

This m eans that experim entally you need only determ ine a few points of equilibrium com position to predict the entire phase diagram .

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Invariant Phase Equilibria

There are three degrees of freedom for a binary system . If three phases m eet the

conditions are invariant. This m eans that the phase com position acts like a pure

com ponent. An azeotrope or congruent m elting point splits the phase diagram into two

phase diagram s. There are several types of “invariant reactions”.

Azeotrope

Congruent m elting point

Hex to Cubic Phase transition of Y2O 3

Peritectic Reaction

Peritectoid = 3 solid phases

Eutectic Reaction

Eutectoid = liquid & 2 solid phases

M onotectic Reaction L1 => S + L2

Syntectic Reaction L1 + L2 => S

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Stoichiom etric Interm ediate Phase, CaZrO 3

Stoichiom etric Interm ediate Phase

Only stable at one exact com position w here it has a very low free energy

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The intermediate phases in the system Si–Ti display also a variety

of other features. W hile TiSi2 and Ti5Si3 are congruently m elting phases (melt and crystal have same composition), Ti5Si4 and TiSi m elt incongruently . Finally, Ti3Si decomposes to b-Ti and Ti5Si3 at T = 1170 K, in a peritectoid reaction (3 solid phases) while b-Tidecomposes eutectoidally (L and 2S) on cooling forming a-Ti + Ti3Si

at T = 862 K.

Interm ediate Phases

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Antimony

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M onotectic Phase Diagram

L1 => L2 + a

Self-lubricating bearingsNano-structure Bi

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Freezing Point Depression

dµB,Solid = Vm B,Solid dP – Sm B,Solid dT + RT d(lnaB,Solid)

Pure solid in equilibrium with a binary solution following Henry’s Law

Isobaric, pure com ponent B so lnaB,Solid = 0

dµB,Solid = – Sm B,Solid dTfp

Binary solution following Henry ’s Law

dµB,Solution = – Sm B,Solultion dTfp + RTfp d(lnyB,Solution)P,T

For sm all x: e-x = 1 - x + … or ln(1-x) = -xSo for sm all yB,Solution: lnyB,Solution ~ -yA,Solution

So,

Sm B,Solid dTfp = Sm B,Solution dTfp + RTfp dyA,Solution

dyA,Solution = (SmB,Solid - Sm

B,Solution)/(RTfp) dTfp ~ -DSmB/(RTfp) dTfp = -DH m

B/(RTF) (dTfp)/Tfp

yA,Solution = -DH m B/(RTF) ln(Tfp/TF) For sm all x: ln(x) = x – 1

yA,Solution = -DH mB/(RTF) (Tfp/TF - 1) = -DH m

B/(RT2F) DT

Tfp = TF - yA,SolutionRT2F/DH m B

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DT = -RT2FyA,Solution/ DH m B

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Ternary Phase Diagram , xA, xB, xC W ith no grid you can get the com positionFrom the norm als

Go counter clockwise0 at start 1 at corner

Phase point is where parallel lines m eetPlot is isotherm al, isobaric, isochoricxA + xB + xC = 1

This can be read if the plot hasa grid

You can also use intersection lines to get the com position

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31

Predom inance Diagram (W hich species is dom inant)

pH = -log [H +]pCr = -log [Cr]

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