“Phase Equilibria in Materials”

“Phase Equilibria in Materials”

Eun Soo Park

Office: 33-313

Telephone: 880-7221 Email: espark@snu.ac.kr

Office hours: by an appointment

2021 Spring

03.09.2021

1

Gibbs Free Energy as a Function of Temp. and Pressure

- Single component system Chapter 1

= 0 dG

P V S G

T G

T P

 =



 





∂

− ∂

 =



 





∂

∂ ,

Thermondynamics and Phase Diagrams

- Equilibrium

2 1

0

G G G

∆ = − <

- Phase Transformation

Contents for previous class

Lowest possible value of G

V T

H dT

dP

eq

eq

∆

= ∆ ) (

Clausius-Clapeyron Relation

- Classification of phase transition

First order transition: CDD/Second order transition: CCD

3

The First-Order Transitions

G

T S

T

∆S = L/T

T C_P

N P

P T

T S C

,



 





∂

= ∂ Latent heat

Energy barrier

Discontinuous entropy, heat capacity

The Second Order Transition

G

T S

T

∆S=0

Second-order transition

T C_P

No Latent heat

Continuous entropy

∞

 →



 





∂

= ∂

N P

P T

T S C

,

Liquid Undercooled Liquid Solid

<Thermodynamic>

Solidification:

• Interfacial energy ΔT_N

Melting:

• Interfacial energy

SV LV

SL

γ γ

γ + <

No superheating required!

No ΔT_N T_m

vapor

Melting and Crystallization are Thermodynamic Transitions

6 NANO STRUCTURED MATERIALS LAB.

T

7

Contents for today’s class

- Chemical potential ^and Activity

- Gibbs Free Energy in Binary System

Ideal solution and Regular solution

- Binary System mixture/ solution / compound

Hume-Rothery Rules for Alloys

8

Q1: What are binary systems?

“Mixture vs. Solution vs. Compound”

Multi-component system:

9

*

Binary system (two components)

- Mixture ; A – A, B – B ;  the physical combination of two or

more substances on which the identities and boundaries are retained.

A B

* Single component system

: Equilibrium depends on pressure and

: Equilibrium depends on not only pressure and temperature but also composition.

Alluvial mining

Winnowing

wash rice Select egg

10

- Solution ; A – A – A ;  atomic scale mixture/ Random distribution A – B – A Solid solution : substitutional or interstitial

- Compound ; A – B – A – B ;  fixed A, B positions/ Ordered state B – A – B – A

MgCNi3

11

Q2: What is “Alloying”?

Ordered Compounds or Solid Solutions

12 12

Ordered Substitutional and Interstititials Compounds

Substitutional Interstitial

element replaces host atoms element goes into holes in an orderly arrangement in an orderly arrangement

e.g., Ni₃Al (hi-T yield strength),

Al₃(Li,Zr) (strengthening) e.g., small impurities, clays ionic crystals, ceramics.

“Alloying” : atoms mixed on a lattice

Ordered Compounds and Solid Solutions

13

Intermetallic Compounds

Mg₂Pb

Intermetallic compounds form lines - not areas - because stoichiometry (i.e.

composition) is exact.

Antifluorite Structure:

• FCC unit cell with anions occupying FCC sites

• Cations occupying 8

octahedral interstitial sites

14

- Solution ; A – A – A ;  atomic scale mixing, Random distribution A – B – A

Two Possibilities for Solid Solutions: B atoms in A atoms

Substitutional Interstitials

‘new element replaces host atoms’ ‘new element goes in holes’

15

Q3: “Solution vs. Intermetallic compound”?

ruthenium 루테늄 《백금류의 금속 원소;기호 Ru, 번호 44》 16

Solid Solution vs. Intermetallic Compound

Crystal structure

Surface

Pt_0.5Ru_0.5 – Pt structure (fcc) PtPb – NiAS structure

solid solution and ordered compound Alloying: atoms mixed on a lattice

17

Solid Solution vs. Intermetallic Compounds

Pt_0.5Ru_0.5 – Pt structure (fcc) PbPt – NiAS structure

: in order to introduce some of the basic concepts of the thermodynamics of alloys

Assumption: a simple physical model for “binary solid solutions”

18

• Solid solution of B in A plus particles of a new phase (usually for a larger amount of B)

Second phase particle --different composition --often different structure.

Particles of New Phase in Solid-Solution Alloys

Solid Solution phase: B atoms in A

19

c. Mechanism of Precipitation

(1)

(2) (3)

(1)

(2)

(3)

(1) (2) (3)

Atomic diffusion Precipitate

Matrix atom

Alloying atom

Composition

Temperature

5) Microstructure control : ② Secondary phase control

20

Q4: How can we classify “Solubility”?

21

Solubility

• Unlimited Solubility

– Hume Rothery’ Conditions

• Similar Size

• Same Crystal Structure

• Same Valance

• Similar Electronegativity

– Implies single phase

• Limited Solubility

– Implies multiple phases

• No Solubility

- oil and water region

22

Cu-Ag Alloys Cu-Ni Alloys

complete solid solution limited solid solution

23

* Complete immiscibility of two metals does not exist.

: The solubility of one metal in another may be so low (e.g. Cu in Ge <10^-7 at%.)

that it is difficult to detect experimentally, but there will always be a measure of solubility.

24

25

26

27

Q5: Can we roughly estimate

what atoms will form solid solutions?

“Hume-Rothery Rules”

28

Hume-Rothery Rules for Alloys (atoms mixing on a lattice)

Will mixing 2 (or more) different types of atoms lead to a solid-solution phase?

Empirical observations have identified 4 major contributors through :

+1 +2

Atomic Size Factor , Crystal Structure, Electronegativity, Valences

29

Hume-Rothery Rules for Mixing

Empirical rules for substitutional solid-solution formation were identified from experiment that are not exact, but give an expectation of formation.

Briefly,

1) Atomic Size Factor The 15% Rule

If "size difference" of elements are greater than ±15%, the lattice distortions (i.e. local lattice strain) are too big and solid-solution will not be favored.

DR%= < ±15% will not disallow formation.

2) Crystal Structure Like elemental crystal structures are better

For appreciable solubility, the crystal structure for metals must be the same.

3) Electronegativity DE ~ 0 favors solid-solution.

The more electropositive one element and the more electronegative the other, then "intermetallic compounds" (order alloys) are more likely.

4) Valences Higher in lower alright. Lower in higher, it’s a fight.

A metal will dissolve another metal of higher valency more than one of lower valency.

r_solute − r_solvent

r_solvent x100%

30

Is solid-solution favorable, or not?

• Cu-Ni Alloys

Rule 1: r_Cu = 0.128 nm and r_Ni= 0.125 nm.

DR%= = 2.3% favorable √

Rule 2: Ni and Cu have the FCC crystal structure. favorable √ Rule 3: E_Cu = 1.90 and E_Ni= 1.80. Thus, DE%= -5.2% favorable √ Rule 4: Valency of Ni and Cu are both +2. favorable √

Expect Ni and Cu forms S.S. over wide composition range.

At high T, it does (helpful processing info), but actually phase separates at low T due to energetics (quantum mechanics).

Hume-Rothery Empirical Rules in Action

r_solute

− r

r_solvent x100%

31

Is solid-solution favorable, or not?

• Cu-Ag Alloys

Rule 1: r_Cu = 0.128 nm and r_Ag= 0.144 nm.

DR%= = 9.4% favorable √

Rule 2: Ag and Cu have the FCC crystal structure. favorable √ Rule 3: E_Cu = 1.90 and E_Ni= 1.80. Thus, DE%= -5.2% favorable √ Rule 4: Valency of Cu is +2 and Ag is +1. NOT favorable

Expect Ag and Cu have limited solubility.

In fact, the Cu-Ag phase diagram (T vs. c) shows that a solubility of only 18% Ag can be achieved at high T in the Cu-rich alloys.

r_solute − r_solvent

r_solvent x100%

Hume-Rothery Empirical Rules in Action

32

Cu-Ag Alloys Cu-Ni Alloys

complete solid solution limited solid solution

33

(1) Thermodynamic : high entropy effect (2) Kinetics : sluggish diffusion effect (3) Structure : severe lattice distortion effect (4) Property : cocktail effect

Major element 2

Major element 3 Major

element 4, 5..

Major element 1

Minor element 2 Minor element 3

Major element

Traditional alloys

Minor element 1

Conventional alloy system High entropy alloy system

Yong Zhang et al., Adv. Eng. Mat. P534-538, 2008

Ex) 304 steel - Fe74Cr18Ni8 Ex) Al²⁰Co²⁰Cr²⁰Fe²⁰Ni²⁰

Severe lattice distortion → Sluggish diffusion & Thermal stability

High entropy alloy (HEA)

34

Q6: How to calculate

“Gibbs Free Energy of Binary Solutions”?

G

= G

+ ΔG

J/mol

35

* Composition in mole fraction X_A, X_B X_A+ X_B = 1

Step 1. bring together X_A mole of pure A and X_B mole of pure B

Step 2. allow the A and B atoms to mix together to make a homogeneous solid solution.

2) Gibbs Free Energy of Binary Solutions

: binary solid solution/ a fixed pressure of 1 atm Binary Solutions

36

Gibbs Free Energy of The System

In Step 1

- The molar free energies of pure A and pure B pure A;

pure B;

;X

, X

(mole fraction

, (T P G_A

) , (T P G_B

G

= X

G

+ X

G

J/mol

Free energy of mixture

1.3 Binary Solutions

37

Gibbs Free Energy of The System

In Step 2 G

= G

+ ΔG

J/mol

Since G

and G

And putting

∆G_mix = ∆H_mix - T∆S_mix

∆H_mix : Heat of Solution i.e. heat absorbed or evolved

during step 2

∆S_mix : difference in entropy

between the mixed and unmixed state

How can you estimate ΔH

and ΔS

?

1.3 Binary Solutions

38

Q7: How can you estimate

“ΔG _mix of ideal solid solution”?

Gibbs Free Energy of Binary Solutions

) ln

ln

(

mix

RT X X X X

G = +

∆

ΔG

= -T ΔS

39

Mixing free energy, ΔG

Assumption 1; ∆ H

=0 :

; A & B = complete solid solution ( A,B ; same crystal structure)

; no volume change

- Ideal solution

w k

S = ln

 thermal; vibration ( no volume change )

 Configuration; number of distinguishable ways of arranging the atoms

ΔG

= -T ΔS

∆G

= ∆ ^H

- T ∆ ^S

Entropy can be computed from randomness by Boltzmann equation, i.e.,

=

+

S S S

1.3 Binary Solutions

40

)]

ln (

) ln

( ) ln

[(

,

,

1

! ln

!

)!

ln (

0 0

0

o B o

B o

B o

A o

A o

A o

o o

B A

B B

A A

B A

B before A

after mix

N X N

X N

X N

X N

X N

X N

N N

k

N N

N N

X N

N X N

N k N

N k N

S S

S

−

−

−

−

−

=

= +

=

=

→

+ −

=

−

=

∆

N N

N

N ! ≈ ln − ln

Excess mixing Entropy

) , _(

_

!

!

)!

(

) _

_(

_

1

B A B

A B A

config config

N N solution after

N N

N w N

pureB pureA

solution before

w

+ →

=

→

=

If there is no volume change or heat change,

Number of distinguishable way of atomic arrangement

kN

R =

using Stirling’s approximation and

1.3 Binary Solutions

Since we are dealing with 1 mol of solution,

) ln

ln

( X

X

X

X

R +

−

=

(the universal gas constant)

41

Excess mixing Entropy

) ln

ln

(

mix

R X X X X

S = − +

∆

) ln

ln

(

mix

RT X X X X

G = +

∆

ΔG

= -T ΔS

1.3 Binary Solutions

Fig. 1.9 Free energy of mixing for an ideal solution

42

Q8: How can you estimate

“Molar Free energy for ideal solid solution”?

Gibbs Free Energy of Binary Solutions

G₂ = G₁ + ΔG_mix G = X_AG_A + X_BG_B + RT(X_AlnX_A + X_BlnX_B)

43

Since ΔH_mix = 0 for ideal solution,

Compare G_solution between high and low Temp.

G₂ = G₁ + ΔG_mix

G₁ ΔG_mix

1) Ideal solution

1.3 Binary Solutions

Fig. 1.10 The molar free energy (free energy per mole of solution) for an ideal solid solution.

A combination of Figs. 1.8 and 1.9.

+ =

G = X

G

+ X

G

+ RT(X

lnX

+ X

lnX

)

44

Q9: How the free energy of a given phase will change when atoms are added or removed?

“ Chemical potential ”

Gibbs Free Energy of Binary Solutions

45

Chemical potential

= µ

dG' dn

The increase of the total free energy of the system by the increase of very small quantity of A, dn

, will be proportional to μ

.

dn_A~ small enough

(∵ µ_A depends on the composition of phase)

(T, P, n

: constant )

G

= H-TS

For A-B binary solution, dG' ^{= µ}

dn

^{+ µ}

dn

= − + + µ

+ µ

dG' SdT VdP dn dn

For variable T and P

1) Ideal solution

1.3 Binary Solutions

µ

:

 ∂ 

µ = ^A ∂ ^A T, P, n_B

G' n

 ∂ 

µ = ^B  ∂ ^B T, P, n_A

G' n

46

47

Chemical potential

= µ

dG' dn

The increase of the total free energy of the system by the increase of very small quantity of A, dn

, will be proportional to μ

.

dn_A~ small enough

(∵ µ_A depends on the composition of phase)

(T, P, n

: constant )

G

= H-TS

For A-B binary solution, dG' ^{= µ}

dn

^{+ µ}

dn

= − + + µ

+ µ

dG' SdT VdP dn dn

For variable T and P

1) Ideal solution

1.3 Binary Solutions

µ

:

 ∂ 

µ = ^A ∂ ^A T, P, n_B

G' n

 ∂ 

µ = ^B  ∂ ^B T, P, n_A

G' n

48

Q10: “Correlation between chemical potential and free energy”?

Gibbs Free Energy of Binary Solutions

49

−1

= µ

+ µ

G X X Jmol

A A B B

dG = µ dX + µ dX

B A

B

dG

dX = µ − µ

A B

B

dG µ = µ − dX

( ¹ )

B A B B

B

B A A B B

B

B A

B

B B

B

G dG X X

dX

X dG X X

dX dG X dX

dG X

dX

 

= µ −   + µ

 

= µ − + µ

= µ −

= µ − −

B A

B

G dG X µ = + dX

For 1 mole of the solution (T, P: constant )

1) Ideal solution

Correlation between chemical potential and free energy

50

−1

= µ

+ µ

G X X Jmol

1) Ideal solution For 1 mole of the solution

Correlation between chemical potential and free energy

G_A G_B

μ_A

μ_B

X_A X_B X_B

B A

B

G dG X µ = + dX

) 1

(

B

B

X

dX

G + dG − µ =

B B

A

B

dX

X dG

X )

( +

−

= µ

G

A B C

X

A B

B

dG µ = µ − dX

CB DC

DA − −

=

D

X

DC DA +

=

51

A B

B

dG µ = µ − dX

−1

= µ

+ µ

G X X Jmol

B A

B

G dG X

µ = + dX

For 1 mole of the solution (T, P: constant )

Correlation between chemical potential and free energy

1) Ideal solution

Fig. 1.11 The relationship between the free energy curve for a solution and the chemical potentials of the components.

52

1) Ideal solution 1.3 Binary Solutions

μ_B’

μ_A’ X_B’

Fig. 1.12 The relationship between the free energy curve and Chemical potentials for an ideal solution.

= µ

+ µ

G X X

B B

B A

A

A

RT X X G RT X X

G ln ) ( ln )

( + + +

=

53

Contents for today’s class

- Chemical potential ^and Activity

- Gibbs Free Energy in Binary System - Binary System

Ideal solution

Regular solution

mixture/ solution / compound

G

= G

+ ΔG

J/mol G

= X

G

+ X

G

J/mol

(∆H

=0) ^{∆ G}

^{= RT ( X}

^{ln X}

^{+ X}

^{ln X}

⁾

∆

=

ε ε = ε −

1 ε + ε

H P where ( )

2

) ln

ln

(

B A B

B A

A

G X G X X RT X X X X

X

G = + + Ω + +

 ∂ 

µ = ^A  ∂ ^A T, P, n_B

G' n

• μ

= G

+ RTlna

dn_A~ small enough (∵µ_A depends on the composition of phase)

“Phase Equilibria in Materials”

Eun Soo Park

Office: 33-313

Telephone: 880-7221 Email: espark@snu.ac.kr

Office hours: by an appointment

2021 Spring

03.11.2021

1

2

Contents for previous class

- Chemical potential ^and Activity

- Gibbs Free Energy in Binary System - Binary System

Ideal solution

Regular solution

mixture/ solution / compound

G

= G

+ ΔG

J/mol G

= X

G

+ X

G

J/mol

(∆H

=0) ^{∆ G}

^{= RT ( X}

^{ln X}

^{+ X}

^{ln X}

⁾

∆

=

ε ε = ε −

1 ε + ε

H P where ( )

2

) ln

ln

(

B A B

B A

A

G X G X X RT X X X X

X

G = + + Ω + +

 ∂ 

µ = ^A  ∂ ^A T, P, n_B

G' n

• μ

= G

+ RTlna

dn_A~ small enough (∵ ^µA depends on the composition of phase)

3

Solid Solution vs. Intermetallic Compounds

Pt_0.5Ru_0.5 – Pt structure (fcc) PbPt – NiAS structure

*

Binary System (two components)

: Equilibrium depends on not only pressure and temperature but also composition.

- atomic scale mixture/ Random distribution on lattice - fixed A, B positions/ Ordered state

4

• Solid solution:

– Crystalline solid

– Multicomponent yet homogeneous

– Impurities are randomly distributed throughout the lattice

• Factors favoring solubility of B in A (Hume-Rothery Rules) – Similar atomic size: ∆r/r ≤ 15%

– Same crystal structure for A and B

– Similar electronegativities: |χ_A – χ_B| ≤ 0.6 (preferably ≤ 0.4) – Similar valence

• If all four criteria are met: complete solid solution

• If any criterion is not met: limited solid solution

Empirical rules for substitutional solid-solution formation were identified from experiment that are not exact, but give an expectation of formation.

Hume-Rothery Rules for Mixing

Cu-Ag Alloys Cu-Ni Alloys

complete solid solution limited solid solution

: in order to introduce some of the basic concepts of the thermodynamics of alloys

Assumption: a simple physical model for “binary solid solutions”

6

Chemical potential

= µ

dG' dn

The increase of the total free energy of the system by the increase of very small quantity of A, dn

, will be proportional to μ

.

dn_A~ small enough

(∵ ^µA depends on the composition of phase)

(T, P, n

: constant )

G

= H-TS

For A-B binary solution, dG' ^{= µ}

dn

^{+ µ}

dn

= − + + µ

+ µ

dG' SdT VdP dn dn

For variable T and P

1) Ideal solution

1.3 Binary Solutions

µ

:

 ∂ 

µ = ^A ∂ ^A T, P, n_B

G' n

 ∂ 

µ = ^B  ∂ ^B T, P, n_A

G' n

B B

B A

A

A

RT X X G RT X X

G ln ) ( ln )

( + + +

=

1) Ideal solution 1.3 Binary Solutions

μ_B’

μ_A’ X_B’

Fig. 1.12 The relationship between the free energy curve and

Chemical potentials for an ideal solution. 7

Contents for today’s class

- Equilibrium in heterogeneous system - Binary Solid Solution

Real solution

Ideal solution and Regular solution

- Ordered phases: SRO & LRO, superlattice, Intermediate phase (intermetallic compound) - Clustering

: Chemical potential

Activity

8

Q1: What is “Regular Solution”?

∆Gmix = ∆Hmix - T∆Smix

1.3 Binary Solutions

9

Ideal solution : ∆H

= 0

Quasi-chemical model assumes that heat of mixing, ∆H

, is only due to the bond energies between adjacent atoms.

Structure model of a binary solution

Regular Solutions

∆Gmix = ∆Hmix - T∆Smix

1.3 Binary Solutions

Assumption: the volumes of pure A and B are equal and do not change during mixing so that the interatomic distance and bond energies are independent of composition.

This type of behavior is exceptional in practice and usually mixing is endothermic or exothermic.

Fig. 1.13 The different types of interatomic bond in a solid solution.

10

Q2: How can you estimate

“ΔH _mix of regular solution”?

Gibbs Free Energy of Regular Solutions

∆H

= ΩX

X

where Ω = N

z ε

11

Bond energy Number of bond A-A ε

P

B-B ε

P

A-B ε

P

=

ε +

ε +

ε

E P P P

∆

=

ε ε = ε −

1 ε + ε

H P where ( )

2

Internal energy of the solution

Regular Solutions

1.3 Binary Solutions

12

∆ H

= 0

Completely random arrangement

ideal solution

AB

=

a

P N zX X bonds per mole N : Avogadro ' s number

z : number of bonds per atom

∆H

= ΩX

X

where Ω = N

z ε

∆ H

= P

ε

Regular Solutions

= 0

ε ( )

2 1

BB AA

AB

ε ε

ε = +

↑

→

< 0 P

ε ε ^{> 0 → P}

^↓

≈ 0 ε

Ω >0

1.3 Binary Solutions

(1) (2)

(3)

Fig. 1.14 The variation of ΔHmixwith composition for a regular solution.

13

Q3: How can you estimate

“Molar Free energy for regular solution”?

) ln

ln

(

B A B

B A

A

G X G X X RT X X X X

X

G = + + Ω + +

G₂ = G₁ + ΔG_mix ∆H_mix -T∆S_mix

14

Gibbs Free Energy of Regular Solutions

Regular Solutions

Reference state

Pure metal

G

= G

= 0

) ln

ln

(

B A B

B A

A

G X G X X RT X X X X

X

G = + + Ω + +

G₂ = G₁ + ΔG_mix

∆G_mix

=

∆H_mix -T∆S_mix

15

Q4: How can you calculate

“critical temperature, T _c ”?

16

Gibbs Free Energy of Regular Solutions

17

ε > 0, ∆H

> 0 / ∆H

~

Regular Solutions

∆G_mix

=

∆H_mix -T∆S_mix

Reference state

Pure metal

G

= G

= 0

(a)

(b)

19

where,

: energy term

Low temp.

High temp.

* Variation of free energy with composition for a homogeneous solution with ΔH_mix > 0

→The curves with kT/C < 0.5 show two minima, which approach

each other as the temperature rise.

20

where,

Taking the curve kT/C = 0.25 as an example, we can substitute C=4kT in upper eq. to obtain

∆H_mix ∆S_mix

* Free energy curve exhibit two minima at X_A

= 0.02 and X

=0.98

where,

Low temp.

High temp.

* The curves with kT/C < 0.5 show two minima, which approach each other as the temperature rise.

22

where,

: energy term

Low temp.

High temp.

* With kT/C ≥ 0.5 there is a

continuous fall in free energy from X_A=0 to X_A=0.5 and X_A=1.0 to

X_A=0.5. The free energy curve thus assumes the characteristic from one associates with the formation of homogeneous solutions.

* The curves with kT/C < 0.5 show two minima, which approach

each other as the temperature rise.

23

where,

: solubility limit

Figure. 21. Solubility curve obtained from eqn. (100)

by plotting the concentration A in two-existing phases as a function of kT/C.

The solubility of the components in each other increases with temperature until a temperature is reached where the components are completely miscible (soluble) in each other.

The temperature at which complete miscibility occurs is called the critical temperature, T_c.

25

26

ε > 0, ∆H

> 0 / ∆H

~

Incentive Homework1:

please find and summary models for asymmetric miscibility gaps as a ppt file.

Ref. Acta Meter. 1 (1953) 202/ Acta Meter. 8 (1960) 711/ etc.

27

ε > 0, ∆H

> 0 / ∆H

~

kJ/mol

27

Q5: “Correlation between chemical potential and free energy”?

Gibbs Free Energy of Binary Solutions

28

−1

= µ

+ µ

G X X Jmol

For 1 mole of the solution (T, P: constant )

= E+PV-TS

G

= H-TS

2) regular solution

A B B

A B

A B A B

AX X X X X X X X X

X = ( + ) = ² + ²

B B

B B

A A

A A

X RT

X G

X RT

X G

ln )

1 (

ln )

1 (

2 2

+

− Ω

+

=

+

− Ω

+

= µ µ

) ln

ln

(

B A B

B A

A

G X G X X RT X X X X

X

G = + + Ω + +

) ln

) 1

( (

) ln

) 1

(

( _{A A} ² _{A B B B} ² _B

A G X RT X X G X RT X

X +Ω − + + +Ω − +

=

µ = +

µ = +

A A A

B B B

G RT ln X G RT ln X

Correlation between chemical potential and free energy

Ideal solution

복잡해졌네 --;;

Regular solution

29

Q6: What is “activity”?

Gibbs Free Energy of Binary Solutions

30

Activity, a : effective concentration for mass action

ideal solution

μ

= G

+ RTlna

μ

= G

+ RTlna

µ = +

µ = +

A A A

B B B

G RT ln X G RT ln X

B B

B B

A A

A

A = G + Ω (1 − X )² + RT ln X

µ

µ

γ =

B

a X

)

1 ( )

ln(

B

B

X

RT X

a = Ω −

31

Degree of non-ideality

> 1

A

XA

1

<

A

X

a

Degree of non-ideality

32

Variation of activity with composition (a) a

, (b) a

a

a

Line 1 : (a) a_B=X_B, (b) a_A=X_A Line 2 : (a) a_B<X_B, (b) a_A<X_A Line 3 : (a) a_B>X_B, (b) a_A>X_A

ideal solution…Rault’s law ΔHmix<0

ΔHmix>0

B in A

random mixing

A in B

random mixing

33

Q7: “Chemical equilibrium of multi-phases”?

Gibbs Phase Rule

Gibbs Free Energy of Binary Solutions

34

35

36 2

37

Degree of freedom

The Gibbs Phase Rule

= the number of variables – the number of constraints

38

The Gibbs Phase Rule

In chemistry, Gibbs' phase rule describes the possible number of degrees of freedom (F) in a closed system at equilibrium, in terms of the number of separate phases (P) and the number of chemical components (C) in the system. It was deduced from thermodynamic principles by Josiah Willard Gibbs in the 1870s.

In general, Gibbs' rule then follows, as:

F = C − P + 2 (from T, P).

From Wikipedia, the free encyclopedia

1.5 Binary phase diagrams

39

2

2 3 2 1

1

1

1 single phase F = C - P + 1

= 2 - 1 + 1

= 2

can vary T and composition independently

2 ^{two phase}

F = C - P + 1

= 2 - 2 + 1

= 1

can vary T or composition

3 eutectic point F = C - P + 1

= 2 - 3 + 1

= 0

can’t vary T or composition For Constant Pressure,

P + F = C + 1

The Gibbs Phase Rule

40

The Gibbs Phase Rule