Binary solid mixtures - uni-greifswald

©Matthias Eschrig

Binary solid mixtures

©Matthias Eschrig

What happens when we cool down a mixture of chemical elements?

• At high temperatures, in the gaseous phase, all atoms mix, resulting in a single phase.

• In liquid phase they often mix perfectly, but can occasionally also show partial

immiscibility (like for example “emulsions” like milk-oil mixtures).

• In solid phase often phase separation between solution of element B in element A (α-

phase) and solution of element A in element B (β-phase) exists – a binary alloy.

• The phase transition from liquid to solid phase in such mixtures shows a characteristic

phase diagram, in which three components react with each other (e.g. liquid, α-phase,

β-phase).

• These reactions are classified into eutectic, peritectic and monotectic.

©Matthias Eschrighttp://resource.npl.co.uk/mtdata/phdiagrams/png/cuni.png

http://resource.npl.co.uk/mtdata/phdiagrams/png/nipd.png

Phase diagrams of two-component (“binary”) alloy systems:

often characterized by limited solubility of each of the components in the other

Solubility phase diagrams:

example: Copper-Nickel alloy example: Nickel-Palladium alloy

Data source: National Physical Laboratory (NPL) online service:

liquid Ag/Pt (L)liquid Cu/Ni (L)

Cu/Ni solid solution (S)

solid phases (below critical point):

Cu-rich (α) Ni-rich (β)

L+S

α + β


Ni-rich (α) Pd-rich (β)

αβ

βα

α + β

Ni/Pd solid solution (S) critical

point

critical

point

L+SL+S

liquidus

solidus

solvus

solidus

liquidus

solidus

liquidus

liquid Cu/Ni (L)

Cu/Ni solid solution (S)


Cu-rich (α) Ni-rich (β)

L+S

α + β

l

αβ

critical

point

liquidus

solidus

solvus

©Matthias Eschrighttp://resource.npl.co.uk/mtdata/phdiagrams/png/cuni.pngData source: National Physical Laboratory (NPL) online service:

lever rule for relative

fraction of phases

wS wL

L

S

wL+wS=1

L

L

S

S

α

β

©Matthias Eschrighttp://resource.npl.co.uk/mtdata/phdiagrams/png/agpt.png

http://resource.npl.co.uk/mtdata/phdiagrams/png/agcu.png



Eutectic phase diagram

example: Silver-Copper alloy

Peritectic phase diagram

example: Silver-Platinum alloy


liquid Ag/Pt (L)

α + β

L+β

L+α peritectic

point

peritectic transition L + β → α

liquid Ag/Cu (L)

α + β

solid phases:

Ag-rich (α) Cu-rich (β)

L+βL+α

eutectic

point

eutectic transition: L → α + β

solid phases:

Ag-rich (α) Pt-rich (β)

αβ

β

α

liquid Ag/Cu (L)

α + β

solid phases:


L+βL+α

eutectic

point


αβ

©Matthias Eschrighttp://resource.npl.co.uk/mtdata/phdiagrams/png/agcu.png


L

α/β

liquid Ag/Cu (L)

α + β

solid phases:


L+βL+α

eutectic

point


αβ

©Matthias Eschrighttp://resource.npl.co.uk/mtdata/phdiagrams/png/agcu.png


L

α/β

β

L

β

α/β

β



liquid Ag/Pt (L)

α + β

L+β

L+α peritectic

point


solid phases:


β

α

L

L

β

α

β

α



liquid Ag/Pt (L)

α + β

L+β

L+α peritectic

point


solid phases:


β

α

L

L

β

β α

β α

©Matthias Eschrighttp://resource.npl.co.uk/mtdata/phdiagrams/png/agpb.png

http://resource.npl.co.uk/mtdata/phdiagrams/png/cumg.png



Monotectic phase diagram

example: Copper-Lead alloy

More complex phase diagrams

example: Copper-Magnesium alloy


Liquid Cu/Mg

γ + δ

α

L+sAg

peritectic

point

e1: L1 → α + γ,

m: L2 → γ + L3,

e2: L3 → δ + β

Liquid Cu/Pb

α + β

αL1+L2

L2+α

monotectic

point

monotectic transition: L1 → α + L2

L1

L2

L1+α

solid phases:

Cu-rich (α) Pb-rich (β)solid phases:

Cu-rich (α) Mg-rich

(β)

β

γ

β

L1+α L1

L2

L2+γ

L1+γ

L3 L3+β

δ

L3+δ

δ + β

α + γ

e1

me2

congruent

melting points

criticalpoint

Liquid Cu/Mg

γ + δ

α

L+sAg

peritectic

point

solid phases:

Cu-rich (α) Mg-rich (β)

γ

β

L1+α L1

L2

L2+γ

L1+γ

L3L3+β

δ

L3+δ

δ + β

α + γ

e1

me2

congruent

melting points

©Matthias Eschrighttp://resource.npl.co.uk/mtdata/phdiagrams/png/cumg.png


Stoichiometric

(intermetallic)

compound

CuMg2

Non-stoichiometric

compound

Laves C15 phase

(MgCu2)

©Matthias Eschrig

1.) Gibbs free enthalpy of a real binary mixture at constant pressure:

G(T,xB)/Ν = xA μA,0(T)+xB μB,0(T) + kBT (xA ln xA + xB ln xB) + xA xB [a(T)+(xA-xB) b(T)]

(Note that xA=1−xB)

Interpolates linearly

Between pure

compound Gibbs FE’s

Contribution due to

“mixing entropy”; this

term is always negative.

“Excess enthalpy”; quantifies

deviation from ideal behavior.

Can be positive or negative.

a and b are functions of T

The excess enthalpy is due to the fact that the average energy for interactions between A and B

atoms is not the same as that for A-A and B-B interactions.

In so-called ideal mixtures it is negligible (this defines ideal mixtures). This term also is zero in

the pure compounds. If the attractive interactions between A-B are on average weaker than that

for A-A and B-B, this term is positive.

The second term is due to the mixing of atoms A and B. It is zero in the pure compounds and

negative in between.

The first term is a linear function of xB that interpolates between the Gibbs free enthalpies of

the pure compounds.

Thermodynamics background: Consider two chemical components, A and B (e.g. A=Ag, B=Cu).

For each phase of a binary mixture the Gibbs free energy is given as:

©Matthias Eschrig

The construction of the lower convex hull to two curves:

Find the set of all tangents to the two curves that fulfill the condition to lie entirely below

both curves. Then find the envelope for these tangents (the “caustic”). The picture above

illustrates the procedure.

In our case it is the correct construction to find the equilibrium state between two phases from

the two Gibbs free energy curves of each phase. The red and blue curves will e.g.

correspond to the liquid and solid phases, and the green line to the coexistence region. It is

defined as a tangent to both the red and blue curves. The envelope curve (red+green+blue)

defines the thermodynamic stable state.

©Matthias Eschrig

Gibbs free energy for ideal mixture:

Ideal mixtures are mixtures where the excess enthalpy term can be neglected:

G(T,xB)/N = xA μA,0(T)+xB μB,0(T) + kBT (xA ln xA + xB ln xB)

kBT (xA ln xA + xB ln xB)

xA μA,0(T)+xB μB,0(T)

G(T,xB)/N

0 1XB

A BA/B mixture

T=const.

Let us fix T and study the qualitative dependence on xB:

0

©Matthias Eschrig


There is a version for the liquid state, GL(T,xB)/NL, and a version for the solid state, GS(T,xB)/NS .

Each of these versions of this equation has its own set of parameters.

When lowering temperature, the chemical potentials μA,0S(T) and μB,0

S(T) for the solid increase

less rapidly than μA,0L(T) and μB,0

L(T) for the liquid, as the pure solid phases must have a lower

Gibbs free energy at low temperatures than the pure liquid phases.

The mixing entropy term decreases in magnitude with decreasing temperature (it is ~T ).

NA[μA,0S(T)-μA,0

L(T)]=(-59725 + 48.35 T) J/mol

NA[μB,0S(T)-μB,0

L(T)]=(-75270 + 48.10 T) J/mol

μA,0L(T)=μB,0

L(T)

In the following pictures we take a model material with the following parameters (all

parameters are related to 1 mol, i.e. to Avogadro’s number NA=6.022x1023 mol-1):

Ideal mixtures are mixtures where the excess enthalpy term can be neglected:

G(T,xB)/N = xA μA,0(T)+xB μB,0(T) + kBT (xA ln xA + xB ln xB)

kB=1.38x10-23 J/K

NAkB=R=8.314 J/(mol K)

mixing entropylinear term

0 0.2 0.4 0.6 0.8 0

0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1400

0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1400

0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1400

©Matthias Eschrig

T=1280oC


LL L S

L+SL+S

S

solidification starts

T=1340oC

T/oC

XB0 1

coexistence region:

separation into two phases

Red: Gibbs free energy of solid

Blue: Gibbs free energy of liquid

T=1220oC

In the following series of graphs (next three slides) the temperature (yellow line) is gradually lowered:

0 0.2 0.4 0.6 0.8 0

0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1400

0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1400

0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1400

©Matthias Eschrig

LL L S

L+S


T/oC

XB0 1 S

SL+S L+S



T=1160oC T=1100oC T=1040oC

0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1400

0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1400

0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1400

©Matthias Eschrig

T=980oC T=920oC

L S SS

L+S


0 0.2 0.4 0.6 0.8 1

T/oC

XB0 1

T=860oC

solidification about

to be finished

solidification is finished

no liquid left

“liquidus line”

“solidus line”



©Matthias Eschrig

2.) Composition and substance fraction of phases:

0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1400

x1x2x3x4

T1

T3

T2

A) What is the composition of the phases?

L

α

L α

T1

T2

T3

x

x x1

x3 x2

x4 x

above T1 there is only

liquid with composition x

below T3 there is only

solid with composition x

Between T1 and T3 the composition of the liquid changes from x to x4 and the composition

of the solid changes from x1 to x.

©Matthias Eschrig

0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1400

x2x3

T2

B) What is the fraction of each of the two phases?

L

α

x

sαsLwα + wL = 1

wα sα+ wL sL= sα + sL

wα= sL/(sα+sL)=(x-x3)/(x2-x3)

wL= sα/(sα+sL)=(x2-x)/(x2-x3)

“Lever rule”:

2.) Composition and substance fraction of phases:

©Matthias Eschrig

In real mixtures the excess term becomes important:

G(T,xB)/N = xA μA,0(T)+xB μB,0(T) + kBT (xA ln xA + xB ln xB) + xA xB [a(T)+(xA-xB) b(T)]

Again, there is a version pf this equation for the liquid phase, GL(T,xB)/NL, and a version for the

solid phase, GS(T,xB)/NS. Each of these versions has its own set of parameters.

NA[μA,0S(T)-μA,0

L(T)]=(-59725 + 48.35 T ) J/mol

NA[μB,0S(T)-μB,0

L(T)]=(-75270 + 48.10 T ) J/molμA,0

L(T)=μB,0L(T)

In the following pictures we take a model material with the following parameter set:

3.) Gibbs free energy leading to a miscibility gap:

aS(T)=(20719.2 – 5.5068 T ) J/mol

bS(T)=(-3597.6 + 1.035 T ) J/mol

aL(T)=(9102.6 – 1.5222 T ) J/mol

bL(T)=(-1455 + 0.5676 T ) J/molNAkB=R=8.314 J/(mol K)

mixing entropy excess enthalpylinear term

0 0.2 0.4 0.6 0.8 0

©Matthias Eschrig

1 0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

12001 0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

12001 0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1200T=1130.7oC

L L L S

L+S

T/oC

XB0 1



T=990.7oCT=1060.7oC


0 0.2 0.4 0.6 0.8 0

©Matthias Eschrig

1 0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

12001 0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

12001 0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1200

SSSSS LL

L+S L+S

T/oC

XB0 1



T=780.7oCT=850.7oCT=920.7oC

0 0.2 0.4 0.6 0.8 1

©Matthias Eschrig

1 0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

12001 0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

12001 0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1200

T=710.7oC

S α β βαα+β α+β

miscibility gap miscibility gap

Mixture of two phases: α and β

T/oC

XB0 1



T=640.7oC T=570.7oC

©Matthias Eschrig

And again, there is a version for the liquid state, GL(T,xB)/NL, and a version for the solid state,

GS(T,xB)/NS.

NA[μA,0S(T)-μA,0

L(T)]=(-59725 + 48.35 T ) J/mol

NA[μB,0S(T)-μB,0

L(T)]=(-75270 + 48.10 T ) J/molμA,0

L(T)=μB,0L(T)

In the following pictures we take the parameter set appropriate for reproducing the eutectic

phase diagram for an Ag-Cu system:

aS(T)=(34532 – 9.67 T ) J/mol

bS(T)=(-5996 + 1.725 T ) J/mol

aL(T)=(15171 – 2.537 T ) J/mol

bL(T)=(-2425 + 0.946 T ) J/mol

4.) Gibbs free energy for eutectic Ag-Cu system:

Again we use the following expression for the Gibbs free energy:

G(T,xB)/N = xA μA,0(T)+xB μB,0(T) + kBT (xA ln xA + xB ln xB) + xA xB [a(T)+(xA-xB) b(T)]

NAkB=R=8.314 J/(mol K)

mixing entropy excess enthalpylinear term

©Matthias Eschrig

0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

12001 0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1200

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1

1 0 0.2 0.4 0.6 0.8 1

1 0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1

1 0 0.2 0.4 0.6 0.8 1

1 0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1

1 0 0.2 0.4 0.6 0.8 1

600

800

1000

1200

1200

Transition from ideal to eutectic phase diagram with increasing excess enthalpy(obtained by scaling the excess enthalpy part of the Gibbs free energy for both solid and

liquid by the six factors 0.4, 0.5, 0.6, 0.7, 0.8, 0.9 for the six graphs shown)

pre

ssu

re

0 0.2 0.4 0.6 0.8 0

©Matthias Eschrig

4.) Gibbs free energy for eutectic Ag-Cu system:

0 0.2 0.4 0.6 0.8 1500

600

700

800

900

1000

1100

1200

0 0.2 0.4 0.6 0.8 1500

600

700

800

900

1000

1100

1200

0 0.2 0.4 0.6 0.8 1500

600

700

800

900

1000

1100

1200

T=1130.7oC T=1060.7oC T=990.7oC

L L L ββ

L+β L+βRed: Gibbs free energy of solid


T/oC

XB0 1


0 0.2 0.4 0.6 0.8 0

©Matthias Eschrig

Gibbs free energy for eutectic Ag-Cu system:

0 0.2 0.4 0.6 0.8 1500

600

700

800

900

1000

1100

1200

0 0.2 0.4 0.6 0.8 1500

600

700

800

900

1000

1100

1200

0 0.2 0.4 0.6 0.8 1500

600

700

800

900

1000

1100

1200

L L βββ

ααα

LL+βL+β

L+α L+α

L+α L+β



T/oC

XB0 1

T=920.7oC T=850.7oC T=780.7oC

0 0.2 0.4 0.6 0.8 1

©Matthias Eschrig

Gibbs free energy for eutectic Ag-Cu system:

0 0.2 0.4 0.6 0.8 1500

600

700

800

900

1000

1100

1200

0 0.2 0.4 0.6 0.8 1500

600

700

800

900

1000

1100

1200

0 0.2 0.4 0.6 0.8 1500

600

700

800

900

1000

1100

1200

T=640.7oC T=570.7oC

α βα+βα+β α+βα αβ β



T/oC

XB0 1

T=710.7oC

©Matthias Eschrig

0 0.2 0.4 0.6 0.8 1500

600

700

800

900

1000

1100

1200

solv

us

line

solv

us

line

Limit of solubility is defined by the solvus line:

eutectic

isotherm

eutectic

point

eutectic

concentrationEutectic reaction: L → α+β

(takes place along eutectic isotherm)

eutectic

temperature

©Matthias Eschrig

0 0.2 0.4 0.6 0.8 1500

600

700

800

900

1000

1100

1200

L

α

βL+βL+α

α+β

TE

xx1x2xE

Fractions of the phases during eutectic reaction:

Just above TE:

α : x1 , wα=(xE-x)/(xE-x1)

L : xE , wL=(x-x1)/(xE-x1)

Just below TE:

α : x1 , wα=(x2-x)/(x2-x1)

β : x2 , wβ=(x-x1)/(x2-x1)

α

αα/β In eutectic mixture just

below TE :

α : x1 , wEα=(x2-xE)/(x2-x1)

β : x2 , wEβ=(xE-x1)/(x2-x1)

Total α below TE = primary α formed just above TE + liquid present just above TE times fraction

of α in eutectic mixture just below TE : (x2-x)/(x2-x1)= (xE-x)/(xE-x1)+(x-x1)/(xE-x1) * (x2-xE)/(x2-x1)

i.e., wα (below TE)= wα (above TE)+ wL (above TE)* wEα (below TE)

Example of how to calculate the fractions when crossing TE from above using the lever rules:

©Matthias Eschrig

Classification of various types of binary phase diagrams:

L1

L2 S

LS2S1

S1

S2L

L1 L2S

L S1S2

monotectic

eutectic

catatectic

(metatectic)

syntectic

peritectic

Eutectic type:

Peritectic type:

A→B+C

under cooling

A+B→C

under cooling

Schematic reaction types at critical lines

(slopes of phase boundaries can be deformed)

S2 S3

S1

eutectoid:

S1 S2S3

peritectoid:

L1→L2+S

L→S1+S2

S1→L+S2

L1+L2→S

L+S1→S2

S1→S2+S3

S1+S2→S3

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