II. Synthetic Aspects Eutectic Mixtures A.R. West, Basic Solid State Chemistry, pp. 256-280 A and B...

II. Synthetic Aspects Eutectic Mixtures A.R. West, Basic Solid State Chemistry, pp. 256-280

A and B form a homogeneous mixture in the liquid phase;A and B form a heterogeneous mixture in the solid phase;

Bivariant (F = 2)

Univariant (F = 1)

Univariant (F = 1)

LIQUID: F = 2; T and xB

A + LIQUID: F = 1; T or xB

B + LIQUID: F = 1; T or xB

A + B: F = 1; T

Hand-Outs: 11



Tf(B)

Tf(A)

x: F = 0; melting pt. of A

z: F = 0; melting pt.of B

y: F = 2 3 + 1 = 0; Invariant (Eutectic Point)

0.7 0.30.7 A(s) 0.3 B(s) A B (l)

LIQUID: F = 2; T and xB

A + LIQUID: F = 1; T or xB

B + LIQUID: F = 1; T or xB

A + B: F = 1; T

Hand-Outs: 11

Bivariant (F = 2)

Univariant (F = 1)

Univariant (F = 1)

A B



Take a 50% A: 50% B mixture and heat it into the LIQUID region.

On cooling,

Hand-Outs: 11




On cooling, B will precipitate with T reaches the liquidus curve.

Hand-Outs: 11





Further cooling precipitates more B, while LIQUID follows liquidus curve (richer in A) until eutectic point.

Hand-Outs: 11





Further cooling precipitates more B, while LIQUID follows liquidus curve (richer in A) until eutectic point.

Further cooling gives A(s) + B(s).

Hand-Outs: 11

0

0 2

Ideal Solution : ln

Regular Solution : ln

A A A

A A A B

RT x

RT x wx

II. Synthetic Aspects Solid Solutions A.R. West, Basic Solid State Chemistry, pp. 256-280

Complete immiscibility in the solid state is not possible, although the solubility of onecomponent into another can be immeasurably small. We identify solid solutions on aphase diagram as follows:

B dissolves in A (A ss)

Solubility increasesuntil “m”, thenrapidly decreases.

C = 2, P = 1,F = 2

• B replaces A atoms in structure;• B occupy interstitial sites

Reference: http://rkt.chem.ox.ac.uk/lectures/liqsolns/regular_solutions.html

Hand-Outs: 11

II. Synthetic Aspects Phase Diagrams A.D. Pelton, Prog. Solid State Chem. 10, 119-155.

Two-Component System, A and B,with no compounds. Use regularsolution model for both solid and liquid.

TfA = 800 K; TfB = 1200 KsfA = sfB = 10 J/Kmol

Solid

g(s)(T,x) = RT[(1x) ln(1x) + x ln x] + s x(1x)

Liquid

g(l)(T,x) = RT[(1x) ln(1x) + x ln x] + l x(1x)+ (1x) (TfAT) sfA + x (TfBT) sfB

Plot g(s)(T,x) and g(l)(T,x);Use common tangent approach

() < 0: A-B more attractive than A-A, B-B

() > 0: A-B less attractive than A-A, B-B

Free Energyof Mixing

Enthalpyof Mixing

Hand-Outs: 12-13


x

0.0 0.2 0.4 0.6 0.8 1.0

g mix

ing

(kJ/

mol

)

-14

-12

-10

-8

-6

-4

-2

0 T = 1400 K

Liquid

Solid

x

0.0 0.2 0.4 0.6 0.8 1.0

g mix

ing

(kJ/

mol

)

-8

-6

-4

-2

0

2

4

T = 1000 K

Liquid

Solid

x

0.0 0.2 0.4 0.6 0.8 1.0

g mix

ing

(kJ/

mol

)

-4

-2

0

2

4

6

8

T = 600 K

Liquid

Solid

ExampleTfA = 800 K; TfB = 1200 K

Si-Ge

Ideal Solution Behavior in Solid and Liquid

(s) = 0 kJ/mol (l) = 0 kJ/mol

Hand-Outs: 14


Two-Component System, A and B,with no compounds. Use regularsolution model for both solid and liquid.

TfA = 800 K; TfB = 1200 KsfA = sfB = 10 J/Kmol

Solid

g(s)(T,x) = RT[(1x) ln(1x) + x ln x] + s x(1x)

Liquid

g(l)(T,x) = RT[(1x) ln(1x) + x ln x] + l x(1x)+ (1x) (TfAT) sfA + x (TfBT) sfB

Plot g(s)(T,x) and g(l)(T,x);Use common tangent approach

() < 0: A-B more attractive than A-A, B-B

() > 0: A-B less attractive than A-A, B-B

Free Energyof Mixing

Enthalpyof Mixing

Hand-Outs: 12-13

x

0.0 0.2 0.4 0.6 0.8 1.0

g mix

ing

(kJ/

mol

)

-8

-6

-4

-2

0

2

4

T =1000 K

Liquid

Solid


ExampleTfA = 800 K; TfB = 1200 K

Bi-Cd

Ideal Solution Behavior in LiquidRegular Solution Behavior in Solid

(s) = +15 kJ/mol (l) = 0 kJ/mol

x

0.0 0.2 0.4 0.6 0.8 1.0

g mix

ing

(kJ/

mol

)

-3

-2

-1

0

1

2

3

4

5

T = 750 K

Liquid

Solid

x

0.0 0.2 0.4 0.6 0.8 1.0

g mix

ing

(kJ/

mol

)

-2

0

2

4

6

8

T = 500 K

Liquid

Solid

Hand-Outs: 14

II. Synthetic Aspects Peritectic Reactions A.R. West, Basic Solid State Chemistry, pp. 256-280

AB is a compound in each phase diagram:

3At : AB(s) AB(l)T

At y:

C = 1P = 2F = 0

AB melts congruently.

At x:

C = 2P = 3F = 0

???

Hand-Outs: 15



3At : AB(s) AB(l)T

At y:

C = 1P = 2F = 0


At x:

C = 2P = 3F = 0

2 0.3 0.7At : AB(s) A(s) + A B (l)

0.57; 1.43

T a b

a b

AB(s) + liquid

At p: A0.4B0.6 aA(s) + bA0.3B0.7(l)Just above T2

a = 0.15; b = 0.85

At T2:A(s) + Liq.(x) AB(s) + Liq.(x)

AB melts incongruently.

Hand-Outs: 15



3At : AB(s) AB(l)T

At y:

C = 1P = 2F = 0


At x:

C = 2P = 3F = 0

2 0.3 0.7At : AB(s) A(s) + A B (l)

0.57; 1.43

T a b

a b

AB melts incongruently.

AB(s) + liquid

At p: A0.4B0.6 aA(s) + bA0.3B0.7(l)Just above T2

a = 0.15; b = 0.85

At T2:A(s) + Liq.(x) AB(s) + Liq.(x)

To obtain “pure” AB(s)

Hand-Outs: 15

II. Synthetic Aspects Phase Diagrams ASM International's Binary Alloy Phase Diagrams

Mg-Sn

Hand-Outs: 16


Mg-Sn

Mg2Sn + L’

Mg2Sn + -Sn

Mg2Sn + Mg(Sn)

Mg2Sn + L

Hand-Outs: 16


Eu-Ge

Eu3Ge+ EuGe

Hand-Outs: 16

Gd-Ge


Hand-Outs: 16

Flux (mp, C) Crystal

Al (660.4) TiB2, ZrB2

LnAlB4

LnB4, UB4, NpB4

LnB6

LnBe13, UBe13

Si(a), Ge(a)

Bi (271.5) UPt3

UAl3

GaP(a)ZnSiP2, CdSiP2

Cu (1083.5) LnRh4B4

LnCu2Si2

V3Si

Ga (29.8) Si(a), Ge(a), GaSb(a)

Sn (232.0) LnFe4P12

II. Synthetic Aspects Fluxes

Metal Container

Alkali Metals Ta, steel

Alkaline Earth Metals Ta, graphite for Ba, steel

Al, Ga Alumina, MgO, BeO

Mg MgO, Ta, graphite or steel

Mn Alumina

Cu, Ag, Au Graphite, MgO, alumina, Ta

Fe, Co, Ni Alumina, zirconia

Zn, Cd, Hg Alumina

In Alumina, Ta

Ln Ta, Mo, W, BeO

Bi, Sn Alumina, silica, graphite

Sb Silica, graphite

Handbook on the Physics and Chemistry of the Rare Earths, 12, 53-70 (1989)Phil. Mag. B 65, 1117-1123 (1992)

Separating Flux from Product:(a) Centrifugation – plug of silica wool above crucible; metal screen;(b) Chemical Etching – NaOH for Al(l), HCl for Ga(l) or In(l), dilute HCl for Sn(l)

Liquid more densethan Solid

Hand-Outs: 17

II. Synthetic Aspects Fluxes: Ce-Sb

How would you prepare? CeSb2 CeSb Ce2Sb Ce5Sb3

Hand-Outs: 18

II. Synthetic Aspects Fluxes: Borocarbide Superconductors (LnNi2B2C)

Ca. 1 g crystal of LuNi2B2C

(Canfield, et. al. , Physics Today, October, 1998)

C

B

Ni

Ln = Y, Gd-Lu

Ni-B

Ni-Y

Tc: 6-19 K

Hand-Outs: 19

II. Synthetic Aspects Containers

• Inert to reactant mixtures;• Compatible expansion/compression coefficients;• (with gaseous species) Constructed to withstand internal pressures.

Often, secondary containers used to prevent oxidation of primary container.

Fused Silica: More active metals (Li, Na, Ca, some RE) – reduce SiO2 to oxidesand silicides (esp. Zr, Nb, Ta, Mo);

At high T, reduction of water on silica or interstitial H in metals – transport silica and contaminate product:

SiO2(s) + H2(g) SiO(g) + H2O(g)

Basic oxides (e.g., CaO) will form silicates: e.g., CaSiO3

Hand-Outs: 20

II. Synthetic Aspects Containers

Ceramics: Yttrium-stabilized zirconia (Y:ZrO2); Alumina (Al2O3) allowhigher temperatures, but!!!

4 Al(l) + Al2O3(s) 3 Al2O(g) ca. 1200 K

Metals: Reactive containers (Cu, Ni, Co, Ti, Nb, Ta, Mo) – effectivereductants for low oxidation state products (cluster compounds)

NbI4(s) Nb6I11(s)

Nb, Ta: versatile (ductile, strong, easy to weld)Ta: permeable to hydrogen, can withstand ca. 30 atm.Mo, W: more inert, more difficult to work/seal

Inert containers (Pt, Au) – useful for hydrothermal reactions; do not react with oxygen (esp. Au)

Pt: transported by chlorine, oxygen at higher temps, CO(g)Pt, Au: early metals form brittle compounds – Cs-Au, Re-Au, Zr-Pt

HeatNb Container

Hand-Outs: 20

II. Synthetic Aspects Eutectic Mixtures A.R. West, Basic Solid State Chemistry, pp. 256-280 A and B...

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