Chapter 6 Interpretation of Phase Diagrams Phase diagrams summarize in graphical form the ranges of temperature (or pressure) and composition over which.

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Presentation transcript:

Chapter 6 Interpretation of Phase Diagrams Phase diagrams summarize in graphical form the ranges of temperature (or pressure) and composition over which phases or mixtures of phases are stable under conditions of thermodynamic equilibrium. Phase diagram contains information of compound’s composition, solid solution, phase transition and melting temperature.

Phase rule: P + F = C + 2 P: number of phases C: number of components F: degree of freedom Total number of degree of freedom: (C-1)P + 2 Number of relations: C(P-1) Net degree of freedom: F = (C-1)P + 2 – C(P-1) = C – P + 2 P-1 relations:  1 =  2 =  3 =  4 =  5 = …….=  P  i : free energy PV = nRT 3 variables, 1 equation C1 + C2 + …..+ Cc = 100% One needs P-1 equations to describe the equilibrium between phases

P: number of phases How to distinguish phases? crystalline phases: MgSiO 3, Mg 2 SiO 4 polymorphism  -  -  - solid solution ( one phase ) defects ordered defects: distinguished phases disordered defects: one phase (considered as solid solution) WO 3-x (W n O 3n-1 : W 20 O 59, W 19 O 56 ) liquid (miscibility or immiscibility) gas: one phase

C: number of components a) CaO-SiO 2 : two components b) MgO (heating below mp): one component c) “FeO” appears to be one component, but two components ∵ 3Fe 2+  2Fe 3+ + Fe o ∴ FeO is in fact Fe 1-x O (wüstite) + x Fe

Phase diagram of CaO-SiO 2

F: degree of freedom Independent variable to describe a system For example: boiling water P = 2, C = 1 F = C – P + 2 = = 1 So only temperature or pressure is enough to describe the system, that is, T and P are dependent variables.

Thermodynamically stable or kinetically stable

How do you know a compound is thermodynamically stable or kinetically stable? eg. 1. Ca 3 SiO 5 is prepared above 1300 o C. At 1100 o C, Ca 3 SiO 5  CaO + Ca 2 SiO 4 Ca 3 SiO 5 is thermodynamically stable above 1300 o C but is kinetically stable at 1100 o C. eg. 2. At ambient temperature and pressure, graphite is thermodynamically stable and diamond is kinetically stable. Usually metastable products can be obtained by quenching the reaction before it reaches equilibrium.

One component system F = C –P +2 = 3 – P One phase, P = 1, F = 2 It takes two independent variables to describe the system. Ex. Ideal gas law: PV = nRT Two phases, F = 1, e.g. Boiling water. Need to know P or T. Three phases: no variables.

One component system BE: gives the change of transition temperature with pressure. (F = 1) FC: change of melting point of polymorph Y with pressure. (F = 1) AB, BC: sublimation curves for X and Y. (F = 1) CD: vapour pressure curve for the liquid. (F = 1) Points B, C are triple points. (F = 0) In area X, Y etc, F = 2

Phase diagram of water

Phase diagram of SiO 2  -quartz   -quartz   -tridymite   -cristobalite  liquid SiO o C870 o C1470 o C1710 o C At 500 bar  -quartz   -quartz  liquid SiO 2 ∵  -tridymite and  -cristobalite have lower density than quartz At 1600 bar However, many metastable phases can be obtained by quenching.

Condensed System For most systems and applications of interest in solid state chemistry, the condensed phase rule is applicable, pressure is not a variable and the vapour phase is not important. Condensed Phase Rule: P + F = C + 1 e.g. SiO 2 (Fig 6.5) 1 + F = => F = 1 So that temperature is the only factor for the change of polymorphs

Eutectic Binary System y: invariant point xyz: liquidus curve cyd: solidus curve In order to determine the relative amounts of two phases in a mixture, the level rule is used What happens if the system is heated from points e and f ? eutectic point

Level Rule (amount of liquid) x (distance of hf) = (amount of B) x (distance of Bf) (amount of B) = hf/Bh (amount of liquid) = Bf/Bh

(amount of B) = hf/Bh (amount of liquid) = Bf/Bh Composition of liquid = h BLiquid

Amount of liquid in varies T T1: 0.43 (43% liq.) T2: liq in f = Bf/Bj = 0.53 T3: 0.71 T4: melt complete

Eutectic Reaction Reactions at f point: T > T1 : 57% B (43% liq.) T < T1 : 70% B (30% A)  mixture of A & B crystallized The reaction described above are those that should occur under equilibrium conditions.

A method to lower melting point From the eutectic binary system, it can be considered that B is added to lower the melting point of A. For example, NaCl is added to lower the melting point of ice (to –21 o C)

Binary systems with the formation of compounds Point x: peritectic point. Compound AB melts congruently or incongruently. Describe what happens when system is cooled at composition n ?

Non-equilibrium products Describe what happens when system is cooled at composition n ? When the system is cooled at composition n, one should get AB + B (equilibrium product) But if A + L  AB + L is very slow, one will get A + AB + B ( non-equilibrium products)

Phase diagram of CaO-SiO 2 system

Immiscible Liquids

Formation of metastable 2-liquid system It happens when “2-liquid  B + liquid” is very slow upon cooling

Binary system with complete range of solid solution

Cored (phenomenon of non-equilibrium) When the system is cooled at composition b, the central part that forms first may have composition a and on moving out radially from the centre the crystal becomes increasingly rich in B. Still forms a single crystal (the same structure type of A and B) but with concentration gradient.

What forms solid solution? A and B have the same structure type and ions in them have similar size. e. g. CaAl 2 Si 2 O 8 and NaAlSi 3 O 8 in plagioclase feldspar system

The plagioclase feldspar system

Glass formation Recently, there has been much scientific and technological interest in glassy semiconductors and metals, materials with unusual electrical and mechanical properties. If the liquid is cool rapidly (quenched) to room temperature, there may not be time for any crystallization to occur and a glass forms.

Partial Solid Solution If ions are quite different in size  complete solid solution may not be possible. e.g. forsterite (Mg 2 SiO 4 ) vs willemite (Zn 2 SiO 4 )

Partial Solid Solution

A slightly complicated partial solid solution

A complicated Partial Solid Solution

Binary system with polymorph transitions

example

Another example

The Fe-C diagram