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Chapter 6. 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.
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Chapter 6 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. Interpretation of Phase Diagrams
Phase rule:P + F = C + 2 P: number of phases C: number of components F: degree of freedom PV = nRT 3 variables, 1 equation 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 C1 + C2 + …..+ Cc = 100% P-1 relations: m1 = m2 = m3 = m4 = m5 = …….= mP mi: free energy One needs P-1 equations to describe the equilibrium between phases
P: number of phases How to distinguish phases? lcrystalline phases: MgSiO3, Mg2SiO4 lpolymorphism a- b- g- lsolid solution ( one phase ) ldefects ordered defects: distinguished phases disordered defects: one phase (considered as solid solution) WO3-x (WnO3n-1: W20O59, W19O56) lliquid (miscibility or immiscibility) lgas: one phase
C: number of components • CaO-SiO2 : two components • MgO (heating below mp): one component • “FeO” appears to be one component, but two components ∵ 3Fe2+ 2Fe3+ + Feo ∴ FeO is in fact Fe1-xO (wüstite) + x Fe
F: degree of freedom Independent variable to describe a system For example: boiling water P = 2, C = 1 F = C – P + 2 = 1 - 2 + 2 = 1 So only temperature or pressure is enough to describe the system, that is, T and P are dependent variables.
How do you know a compound is thermodynamically stable or kinetically stable? eg. 1. Ca3SiO5 is prepared above 1300oC. At 1100oC, Ca3SiO5 CaO + Ca2SiO4 Ca3SiO5 is thermodynamically stable above 1300oC but is kinetically stable at 1100oC. 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.
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) One component system 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 SiO2 At 1600 bar • a-quartz b-quartz • liquid SiO2 ∵b-tridymite and b-cristobalite have lower density than quartz However, many metastable phases can be obtained by quenching. At 500 bar a-quartz b-quartz b-tridymite b-cristobalite liquid SiO2 573oC 870oC 1470oC 1710oC
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. SiO2 (Fig 6.5) 1 + F = 1 + 1 => 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 eutectic point What happens if the system is heated from points e and f ? In order to determine the relative amounts of two phases in a mixture, the level rule is used
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
Liquid B (amount of B) = hf/Bh (amount of liquid) = Bf/Bh Composition of liquid = h
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 oC)
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 compositionn ?
Non-equilibrium products Describe what happens when system is cooled at compositionn ? 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)
Formation of metastable 2-liquid system It happens when “2-liquid B + liquid” is very slow upon cooling
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. CaAl2Si2O8 and NaAlSi3O8 in 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 (Mg2SiO4) vs willemite (Zn2SiO4)
