1. II. Synthetic Aspects Preparation vs. Separation; Quality CriteriaCheetham & Day, pp. 1-38 Incomplete conversion of reactants into products – filtration, distillation, … Unwanted side reactions Metal source is really MH n, e.g., Ba is really BaH <0.20 Cu screen for filtering products Minimize impurities Clean gaseous reagents (phosphines often added to inert gases) Sublime/Distill elements/reactants to insure purity Sound quantitative analysis (bulk measurements) X-ray diffraction, Electron microscopy, Microprobe analysis, X-ray fluorescence, X-ray photoelectron spectroscopy, ICP-Mass Spectrometry Hand-Outs: 4

2. II. Synthetic Aspects Quality CriteriaCheetham & Day, pp. 1-38 Some Historical Problems: (a)  -Tungsten:Really W 3 O (b) Ti 2 S: (prepared in graphite crucible) really Ti 2 SC (c) Sr 3 Sn: (nonmetallic behavior) really Sr 3 SnO (d) In 5 S 4 : (prepared in a Sn flux; could not be prepared pure in binary mixture) really In 4 SnS 4 (e) NbF 3 : really NbO x F 3  x C Ti 2 S “[In 5 ] 8+ ” = [SnIn 4 ] 8+ Hand-Outs: 4

3. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Gaining an understanding of the equilibrium state of a system; Reading phase diagrams for synthetic or other chemical use.

4. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Gibbs Phase Rule:F = C  P + 2 F = # degrees of freedom (independent intensive variables) among T, p, composition. Composition given as mole fractions, x i ( ) (i = 1, …, C), ( = 1, …, P)

5. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Gibbs Phase Rule:F = C  P + 2 F = # degrees of freedom (independent intensive variables) among T, p, composition. Composition given as mole fractions, x i ( ) (i = 1, …, C), ( = 1, …, P) P = # phases present at equilibrium; are physically distinct, separable (in principle) A phase is uniform throughout, not only in chemical composition but in physical state. E.g., H 2 O(s) vs. H 2 O(l), MgSiO 3 (s) vs. Mg 2 SiO 4 (s),  -Fe(s) vs.  -Fe(s). A homogeneous mixture is a single phase: N 2 (g) and O 2 (g), NaCl(aq) A solid solution is single phase:  -Al 2 O 3 / Cr 2 O 3, Nb 3  x Cl 8 = (Nb 3 ) 1  x (Nb 2 ) x Cl 8

6. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Gibbs Phase Rule:F = C  P + 2 F = # degrees of freedom (independent intensive variables) among T, p, composition. Composition given as mole fractions, x i ( ) (i = 1, …, C), ( = 1, …, P) P = # phases present at equilibrium; are physically distinct, separable (in principle) A phase is uniform throughout, not only in chemical composition but in physical state. E.g., H 2 O(s) vs. H 2 O(l), MgSiO 3 (s) vs. Mg 2 SiO 4 (s),  -Fe vs.  -Fe. A homogeneous mixture is a single phase: N 2 (g) and O 2 (g), NaCl(aq) A solid solution is single phase:  -Al 2 O 3 / Cr 2 O 3, Nb 3  x Cl 8 = (Nb 3 ) 1  x (Nb 2 ) x Cl 8 C = # constituents that can undergo independent variation in different phases; minimum number of constituents needed to completely describe the composition of the system

7. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Gibbs Phase Rule:F = C  P + 2(Simple Derivation) F = # variables  # of restraints (equations between variables) From Gibbs Free Energy:

8. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Gibbs Phase Rule:F = C  P + 2(Simple Derivation) F = # variables  # of restraints (equations between variables) From Gibbs Free Energy: Variables are T,p x 1 (  ) x 1 (  ) x 1 (  )  x 2 (  ) x 2 (  ) x 2 (  )  ……… # Variables = (C)(P) + 2 C P

9. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Gibbs Phase Rule:F = C  P + 2(Simple Derivation) F = # variables  # of restraints (equations between variables) From Gibbs Free Energy: Variables are T,p x 1 (  ) x 1 (  ) x 1 (  )  x 2 (  ) x 2 (  ) x 2 (  )  ……… Equations are x 1 (  ) + x 2 (  ) + x 3 (  ) +  = 1 x 1 (  ) + x 2 (  ) + x 3 (  ) +  = 1   1 (  ) =  1 (  )  2 (  ) =  2 (  )   C (  ) =  C (  )  1 (  ) =  1 (  )  2 (  ) =  2 (  )   C (  ) =  C (  )   1 (P  1) =  1 (P)  2 (P  1) =  2 (P)   C (P  1) =  C (P) # Variables = (C)(P) + 2 # Equations = P + (C)(P  1) Chemical Potentials F = # Variables  # Equations = (CP + 2)  (P +CP  C) P P  1 C

10. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Determining C:C = S  R   S = # chemical species identified at equilibrium (unrestricted) R = # independent net chemical equations (reactions)  = # restraints imposed by the preparation Examples: (a)Solid-Gas Equilibria: Consider an equilibrium mixture of BaCO 3 (s), BaO(s), CO 2 (g). How many independent intensive variables (degrees of freedom F) are there for any given temperature?

11. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Determining C:C = S  R   S = # chemical species identified at equilibrium (unrestricted) R = # independent net chemical equations (“reactions”)  = # restraints imposed by the preparation Examples: (a)Solid-Gas Equilibria: Consider an equilibrium mixture of BaCO 3 (s), BaO(s), CO 2 (g). How many independent intensive variables (degrees of freedom F) are there for any given temperature? # Phases = P = 3: BaCO 3 (s), BaO(s), CO 2 (g) (2 distinct solid phases + gas phase) Since F = C  P + 2, then what is C ? BaCO 3 (s), BaO(s) and CO 2 (g) are species:S = 3 Independent Net Reactions R = ?? Set up species-by-element matrix and use row reduction… (General)  = 0 (No restraints, in general)

12. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Examples: (a)Solid-Gas Equilibria:BaCO 3 (s), BaO(s) and CO 2 (g) are species Independent Net Reactions – set up species-by-element matrix and use row reduction: BaOCO 2 BaCO 3 Ba101 C011 O123 C = S  R   = ??? S = # chemical species identified at equilibrium (unrestricted) R = # independent net chemical equations (reactions)  = # restraints imposed by the preparation

13. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Examples: (a)Solid-Gas Equilibria:BaCO 3 (s), BaO(s) and CO 2 (g) are species Independent Net Reactions – set up species-by-element matrix and use row reduction: BaOCO 2 BaCO 3 Ba101 C011 O123 BaOCO 2 BaCO 3 Ba101 C011 O022 Subtract Row 1 from Row 3 C = S  R   = ??? S = # chemical species identified at equilibrium (unrestricted) R = # independent net chemical equations (reactions)  = # restraints imposed by the preparation

14. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Examples: (a)Solid-Gas Equilibria:BaCO 3 (s), BaO(s) and CO 2 (g) are species Independent Net Reactions – set up species-by-element matrix and use row reduction: BaOCO 2 BaCO 3 Ba101 C011 O123 BaOCO 2 BaCO 3 Ba101 C011 O022 BaOCO 2 BaCO 3 Ba101 C011 O000 Subtract Row 1 from Row 3 Divide Row 3 by 2; Subtract Row 2 from Row 3 C = S  R   = ??? S = # chemical species identified at equilibrium (unrestricted) R = # independent net chemical equations (reactions)  = # restraints imposed by the preparation Matrix has Rank = 2

15. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Examples: (a)Solid-Gas Equilibria:BaCO 3 (s), BaO(s) and CO 2 (g) are species (S = 3) Independent Net Reactions – set up species-by-element matrix and use row reduction: BaOCO 2 BaCO 3 Ba101 C011 O123 BaOCO 2 BaCO 3 Ba101 C011 O022 BaOCO 2 BaCO 3 Ba101 C011 O000 Subtract Row 1 from Row 3 Divide Row 3 by 2; Subtract Row 2 from Row 3 C = S  R   = ??? S = # chemical species identified at equilibrium (unrestricted) R = # independent net chemical equations (reactions)  = # restraints imposed by the preparation Matrix has Rank = 2 R = S  Rank = 3  2 = 1

16. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Examples: (a)Solid-Gas Equilibria:BaCO 3 (s), BaO(s) and CO 2 (g) are species (S = 3) Independent Net Reactions – set up species-by-element matrix and use row reduction: BaOCO 2 BaCO 3 Ba101 C011 O123 BaOCO 2 BaCO 3 Ba101 C011 O022 CBaCO 3 BaO101 CO 2 011 O000 Subtract Row 1 from Row 3 Divide Row 3 by 2; Subtract Row 2 from Row 3 C = S  R   = 3  1  0 = 2 S = # chemical species identified at equilibrium (unrestricted) R = # independent net chemical equations (reactions)  = # restraints imposed by the preparation R = S  Rank = 3  2 = 1 Switch “Basis” BaCO 3 (s) BaO(s) + CO 2 (g)

17. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Examples: (a)Solid-Gas Equilibria:Equilibrium mixture of BaCO 3 (s), BaO(s) and CO 2 (g) C = 2,P = 3,Therefore, F = C  P + 2 = 2  3 + 2 = 1 (univariant) BaCO 3 (s) BaO(s) + CO 2 (g) K p = p(CO 2 ) = p = exp(  G/RT); (p is a function of T) Another view:Variables (4): Equations (3): F = # Variables  # Equations = 1

18. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Examples: (a)Solid-Gas Equilibria:Equilibrium mixture of BaCO 3 (s), BaO(s) and CO 2 (g) C = 2,P = 3,Therefore, F = C  P + 2 = 2  3 + 2 = 1 (Univariant) BaCO 3 (s) BaO(s) + CO 2 (g) K p = p(CO 2 ) = p = exp(  G/RT); (p is a function of T) Arbitrary Restraint Created by Preparation? (i) If equilibrium is established by using just BaCO 3 (s)… There is no new restraint involving the intensive variables, Therefore,  = 0. Then, C = 2, F = 2  3 + 2 = 1 (Univariant) Let n 0 = # moles BaCO 3 (s) placed in the container. Using mass balance: Ba: n 0 = n(BaCO 3,s) + n(BaO,s) C: n 0 = n(BaCO 3,s) + n(CO 2,g)  n(BaO,s) = n(CO 2,g) O: 3n 0 = 3n(BaCO 3,s) + n(BaO,s) + 2n(CO 2,g) BUT: x(CO 2,g) = 1 (it is the only gaseous species)

19. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Examples: (a)Solid-Gas Equilibria:Equilibrium mixture of BaCO 3 (s), BaO(s) and CO 2 (g) C = 2,P = 3,Therefore, F = C  P + 2 = 2  3 + 2 = 1 (Univariant) BaCO 3 (s) BaO(s) + CO 2 (g) K p = p(CO 2 ) = p = exp(  G/RT); (p is a function of T) Arbitrary Restraint Created by Preparation? (ii) If equilibrium is established by using just BaCO 3 (s) and in a piston which equilibrates with the external atmospheric pressure, say 1.00 atm. This is a new restraint on the total pressure, p TOT = p(CO 2 ) = 1.00 atm Therefore,  = 1. Then, C = 1, F = 1  3 + 2 = 0 (Invariant) NOTE: This is still a 2-component system with a physical restraint.

20. II. Synthetic Aspects Thermodynamic Issues H.F. Franzen, Physical Chemistry of Solids, Ch. 7 Examples: (b)Solid-Gas Equilibria: FeSO 4 (s) is heated to 929 K in an evacuated container. At equilibrium, FeSO 4 (s), Fe 2 O 3 (s), SO 2 (g), SO 3 (g), and O 2 (g) are present. How many independent intensive variables (F) are there in this system? Write down the equations relating any of the intensive variables with each other.