II. Synthetic Aspects Preparation vs. Separation; Quality CriteriaCheetham & Day, pp. 1-38...

II. Synthetic Aspects Preparation vs. Separation; Quality Criteria Cheetham & Day, pp. 1-38

• Incomplete conversion of reactants into products – filtration, distillation, …• Unwanted side reactions

Metal source is really MHn, e.g., Ba is really BaH<0.20

Cu screen for filtering products

• Minimize impuritiesClean 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

II. Synthetic Aspects Quality Criteria Cheetham & Day, pp. 1-38

Some Historical Problems:

(a) -Tungsten: Really W3O

(b) Ti2S: (prepared in graphite crucible)really Ti2SC

(c) Sr3Sn: (nonmetallic behavior) really Sr3SnO

(d) In5S4: (prepared in a Sn flux; could not be prepared pure in binary mixture) really In4SnS4

(e) NbF3: really NbOxF3x

CTi2S

“[In5]8+” = [SnIn4]8+

Hand-Outs: 4

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.

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, xi

() (i = 1, …, C), ( = 1, …, P)




() (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., H2O(s) vs. H2O(l), MgSiO3(s) vs. Mg2SiO4(s), -Fe(s) vs. -Fe(s). A homogeneous mixture is a single phase: N2(g) and O2(g), NaCl(aq) A solid solution is single phase: -Al2O3 / Cr2O3, Nb3xCl8 = (Nb3)1x(Nb2)xCl8




() (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., H2O(s) vs. H2O(l), MgSiO3(s) vs. Mg2SiO4(s), -Fe vs. -Fe. A homogeneous mixture is a single phase: N2(g) and O2(g), NaCl(aq) A solid solution is single phase: -Al2O3 / Cr2O3, Nb3xCl8 = (Nb3)1x(Nb2)xCl8

C = # constituents that can undergo independent variation in different phases; minimum number of constituents needed to completely describe the composition

of the system


Gibbs Phase Rule: F = C P + 2 (Simple Derivation)

F = # variables # of restraints (equations between variables)

From Gibbs Free Energy: 1

, , molar free energy, entropy, volume

C

i ii

dg s dT v dp dx

g s v




From Gibbs Free Energy:

Variables are T, px1

() x1() x1

() x2

() x2() x2

()

… … …

# Variables = (C)(P) + 2

1


C

i ii

dg s dT v dp dx

g s v

C

P




From Gibbs Free Energy:

Variables are T, px1

() x1() x1

() x2

() x2() x2

()

… … …

Equations are x1() + x2

() + x3() + = 1

x1() + x2

() + x3() + = 1

1(a) = 1

(b) 2(a) = 2

(b) C(a) = C(b) 1

(b) = 1(c) 2

(b) = 2(c) C(b) = C(c)

1

(P1) = 1(P) 2

(P1) = 2(P) C(P1) = C(P)

# Variables = (C)(P) + 2

# Equations = P + (C)(P 1)

1


C

i ii

dg s dT v dp dx

g s v

ChemicalPotentials

F = # Variables # Equations = (CP + 2) (P +CP C)

P

P 1

C


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 BaCO3(s), BaO(s), CO2(g).

How many independent intensive variables (degrees of freedom F) are there for anygiven temperature?


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 BaCO3(s), BaO(s), CO2(g).

How many independent intensive variables (degrees of freedom F) are there for anygiven temperature?

# Phases = P = 3: BaCO3(s), BaO(s), CO2(g) (2 distinct solid phases + gas phase)

Since F = C P + 2, then what is C ?

BaCO3(s), BaO(s) and CO2(g) are species: S = 3Independent Net Reactions R = ??

Set up species-by-element matrix and use row reduction… (General)

= 0 (No restraints, in general)


Examples:(a) Solid-Gas Equilibria: BaCO3(s), BaO(s) and CO2(g) are species

Independent Net Reactions – set up species-by-element matrix and use row reduction:BaO CO2 BaCO3

Ba 1 0 1

C 0 1 1

O 1 2 3

C = S R = ??? S = # chemical species identified at equilibrium (unrestricted) R = # independent net chemical equations (reactions) = # restraints imposed by the preparation




Ba 1 0 1

C 0 1 1

O 1 2 3

BaO CO2 BaCO3

Ba 1 0 1

C 0 1 1

O 0 2 2

Subtract Row 1 from Row 3





Ba 1 0 1

C 0 1 1

O 1 2 3

BaO CO2 BaCO3

Ba 1 0 1

C 0 1 1

O 0 2 2

BaO CO2 BaCO3

Ba 1 0 1

C 0 1 1

O 0 0 0


Divide Row 3 by 2;Subtract Row 2 from Row 3


Matrix has Rank = 2


Examples:(a) Solid-Gas Equilibria: BaCO3(s), BaO(s) and CO2(g) are species (S = 3)


Ba 1 0 1

C 0 1 1

O 1 2 3

BaO CO2 BaCO3

Ba 1 0 1

C 0 1 1

O 0 2 2

BaO CO2 BaCO3

Ba 1 0 1

C 0 1 1

O 0 0 0




Matrix has Rank = 2

R = S Rank = 3 2 = 1


Examples:(a) Solid-Gas Equilibria: BaCO3(s), BaO(s) and CO2(g) are species (S = 3)


Ba 1 0 1

C 0 1 1

O 1 2 3

BaO CO2 BaCO3

Ba 1 0 1

C 0 1 1

O 0 2 2

Ba C BaCO3

BaO 1 0 1

CO2 0 1 1

O 0 0 0



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”

BaCO3(s) BaO(s) + CO2(g)


Examples:(a) Solid-Gas Equilibria: Equilibrium mixture of BaCO3(s), BaO(s) and CO2(g)

C = 2, P = 3, Therefore, F = C P + 2 = 2 3 + 2 = 1 (univariant)


Kp = p(CO2) = p = exp(G/RT); (p is a function of T)

Another view: Variables (4):

Equations (3): F = # Variables # Equations = 1

2

GAS BaO,sCO BaO, , ,T p x x

2 3 2

GAS BaO,sCO BaO BaCO BaO CO1; 1;x x



C = 2, P = 3, Therefore, F = C P + 2 = 2 3 + 2 = 1 (Univariant)



Arbitrary Restraint Created by Preparation? (i) If equilibrium is established by using just BaCO3(s)…

There is no new restraint involving the intensive variables,

Therefore, = 0. Then, C = 2, F = 2 3 + 2 = 1 (Univariant)

2

GAS BaO,sCO BaO, , ,T p x x

Let n0 = # moles BaCO3(s) placed in the container. Using mass balance:

Ba: n0 = n(BaCO3,s) + n(BaO,s)C: n0 = n(BaCO3,s) + n(CO2,g) n(BaO,s) = n(CO2,g)O: 3n0 = 3n(BaCO3,s) + n(BaO,s) + 2n(CO2,g)

BUT: x(CO2,g) = 1 (it is the only gaseous species)



C = 2, P = 3, Therefore, F = C P + 2 = 2 3 + 2 = 1 (Univariant)



Arbitrary Restraint Created by Preparation? (ii) If equilibrium is established by using just BaCO3(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, pTOT = p(CO2) = 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.


Examples:(b) Solid-Gas Equilibria: FeSO4(s) is heated to 929 K in an evacuated container.

At equilibrium, FeSO4(s), Fe2O3(s), SO2(g), SO3(g), and O2(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.

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