Notation con-vention

Let G' stand for total free energy and in a similar fashion S', V', H', etc.

Then we will let = G'/n represent the free energy per mole, n, and in a similar fashion. S', V', H', etc.

Partial Molar & Excess Quantities

G

Consider a solution of volume V’ containing n, moles of a component 1, n2

moles of a component 2 and etc.

,2,1 nnVV

2

,,,21

,,,13232

dnn

Vdn

n

VVd

nnPTnnPT

def

jnPTin

VV

,,

Partial molar volume of component i

iiG Where the def the chemical potential of i

2211 dnVdnVVd

2211 dnHdnHHd

2211 dnGdnGGd

2211 dndnGd

A

Integrating the set of equation A ,

2211 nnG

2211 nVnVVB

etc

22221111 dndndndnGd

02211 dndn etc

Can also be written as

02211 dxdx

Gibbs-Duhem eqns.

Differentiating the set B

and comparing and A B

*In the derivation of the Gibbs phase rule the Gibbs-Duhem eqns.

play an important rule in restricting the allowed variations.

Activity and the activity coeffi-cient

Def : activity

0i

ii p

pa

Ideal Solution

Let Poi correspond to the equilibrium vapor pressure of the pure component, i.

ai

1.0

xi 1

Raoult’s Law

Raoult’s law iii pxp

by def of aiii xa

Regular Solution

Henry’s law

by def of ai

gi is known as the activity coeff. (* g = 1 for ideal solution)

gi is independent of Xi only when Xi is “small”

ai

1.0

xi 1.0Henry’s law

pos. and neg. deviations (g > or < 1)

from ideality

ii bxp

iiiii

ii xx

p

b

p

bxa

Consider a regular solution of components A and B

in a Regular Solution the solute will often follow Henry’s law and

the solvent Raoults law when the solution is “dilute”.

< 0ai

1.0

xB 1.0

Henry’s law

Raoult’s Law

Partial molar Quantities in Ideal and regular solution

Grouping terms in and comparing coefficients of XA and XB in

and

AA

B

AAA xRT ln0

BBB xRT ln0

Where AA G0 and

BB G0

Ideal Solution

ln lnA A B B A A B BG x G x G RT x x x x A

We can also write the Gibbs potential as

A A B BG x x B

Regular Solutions:

0 0 ln lnA A B B A B A A B BG x x x x RT x x x x A

Using the identity ABBBABABA xxxxxxxxxxA

22

Again comparing and A B

AAAA xRTx ln1 20

BBBB xRTx ln1 20

This can be written in the form:

AAA aRT ln0

BBB aRT ln0

so that in the regular solution model:

2

01exp AA

A

AA x

RTx

p

pa

2

01exp BB

B

BB x

RTx

p

pa

Summarizing

Ideal Solution

ii xa

Regular Solution

21exp iii xRT

xa

How are the partial molar quantities related to the molar quantities ?

Consider :

BBAA dxdxdG

ABBdx

dG

Since xA + xB = 1BA dxdx

A

BAB dx

dG

and substituting for A from A

BBB dx

dGxG 1

And similarly

A BB

dGG x

dx

Also usingA A B BG x x

B

BBA x

xG

1

Graphical Interpretation

*

*** )()(BxxB

BBBA dX

dGxxGx

*

)1()()( ***

BxxBBBBB dx

dGxxGx

Given G as a function of xB as above, the chemical potential @ a composi-

tion xB* can be obtained graphically by extrapolating the tangent of the G

curve @ xB= xB* to xB= 0 and xB= 1

*B

B X

dG

dx

0 1

G (XB*)

ΔGmix

X*B

G

XB

A

A

B

B

Chemical Equilibrium

aA + bB + … xX + yY + …

A, B … reactantsX, Y … products

} standard states

a, b, …., x, y,…. Stoichiometric Coefficients

Equilibrium is defined by the condition

0 BAyx bayxGd

Since iii aRT ln0

0 0 0 0

y A BxdG x y a b

( ln ln ) ( ln ln ) 0x y A BRT x a y a RT a a b a

≣ standard free energy change /mole

0 ln lnx yX Ya bA B

a aRT RT K

a a

K is the equilibrium constant for the reaction.

Solving for

The equilibrium constant is defined by

When components are not in standard state:

bB

aA

yY

xX

aa

aaRT ln0

)exp(RT

K

Equilibrium in multiphase Solutions

Consider 2 phases containing the same component i in α and β

The component i has and activity aαi and aβ

i .

Imagine and infinitesimal amount of i is transferred from α to β so that

the compositions have not been altered.

The reaction is:

)()( ii aiai

The chemical potential of i in each phase is

lnoi i iRT a

lnoi i iRT a

i

i

a

aRTG ln if

ii aa => spontaneous (ΔG < 0)

Equilibrium is defined by ΔG = 0 or

ii aa

ii ;

In general for a multicomponent system with different phases

present, equilibrium is defined by:

Equal P & T for all phases

pA A A A

pB B B B

pC C C C

…

The free energy change for the reaction at const T, P is,

The Gibbs Phase Rule

Equilibrium for different phases, , , , …p in contact with one another is given by the conditions;

pp p p p A pT T T T B

pA A A A

pB B B B

pC C C C

…

C

Equilibrium will in general not be maintained if the parameters are arbi-trarily varied. The Gibbs phase rule restricts the manner in which the parameters can be varied such that equilibrium is maintained.

The composition of a given phase is set by the additional condition.

... 1P P P

A B Cx x x C - 1 composition variables per phase

C components present

p phases present (p-1) eqns. for each component

For each phase there are (C – 1) + 2 = C + 1 independent variables ordegrees of freedom.

CThe set of equations represent a system of coupled linear equations. There are ( P – 1) equations for each component and C components so we have (P – 1)(C+2) equations.

T, P

P (C +1) unknowns (P – 1)(C+2) equations

In order for a solution to these equations to exist, # unknowns # of equations

P (C +1) (P – 1)(C+2)

P C + 2

The Gibbs Phase Rule

P C + 2 For a system of C independent components not morethan C + 2 phases can co-exist in equilibrium.

Trivial Example: C = 1, P = 3 solid, liquid vapor

If P is less than C + 2 then C + 2 – P variables can take on arbitrary values(degrees of freedom) without disturbing equilibrium.

def. Thermodynamic degrees of freedom, f

2 0f C P Gibbs Phase Rule

The Gibbs Phase Rule

Geometrical Interpretation of Equilibrium in Multi-component Systems.

Consider a 2 phase binary : GpTxxGG BA ; ) , , ,(

AA BB } Equilibrium

x

x

x

10

G

xB →

ab

A A

B B

The intercepts of the common tangent to the free energy curve give the chemical potentials defining the heterogeneous equilibrium.

xxB 0

xxx B 1 Bxx

- single phase; a

- 2 phase; + field

@ T and P - single phase; b

Composition x* is in the 2 phase α + β field.Consider a simple rule of mixtures mass balance :

xfxfx *

fi is the fraction of alloy composed of i

1 ff

xfxfx )1(*

xx

xxf

*

xx

xxf

*

These formulas are analogous to mass balance for a lever with ful-crum @ x*. => “ Lever Rule”

Binary Phase Diagrams

xB →

A B

T1

G

S

l

xB →

A B

Tmp(A)

G

S

l

xB →

A B

T2

G Sl

xB →

G

A B

Tmp(B)

S

l

A and B complete miscibility in both solid and liquid

xB →

A B

T3

G

S

l

A B

T1

T2

T3

xB →x*

b

c

xs xl

At temp T2 x* is composed of some fl with composition c and fs at composition b:

* :s s l lx f x f x sb x

lc x

Lever Rule

sl

sl

sl

ls xx

xxf

xx

xxf

**

; Ag-AuSi-Ge

System with a miscibility gap

A B

T > Tc

G

xB → xB →

A B

T = Tc

G

xB →

A B

T << Tc

G

g h

G

xB →

A B

T < Tc

ef

2cTR

A B

T < T’c

T < <T’c

xB →

Tc

e f

g h+a a’’

• The miscibility gap is the region where the overall composition exceeds the

solubility limit.

• The solid solution a is most stable as a mixture of two phases a’ + a’’.

• Usually a, a’, and a’’ have the same crystal structure. Cu –Pb, Au-Ni, Fe-

Sn, Cr-W, NaCl-KCl, TiO2-SiO2.

G

T1

XB

Solid

Liquid

T2

a b c d

T3

fe

Negative Curvature

Negative Curvature

ab c

d

fe

a

+a a’’T3

T2

T1liquid

A BXB

Free energy curves and phase diagram for ∆Hsmix> ∆HL

mix= 0.

• The A and B atoms dislike each other.

• Note that the melting point of the alloy is less than that of either

of the pure phases.

xB →

A Ba’

a

liquid

Phase diagram for ΔHSmix <

ΔHlmix < 0

Phase diagram for ΔHSmix

<< ΔHlmix < < 0

Since ΔHSmix < 0, a maximum melting point mixture may appear.

Ordered AlloyFormation

B

liquid

xB →

A

+a b b+ g

a b

g

Ordered Alloy Formation

Eutectic Alloy; ΔHSmix >> ΔHl

mix > 0

A B

T1

Gsolid

l

xB → A B

T3

solid

l

lαa1

l+ a1

l+ αa2

a2

A B

T4

solidl

A B

T5

solid

l

A B

T3

T4

Tb

T2

T1

T5

xB →

T

TA

A B

T2

solid

l

la1

l+ a1

a1l+a1

l+a2

a2

a1a1+a2

a2

l

a1 a2

l+ a1

l+ a 2

a1+a2

• If Hsmix >> 0, the miscibility gap can extend into the liquid phase (T2, T3, T4)

resulting in a simple eutectic phase diagram. • A similar result can occur if the A and B components have a different crystal

structure.

A B

T2

l

α

β

l

α l+ α l+ β β

T4

l β

α

α βα+ β

A B

A B

T1

Gl

xB → β

β

α

α l+ α lA B

T3

lβ

α

α l+ α l+ β β

A B

α

l+ α

l+ β

α +β

β

T2

T3

Tb

T1

T4

xB →

T

TAl

Eutectic phase diagram where each solid phase has a different crystal structure.

A B

T1

liquid

βα

γ

A B

T2

βα

γ

α

α +β+l

l

l

A B

T3

βα

γ

α +β

l

lα β β+l

A B

T4

βα

α +β

l

l+γα β β+l

γ

l

A B

T5

βα

α +β

α β β+l+ γ

l

A B

T3T4

T2

T1

T5

Tl+α

liquid

α +β

β

α

β+γγ

l+ γ

l+β

P Q

A B

T3T4

T2

T1

T5

Tl+α

liquid

α +β

β

α

β+γγ

l+ γ

l+β

P Q

A B

T3T4

T2

T1

T5

Tl+α

liquid

α +β

β

α

β+γγ

l+ γ

l+β

P Q

A B

T3T4

T2

T1

T5

Tl+α

liquid

α +β

β

α

β+γγ

l+ γ

l+β

P Q

A B

T3T4

T2

T1

T5

Tl+α

liquid

α +β

β

α

β+γγ

l+ γ

l+β

P Q

A B

T3T4

T2

T1

T5

Tl+α

liquid

α +β

β

α

β+γγ

l+ γ

l+β

P Q

A B

T3T4

T2

T1

T5

Tl+α

liquid

α +β

β

α

β+γγ

l+ γ

l+β

P Q

A B

T3T4

T2

T1

T5

Tl+α

liquid

α +β

β

α

β+γγ

l+ γ

l+β

P Q

A B

T3T4

T2

T1

T5

Tl+α

liquid

α +β

β

α

β+γγ

l+ γ

l+β

P Q

A B

T3T4

T2

T1

T5

Tl+α

liquid

α +β

β

α

β+γγ

l+ γ

l+β

P Q

A B

T3T4

T2

T1

T5

Tl+α

liquid

α +β

β

α

β+γγ

l+ γ

l+β

P Q

A B

T3T4

T2

T1

T5

Tl+α

liquid

α +β

β

α

β+γγ

l+ γ

l+β

P Q

The derivation of a complex phase diagram showing the formation of stable

intermediate phases(β). At a composition indicated by the red line, just above

T2, (green line) a solid at composition P is in equilibrium with a liquid at

composition Q. At a temperature just below T2 ( not shown ) the two phases

in equilibrium are solid b and liquid . The following peritectic reaction oc-

curs on cooling: l + α → β