Chapter10 Gibbs Free Energy-Composition

Curves and Binary Phase Diagrams

§10-1. Introduction

§10-2. Gibbs Free Energy Curve, ∆ ( )B

M XG

§10-3. ∆ ( )B

M XG of a Regular Solution

§10-4. Criteria For Phase Stability in Regular Solutions

§10-5. Standard States and Two-Phase Equilibrium

§10-6. Binary Phase Diagrams with Liquid and Solid Exhibiting

Regular Solution.

§10-7. Eutectic Phase Diagrams and Monotectic Phase Diagram

§10-1. Introduction 1. At constant T, P

(1) Stable (equilibrium) state ⇔ G = minG

(2) When phases coexist, ( a、ß、?… )

α

iG =β

iG =γ

iG =… … .

2. Isobaric binary phase diagrams can be determined from ∆ ( )iM XG curves at

different T for each phase. e.g : Si-Ge isomorphous phase diagram.

T > ( )SiTm : Liquid T ≤ ( )SiTm or liquidus line: solid phase coexists with liquid phase

( )TX si 、 ( )TX i

λ can be determined.

§10-2. Gibbs Free Energy Curve, ( )BM XG

MG∆ =RT（ AX ln Aa + BX ln Ba ）

∵ id,MG∆ =RT（ AX ln AX + BX ln BX ）

∴ MG∆ = id,MG∆ +RT（ AX ln Aγ + BX ln Bγ ）

Figure 10.1 I . ∆ id,MG ＜0

II . iγ ＞1 III. iγ ＜1

＊When BX = °BX ，tangent intercept at BX =1 is

MBG∆

∴∣ =BbM

BG∆ =RT ln Ba (II)∣<∣ =BaM

BG∆ = RT ln Ba (I)∣

<∣ =BcM

BG∆ = RT ln Ba (III)∣

∴ Bγ (II) > =γidB 1 > Bγ (III)

＊ 0X i → ， ia 0→ ， RTG i =∆ ln ia −∞→

§10-3. ∆ ( )BM XG of a Regular Solution

∵Regular Solution, 0Sxs =

id,MM GG ∆−∆ = xsG = ∆ MH =RT α AX BX =Ω AX BX

＊ For ∆ MH ＜0， MG∆ is more negative than id,MG∆ ＜0

a homogeneous solution is stable for all BX .

＊ For ∆ MH ＞0， a＞0

RTGM∆－

RTG id,M∆

=RTHM∆＝a AX BX

ideal solution RTG id,M∆

=－R

S id,M∆= AX ln AX + BX ln BX

Figure 10.2 I : RTG id,M∆

（a=0）

When a ↑ ⇒ part of RTGM∆

curve is convex upward.

＊ When all ∆ ( )BM XG curve is 〝convex downward〞.

Figure 10.3 at any specific °BX , ∆ MG ( )°

BX =d.

if it is mixed by any other two different compositions ( ) cba →+

∆ MG ( )c ＞ ∆ MG ( )d

i.e. no phase separation.

＊ When part of ∆ ( )BM XG curve is 〝convex upward〞.

i. m＜ BX ＜q

∆ MG is minimized when two solutions（m, q）coexist

e.g. →γ m＋q

ii. m, q are the common tangent of ∆ MG curve.

∴

==

)qsolution(G)msolution(G)qsolution(G)msolution(G

BB

AA

∴ subtracting °AG and °

BG

then

==

)q()m()q()m(a

BB

AA

aaa

i.e. solutions m, q are in equilibrium, they coexist.

a ↑ ⇒ clustering causes phase separation curve AmqB represents the equilibrium state of the system, curve mnopq has no physical significance.

§10-4. Criteria For Phase Stability in Regular Solutions ＊ Given T, a critical crα occurs ,

α>αα<α

.occursseparationphase,.stableissolutionogeneoushom,

cr

cr

＊ When α = crα , 0XG

B

M

=∂∆∂

, 0X

G2

B

M2

=∂

∆∂, 0

X

G3

B

M3

=∂

∆∂

Figure 10.4

∴ MG∆ =RT( BBAA XlnXXlnX + )+RT BA XXα

∴ =∂∆∂

B

M

XG

RT [ lnA

B

XX

+α ( BA XX − )]

=∂

∆∂2

B

M2

X

GRT( α−+ 2

X1

X1

BA

)

=∂

∆∂3

B

M3

X

G( 2

B2

A X

1

X

1− )

From 0X

G3

B

M3

=∂

∆∂, ∴ 5.0XX BA ==

Then from 0X

G2

B

M2

=∂

∆∂, =α 2cr =α

∵Ω =RT α

∴For a given positive Ω ( 0>Ω ) , R2R

Tcr

crΩ

=α

Ω=

＊ Given fixed Ω >0

When

>α<<α>

.occursseparationphase,2,TT.solutionogeneoushom,2,TT

cr

cr

∵ MG∆ = )XlnXXlnX(RT BBAA + + BAXXRTα

(1) (2) RT α =Ω =constant

↓⇒ negativelessis)1(termT:)1(termTof.indepis)2(term

⇒≤ crTT eventually , at ,5.0X B = MG∆ is positive

MG∆ ( iX ) is convex upward.

＊ Miscibility curve bounding two-phase region in phase diagram is the locus of common tangent compositions .

Figure 10.5 (a),(b) for Ω =16630J , crT =1000K

＊ Ba )X( B at different T : Figure 10.5 (c)

(1) at T= crT , ,5.0X B = inflexion occurs (i).

i.e. B

B

Xa

∂∂

=0 and 2

2

B

B

X

a∂

∂=0

(2) crTT < , Ba )X( B has a maximum, and minimum point.

e.g. Figure 10.6 points b, c where B

B

Xa

∂∂

=0 and 2B

MB

2

X

G

∂

∆∂=0

⇒ spinodal compositions.

＊ Ba )X( B curve between∩bc , (

B

B

Xa

∂∂

)<0 ,

this violates the stability criterion : ↑BX , (B

B

Xa

∂∂

)↑

⇒ curve ∩bc has no physical significance.

⇒ horizontal ad is actual constant activity in the two-phase region.

Points a, d are the common tangent of the MG∆ ( BX ) curve.

§10-5. Standard States and Two-Phase Equilibrium 1. Standard states: pure component in its stable state at specific T, P. ＊ ∵Standard state changes with T.

∴Choice of standard state is based on convenience.

2. Consider phase diagram in Figure 10.7 and )B(m)A(m TTT <<

Choose standard states: ολ)(AG = 0, ο

)S(AG = 0

∴ ο)S(AG － ο

λ)(AG =－ ο)A(mG∆ =－( οο

)A(m)A(m STH ∆−∆ )

∆ H= ∆ οH + ∫ ∆ dTCP , ∆ S= ∆ οS + ∫∆

dTTCP

∆∆=

∆=∆∴=∆=

.Tof.indepareS,H,CCif

STH0G,TTat

)A(m)A(m)(A,P)S(A,P

)A(mm)A(m)A(m)A(m

οολ

οοο

∴ ο)A(mG∆ = ο

)A(mH∆

−

)A(m

)A(m

T

TT

∴point c, ο)s(AG is more positive, when T↑ ( T> )A(mT )

line cb is Gibbs free energy of unmixed

BsolidAsolid

∆ G=－ ο)A(mA GX ∆⋅

similarly, ολ)(BG － ο

)S(BG = ο)B(mG∆ = ο

)B(mH∆ －T ο)B(mS∆ ≅ ο

)B(mH∆

−

)A(m

)A(m

T

TT

line ab , G∆ = BX ο)B(mG∆⋅

∆

∆∩

∩

M)S(

M)(

G,solutionsolidiscfbcurve

G,solutionliguidtheisaedcurve λ

＊ Assume : Ideal solution for solid and liquid solutions.

∴

∆−+=∆

∆++=∆

)2......(GX)XlnXXlnX(RTG

)1......(GX)XlnXXlnX(RTG

)A(mABBAAM

)S(

)B(mBBBAAM

)(ο

ολ

＊ Common tangent positions e, f are the compositions of liquid and solid solutions in equilibrium.

＊ When T↓ , ↓ca and ↑bd , positions of common tangent shift to left.

?)T(X )(A =λ , ?)T(X )s(A =

For equilibrium between solid and liquid

∆=∆

∆=∆

)4......(GG

)3......(GGM

)(BM

)s(B

M)A(

M)s(A

λ

λ

∵ M)(AG λ∆ = M

)(AG λ∆ + )(BX λ)(A

M)(

X

G

λ

λ

∂

∆∂

From (1) )(A

M)(

X

G

λ

λ

∂

∆∂=RT( ln )(AX λ －ln )(BX λ )－ ο

)B(mG∆

∴ M)(AG λ∆ =RT ln )(AX λ … … (5)

Similarly, M)S(AG∆ = M

)S(G∆ + )S(BX)S(A

M

X

G

∂

∆∂

From (2) : ∴ M)S(AG∆ =RT ln )S(AX － ο

)A(mG∆ … … (6)

From (3),(5),(6), ∴RT ln )(AX λ = RT ln )S(AX － ο)A(mG∆

From (1),(2),(4), RT ln )S(BX = RT ln )(BX λ + ο)B(mG∆

∴

∆−⋅=

∆−⋅=

)RT

Gexp(XX

)7)......(RT

Gexp(XX

)B(m)S(B)(B

)A(m)S(A)(A

ο

λ

ο

λ

or (1－ )(AX λ )=(1－ )S(AX )exp(－RT

G )B(mο∆

)… … (8)

From (7),(8)

∆−−

∆−

∆−⋅

∆−−

=

∆−−

∆−

∆−−

=

)RT

Gexp()

RT

Gexp(

)RT

Gexp()]

RT

Gexp(1[

X

)RT

Gexp()

RT

Gexp(

)RT

Gexp(1

X

)B(m)A(m

)A(m)B(m

)(A

)B(m)A(m

)B(m

)S(A

οο

οο

λ

οο

ο

If )(i,PC λ = )S(i,PC

ο)i(mG∆ ο

)i(mH∆≅

−

)i(m

)i(m

TTT

∴Given )A(mT , )A(mT , ο)A(mH∆ , ο

)B(mH∆ and

assuming ideal solution model for both liquid and solid solutions.

The solidus and liquidus lines in isomorphous phase diagram can be calculated.

Example: Ge-Si Figure 10.8 ＊Positions of points of double tangency are not influenced by choice of standard

states; they are determined only by T, ο)A(mG∆ , ο

)B(mG∆ .

See. Eq.(1),(2) and Figure 10.9

＊Values of activity, ia , will depend on the choice of standard state of i and T.

See. Figure 10.7 (a)(b)(c):

Consider Ba : )B(m)A(m TTT <<

If pure B(s) is chosen as standard state,

i.e. ο)s(BG =0 , ο

λ)(BG >0

∴ )s(Ba =1, at 1XB = . i.e. gn =1

point m is )(Ba λ of pure liquid B

∴ )(Ba λ =qnmn

<1

∵

+=

+=

)B.t.r.wln(RTGG

)B.t.r.wln(RTGG

)(B)(BB

)S(B)s(BB

aa

λο

λ

ο

But BG should be independent of choice of standard state.

∴ ολ)(BG － ο

)s(BG = ο)B(mG∆ =RT ln [ ]

)(B

)s(B

B.t.r.w

B.t.r.w

aa

λ

Since T< )B(mT , ∴ ο)B(mG∆ >0 , ∴ [ ]

)(B

)s(B

B.t.r.w

B.t.r.w

aa

λ

>1

∴)(B

)s(B

B.t.r.w

B.t.r.w

aa

λ

>1

i.e. )s(Ba > )(Ba λ

)(B

)s(B

B.t.r.w

B.t.r.w

aa

λ

=exp(RT

G )B(mο∆

)

Similarly, iliquid.t.r.w

isolid.t.r.w

i

i

aa

=exp(RT

G )i(mο∆

)=exp

−∆

)i(m

)i(m)i(m RTT

TTHο

∴In Figure 10.7(c) , Ba ( BX )

statedardtansas)(Bonbasedisjklmline

statedardtansas)s(Bonbasedisjihgline

λ

these two lines vary in a constant ratio of exp(RT

G )B(mο∆

)

＊ Similar discussions can be applied to Aa , Figure 10.7(d)

＊ When T↓ , since T< )B(mT

∴ ο)B(mG∆ >0 , and ο

)B(mG∆ ↑

[)(B

)s(B

aa

λ

] ↑ , i.e.

mngn

↑ or

gnmn

↓

When T↑ , ο)B(mG∆ ↓ ; at T= )B(mT , ο

)B(mG∆ =0 , )s(Ba = )(Ba λ

Point g and point m coincide.

§10-6. Binary Phase Diagrams with Liquid and Solid

Exhibiting Regular Solution.

1. K800T )A(m = , K1200T )B(m = Figure 10.10

−=∆

−=∆

T1012000G

T108000G

)B(m

)A(m

)ideal(0)gular(ReKJ20

s =Ω−=Ωλ

(1) )B(m)A(m TK1000TT <=<

Standard states : ο)s(BG =0 , ο

λ)(AG =0

∴

Ω+++∆⋅−=∆

Ω+++∆⋅=∆

BASBBAA)A(mAMS

BABBAA)B(mBM

XX)XlnXXlnX(RTGXG

XX)XlnXXlnX(RTGXGο

λο

λ

(2) T= )A(mT =800K , Figure 10.10(b)

(3) 480K<T=600K< )A(mT =800K , Figure 10.10(c)

Standard states : ο)s(AG , ο

)s(BG

∴

+=∆Ω+++∆+∆=∆

)XlnXXlnX(RTGXX)XlnXXlnX(RTGXGXG

BBAAMS

BABBAA)A(mA)B(mBM

λοο

λ

(4) T=480K , MSG∆ = MG λ∆ at BX =0.41

(5) T<480K , MSG∆ < MG λ∆

2.

=

=

K1200T

K800T

)B(m

)A(m

T1012000GT10800G

)B(m

)A(m

−=∆−=∆

+=Ω−=Ω

)gular(ReKJ10)gular(ReKJ20

S

λ

Figure 10.11

＊ For solid solution : crT =R2SΩ

= 601314.82

10000=

×(K)

But MSG∆ = MG λ∆ , at T=661K , 35.0X B =

＊ When ↓Ω λ and SΩ ↑

↑crT and point ( MSG∆ = MG λ∆ ) ↓

Eventually, these two points merge, then eutectic system occurs.

3. Same conditions , but

+=Ω+=Ω

)gular(ReKJ30)gular(ReKJ20

S

λ

Figure 10.12 ＊ Critical temperatures for solid and liquid solutions.

)(crT λ =R2λΩ

=1203K

)S(crT =R2SΩ

=1804K > )(crT λ

＊

α+α→α+→

/

/12

:Eutectic:Monotectic

λλλ

are shown.

4. Figure 10.13 Different ,λΩ SΩ

＊ (a) → (d), λΩ ↑ ⇒ ↑eutecticT , (e) liquid is unstable ⇒ monotectic.

§7. Eutectic Phase Diagrams and Monotectic Phase

Diagram

1. Complete solid solubility of components A, B:

valenceSimilar)4(ativityelectronegSimilar)3(

sizeatomicComparable)2(structurecrystalSame)1(

if any one condition is not met⇒ miscibility gap.

2. A-B system with two terminal solid solutions. ( βα, )

e.g. eutectic phase diagram. Figure 10.14

3. Solid solubility in βα, phases are extremely small.

＊ When solid solubility↓ ⇒ MG α∆ is compressed toward 1X A =

i.e. it coincides with vertical axis.

Figure 10.15 ＊

statesolidinityimmiscibilCompletestateliquidinymiscibilitComplete

Figure 10.16 Liquidus curves can be calculated assuming that liquid solution

is an ideal solution.

Aa 1XatA.t.r.w A)S( = Aa =1

Consider at T< )A(mT , ( pure )s(A + liquid ) coexist.

∴ )(A)s(A GG λο = = ο

λ)(AG +RT ln Aa )(A.t.r.w λ

∴ οολ )s(A)(A GG − = ο

)A(mG∆ =－RT ln Aa

assuming ideal liquid solution Aa = )(AX λ

∴ ο)A(mG∆ =－RT ln )(AX λ

i.e. )(AX λ =exp(RT

G )A(mο∆

− )

If ο)A(mG∆ = f (T) is known ⇒ )(AX λ =g (T) can be obtained.

e.g. Bi-Cd phase diagram Figure 10.17

(3.1) Bi

×+×+=

×+==∆

=

−−

−

)K/J(T101.21T1015.620C

)K/J(T106.228.18C)J(10900H

K544T

253)(Bi,P

3)s(Bi,P

)Bi(m

)Bi(m

λ

ο

∴ At T= )Bi(mT =544K, ο)Bi(mG∆ = 0 = ο

)Bi(mH∆ － ⋅mT ο)Bi(mS∆

∴ ο)Bi(mS∆ =

)Bi(m

)Bi(m

T

Hο∆=20.0 (J/K)

ο)Bi(mG∆ (T) = ? T )Bi(mT≠

∴ Bi,PC∆ = )(Bi,PC λ － )s(Bi,PC =1.2－16.45 310 −× T+21.1 25 T10 −×

ο)Bi(mG∆ = ο

544),Bi(mH∆ + ∫ ∆T

544 Bi,P dTC －T( ο544),Bi(mS∆ + ∫

∆T

544

Bi,P dTT

C)

=16560－23.79T－1.2T lnT+8.225 23 T10 −× －10.55× 15 T10 −

ideal solution : －RT ln =)(BiX λο

)Bi(mG∆

∴

ln =)(BiX λ 2

54

T

10269.1T10892.9Tln144.0861.2

T1992 ×

+×−++− − … … (1)

∴ )(BiX λ (T) is line(i) in Fig. 10.17

(3.2) Cd

=×+=

=∆=∆=

−

)K/J(7.29C)K/J(T103.122.22C

K/J77.10S,J6400HK594T

)(Cd,P

3)s(Cd,P

)Cd(m)Cd(m

)d(m

λ

οο

∴ ο)Cd(mG∆ = ο

)Cd(mH∆ + ∫ ∆T

594 Cd,P dTC －T( ο)Cd(mS∆ + ∫

∆T

594

)Cd(,P dTT

C)

ο)Cd(mG∆ =－RT ln )(CdX λ ,

∗∗

ilitylubsosolidneglectsolutionliquididealassume

∴ ln )(CdX λ =T495−－4.489+0.90 lnT－ T10397.7 4−× … … (2)

(3.3) ∴Eutectic temperature:

=−∗=∗

)T(gX1)2()T(fX)1(

)(Bi

)(Bi

λ

λ

∴ 1－f (T)=g (T) ⇒ T= eT =406 K

Actual : eT =419K＞ ( ) .cale K406T =

∵Caculation is based on “ideal” solution.

At T=419K ,

=∆

=∆

J1898G:)2(from

J2482G:)1(from

)Cd(m

)Bi(mο

ο

∵

=∆=∆

lnRTGlnRTG

)Cd(m

)Bi(mο

Cd

Bi

aa

∴

Cd

Bi

aa

580

490

.

.

=

= , Actual eutectic composition is 45.0X,55.0X BiCd ==

∴

==γ

==γ

051

091

.Xa

.Xa

Cd

CdCd

Bi

BiBi

＞1.0

∴ positive deviation from ideality ! ⇒ .)cal(ee TT >

4. When positive deviation from ideal liquid solution increase , 0G XS > , ↑XSG ⇒ Ω > crΩ (1) Liquidus curve is not a monotonic. Max. and min. of curve

forms.

(2) Liquid miscibility gap forms. ⇒ as conditions (1),(2) merge, monotectic system appears.

＊ Assuming liquid solution is regular.

∴－ ο)A(mG∆ =RT ln Aa =RT ln AX +RT ln Aγ

∵regular solution : 2A

A

)X1(

ln

−

γ=α =

RTΩ

∴ ο)A(mG∆ =RT ln AX +RT α 2

A )X1( −

ο)A(mG∆ =RT ln AX +Ω 2

A )X1( −

e.g. )A(mT =2000K, ο)A(mH∆ =10KJ

∴－ ο)A(mG∆ =10000+5T=RT ln AX +Ω 2

A )X1( −

Liquidus curve )(AX λ for different Ω is shown in Figure 10.18

∴

Ω > crΩ =25.3KJ

max. and min. appears⇒ part of curve has no physical meaning.

At Ω = crΩ

==

5.0XK1413T

A

cr

(AdX

dT)=0,

2A

2

dXTd

=0

Proof :

∵)(A

)S(A

A.t.r.wa

A.t.r.w

λ

a = exp[

RT

G )A(mο∆

] = exp[RT

H )A(mο∆

()A(m

)A(m

T

TT −)]

pure )S(A is standard state. ∴ Aa =1 w.r.t. )S(A

∴)l(Aa

1=exp[ −

∆

RT

H )A(mο

)A(m

)A(m

RT

Hο∆]

∴ln Aa =－RT

H )A(mο∆

+)A(m

)A(m

RT

Hο∆

dT

alnd A = 2)A(m

RT

Hο∆, ∴d ln Aa =

A

A

ada

= 2)A(m

RT

Hο∆dT

AdX

dT=

A)A(m aHRT

⋅∆ ο

2

A

A

dXda⋅

2

2

AdXTd

=(A)A(m aH

RT⋅∆ ο

2

A

A

dXda

⋅ )AdX

dT－(

ο)A(mH

RT∆

2

A

A

dXda

⋅ )． 2

1

Aa．(

A

A

dXda

)

+A)A(m aH

RT⋅∆ ο

2

． 2

2

A

A

dXad

∵ =Ω crΩ , T= crT , A

A

dXda

=0 and 2

2

A

A

dXad

=0

∴AdX

dT=0 and

2A

2

dXTd

=0

＊ When Ω =30KJ , for regular liquid solution

crT =R2

Ω=

231.830000

×=1804K Figure 10.19.

Ex: Cs-Rb phase diagram, Figure 10.20

=

=

C5.39T

C4.28T

)Rb(m

)Cs(mο

ο

,

−=∆

−=∆

)J(T05.72200G

)J(T95.62100G

)Rb(m

)Cs(mο

ο

Compare with theory assuming:

=Ω ?,regular:solutionsolidideal:solutionliquid

S

Sol: From phase diagram : liquidus and solidus curves touch each other

at

==

C7.9T47.0XRb

ο

when

<<

m(Rb)

m(Cs)

TTTT

Standard states : ο)s(CsG =0 , ο

)s(RbG =0

∴

Ω++=∆

∆⋅+∆⋅++=∆

CsRbSCsCsRbRbM

)s(

)Cs(mCs)Rb(mRbCsCsRbRbM

)(

XX)XlnXXlnX(RTG

GXGX)XlnXXlnX(RTG οολ

at

====

53.0X,47.0XK7.282C7.9T

CsRb

ο

M)(G λ∆ = M

)s(G∆ , ∴ sΩ =668(J)

c.p. When T=20 Cο =293K

common tangents to M)(G λ∆ and M

)s(G∆

==

13.0X10.0X

)(Rb

)s(Rb

λ

==

81.0X75.0X

)s(Rb

)(Rb λ

Very good fitting!! Figure 10.21