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Chapter10 Gibbs Free Energy-Composition
Curves and Binary Phase Diagrams
§10-1. Introduction
§10-2. Gibbs Free Energy Curve, ∆ ( )BM XG
§10-3. ∆ ( )BM 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 iλ 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)∣ 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
-
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<
.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
∂∂
)
-
∴ 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, ο λ)(BG －ο
)S(BG =ο
)B(mG∆ =ο
)B(mH∆ －Tο
)B(mS∆ ≅ο
)B(mH∆
−
)A(m
)A(m
T
TT
line ab , G∆ = BXο
)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 0
∴ )s(Ba =1, at 1XB = . i.e. gn =1
point m is )(Ba λ of pure liquid B
∴ )(Ba λ = qnmn
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
-
Figure 10.11
＊ For solid solution : crT = R2SΩ = 601
314.8210000
=×
(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
∴ οο λ )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∆ － ⋅mTο
)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 =1.2－16.45310 −× T+21.1 25 T10 −×
ο )Bi(mG∆ =ο
544),Bi(mH∆ + ∫ ∆T
544 Bi,PdTC －T( ο 544),Bi(mS∆ + ∫
∆T544
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 25
4
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,PdTC －T( ο )Cd(mS∆ + ∫
∆T594
)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 α2
A )X1( −
ο )A(mG∆ =RT ln AX +Ω2
A )X1( −
e.g. )A(mT =2000K, ο
)A(mH∆ =10KJ
∴－ ο )A(mG∆ =10000+5T=RT ln AX +Ω2
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 =－ RTH )A(m
ο∆+
)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
⋅∆ ο2
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
⋅ )． 21
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 = R2Ω
=231.8
30000×
=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