MATSCI 204 Thermodynamics and Phase Equilibria ... - PDFKUN.COM
MATSCI 204 Thermodynamics and Phase Equilibria Winter 2013 Chapter #5 Practice problems Problem 4 a-Assuming that you are cooling an equimolar liquid Au-Bi solution reversibly from 1200°C, describe the sequence of events (i.e. phases that form and disappear at what temperatures and all phase present as you cool) until you reach 0°C. b-Sketch and label Gibbs free energy vs. composition diagrams for the Au-Bi system at 371°C, 247°C and 241°C.
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Solution a- At 1200°C, a 50% at Au-Bi solution is homogeneous. As we cool, the first phase transformation occurs at 530°C, where pure Au is in equilibrium with the liquid. The two-phase equilibrium persists until 371°C, where Au disappears and the liquid is in equilibrium with a Au3Bi intermetallic compound. At 241°C, the system incurs in a triple point, where Au3Bi, liquid Au-Bi solution and Bi are in equilibrium. Upon further cooling until 116°C, the system is in a two-phase Au3Bi/Bi field. At T0.33
3
Problem: PtSi is an ohmic contact used in Si microelectronic devices. It is one of several silicide compounds that can form in the Pt-Si binary system (see attached). Note: the dashed line near XSi = 0 is the estimated solubility of Si in the Pt-Si fcc alloy. a) Sketch the activity of Pt as a function of composition at 1000°C. Label your sketch fully and be as quantitative as possible. b) Assume that the liquid behaves as a regular solution with an enthalpy of mixing of 1000J/mol at XSi = 0.5. The heat of fusion of Si is 50.2 kJ/mol and that of Pt is 19.7 kJ/mol(Tm,Pt = 1768°C). Determine the Gibbs free energy change at 1210°C for the reaction: Pt(l) + Si(l) = PtSi(s).
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Solution Plot of activity of Pt as a function of composition @ 1000 oC.
atomic % Si
L+ Pt2Si
L+ Si fcc +L
fcc Pt(Si)
L
L
aPt
L L+ PtSi
Ideal solution
0.02
0.21
0.28 0.33
0.41
0.65 0.5
0.67
0.38
N.B. Activity of Pt drops abruptly to zero @ xSi
0.
b) 0S We are asked to calculate GPtSi " GPt0L " GSi0L From the phase diagram, @ T = 1210 oC, PtSi(s) undergoes congruent melting.
!
5
1. 0
1 PtSi L at x=0.5 with the consistent choice of reference "Gmix = "Gmix 2 L states (as discussed in class), where "Gmix is the free-energy of mixing of the liquid Pt-Si solution. The factor or ½ comes from the fact that at x=0.5, we have 0.5 moles of Pt and 0.5 moles of Si, which form 0.5 moles of PtSi. ! We choose pure solid Pt and Si as reference states: PtSi 0S ! the free-energy of formation of PtSi from pure solid "Gmix = GPtSi # GPt0S # GSi0S is components. L "Gmix = x Si (GSiL # GSi0S ) + x Pt (GPtL # GPt0S ) At the congruent melting point: 1 0S (1) GPtSi " GPt0S " GSi0S ) = x Si (GSiL " GSi0S ) + x Pt (GPtL " GPt0S ) ( 2 The liquid solution is a regular solution with a given "H mix , therefore: x Si (GSiL " GSi0L ) + x Pt (GPtL " GPt0L ) = #H mix + RT ( x Si ln x Si + x Pt ln x Pt ) This is equivalent to
! !
! ! !
!
!
We now introduce GPt0S and GSi0S : ! Si0L $S ) = #H mix + RT ( x Si ln x Si + x Pt ln x Pt ) x Si (GSiL " GSi0S + #GSi0L $S ) + x Pt (GPtL " GPt0S + #G Plugging this result in Eq. 1: 1 0S ! 0S GPtSi " GPt " GSi0S!) = #H mix + RT ( x Si ln x Si + x Pt ln x Pt ) " x Si#GSi0L $S " x Pt #GPt0L $S ( 2 Moreover, 1 0S 1 0S #GPt0L $S #GSi0L $S 0S 0S 0L 0L G " G " G = G " G " G " ( PtSi Pt Si ) 2 ( PtSi Pt Si ) 2 " 2 2 Since at the congruent melting point, xSi=xPt=0.5, We have: 1 0S #GPt0L $S #GSi0L $S GPtSi " GPt0L " GSi0L ) " " = ( 2 2 2 = #H mix + RT (0.5ln0.5 + 0.5ln0.5) " 0.5#GSi0L $S " 0.5#GPt0L $S As a result, the free-energy of PtL+SiL=PtSiS is: 0S GPtSi " GPt0L " GSi0L = 2[#H mix + RT (0.5ln0.5 + 0.5ln0.5)] =-15.09 kJ/mol
!
!
6
(2)
Problem 4 a) A homogeneous liquid of composition xSiO2=0.6 is cooled slowly from 2500°C to rom temperature such that at any temperature during cooling the sample may be assumed to be at equilibrium. List the phases present during each stage of the cooling process and the temperature ranges over which they are observed. Be sure to indicate the temperatures at which phase boundaries are crossed during cooling of this sample and describe which new phases are fomed (and which old phases may disappear) at each of these temperatures. b) The kinetics of formation of Zircon, ZrSiO2, are notoriously slow. Repeat question b, assuming that the cooling process is performed more quicky, such that zircon does not form. c) Sketch the variation of the activity of SiO2 across the composition range from pure ZrO2 to pure SiO2 at 2300°C. Indicate regions of to phase and single phase stability and be quantitative in locating phase boundaries.
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Solution Part a) Equilibrium cooling: ZrO2/SiO2 system; XSiO2 = 0.6, Tinit = 2500 0C. i) ii) iii) iv) v) vi) vii) viii)
2500 oC > T > 2336 oC T = 2336 oC 2336 oC > T > 2250 oC T = 2250 oC 2250 oC> T > 1687 oC T = 1687 oC 1687 oC > T > 1676oC T = 1676 oC 1676 oC > T > 1470 oC T = 1470 oC 1470 oC > T > 867 oC T = 867 oC 867 oC > T > 573 oC T = 573 oC 573oC > T > 25 oC
Liquid(L) L → L1 + L2 (phase separation) L1 + L2 L1 → ZrO2 (tet) + L2 (monotectic rxn) ZrO2(tet) + L2 L2→ZrO2(tet)+SiO2(crist) (eutectic) ZrO2(tet) + SiO2 (crist) ZrO2(tet) + SiO2(crist)→ZrSiO4(peritectoid) ZrSiO4 + SiO2 (crist) SiO2(crist)→SiO2(trid) (pseudomorphic trans.) ZrSiO4 + SiO2 (trid) SiO2(trid) →SiO2(H-qtz) (pseudomorphic trans.) ZrSiO4 + SiO2 (H-qtz) SiO2(H-qtz)→SiO2(L-qtz) (pseudomorphic trans.) ZrSiO4 + SiO2 (L-qtz)
Part b) Till T = 1676 oC, the phases present & the phase transformations are same as that of Part (a). 1676 oC > T > 1470 oC T = 1470 oC
ZrO2 (tat) + SiO2 (crest) SiO2 (crist) →SiO2 (trid)
1470 oC > T > 1170 oC T = 1170 oC
ZrO2 (tet) + SiO2 (trid) ZrO2 (tet) →ZrO2 (mon)
1170 oC > T > 867 oC T = 867 oC
ZrO2 (mon) + SiO2 (trid) SiO2 (trid) →SiO2 (H-quartz)
867 oC > T > 573 oC T = 573 oC
ZrO2 (mon) + SiO2 (H-quartz) SiO2 (H-quartz) →SiO2 (L-quartz)
573 oC > T > 25 oC
ZrO2(mon) + SiO2 (L-quartz)
Part c) At T = 2300 oC, the following phase fields exist. ZrO2 (cub) + L1 L1
0<x 0 at T=T1)
!
L #S "gFeO =
!
Using the first condition (derivatives must be equal at xS=0.5 and xL) we obtain:
! S $L ! + % x (. #h S $L # 0.5 & #hFeO " T1 " Tm,FeO ) = RT-ln' L *0 + MnO (Tm,MnO " T1 ) because ln% ()0 ( $1" 0.5 ' Tm,FeO , &1" x L )/ Tm,MnO we extract: T1 = !
S #L "hFeO Tm,FeO
S #L S #L "hFeO $ "hMnO + % x (. "h S #L $ MnO + RT-ln' L *0 Tm,MnO , & 1$ x L )/
!
The second condition gives:
RT !1 [ x L ln(x L ) + (1" x L )ln(1" x L )] + x L = RT ln(0.5) + 0.5
!
S $L #hMnO (Tm,MnO " T1) = Tm,MnO
S $L #hFeO #h S $L T1 " Tm,FeO ) " FeO (T1 " Tm,FeO ) % (0.5 " x L ) ( Tm,FeO Tm,FeO
These two equations can be put through Mathematica to find xL and T1. I calculated T1 as a function of xL with the first equation and then found the root of the second by plotting it. With this approximate method, I obtain 0.34<xL Au3.6Sn0.4 compositions: Au is 5% Sn in 95% Au; L is 79% Au and 21% Sn • @521°C Au3.6Sn0.4+L -> Au0.85Sn0.15 compositions: L is 78% Au and 22% Sn • @309°C AuSn+L ->AuSn2 compositions: L is 29% Au and 71% Sn • @252°C AuSn2+L -> AuSn4 compositions: L is 12% Au and 88% Sn Eutectics: • @280°C L -> Au0.85Sn0.15+AuSn compositions: L is 71% Au and 29% Sn; Au0.85Sn0.15 is 82% Au and 18% Sn (i.e. it’s a solid solution of Au0.85Sn0.15 and excess Sn) • @217°C L -> AuSn4 + Sn compositions: L is 6% Au and 94% Sn Peritectoid: • @190°C Au0.85Sn0.15 + AuSn -> Au5Sn compositions: Au0.85Sn0.15 is 86% Au and 14% Sn (it’s nonstoichiometric with an excess of Au) Eutectoid: • @55°C AuSn4 -> Sn + AuSn2 AuSn has a congruent melting point and therefore can be cast directly from the melt.
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Problem Consider the attached phase diagram of the binary system Ni/Al. List all the eutectics, peritectics, eutectoids, peritectoids and congruent melting points. In your list, write the reaction that occurs upon cooling and the reaction temperature.
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Answer Congruent: 1638°C: L->NiAl Eutectics: 640°C: L->Al+NiAl3; 1385°C: L->Ni3Al+Ni Peritectics: 1133°C: L+NiAl->Ni2Al3; 1395°C: L+NiAl->Ni3Al; 854°C: L+Ni2Al3->NiAl3 Peritectoid: 700°C: NiAl+Ni3Al->Ni5Al3
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Problem 4 Consider the following phase diagram of a binary system A/B. You can consider AB as a line compound.
We will work at the T=1300K isotherm 1- First assume that all solutions (solid and liquid) are ideal. - plot the activity of A as a function of xB. Be quantitative wherever possible. (pick the equilibrium state of A at T=1300K as the reference state) - calculate the activity of A in AB when AB is in equilibrium with the A-rich liquid solution and with the B-rich liquid solution as well. - estimate the molar melting enthalpy of pure A Δhm at its melting point. Explain the approximations you make.
!
2- The terminal B-rich solution now obeys the dilute solution model for solvent and solute throughout the whole composition range at 1300K. Henry’s coefficient for A is 0.25. Furthermore the liquid solution in equilibrium with the terminal B-rich solid is modeled as a regular solution with a molar enthalpy of mixing of the form "hmix = #x A x B . - calculate α. - what’s the activity of A in AB when AB is in equilibrium with the B-rich liquid solution (i.e. right side of the AB line compound)? 3- The liquid in equilibrium with the terminal A-rich solid is not an ideal solution either. Assume it’s a regular solution and the Δgmix at xB=0.15 is -7 kJ/mol. 23
Furthermore, the activity coefficient of B at xB=0.15 is equal to 0.4. What is the activity of A in AB when it’s in equilibrium with the A-rich liquid solution (i.e. left side of the AB line compound)?
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Answer 1 With the stated reference state and using all the information (i.e. that all solutions are ideal), we conclude that the a vs. x lines for the two liquid solutions are both on the diagonal, which fixes all points:
! !
!
As a result, the activity of A in AB on the A-rich side is 0.85 and that of A on the B-rich side is 0.45. The Δhm of A is obtained from the intercept of the activity of A in the B-rich terminal solid solution with the y-axis at xB=0 (or xA=1). Due to the choice of reference states, the $ #g m ' intercept is exp&" A ) . Since the solution is ideal, we conclude that: % RT ( % $g m ( aA = x A " exp'# A * . And since aA=0.3 at xA=0.1, we get & RT ) $a ' ! "gAm = #RT ln& A ) = #RT ln( 3) % xA ( Further, using the Turnbull approximation, since we are at T
1
Solution a- At 1200°C, a 50% at Au-Bi solution is homogeneous. As we cool, the first phase transformation occurs at 530°C, where pure Au is in equilibrium with the liquid. The two-phase equilibrium persists until 371°C, where Au disappears and the liquid is in equilibrium with a Au3Bi intermetallic compound. At 241°C, the system incurs in a triple point, where Au3Bi, liquid Au-Bi solution and Bi are in equilibrium. Upon further cooling until 116°C, the system is in a two-phase Au3Bi/Bi field. At T0.33
3
Problem: PtSi is an ohmic contact used in Si microelectronic devices. It is one of several silicide compounds that can form in the Pt-Si binary system (see attached). Note: the dashed line near XSi = 0 is the estimated solubility of Si in the Pt-Si fcc alloy. a) Sketch the activity of Pt as a function of composition at 1000°C. Label your sketch fully and be as quantitative as possible. b) Assume that the liquid behaves as a regular solution with an enthalpy of mixing of 1000J/mol at XSi = 0.5. The heat of fusion of Si is 50.2 kJ/mol and that of Pt is 19.7 kJ/mol(Tm,Pt = 1768°C). Determine the Gibbs free energy change at 1210°C for the reaction: Pt(l) + Si(l) = PtSi(s).
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Solution Plot of activity of Pt as a function of composition @ 1000 oC.
atomic % Si
L+ Pt2Si
L+ Si fcc +L
fcc Pt(Si)
L
L
aPt
L L+ PtSi
Ideal solution
0.02
0.21
0.28 0.33
0.41
0.65 0.5
0.67
0.38
N.B. Activity of Pt drops abruptly to zero @ xSi
0.
b) 0S We are asked to calculate GPtSi " GPt0L " GSi0L From the phase diagram, @ T = 1210 oC, PtSi(s) undergoes congruent melting.
!
5
1. 0
1 PtSi L at x=0.5 with the consistent choice of reference "Gmix = "Gmix 2 L states (as discussed in class), where "Gmix is the free-energy of mixing of the liquid Pt-Si solution. The factor or ½ comes from the fact that at x=0.5, we have 0.5 moles of Pt and 0.5 moles of Si, which form 0.5 moles of PtSi. ! We choose pure solid Pt and Si as reference states: PtSi 0S ! the free-energy of formation of PtSi from pure solid "Gmix = GPtSi # GPt0S # GSi0S is components. L "Gmix = x Si (GSiL # GSi0S ) + x Pt (GPtL # GPt0S ) At the congruent melting point: 1 0S (1) GPtSi " GPt0S " GSi0S ) = x Si (GSiL " GSi0S ) + x Pt (GPtL " GPt0S ) ( 2 The liquid solution is a regular solution with a given "H mix , therefore: x Si (GSiL " GSi0L ) + x Pt (GPtL " GPt0L ) = #H mix + RT ( x Si ln x Si + x Pt ln x Pt ) This is equivalent to
! !
! ! !
!
!
We now introduce GPt0S and GSi0S : ! Si0L $S ) = #H mix + RT ( x Si ln x Si + x Pt ln x Pt ) x Si (GSiL " GSi0S + #GSi0L $S ) + x Pt (GPtL " GPt0S + #G Plugging this result in Eq. 1: 1 0S ! 0S GPtSi " GPt " GSi0S!) = #H mix + RT ( x Si ln x Si + x Pt ln x Pt ) " x Si#GSi0L $S " x Pt #GPt0L $S ( 2 Moreover, 1 0S 1 0S #GPt0L $S #GSi0L $S 0S 0S 0L 0L G " G " G = G " G " G " ( PtSi Pt Si ) 2 ( PtSi Pt Si ) 2 " 2 2 Since at the congruent melting point, xSi=xPt=0.5, We have: 1 0S #GPt0L $S #GSi0L $S GPtSi " GPt0L " GSi0L ) " " = ( 2 2 2 = #H mix + RT (0.5ln0.5 + 0.5ln0.5) " 0.5#GSi0L $S " 0.5#GPt0L $S As a result, the free-energy of PtL+SiL=PtSiS is: 0S GPtSi " GPt0L " GSi0L = 2[#H mix + RT (0.5ln0.5 + 0.5ln0.5)] =-15.09 kJ/mol
!
!
6
(2)
Problem 4 a) A homogeneous liquid of composition xSiO2=0.6 is cooled slowly from 2500°C to rom temperature such that at any temperature during cooling the sample may be assumed to be at equilibrium. List the phases present during each stage of the cooling process and the temperature ranges over which they are observed. Be sure to indicate the temperatures at which phase boundaries are crossed during cooling of this sample and describe which new phases are fomed (and which old phases may disappear) at each of these temperatures. b) The kinetics of formation of Zircon, ZrSiO2, are notoriously slow. Repeat question b, assuming that the cooling process is performed more quicky, such that zircon does not form. c) Sketch the variation of the activity of SiO2 across the composition range from pure ZrO2 to pure SiO2 at 2300°C. Indicate regions of to phase and single phase stability and be quantitative in locating phase boundaries.
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Solution Part a) Equilibrium cooling: ZrO2/SiO2 system; XSiO2 = 0.6, Tinit = 2500 0C. i) ii) iii) iv) v) vi) vii) viii)
2500 oC > T > 2336 oC T = 2336 oC 2336 oC > T > 2250 oC T = 2250 oC 2250 oC> T > 1687 oC T = 1687 oC 1687 oC > T > 1676oC T = 1676 oC 1676 oC > T > 1470 oC T = 1470 oC 1470 oC > T > 867 oC T = 867 oC 867 oC > T > 573 oC T = 573 oC 573oC > T > 25 oC
Liquid(L) L → L1 + L2 (phase separation) L1 + L2 L1 → ZrO2 (tet) + L2 (monotectic rxn) ZrO2(tet) + L2 L2→ZrO2(tet)+SiO2(crist) (eutectic) ZrO2(tet) + SiO2 (crist) ZrO2(tet) + SiO2(crist)→ZrSiO4(peritectoid) ZrSiO4 + SiO2 (crist) SiO2(crist)→SiO2(trid) (pseudomorphic trans.) ZrSiO4 + SiO2 (trid) SiO2(trid) →SiO2(H-qtz) (pseudomorphic trans.) ZrSiO4 + SiO2 (H-qtz) SiO2(H-qtz)→SiO2(L-qtz) (pseudomorphic trans.) ZrSiO4 + SiO2 (L-qtz)
Part b) Till T = 1676 oC, the phases present & the phase transformations are same as that of Part (a). 1676 oC > T > 1470 oC T = 1470 oC
ZrO2 (tat) + SiO2 (crest) SiO2 (crist) →SiO2 (trid)
1470 oC > T > 1170 oC T = 1170 oC
ZrO2 (tet) + SiO2 (trid) ZrO2 (tet) →ZrO2 (mon)
1170 oC > T > 867 oC T = 867 oC
ZrO2 (mon) + SiO2 (trid) SiO2 (trid) →SiO2 (H-quartz)
867 oC > T > 573 oC T = 573 oC
ZrO2 (mon) + SiO2 (H-quartz) SiO2 (H-quartz) →SiO2 (L-quartz)
573 oC > T > 25 oC
ZrO2(mon) + SiO2 (L-quartz)
Part c) At T = 2300 oC, the following phase fields exist. ZrO2 (cub) + L1 L1
0<x 0 at T=T1)
!
L #S "gFeO =
!
Using the first condition (derivatives must be equal at xS=0.5 and xL) we obtain:
! S $L ! + % x (. #h S $L # 0.5 & #hFeO " T1 " Tm,FeO ) = RT-ln' L *0 + MnO (Tm,MnO " T1 ) because ln% ()0 ( $1" 0.5 ' Tm,FeO , &1" x L )/ Tm,MnO we extract: T1 = !
S #L "hFeO Tm,FeO
S #L S #L "hFeO $ "hMnO + % x (. "h S #L $ MnO + RT-ln' L *0 Tm,MnO , & 1$ x L )/
!
The second condition gives:
RT !1 [ x L ln(x L ) + (1" x L )ln(1" x L )] + x L = RT ln(0.5) + 0.5
!
S $L #hMnO (Tm,MnO " T1) = Tm,MnO
S $L #hFeO #h S $L T1 " Tm,FeO ) " FeO (T1 " Tm,FeO ) % (0.5 " x L ) ( Tm,FeO Tm,FeO
These two equations can be put through Mathematica to find xL and T1. I calculated T1 as a function of xL with the first equation and then found the root of the second by plotting it. With this approximate method, I obtain 0.34<xL Au3.6Sn0.4 compositions: Au is 5% Sn in 95% Au; L is 79% Au and 21% Sn • @521°C Au3.6Sn0.4+L -> Au0.85Sn0.15 compositions: L is 78% Au and 22% Sn • @309°C AuSn+L ->AuSn2 compositions: L is 29% Au and 71% Sn • @252°C AuSn2+L -> AuSn4 compositions: L is 12% Au and 88% Sn Eutectics: • @280°C L -> Au0.85Sn0.15+AuSn compositions: L is 71% Au and 29% Sn; Au0.85Sn0.15 is 82% Au and 18% Sn (i.e. it’s a solid solution of Au0.85Sn0.15 and excess Sn) • @217°C L -> AuSn4 + Sn compositions: L is 6% Au and 94% Sn Peritectoid: • @190°C Au0.85Sn0.15 + AuSn -> Au5Sn compositions: Au0.85Sn0.15 is 86% Au and 14% Sn (it’s nonstoichiometric with an excess of Au) Eutectoid: • @55°C AuSn4 -> Sn + AuSn2 AuSn has a congruent melting point and therefore can be cast directly from the melt.
20
Problem Consider the attached phase diagram of the binary system Ni/Al. List all the eutectics, peritectics, eutectoids, peritectoids and congruent melting points. In your list, write the reaction that occurs upon cooling and the reaction temperature.
21
Answer Congruent: 1638°C: L->NiAl Eutectics: 640°C: L->Al+NiAl3; 1385°C: L->Ni3Al+Ni Peritectics: 1133°C: L+NiAl->Ni2Al3; 1395°C: L+NiAl->Ni3Al; 854°C: L+Ni2Al3->NiAl3 Peritectoid: 700°C: NiAl+Ni3Al->Ni5Al3
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Problem 4 Consider the following phase diagram of a binary system A/B. You can consider AB as a line compound.
We will work at the T=1300K isotherm 1- First assume that all solutions (solid and liquid) are ideal. - plot the activity of A as a function of xB. Be quantitative wherever possible. (pick the equilibrium state of A at T=1300K as the reference state) - calculate the activity of A in AB when AB is in equilibrium with the A-rich liquid solution and with the B-rich liquid solution as well. - estimate the molar melting enthalpy of pure A Δhm at its melting point. Explain the approximations you make.
!
2- The terminal B-rich solution now obeys the dilute solution model for solvent and solute throughout the whole composition range at 1300K. Henry’s coefficient for A is 0.25. Furthermore the liquid solution in equilibrium with the terminal B-rich solid is modeled as a regular solution with a molar enthalpy of mixing of the form "hmix = #x A x B . - calculate α. - what’s the activity of A in AB when AB is in equilibrium with the B-rich liquid solution (i.e. right side of the AB line compound)? 3- The liquid in equilibrium with the terminal A-rich solid is not an ideal solution either. Assume it’s a regular solution and the Δgmix at xB=0.15 is -7 kJ/mol. 23
Furthermore, the activity coefficient of B at xB=0.15 is equal to 0.4. What is the activity of A in AB when it’s in equilibrium with the A-rich liquid solution (i.e. left side of the AB line compound)?
24
Answer 1 With the stated reference state and using all the information (i.e. that all solutions are ideal), we conclude that the a vs. x lines for the two liquid solutions are both on the diagonal, which fixes all points:
! !
!
As a result, the activity of A in AB on the A-rich side is 0.85 and that of A on the B-rich side is 0.45. The Δhm of A is obtained from the intercept of the activity of A in the B-rich terminal solid solution with the y-axis at xB=0 (or xA=1). Due to the choice of reference states, the $ #g m ' intercept is exp&" A ) . Since the solution is ideal, we conclude that: % RT ( % $g m ( aA = x A " exp'# A * . And since aA=0.3 at xA=0.1, we get & RT ) $a ' ! "gAm = #RT ln& A ) = #RT ln( 3) % xA ( Further, using the Turnbull approximation, since we are at T
