H.2 Euler-Maclaur in Sum Formula
10.3 Phase Diagram for an Ideal Solid and an Ideal Liquid
As we have seen above,g org for an ideal binary solution is a convex function ofXB, and this could happen either below or above the melting points of the pure components.
The interesting question is: What happens for temperatures between the melting points of the pure elements? For example, Si melts at 1685 K and Ge melts at 1210.4 K. Moreover, Ni melts at 1736 K and Cu melts at 1356.5 K.
The answer is that there is a miscibility gap and phase separation into a composite, part solid and part liquid. We analyze this situation as an application of the ideal solution model, and arbitrarily takeAto be the component having the higher melting point,TA. ThusTA>TB, whereTBis the melting point of pureB. For the ideal liquid, we have
μLA=μ0LA (T,p)+RTln(1−XB); μLB=μ0LB (T,p)+RTlnXB, (10.36) and for the ideal solid
μSA=μ0SA (T,p)+RTln(1−XB); μSB=μ0SB(T,p)+RTlnXB, (10.37) where the superscripts L and S denoting liquid and solid have been added to Eq. (10.34).
For a temperatureT that satisfiesTA >T > TB, we know thatμ0LA (T,p) > μ0SA (T,p)and μ0LB (T,p) < μ0SB (T,p). Therefore, graphs ofgL=μLA(1−XB)+μLBXBandgS=μSA(1−XB)+ μSBXBresult in two curves that cross, as shown inFigure 10–4. By applying the common
μ0LA μ0SA
μSA=μLA
gS
gL μ0SB
μ0LB
μSB=μLB
0.0 XBS XBL 1.0
FIGURE 10–4 Curves ofgLfor an ideal liquid solution andgSfor an ideal solid solution versusXBfor a temperature Tbetween the melting points of pureAand pureB. The common tangent construction applies, with tangency at XBSandXBL. As the temperature changes, the curves shift, resulting in a change of the points of common tangency.
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tangent construction toFigure 10–4, we see that there is a miscibility gap forXBS<XB<XBL. Here,XBSandXBLare the compositions at which common tangency occurs for the value of T corresponding to the figure. As T varies, these curves shift and we can trace out the miscibility gap in theXB,Tplane. The result is a “lens type” binary phase diagram, such as plotted inFigure 10–5.
10.3.1 Equations for the Miscibility Gap
Equations to determine the values ofXBSandXBLthat correspond to the boundary of the miscibility gap in theXB,Tplane can be determined by equating chemical potentials:
μLA(T,p,XBL)=μSA(T,p,XBS); μLB(T,p,XBL)=μSB(T,p,XBS). (10.38) Substitution of Eqs. (10.36) and (10.37) into Eq. (10.38) gives:
μ0LA (T,p)−μ0SA (T,p)=RTln
1−XBS 1−XBL
; μ0LB (T,p)−μ0SB (T,p)=RTln
XBS XBL
. (10.39)
In order to proceed further, we need a model to evaluate the chemical potential differences on the left of Eq. (10.39). We write these in the forms
μA:=μ0LA (T,p)−μ0SA (T,p)=hA−TsA;
μB:=μ0LB (T,p)−μ0SB (T,p)=hB−TsB, (10.40) wherehAandhBare enthalpy differences (liquid minus solid) andsAandsBare the corresponding entropy differences. We assume as an approximation that these enthalpy and entropy differences are constants that we relate by requiringμA =0 atT =TAand μB = 0 atT = TB. This giveshA = TAsAandhB = TBsB, soEqs. 10.40become approximately
μA=hA(1−T/TA); μB=hB(1−T/TB). (10.41) We recognizehAandhBas the respective latent heats of fusion per mole and observe that Eqs. (10.41) have the expected algebraic signs, that is,μA > 0 andμB < 0 for TA > T > TB. An alternative view of Eqs. (10.41) is to assume that they are the leading terms in expansions in the variables(T−TA)/TAand(T−TB)/TB.
Substitution of Eqs. (10.41) into Eqs. (10.39) shows thatXBS<XBLas expected and 1−XBL
1−XBS =exp
−hA R
1 T − 1
TA ≡EA(T); XBS
XBL =exp hB
R 1
T − 1
TB ≡EB(T). (10.42)
ForTA>T >TB, we note that 0<EA(T) <1 and 0<EB(T) <1. Solving Eq. (10.42) forXBS andXBLgives
0.0 0.2 0.4 0.6 0.8 1.0 1200 K
1400 K 1600 K TA
Solid (crystal)
Liquid XBS(T)
XBL(T)
S + L
XB
TB
FIGURE 10–5 Computed phase diagram for an ideal solid solution and an ideal liquid solution for physical constants that resembleA=Si andB=Ge. The plot showsXBSandXBLas functions ofTaccording to Eq. (10.43).TA=1685 K andTB=1210.4 K.
XBS=EB(T) 1−EA(T)
1−EA(T)EB(T); XBL= 1−EA(T)
1−EA(T)EB(T). (10.43) We observe from Eq. (10.43) that XBS and XBL increase from 0 to 1 with XBS ≤ XBL as T decreases fromTAtoTB.
In order to make a plot of Eq. (10.43) we need some numerical values of the physical constants. IfA were Si andBwere Ge, then hA/(RTA) = 3.59 andhB/(RTB) = 3.14.
Figure 10–5shows a plot of the resulting phase diagram. There is a miscibility gapwith a “lens shape” connecting the melting points of pureAandB. Above the miscibility gap, the liquid solution is stable, and below it the solid solution is stable. The curveXBS(T)is called thesolidusandXBL(T)is called theliquidus. For a point within the miscibility gap, a homogeneous solution is unstable, so the corresponding equilibrium state is a composite, consisting of part liquid solution and part solid solution. The amounts of solid and liquid in this composite are governed by the lever rule.
Example Problem 10.2. For the phase diagram depicted inFigure 10–5, what is the mole fraction of solid in equilibrium with liquid atT =1600 K if the overall composition isXB=0.22 mole fraction? By how much does the chemical potential ofAin this solid differ from that of pure solidAatT =1600 K?
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Solution 10.2. AtT =1600 K, the compositions at the solidus and liquidus are estimated to be 0.13 and 0.28. By the lever rule, the mole fraction of solid is(0.28−0.22)/(0.28−0.13)=0.4. From the first of Eqs. (10.37) we obtainμSA−μ0SA =RTln(1−XB)=3200 ln(1−0.13)= −456 cal/mol.
Example Problem 10.3. For a very dilute ideal solution XB 1, develop approximate formulae for the distribution coefficientk:=XBS/XBLand the slope dXBL/dT.
Solution 10.3. From the second of Eqs. (10.42) we can approximateT = TAto obtaink = EB(TA)=constant<1. Then we take the derivative of the first of Eqs. (10.42) with respect toT and evaluate the result atXBS=XBL=0 andT=TAto obtain
d(XBL−XBS)
dT = −hA
RTA2. (10.44)
Then use ofkto eliminateXBSgives dXBL
dT = − hA
(1−k)RTA2. (10.45)
This result is related to J. H. van’t Hoff’s law of freezing point lowering for a dilute solid solution [22, p. 235]. Similar formulae can be obtained at the other end of the phase diagram forXA1.
The results arek:=XAS/XAL=1/EA(TB)=constant>1 and dXAL/dT=hB/[(k−1)RTB2].