Solid-liquid equilibria (SLE) calculations begin just as VLE and LLE calculations, by equating fugacities. From Eqn. 11.13, 
. The next step is to equate 
and derive an equation to solve for temperature or composition depending on the problem statement. We have deliberately avoided substituting 
, however, as we did for VLE and LLE. This is because pure components below their melting temperature do not exist as liquids, so the vapor pressure is not relevant. This peculiarity necessitates a derivation to characterize a hypothetical liquid state. The treatment of the solid phase is usually much simpler, however, because most solids exist in a practically pure state. The purity of crystallized solid phases is one of the reasons for the prevalence of crystallization and filtration in the pharmaceutical industry.
Pressure Effects
For SLE, as for LLE, pressure changes usually have very small effects on the equilibria unless the pressure changes are large (10 to 100 MPa), because the enthalpies and entropies of condensed phases are only weakly pressure-dependent. Since dG = RT d ln f = dH – T dS = V dP for a pressure change at constant T, the Gibbs energy and fugacity exhibit only small changes with pressure when the enthalpy and entropy exhibit small changes. (Recall that the Poynting correction factor is usually very near one.) In a mixture of liquids, the analysis must be done with partial molar enthalpies and partial molar entropies; however, these properties also depend only weakly on pressure. In the following subsections, we calculate properties at the triple-point, or other low pressures, and use the results at atmospheric pressure without pressure correction due to the weak pressure dependence.
Pressure effects on SLE are usually neglected.
SLE in a Single Component System
To begin our discussion, we will consider a single component. At 1 bar, water freezes at 0°C. Ice exists below this temperature and liquid exists above. From the principles of thermodynamics, the Gibbs energy is minimized at constant pressure and temperature; therefore, above 0°C, the Gibbs energy of liquid water must be lower than the Gibbs energy of solid water. In order to express this concept quantitatively, we must consider how the Gibbs energies of each of these phases changes with temperature.
The effect of temperature on the Gibbs energy of any phase may be determined most easily at constant pressure. We may write dG =–S dT + V dP, and recognize using the concepts of Chapter 6
(∂G/∂T)P = –S
The temperature dependence of Gibbs energy is then dependent on the entropy of the phase. Entropy is a positive quantity; therefore, the Gibbs energy of any phase must decrease with increasing temperature. However, the entropy of a liquid phase is greater than the entropy of a solid phase; thus, the Gibbs energy of a liquid phase decreases more rapidly as the temperature increases. Since the Gibbs energies of the phases are equal at the freezing temperature, the Gibbs energy of the liquid will lie below the Gibbs energy of the solid above the freezing temperature (see Fig. 14.8).1 Below the freezing temperature, we can still extrapolate the Gibbs energy of the liquid, but we must refer to this as a hypothetical liquid because it does not exist as a stable phase. The portion of the solid Gibbs energy curve above the melting temperature represents the Gibbs energies of a hypothetical solid, and a melting process will occur spontaneously at constant temperature and pressure because the ΔGfus for the process is negative. Alternatively, an equilibrium solid will not be formed above the melting temperature because ΔGfus is positive. A discussion for behavior below the melting temperature is not presented, although the ideas are similar. Melting of a solid below the freezing temperature requires a ΔGfus > 0 at constant temperature and pressure. The mixing process balances this positive Gibbs energy change to create mixtures below the normal melting temperature.
Figure 14.8. Illustration of Gibbs energies for pure SLE.
ΔGfus > 0 below the normal melting temperature.
The Calculation Pathway for Mixtures
We know that solids dissolve in liquid mixtures well below their normal melting temperatures. Sugar and salt both dissolve in water at room temperature, although the pure solid melting temperatures are far above room temperature. Also, salt is spread on the highways in the winter to lower the temperature at which solid ice forms. We may choose to address several problems such as: 1) How much sugar may be dissolved in water before the solubility limit is reached? 2) In a water/salt solution, at what temperature will a solid form, and will the crystals be water or salt? (Salt is introduced as a practical example, although rigorous treatment of the problem involves electrolyte thermodynamics.) 3) How may a solvent or solvent mixture be modified to regulate crystallization? In order to deal with these concepts mathematically, we use state properties to calculate thermodynamic changes along convenient pathways that involve hypothetical steps.
Let us consider a practical example of dissolving naphthalene (2) in n-hexane (1) at 298 K. Since the normal melting temperature for pure naphthalene is 353.3 K, how can we explain the phenomenon that the naphthalene dissolves in hexane? First, recall that the naphthalene will dissolve in the n-hexane if the total Gibbs energy of the system (n-hexane and naphthalene) decreases upon dissolution. Thus, more and more solid may be added to the liquid solution until any further addition causes the total Gibbs energy to increase rather than decrease. This method of calculating equilibrium is fairly tedious to apply, and a preferred method is used that gives identical results: The solubility limit is reached when the chemical potential of the naphthalene in the liquid is the same as the chemical potential of the pure solid naphthalene. Therefore, we solve the problem by equating the chemical potentials for naphthalene in the liquid and solid phases.
The equilibrium can be written as 
or, recognizing the notational definitions 
and 
, therefore,
Next, envision the hypothetical pathway shown in Fig. 14.9 to calculate the chemical potential difference given by Eqn. 14.14, which consists of two primary steps.
Figure 14.9. Illustration of the two-step process for calculating solubility of solids in liquids. Overall, 
. Note that the Gibbs energy goes up in Step 1 to create liquid, below the normal melting Tm, but the Gibbs energy goes down when the liquid is mixed.
Step 1. Naphthalene is melted to form a hypothetical liquid at 298 K. The Gibbs energy change for this step is positive as discussed above. The Gibbs energy change is:
where the superscript hypL indicates a hypothetical liquid.
Step 2. The hypothetical liquid naphthalene is mixed with liquid n-hexane. If the solution is nonideal, the Gibbs energy change for component 2 is
The Gibbs energy change for this step is always negative if mixing occurs spontaneously, and must be large enough to cancel the Gibbs energy change from step 1.
Solubility is determined by a balance between the positive ΔGfus and the negative Gibbs energy effect of mixing.
Then clearly, from Fig. 14.9 and Eqns. 14.14 through 14.16,
or
where T is 298 K for our example. Relations for the activity coefficients in the right-hand side of the equation have been developed in previous chapters. In the next subsection, the calculation of 
is explained.
Formation of a Hypothetical Liquid
The Gibbs energy change for step 1 is most easily calculated using the entropies and enthalpies. For an isothermal process, we can write:
dG = dH – T dS
Continuing with the example of dissolving naphthalene at 298 K, the Gibbs energy change for melting naphthalene at 298 K can be calculated from the enthalpy and entropy of fusion,
where T = 298 K. ΔH298fus can be calculated by determining the enthalpies of the liquid and solid phases relative to the normal melting temperature, where the heat of fusion, 
. The enthalpy of solid at 353.3 K relative to solid at 298 K (step 1(a) of Fig. 14.9) is
The enthalpy of the liquid at 298 K relative to liquid at 353.3 K (step 1(c) of Fig. 14.9) is:
Thus, the 
for melting is
which we can generalize to
A similar derivation for the entropy gives
which we can generalize to
In addition to these relationships, at the normal melting temperature, since 
,
where Tm = 353.3 K. Combining the results and neglecting the integrals (which are nonzero, but essentially cancel each other), we have
where for our example, T = 298.15 and Tm = 353.3. This is the that we desire for Eqn. 14.17.
Criteria for Equilibrium
In general, combining Eqn. 14.22 with Eqn. 14.17, we arrive at the equation for the solubility of component 2,
Solubility equation for crystalline solids.
where heat of fusion is at the normal melting temperature of 2, and heat capacity integrals are neglected.
Example 14.8. Variation of solid solubility with temperature
Estimate the solubility of naphthalene in n-hexane for the range T = [298, 350K] using the SSCED model. Plot log10(xN) versus 1000/T.
Solution
From Appendix E, Tm,2 = 353.3 K and ΔHfus = 18,800J/mol.
We can begin at 298 K, assuming an ideal solution. Then
xN = exp[(–18800/8.314)·(1/298 – 1/353.3)] = 0.305.
Starting with xN = 0.305 as an initial guess,
ΦN = 0.305·130.6/(0.695·130.3 + 0.305·130.6) = 0.306.
Noting that δ1′ = 14.93, δ2′ = 19.19 and k12 = 0.0052 => γ2 = 1.693.
xN = 0.305/1.693 = 0.1802. Iterating on xN to achieve consistency, xN = 0.135. Repeating this procedure at other temperatures gives Fig. 14.10.
Figure 14.10. Freezing curve for the system n-hexane(1) + naphthalene(2). Experimental data of H.L. Ward, 1926. J. Phys. Chem., 30:1316.
Hexane also dissolves in a hexane–naphthalene solution below its melting temperature. The general relationship for solving SLE can be written as:
where the heat of fusion is for the pure ith component at its normal melting temperature, Tm,i. Note that Eqns. 14.23 and 14.24 may be used to determine crystallization temperatures at specified compositions.
Example 14.9. Eutectic behavior of chloronitrobenzenes
Fig. 14.11 illustrates application to the system o-chloronitrobenzene (1) + p-chlorornitrobenzene (2). The compounds are chemically similar; thus, the liquid phase may be assumed to be ideal, and the activity coefficients may be set to 1. The two branches represent calculations performed from Eqns. 14.23 and 14.24, each giving one-half the diagram. The curves are hypothetical below the point of intersection. This temperature at the point of intersection of the two curves is called the eutectic temperature, and the composition is the eutectic composition.
Figure 14.11. Freezing curves for the system o-chloronitrobenzene(1) + p-chloronitrobenzne(2).
Example 14.10. Eutectic behavior of benzene + phenol
In most systems, an activity coefficient model must be included. Fig. 14.12 shows an example where the ideal solution model is not a good approximation, and the activity coefficients are modelled with the UNIFAC activity coefficient model. To solve for solubility given a temperature, the following procedure may be used (taking component 2, for example):
1. Assume the γ2 = 1.
2. Solve Eqn. 14.23 for x2.
3. At this value of x2, determine γ2 from the activity model.
4. Return to step 2, including the value of γ2 in Eqn. 14.23, iterating to converge.
Figure 14.12. Freezing curves for the system benzene(1) + phenol(2). Solid line, UNIFAC prediction; dashed line, ideal solution prediction; squares, Tsakalotos, D., Guye, P. 1910. J. Chim. Phys. 8:340; circles, Hatcher, W., Skirrow, F. J. Am. Chem. Soc., 1917. 39:1939. Based on figure of Gmehling, J., Anderson, T., Prausnitz, 1978. J. Ind. Eng. Chem Fundam. 17:269.
This is a relatively stable iteration, and Excel Solver can iterate on activity by adjusting composition, bypassing the successive substitution method above.
A common procedure when crystallizing products of a pharmaceutical process is to add an antisolvent. The SSCED model is especially convenient for this kind of application because: (1) The concentrations are generally small for crystalline products left in solution; and (2) the infinite dilution activity coefficient of the precipitate involves a simple average of the solvent and antisolvent properties. With these concepts in mind, it is straightforward to tailor a solvent to achieve a target composition at a given temperature.
Example 14.11. Precipitation by adding antisolvent
Ephedrine is a commonly used stimulant, appetite suppressant, and decongestant, related to pseudoephedrine. It can be extracted from the Chinese herb, ma huang. Ephedrine is to be crystallized from ethanol at 278 K by adding water as an antisolvent.
a. Estimate the mole fraction of water needed to reduce the concentration of ephedrine in solution to 0.1mol% using the SSCED model.
b. Yalkowsky and Valvani (1980) have suggested that ΔSfus = 56.5 J/mol-K for rigid molecules.a Evaluate this relation in comparison to the estimated value of ΔHfus = 25kJ/mol.
Additional data for ephedrine are Mw = 165.2; Tm = 313K; and ΔHfus = 25kJ/mol.
Solution
a. From Eqn. 14.25, Tm,i = 313 K and ΔHfus = 25,000J/mol. We can begin by solving for the target value of the activity coefficient, noting that the concentration of drug is practically infinitely dilute. Then using xi = 1E-3 to approximate infinite dilution,
The solution requires iteration using Eqn. 12.55. As the mole fraction of water is increased, the activity coefficient of ephedrine increases because water is the antisolvent.
Since we have worked the problem before, “Guessing” a value of x1 = 0.2102,
Φ1 = 0.2102(58.3)/(0.2102·58.3 + 0.7898·18.0) = 0.4627
<δ’> = 0.4627(18.68) + 0.5373(27.94) = 23.66
From Eqn. 12.51,
k12 = 0.0318; k13 = 0.0028; k23 = 0.0571;
=> <k3m> = 0.4627(18.68)0.0028 + 0.5373(27.94)0.0571 = 0.8808;
Similarly, <k1m> = 0.4779; <k2m> = 0.2752;
<<kmm>> = 0.4627(18.68)0.4779 + 0.5373(27.94)0.2752 = 8.262.
By Eqn. 12.55, RTlnγ3 = 172.3((16.36 – 23.66)2 + 2(16.36)0.8808 – 8.262) => γ3 = 245. So the solution should be 79 mol% water. Good “guess!”
b. With ΔSfus = 56.5 and Tm = 313K, ΔHfus = 56.5(313) = 17,700 J/mol, 29% lower than 25,000. The rule does not appear to apply to this compound.
a. Connors, K.A. 2002. Thermodynamics of Pharmaceutical Systems: An Introduction for Students of Pharmacy, Hoboken, NJ: Wiley, p. 129.
A special feature of Example 14.11 is the way it shows how to tailor a solvent to achieve a particular environment for a target solute. A similar approach could be applied to compatibilizing a liquid solvent to avoid LLE. For example, how much methanol should be added to isooctane to reduce the activity coefficient of water below a value of 7.4? This is the calculation behind “dry gas,” used to dissolve water from gas tanks.
SLE with Solid Mixtures
So far, we have only covered phase behavior in systems where the solids are completely immiscible in each other. Fig. 14.13 illustrates a case where the solids form solid solutions, and Fig. 14.14 illustrates behavior where compounds are formed in the solid complexes. Also, the case of wax precipitation from petroleum results in a range of n-C20 to n-C35 straight chain alkanes being mixed in the solid phase. Paraffin wax that you can purchase in the grocery store is primarily composed of n-C20 to n-C35 straight chain alkanes. For the case of liquid and solid mixtures in equilibrium, the derivation of the equilibrium relationship can be modified by adding a step for “unmixing” of solid solutions to the schematic of Fig. 14.9 on page 558. This step is analogous to a reversal of step 2 of the diagram except involving solid solutions and pure solids rather than liquid solutions and pure liquids. For each component in the mixtures:
Figure 14.13. Freezing curves for the Azoxybenzene(1) + azobenzene(2) system illustrating a system with solid-solid solubility. Based on Hildebrand, J.H., Scott, R.L., Solubility of Nonelectrolytes, New York, NY: Dover, 1964.
Figure 14.14. Solid-liquid and vapor-liquid behavior for the ammonia(1) + water(2) system at 1.013 bar. NH3 and H2O form two crystals in the stoichiometries: (α) NH3·H2O;(β) 2NH3·H2O. (Based on Landolt-Börnstein, 1960. II/2a:377.)
Thus, we can recognize an SLE K-ratio on the left-hand side, 
,
and we can recognize Eqn. 14.24 as a simplification where for a pure solid, 
.
Petroleum Wax Precipitation
An especially difficult problem in the recovery of natural gas is the clogging of pipes caused by small amounts of wax that accumulate over time. In the Gulf of Mexico, natural gas at the bottom of the well can be 250 bars and 100°C, but it must be reduced to 100 bars to be permitted in the pipeline, and the sea floor can drop to 5°C. The reduction in pressure and temperature results in a loss of carrying power and the small amounts of heavy liquid hydrocarbons can condense, eventually coating the walls with viscous liquid. After the liquid has formed, further cooling can cause solid wax to deposit on the walls of the pipe. These deposits cause constrictions and larger pressure drops that lead to more deposits, and so forth.
Naturally occurring petroleum co-produced with natural gas is a complex mixture of hundreds of individual components. Rather than attempt to specify the identity and composition of every component, it is conventional to collect several fractions of the original according to the ranges of their molecular weights. Hansen et al.2 provide the data in the first four columns of Table 14.1 for the composition, mass density, and molecular weight of several fractions from a typical petroleum stream. This kind of data is typically collected by distilling the initial sample and collecting fractions over time. As the lower molecular weight species are removed, the boiling temperature rises and the distillate collected over each particular temperature range is stored in a separate container. The weight of each fraction relative to the weight of the initial sample gives the composition of that species fraction. The mass of each fraction divided by its volume gives the density. And the average molecular weight can be characterized by gas chromatography or through correlations with viscosity. Note that the molecular weight for any particular species is not necessarily equal to the molecular weight for the corresponding saturated hydrocarbon. This is an indication that olefins, naphthenics, and aromatics are present in significant compositions. The objective of this example problem is to treat the data of Hansen et al. as characteristic of a gas condensate and compute the fractions of the stream that form solids at each temperature.
Table 14.1. Summary of Data for Wax Fractions and Calculations of the Precipitate Composition as Calculated by Example 14.12
The fusion (melting) temperatures and heats of fusion for n-paraffins can be calculated according to the correlations of Won.3
Noting that each species fraction can contain many species besides the n-paraffins that are responsible for practically all wax formation, it is necessary to estimate the portion of each species fraction that can form a wax. Since the densities of n-paraffins are well known, it is convenient to use the difference between the observed density and the density of an n-paraffin of the same molecular weight as a measure of the n-paraffin content. The correlations for estimating the percentage of wax-forming components in the feed (ziW) are taken from Pedersen.4
where 
is the species overall mole fraction in the initial sample.
is the portion of that fraction which is wax-forming (i.e., n-paraffin).
Example 14.12. Wax precipitation
Use the data from the first four columns of Table 14.1 and correlations for wax to estimate the solid wax phase amount and the composition of the solid as a function of temperature. Use your estimates to predict the temperature at which wax begins to precipitate. Hansen et al. give the experimental value as 304 K.
Solution
This problem is basically a multicomponent variation of the binary solid-liquid equilibrium problems discussed above. The main difference is that the solid phase is not pure. We can adapt the algorithm as follows.
Assuming ideal solution behavior for both the solid and liquid phases, we define 
, and as before, we assume the difference in heat capacities between liquid and solid is negligible relative to the heat of fusion,
which is independent of the compositions of the liquid and solid phases because of the ideal solution assumptions. The solid solution mole fraction is given by 
. Compare this method to the vapor-liquid calculations using the shortcut K-ratio in Chapter 9. This is a liquid-solid freezing temperature analog to the vapor-liquid dew-temperature procedure. The liquid mole fractions are given by the ziW values in the table below. All that remains is to guess values of T, which changes all Ki until 
. Hand calculations would be easy with a couple of components, but spreadsheets are recommended for a multicomponent mixture. Using Solver for spreadsheet Wax.xlsx distributed with the textbook software gives T = 320.7 K. Intermediate results are tabulated in Table 14.1. The T is slightly higher than the experimental value, but reasonably accurate considering the complex nature of the petroleum fractions and their variabilities from one geographic location to another.
Solid-Liquid Equilibria Summary
Phase equilibrium involving solids is an extension of previous modeling concepts. Like liquid-liquid equilibria, the condensed phase fugacities are quite insensitive to pressure, so the partition coefficients are simply functions of temperature. The main difference is that the heats of fusion are used to relate the component fugacities in their various states of matter. In multicomponent mixtures, the solid-liquid procedures for calculation are analogs of the vapor-liquid procedures, where the partition coefficients are calculated in a different manner.
14.11. Summary
The presentations of LLE and SLE have been brief, but have opened a broad new frontier of phase behavior analysis: multiphase equilibrium. You might find it incredible to see how many phases and peculiar behaviors can be observed with just “oil,” water, and special third components known as surfactants. (Soap is an example of a surfactant.) The short introduction here is a branching point that barely scratches the surface of such oleic and aqueous systems.5 Such molecules are more complicated than we can represent with the simple models here because of the way that they organize to make films and micelles. Far down this road you may begin to understand the forces that hold cell membranes and living organisms together. At the more practical engineering level, you should be able to perform preliminary designs of liquid extraction and crystallization equipment with little more thermodynamical background than has been presented here.
This kind of breadth is possible with such short introductions because the fundamentals have been laid out previously. The key equation in both LLE and SLE is familiar from VLE (Eqn. 11.13):
The liquid phases split because the Gibbs excess energy becomes so large that the stability limit is exceeded. In other words, the fugacities are so high that the components must escape each other, even if their volatilities are too low for VLE. An entropic penalty must be paid, but a highly unfavorable energy of interaction may more than compensate. A convenient guideline is,
Suspect LLE if 
, in which case 
is a good initial guess.6
With this guideline, LLE computation is a simple coincidence that may occur occasionally during VLE computations. To paraphrase Pasteur, its observation presents no problem if we know how to recognize it. The procedure for calculating Gibbs energy is the same as for VLE. The calculation of the phase distributions requires a flash calculation in terms of liquid-liquid K-ratios. But the flash algorithm has been discussed before and liquid-liquid flash calculations are hardly different from vapor-liquid flash calculations. The introduction to the prospect of LLE has reminded us of the need to check for stability, however. We first encountered the concept of stability in contemplating the critical points of pure fluids. With the prospect of multiple phases, we begin to realize the need to explore critical phase behavior in a systematic fashion. We have treated that problem for binary and ternary mixtures but have not generalized to multicomponent systems. The generalization of critical phase behavior requires a fair amount of calculus and matrix manipulation,7 but leads to a much deeper understanding of stability behavior than we have provided here. As a practical matter, activity coefficient models must be used carefully because the precision of the models is usually inferior to that of VLE predictions for the same systems. In particular, the temperature dependence of LLE is usually not predicted well. Frequently, the miscibility increases with temperature more quickly than predicted by the activity coefficients even with more sophisticated models like UNIQUAC or UNIFAC. Temperature-dependent parameters can improve the fit, but have little theoretical basis, making extrapolations more tenuous than for VLE.
Important Equations
For LLE, the key equation comes from setting the two expressions for fiL equal and canceling fio,
The iteration for LLE is a little tricky because it relies on the γ’s being large but they get smaller as the iteration proceeds. If you guess a composition too close to equimolar, you might miss the LLE.
For SLE, the key equation is (Eqn. 14.23),
If γi >> 1 then T → Tm. (It is nonsense if the computations yield a value of T > Tm.) Otherwise, the solubility may be significant at T < Tm. This observation suggests changing the solvent (by adding anti-solvent to increase γi), in addition to simply cooling. This is a common technique in pharmaceutical crystallization, among other applications. SLE computation requires iteration if γi≠ 1, because the liquid composition must be guessed in order to estimate γi. But iterative problems like this are familiar from VLE experience and present no problem.
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