lectures20-21 - Phase equilibrium involving mixtures...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Phase equilibrium involving mixtures Suppose we mix two liquids. What is the composition of the vapor that is in equilibrium with the liquid mixture? What is the boiling point of the mixture? To answer this kind of questions, we need to write down the equilibrium conditions for mixtures. At constant P and T these conditions arise from the requirement for G to be minimal. Consider equilibrium between two phases, e.g., liquid and gas. Consider one particular component of the mixture and suppose that dn moles of this component go from liquid top gas. The resulting change in G is dG = dnµi ( g ) − dnµi (l ) where µi (l ( g )) is the chemical potential of this component in the liquid (or gas) mixture. We see that if the two chemical potentials are not equal the system can always reduce its Gibbs energy by transferring some molecules from one phase to the other. So in equilibrium we will have µi ( g ) = µi (l ) In general, if we have a system that involves several different phases of several compounds mixed with one another and coexisting with each other at some (P,T) then the equilibrium condition is that the chemical potential of each component is the same everywhere in the system. We have already seen an example of this rule: phase equilibrium of a pure chemical. Our new rule is more general but it can be derived in exactly the same way from the requirement for the free energy G to be minimum. Below, we will see a few other examples of this rule. 1. Depression of the freezing point. Salt is often used to melt ice on roads in the winter. To understand the physical basis of this method, consider phase equilibrium between the solution of salt in water and ice. First note that water freezes out of solution. Therefore the chemical potential of pure water in the form of ice should be equal to that of liquid water mixed with salt. 0 0 µ water ( P, T ) + RT ln(1 − xNaCl ) = µice ( P, T ) where we have used the fact that xwater+xNaCl=1. Let T0 be the normal melting point of ice. Then 0 0 µ water ( P, T0 ) = µice ( P, T0 ) Let T = T0+ΔT and assume that ΔT is small and that xNaCl<<1. Then we can write 0 0 µ water ( P, T ) ≈ µ water ( P, T0 ) − S water ΔT and a similar equation for the chemical potential of ice. The equilibrium condition becomes 0 0 µ water ( P, T0 ) − S water ΔT − RTxNaCl = µice ( P, T0 ) − Sice ΔT or ΔT = − RTxNaCl RT 2 xNaCl =− S water − Sice ΔH melting This is a negative number meaning that the freezing point is decreased. Boiling point elevation If salt is added to water, this increases its boiling point. Assume there is no salt in the vapor above the boiling solution of NaCl in water. The chemical potential of water in the solution and in vapor should be the same: 0 0 µwater ( P, T ) + RT ln(1 − xNaCl ) = µvapor ( P, T ) Repeating what we did when considering the freezing point of the NaCl solution in water, we find ΔT = − RTxNaCl RT 2 xNaCl = > 0 S water − Svapor ΔH vaporization We see that, indeed, the change in the temperature is positive Osmotic pressure Consider a mixture of A and B separated from pure A by a semipermeable membrane that is transparent to A but not permeable for B. A will want to equalize its chemical potential to both sides of the membrane. The chemical potential is: 0 µ A = µ A + RT ln xA where xA=1 for the pure A and xA<1 for the mixture. It is clear that A will want to flow into the mixture thereby increasing its concentration there. This flow can be compensated for by an increase in the pressure of the mixture, ΔP. The equilibrium condition is: 0 0 µ A ( P + ΔP, T ) + RT ln xA = µ A ( P, T ) Assuming that the concentration of B is small and that ΔP is small, we have ln(xA) = ln(1 ­xB) ~  ­xB and 0 0 µ A ( P + ΔP, T ) − µ A ( P, T ) = VAΔP = RTxB or ΔP = RTxB/vA = RT nB/V where V≈nA VA is the total volume of the mixture. Vapor ­liquid equilibrium for binary liquid mixtures Consider an ideal solution that consists of A and B, with mole fractions xA and xB =1 ­xA. What is the composition of the vapor that is in equilibrium with this solution? That is, what are the partial pressures PA and PB of the two gases in the vapor phase? Let the total pressure be P. The chemical potential of A in the liquid phase is l l µ A ( P, T ) + RT ln xA where µ A ( P, T ) is the chemical potential of the pure liquid. The chemical potential of the A vapor, assuming that it is an ideal gas, can be written as: g µ A ( P, T ) + RT ln( PA / P) so that l g µ A ( P, T ) + RT ln xA = µ A ( P, T ) + RT ln( PA / P) (1) Now consider the equilibrium of pure A liquid with its vapor at temperature T. Let PA* be the equilibrium pressure. Then we have: l g g µ A ( PA* , T ) = µ A ( PA* , T ) = µ A ( P, T ) + RT ln( PA* / P) l Neglecting the change in the molar volume of liquid, we have µ A ( P* , T ) = l µ A ( P, T ) + VAl (PA* − P) , so that l l g µ A ( P, T ) + VA (PA* − P) = µ A ( P, T ) + RT ln( PA* / P) (2) Subtracting (2) from (1), we get RT ln x A − VAl ( PA − P ) = RT ln( PA / PA ) * * (3) In most textbooks that derive this expression the second term on the lhs is missing. This term can actually be neglected in many cases. Here is why: for an ideal gas (vapor) P V = RT. Since the molar volume of a liquid is much smaller than that of a gas then the second term in Eq. 3 is much less than RT and therefore it would typically be much smaller than the other two terms. Neglecting it, we find: PA(T)= PA*(T) xA This is Raoult’s law. In particular, if xA is close to 1 then PA is close to P*. The dependence of the vapor pressure PA*(T) on T is given by the liquid ­vapor coexistence curve for pure A and can be found by using the Clausius ­Clapeyron equation. However the assumption that the escape probability of A is not affected by the presence of B is a very strong one. The escape probability is determined by how strongly A is attracted to its neighbors in the liquid. If AB, BB, and AB interactions are the same then, indeed, the escape probability is independent of the amounts of A and B. As we have discussed before, this is equivalent to assuming that A and B form an ideal solution. For real solutions, things are much more complicated but the limit of a very dilute solution, xB = 1, can be understood. In this limit, an A molecule on the surface will mostly only see A molecules and so the probability of it escaping is the same as for pure A, regardless of the magnitude of the A ­B interactions. In this limit, Raoult’s law is valid again for the solvent A: * PA = PA xA For the solute B, the probability of a molecule to escape will be mostly determined by its interactions with A, not B, and so it will be different from the case of pure B. Since the number of molecules B that are ready to escape is still proportional to xB, we still can write PB = kH xB but the constant kH is no longer equal to the vapor pressure of pure B and it generally depends on the properties of A and B. The above formula is knows as Henry’s law and kH is Henry’s constant. Finally, we mention that, when the solution is non ­ideal, it is common to write µi ( P, T , { xi }) = µi0 ( P, T ) + RT ln ai , where ai is called activity. Then all of the equations we have derived for ideal solutions will look the same provided that xi’s are replaced by ai’s. The catch, of course, is that, while xi’s are generally known, ai’s are not. ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online