“Liquid Solutions I- Liquid/Liquid” Prof. Gianluigi Veglia Partial Molar Quantities: when 1+1≠2. We are getting ready to talk about mixtures, especially liquid solutions: benzene and toluene, methanol and water, sugar and water etc. We begin by introducing the subject of partial molar quantities, beginning with partial molar volume. The partial molar volume of a component j in a solution containing an arbitrary number of components is : ',, nPT j j n V V ∂ ∂ = The unit of the partial molar volume is L or cm 3 per mole. Thus the partial molar volume is an intensive quantity that depend the temperature, pressure and composition. What is the physical meaning of the partial molar volume? If I had an infinite quantity of some solution and were to add j to the solution molecule by molecule (infinite volume ensures constant composition) and were to measure the volume change with added j that would be the partial molar volume. What good is the partial molar volume? Let’s begin writing the volume of a solution as a function of T, P and the number of molecules of all components: Then ( ) j nnnPTVV ,...,,, 21 = j nPT jnTnP dn n V dP P V dT T V dV i jj ∑ ∂ ∂ + ∂ ∂ + ∂ ∂ = ,, ,, j j nTnP dnVdP P V dT T V dV jj ∑ + ∂ ∂ + ∂ ∂ = ,, Let’s suppose that we prepare the solution under conditions of constant T and P. Then we can write: Instead of expression V in terms of n 1 , n 2 , …, we will often find it more convenient to express V in terms of χ 1 , χ 2 ,… (the mole fraction of each component) and n tot , the total number of moles of stuff present: j j dnVdV ∑ = tot j i j j n n n n == ∑ χ jjtot nn =χ at constant T and P. Now let’s imagine a change at constant T and P and composition. That is I change the total number of moles of stuff, but not the composition (e.g., I might add 5 ml of a 0.5 M sugar solution to a 500 ml of 0.5 M sugar solution). Now the expression for dV becomes: At constant T, P and composition. Let’s look at V in another way. It only stands to reason that V is a function T, P, χ 1 , χ 2 ,…and also n tot . ( ) jtot j ndVdV χ ∑ = totj j dnVdV χ ∑ = There has to be some intensive quantity, f(T,P, χ i ), which is the volume of a mole of solution that V, the total solution volume is: What is this function? At constant T, P and composition: At constant T, P and composition. We now have two different expressions for dV, which we equate: ),,( itot PTfnV χ⋅= toti dnPTfdV ⋅= ),,( χ totj j toti dnVdnPTf χχ ∑ =⋅),,( Therefore: We can also write: So the volume of a solution is simple equal to the sum of the partial molar volume of each component multiplied by the number of moles of the component. j j i VPTf χχ ∑ =),,( j j totitot VnPTfn χχ ∑ =),,( j j totitot VnPTfnV χχ ∑ == ),,( j j nVV ∑ = How might one measure partial molar volumes? Suppose one were to dissolve various quantities of a solute (for instance MgSO 4 ) in a fixed quantity of solvent (say one mole of water) under conditions of constant T and P. One could measure the volume of each solution. Thus, we can infer that the molar volume of magnesium sulfate from the slope of the plot V versus number of moles. Note that the molar volume is negative at low concentrations and becomes positive at high concentrations. Partial molar volumes are reported as a function of the mole fraction. In a binary (two components) solution: Water Ethanol: Here is an example of partial molar volume data for a solution of water ethanol, reported as a function of the mole fraction of ethanol. Note here that whenever Increases decreases BA A A nn n + =χ BA B B nn n + =χ OH V 2 OHHCV 52 OHHC V 52 Why this relationship? Is it an accident? No! Imagine to have a solution of A and B. We write the volume as a function of T, P, n A , n B : A. At constant pressure and temperature. We have another expression for V: at constant pressure and temperature. B B A A dnVdnVdV += B B A A VnVnV += Differentiating: B. Since the two expression A and B for dV must be equal: Or in terms of mole fractions: B BB BA AA A dnVVdndnVVdndV +++= ( ) ( ) A A B BB B A A dnVdnVVdnVdndV +++= ( ) 0=+ B B A A VdnVdn A B A B Vd n n Vd −= A B A B VdVd −= χ χ This is great because it means that we need to measure only the partial molar volume of A as a function of the composition for a to obtain the change in the partial molar volume of B. We can generalize the above expression to a solution containing an arbitrary number of components: This expression valid at constant pressure and temperature is called Gibbs-Duhem equation, and it is one of the most important equations of thermodynamics because it expresses the interrelationship between the partial molar volumes. 0= ∑ i i Vdn Other state functions (S, U, H, etc.) can be expressed as partial molar quantities, i.e.: We can think of the chemical potential as another name for we might otherwise call the partial molar free energy. Other state functions (S, U, H, etc.) can be expressed as partial molar quantities, i.e.: ',, nPT j j n S S ∂ ∂ = ',, nPT j j j n G G ∂ ∂ ==µ All of the expressions derived above for the partial molar volume have analogies in the other partial molar terms, e.g.: Or Gibbs-Duhem equation. jjjjtot nGnG µ ∑∑ == 0== ∑∑ jjj j dndnG µ In addition, the expressions that relate changes in the state function also relate changes in analogous partial molar quantities, e.g.: We can obtain the following expressions: From those expressions we can obtain that VdPSdTdG +−= S T G P −= ∂ ∂ V P G T = ∂ ∂ ; ',nP j j T S ∂ ∂ = µ ',nT j j P V ∂ ∂ = µ ; Let’s consider a binary, liquid solution of A and B in equilibrium with vapor phase: At equilibrium we know that the chemical potentials must be equal: We know )()( gl AA µµ = )()( gl BB µµ = 0 0 ln)()()( P P RTggl A A AA +== µµµ 0 0 ln)()()( P P RTggl B B BB +== µµµ Now let’s consider a pure liquid for which we designate the chemical potential (A=B): Subtracting the equations: )( * lAµ )( * lBµ and 0 * 0* ln)()( P P RTgl A AA +=µµ 0 0 ln)()( P P RTgl A A A +=µµ 0 * 0* ln)()( P P RTgl A AA +=µµ A A A A P P RTll * * ln)()( =−µµ A A A A P P RTll * * ln)()( +=µµ We still are not quite where we want to be in terms of expressions for µ A (l) and µ B (l). In fact, we want the chemical potentials in terms of the composition of the solution, not of the gas phase! To get expressions for µ A (l) and µ B (l), we need to develop some simple models of solutions. We then will consider how to account for the deviations from those models. We will develop two models: the ideal solution and the ideal-dilute solution. We begin with the ideal solution model. Consider a solution of A and B in which A and B as so similar that A can be interchanged with B and with A without the solution energy changing: Examples of such solutions might be benzene/toluene, CH 3 OH/CH 3 OD, hexane/heptanes. The intermolecular interactions between A-A and B-B, and A-B are so similar that mixing does not change the internal energy. Experience has shown that the vapor pressure of A (P A ) in such solution is just equal to χ A P * A , where χ A is the mole fraction of A in solution and P * A is the vapor pressure of pure A. This makes sense because the forces holding A in solution are identical to those holding A in pure A. But the amount of A available for evaporation is reduced. So: This is called Raoult’s law for ideal solutions: * AAA PP χ= * * * ln)()( A AA A A P P RTll χ µµ += A A A RTll χµµ ln)()( * += Calculate ∆G and ∆S for mixing 1 mole of benzene and 2 moles of toluene at 300K. initialfinalmix GGG −=∆ ∑∑ −=∆ )()( finalj j finalj jmix GnGnG [ ] [ ])()()()( initialninitialnfinalnfinalnG toltolbenbentoltolbenbenmix µµµµ −−−=∆ [ ] [ ])ln()ln( **** tolbenben toltoltolbenbenmix RTnRTnG µχµµχµ −+−−+=∆ toltolbenbenmix RTnRTnG χχ lnln +=∆ [ ] toltolbenbentotalmix RTnG χχχχ lnln +=∆ JG mix 4760−=∆ Let’s look at the entropy: This example shows that A. benzene and toluene are completely miscible B. The entropy of mixing is relatively small (A~B) C. The variation of enthalpy is zero! D. If you compare the expressions for mixing an ideal solutions, they are similar to those of mixing ideal gases. ∆ −=∆ dT G dS mix mix [ ] toltolbenbentotalmix RnS χχχχ lnln +−=∆ 1 9.15 − =∆ JKS mix 0≅∆−∆=∆ STGH mixmix Unfortunately, many solutions are made up with dissimilar components (e.g., sugar/water, CHCl 3 and acetone etc. We cannot longer assume that A and B are interchangeable. How do we handle this situation? We first define two things: the solvent (A, the majority component) and the solute (I, everything else). At very low solute concentrations, the number of solute-solvent interactions is low and Raoult’s law is applicable for the solvent! The solutes are a different story. However, solute molecules interact almost exclusively with the solvent molecules. Their vapor pressures are calculated using Henry’s law: Where k i is determined experimentally. So we have the chemical potentials in a solution for which the solvent obeys Raoult’s law and the solute obeys to Henry’s law. * AAA PP χ= iii kP χ= iii kP χ= This solution is called ideal-dilute solution and it applies to solutions of dissimilar components in which the solute concentrations is so low that solvent interact only with solvent: For solute: A A A RTll χµµ ln)()( * += * * ln i ii i i P k RT χ µµ += ii i i i i i RT RT P k RT χµ χµµ ln lnln # * * += ++= Ideal solutions: For the solvent: Ideal Dilute-solutions: A A A RTll χµµ ln)()( * += A A A RTll χµµ ln)()( * += iii RT χµµ ln # += We will show that the components of an ideal solution obey to Raoult’s Law at all concentrations. Consider a liquid mixture of two components: A and B in equilibrium with the gas phase: 2 components 2 phases F=C-P+2=2 In order to fully define our system we need to specify: T or P and χ A Let’s consider how the vapor pressure of the system varies with the mole fraction of A in the liquid phase (χ A (l) ) at constant temperature (Note that α=gas and β=liquid). If the solution is ideal, the Raoult’s law holds and: Since *)(*)( B l BA l Atot BAtot PPP PPP χχ += += ( ) ( ) ( ) ** 1 1 B l AA l Atot BA PPP χχ χχ −+= =+ ( ) ( ) *** B l ABAtot PPPP +−= χ Therefore the plot for the total pressure for the liquid will have a straight line. For instance, in the case of benzene toluene: Now, what is the composition of the gas phase in a solution having that vapor pressure? Since After substituting, simplifying and solving for P tot : ( ) ( ) ( ) ( ) *** B l ABA g A tot A g A PPPP P +− == χ χ χ ( ) ( ) *** B l ABAtot PPPP +−= χ ( ) ( ) ( ) ** * BA Btot l A PP PP − − =χ ( ) ( )g AABA BA tot PPP PP P χ *** ** −+ = Let’s plot this in the Pressure/composition diagram and we have our first two-component phase diagram: How to interpret our phase diagram? It can be interpreted much more broadly than the derivation of two pressure curves implies. The phase diagram shows the stable configuration of a mixture of A and B and overall composition of χ A . In the one-phase regions (only gas or liquid present) the phase rule tells us that F=C-P+2=3 (2 components, 1 phase). In the two-phase regions F=2. In both cases, one degree of freedom is taken up by temperature. What are the compositions of the two phases if one is in the l+g region? One finds them by constructing tie lines. Suppose one is the two phase region, with a system of overall χ A . What are the relative amounts of the two phases formed? Let’s say that some fraction f (a mole fraction) of the material is in the liquid phase. Then (1-f) is in the vapor phase. Since the overall composition is χ A : ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ] ( ) ( ) [ ] A g A l AA AA g A l A A g A l A ff ffff ff χχχχ χχχχ χχχ −−=−⋅ ⋅−+⋅=⋅−+⋅ =⋅−+⋅ 1 11 1 Reworking: This is called the lever rule: the closer the overall composition is to one of the stable phases, the more of the system partitions into the stable phase. Alternatively, you can find the following nomenclature: ( ) [ ] ( ) [ ] A g A gl AA l ff χχχχ −=− )()( ββαα lnln = Most of you are familiar with the fractional distillation column and know that the repeated condensation and re- evaporation occur along the entire column and the vapor becomes richer and richer of the more volatile component as it moves up the column. The phase diagram derived above applies to an ideal solution. Moreover, it shows how a vapor pressure varies with composition at constant temperature. Often, however, we are interested in how things change as a function of temperature at constant pressure. For instance, how does the boiling point change with composition? Everything we said P vs χ A diagrams applies here as well: the lever rule, the determination of the phase composition etc. We simply are plotting a different region in the three dimensional space: (F=C-P+2=4-P) The phase diagram tells us how a fractional distillation process is expected to proceed. The gas phase is enriched in the lower boiling substance. Therefore, if one condenses said gas-phase, one has taken the first step toward purification. The efficiency of a fractionation column is expressed in terms of the number of theoretical plates: the number of effective vaporization and condensation steps that are required to achieve a condensate of given composition from a given distillate. To achieve a good purification in figure A, we must have a column with three theoretical plates. To achieve the same separation for a system that behaves as in B we must have a column with 5 theoretical plates! Although some solutions exhibit behavior very much (similar) like that of an ideal solution, and some exhibit behavior very much like that of an ideal-dilute solution , most solutions deviate from these idealized behaviors. The situation is even more extreme that it is for a gas, which usually approximates the ideal gas except at very high pressures. Solutions (real ones) can deviate markedly from idealized ones. To deal with real solutions, we introduce the concept of activity. Activities correct for deviations from ideal or ideal-dilute solutions. Let’s consider the correction to the ideal solution model: In a real solution: Where a is the activity and γ A is called activity coefficient. AA A A A A A RT aRT χγµµ µµ ln ln * * += += Since all the solutions behave ideally in the limit of infinitively low concentration (i.e. in the limit of pure A), we have If we start with the ideal-dilute model (which is more suitable for solutions, e.g. sugar/water), we have: Note that this expression is valid for the solvent. Since all the solutions behave ideally in the limit of infinitively low concentration (i.e. in the limit of pure A), we have 1lim 1 = → A A γ χ AA A A A A A RT aRT χγµµ µµ ln ln * * += += 1lim 1 = → A A γ χ For the solute with For solutions of low concentrations, it is often more convenient to express solute concentration as molality (mol kg -1 ) at low concentration (n i <