Computing Liquid-Vapor Phase Diagrams for non-ideal binary mixtures by Franklin Chen Department of Natural and pplied Science University of Wisconsin-Green Bay Green Bay, WI 543 chenf@uwgb.edu Goal The goal of this document is to introduce the mathematical models that are used to describe vapor-liquid equilibrium of binary mixtures. This document focuses on the solubility parameter theory approach to calculate the activity coefficients of non-ideal mixtures. Prerequisites. Experience with concepts and practices with nd Law of thermodynamics, chemical potentials, and equilibrium constants.. Moderate skill with Mathcad. Performance Objectives t the end of this exercise you will be able to:. relate phase rule to liquid-phase and vapor-phase compositions of binary mixtures at equilibrium;. explain why azeotrope arises only in the non-ideal mixtures; 3. calculate activity coefficients using solubility parameter theory and the variation of the solubility parameter approaches; 4. construct a phase diagram of a given non-ideal binary mixture. chenvle.mcd Page
Chemical potential, activity, and activity coefficient Chemical potential is a partial molar quantity. The aoult's Law allows us to write µ 0 µ T ln X Eq. [] when the mixture is an ideal mixture. chemical equilibrium in the ideal case: a bb cc dd Eq. [] The equilibrium constant and the standard free energy change is related as: G 0 K T ln K X X c C a X X d D b B Eq. [3] µ 0 µ T ln a Eq. [4] Exercise How do you write the equilibrium constant for the equilibrium reaction shown in Eq.[] when chemical potential is written as Eq.[4]? What is the definition of this new standard state? We need to invent another thermodynamic entity, called the activity coefficient γ, to relate the non-ideal case to the ideal case: a γ X Eq. [5] Exercise : Can activity coefficient be greater than? chenvle.mcd Page
Exercise 3: The following example allows us to calculate the activity coefficients of chlorobenzene in equilibrium with a chlorobenzene/-nitropropane solution at 75 o C are listed as follows: x chlorobenzene : ( 0.9 0.89 0.460 0.69.00 ) p chlorobenzene : ( 9.0 4.9 6.4 86.4 9 ) units in torr i : 0,.. 4 a chlorobenzenei : a chlorobenzene T T p chlorobenzene i 9 ( 0.6 0.35 0.54 0.76 ) There are two kinds of subscripts. s for a subscript to designate a name, we use period "." after the name. If we want an entity from a vector, we use "[". In Mathcad, the vector entity start with 0. This is why we define i0, (;) 4. The semi-colon will appear as ".." in the document. You should be able to write a vector for the activity coefficients for the chlorobenzene at various composition in this mixture. t what mole fraction of chlorobenzene in the mixture would its activity coefficient be equal to? re the activity coefficients at other mixture conditions greater than or less than zero? Plot real and ideal vapor pressures of chlorobenzene against mole fractions of chlorobenzenes and explain the deviations of real vapor pressure from ideal vapor pressure. p idealchloroi : x chlorobenzene T 9 i 0 9.5 Vapor pressure torr 65 37.5 0 0 0.5 0.5 0.75 Ideal eal Mole fraction of chlorobenzene chenvle.mcd Page 3
Phase rule, Vapor and liquid phase compositions at equilibrium You may recall that Gibbs phase rule states that degree of freedom, f, is realted to the number of the component, c; number of phases, p, by the following relation: fc-p Eq. [6] Exercise 4 Consider the vapor-phase equilibrium of a binary mixture, how would you assign c and p? What is the degree of freedom when both phases are present at constant temperature and pressure? s for a binary mixture, c. p when both liquid and vapor are present; p for either vapor phase or liquid phase. t constant p and T, and when both vapor and liquid phases are present, We are now ready to construct vapor-liquid diagrams in which the y-axis is the boiling point of the mixture while the x-axis is either the mole fraction of the liquid phase, or the mole fraction of the vapor phase. Exercise 5 What is the definition of a boiling point? What is the total vapor pressure when the mixture is at the boiling point? Boiling point of a liquid is the temperature at which the total vapor pressure of the liquid is equal to the external pressure. Since boiling points vary with the compositions, and the total vapor pressures of the mixture are also function of temperature, we need to construct vapor pressure versus temperature data. Some of these data at selected temperatures are available in the CC Chemistry Handbook. We still need to construct regression equations to interpolate. Chemical engineers are familiar with the ntoine equation which has the form of log P-b/(tC), where, B, and C are constants depending upon the component, and t is the temperature in Celcius. Here, we will use Clausius-Clapeyron equation, which you have learned from your General Chemistry. ln p p vap H T T Eq. [7] where p and T are the reference point vapor pressure and temperature, vap H, is the gas constant. One good reference point is the normal boiling points whose values are available in most CC Handbook references. Exercise 6: What are the assumptions we made for the Clausius-Clapeyron equation? chenvle.mcd Page 4
There are assumptions: )Molar volume of a liquid is insignificant when compared to that of a gas.) Heat of vaporization is a constant from T to T. The following reference gives the triple point and normal boiling point data for toluene (http://www.nist.gov/srd/pdf%0files/jpcd37.pdf) Goodwin.D., J. Phy. Chem. ef. Data, 8, 565-636. Toluene, triple point: 78.5K, 4.36*0-7 bar. boiling point: 383.764K,.035 bar. Heat of vaporization at 388 K is 33. kj/mol. s for benzene, the reference is http://www.nist.gov/srd/pdf%0files/jpcd350.pdf, Goodwin,.D., J Phy. Chem. ef. Data, 7, 54 (988) Benzene, triple point: 78.65K, 0.04785 bar. boiling point: 353.4K,.035 bar. Heat of vaporization at 300K is 33.737 kj/mol We are going to set up vapor pressure data for both toluene and benzene from 300 o K to 400 o K at 0 o K interval : 8.34 joule K mole bar : 0 5 Pa kj : 0 3 joule Exercise 7: We choose our reference points for toluene and benzene as their normal boiling points. Explain why these fixed points are chosen as reference points? T tolu : 383.76 K p tolu :.03 bar T benz : 353.4 K p benz :.03 bar H vaptolu : 3.34 kj mole H vapbenz : 33.737 kj mole j : 0,.. 00 T j : ( 350 j 0.7) K Ideal Case: p toluj p tolu exp H vaptolu : p benzj p benz exp H vapbenz : T tolu T j T benz T j Exercise 8: Let us assume benzene and toluene form an ideal solution. Is this a good assumption? Why and why not? chenvle.mcd Page 5
s for the ideal case, the strategy of calculation is set Tj as the boiling temperature. We will then use both the aoult's law and Dalton's law to find both Xj and Yj such that p benzj p toulj p ext.03 bar p ext :.03 bar p ext p benzj p toluj X j X j : Y j : p toluj p benzj p ext Block # Liquid Vapor Equilibrium of Benzene Toluene Mixture 390 377.5 b.p. (K) 365 35.5 340 0 0.5 0.5 0.75 Liquid Vapor Mole Fraction of Toluene Exercise 9: This diagram correctly predict the b.p. of pure benzene which is at 353oK,and b.p. of toluene which is at 383.76K. Practice yourself using the triple point of the toluene as the reference point. What do you see? You can also practice to construct VLE diagram using the ntoine Equation with the ntoine constants shown as follows: Benzene: 6.90563, B.03, C0.79; Toluene: 6.95466, B344.80, C9.48 ntoine Equation: Log0 (p)-b/(tc) where p is in torr, t in Centigrade. Discuss the strength and the weakness of ntoine Equation. chenvle.mcd Page 6
Now, in the ideal case plot -Y i vs (-X j ) (benzene vapor phase composition against benzene liquid phase composition), what do you see? Would this plot convince you that there will be no azeotrope when (-Y i) >(-X i ) whether -X-->0 or -X-->? Y j 0.5 0 0 0.5 X j Non-ideal Case comes from enthalpy and entropy changes induced by interactions. The effects of interactions is treated in terms of a contact free energy change, G contact m 0 ideal ( µ ) T ln X µ Eq. [8] 0 E ( µ µ ) T lnγ Eq [9] Different models propose different expressions for the excess chemical potential.hildebrand estimated the enthalpy of mixing, H contact m H contact m V m φ ( δ ) φ δ Eq.[0] [( H T )/V ] / vap δ Eq [] where V m is the molar volume of the mixture, δ and δ are the solubility parameters of components, and respectively, V is the molar volume.φ and φ are the component volume fractions of components and. Exercise 0: Why do we substract T from vap H in Eq. []? The definition of the solubility parameter is actually: chenvle.mcd Page 7
δ ( U /V ) / U and H are related as U H-T Exercise : Calculate the solubility parameters for toluene and benzene using the following parameters. Toluene: vap H38 kj/mol; V06.75 ml/mole Benzene: vap H33.737 kj/mol; V89.786 ml/mole δ toluene : δ benzene : ( 38 kj mole 300 K) 06.75 ml mole 0.5 33.737 kj mole 0.5 300 K 89.786 ml mole δ toluene 8.78 0 3 joule 0.5 m.5 δ benzene 8.654 0 3 joule 0.5 m.5 Two solubility parameter numbers are really close. The enthalpy of interaction between toluene and benzene is approximately zero. The mixture is close to ideal mixture. Van Laar Theory ewrite Eq.[0], we have: H contact m V m φ φ ) where X and X are the mole fractions of components and, molar volumes for components and. ( δ δ ) ( X V X V φ φ ( δ δ ) _ V _ and V Eq [] are partial Taking partial differentiation of Eq.[] against n, the number of moles of component,we have T lnγ V φ ( δ δ ) Eq [3] Equation [3] is derived from: 0 E ( µ µ ) T ln γ nonideal ideal ( G G) m n m n contact m H chenvle.mcd Page 8
Similarly, we have ( δ δ ) T lnγ V φ Eq [4] If we define Van Laar Coefficients as: V T V T ( δ δ ) ( δ δ ) Eq.[5] Then, we can show that g H contact m Eq [6] T ( X X ) X X g lnγ X lnγ g X X X X X Eq [7] This is because: contact m n H T lnγ lthough Van Laar theory is not different from the solubility parameter theory in principle, it does provide an algorithm to calculate the activity coefficients for a non-ideal mixture if azeotrope data of the mixtures are available. The tasks are to determine and, two Van Laar coefficients. Once these two coefficients are found, Eq.[7] allows us to calculate γ and γ at any compositions. Looking at Eq.[5], and can be calculated from solubility parameters and partial molar volumes of the components. Solubility parameters can be calculated from heat of vaporization, and molar volume. Partial molar volume can be calculated from density-composition data. In principle, they can be calculated, although the calculations may be quite tedious. If we know the azeotrope point at one known external pressure, then Eq.[6] and Eq[7] allow us to calculate γ and γ at any compositions. These data, in turn, allow us to calculate azetrope points at any other external pressures. We shall use water-propanol binary system as the example. Before we discuss the calculations of γ and γ at the azeotrope conditions, we shall simplify the expressions in the brackets of Eq[7] chenvle.mcd Page 9
Exercise : Show that x lnγ x lnγ Eq [8] Before we proceed to prove Eq[8], we shall see the utility of Eq[8]. This relation suggests that and can be obtained from Eq[7] using the relation of Eq[8]. ln( γ ) ln( γ ) ln ln( γ ) ln ln ln( γ ) ln ( γ ) ( γ ) ( γ ) ( γ ) x x Eq [9] Eq[9] indicates that and can be calculated for a given x, x, γ, and γ. Once and are determined, Eq[7] can the be used to determine γ and γ at any compositions We now shall proceed to prove Eq [8] In Eq[7], multiplying Eq[7 b] with x and Eq[7 a] with x and take the ratio for these expressions. We have: x lnγ x lnγ x Eq [0] Exercise 3 In a binary mixture, x /x are liquid compositions and y /y are vapor compositions of component and. t azeotropic composition, which one of the following is correct? (a) x x ; y y ; (b) x y, x y s for the n-propanol/water binary system, x propanol y propanol 0.4 for p ext.03 bar; Water, T b 373.5K, vap H44.06 kj/mol at 98 K Propanol, T b 370.3 K vap H47.5 kj/mol Molar volume for water is 8mL/mole, n-propanol 74.84 ml/mole eferences: Majer,V., and Svoboda,V., [985] Lee and Shen[003] Eq [] Exercise 4: Given the data in Eq[], calculate solubility parameters for water and propanol. chenvle.mcd Page 0
( ) 44.06 kj mole 0.5 98 K δ water : mole δ water 48.038 joule 0.5 m.5 8 ml 0 3 47.5 kj mole 0.5 370.3 K δ propanol : mole δ propanol 4.363 joule 0.5 m.5 74.84 ml 0 3 Eq [] The difference is big eneough to treat the case with non-linear approach. That means we need to calculate activity coefficients for water and propanol at every compositions. Propanol Mole Fraction at azetrope point T az : 360.95 X az : 0.4 p ext : bar Enter your azeotrope composition data (acetone mole fractions from water-n-propanol mixtures are used here as an example) chenvle.mcd Page
T propanol : T water : 370.3 K 373.5 K H vappropanol : H vapwater : 47.5 kj mole 44.06 kj mole p propanol : p water :.03 bar.03 bar T az : 360.95 K p wateraz p water exp H vapwater : p propanolaz p propanol exp H vappropanol : p propanolaz 0.679 bar T water T az T propanol T az p wateraz 0.67 bar Eq [3] p wateraz p propanolaz.306 bar Exercise 4: Why the sum of the partial pressures of propanol and water are greater the total external pressure which is.03 bar? These are the partial pressures that the pure substances would have at the azeotrope if they acted as ideal gases. The activity coefficients at the azeotrope are thus p 0az : p propanolaz p 0az : p wateraz ecall that a(p ext y az )/p*(p ext x az )/p*; γa/x az; γp ext /p*p ext /p az Eq [4] p ext γ az : p 0az γ az.47 p ext γ az : p 0az γ az.595 Eq [5] Using the van Laar equation (Eq[7]), one can calculate the van Laar constants for this system. X az : X az Eq [6] chenvle.mcd Page
: log γ az ( ) ( ) ( ) X az log γ az X az log γ az ( ) : log γ az ( ) ( ) X az log γ az X az log γ az.94 0.59 Eq [7] Exercise 5: Can and be negative numbers? Why and why not? By definition, ~~( vap H - vap H )/T ( 47.5 kj mole 44.0 kj mole ).63 360 K Eq [8] The and values obtained are therefore reasonable. The calculations of γ and γ depend on the values of and. You may want to write different and values after the water-propanol azeotrope studies are complete to get a feel about the dependence of these two van Laar constants on the azeotrope composition. The van Laar constants are now used to determine the activity coefficients at each mole fraction. The process so far is that the azeotrope composition allowed the determination of the van Laar constants and now that we have the van Laar constants we can determine the activity coefficients for all other compositions. i : 0.. 30 j : 0.. 9 XN i : i 30 log( e) XN j XN j γ j 0 : log( e) XN j ( XN j ) γ 0 : j γ : 30 γ : 0 Block # The algorithm of Block # is straightforward, except you need to pay attention the indices i and j such that you don't want XNj in calculating γj, and XN j 0 in calculating γ j. chenvle.mcd Page 3
To complete the phase diagram, we must find the temperatures at which a solution of mole fraction X will boil. This calculation is performed by using the temperature-dependent values of the partial pressures from the Clausius Clapeyron equation and the activity coefficients from the van Laar equation. t the boiling point, the sum of the partial pressures must equal the external pressure. We would not the algorithm of Block #, because now we have several vectors to consider simultaneously. Can you tell what are these vectors? MTHCD uses a function in which an initial guess is made, and a Leavenberg-Marquardt algorithm modifies the guess, minimizing the difference on either side of the sign. You should refer to Mathcad esource Center " Solving a Non-linear System Equation for help. s an initial guess, try 350K., which is slightly below the boiling points of either of the two species. Let us simplify the notation as follows: p p : p propanol T p : T propanol p w : p water H p : H vappropanol H w : H vapwater T w : T water Eq [9] Tn : 350 K Given Beginning of the sove block. p p exp H p XN γ p w exp H w ( XN) γ p ext T p Tn T w Tn Eq [30] Temp( XN, γ, γ) : Find( Tn) Note here we use Ctr for the equal sign on the equation above. XN, γ, and γ are vectors. End of the solve block #3. Exercise 6: What is the physical meaning of Eq.[30]? chenvle.mcd Page 4
Tn : Temp( XN, γ, γ) This is a vector of the temperatures corresponding to the liquid mole fraction vector XN. Type Tn in the space to the right to see all of the components of the vector. YN is a vector containing the vapor-phase mole fractions. It is found directly from Dalton;s Law of Partial Pressures. YN i : γ i XN i p p exp H p T p Tn i γ i XN i p p exp H p γ i ( XN i ) p w exp H w T p Tn i T w Tn i Eq [3] chenvle.mcd Page 5
Exercise 7: Explain the physical meanings of the numerator and the denominator of Eq.[3] Exercise 8: Plot the non-ideal liquid-vapor phase diagram for water-propanol system. NON-IDEL LIQUID-VPO PHSE DIGM 380.0 Tn i Tn i Temp 350 0 XN, YN i i Mole Fraction of propanol chenvle.mcd Page 6
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Now plot activity coefficients as a function of mole fraction. 3.5 ctivity Coefficients 3 γ i.5 γ i.5 0 0.5 0.5 0.75 XN i eferences Gentilcore,M., and Healthcare, T., (004) pply Solubility Theory for Process Improvement, www. cepmagzine.org; 38-4 Lee,L.S., and Shen, H.C., Ind. Eng. Chem. es; 4, 5905-594(003) Majer,V., and Svoboda,V., Enthalpies of vaporization of organic compounds: a critical review and data compilation.international Union of Pure and pplied Chemistry Chemical Data Series, 985, N 3, 300 pp Young,S., Computing a Liquid-Vapor Phase Diagram; Symbolic Mathematics Documents for Physical Chemistry http://bluehawk.monmouth.edu/~tzielins/mathcad/index.htm, 996 chenvle.mcd Page 8
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