Introductory Chemical Engineering Thermodynamics By J.R. Elliott and C.T. Lira Chapter 11 - Activity Models NONIDEAL SOLUTIONS When a solution does not follow the ideal solution approximation we can apply an EOS or the "correction factor", γi, yielding the general expression for K-ratio γ iL Pi vap ϕ isat exp[V ( P − Pi vap ) / RT ] Ki = ϕi γ iV P We refer to this "correction factor" as the activity coefficient. To derive the thermodynamic meaning of the activity coefficient, note: E is ∆G G G G G ≡ − = − ∑ x i i + ln( x i nRT nRT nRT nRT nRT ) γi ≡ fi /xi fi° where fi° ≡ f at T and P x i Gi x i ( µ i − Gi ) f$i G ∑ − =∑ = ∑ xi ln( o ) = ∑ xi ln( x i γ i ) nRT RT RT fi E ∆G G ∑ xi Gi − x ln( x ) = x ln( x γ ) − x ln( x ) = x ln( γ ) ≡ − ∑ i i ∑ i i i ∑ i i ∑ i i nRT nRT RT E ∆G = ∑ ni ln( γ i ) RT Hence we see that the activity coefficient gives a correction to the ideal solution estimate of the Gibbs energy, component by component. Letting Elliott and Lira: Chapter 11 - Activity Models Slide 1 Activity coefficients as derivatives Show that expressions for all the activity coefficients can be derived once a single expression for the Gibbs excess energy is available. ∆G Given: RT = E ∂( ∆G E / RT ) = ∂n j ∑n ∂( ∆G / RT ) ln γ = j Prove: ∂n j E i ln( γ i ) ∂ni ∂ ln γ i ∑ ln γ i ∂n + ∑ ni ∂n j j ∂n i ∂ni 0 if i ≠ j = ∑ ln γ i ∂n = ln γ j ∂n j 1if i = j ⇒ j As for the second sum, we must show that it goes to zero. By definition, RTd ln γ i ≡ dµ i ⇒ ∑ ni ( ∂ ln γ i / ∂n j ) = ∑ ni ( ∂µ i / ∂n j ) / RT But, Gibbs-Duhem Therefore ∑ n ( ∂µ ∑ n ( ∂ ln γ i i i i ) / ∂n j = 0 ) / ∂n j = 0 Gibbs-Duhem for activity coefficients ∂( ∆ G / RT ) E = ln γ j Combining these results, So, G (T,P,x), → γ’s. ∂n j E Elliott and Lira: Chapter 11 - Activity Models Slide 2 Example. Activity Coefficients by the 1-Parameter Margules Equation Perhaps the simplest expression for the Gibbs excess function is the 1-Parameter Margules (also known as the two-suffix Margules). E ∆G A x1 x2 = nRT RT Derive the expressions for the activity coefficients from this expression. Solution: E An2 n1 ∆G = RT RT n E ∂ ( ∆G / RT ) An2 1 n1 A n2 n1 A x 2 ( 1 − x1 ) = − = = − 1 ∂n1 RT n n 2 RT n n RT A 2 x ⇒ ln γ 1 = RT 2 Elliott and Lira: Chapter 11 - Activity Models Slide 3 Example. VLE prediction using UNIFAC activity coefficients The isopropyl alcohol (IPA) + water (w) system is known to form an azeotrope at atmospheric pressure and 80.37°C (xw = 0.3146) (cf.Perry’s 5ed, p13-38). Use UNIFAC to estimate the conditions of the azeotrope. Solution: We will need the following data, Compo UNIFAC Groups ANTA ANTB ANTC Tmin Tmax water 1-H2O 8.87829 2010.33 252.636 -26 83 IPA 2-CH3; 1-CH, 1-OH 8.07131 1730.63 233.426 1 100 Entering the mole fractions and 80.37°C ⇒ γw = 2.1108; γipa =1.0886 vap T yw Pipa Pwvap ∑ x i Pi vap 80.37 695 360 757 0.3158 82.50 760 395 829 0.3164 80.46 697 361 760 0.3158 Since 0.3158 ≠ 0.3146, we did not find the azeotrope yet. Try xw = 0.3168 ⇒ γw = 2.1053; γipa =1.0898 T Pwsat Pipasat ΣxiPisat yw 80.46 697 361 760 0.3168 Since xw = 0.3168 = yw this must be the composition of the azeotrope estimated by UNIFAC. UNIFAC seems to be fairly accurate for this mixture. Also note that T vs. x is fairly flat near an azeotrope. Elliott and Lira: Chapter 11 - Activity Models Slide 4 "Regular" Solutions The energetics of mixing are described by the van der Waals equation with quadratic mixing rules, but we circumvent the iterative determination of the density by assuming a molar average for the volume of mixing. U − U ig − ρ −1 = x x a = x i x j a ij ∑ ∑ ∑ ∑ i j ij VRT RT RT V = ΣxiVi according to "regular solution theory," − ∑ ∑ x i x j a ij U − U ig = ∑ x i Vi For the pure fluid, taking the limit as xi→1, − a ii U − U ig = ⇒ U − U ig = − ∑ x i a ii / Vi i is Vi For a binary mixture, subtracting the ideal solution result to get the excess energy gives, a 11 a 22 x12 a 11 + 2 x 1 x 2 a 12 + x 22 a 22 E U = x1 + x2 − V2 x 1V1 + x 2V2 V1 ( ) ( ) ( ) Elliott and Lira: Chapter 11 - Activity Models Slide 5 Collecting a common denominator a 11 a ( x 1V1 + x 2V2 ) + x 2 22 ( x 1V1 + x 2V2 ) − ( x 12 a 11 + 2 x 1 x 2 a 12 + x 22 a 22 ) V2 V1 UE = x 1V1 + x 2V2 V V x12 a11V1 + x1 x2 a11 2 + + x22 a22V1 + x1 x2 a22 1 − ( x12 a11 + 2 x1 x2 a12 + x22 a22 ) V1 V2 UE = x1V1 + x2V2 V V VV x1 x2 a11 2 + x1 x2 a22 1 − 2 x1 x2 a12 2 1 V1V2 V1 V2 UE = x1V1 + x2V2 x1 Scatchard and Hildebrand now make an assumption which is very similar to assuming kij=0 in an equation of state. Setting a12= a11 a 22 , and collecting terms in a slightly subtle way, 2 a 11 a 22 x x VV a a a a x x VV U E = 1 2 1 2 112 + 222 − 2 112 222 = 1 2 1 2 − x1V1 + x 2V2 V1 V2 V2 V1 V2 x 1V1 + x 2V2 V1 and finally, defining a term called the "solubility parameter" 2 U E = Φ 1Φ 2 ( δ 1 − δ 2 ) ( x1V1 + x 2V2 ) Φ i ≡ xiVi / ∑ xiVi is known as the " volume fraction" where δ i ≡ a ii / Vi is known as the " solubility parameter" Elliott and Lira: Chapter 11 - Activity Models Slide 6 Solubility Parameters in (cal/cc)½ To estimate the value of δi, Scatchard and Hildebrand suggested that experimental data near typical conditions be used instead of the critical point. δ i ≡ ∆Uvap / Vi (Note the units on the "a" parameter and the way Vi moves inside.) By scanning the tables for the values of solubility parameters, we can quickly estimate whether the ideal solution will be accurate or not. Alkanes Olefins Napthenics Aromatics n-pentane 7.0 1-pentene 6.9 cyclopentane 8.7 benzene 9.2 n-hexane 7.3 1-hexene 7.4 cyclohexane 8.2 toluene 8.9 n-heptane 7.4 1,3 butadiene 7.1 Decalin 8.8 ethylbenzene 8.8 n-octane 7.6 styrene 9.3 n-nonane 7.8 n-propylbenzene 8.6 n-decane 7.9 anthracene 9.9 phenanthrene 9.8 naphthalene 9.9 Turning to the free energy, with the elimination of excess entropy and excess volume at constant pressure, we have, 2 ∆G E = U E = Φ 1Φ 2 ( δ 1 − δ 2 ) ( x 1V1 + x 2V2 ) And the resulting activity coefficients are RT ln γ 1 = v1Φ ( δ 1 − δ 2 ) 2 2 2 RT ln γ 2 = v 2 Φ (δ 1 − δ 2 ) Elliott and Lira: Chapter 11 - Activity Models 2 1 2 Slide 7 More Solubility Parameters in (cal/cc)½ For oxygenated hydrocarbons and amines, the solubility parameters tend to be larger. This is largely a reflection of the higher heats of vaporization resulting from hydrogen bonding, but also from the polar moments typical of these components. Alcohols Amines Ethers Ketones water 23.4 ammonia 16.3 dimethyl ether 8.8 acetone 9.9 methanol 14.5 methyl amine 11.2 diethyl ether 7.4 2-butanone 9.3 ethanol 12.5 ethyl amine 10.0 dipropyl ether 7.8 2-pentanone 8.7 n-propanol 10.5 pyridine 14.6 furan 9.4 2-heptanone 8.5 n-butanol 13.6 THF 9.1 n-hexanol 10.7 n-dodecanol 9.9 We can also obtain a compromise by assuming a12= a a (1-kij) where kij is an adjustable parameter also called the binary interaction coefficient The activity coefficient expressions become 2 2 RT ln γ 1 = V1Φ 22 ( δ 1 − δ 2 + 2 k 12 δ 1 δ 2 ) ; RT ln γ 2 = V2 Φ 12 (δ 1 − δ 2 + 2 k 12 δ 1δ 2 ) 11 22 Elliott and Lira: Chapter 11 - Activity Models Slide 8 Example. VLE Predictions using regular solution theory Benzene and cyclohexane are to be separated by distillation at 1 bar. Use regular solution theory to predict whether an azeotrope should be expected for this mixture. Tc (K) Pc (bar) Vi(cc/mol) ω δ(cal/cc)½ Benzene 562.2 48.98 0.211 89 9.2 Cyclohexane 553.5 40.75 0.215 109 8.2 Solution: Consider y vs. x at x =0.01 and 0.99. If yB >xB at xB =0.01 and yB <xB at xB =0.99, then yB =xB (i.e. there is an azeotrope) somewhere in between. If y >x or y<x for all xB, then there is no azeotrope. Given xB and P, we should perform bubble point temperature calculations. At xB =0.99, guess T=350K ⇒ ΦB = 0.99(89)/[0.99(89)+0.01(109)] = 0.9878 PBsat= 48.98*10**[7/3*1.211*(1-562.2/350)]= 0.9481 bar PCsat = 40.75*10**[7/3*1.215*(1-553.5/350)]= 0.9158 bar lnγB = 89/1.987(350) (1-.9878)2(9.2-8.2)2= 0.00001911 ⇒ γB = 1.00002 lnγC = 109/1.987(350) (.9878)2 (9.2-8.2)2 = 0.1529 ⇒ γC = 1.1652 Σyi = Σxiγi Pisat/P = 0.99(0.9481)1.00002+0.01(0.9158)1.1652 = 0.9493 ⇒yB =0.9887 Guess T=353K ⇒ PBsat = 1.036; PCsat = 0.9997; γB=1.00; γC =1.1652*353/350=1.1752 Elliott and Lira: Chapter 11 - Activity Models Slide 9 Σyi = Σ xiγi Pisat/P = 0.99(1.036)1.00 + 0.01(0.9997)1.1752 = 1.0374 ⇒yB =0.9887 T≈350+3*(1-0.9493)/(1.0374-0.9493)=351.73 Guess T=351.73K⇒PBsat=0.9981;PCsat=0.9634;γB=1.0;⇒γC=1.1652*351.73/350=1.1710 Σ yi = 0.99(0.9981)1.0 + 0.01(0.9634)1.1710 = 0.99944 ⇒yB =0.9887 < 0.99 At xB =0.01, guess T=353K ⇒ΦB = 0.01(109)/[0.01(89)+0.99(109)] = 0.0082 lnγC = 109/1.987(353) (1-.0082)2(9.2-8.2)2 ≈ 0 ⇒ γC = 1.00 lnγB = 89/1.987(353) (.0082)2(9.2-8.2)2 = 0.1248 ⇒ γB = 1.1330 Σ yi = Σ xiγi Pisat/P = 0.01(1.036)1.1330 + 0.99(0.9997)1.00 = 1.0014 ⇒yB=.0138 Therefore, (yB- xB) changes sign between 0.01-0.99 ⇒ AZEOTROPE. NOTES: 1. γ is a strong function of composition but weak w.r.t. Temperature. 2. γi(xi→1) ≈ 1.00; γi(xi→0) = γimax 3. If Σ yi ε [0.95,1.05], then yi= xiγi Pisat/(PΣyi ) is an accurate estimate. 4. If PBsat ≈ PCsat then a small non-ideality can cause an azeotrope. Elliott and Lira: Chapter 11 - Activity Models Slide 10 Van Laar’s Equations The regular solution equations can easily be rearranged into the van Laar form by writing two adjustable parameters, A12 and A21. A12 V1 V 2 V = A12 = 1 (δ 1 − δ 2 ) ; A21 = 2 (δ 1 − δ 2 )2 ; RT A V2 RT 21 NOTE: Do NOT estimate A12 and A21 from δ1 and δ2. This how we rename this particular grouping of parameters to obtain two adjustable parameters, A12 and A21. ∆G E U E A12 A21 x1x2 = = RT RT RT ( x1 A12 + x2 A21 ) Giving expressions for the activity coefficients, A12 A21 ln γ 1 = ln γ = 2 2 2 A12 x1 ; A21 x 2 (11.28) 1 + 1 + A21 x 2 A12 x1 The point of van Laar theory is to use experimental data for mixtures to estimate the values of A12 and A21. These equations can be rearranged to obtain A12 and A21 from γ1 and γ2 given any one VLE point. x 2 ln γ 2 A12 = ln γ 1 1 + x1 ln γ 1 2 x1 ln γ 1 A21 = ln γ 2 1 + x 2 ln γ 2 Elliott and Lira: Chapter 11 - Activity Models 2 (11.29) Slide 11 Example. Application of the Van Laar equation A particularly useful data point for VLE is the azeotrope because 1) x1=y1 ⇒ γ1 = P/P1sat; γ2 = P/P2sat 2) Many tables of known azeotropes are commonly available 3) The location of an azeotrope is very important for distillation design. Consider the benzene(1)+ethanol(2) system which exhibits an azeotrope at 760 mmHg and 68.24 °C containing 44.8 mol% Ethanol. Calculate the composition of the vapor in equilibrium with an equimolar liquid solution at 760 mmHg given the Antoine constants log P1sat = 6.8975 - 1206.35/(T+220.24) log P2sat = 8.1122 - 1592.86/(T+226.18) Solution: at T = 68.24°C, P1sat = 519.6 mmHg; P2sat = 503.4 mmHg γ1 = 760/519.6 = 1.4627; γ2 = 760/503.4 = 1.5097 x1 = 0.552 ; x2 = 0.448 2 2 x 2 ln γ 2 x1 ln γ 1 A12 = ln γ 1 1 + A21 = ln γ 2 1 + x1 ln γ 1 x 2 ln γ 2 = 1.3424 ; = 1.8814 Elliott and Lira: Chapter 11 - Activity Models Slide 12 Now consider x1 = x2 = 0.5 A12 A21 ln γ 1 = ln γ 2 = 2 2 A12 x1 ; A21 x 2 1 + A x 1 + A x 21 2 12 1 γ1 = 1.580; γ2=1.386 Problem statement ⇒ bubble point temperature is required Guess T=60°C ⇒ P1sat = 391.5 mmHg; P2sat = 351.9 mmHg yi = xi γi P1sat /P ⇒ y1 = 0.407; y2 = 0.321; Σyi = 0.728 ⇒ T guess is too low. at T = 68.24°C, P1sat = 519.6 mmHg; P2sat = 503.4 mmHg yi = xi γi Pisat /P ⇒ y1 = 0.540; y2 = 0.459; Σyi = 0.999 ⇒ T guess is practically Taz. Elliott and Lira: Chapter 11 - Activity Models Slide 13 Free volume and Flory-Huggins Theory The volume occupied by one molecule is not accessible to the other molecules. When we mix two components, each component’s entropy increases according to how much more space it has: ∆Si = Ni k ln(V f m / V f i ) V f m = the free volume of the mixture where V f i = the free volume in the ith pure component It is customary to assume that the fraction of free volume in any component is the same. V f i = Nivi vf where vi = volume of the ith species vf = universal fraction of free volume The entropy may be taken as that of a perfect gas composed of the same number of molecules confined to a volume equal to the free volume. V fm V fm ∆S = x1 ln( ) + x1 ln( ) V f2 V f1 Nk n v + n2 v 2 n v + n2 v 2 ∆S = x1 ln( 1 1 ) + x 2 ln( 1 1 ) = − ∑ x i ln Φ i Nk n1 v1 n2 v 2 ∆S E = − ∑ x i ln Φ i + ∑ x i ln x i = − ∑ x i ln( Φ i / x i ) Nk Elliott and Lira: Chapter 11 - Activity Models Slide 14 For a binary solution, (δ 1 −δ2) Φ Φ ∆G H ∆S = − = x1 ln 1 + x 2 ln 2 + Φ1Φ2 NkT NkT Nk x1 x2 RT v1 2 2 ln γ 1 = ln(Φ1 / x1 ) + (1 − Φ1 / x1 ) + Φ2 ( δ 1 − δ 2 ) RT v2 2 2 ln γ 2 = ln(Φ2 / x2 ) + (1 − Φ2 / x 2 ) + Φ1 ( δ 1 − δ 2 ) RT E E E Elliott and Lira: Chapter 11 - Activity Models 2 ( x1v1 + x 2 v 2 ) Slide 15 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 0.975 1 10 0 0.07 0.14 0.2 0.26 0.32 0.38 0.43 0.47 0.52 0.55 0.58 0.61 0.62 0.62 0.6 0.57 0.51 0.41 0.26 0.15 0 V1/V2 100 1000 0 0 0.18 0.29 0.36 0.59 0.53 0.87 0.7 1.16 0.87 1.44 1.03 1.72 1.19 1.99 1.34 2.25 1.48 2.51 1.62 2.76 1.75 3 1.86 3.23 1.96 3.44 2.04 3.63 2.1 3.8 2.11 3.92 2.07 3.98 1.93 3.92 1.55 3.59 1.13 3.08 0 0 4 V2/V1=1000 3.5 3 Excess Entropy/Nk x1 2.5 V 2/V1= 100 2 1.5 1 V2/V1=10 0.5 0 -0.5 0 0.2 0.4 0.6 0.8 1 x1 Elliott and Lira: Chapter 11 - Activity Models Slide 16 Example. Combinatorial contribution to the activity coefficient Consider the case when 1 g of benzene is added to 1g of pentastyrene to form a solution. Estimate the activity coefficient of the benzene in the pentastyrene if δps = δb =9.2 and Vps and Vb are estimated using the "R" parameters from UNIQUAC/UNIFAC. Solution: Since δps = δb =9.2, we can ignore the residual contribution. Therefore, ln γ b = ln(Φb / x b ) + (1 − Φb / x b ) Benzene is comprised of 6(ACH) groups @ 0.5313 R-units per group ⇒ Vb ~3.1878 Pentastyrene is 25(ACH)+1(ACCH2)+4(ACCH)+4(CH2)+1(CH3) 25*0.5313+1.0396+4*0.8121+4*0.6744+0.9011⇒ Vps ~21.17 Mb = 78 and Mps = 522 ⇒ xb = 0.8696 Φb = 0.8696(3.1878)/[0.8696(3.1878)+0.1304(21.17)] = 0.5010 (Note: The volume fraction is very close to the weight fraction.) ln γ b = ln(0.5010 / 0.8696) + (1 − 0.5010 / 0.8696) = −0.1275 ⇒ γ b = 0.8803 Note: The activity of benzene is soaked up like a sponge if there is no energetic contribution. Elliott and Lira: Chapter 11 - Activity Models Slide 17 Example. Polymer mixing Suppose 1g each of two different polymers (polymer A and polymer B) is heated to 127°C and mixed as a liquid. Estimate the activity coefficients of A and B using Scatchard-Hildebrand theory combined with the Flory-Huggins combinatorial term. MW V δ(cal/cc)½ A 10,000 1,540,000 9.2 B 12,000 1,680,000 9.3 Solution: xA = (1/10,000)/(1/10,000+1/12,000) = .5455; xB = .4545 ΦA = 0.5455(1.54)/[0.5455(1.54)+0.4545(1.68)] = 0.5238; ΦB = 0.4762 lnγ A = ln(0.5238/0.5455) + (1 - 0.5238/0.5455) + 1.54E6(9.3 - 9.2)2 (0.4762)2 /1.987(400) = -.0008 + 4.395 ⇒ γA = 81 lnγ B = ln(0.4762/ 0.4545) + (1 - 0.4762/0.4 545) + 1.68E6(9.3 - 9.2) 2 (0.5238) 2 /1.987(400 ) = +.0008 + 5.800 ⇒ γB = 330 Note: These high γ‘s actually lead to LLE discussed below. Elliott and Lira: Chapter 11 - Activity Models Slide 18 Local Composition Theory Define a local mole fraction by: xij ≡Nij/Ncj Nij = number of "i" atoms around a "j" atom Ncj = ∑ N ij i The local mole fraction can be related to the bulk mole fraction by N i σij3 Rij xij = g ij 4πrij2 drij ∫ VNc j 0 where rij = r/σij Rij = "neighborhood" Further, we can write xij x jj = 3 ij Nc j N j 3jj Nc j N i Noting ⇒ ∫ gij 4 ∫ g jj 4 rij2 drij r jj2 dr jj ≡ xi Ω ij xj ∑ xij = 1 = ∑ xi Ωij x jj / x j = x jj / x j ∑ xi Ωij i 1 = x jj / x j ∑ xi Ωij ⇒ i xj x jj = ∑ xi Ωij ⇒ xij = i Elliott and Lira: Chapter 11 - Activity Models xi Ωij ∑ xk Ωkj k Slide 19 Example 11.12(p383). Compute the local compositions for the following lattice based on rows and columns away from the edges. O X X O X X O O X O X X O O X X X O O O X O X O O# #X’S #O’S O X 2 3 0 3 3 0 4 2 0 5 1 1 6 1 0 7 0 3 X X O X 1 3 2 X 8 2 1 X O 9 2 = 17 1 = 8 xxo = 17/25; xo = 9/22; Ωxo = (17/8)*(9/13) = 1.47 Elliott and Lira: Chapter 11 - Activity Models Slide 20 Obtaining the Free energy from the local compositions Recalling the energy equation for mixtures, N A uij U − U ig ρ = x x gij N A 4πr 2 dr ∑ ∑ i j∫ RT RT 2 We would like to specify some (uij)avg ≡ εij such that N A uij N A ε ij 2 2 ∫ RT gij N A 4πr dr = RT ∫ gij N A 4πr dr ⇒ ni N A σ ij3 N A ε ij U − U ig 1 g ij 4πrij2 drij = ∑∑x j ∫ V RT RT 2 Substituting Ncj, Λij, and xij into the energy equation for mixtures ( U − U ig ) = 21 ∑ x j Nc j ∑ xij εij j i ~(11.77) If we assume that Ncj = Nci ≡ z where z is assumed to be the same coordination number for all the components, U E = 1 2 ∑x j j Nc j ∑ xij ( ij i jj ); UE = 1 2 ∑ x j Nc j ∑ j i xi Ω ij ∑x Ω i ( ij - ij ) ij (11.80) k Elliott and Lira: Chapter 11 - Activity Models Slide 21 Obtaining the Free energy from the local compositions A = U - TS ⇒ A/RT = U/RT - S/R T ∂U TU T ∂S Cv U T Cv U ∂ ( A / RT ) = − − = − − = − T V RT ∂T V RT 2 R ∂T V ∂T R RT R T RT −U E dT AE =∫ + C where C is an integration constant. Recall the analogous RT RT T E 2 procedure for regular solutions (i.e. U = Φ 1Φ 2 (δ 1 − δ 2 ) ( x1V1 + x2V2 ) ) isindependent of temperature, so it can be factored out of the intgral, and A E U E − dT UE = +C = +C 2 ∫ RT R RT T For local composition theory, we just need to repeat this complete procedure but E recognize that U can be a function of temperature. In local composition theory, the temperature dependence shows up in Ωij. We assume, Ωjj = Bij exp[-AijNcj /2RT] where Ajj = ( εij - εjj ) (Note: Aij ≠Aji even though εij = εji ) the integration with AE = −∑ x j ln ∑ xi Ω ij + C respect to T becomes very simple. Then, RT j i Elliott and Lira: Chapter 11 - Activity Models Slide 22 Wilson’s equation Ncj =2 for all j at all ρ; Bij = Vj/Vi ; C = 0 GE =− RT ∑ j x j ln ∑ i xi Λ ji ⇒ GE = −∑ n j ln ∑ ni Λ ji - ln (n ) RT i j Taking the last term first: + ∑ n j [ln (n )] = n ln(n); j ∂ ∑ n j ln ∑ ni i j ∂n k ∂ ( n ln n) 1 = ln n + n ∂n k n ji = ln ∑ ni i ji ∂ G E / RT 1 jk = ln n + n − ln ∑ ni ki − ∑ n j ln γ k = ∂n k n i j ∑ ni i { jk n − ∑ j n ki j ∑ i i } Elliott and Lira: Chapter 11 - Activity Models ji = 1 − ln ∑ xi i jk x − ∑ ki j j ∑ xi i Slide 23 ji UNIFAC and UNIQUAC Abrams, et al. (1975), Maurer and Prausnitz (1978), Fredenslund et al. (1975) Ncj =qj for all j at all ρ; C = Σxiln(Φi/xi) -5Σqixiln(Φi/θi) x j rj xjqj qi Φj ≡ θj ≡ B ≡ r = n r q = n q ij where ∑ xjqj ∑ xi ri ; ∑ xi qi ; j ∑ kj kj ; j ∑ kj kj ; k i GE =− RT ∑ j k j i q j x j ln ∑ i xi Ω ij + ∑ x j ln(Φ j /x j ) -5∑ q j x j ln(Φ j /θ j ) j j RES ln γ k = ln γ COMB + ln γ k k ln γ kCOMB = ln (Φ k / xk ) - (1 − Φ k / xk ) - 5qk [ln(Φ k / θ k ) − (1 − Φ k / θ k )] ln γ kRES = qk 1 − ln ∑ i xi Ω ik − ∑ j x j Ω kj xi Ω ij i ∑ Elliott and Lira: Chapter 11 - Activity Models Slide 24 Example. Application of Wilson’s equation to VLE For the binary system n-pentanol(1)+n-hexane(2), the Wilson equation constants are A12 = 1718 cal/mol A21 = 166.6 cal/mol Assuming the vapor phase to be an ideal gas, determine the composition of the vapor in equilibrium with a liquid containing 20 mole percent n-pentanol at 30xC. Also calculate the equilibrium pressure. Given: P1sat= 3.23 mmHg; P2sat = 187.1 mmHg Solution From CRC, ρ1 = 0.8144 g/ml (1mol/88g) ⇒ V1 = 108 cm3/mol ρ2 = 0.6603 g/ml (1mol/86g) ⇒ V2 = 130 cm3/mol Note: ρ1 and ρ2 are functions of T but ρ1/ρ2 ≈ const. V2/V1 = 1.205 Λij = Vj /Vi exp(-Aij/RT) Λ12 = 1.205 exp(-1718/1.987/303) = 0.070 Λ21 = 1/1.205 exp(-166.6/1.987/303) = 0.625 Elliott and Lira: Chapter 11 - Activity Models Slide 25 The activity coefficients from the Wilson equation are: x1Λ11 x2 Λ 21 ln γ 1 = 1 − ln( x1Λ11 + x2 Λ12 ) − − x1Λ11 + x2 Λ12 x1Λ 21 + x2 Λ 22 x1 Λ 12 x 2 Λ 22 ln γ 2 = 1 − ln( x1 Λ 21 + x 2 Λ 22 ) − − x1 Λ 11 + x 2 Λ 12 x1Λ 21 + x 2 Λ 22 Noting that Λ11= Λ22 =1, we can rearrange for binary mixtures to obtain the slightly simpler relations: ln γ 1 = 1 − ln( x1 Λ 11 + x 2 Λ 12 ) + x 2 Q ln γ 2 = 1 − ln( x1 Λ 21 + x 2 Λ 22 ) − x1Q Λ 12 Λ 21 Q = − where x1 + x 2 Λ 12 x1 Λ 21 + x 2 Q = 0.070/(0.2+0.8*0.070) - 0.625/(0.8+0.2*0.625) = -0.4022 ln γ 1 = 1 − ln(0.2 + 0.8 * 0.070) + 0.8Q = 1.0408 ⇒ γ1= 2.824 ln γ 2 = 1 − ln(0.8 + 0.20*.625) − 0.2Q = 0.1584 ⇒ γ2= 1.172 + x2 γ2 P2sat P = (y1+y2)P = x1γ1 P1sat = 0.2*2.824*3.23 + 0.8*1.172*187.1 = 177.2 mmHg y1 = x1γ1 P1sat /P = 0.2*2.824*3.23/177.2 = 0.0103 Elliott and Lira: Chapter 11 - Activity Models Slide 26 Question: What value for Ωij is implied by the van der Waals EOS? Z= aρ 1 − 1 − bρ RT b = Σxibi is reasonable. As for "a", we must carefully consider how this term relates to the energy of mixing: N A uij U − U ig aρ N A ρ =− = xi x j ∫ g ij 4πr 2 dr ∑ ∑ 2 RT RT RT Comparing to the result for pure fluids a ii = − NA 2 ∫ U − U ig N ρ = − A ∑ ∑ xi x j aij ⇒ a = ∑ ∑ xi x j aij N A uii gii 4πr dr ⇒ RT RT 2 where aij ≡ − ⇒ Ω ij ≡ − NA 2ε ij − NA 2ε jj NA 2 2 N u g 4π r dr where we set aij= aii a jj (1 - kij), A ij ij ∫ 2 π N u g r dr 4 A ij ij ∫ 2 ∫ N Au jj g jj 4πr dr = aij ε jj a jj ε ij ~ σ ij3 σ 3jj Elliott and Lira: Chapter 11 - Activity Models Slide 27