Introductory Chemical Engineering Thermodynamics Chapter 11

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 L P vap  sat exp[VP (− Pvap ) / RT ] K = i i  i i  i V ϕ  γi P i We refer to this "correction factor" as the activity coefficient. To derive the thermodynamic meaning ofϕ the activity coefficient, note: ∆G EisG G G  G  ≡−=−∑ x  i +ln(x ) nRT nRT nRT nRT i nRT i  ° ° Letting γi ≡ fi /xi fi where fi ≡ f at T and P $ ∑ xG xG()µ − f G −=ii ii i==i γ ∑∑∑xi ln( o )ln()xxiii nRT RT RT f i ∆G E G ∑ xG ≡−ii −∑∑xxln( ) = xx ln(γγ ) − ∑∑ xx ln( ) = x ln( ) nRT nRT RT ii iiiii ii ∆G E = ∑n ln(γ ) RT ii Hence we see that the activity coefficient gives a correction to the ideal solution estimate of the Gibbs energy, component by component.

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 E ∂(/)∆GRTE = ∑n ln(γ ) ln γ = Given: ii Prove: j ∂ RT n j ∂(/)∆GRTE  ∂n  ∂γln  = ∑∑ln γ  i  + n  i  ∂ i ∂ i ∂ n j  n j   n j  ∂  ≠   ni 0if i j ∂ ni =  ⇒ ∑ lnγ   = ln ∂n 1if i = j i  n  j j ∂ j γ As for the second sum, we must show that it goes to zero. γµ≡⇒ ∂γ∂∑∑()() ∂µ∂ = By definition, RTdlnii d n i ln ij / n n iij / n / RT ∑ ()∂µ ∂ = But, Gibbs-Duhem nnii/0 j ∑ ()∂γ∂ = Therefore nniijln / 0 Gibbs-Duhem for activity coefficients ∂(/)∆GRTE ln γ = E Combining these results, j ∂ So, G (T,P,x), → γ's . n j

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). ∆G E A = xx nRT RT 12 Derive the expressions for the activity coefficients from this expression. Solution: ∆ E An n G = 21 RT RT n ∂(/)∆GRTE An 1 n  A n  n  A =−21=−2111=−xx() ∂  2    21 n1 RT n n  RT n  n  RT A ⇒=lnγ x 2 12RT

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 vap vap T Pipa Pw ∑ xPi i yw 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 sat sat Σ sat TPw Pipa xiPi 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. UU− ig  −ρ −1   = ∑∑∑ xxa = ∑ xxa  RT RT ijijVRT ijij V = ΣxiVi according to "regular solution theory," −∑∑ xxa ()UU−=ig ijij ∑ xVii For the pure fluid, taking the limit as xi→1, −a ()UU−=ig ii ⇒−()UUig =−∑ xaV/ i is iii i Vi For a binary mixture, subtracting the ideal solution result to get the excess energy gives, a a  xa2 ++2 xxa xa2  UxE =+11 x 22 − 1 11 1 2 12 2 22  1 2 + V1 V2  xV11 xV 2 2 

Elliott and Lira: Chapter 11 - Activity Models Slide 5 Collecting a common denominator a a x 11 ()()(xV++ xV x 22 xV +− xV x2 a +2 x x a + x2 a ) 1 V 11 2 2 2V 11 2 2 1 11 1 2 12 2 22 U E = 1 2 + xV11 xV 2 2 V V xaV2 ++++−++ xxa 2 xa2 V xxa 1 ()xa2 2 xxa xa2 1 11 1 1 2 11 V 2 22 1 1 2 22 V 1 11 1 2 12 2 22 U E = 1 2 + xV11 xV 2 2 V V VV xxa 2 +−xxa 1 2xxa 21 1211V 1222V 1212VV U E = 1 2 12 + xV11 xV 2 2 Scatchard and Hildebrand now make an assumption which is very similar to assuming aa kij=0 in an equation of state. Setting a12= 11 22 , and collecting terms in a slightly subtle way, 2 xxVV  a a a a  xxVV  a a  U E = 1212  11 +−22 2 11 22  = 1212  11 − 22  +  2 2 2 2  +   xV11 xV 2 2V1 V2 V1 V2  xV11 xV 2 2 V1 V2  and finally, defining a term called the "solubility parameter" E =−ΦΦ δδ2 + UxVxV12() 1 2 ()11 2 2 Φ ≡ where iiiiixV/∑ xV is known as the " volume fraction" δ ≡ iiiiaV/ 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 ≡∆ iiUvap/ V (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, ∆ΦΦEE==()δδ −2 + GU12 1 2 () xVxV11 2 2 And the resulting activity coefficients are 2 2 γδδ=−Φ2 ()γδδ=−Φ2 () RTln 112 v 12 RTln 221 v 12

Elliott and Lira: Chapter 11 - Activity Models 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 aa a12= 11 22 (1-kij) where kij is an adjustable parameter also called the binary interaction coefficient The activity coefficient expressions become γδδδδ=−+Φ 2 ()2 γδδδδ=−+Φ 2 ()2 RTln 112 V122 k 1212; RTln 221 V122 k 1212

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) ω Vi(cc/mol) 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 x or y

⇒ sat sat γ γ Guess T=353K PB = 1.036; PC = 0.9997; B=1.00; C =1.1652*353/350=1.1752

Elliott and Lira: Chapter 11 - Activity Models Slide 9 Σ Σ γ sat ⇒ yi = xi i Pi /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 ⇒ sat sat γ ⇒γ Guess T=351.73K PB =0.9981;PC =0.9634; B=1.0; C=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 Σ γ sat ⇒ yi = xi i Pi /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 Σ ε γ sat Σ 3. If yi [0.95,1.05], then yi= xi i Pi /(P yi ) is an accurate estimate. sat sat 4. If PB ≈ PC 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. V 2 V δ A V =−1 ()δδ = 2 ()−δ 2 12 = 1 A12 12; A21 1 2 ; RT RT A21 V2

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. ∆ E E G = U = A12 A21 x1x2 + RT RT RT (x1A12 x2 A21) Giving expressions for the activity coefficients, A A γ = 12 γ = 21 ln 1 2 ln 2 2  Ax ;  Ax (11.28) 1+ 12 1  1+ 21 2      Ax21 2 Ax12 1 The point of van Laar theory is to use experimental data for mixtures to estimate the γ values of A12 and A21. These equations can be rearranged to obtain A12 and A21 from 1 and γ2 given any one VLE point. 2 2  x ln   x ln  A =+lnγ 1 22 A =+lnγ 1 11 12 1 γ 21 2 γ (11.29)  x11ln   x22ln  γ γ

Elliott and Lira: Chapter 11 - Activity Models Slide 11 Example. Application of the Van Laar equation A particularly useful data point for VLE is the azeotrope because ⇒ γ sat γ sat 1) x1=y1 1 = P/P1 ; 2 = P/P2 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 sat log P1 = 6.8975 - 1206.35/(T+220.24) sat log P2 = 8.1122 - 1592.86/(T+226.18) Solution: sat sat at T = 68.24°C, P1 = 519.6 mmHg; P2 = 503.4 mmHg γ1 = 760/519.6 = 1.4627; γ2 = 760/503.4 = 1.5097 x1 = 0.552 ; x2 = 0.448 2 2  x ln   x ln  A =+lnγ 1 22 A =+lnγ 1 11 12 1 γ  21 2  γ  x11ln x22ln = 1.3424 γ ; = 1.8814 γ

Elliott and Lira: Chapter 11 - Activity Models Slide 12 Now consider x1 = x2 = 0.5 A A γ = 12 γ = 21 ln 1 2 ln 2 2  Ax ;  Ax 1+ 12 1  1+ 21 2      Ax21 2 Ax12 1 γ1 = 1.580; γ2=1.386

Problem statement ⇒ bubble point temperature is required sat sat Guess T=60°C ⇒ P1 = 391.5 mmHg; P2 = 351.9 mmHg γ sat ⇒ Σ ⇒ yi = xi i P1 /P y1 = 0.407; y2 = 0.321; yi = 0.728 T guess is too low. sat sat at T = 68.24°C, P1 = 519.6 mmHg; P2 = 503.4 mmHg γ sat ⇒ Σ ⇒ az yi = xi i Pi /P y1 = 0.540; y2 = 0.459; yi = 0.999 T guess is practically T .

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: ∆ V /V Si = Ni k ln( ffmi) V where fm = the free volume of the mixture V fi = 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 fi = 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.

∆S V f V f =+x ln(mm )x ln( ) Nk 11V V f12f ∆ nv+ nv nv+ nv S = 11 2 2 + 11 2 2 =− Φ x1 ln( )x2 ln( )∑ xii ln Nk nv11 nv22 ∆S E =−∑∑xxxxxlnΦΦ + ln =− ∑ ln( / ) Nk ii ii i ii

Elliott and Lira: Chapter 11 - Activity Models Slide 14 For a binary solution, 2 ∆∆G EEH S E ΦΦ ()δδ− =−=1 +2 +ΦΦ 12 + x1 lnx2 ln12 (xv11 xv 2 2 ) NkT NkT Nk x1 xRT2 v 2 lnγδδ=+−+− ln(ΦΦΦ /xx ) (1 / ) 1 2 () 11111RT 2 12 v 2 lnγδδ=+−+− ln(ΦΦΦ /xx ) (1 / ) 2 2 () 22222RT 1 12

Elliott and Lira: Chapter 11 - Activity Models Slide 15 x1 V1/V2 10 100 1000 0000 0.05 0.07 0.18 0.29 4 0.1 0.14 0.36 0.59 V2/V1=1000 0.15 0.2 0.53 0.87 3.5 0.2 0.26 0.7 1.16 3 0.25 0.32 0.87 1.44 0.3 0.38 1.03 1.72 2.5 0.35 0.43 1.19 1.99 V2/V1=100 0.4 0.47 1.34 2.25 2 0.45 0.52 1.48 2.51 0.5 0.55 1.62 2.76 1.5 0.55 0.58 1.75 3 0.6 0.61 1.86 3.23 1 0.65 0.62 1.96 3.44 Excess Entropy/Nk V2/V1=10 0.7 0.62 2.04 3.63 0.5 0.75 0.6 2.1 3.8 0.8 0.57 2.11 3.92 0 0.85 0.51 2.07 3.98 0 0.2 0.4 0.6 0.8 1 0.9 0.41 1.93 3.92 -0.5 0.95 0.26 1.55 3.59 0.975 0.15 1.13 3.08 x1 1000

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, γ =+−ΦΦ lnbbbbb ln( / xx)(1 / ) 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.) γ = + − = − ⇒ γ = ln b ln(0.5010/ 0.8696) (1 0.5010/ 0.8696) 0.1275 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 γ 2 2 ln A = ln(0.5238/0.5455) + (1- 0.5238/0.5455) + 1.54E6(9.3- 9.2) (0.4762) /1.987(400) = -.0008 + 4.395 ⇒ γA = 81 γ 2 2 ln B = ln(0.4762/0.4545) + (1- 0.4762/0.4545) + 1.68E6(9.3 - 9.2) (0.5238) /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 ∑ N Ncj = ij i The local mole fraction can be related to the bulk mole fraction by N σ 3 Rij = iij∫ 2 xij grdrij 4 ij ij VNc j 0 π where rij = r/σij Rij = "neighborhood" Further, we can write 2 x Nc N 3 ∫ g 4 r dr ij = j i ij ij ij ij ≡ xi Ω 3 2 ij x jj Nc j N j jj ∫ g jj 4 rjjdrjj x j = = Ω = Ω Noting ∑ xij 1 ∑ xi ij x jj / x j x jj / x j ∑ xi ij i x Ω x j = i ij 1= x / x x Ω ⇒ = x Ω xij ⇒ jj j ∑ i ij ∑ i ij ⇒ ∑ x Ω i x jj i k kj k

Elliott and Lira: Chapter 11 - Activity Models Slide 19 Example 11.12(p383). Compute the local compositions for the following lattice based on rows and columns away from the edges.

OOXO XOXX XXXOXO OXO X OX OXO X XOOX OX X X O

O#123456789 #X’S333211022 = 17 #O’S200010311 = 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, UU− ig  ρ Nu   = ∑∑ xx ∫ Aij g Nrdr4π 2  RT  2 ij RT ij A We would like to specify some (uij)avg ≡ εij such that ε NuAij N Aij ∫∫ gNrdr44π 22= gNrdrπ ⇒ RT ij A RT ij A ig σε3 UU−  1 nNiAijAijN   = ∑∑ x ∫ grdr4π 2  RT  2 j V RT ij ij ij Substituting Ncj, Λij, and xij into the energy equation for mixtures ()UU−=ig 1 ∑∑ xNcxε 2 j jijij ~(11.77) j i If we assume that Ncj = Nci ≡ z where z is assumed to be the same coordination number for all the components, Ω E xi ij E = 1 ()U = 1 ∑ x Nc ∑ ( - ) U 2 ∑ x j Nc j ∑ xij ij - jj ; 2 j j Ω ij ij (11.80) j i j i ∑ xi ij 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 ∂ (/ART ) T ∂U  TU T ∂S  Cv U T Cv U T  =   −−  =−− =−  ∂   ∂  2 ∂  T VVRT T RT R T VR RT R T RT A EE−U dT = ∫ + C where C is an integration constant. Recall the analogous RT RT T δ E = Φ Φ −δ 2 + procedure for regular solutions (i.e. U 1 2 ( 1 2 ) (x1V1 x2V2 ) ) isindependent of temperature, so it can be factored out of the intgral, and A EEU −dT U E = ∫ +=C +C RT R T 2 RT 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 ln∑ x Ω  + C respect to T becomes very simple. Then, j  i ij  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 E   E     G = −  Λ  G = −   Λ  () ∑ x j ln∑ xi ji  ⇒ ∑n j ln∑ni ji  - ln n RT j  i  RT j   i   Taking the last term first: ∂(n ln n)  1  + ∑ n []ln()n = n ln(n); = ln n + n  j ∂ j nk  n     ∂   ∑ n j ln ∑ ni ji     j  i      = ln∑ n  − ∑ n  jk  ∂ i ki j nk  i  j  ∑ ni ji   i      ∂{}G E / RT  1          lnγ = = ln n + n  − ln∑ n  − ∑ n  jk  = 1− ln∑ x  − ∑ x  jk  k ∂ i ki j i ki j nk  n   i  j  ∑ ni ji   i  j  ∑ xi ji   i   i 

Elliott and Lira: Chapter 11 - Activity Models Slide 23 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 r x q q Φ ≡ jj θ ≡ jj ≡ i j j rnr= ∑ qnq= ∑ Bij whereγ ∑ xr ; ∑ xq ; jkjkj; jkjkj; ∑ xq ii ii k k jj i i j E   G = −  Ω  + ()Φ Φ θ ( ) ∑ q j x j ln∑ xi ij  ∑ x j ln j /x j -5∑ q j x j ln j / j RT j  i  j j γ =+γ COMB γ RES θ lnk lnk ln k COMB = ()()Φ − Φ []Φ − ()()− Φ θ ln k ln k / xk - 1 k / xk - 5qk ln k / k 1 k / k      x Ω  lnγ RES = q 1 − ln∑ x Ω  − ∑ j kj k k   i ik  Ω    i  j ∑ xi 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. sat sat Given: P1 = 3.23 mmHg; P2 = 187.1 mmHg ρ ⇒ Solution From CRC, 1 = 0.8144 g/ml (1mol/88g) V1 = 108 cm3/mol ρ ⇒ 2 = 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: x Λ x Λ lnγ =1− ln(x Λ + x Λ ) − 1 11 − 2 21 1 1 11 2 12 Λ + Λ Λ + Λ x1 11 x2 12 x1 21 x2 22 x Λ x Λ lnγ = 1− ln(x Λ + x Λ ) − 1 12 − 2 22 2 1 21 2 22 Λ + Λ Λ + Λ x1 11 x2 12 x1 21 x2 22 Noting that Λ11= Λ22 =1, we can rearrange for binary mixtures to obtain the slightly simpler relations: γ =−ΛΛ + ln111121221 ln(xx)+ xQ γ =−ΛΛ + − ln212122211 ln(xx) xQ Λ Λ Q = 12 − 21 where + Λ Λ + xx1212 x1212 x Q = 0.070/(0.2+0.8*0.070) - 0.625/(0.8+0.2*0.625) = -0.4022 γ =− + ⇒ γ ln1 1020800700 ln( . . * . ) + .8Q = 1.0408 1= 2.824 γ =− + − ⇒ γ ln2 1 ln( 08 . 0 . 20 *. 625 ) 0 . 2Q = 0.1584 2= 1.172 γ sat γ sat P = (y1+y2)P = x1 1 P1 + x2 2 P2 = 0.2*2.824*3.23 + 0.8*1.172*187.1 = 177.2 mmHg γ sat y1 = x1 1 P1 /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? 1 aρ Z = − 1− bρ RT b = Σxibi is reasonable. As for "a", we must carefully consider how this term relates to the energy of mixing: ig UU− aρ N ρ NuAij =− = A ∑∑ xx ∫ grdr4π 2 RT RT 2 ij RT ij

Comparing to the result for pure fluids UU− ig N ρ =−N A π 2 =−A ⇒ = aNugrdr∫ 4 ⇒ ∑∑∑ xxaijij a∑ xxa ijij ii 2 Aiiii RT RT π aNugrdr≡−N A 4π 2 where ij 2 ∫ Aijij where we set aij= aaii jj (1 - kij), N ε − A π 2 3 ε ∫ N Auij gij 4 r dr a σ Ω ≡ 2 ij = ij jj ij ij ε ~ ⇒ − N A 2 a σ 3 ε ∫ N Au jj g jj 4 r dr jj ij jj 2 jj

Elliott and Lira: Chapter 11 - Activity Models Slide 27