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Introductory Chemical Engineering Thermodynamics Chapter 11

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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 ϕ sat exp[VP (− Pvap ) / RT ] K = i i i i i γ V ϕ i P i We refer to this "correction factor" as the activity coefficient. To derive the thermodynamic meaning of 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 ∂n 0if i ≠ j ∂n i = ∑ lnγ i = lnγ ∂ = ⇒ i ∂ j n j 1if i j n j 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 2V1 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 <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 sat PB = 48.98*10**[7/3*1.211*(1-562.2/350)]= 0.9481 bar sat PC = 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 Σ Σ γ sat ⇒ yi = xi i Pi /P = 0.99(0.9481)1.00002+0.01(0.9158)1.1652 = 0.9493 yB =0.9887 ⇒ 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.
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