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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", γ i, yielding the general expression for K-ratio   γ ϕ exp[ ( − ) / ] L P vap sat V P P vap RT i i i i = K   i γ V ϕ P   i i 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  ∑ ≡ − = − i + x ln( x )   i i nRT nRT nRT nRT nRT   Letting γ i ≡ fi /xi fi ° where fi ° ≡ f at T and P ∑ x G µ − � G x ( G ) f ∑ ∑ ∑ i i − = i i i = i = γ x ln( ) x ln( x ) i i i i o nRT RT RT f i ∑ x G E ∆ G G ∑ ∑ ∑ ∑ i i ≡ − − = γ − = γ x ln( x ) x ln( x ) x ln( x ) x ln( ) i i i i i i i i i nRT nRT RT ∆ G E = ∑ γ n ln( ) i i RT 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. E ∂ ∆ E ∆ G ( G / RT ) = ∑ γ = γ Prove : ln n ln( ) Given : j i i ∂ n RT j     E ∂ ∂ γ ∂ ∆ G n ln ( / RT ) ∑ ∑ = γ i + i ln   n   i i ∂ ∂ ∂ n n n         j j j 0 ≠   if i j ∂  n ∂ n ∑ i = i ln γ  = ln γ   ⇒ 1 = ∂ if i j i ∂ j n n    j j As for the second sum, we must show that it goes to zero. ( ) ( ) ∑ ∑ ln γ ≡ µ ⇒ ∂ ln γ / ∂ = ∂µ / ∂ / By definition, RTd d n n n n RT i i i i j i i j ( ) ∑ / 0 ∂µ ∂ = n n But, Gibbs-Duhem i i j ( ) ∑ ln / 0 Gibbs-Duhem for activity coefficients ∂ γ ∂ = n n Therefore i i j E ∂ ∆ ( G / RT ) γ = Combining these results, ln So, G E (T,P,x) , → γ ’s. j ∂ 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). E ∆ G A = RT x x 1 2 nRT Derive the expressions for the activity coefficients from this expression. Solution: E ∆ G An n = 2 1 RT RT n E 1 ∂ ∆ G An n n n ( / RT )   A   A 1 1 = 2 − 1  = 2 − 1  = − RT x ( x )   2 1 2 ∂ n RT n n RT n n     1 A ln γ 1 2 ⇒ = RT x 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 ⇒ γ w = 2.1108 ; γ ipa =1.0886 ∑ T yw P vap vap x P vap P ipa i i w 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 ⇒ γ w = 2.1053; γ ipa =1.0898 T sat sat sat yw Σ x i P i P w P ipa 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.   1 − ig  = − ρ = − U U ∑ ∑ ∑ ∑ x x a x x a  i j ij i j ij RT RT VRT   V = Σ xiVi according to "regular solution theory," ∑ ∑ = − x x a ( ) i j ij − ig U U ∑ x V i i For the pure fluid, taking the limit as xi → 1, = − a ( ) ( ) = − ∑ − ig ⇒ − ig U U ii U U x a / V i ii i V i is i For a binary mixture, subtracting the ideal solution result to get the excess energy gives,   2 2 2 + + x a x a x a x x a x a E = 11 22 1 11 1 2 12 2 22 + − U   1 2 + V V x V x V   1 2 1 1 2 2 Elliott and Lira: Chapter 11 - Activity Models Slide 5

Collecting a common denominator x a x a 11 22 2 2 2 + + + − + + ( x V x V ) ( x V x V ) ( x a x x a x a ) 1 1 1 2 2 2 1 1 2 2 1 11 1 2 12 2 22 V V E = 1 2 U + x V x V 1 1 2 2 x x a V V 2 2 2 2 2 + 2 + + + 1 − + + x a V x a V x x a ( x a x x a x a ) 1 11 1 1 2 11 2 22 1 1 2 22 1 11 1 2 12 2 22 V V E = 1 2 U + x V x V 1 1 2 2 x x a V V V V 2 2 + 1 − 2 1 x x a x x a 1 2 11 1 2 22 1 2 12 V V VV E = 1 2 1 2 U + x V x V 1 1 2 2 Scatchard and Hildebrand now make an assumption which is very similar to assuming k ij =0 in an equation of state. Setting a 12 = a a 22 , and collecting terms in a slightly 11 subtle way, 2     a a x x V V a a a a x x V V E = 11 22 1 2 1 2 11 + 22 − 2 11 22  = 1 2 1 2 − U        + 2 2 2 2 + x V x V x V x V V V V V V V     1 1 2 2 1 2 1 2 1 1 2 2 1 2 and finally, defining a term called the "solubility parameter" E = ( ) 2 Φ Φ δ − δ + U ( x V x V ) 1 2 1 2 1 1 2 2 ∑ Φ i ≡ where x V / x V is known as the " volume fraction" i i i i δ i ≡ a / V is known as the " solubility parameter" ii i Elliott and Lira: Chapter 11 - Activity Models Slide 6

Solubility Parameters in (cal/cc)½ To estimate the value of δ i , Scatchard and Hildebrand suggested that experimental data near typical conditions be used instead of the critical point. (Note the units on the " a " parameter and the way Vi moves inside.) δ i ≡ ∆ Uvap V / i 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 ∆ = = Φ Φ δ − δ + G E U E ( x V x V ) 1 2 1 2 1 1 2 2 And the resulting activity coefficients are ( ) ( ) 2 2 ln γ 2 ln γ 2 = Φ δ − δ = Φ δ − δ RT v RT v 1 1 2 1 2 2 2 1 1 2 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 a 12 = a a 22 (1- k ij ) 11 where kij is an adjustable parameter also called the binary interaction coefficient The activity coefficient expressions become ( ) ( ) 2 2 2 2 2 2 ln γ = Φ δ − δ + δ δ ln γ = Φ δ − δ + δ δ RT V k ; RT V k 1 1 2 1 2 12 1 2 2 2 1 1 2 12 1 2 Elliott and Lira: Chapter 11 - Activity Models Slide 8