U = ig U U x x a Flory, Flory-Huggins - - - - - PowerPoint PPT Presentation

Outline 11.4 Van der Waals perspective o Bubble/Dew Calculations using MRL Regular Solutions (V E = 0, S E = 0) Making the function for G E realistic: energetics of mixing are described by the same energy o 11.4 Van der Waals perspective equation for mixtures that we previously developed in Van Laar discussing the simple basis for mixing rules. Scatchard/Hildebrand U ig a ρ ∑ ∑ U – – − = − ρ ig U U x x a Flory, Flory-Huggins - - - - - - - - - - - - - - - - - - = - - - - - - - - - pg 322 i j ij RT RT o 11.5 Theory - Skip Noting that 1/ ρ = V = Σ x i V i according to “regular o 11.6 Local Composition Theory ∑ ∑ solution theory,” a = x i x j a ij Wilson ∑ ∑ – x i x j a ij UNIQUAC ( U ig ) U – = - - - - - - - - - - - - - - - - - - - - - - - - - - - - UNIFAC ∑ x i V i For the pure fluid, taking the limit as x i → 1, 1 2 van Laar a ii x i a ii U ig U is U ig ( ) i ⇒ ( ) ∑ U – = – - - - - - – = – - - - - - - - - - - V i V i 2 For a binary mixture, subtracting the ideal solution F a a I x x V V ≡ − ⇒ = 11 22 E 1 2 1 2 Q U x V Q result to get the excess energy gives, G J + V V x V H 1 2 K 1 1 2 2 A A x x  2 a 11 2 a 22  11.27 = = a 11 a 22 x 1 + 2 x 1 x 2 a 12 + x 2 E E 12 21 1 2 G U U E   + = x 1 - - - - - - - + x 2 - - - - - - - – - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ( x A x A )   V 1 V 2 x 1 V 1 + x 2 V 2 1 12 2 21 A 12 ln γ 2 11.28 γ 1 ln = - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 A 12 x 1 exchange subscripts 1 + - - - - - - - - - - - - - A 21 x 2 3 4

Fitting Van Laar to a single experiment: Infinite dilution can be used. See example 11.7 γ 2 x 2 ln 2 γ 1 1 A 12 = ln + - - - - - - - - - - - - - - - γ 1 ln x 1 ∞ γ 1 ln = A 12 ln γ i ∞ γ 2 γ 1 ln ln 2 x 1 γ 2 1 A 21 = ln + - - - - - - - - - - - - - - - γ 2 x 2 ln Azeotrope point can be used. See example 11.6 0 x i γ i P i sat y i P x 1 = P y i P γ i = - - - - - - - - - - - - - = - - - - - - - - - sat sat x i P i P i 5 6 Scatchard Hildebrand d i ∫ a / V is known as the "solubility parameter" ii i a 12 = a a ∆ ∆ − 11 22 vap vap U H RT δ i ≡   = x 1 x 2 V 1 V 2 - a 11 a 22 2 a 11 - a 22 U E   V V = - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - + - - - - - - - – - - - - - - - - - - - - - i i x 1 V 1 + x 2 V 2  2 2 2 2  V 1 V 2 V 1 V 2 This is a predictive technique valid for nonpolar 2 substances. See table 11.1 for parameters.   x 1 x 2 V 1 V 2 a 11 a 22 U E   = - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - – - - - - - - - - - - -   x 1 V 1 + x 2 V 2 V 1 V 2 Also can make adjustable. ( ) = = Φ Φ δ − δ + 2 E E a 12 = a 1 a 2 1 – k 12 G U f ( x V x V ) a 1 2 1 2 1 1 2 2 = = ΦΦ δ − δ + 2 E E G U f ( xV x V ) 2 ) 2 γ 1 V 1 Φ 2 [ ( δ 1 δ 2 2 k 12 δ 1 δ 2 ] a RT ln = – + 1 2 1 2 1 1 2 2 Â F i ∫ x V / x V is known as the "volume fraction" i i i i 7 8

Van Laar and Scatchard-Hildbrand Flory’s Equation Recall Ideal Solution result (pg 95) 355 n 1 permit mixing n 2 350 345 V T V T T(K) ∆ S 1 = n 1 R ln - - - - - - = n tot x 1 R ln - - - - - - Scatchard-Hildebrand i i 340 V 1 V 1 k ij =-0.038 335 V T V T van Laar ∆ S 2 = n 2 R ln - - - - - - = n tot x 2 R ln - - - - - - 330 i i 0 0.2 0.4 0.6 0.8 1 V 2 V 2 x,y MeOH methanol + benzene 9 10   ( )λ ( )λ V T V T x 1 V 1 + x 2 V 2 x 1 V 1 + x 2 V 2     ∆ ∆ ∆   ∆ S = S 1 + S 2 = n tot R x 1 ln - - - - - - + x 2 ln - - - - - - S = nR x 1 ln - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - + x 2 ln - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -     x 1 V 1 λ x 2 V 2 λ   i i V 1 V 2 is ∆ ( ) = ln + ln S nR x 1 x 1 x 2 x 2 free volume (not occupied) is S E S is S is ( ) ( ) ∆ ∆ ∑ ∑ = S – = S – x i S i – – x i S i = S – S n i V i λ V f i = , athermal ( )λ V f = n 1 V 1 + n 2 V 2 Φ Φ = − = + E E E 1 2 G H TS RT x ( ln x ln ) 1 2 x x 1 2 11 12

Flory-Huggins Model 11.6 Local composition models (nonrandom) Common Features Φ 1 Φ 2 o Lattice Model - Fixed number of neighbors = 10   G E Φ 1 Φ 2 x 1 ( )χ RT = RT x 1 ln - - - - - - + x 2 ln - - - - - - + + x 2 R o Uses G E = A E approximation (good)   x 1 x 2 o Local Composition Flory Scatchard-Hildebrand x 21 - 2’s around 1 1 V 1 δ 1 ( δ 2 ) 2 2 – x 11 - 1’s around 1 V 2 χ = - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = - - - - - - R RT V 1 1 x 12 - 1’s around 2 χ athermal = 0 x 22 - 2’s around 2 1 2 However χ usually nonzero even with H E = 0 (compensate for error in Flory) 13 14 Wilson’s equation x 21 x 2 x 12 x 1 - Ω 21 - Ω 12 - - - - - - - = - - - - - - - - - - - = - - - - V 1 – A   x 11 x 1 x 22 x 2 Ω 12 Λ 21 21 = = - - - - - - exp - - - - - - - - - - -   V 2 RT o Two-fluid Theory ( ) ( ) 1 2 ( U ig ) ( U ig ) ( U ig ) U – = x 1 U – + x 2 U – integrate U E to get A E = G E (Eqn 11.85) differentiate G E to get γ expressions Λ 12 Λ 21 - x 1 x 2 Ω 21 ε 21 ( ε 11 ) x 2 x 1 Ω 12 ε 12 ( ε 22 )   N A z – – γ 1 ( x 2 Λ 12 ) U E ln = – ln x 1 + + x 2 - - - - - - - - - - - - - - - - - - - - - - - - - – - - - - - - - - - - - - - - - - - - - - - - - - - = - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -   x 2 Λ 12 x 1 Λ 21 x 2 Ω 21 x 1 Ω 12 x 1 + + x 2 2 x 1 + + x 2 A E U E – - T d ∫ - - - - - - - = - - - - - - - - - - - - - - - + C RT RT T 15 16

UNIQUAC A z ε ij ( ε jj ) – – q i N q i – a q i     Ω ij ij - τ ij = - - - - exp - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = - - - - exp - - - - - - - - - = - - -     q j 2 RT q j T q j correction to Flory q = surface area of molecule Flory ν k R k ∑ r i E G volume (look up) = Φ Φ } ln / + x ln / x x x groups a f a f comb 1 1 1 2 2 2 ν k Q k ∑ RT q i surface area (look up) − Φ θ + Φ θ groups 5 q x ln / q x ln / a f a f θ i ( ) ⁄ ( ) 1 1 1 1 2 2 2 2 } ∑ x i q i x i q i surface area fractions resid − θ + θ τ − θ τ + θ ln( ) ln( ) q x q x – a 1 1 1 2 21 2 2 1 12 2   parameter τ ij ij = exp - - - - - - - - -   T 17 18 Φ 1 Φ 1 Φ 1 Φ 1     Group parameters for the UNIFAC and UNIQUAC equations. γ 1 ln = ln - - - - - - + 1 – - - - - - - – 5 q 1 ln - - - - - - + 1 – - - - - - - AC in the table means aromatic carbon. (DIFFERS SLIGHTLY FROM TEXT)     θ 1 θ 1 x 1 x 1 Main Sub- R(rel.vol.) Q(rel.area) Example θ 2 τ 12 θ 1 Group group CH2 CH3 0.9011 0.8480 ( θ 1 θ 2 τ 21 ) + q 1 1 – ln + – - - - - - - - - - - - - - - - - - - - - - - - - – - - - - - - - - - - - - - - - - - - - - - - - - CH2 0.6744 0.5400 n-hexane: 4 CH2+2 CH3 θ 1 θ 2 τ 21 θ 1 τ 12 θ 2 + + CH 0.4469 0.2280 isobutane: 1CH+3 CH3 C 0.2195 0 neopentane: 1C+ 4 CH3 C=C CH2=CH 1.2454 1.1760 1-hexene: 1 CH2=CH+3 CH2+1 CH3 Φ 2 Φ 2 Φ 2 Φ 2     CH=CH 1.1167 0.8670 2-hexene: 1 CH=CH+2 CH2+2 γ 2 ln = ln - - - - - - + 1 – - - - - - - – 5 q 2 ln - - - - - - + 1 – - - - - - -     CH3 θ 2 θ 2 x 2 x 2 CH2=C 1.1173 0.9880 CH=C 0.8886 0.6760 θ 1 τ 21 θ 2 C=C 0.6605 0.4850 ( θ 1 τ 12 θ 2 ) + q 2 1 – ln + – - - - - - - - - - - - - - - - - - - - - - - - - – - - - - - - - - - - - - - - - - - - - - - - - - ACH ACH 0.5313 0.4000 benzene: 6ACH θ 1 θ 2 τ 21 θ 1 τ 12 θ 2 + + AC 0.3652 0.1200 benzoic acid: 5ACH+1AC+1COOH ACCH2 ACCH3 1.2663 0.9680 toluene: 5ACH+1ACCH3 ACCH2 1.0396 0.6600 ethylbenzene: 5ACH+1ACCH2+1CH2 ACCH 0.8121 0.3480 OH OH 1.0000 1.2000 n-propanol: 1OH+1 CH3+2 CH2 19 20