Introductory Chemical Engineering Thermodynamics By J.R. Elliott - - PowerPoint PPT Presentation

Introductory Chemical Engineering Thermodynamics

By J.R. Elliott and C.T. Lira

Chapter 11 - Activity Models

Elliott and Lira: Chapter 11 - Activity Models Slide 1 NONIDEAL SOLUTIONS When a solution does not follow the ideal solution approximation we can apply an EOS

• r the "correction factor", γi, yielding the general expression for K-ratio

K P P V P P RT

i i L i vap i V i sat i vap i

= −       γ γ ϕ ϕ exp[ ( ) / ]

We refer to this "correction factor" as the activity coefficient. To derive the thermodynamic meaning of the activity coefficient, note:

∆G nRT G nRT G nRT G nRT x G nRT x

E is i i i

≡ − = − +      

∑

ln( )

Letting γi ≡ fi /xi fi° where fi° ≡ f at T and P G nRT x G RT x G RT x f f x x

i i i i i i i i

• i

i i

− = − = =

∑ ∑ ∑ ∑

( ) ln(

• )

ln( ) µ γ ∆G nRT G nRT x G RT x x x x x x x

E i i i i i i i i i i i

≡ − − = − =

∑ ∑ ∑ ∑ ∑

ln( ) ln( ) ln( ) ln( ) γ γ ∆G RT n

E i i

= ∑ ln( ) γ Hence we see that the activity coefficient gives a correction to the ideal solution estimate

• f the Gibbs energy, component by component.

Elliott and Lira: Chapter 11 - Activity Models Slide 2 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. Given:

∆G RT n

E i i

= ∑ ln( ) γ

Prove: ln

( / ) γ ∂ ∂

j E j

G RT n = ∆

∂ ∂ γ ∂ ∂ ∂ γ ∂ ( / ) ln ln ∆G RT n n n n n

E j i i j i i j

=         +        

∑ ∑

   = ≠ = j i if j i if n n

j i

1 ∂ ∂ ⇒

ln ln γ ∂ ∂ γ

i i j j

n n

∑

      =

As for the second sum, we must show that it goes to zero. By definition,

( ) ( )

RTd d n n n n RT

i i i i j i i j

ln ln / / / γ µ ∂ γ ∂ ∂µ ∂ ≡ ⇒ =

∑ ∑

But, Gibbs-Duhem

( )

n n

i i j

∑

= ∂µ ∂ / Therefore

( )

n n

i i j

∑

= ∂ γ ∂ ln / 0 Gibbs-Duhem for activity coefficients Combining these results, ln

( / ) γ ∂ ∂

j E j

G RT n = ∆

So, GE(T,P,x), → γ’s.

Elliott and Lira: Chapter 11 - Activity Models Slide 3

• 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 nRT A RT x x

E

=

1 2

Derive the expressions for the activity coefficients from this expression. Solution:

∆G RT An RT n n

E

=

2 1

∂ ∂ ( / ) ( ) ∆G RT n An RT n n n A RT n n n n A RT x x

E 1 2 1 2 2 1 2 1

1 1 1 = −       = −       = −

⇒ = lnγ 1

2 2

A RT x

Elliott and Lira: Chapter 11 - Activity Models Slide 4

• 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 T

P

ipa vap

P

w vap

x P

i i vap

∑

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 T Pw

sat

Pipa

sat

ΣxiPi

sat

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 5 "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 RT RT x x a VRT x x a

ig i j ij i j ij

−       = − = −

∑ ∑ ∑ ∑

ρ 1 V = ΣxiVi according to "regular solution theory,"

( )

U U x x a x V

ig i j ij i i

− = −

∑ ∑ ∑

For the pure fluid, taking the limit as xi→1,

( ) ( )

U U a V U U x a V

ig i ii i ig is i ii i

− = − ⇒ − = −∑ / For a binary mixture, subtracting the ideal solution result to get the excess energy gives, U x a V x a V x a x x a x a x V x V

E =

+ − + + +      

1 11 1 2 22 2 1 2 11 1 2 12 2 2 22 1 1 2 2

2

Elliott and Lira: Chapter 11 - Activity Models Slide 6 Collecting a common denominator

U x a V x V x V x a V x V x V x a x x a x a x V x V

E =

+ + + − + + +

1 11 1 1 1 2 2 2 22 2 1 1 2 2 1 2 11 1 2 12 2 2 22 1 1 2 2

2 ( ) ( ) ( )

U x a V x x a V V x a V x x a V V x a x x a x a x V x V

E =

+ + + + − + + +

1 2 11 1 1 2 11 2 1 2 2 22 1 1 2 22 1 2 1 2 11 1 2 12 2 2 22 1 1 2 2

2 ( )

U

x x a V V x x a V V x x a V V VV x V x V

E =

+ − +

1 2 11 2 1 1 2 22 1 2 1 2 12 2 1 1 2 1 1 2 2

2

Scatchard and Hildebrand now make an assumption which is very similar to assuming kij=0 in an equation of state. Setting a12= a a

11 22 , and collecting terms in a slightly

subtle way, U x x V V x V x V a V a V a V a V x x V V x V x V a V a V

E =

+ + −         = + −        

1 2 1 2 1 1 2 2 11 1 2 22 2 2 11 1 2 22 2 2 1 2 1 2 1 1 2 2 11 1 22 2 2

2 and finally, defining a term called the "solubility parameter"

( )

U x V x V

E =

− + Φ Φ

1 2 1 2 2 1 1 2 2

δ δ ( ) where

Φi

i i i i

x V x V ≡

∑

/ is known as the " volume fraction"

δi

ii i

a V ≡ / is known as the " solubility parameter"

Elliott and Lira: Chapter 11 - Activity Models Slide 7 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.

δi

i

Uvap 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,

( )

∆ Φ Φ G U x V x V

E E

= = − +

1 2 1 2 2 1 1 2 2

δ δ ( ) And the resulting activity coefficients are

( )

RT v lnγ δ δ

1 1 2 2 1 2 2

= − Φ

( )

RT v lnγ δ δ

2 2 1 2 1 2 2

= − Φ

Elliott and Lira: Chapter 11 - Activity Models Slide 8 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

11 22 (1-kij)

where kij is an adjustable parameter also called the binary interaction coefficient The activity coefficient expressions become

( )

RT V k ln γ δ δ δ δ

1 1 2 2 1 2 12 1 2 2

2 = − + Φ ;

( )

RT V k ln γ δ δ δ δ

2 2 1 2 1 2 12 1 2 2

2 = − + Φ

Elliott and Lira: Chapter 11 - Activity Models Slide 9

• 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 PB

sat= 48.98*10**[7/3*1.211*(1-562.2/350)]= 0.9481 bar

PC

sat = 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 Pi

sat/P = 0.99(0.9481)1.00002+0.01(0.9158)1.1652 = 0.9493 ⇒yB =0.9887

Guess T=353K ⇒ PB

sat = 1.036; PC sat = 0.9997; γB=1.00; γC =1.1652*353/350=1.1752

Elliott and Lira: Chapter 11 - Activity Models Slide 10 Σyi = Σ xiγi Pi

sat/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⇒PB

sat=0.9981;PC sat=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 Pi

sat/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 Pi

sat/(PΣyi ) is an accurate estimate.

• 4. If PB

sat ≈ PC sat then a small non-ideality can cause an azeotrope.

Elliott and Lira: Chapter 11 - Activity Models Slide 11 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.

( )

A V RT

12 1 1 2 2

= − δ δ ;

( )2

2 1 2 21

δ δ − = RT V A

; A A V V

12 21 1 2

= 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.

) (

21 2 12 1 2 1 21 12

A x A x x x RT A A RT U RT G

E E

+ = = ∆

Giving expressions for the activity coefficients, lnγ 1

12 12 1 21 2 2

1 = +       A A x A x ; lnγ 2

21 21 2 12 1 2

1 = +       A A x A x (11.28) 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. A x x

12 1 2 2 1 1 2

1 = +       ln ln ln γ γ γ A x x

21 2 1 1 2 2 2

1 = +       ln ln ln γ γ γ (11.29)

Elliott and Lira: Chapter 11 - Activity Models Slide 12

• Example. Application of the Van Laar equation

A particularly useful data point for VLE is the azeotrope because 1) x1=y1 ⇒ γ1 = P/P1

sat;

γ2 = P/P2

sat

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 P1

sat = 6.8975 - 1206.35/(T+220.24)

log P2

sat = 8.1122 - 1592.86/(T+226.18)

Solution: at T = 68.24°C, P1

sat = 519.6 mmHg; P2 sat = 503.4 mmHg

γ1 = 760/519.6 = 1.4627; γ2 = 760/503.4 = 1.5097 x1 = 0.552 ; x2 = 0.448 A x x

12 1 2 2 1 1 2

1 = +       ln ln ln γ γ γ A x x

21 2 1 1 2 2 2

1 = +       ln ln ln γ γ γ = 1.3424 ; = 1.8814

Elliott and Lira: Chapter 11 - Activity Models Slide 13 Now consider x1 = x2 = 0.5 lnγ 1

12 12 1 21 2 2

1 = +       A A x A x ; lnγ 2

21 21 2 12 1 2

1 = +       A A x A x γ1 = 1.580; γ2=1.386 Problem statement ⇒ bubble point temperature is required Guess T=60°C ⇒ P1

sat = 391.5 mmHg; P2 sat = 351.9 mmHg

yi = xi γi P1

sat /P ⇒ y1 = 0.407; y2 = 0.321; Σyi = 0.728 ⇒ T guess is too low.

at T = 68.24°C, P1

sat = 519.6 mmHg; P2 sat = 503.4 mmHg

yi = xi γi Pi

sat /P ⇒ y1 = 0.540; y2 = 0.459; Σyi = 0.999 ⇒ T guess is practically Taz.

Elliott and Lira: Chapter 11 - Activity Models Slide 14 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

V

f f

m i

/

) where

V 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 Nk x V V x V V

f f f f

m m

= +

1 1

1 2

ln( ) ln( ) ∆ Φ S Nk x n v n v n v x n v n v n v xi

i

= + + + = −∑

1 1 1 2 2 1 1 2 1 1 2 2 2 2

ln( ) ln( ) ln ∆ Φ Φ S Nk x x x x x

E i i i i i i i

= − + = −

∑ ∑ ∑

ln ln ln( / )

Elliott and Lira: Chapter 11 - Activity Models Slide 15 For a binary solution,

( )

∆ ∆ Φ Φ Φ Φ G NkT H NkT S Nk x x x x RT x v x v

E E E

= − = + + − +

1 1 1 2 2 2 1 2 1 2 2 1 1 2 2

ln ln ( ) δ δ

( )

ln ln( / ) ( / ) γ δ δ

1 1 1 1 1 1 2 2 1 2 2

1 = + − + − Φ Φ Φ x x v RT

( )

ln ln( / ) ( / ) γ δ δ

2 2 2 2 2 2 1 2 1 2 2

1 = + − + − Φ Φ Φ x x v RT

Elliott and Lira: Chapter 11 - Activity Models Slide 16

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

• 0.5

0.5 1 1.5 2 2.5 3 3.5 4 0.2 0.4 0.6 0.8 1 V2/V1=10 V2/V1=100 V2/V1=1000

Elliott and Lira: Chapter 11 - Activity Models Slide 17

• 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 ln( / ) ( / ) γ b

b b b b

x x = + − Φ Φ 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.)

8803 . 1275 . ) 8696 . / 5010 . 1 ( ) 8696 . / 5010 . ln( ln = ⇒ − = − + =

b b

γ γ

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 18

• 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 ) /1.987(400 (0.4762) 9.2)

• 1.54E6(9.3

+ 455) 0.5238/0.5

• (1

+ 0.5455) ln(0.5238/ = ln

2 2 A

γ = -.0008 + 4.395 ⇒ γA = 81

) /1.987(400 (0.5238) 9.2)

• 1.68E6(9.3

+ 545) 0.4762/0.4

• (1

+ 0.4545) ln(0.4762/ = ln

2 2 B

γ

= +.0008 + 5.800 ⇒ γB = 330 Note: These high γ‘s actually lead to LLE discussed below.

Elliott and Lira: Chapter 11 - Activity Models Slide 19 Local Composition Theory Define a local mole fraction by: xij ≡Nij/Ncj Nij = number of "i" atoms around a "j" atom Ncj = Nij

i

∑

The local mole fraction can be related to the bulk mole fraction by x N VNc g r dr

ij i ij j ij R ij ij

ij

=

∫

σ π

3 2

4 where rij = r/σij Rij = "neighborhood" Further, we can write

ij j i jj jj jj ij ij ij jj j j ij i j jj ij

x x dr r g dr r g N Nc N Nc x x Ω ≡ =

∫ ∫

2 2 3 3

4 4

Noting

ij i j jj j jj ij i i ij

x x x x x x x Ω = Ω = =

∑ ∑ ∑

/ / 1

⇒

ij i i jj j ij i i j jj

x x x x x x Ω = ⇒ Ω =

∑ ∑

/ 1

⇒

∑

Ω Ω =

k kj k ij i ij

x x x

Elliott and Lira: Chapter 11 - Activity Models Slide 20 Example 11.12(p383). Compute the local compositions for the following lattice based

• n rows and columns away from the edges.

O O X O X O X X X X X O X O O X O X O X O X O X X O O X O X X X O O# 1 2 3 4 5 6 7 8 9 #X’S 3 3 3 2 1 1 2 2 = 17 #O’S 2 1 3 1 1 = 8 xxo = 17/25; xo = 9/22; Ωxo = (17/8)*(9/13) = 1.47

Elliott and Lira: Chapter 11 - Activity Models Slide 21 Obtaining the Free energy from the local compositions Recalling the energy equation for mixtures, U U RT x x N u RT N r dr

ig i j A ij ij A

−       =

∑ ∑ ∫

ρ π 2 4

2

g We would like to specify some (uij)avg ≡ εij such that N u RT N r dr N RT N r dr

A ij ij A A ij ij A

∫ ∫

= g g 4 4

2 2

π ε π ⇒

U U RT x n N V N RT g r dr

ig j i A ij A ij ij ij ij

−       =

∑ ∑ ∫

1 2 4

3 2

σ ε π Substituting Ncj, Λij, and xij into the energy equation for mixtures

( ) U U x Nc x

ig j j ij i

− = ∑

∑

1 2 j ij

ε ~(11.77) If we assume that Ncj = Nci ≡ z where z is assumed to be the same coordination number for all the components,

( )

jj ij i ij j j j E

• x

Nc x U

∑ ∑

= 2

1

;

) (

2 1 ij ij i k ij i ij i j j j E

• x

x Nc x U

∑∑ ∑

Ω Ω =

(11.80)

Elliott and Lira: Chapter 11 - Activity Models Slide 22 Obtaining the Free energy from the local compositions A = U - TS ⇒ A/RT = U/RT - S/R T A RT T T RT U T TU RT T R S T Cv R U RT T R Cv T U RT

V V V

∂ ∂ ∂ ∂ ∂ ∂ ( / )       =       − −       = − − = −

2

A RT U RT dT T C

E E

= − +

∫

where C is an integration constant. Recall the analogous procedure for regular solutions (i.e.

) ( ) (

2 2 1 1 2 2 1 2 1

V x V x U E + − Φ Φ = δ δ

) isindependent

• f temperature, so it can be factored out of the intgral, and

A RT U R dT T C U RT C

E E E

= − + = +

∫

2

For local composition theory, we just need to repeat this complete procedure but recognize that U

E 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 respect to T becomes very simple. Then,

C x x RT A

i i j j E

+ ln

ij

       Ω − =

∑ ∑

Elliott and Lira: Chapter 11 - Activity Models Slide 23 Wilson’s equation Ncj =2 for all j at all ρ; Bij = Vj/Vi ; C = 0

ln

ji 

       Λ − =

∑ ∑

i i j j E

x x RT G

⇒

( )

                Λ − =

∑ ∑

n n n RT G

i i j j E

ln

• ln

ji

Taking the last term first:

( ) [ ]

      + = ∂ ∂ = + ∑ n n n n n n n n n n

k j j

1 ln ) ln ( ); ln( ln

∑ ∑ ∑ ∑ ∑

          −       = ∂             ∂

j ji i i jk j ki i i k ji i i j j

n n n n n n ln ln

{ }

∑ ∑ ∑ ∑ ∑ ∑

          −       − =           −       −       + = ∂ ∂ =

j ji i i jk j ki i i j ji i i jk j ki i i k E k

x x x n n n n n n n RT G ln 1 ln 1 ln / lnγ

Elliott and Lira: Chapter 11 - Activity Models Slide 24 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) where Φ j

j j i i i

x r x r ≡ ∑

; θ j

j j i i i

x q x q ≡ ∑

; r

n r

j kj kj k

= ∑

; q

n q

j kj kj k

= ∑

; B

q x q

ij i j j j

≡ ∑

( ) ( )

∑ ∑ ∑ ∑

Φ Φ +         Ω − =

j j j j j j j j j ij i i j j j E

/ x q

• /x

x x x q RT G θ ln 5 ln ln

ln ln ln γ γ γ

k k COMB k RES

= +

( ) ( ) ( ) ( ) [ ]

k k k k k k k k k COMB k

q x x θ θ γ / 1 / ln 5

• /

1

• /

ln ln Φ − − Φ Φ − Φ =

          Ω Ω −         Ω − =

∑∑ ∑

j i ij i kj j ik i i k RES k

x x x q ln 1 lnγ

Elliott and Lira: Chapter 11 - Activity Models Slide 25

• 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: P1

sat= 3.23 mmHg; P2 sat = 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 26 The activity coefficients from the Wilson equation are:

22 2 21 1 21 2 12 2 11 1 11 1 12 2 11 1 1

) ln( 1 ln Λ + Λ Λ − Λ + Λ Λ − Λ + Λ − = x x x x x x x x γ

22 2 21 1 22 2 12 2 11 1 12 1 22 2 21 1 2

) ln( 1 ln Λ + Λ Λ − Λ + Λ Λ − Λ + Λ − = x x x x x x x x γ Noting that Λ11= Λ22 =1, we can rearrange for binary mixtures to obtain the slightly simpler relations: ln ln( γ 1

1 11 2 12 2

1 = − + x x x Q Λ Λ ) + ln ln( γ 2

1 21 2 22 1

1 = − + − x x x Q Λ Λ )

where Q x x x x = + − + Λ Λ Λ Λ

12 1 2 12 21 1 21 2

Q = 0.070/(0.2+0.8*0.070) - 0.625/(0.8+0.2*0.625) = -0.4022

ln ln( . . * . ) γ 1 1 0 2 08 0 070 = − + + .8 Q = 1.0408 ⇒ γ1= 2.824 ln ln( . . *. ) . γ 2 1 08 0 20 625 0 2 = − + − Q = 0.1584 ⇒ γ2= 1.172

P = (y1+y2)P = x1γ1 P1

sat

+ x2 γ2 P2

sat

= 0.2*2.824*3.23 + 0.8*1.172*187.1 = 177.2 mmHg y1 = x1γ1 P1

sat /P = 0.2*2.824*3.23/177.2 = 0.0103

Elliott and Lira: Chapter 11 - Activity Models Slide 27 Question: What value for Ωij is implied by the van der Waals EOS?

Z b a RT = − − 1 1 ρ ρ

b = Σxibi is reasonable. As for "a", we must carefully consider how this term relates to the energy of mixing:

U U RT a RT N x x N u RT g r dr

ig A i j A ij ij

− = − =

∑ ∑ ∫

ρ ρ π 2 4

2

Comparing to the result for pure fluids a N u g r dr

ii N A ii ii

A

= −

∫

2 2

4π ⇒

U U RT N RT x x a a x x a

ig A i j ij i j ij

− = − ⇒ =

∑ ∑ ∑ ∑

ρ

where a

N u g r dr

ij N A ij ij

A

≡ −

∫

2 2

4π

where we set aij= a a

ii jj (1 - kij),

⇒

3 3 2 2 2 2

~ 4 4

jj ij ij jj jj ij jj jj A N ij ij A N ij

a a dr r g u N dr r g u N

jj A ij A

σ σ ε ε π π

ε ε

= − − ≡ Ω

∫ ∫