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Introductory Chemical EngineeringThermodynamics

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

Chapter 11 - Activity Models

Elliott and Lira: Chapter 11 - Activity Models Slide 1

NONIDEAL SOLUTIONSWhen a solution does not follow the ideal solution approximation we can apply an EOSor the "correction factor", γi, yielding the general expression for K-ratio

KP

P

V P P RTi

iL

ivap

iV

isat

ivap

i

=−

γ

γϕ

ϕexp[ ( ) / ]

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

nRT

G

nRT

G

nRT

G

nRTx

G

nRTx

E is

ii

i≡ − = − +

∑ ln( )

Letting γi ≡ fi /xi fi° where fi° ≡ f at T and P

G

nRT

x G

RT

x G

RTx

f

fx xi i i i i

ii

io i i i− =

−= =∑ ∑ ∑ ∑( )

ln(\$

) ln( )µ

γ

∆G

nRT

G

nRT

x G

RTx x x x x x x

Ei i

i i i i i i i i i≡ − − = − =∑ ∑ ∑ ∑ ∑ln( ) ln( ) ln( ) ln( )γ γ

∆G

RTn

E

i i= ∑ ln( )γ

Hence we see that the activity coefficient gives a correction to the ideal solution estimateof the Gibbs energy, component by component.

Elliott and Lira: Chapter 11 - Activity Models Slide 2

Activity coefficients as derivativesShow that expressions for all the activity coefficients can be derived once a singleexpression for the Gibbs excess energy is available.

Given: ∆G

RTn

E

i i= ∑ ln( )γ Prove: ln( / )γ ∂

∂j

E

j

G RT

n= ∆

∂∂

γ∂∂

∂ γ∂

( / )ln

ln∆G RT

n

n

nn

n

E

ji

i

ji

i

j

=

+

∑ ∑

=≠

=jiif

jiif

n

n

j

i

1

0

∂∂

⇒ ln lnγ∂∂

γii

jj

n

n∑

=

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

By definition, ( ) ( )RTd d n n n n RTi i i i j i i jln ln / / /γ µ ∂ γ ∂ ∂µ ∂≡ ⇒ =∑ ∑But, Gibbs-Duhem ( )n ni i j∑ =∂µ ∂/ 0

Therefore ( )n ni 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 EquationPerhaps the simplest expression for the Gibbs excess function is the 1-ParameterMargules (also known as the two-suffix Margules).∆G

nRT

A

RTx 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

RTx x

E

1

2 12

2 12 1

11 1= −

= −

= −

⇒ =lnγ 1 22A

RTx

Elliott and Lira: Chapter 11 - Activity Models Slide 4

Example. VLE prediction using UNIFAC activity coefficientsThe isopropyl alcohol (IPA) + water (w) system is known to form an azeotrope atatmospheric 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 Tmaxwater 1-H2O 8.87829 2010.33 252.636 -26 83IPA 2-CH3; 1-CH, 1-OH 8.07131 1730.63 233.426 1 100Entering the mole fractions and 80.37°C ⇒ γw = 2.1108; γipa =1.0886 T Pipa

vap Pwvap x Pi i

vap∑ yw80.37 695 360 757 0.315882.50 760 395 829 0.316480.46 697 361 760 0.3158Since 0.3158 ≠ 0.3146, we did not find the azeotrope yet.Try xw = 0.3168 ⇒ γw = 2.1053; γipa =1.0898 T Pw

sat Pipasat ΣxiPi

sat yw80.46 697 361 760 0.3168Since xw = 0.3168 = yw this must be the composition of the azeotrope estimated byUNIFAC. UNIFAC seems to be fairly accurate for this mixture. Also note that T vs. x isfairly flat near an azeotrope.

Elliott and Lira: Chapter 11 - Activity Models Slide 5

"Regular" SolutionsThe 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 amolar average for the volume of mixing.

U U

RT RTx x a

VRTx x a

ig

i j ij i j ij

= − = −∑∑ ∑∑ρ 1

V = ΣxiVi according to "regular solution theory,"

( )U Ux x a

x Vig i j ij

i i

− =− ∑∑

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

( ) ( )U Ua

VU U x a Vig

i

ii

i

ig

is i ii i− =−

⇒ − = −∑ /

For a binary mixture, subtracting the ideal solution result to get the excess energy gives,

U xa

Vx

a

V

x a x x a x a

x V x VE = + −

+ ++

1

11

12

22

2

12

11 1 2 12 22

22

1 1 2 2

2

Elliott and Lira: Chapter 11 - Activity Models Slide 6

Collecting a common denominator

U

xa

Vx V x V x

a

Vx V x V x a x x a x a

x V x VE =

+ + + − + +

+

111

11 1 2 2 2

22

21 1 2 2 1

211 1 2 12 2

222

1 1 2 2

2( ) ( ) ( )

U

x a V x x aV

Vx a V x x a

V

Vx a x x a x a

x V x VE =

+ + + + − + +

+

12

11 1 1 2 112

122

22 1 1 2 221

212

11 1 2 12 22

22

1 1 2 2

2( )

Ux x a

V

Vx x a

V

Vx x a

V V

V V

x V x VE =

+ −

+

1 2 112

11 2 22

1

21 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 a11 22 , and collecting terms in a slightlysubtle way,

Ux 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

VE =

++ −

=

+−

1 2 1 2

1 1 2 2

11

12

22

22

11

12

22

22

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 VE = − +Φ Φ1 2 1 2

2

1 1 2 2δ δ ( )

where Φ i i i i ix V x V≡ ∑/ is known as the " volume fraction"

δ i ii ia 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 datanear typical conditions be used instead of the critical point.δi iUvap 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 estimatewhether the ideal solution will be accurate or not.Alkanes Olefins Napthenics Aromaticsn-pentane 7.0 1-pentene 6.9 cyclopentane 8.7 benzene 9.2n-hexane 7.3 1-hexene 7.4 cyclohexane 8.2 toluene 8.9n-heptane 7.4 1,3 butadiene 7.1 Decalin 8.8 ethylbenzene 8.8n-octane 7.6 styrene 9.3n-nonane 7.8 n-propylbenzene 8.6n-decane 7.9 anthracene 9.9

phenanthrene 9.8naphthalene 9.9

Turning to the free energy, with the elimination of excess entropy and excess volume atconstant pressure, we have,

( )∆ Φ ΦG U x V x VE E= = − +1 2 1 2

2

1 1 2 2δ δ ( )And the resulting activity coefficients are

( )RT vlnγ δ δ1 1 22

1 2

2= −Φ ( )RT vlnγ δ δ2 2 1

21 2

2= −Φ

Elliott and Lira: Chapter 11 - Activity Models Slide 8

More Solubility Parameters in (cal/cc)½For oxygenated hydrocarbons and amines, the solubility parameters tend to be larger.This is largely a reflection of the higher heats of vaporization resulting from hydrogenbonding, but also from the polar moments typical of these components.Alcohols Amines Ethers Ketoneswater 23.4 ammonia 16.3 dimethyl ether 8.8 acetone 9.9methanol 14.5 methyl amine 11.2 diethyl ether 7.4 2-butanone 9.3ethanol 12.5 ethyl amine 10.0 dipropyl ether 7.8 2-pentanone 8.7n-propanol 10.5 pyridine 14.6 furan 9.4 2-heptanone 8.5n-butanol 13.6 THF 9.1n-hexanol 10.7n-dodecanol 9.9

We can also obtain a compromise by assuminga12= a a11 22 (1-kij)where kij is an adjustable parameter also called the binary interaction coefficientThe activity coefficient expressions become

( )RT V kln γ δ δ δ δ1 1 22

1 2 12 1 2

22= − +Φ ; ( )RT V kln γ δ δ δ δ2 2 1

21 2 12 1 2

22= − +Φ

Elliott and Lira: Chapter 11 - Activity Models Slide 9

Example. VLE Predictions using regular solution theoryBenzene and cyclohexane are to be separated by distillation at 1 bar. Use regular solutiontheory to predict whether an azeotrope should be expected forthis mixture.

Tc (K) Pc (bar) ω Vi(cc/mol) δ(cal/cc)½Benzene 562.2 48.98 0.211 89 9.2Cyclohexane 553.5 40.75 0.215 109 8.2Solution: Consider y vs. x at x =0.01 and 0.99. If yB >xB at xB =0.01 and yB <xB atxB =0.99, then yB =xB (i.e. there is an azeotrope) somewhere in between. If y >x or y<xfor all xB, then there is no azeotrope. Given xB and P, we should perform bubble pointtemperature calculations.At xB =0.99, guess T=350K ⇒ ΦB = 0.99(89)/[0.99(89)+0.01(109)] = 0.9878PB

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

sat = 40.75*10**[7/3*1.215*(1-553.5/350)]= 0.9158 barlnγB = 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 = 1.1652Σyi = Σxiγi Pi

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

Guess T=353K ⇒ PBsat = 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 Pisat/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)=351.73Guess T=351.73K⇒PB

sat=0.9981;PCsat=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 < 0.99At 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 = 1.1330Σ yi = Σ xiγi Pi

sat/P = 0.01(1.036)1.1330 + 0.99(0.9997)1.00 = 1.0014 ⇒yB=.0138Therefore, (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 Pisat/(PΣyi ) is an accurate estimate.

4. If PBsat ≈ PC

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

Elliott and Lira: Chapter 11 - Activity Models Slide 11

Van Laar’s EquationsThe regular solution equations can easily be rearranged into the van Laar form by writingtwo adjustable parameters, A12 and A21.

( )AV

RT121

1 2

2= −δ δ ; ( )221

221 δδ −=

RT

VA ;

A

A

V

V12

21

1

2

=

NOTE: Do NOT estimate A12 and A21 from δ1 and δ2. This how we rename thisparticular grouping of parameters to obtain two adjustable parameters, A12 and A21.

)( 212121

212112

AxAx

xx

RT

AA

RT

U

RT

G EE

+==∆

Giving expressions for the activity coefficients,

lnγ 112

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 thevalues of A12 and A21. These equations can be rearranged to obtain A12 and A21 from γ1and γ2 given any one VLE point.

Ax

x12 12 2

1 1

2

1= +

ln

ln

lnγ

γγ

Ax

x21 21 1

2 2

2

1= +

ln

ln

lnγ

γγ (11.29)

Elliott and Lira: Chapter 11 - Activity Models Slide 12

Example. Application of the Van Laar equationA particularly useful data point for VLE is the azeotrope because1) x1=y1 ⇒ γ1 = P/P1

sat; γ2 = P/P2sat

2) Many tables of known azeotropes are commonly available3) The location of an azeotrope is very important for distillation design.Consider the benzene(1)+ethanol(2) system which exhibits an azeotrope at 760 mmHgand 68.24 °C containing 44.8 mol% Ethanol. Calculate the composition of the vapor inequilibrium with an equimolar liquid solution at 760 mmHg given the Antoine constantslog 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; P2sat = 503.4 mmHg

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

Ax

x12 12 2

1 1

2

1= +

ln

ln

lnγ

γγ A

x

x21 21 1

2 2

2

1= +

ln

ln

lnγ

γγ

= 1.3424 ; = 1.8814

Elliott and Lira: Chapter 11 - Activity Models Slide 13

Now consider x1 = x2 = 0.5

lnγ 112

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 requiredGuess T=60°C ⇒ P1

sat = 391.5 mmHg; P2sat = 351.9 mmHg

yi = xi γi P1sat /P ⇒ y1 = 0.407; y2 = 0.321; Σyi = 0.728 ⇒ T guess is too low.

at T = 68.24°C, P1sat = 519.6 mmHg; P2

sat = 503.4 mmHgyi = 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 TheoryThe volume occupied by one molecule is not accessible to the other molecules. When wemix two components, each component’s entropy increases according to how much morespace it has: ∆Si = Ni k ln(V Vf fm i

/ )

where Vfm = the free volume of the mixture

Vfi = the free volume in the ith pure componentIt is customary to assume that the fraction of free volume in any component is the same. Vfi = Nivi vfwhere vi = volume of the ith species vf = universal fraction of free volumeThe entropy may be taken as that of a perfect gas composed of the same number ofmolecules confined to a volume equal to the free volume.∆S

Nkx

V

Vx

V

Vf

f

f

f

m m= +1 1

1 2

ln( ) ln( )

∆ ΦS

Nkx

n v n v

n vx

n v n v

n vxi i=

++

+= −∑1

1 1 2 2

1 12

1 1 2 2

2 2

ln( ) ln( ) ln

∆ Φ ΦS

Nkx 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 ΦΦ Φ

G

NkT

H

NkT

S

Nkx

xx

x RTx v x v

E E E

= − = + +−

+11

12

2

21 2

1 2

2

1 1 2 2ln ln ( )δ δ

( )ln ln( / ) ( / )γ δ δ1 1 1 1 11

22

1 2

21= + − + −Φ Φ Φx x

v

RT

( )ln ln( / ) ( / )γ δ δ2 2 2 2 22

12

1 2

21= + − + −Φ Φ Φx x

v

RT

Elliott and Lira: Chapter 11 - Activity Models Slide 16

x1 V1/V210 100 1000

0 0 0 00.05 0.07 0.18 0.29

0.1 0.14 0.36 0.590.15 0.2 0.53 0.87

0.2 0.26 0.7 1.160.25 0.32 0.87 1.44

0.3 0.38 1.03 1.720.35 0.43 1.19 1.99

0.4 0.47 1.34 2.250.45 0.52 1.48 2.51

0.5 0.55 1.62 2.760.55 0.58 1.75 3

0.6 0.61 1.86 3.230.65 0.62 1.96 3.44

0.7 0.62 2.04 3.630.75 0.6 2.1 3.8

0.8 0.57 2.11 3.920.85 0.51 2.07 3.98

0.9 0.41 1.93 3.920.95 0.26 1.55 3.59

0.975 0.15 1.13 3.081 0 0 0

x1

Exc

ess

En

tro

py/

Nk

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

0 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 coefficientConsider 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 andVps 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 bx x= + −Φ Φ1Benzene is comprised of 6(ACH) groups @ 0.5313 R-units per group ⇒ Vb ~3.1878Pentastyrene 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 ~21.17Mb = 78 and Mps = 522 ⇒ xb = 0.8696Φb = 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.)

8803.01275.0)8696.0/5010.01()8696.0/5010.0ln(ln =⇒−=−+= bb γγNote: The activity of benzene is soaked up like a sponge if there is no energeticcontribution.

Elliott and Lira: Chapter 11 - Activity Models Slide 18

Example. Polymer mixingSuppose 1g each of two different polymers (polymer A and polymer B) is heated to127°C and mixed as a liquid. Estimate the activity coefficients of A and B usingScatchard-Hildebrand theory combined with the Flory-Huggins combinatorial term. MW V δ(cal/cc)½A 10,000 1,540,000 9.2B 12,000 1,680,000 9.3Solution:xA = (1/10,000)/(1/10,000+1/12,000) = .5455; xB = .4545ΦA = 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 22Aγ

= -.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 22Bγ = +.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 TheoryDefine a local mole fraction by:xij ≡Nij/NcjNij = number of "i" atoms around a "j" atom

Ncj = Niji

∑The local mole fraction can be related to the bulk mole fraction by

xN

VNcg r drij

i ij

jij

R

ij ij

ij

= ∫σπ

3

0

24

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

ijj

i

jjjjjj

ijijij

jjjj

ijij

jj

ij

x

x

drrg

drrg

NNc

NNc

x

xΩ≡=

∫∫

2

2

3

3

4

4

Noting ijijjjjjjijii

ij xxxxxxx Ω=Ω== ∑∑∑ //1

⇒ iji

ijj

jij

iijjj x

x

xxxx Ω=⇒Ω= ∑∑/1 ⇒ ∑ Ω

Ω=

kkjk

ijiij

x

xx

Elliott and Lira: Chapter 11 - Activity Models Slide 20

Example 11.12(p383). Compute the local compositions for the following lattice basedon rows and columns away from the edges.

O O X OX O X XX X X O X O

O X O XO X

O X O XX 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 0 2 2 = 17#O’S 2 0 0 0 1 0 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 compositionsRecalling the energy equation for mixtures,

U U

RTx x

N u

RTN r dr

ig

i jA ij

ij A

= ∑∑ ∫

ρ π2

4 2 g

We would like to specify some (uij)avg ≡ εij such thatN u

RTN r dr

N

RTN r drA ij

ij AA ij

ij A∫ ∫= g g 4 42 2πε

π ⇒

U U

RTx

n N

V

N

RTg r dr

ig

ji A ij A ij

ij ij ij

= ∑∑ ∫

1

24

32

σ επ

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

( )U U x Nc xigj

jij

i

− = ∑ ∑12 j ij ε ~(11.77)

If we assume that Ncj = Nci ≡ z where z is assumed to be the same coordination numberfor all the components,

( )jjiji

ijjj

jE -xNcxU ∑∑= 2

1; )(2

1ijij

ik

iji

ijij

jj

E -x

xNcxU ∑ ∑∑ 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

TA RT

T

T

RT

U

T

TU

RT

T

R

S

T

Cv

R

U

RT

T

R

Cv

T

U

RTV V V

∂∂

∂∂

∂∂

( / )

=

− −

= − − = −2

A

RT

U

RT

dT

TC

E E

= − +∫ where C is an integration constant. Recall the analogous

procedure for regular solutions (i.e. )()( 22112

2121 VxVxU E +−ΦΦ= δδ ) isindependentof temperature, so it can be factored out of the intgral, andA

RT

U

R

dT

TC

U

RTC

E E E

= − + = +∫ 2

For local composition theory, we just need to repeat this complete procedure butrecognize 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, Cxx

RT

A

ii

jj

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−= ∑∑

ii

jj

E

xxRT

G⇒ ( )

Λ−= ∑∑ nnn

RT

G

ii

jj

E

ln- ln ji

Taking the last term first:

( )[ ]

+=

∂∂=+ ∑

nnn

n

nnnnnn

kjj

1ln

)ln();ln(ln

∑ ∑∑∑∑

=

j jii

i

jkjki

ii

k

jii

ij

j

nnn

n

nn

ln

ln

∑ ∑∑∑ ∑∑

−=

+=

∂∂=

j jii

i

jkjki

ii

j jii

i

jkjki

ii

k

E

k xxx

nnn

nnn

n

RTGln1ln

1ln

/lnγ

Elliott and Lira: Chapter 11 - Activity Models Slide 24

UNIFAC and UNIQUACAbrams, 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 jj j

i ii

x r

x r≡

∑ ; θ jj j

i ii

x q

x q≡

∑ ; r n rj kj kjk

= ∑ ; q n qj kj kjk

= ∑ ; Bq

x qij

i

j jj

≡∑

( ) ( )∑∑∑∑ ΦΦ+

Ω−=

jjjjj

jjjjij

ii

jjj

E

/xq -/xxxxqRT

G θln5lnln

ln ln lnγ γ γk kCOMB

kRES= +

( ) ( ) ( ) ( )[ ]kkkkkkkkkCOMBk qxx θθγ /1/ln5-/1- /lnln Φ−−ΦΦ−Φ=

ΩΩ

Ω−= ∑ ∑∑

ji

iji

kjjik

iik

RESk x

x xq ln1lnγ

Elliott and Lira: Chapter 11 - Activity Models Slide 25

Example. Application of Wilson’s equation to VLEFor the binary system n-pentanol(1)+n-hexane(2), the Wilson equationconstants are A12 = 1718 cal/mol A21 = 166.6 cal/molAssuming the vapor phase to be an ideal gas, determine the composition of the vapor inequilibrium with a liquid containing 20 mole percent n-pentanol at 30xC. Also calculatethe equilibrium pressure.Given: P1

sat= 3.23 mmHg; P2sat = 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/molNote: 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:

222211

212

122111

1111221111 )ln(1ln

Λ+ΛΛ−

Λ+ΛΛ−Λ+Λ−=

xx

x

xx

xxxγ

222211

222

122111

1212222112 )ln(1ln

Λ+ΛΛ−

Λ+ΛΛ−Λ+Λ−=

xx

x

xx

xxxγ

Noting that Λ11= Λ22 =1, we can rearrange for binary mixtures to obtain the slightlysimpler relations:ln ln(γ 1 1 11 2 12 21= − +x x x QΛ Λ ) + ln ln(γ 2 1 21 2 22 11= − + −x x x QΛ Λ )

where Q x x x x=

+−

ΛΛ

Λ12

1 2 12

21

1 21 2

Q = 0.070/(0.2+0.8*0.070) - 0.625/(0.8+0.2*0.625) = -0.4022ln ln( . . * . )γ 1 1 0 2 0 8 0 070 0= − + + .8 Q = 1.0408 ⇒ γ1= 2.824ln ln( . . *. ) .γ 2 1 08 0 20 625 0 2= − + − Q = 0.1584 ⇒ γ2= 1.172P = (y1+y2)P = x1γ1 P1

sat + x2 γ2 P2sat

= 0.2*2.824*3.23 + 0.8*1.172*187.1 = 177.2 mmHgy1 = 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?

Zb

a

RT=

−−1

1 ρρ

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

RT

a

RT

Nx x

N u

RTg r dr

igA

i jA ij

ij

− = − = ∑∑ ∫ρ ρ

π2

4 2

Comparing to the result for pure fluids

a N u g r driiN

A ii iiA= − ∫2

24π ⇒U U

RT

N

RTx x a a x x a

igA

i j ij i j ij

− = − ⇒ =∑∑ ∑∑ρ

where a N u g r drijN

A ij ijA≡ − ∫2

24π where we set aij= a aii jj (1 - kij),

⇒ 3

3

22

22

~4

4

jj

ij

ij

jj

jj

ij

jjjjAN

ijijAN

ij a

a

drrguN

drrguN

jj

A

ij

A

σσ

εε

π

π

ε

ε=

−≡Ω

∫∫