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 iL Pi vap  ϕ isat exp[V ( P − Pi vap ) / RT ] 
Ki =


ϕi
γ iV P 

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
≡
−
=
− ∑ x i  i + ln( x i
nRT nRT nRT nRT
 nRT

)

γi ≡ fi /xi fi° where fi° ≡ f at T and P
x i Gi
x i ( µ i − Gi )
f$i
G
∑
−
=∑
= ∑ xi ln( o ) = ∑ xi ln( x i γ i )
nRT
RT
RT
fi
E
∆G
G
∑ xi Gi − x ln( x ) = x ln( x γ ) − x ln( x ) = x ln( γ )
≡
−
∑ i i ∑ i i i ∑ i i ∑ i i
nRT nRT
RT
E
∆G
= ∑ ni ln( γ i )
RT
Hence we see that the activity coefficient gives a correction to the ideal solution estimate
of the Gibbs energy, component by component.
Letting
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
Given: RT =
E
∂( ∆G E / RT )
=
∂n j
∑n
∂( ∆G / RT )
ln
γ
=
j
Prove:
∂n j
E
i
ln( γ i )
 ∂ni 
 ∂ ln γ i 
∑ ln γ i  ∂n  + ∑ ni  ∂n 
 j 

j 

 ∂n i 
∂ni 0 if i ≠ j
=
∑ ln γ i  ∂n  = ln γ j
∂n j 1if i = j ⇒
 j
As for the second sum, we must show that it goes to zero.
By definition, RTd ln γ i ≡ dµ i ⇒ ∑ ni ( ∂ ln γ i / ∂n j ) = ∑ ni ( ∂µ i / ∂n j ) / RT
But, Gibbs-Duhem
Therefore
∑ n ( ∂µ
∑ n ( ∂ ln γ
i
i
i
i
)
/ ∂n j = 0
)
/ ∂n j = 0
Gibbs-Duhem for activity coefficients
∂( ∆ G / RT )
E
=
ln
γ
j
Combining these results,
So,
G
(T,P,x), → γ’s.
∂n j
E
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
x1 x2
=
nRT
RT
Derive the expressions for the activity coefficients from this expression.
Solution:
E
An2 n1
∆G
=
RT
RT n
E
∂ ( ∆G / RT ) An2  1 n1 
A n2  n1 
A
x 2 ( 1 − x1 )
=
−
=
=
−
1
∂n1
RT  n n 2  RT n 
n  RT
A 2
x
⇒ ln γ 1 =
RT 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 ⇒ γw = 2.1108; γipa =1.0886
vap
T
yw
Pipa
Pwvap
∑ x i Pi vap
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
Pwsat
Pipasat
ΣxiPisat
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.
 U − U ig  − ρ
−1
=
x
x
a
=
x i x j a ij
∑
∑
∑
∑
i j ij


VRT
 RT  RT
V = ΣxiVi according to "regular solution theory,"
− ∑ ∑ x i x j a ij
U − U ig =
∑ x i Vi
For the pure fluid, taking the limit as xi→1,
− a ii
U − U ig =
⇒ U − U ig = − ∑ x i a ii / Vi
i
is
Vi
For a binary mixture, subtracting the ideal solution result to get the excess energy gives,
a 11
a 22  x12 a 11 + 2 x 1 x 2 a 12 + x 22 a 22 
E
U = x1
+ x2
−

V2 
x 1V1 + x 2V2
V1

(
)
(
)
(
)
Elliott and Lira: Chapter 11 - Activity Models
Slide 5
Collecting a common denominator
a 11
a
( x 1V1 + x 2V2 ) + x 2 22 ( x 1V1 + x 2V2 ) − ( x 12 a 11 + 2 x 1 x 2 a 12 + x 22 a 22 )
V2
V1
UE =
x 1V1 + x 2V2
V
V
x12 a11V1 + x1 x2 a11 2 + + x22 a22V1 + x1 x2 a22 1 − ( x12 a11 + 2 x1 x2 a12 + x22 a22 )
V1
V2
UE =
x1V1 + x2V2
V
V
VV
x1 x2 a11 2 + x1 x2 a22 1 − 2 x1 x2 a12 2 1
V1V2
V1
V2
UE =
x1V1 + x2V2
x1
Scatchard and Hildebrand now make an assumption which is very similar to assuming
kij=0 in an equation of state. Setting a12= a11 a 22 , and collecting terms in a slightly
subtle way,
2


a 11
a 22
x x VV a
a
a a 
x x VV
U E = 1 2 1 2  112 + 222 − 2 112 222  = 1 2 1 2 
−

x1V1 + x 2V2  V1
V2 
V2
V1 V2  x 1V1 + x 2V2  V1
and finally, defining a term called the "solubility parameter"
2
U E = Φ 1Φ 2 ( δ 1 − δ 2 ) ( x1V1 + x 2V2 )
Φ i ≡ xiVi / ∑ xiVi is known as the " volume fraction"
where
δ i ≡ a ii / Vi 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 i ≡ ∆Uvap / Vi (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,
2
∆G E = U E = Φ 1Φ 2 ( δ 1 − δ 2 ) ( x 1V1 + x 2V2 )
And the resulting activity coefficients are
RT ln γ 1 = v1Φ ( δ 1 − δ 2 )
2
2
2
RT ln γ 2 = v 2 Φ (δ 1 − δ 2 )
Elliott and Lira: Chapter 11 - Activity Models
2
1
2
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
a12= a a (1-kij)
where kij is an adjustable parameter also called the binary interaction coefficient
The activity coefficient expressions become
2
2
RT ln γ 1 = V1Φ 22 ( δ 1 − δ 2 + 2 k 12 δ 1 δ 2 ) ;
RT ln γ 2 = V2 Φ 12 (δ 1 − δ 2 + 2 k 12 δ 1δ 2 )
11 22
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)
Vi(cc/mol)
ω
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
PBsat= 48.98*10**[7/3*1.211*(1-562.2/350)]= 0.9481 bar
PCsat = 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 Pisat/P = 0.99(0.9481)1.00002+0.01(0.9158)1.1652 = 0.9493 ⇒yB =0.9887
Guess T=353K ⇒ PBsat = 1.036; PCsat = 0.9997; γB=1.00; γC =1.1652*353/350=1.1752
Elliott and Lira: Chapter 11 - Activity Models
Slide 9
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)=351.73
Guess T=351.73K⇒PBsat=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 < 0.99
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 Pisat/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 Pisat/(PΣyi ) is an accurate estimate.
4. If PBsat ≈ PCsat then a small non-ideality can cause an azeotrope.
Elliott and Lira: Chapter 11 - Activity Models
Slide 10
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.
A12 V1
V
2
V
=
A12 = 1 (δ 1 − δ 2 ) ;
A21 = 2 (δ 1 − δ 2 )2 ;
RT
A
V2
RT
21
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.
∆G E U E A12 A21
x1x2
=
=
RT
RT
RT ( x1 A12 + x2 A21 )
Giving expressions for the activity coefficients,
A12
A21
ln γ 1 =
ln
γ
=
2
2
2


A12 x1  ;
A21 x 2 
(11.28)
1
+
1
+



A21 x 2 
A12 x1 


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.
 x 2 ln γ 2 
A12 = ln γ 1 1 +

x1 ln γ 1 

2

x1 ln γ 1 
A21 = ln γ 2 1 +

 x 2 ln γ 2 
Elliott and Lira: Chapter 11 - Activity Models
2
(11.29)
Slide 11
Example. Application of the Van Laar equation
A particularly useful data point for VLE is the azeotrope because
1) x1=y1 ⇒ γ1 = P/P1sat; γ2 = P/P2sat
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 P1sat = 6.8975 - 1206.35/(T+220.24)
log P2sat = 8.1122 - 1592.86/(T+226.18)
Solution:
at T = 68.24°C, P1sat = 519.6 mmHg; P2sat = 503.4 mmHg
γ1 = 760/519.6 = 1.4627; γ2 = 760/503.4 = 1.5097
x1 = 0.552
; x2 = 0.448
2
2
 x 2 ln γ 2 

x1 ln γ 1 
A12 = ln γ 1 1 +
A21 = ln γ 2 1 +


x1 ln γ 1 

 x 2 ln γ 2 
= 1.3424
;
= 1.8814
Elliott and Lira: Chapter 11 - Activity Models
Slide 12
Now consider x1 = x2 = 0.5
A12
A21
ln γ 1 =
ln γ 2 =
2
2



A12 x1 ;
A21 x 2 
1 + A x 
1 + A x 


21 2 
12 1 
γ1 = 1.580;
γ2=1.386
Problem statement ⇒ bubble point temperature is required
Guess T=60°C ⇒ P1sat = 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; P2sat = 503.4 mmHg
yi = xi γi Pisat /P ⇒ y1 = 0.540; y2 = 0.459; Σyi = 0.999 ⇒ T guess is practically Taz.
Elliott and Lira: Chapter 11 - Activity Models
Slide 13
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 f m / V f i )
V f m = the free volume of the mixture
where
V f i = 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 f i = 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.
V fm
V fm
∆S
= x1 ln(
) + x1 ln(
)
V f2
V f1
Nk
n v + n2 v 2
n v + n2 v 2
∆S
= x1 ln( 1 1
) + x 2 ln( 1 1
) = − ∑ x i ln Φ i
Nk
n1 v1
n2 v 2
∆S E
= − ∑ x i ln Φ i + ∑ x i ln x i = − ∑ x i ln( Φ i / x i )
Nk
Elliott and Lira: Chapter 11 - Activity Models
Slide 14
For a binary solution,
(δ
1 −δ2)
Φ
Φ
∆G
H
∆S
=
−
= x1 ln 1 + x 2 ln 2 + Φ1Φ2
NkT
NkT
Nk
x1
x2
RT
v1 2
2
ln γ 1 = ln(Φ1 / x1 ) + (1 − Φ1 / x1 ) +
Φ2 ( δ 1 − δ 2 )
RT
v2 2
2
ln γ 2 = ln(Φ2 / x2 ) + (1 − Φ2 / x 2 ) +
Φ1 ( δ 1 − δ 2 )
RT
E
E
E
Elliott and Lira: Chapter 11 - Activity Models
2
( x1v1 + x 2 v 2 )
Slide 15
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
0.975
1
10
0
0.07
0.14
0.2
0.26
0.32
0.38
0.43
0.47
0.52
0.55
0.58
0.61
0.62
0.62
0.6
0.57
0.51
0.41
0.26
0.15
0
V1/V2
100 1000
0
0
0.18 0.29
0.36 0.59
0.53 0.87
0.7 1.16
0.87 1.44
1.03 1.72
1.19 1.99
1.34 2.25
1.48 2.51
1.62 2.76
1.75
3
1.86 3.23
1.96 3.44
2.04 3.63
2.1 3.8
2.11 3.92
2.07 3.98
1.93 3.92
1.55 3.59
1.13 3.08
0
0
4
V2/V1=1000
3.5
3
Excess Entropy/Nk
x1
2.5
V 2/V1= 100
2
1.5
1
V2/V1=10
0.5
0
-0.5
0
0.2
0.4
0.6
0.8
1
x1
Elliott and Lira: Chapter 11 - Activity Models
Slide 16
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 γ b = ln(Φb / x b ) + (1 − Φb / x b )
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.)
ln γ b = ln(0.5010 / 0.8696) + (1 − 0.5010 / 0.8696) = −0.1275 ⇒ γ b = 0.8803
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 17
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
lnγ A = ln(0.5238/0.5455) + (1 - 0.5238/0.5455) + 1.54E6(9.3 - 9.2)2 (0.4762)2 /1.987(400)
= -.0008 + 4.395 ⇒ γA = 81
lnγ B = ln(0.4762/ 0.4545) + (1 - 0.4762/0.4 545) + 1.68E6(9.3 - 9.2) 2 (0.5238) 2 /1.987(400 )
= +.0008 + 5.800 ⇒ γB = 330
Note: These high γ‘s actually lead to LLE discussed below.
Elliott and Lira: Chapter 11 - Activity Models
Slide 18
Local Composition Theory
Define a local mole fraction by:
xij ≡Nij/Ncj
Nij = number of "i" atoms around a "j" atom
Ncj = ∑ N ij
i
The local mole fraction can be related to the bulk mole fraction by
N i σij3 Rij
xij =
g ij 4πrij2 drij
∫
VNc j 0
where rij = r/σij
Rij = "neighborhood"
Further, we can write
xij
x jj
=
3
ij
Nc j N j 3jj
Nc j N i
Noting
⇒
∫ gij 4
∫ g jj 4
rij2 drij
r jj2 dr jj
≡
xi
Ω ij
xj
∑ xij = 1 = ∑ xi Ωij x jj / x j = x jj / x j ∑ xi Ωij
i
1 = x jj / x j ∑ xi Ωij ⇒
i
xj
x jj
= ∑ xi Ωij ⇒
xij =
i
Elliott and Lira: Chapter 11 - Activity Models
xi Ωij
∑ xk Ωkj
k
Slide 19
Example 11.12(p383). Compute the local compositions for the following lattice based
on rows and columns away from the edges.
O
X
X
O
X
X
O
O
X
O
X
X
O
O
X
X
X
O
O
O
X
O
X
O
O#
#X’S
#O’S
O
X
2
3
0
3
3
0
4
2
0
5
1
1
6
1
0
7
0
3
X
X
O
X
1
3
2
X
8
2
1
X
O
9
2 = 17
1 = 8
xxo = 17/25; xo = 9/22; Ωxo = (17/8)*(9/13) = 1.47
Elliott and Lira: Chapter 11 - Activity Models
Slide 20
Obtaining the Free energy from the local compositions
Recalling the energy equation for mixtures,
N A uij
 U − U ig  ρ
=
x
x
gij N A 4πr 2 dr
∑
∑
i j∫


RT
 RT  2
We would like to specify some (uij)avg ≡ εij such that
N A uij
N A ε ij
2
2
∫ RT gij N A 4πr dr = RT ∫ gij N A 4πr dr ⇒
ni N A σ ij3 N A ε ij
 U − U ig  1
g ij 4πrij2 drij
 = ∑∑x j

∫
V
RT
 RT  2
Substituting Ncj, Λij, and xij into the energy equation for mixtures
( U − U ig ) = 21 ∑ x j Nc j ∑ xij εij
j
i
~(11.77)
If we assume that Ncj = Nci ≡ z where z is assumed to be the same coordination number
for all the components,
U
E
=
1
2
∑x
j
j
Nc j ∑ xij ( ij i
jj
);
UE =
1
2
∑ x j Nc j ∑
j
i
xi Ω ij
∑x Ω
i
( ij - ij )
ij
(11.80)
k
Elliott and Lira: Chapter 11 - Activity Models
Slide 21
Obtaining the Free energy from the local compositions
A = U - TS ⇒ A/RT = U/RT - S/R
T  ∂U 
TU T  ∂S 
Cv U
T Cv
U
 ∂ ( A / RT ) 
 =

 −


−
=
−
−
=
−
T

V RT  ∂T V RT 2 R  ∂T V
∂T
R RT R T
RT
−U E dT
AE
=∫
+ C where C is an integration constant. Recall the analogous
RT
RT T
E
2
procedure for regular solutions (i.e. U = Φ 1Φ 2 (δ 1 − δ 2 ) ( x1V1 + x2V2 ) ) isindependent
of temperature, so it can be factored out of the intgral, and
A E U E − dT
UE
=
+C =
+C
2
∫
RT
R
RT
T
For local composition theory, we just need to repeat this complete procedure but
E
recognize that U 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


AE

= −∑ x j ln ∑ xi Ω ij  + C


respect to T becomes very simple. Then, RT
j
 i

Elliott and Lira: Chapter 11 - Activity Models
Slide 22
Wilson’s equation
Ncj =2 for all j at all ρ; Bij = Vj/Vi ; C = 0
GE
=−
RT
∑
j

x j ln 


∑
i

xi Λ ji 


⇒
 


GE


= −∑ n j ln ∑ ni Λ ji - ln (n )


RT
  i

j

Taking the last term first:
+ ∑ n j [ln (n )] = n ln(n);
j


∂ ∑ n j ln  ∑ ni
 i
 j
∂n k
∂ ( n ln n)
1
= ln n + n 
∂n k
n

ji  


= ln ∑ ni
 i



ji 



∂ G E / RT


1
jk
= ln n + n  − ln ∑ ni ki  − ∑ n j 
ln γ k =
∂n k
n
 i
 j
 ∑ ni
 i
{



jk
n
−


∑ j n
ki
 j
∑ i
 i
}
Elliott and Lira: Chapter 11 - Activity Models
ji



 = 1 − ln ∑ xi
 i





jk
x
−

∑
ki
j
 j
 ∑ xi
 i
Slide 23
ji





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)
x j rj
xjqj
qi
Φj ≡
θj ≡
B
≡
r =
n r
q =
n q
ij
where
∑ xjqj
∑ xi ri ;
∑ xi qi ; j ∑ kj kj ; j ∑ kj kj ;
k
i
GE
=−
RT
∑
j
k
j
i

q j x j ln


∑
i

xi Ω ij  +

∑ x j ln(Φ j /x j ) -5∑ q j x j ln(Φ j /θ j )
j
j
RES
ln γ k = ln γ COMB
+
ln
γ
k
k
ln γ kCOMB = ln (Φ k / xk ) - (1 − Φ k / xk ) - 5qk [ln(Φ k / θ k ) − (1 − Φ k / θ k )]
ln γ kRES



= qk 1 − ln 



∑
i

xi Ω ik  −

∑
j

x j Ω kj 
xi Ω ij 
i

∑
Elliott and Lira: Chapter 11 - Activity Models
Slide 24
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: P1sat= 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/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 25
The activity coefficients from the Wilson equation are:
x1Λ11
x2 Λ 21
ln γ 1 = 1 − ln( x1Λ11 + x2 Λ12 ) −
−
x1Λ11 + x2 Λ12 x1Λ 21 + x2 Λ 22
x1 Λ 12
x 2 Λ 22
ln γ 2 = 1 − ln( x1 Λ 21 + x 2 Λ 22 ) −
−
x1 Λ 11 + x 2 Λ 12 x1Λ 21 + x 2 Λ 22
Noting that Λ11= Λ22 =1, we can rearrange for binary mixtures to obtain the slightly
simpler relations:
ln γ 1 = 1 − ln( x1 Λ 11 + x 2 Λ 12 ) + x 2 Q
ln γ 2 = 1 − ln( x1 Λ 21 + x 2 Λ 22 ) − x1Q
Λ 12
Λ 21
Q
=
−
where
x1 + x 2 Λ 12 x1 Λ 21 + x 2
Q = 0.070/(0.2+0.8*0.070) - 0.625/(0.8+0.2*0.625) = -0.4022
ln γ 1 = 1 − ln(0.2 + 0.8 * 0.070) + 0.8Q = 1.0408 ⇒ γ1= 2.824
ln γ 2 = 1 − ln(0.8 + 0.20*.625) − 0.2Q = 0.1584 ⇒ γ2= 1.172
+
x2 γ2 P2sat
P = (y1+y2)P = x1γ1 P1sat
= 0.2*2.824*3.23
+ 0.8*1.172*187.1 = 177.2 mmHg
y1 = x1γ1 P1sat /P = 0.2*2.824*3.23/177.2 = 0.0103
Elliott and Lira: Chapter 11 - Activity Models
Slide 26
Question: What value for Ωij is implied by the van der Waals EOS?
Z=
aρ
1
−
1 − bρ RT
b = Σxibi is reasonable. As for "a", we must carefully consider how this term relates to
the energy of mixing:
N A uij
U − U ig
aρ N A ρ
=−
=
xi x j ∫
g ij 4πr 2 dr
∑
∑
2
RT
RT
RT
Comparing to the result for pure fluids
a ii = −
NA
2
∫
U − U ig
N ρ
= − A ∑ ∑ xi x j aij ⇒ a = ∑ ∑ xi x j aij
N A uii gii 4πr dr ⇒
RT
RT
2
where aij ≡ −
⇒
Ω ij ≡
−
NA
2ε ij
−
NA
2ε jj
NA
2
2
N
u
g
4π
r
dr where we set aij= aii a jj (1 - kij),
A
ij
ij
∫
2
π
N
u
g
r
dr
4
A
ij
ij
∫
2
∫ N Au jj g jj 4πr dr
=
aij ε jj
a jj ε ij
~
σ ij3
σ 3jj
Elliott and Lira: Chapter 11 - Activity Models
Slide 27