Advanced Chemical Engineering Thermodynamics

Class 25

Group Contribution MethodsChapters 6-7 and Appendix F, Prausnitz

• We demonstrated that

• Now we need to remember that

• Thus

• These are the most important relations for the chapter

GE and Activity Coefficients

giE = RT ln [ γi ]

GE = ∑i ni giE

GE = RT ∑i ni ln γi

Expressions for GE

gE = 0 when x1 = 0gE = 0 when x2 = 0

The two-suffix Margules equations were derived from:

gE = A x1 x2

A more complex form for gE as a function of composition can be:

gE = x1 x2 [ A + B (x1 - x2) + C (x1 - x2)2 + D (x1 - x2)3 + … ]

This is the Redlich-Kister expansion

Two-Suffix MargulesThe two-suffix Margules equations were derived from:

gE = A x1 x2

ln γ1 = A/RT x22

ln γ2 = A/RT x12

ln γ1∞ = A/RT

ln γ2∞ = A/RT

The activity coefficients at infinite dilution are:

Wohl’s ExpansiongE / [RT (x1q1 + x2q2)] = 2 a12 z1z2

+ 3 a112 z12z2 + 3 a122 z1z2

2

+ 4 a1112 z12z2 + 4 a1222 z1z2

3

+ 6 a1122 z12z2

2 + …

z1 = x1 q1 / [ x1 q1 + x2 q2 ] z2 = x2 q2 / [ x1 q1 + x2 q2 ]

The parameters ‘q’ are effective volumes or cross sections of the molecules

The parameters ‘a’ are interaction parameters

The contribution to gE of the interaction parameters ‘a’ is weighted by the products that contain the factors ‘z’

Van Laar Equation

gE / RT = 2 a12 x1 x2 q1 q2 / [ x1 q1 + x2 q2 ]

We can calculate the fugacity coefficients of the two components

ln γ2 = B’ / [ 1 + B’/A’ x2/x1 ]2ln γ1 = A’ / [ 1 + A’/B’ x1/x2 ]2

And, at infinite dilution, we obtain

A’ = 2 q1 a12

B’ = 2 q2 a12

ln γ2∞ = B’

ln γ1∞ = A’

Three-Suffix Margules EquationsgE / [RT (x1q1 + x2q2)] = 2 a12 z1z2

+ 3 a112 z12z2 + 3 a122 z1z2

2

+ 4 a1112 z12z2 + 4 a1222 z1z2

3

z1 = x1 q1 / [ x1 q1 + x2 q2 ] z2 = x2 q2 / [ x1 q1 + x2 q2 ]

ln γ1 = A’ x22 + B’ x2

3

ln γ2 = (A’ + 3/2 B’) x12 - B’ x1

3

At infinite dilutionln γ1

∞ = A’ + B’

ln γ2∞ = A’ + 1/2 B’

UNIQUAC

For a binary mixture UNIQUAC states that:

[gE/RT]combinatorial = x1 ln [φ1*/x1] + x2 ln [φ2

*/x2] +

z/2 { x1q1 ln [θ1/φ1*] + x2q2 ln [θ2/φ2

*]}

gE / RT = [ gE / RT ]combinatorial + [ gE / RT ]residual

[gE/RT]residual = - x1 q1’ ln [ θ1

’ + θ2’ τ21 ] - x2 q2

’ ln [ θ2’ + θ1

’ τ12 ]

z is the coordination number

In UNIQUAC z=10

φ* is the segment fraction

θ, and θ’ are area fractions

UNIQUAC

φ1* = x1 r1 / [ x1 r1 + x2 r2 ]

Segment fractions

φ2* = x2 r2 / [ x1 r1 + x2 r2 ]

The parameters ‘r’ are pure-component molecular structure constants

θ1 = x1 q1 / [ x1 q1 + x2 q2 ]Area fractions

For fluids other than water and lower alcohols q = q’

θ2 = x2 q2 / [ x1 q1 + x2 q2 ]

θ1’ = x1 q1

’ / [ x1 q1’ + x2 q2

’ ] θ2’ = x2 q2

’ / [ x1 q1’ + x2 q2

’ ]

Energy parameters

These parameters are given in terms of characteristic energies

τ12 = exp ( -Δu12 / RT ) ≡ exp ( -a12 / RT )

τ21 = exp ( -Δu21 / RT ) ≡ exp ( -a21 / RT )

Parameters, from Table 6.9

Compound r q q’

CCl4 3.33 2.82 2.82

CH3OH 1.43 1.43 0.96

C2H5OH 2.11 1.97 0.92

H2O 0.92 1.40 1.00

Toluene 3.92 2.97 2.97

N-Hexane 4.50 3.86 3.86

*

*

*

Binary Parameters,from Table 6.10

System T/K a12/K a21/K

Acetone/Chloroform 323 -171.71 93.93Acetone/Water 331-368 530.99 -100.71Acetone/Methanol 323 379.31 -108.42

Ethanol/n-octane 348 -123.57 1354.92Ethanol/n-heptane 323 -105.23 1380.30Ethanol/Benzene 350-369 -75.13 242.53Ethanol/CCl4 340-351 -138.90 947.20

Formic acid/Water 374-380 924.01 -525.85N-Hexane/Nitromethane 318 230.64 -5.86

UNQUAC Activity CoefficientsgE / RT = [ gE / RT ]combinatorial + [ gE / RT ]residual

ln γ2 = ln (φ2*/x2) + z/2 q2 ln (θ2/φ2

*) + φ1* (l2 - l1 r2/r1 )

- q2’ ln ( θ2

’ + θ1’ τ12 )

+ θ1’q2

’ [ τ12 / (θ2’ + θ1

’ τ12) - τ21 / (θ1’+θ2

’τ12)]

ln γ1 = ln (φ1*/x1) + z/2 q1 ln (θ1/φ1

*) + φ2* (l1 - l2 r1/r2 )

- q1’ ln ( θ1

’ + θ2’ τ21 )

+ θ2’q1

’ [ τ21 / (θ1’ + θ2

’ τ21) - τ12 / (θ2’+θ1

’τ12)]

l1 = z/2 (r1 - q1) - (r1 - 1) l2 = z/2 (r2 - q2) - (r2 - 1)

Comments

In all the methods we discussed, it is necessary to have some experimental data to obtain the fitting parameters for any binary mixture of interest

When experimental data are not available, an alternative approach is to look at the molecular structure of the compounds and to try obtaining the fitting parameters from the molecular structure of the interacting molecules

ExampleLet’s suppose we have one mixture of acetone and toluene

ExampleLet’s suppose we have one mixture of acetone and toluene

To obtain the interaction parameters, we first look at the chemical structure of both compounds:

CH3

CH3

C=O

Acetone Toluene

CH3CCH

CH CH

CH CH

ExampleLet’s suppose we have one mixture of acetone and toluene

Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions:

CH3

CH3

C=O

Acetone Toluene

CH3CCH

CH CH

CH CH

ExampleLet’s suppose we have one mixture of acetone and toluene

Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions:

CH3

CH3

C=O

Acetone Toluene

CH3CCH

CH CH

CH CH

ExampleLet’s suppose we have one mixture of acetone and toluene

Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions:

CH3

CH3

C=O

Acetone Toluene

CH3CCH

CH CH

CH CH

ExampleLet’s suppose we have one mixture of acetone and toluene

Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions:

CH3

CH3

C=O

Acetone Toluene

CH3CCH

CH CH

CH CH

ExampleLet’s suppose we have one mixture of acetone and toluene

Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions:

CH3

CH3

C=O

Acetone Toluene

CH3CCH

CH CH

CH CH

ExampleLet’s suppose we have one mixture of acetone and toluene

The basic assumption is that each group will behave in any mixture independently on the molecule it is part of

CH3

CH3

C=O

Acetone Toluene

CH3CCH

CH CH

CH CH

ExampleLet’s suppose we have one mixture of acetone and toluene

The basic assumption is that each group will behave in any mixture independently on the molecule it is part ofWe obtain this contribution from available experimental data, and we ‘predict’ the behavior of untested mixtures

CH3

CH3

C=O

Acetone Toluene

CH3CCH

CH CH

CH CH

UNIFACThe Universal Functional Activity Coefficient theory is the best known theory that uses this approachFirst introduced by Fredenslund, Jones, and Prausnitz in 1975

The main advantage of the procedure is that in typical mixtures of non electrolytes the number of group-group interactions is much less than the possible number of molecule-molecule pairs

Another advantage is that correlations are always easier than the experiments they try to substitute

Remember, however, that when no experimental data are used to ‘improve’ the estimation of the interaction parameters, UNIFAC works, at best, as a good first approximation in VLE, but is often poor in LLE predictions

UNIFAC: The Method

ln γi = ln γiC + ln γi

R

UNIFAC: The Method

ln γi = ln γiC + ln γi

R

This is the ‘combinatorial’part, calculated as in the case of UNIQUAC

This is the ‘residual’contribution, which depends on group-group interactions

UNIFAC: The Method

ln γi = ln γiC + ln γi

R

This is the ‘combinatorial’part, calculated as in the case of UNIQUAC

This is the ‘residual’contribution, which depends on group-group interactions

ln γiC = FC (x, φ, θ) As defined for UNIQUAC

ln γiR = FR (X, Q, T, amn)

X = group mole fractionQ = group external surface areaamn = interaction energy between groups n and m

Example: Fig. F4

methanol + water @ 50C

Such good theory-experiment agreement is not common

Example: Fig. F5

n-hexane + methyl ethyl ketone @ 65 C

This is more typical

Appendix FMany binary systems have been studied and the corresponding data have been publishedDIPPR-AIChE provides a continuously-improving compilation

Unfortunately, many other systems have not been reported

UNIFAC, or other group-contribution methods, are in general not very accurate

P/bar

x/n-hexane

L

VUNIFAC

Fig. F-5

Experiments

n-hexane + methyl ethyl ketone

Appendix FLet’s suppose we can obtain only 2 experimental data point

Which ones shall we get?

P/bar

x/n-hexane

L

V

Appendix FLet’s suppose we can obtain only 2 experimental data point

Which ones shall we get?

P/bar

x/n-hexane

L

V

For all binary mixtures, regardless of whether or not they form an azeotrope, binary parameters can be calculated from activity coefficients at infinite dilutionThese often provide the most valuable experimental information

Appendix F: ExampleLet’s suppose we decide to use any functional form for gE of the mixture of interest we like:

gE = F (x, A, B)

x = compositionA,B the two parameters that depend on the specific F we like

Appendix F: ExampleLet’s suppose we decide to use any functional form for gE of the mixture of interest we like:

gE = F (x, A, B)

x = compositionA,B the two parameters that depend on the specific F we like

We saw that, depending on F, we can express the activity coefficient for both compounds as

RT ln γ1 = F1 (x, A, B)RT ln γ2 = F2 (x, A, B)

The specific F1 and F2 will depend on our choice for F

Appendix F: ExampleWe saw that, depending on F, we can express the activity coefficient for both compounds as

RT ln γ1 = F1 (x, A, B)RT ln γ2 = F2 (x, A, B)

A,B are the two parameters we need to implement FIf we measure γ1

∞ and γ2∞, then

lim [F1 (x, A, B) / RT] = ln γ1∞

lim [F2 (x, A, B) / RT] = ln γ2∞

Which is a system of 2 equations in 2 unknowns

x1→0

x2→0

Appendix F: ExampleVan Laar’s equations

ln γ2∞ = B’

ln γ1∞ = A’

Appendix F: ExampleVan Laar’s equations

ln γ2 = B’ / [ 1 + B’/A’ x2/x1 ]2ln γ1 = A’ / [ 1 + A’/B’ x1/x2 ]2

And we can predict the activity coefficients of both components at any composition:

ln γ2∞ = B’

ln γ1∞ = A’

Appendix F: ExampleThree-SuffixMargules equations

ln γ1∞ = A’ + B’

ln γ2∞ = A’ + 1/2 B’

Appendix F: ExampleThree-SuffixMargules equations

And we can predict the activity coefficients of both components at any composition:

ln γ1 = A’ x22 + B’ x2

3

ln γ2 = (A’ + 3/2 B’) x12 - B’ x1

3

ln γ1∞ = A’ + B’

ln γ2∞ = A’ + 1/2 B’