Chapter 5. Activity Coefficient Models. Part 2. Local-composition
models: From Wilson and NRTL to UNIQUAC and UNIFAC
Problem 1. Derivation of the Wilson equation based on the Flory-Huggins model
Derive the Wilson equation for a binary solution, the way the author himself did it
(Wilson1), based on the Flory-Huggins (FH) equation. The purpose is to arrive to the gE
expression for a binary mixture:
gE
5.61
  x1 ln  x1  x212   x2 ln  x2  x121 
RT
starting from the combinatorial term of FH equation (e.g. see Table 4.4), but with the
volume fractions, i being substituted by the local volume fractions zi .
Instructions: Use the definition of the local volume fractions:
x22V2
x11V1
z2 
z1 
5.62
x22V2  x12V1
x11V1  x21V2
together with the mass balances of the local mole fractions ( x21  x11  1, x12  x22  1 ),
see also figure 5.1, and the local mole fraction and Wilson parameter definitions shown
in Tables 5.1 and 5.2.
Problem 2. Properties of the Wilson equation
i. From the expression of Wilson’s equation for the activity coefficient of
multicomponent systems (Table 5.3), derive the expressions for the two activity
coefficients in a binary solution:
 12
 21 
ln  1   ln  x1  x2 12   x2 


 x1  x2 12 x1 21  x2 
5.63
 12
 21 
ln  2   ln  x2  x1 21   x1 


 x1  x2 12 x1 21  x2 
ii. Derive also the expressions for the activity coefficients at infinite dilution:
ln  1   ln 12  1   21
ln  2   ln  21  1  12
5.64a
5.64b
iii. The excess enthalpy is related to the excess Gibbs energy via the equation:
 g E /T
h E  T 2
5.65
P,x
T
Starting from the excess Gibbs energy equation, show that the excess enthalpy for a
binary mixture is given by the equation:


1
 x2 12 
 x1 21 
12  11   x2 
12  22 
h E  x1 
 x1  12 x2 
 x2   21 x1 
5.66
Problem 3. The NRTL equation
i. Which equation do we get from NRTL when the non-randomness factor is equal to
zero (completely random mixture)?
ii. From the expression of NRTL for multicomponent mixtures, show that the activity
coefficient of compound 1 in a binary system is given by:
2
  G

G12 12 
2
21
 

5.67
ln  1  x2  21 
2
  x1  x2G21  x2  x1G12  
What is the value of the activity coefficient at infinite dilution?
iii. Derive the equation for the excess Gibbs energy of a binary mixture from the twofluid framework, following the steps outlined below:
Step 1: Develop first expressions for x12 as a function of G12 and xi and also for x21 as
a function of G21 and xi
Step 2: Starting from basic equation of the two-fluid theory:
5.68
G E  x1 g1  g11   x2 g 2  g 22 
where gii are the Gibbs energies of the pure compounds and the Gibbs energies gi are
given the equations using local compositions:
g1  x11 g11  x21 g 21
5.69
g 2  x22 g 22  x12 g12
prove that for the NRTL the equation from the two-fluid theory is:
5.70
G E  x1 x21 ( g 21  g 11)  x2 x12 ( g12  g 22 )
Step 3: combine the results from step 1 and 2 and derive the expression shown in
Table 5.1. Note that for NRTL, SE and VE are zero, thus the above GE expression is
identical to HE and UE.
Problem 4. Derivation of the Wilson equation based on the two-fluid theory
Derive the Wilson equation for the excess Gibbs energy of a binary solution from the
two-fluid theory based on equations 5.35 and 5.38 (  ij in eq. 5.35 have the same meaning
as  ij ).
Explain the assumptions made in the derivation.
Which model is obtained if all the interaction energies between like and unlike molecules
are the same?
Instructions:
Prove first that, starting from eq. 5.38:
x2 12
x1 21
x21 
x12 
x1  x2 12
x2  x1 21
2
Problem 5. Derivation of the UNIQUAC equation based on the two-fluid theory
Derive the UNIQUAC equation for the excess Gibbs energy of a binary solution from the
two-fluid theory based on equations 5.36, 5.37 and 5.39.
Explain the assumptions made in the derivation. Which model is obtained if all the
interaction energies between like and unlike molecules are the same?
Instructions:
Prove first that, starting from eq. 5.39:
1 12
2 21
12 
21 
2  1 12
1  2 21
Problem 6. Similarities and differences between the three local-composition models
i.
Which are the most important similarities and differences between the three
well-known local composition models (Wilson, NRTL and UNIQUAC) ?
ii.
Write all three models (for a binary mixture) in the form:
ln  i  ln  icomb  ln  ires , where comb and res represent the combinatorial (sizeshape) and residual (energetic) contributions
iii.
Derive the expression for ln  1 for all three models, identifying the
combinatorial and residual contributions. What do you observe ? Compare the
three approaches.
Problem 7. UNIFAC for athermal solutions
In the case of athermal solutions such as systems with alkanes having different sizes e.g.
n-heptane/n-decane, only the combinatorial part of UNIFAC contributes to the activity
coefficient (can you see why?). Table 5.11 provides some experimental data for infinite
dilution activity coefficients of n-heptane in nC20, nC28 and nC36 [n-alkanes with,
respectively, 20, 28 and 36 carbon atoms].
Table 5.11. Experimental data for infinite dilution activity coefficients of n-heptane in
nC20, nC28 and nC36.
System
T (K)
Activity Coefficient of nheptane at infinite dilution
nC7/nC20
353.15
0.899
nC7/nC28
373.15
0.768
nC7/nC36
373.15
0.672
i. Derive the expression of UNIFAC (combinatorial part only) for the infinite dilution
activity coefficient of compound 1 in a binary system:
 r 
 rq 
r 
r q 
ln  1  ln  1   1   1   5q1 ln  1 2   1  1 2 
r2 q1 
 r2 
  r2 q1 
  r2 
3
5.71
ii. Calculate the activity coefficients of n-heptane at infinite dilution in these three
alkanes and compare the results to the experimental data. What do you observe? Use
the r and q values for the groups provided in table 5.12.
Table 5.12. Values of r and q for the groups of problem 7.
Group
R
CH3
0.9011
CH2
0.6744
Q
0.848
0.540
i. Larsen et al. 41 proposed a modified UNIFAC combinatorial term which is given by
the equation:


ln  i  ln i  1  i
xi
xi
5.72
x r 2/3
i  i i 2 / 3
 x j rj
j
Derive for this modified UNIFAC model the expression for the infinite dilution
activity coefficient of compound 1 in a binary system:
2/3
  r 2 / 3
 r1  
1
ln   ln    1    
  r2 
 r2  

1
5.73
and then calculate the activity coefficients of n-heptane at infinite dilution in these
three alkanes using this modified UNIFAC equation. Compare the results to the
experimental data and the calculations from the (original) UNIFAC model (question
ii). Which model performs best?
Problem 8. Combinatorial, free-volume and rotational effects
In an attempt to correct for the underestimation of solvent activities in athermal solutions,
Kontogeorgis et al. (Fluid Phase Equilibria, 1997, 127, 103) have proposed the following
expression which combines combinatorial, free-volume and rotational effects:
 
  fv 
gE
5.74
  xi ln  i    xi ci ln  i 
RT
i
 xi  i
 i 
where  i ,  ifv are the volume and free-volume fractions of compound i and ci is the socalled rotational parameter. Show that the expression for the activity coefficient is given
by the equation:
 
  fv 
   fv 
 
5.75
ln  i   ln i  1  i   ci ln  i   cmix  i  i  , where cmix   xi ci
x1 
xi 
i
 xi
 i 
 xi
Which expressions are obtained in the limiting cases where
i)
all ci values are equal to unity and
ii)
all ci values are equal to unity and the volume and free-volume fractions are
equal to each other.
4
Problem 9. UNIFAC for polymer solutions
Table 5.13 gives experimental values for the weight-fraction based infinite dilution
activity coefficients of various solvents in polyethylene (PE) and polyisobutylene (PIB).
These mixtures can be considered near-athermal. Compare the performance of various
UNIFAC variants including both the original and modified combinatorials as well as
various free-volume versions. What do you observe?
Table 5.13. Experimental values for the weight-fraction based infinite dilution activity
coefficients of various solvents in polyethylene (PE) and polyisobutylene (PIB).
PE +
PIB +
1
1
Hexane (T=423 K)
5.91
Hexane (T=323 K)
6.96
Heptane (T=423 K) 5.26
Hexane (T=423 K)
7.59
Octane (T=423 K)
4.85
Cyclohexane
5.04
(T=323 K)
5