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Chapter 3

UNIFAC and Its Related Calculations

Example 3.1 For the binary system diethylamine (1).n-heptane(2) at 308.15 K,�nd 1 and 2 when x1 = 0:4. The chemical formulas are given as CH3-CH2NH-CH2-CH3 (1) and CH3-(CH2)5-CH3 (2). The �gure below illustrates the chemicalstructure for diethylamine and n-heptane)

H

H

H H

H H

H H

H H

H H

H H

H

H

n-heptane N

H

H

H

H

H

HH

H

H

H

H

diethylamine

H

H

H H

H H

H H

H H

H H

H H

H

H

n-heptane N

H

H

H

H

H

HH

H

H

H

H

diethylamine

Fig. 3.1 Chemical structure for diethylamine (1) and n-heptane (2).

The following table shows the subgroups and their identi�cation numbers k,values of parameters Rk and Qk, and the number of each subgroup in each molecule:Obtain the volume contribution for each molecule as:

ri =Xk

�(i)k Rk;

r1 = (2)(0:9011) + (1)(0:6744) + (1)(1:2070) = 3:6836

r2 = (2)(0:9011) + (5)(0:6744) = 5:1742

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diethylamine (1) n-heptane (2)

Subgroup k Rk Qk �(1)k �

(2)k

CH3 1 0.9011 0.848 2 2CH2 2 0.6744 0.540 1 5CH2NH 33 1.2070 0.936 1 0

In like manner for the surface fractions

qi =Xk

�(i)k Qk;

q1 = (2)(0:848) + (1)(0:540) + (1)(0:936) = 3:1726

q2 = (2)(0:848) + (5)(0:540) = 4:3960

The ri and qi values are molecular properties, independent of composition. Sub-stitute the known values to obtain the eki which are the surface fractions for eachsubgroup in each molecule. The size of this matrix is the number of subgroups timesthe number of components i.e., for this example it is 3� 2.

ek;i =�(i)k Qkqi

e1;1 =�(1)1 Q1q1

=2(0:848)

3:1726= 0:5347

e1;2 =�(2)1 Q1q2

=2(0:848)

4:3960= 0:3858

e2;1 =�(1)2 Q2q1

=1(0:540)

3:1726= 0:1702

e2;2 =�(2)2 Q2q2

=5(0:540)

4:3960= 0:6142

e33;1 =�(1)3 Q3q1

=1(0:936)

3:1726= 0:2951

e33;2 =�(2)3 Q3q2

=0(0:936)

4:396= 0

ek;ik i = 1 i = 2

1 0.5347 0.38582 0.1702 0.614233 0.2951 0.0000

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Determine the overall surface area fraction for each subgroup in the mixturewhich is a vector of size k. The fraction is de�ned as

�k =

Pi

xiqiek;iPj

xjqj

�1 =

Pi

xiqie1;iPj

xjqj=0:4(3:1762)(0:5347) + 0:6(4:3960)(0:3858)

0:4(3:1762) + 0:6(4:3960)= 0:4342

�2 =

Pi

xiqie2;iPj

xjqj=0:4(3:1762)(0:1702) + 0:6(4:3960)(0:6142)

0:4(3:1762) + 0:6(4:3960)= 0:4700

�33 =

Pi

xiqie33;iPj

xjqj=0:4(3:1762)(0:2951) + 0:6(4:3960)(0:0000)

0:4(3:1762) + 0:6(4:3960)= 0:0958

Obtain the interaction parameters matrix for the subgroups from the table. Thisis a square matrix whose elements are determined by the number of subgroups i.e.,for this example it is 3� 3. Notice that ai;i = 0:0

am;k =

1 2 331 a1;1 = 0:0 a1;2 = 0:0 a1;33 = 255:7

2 a2;1 = 0:0 a2;2 = 0:0 a2;33 = 255:7

33 a33;1 = 65:33 a33;2 = 65:33 a33;33 = 0:0

From this matrix, at the given temperature T = 308:15 K, determine the �m;k from

�m;k = exph�am;k

T

i�m;k =

�1;1 = 1 �1;2 = 1 �1;33 = 0:4361

�2;1 = 1 �2;2 = 1 �2;33 = 0:4361

�33;1 = 0:8090 �33;2 = 0:8090 �33;33 = 1

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Determine the �ik. The size of this matrix is number of components times thenumber of subgroups i.e., for this example it is 2� 3.

�ik =Xm

emi�m;k

�1;1 =Xm

em1�m;1 = 0:5347(1) + 0:1702(1) + 0:2951(0:8090) = 0:9436

�1;2 =Xm

em1�m;2 = 0:5347(1) + 0:1702(1) + 0:2951(0:8090) = 0:9436

�1;33 =Xm

em1�m;33 = 0:5347(0:4361) + 0:1702(0:4361) + 0:2951(1) = 0:6024

�2;1 =Xm

em2�m;1 = 0:3858(1) + 0:6142(1) + 0(0:8090) = 1:0000

�2;2 =Xm

em2�m;2 = 0:3858(1) + 0:6142(1) + 0(0:8090) = 1:0000

�2;33 =Xm

em2�m;33 = 0:3858(0:4361) + 0:6142(0:4361) + 0(1) = 0:4361

�i;k =

k

i 1 2 331 0.9436 0.9436 0.60242 1.0000 1.0000 0.4361

Calculate the sk vector (size equal to number of subgroups) from

sk =Xm

�m�m;k

s1 =Xm

�m�m;1 = 0:4342(1) + 0:4700(1) + 0:0958(0:8090) = 0:9817

s2 =Xm

�m�m;2 = 0:4342(1) + 0:4700(1) + 0:0958(0:8090) = 0:9817

s33 =Xm

�m�m;33 = 0:4342(0:4361) + 0:4700(0:4361) + 0:0958(1) = 0:4901

One last step before you are ready to calculate the combinatorial and residual partsof the activity coe¢ cients is to calculate the volume and surface area contributions

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for the molecules i.e.,

Ji =riP

j

xjrj

J1 =r1P

j

xjrj=

3:6836

0:4(3:6836) + 0:6(5:1742)= 0:8046

J2 =r2P

j

xjrj=

5:1742

0:4(3:6836) + 0:6(5:1742)= 1:1302

Li =qiP

j

xjqj

L1 =q1P

j

xjqj=

3:1720

0:4(3:1720) + 0:6(4:3960)= 0:812

L2 =q2P

j

xjqj=

4:3960

0:4(3:1720) + 0:6(4:3960)= 1:1253

Calculate the combinatorial activity coe¢ cients from

ln Ci = 1� Ji + lnJi � 5qi�1� Ji

Li+ ln

JiLi

�ln C1 = 1� J1 + lnJ1 � 5q1(1�

J1L1+ ln

J1L1)

= 1� 0:8046 + ln(0:8046)� 5(3:1720)(1� 0:80460:812

+ ln0:8046

0:812) = �2: 134 7� 10�2

ln C2 = 1� J2 + lnJ2 � 5q2(1�J2L2+ ln

J2L2)

= 1� 1:1302 + ln 1:1302� 5(4:3960)(1� 1:13021:1253

+ ln1:1302

1:1253) = �7: 597 6� 10�3

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Determine the residual part of the activity coe¢ cients from

ln Ri = qi

"1�

Xk

��k�i;ksk

� ek;i ln�i;ksk

�#

ln R1 = q1

"1�

Xk

��k�1;ksk

� ek;1 ln�1;ksk

�#

= 3:1720(1� (0:43420:94360:9817

� 0:5347 ln 0:94360:9817

+0:47000:9436

0:9817� 0:1702 ln 0:9436

0:9817

+0:09580:6024

0:4901� 0:2951 ln 0:6024

0:4901))

= 0:14630

ln R2 = q2

"1�

Xk

��k�2;ksk

� ek;2 ln�2;ksk

�#

= 4:396(1� (0:43421:00000:9817

� 0:3858 ln 1:00000:9817

+0:47001:0000

0:9817� 0:6142 ln 1:0000

0:9817

+0:09580:4360

0:4901� 0:0000 ln 0:4360

0:4901))

= 5: 358 3� 10�2

ln i = ln Ri + ln

Ci

ln 1 = ln R1 + ln

C1 = 0:14630 + (�2: 134 7� 10�2) = 0:124 95

1 = exp(0:124 95) = 1: 133

ln 2 = ln R2 + ln

C2 = 5: 358 3� 10�2 + (�7: 597 6� 10�3) = 4: 598 5� 10�2

2 = exp(4: 598 5� 10�2) = 1: 047 :