Ideal solutions and excess
functions
Part V
problem
• The data in Table 11.2 are experimental values
of VE for binary liquid mixtures of 1,3dioxolane(1) and isooctane(2) at 298.15K and 1
atm.
• (a) Determine from the data numerical values
for a, b, and c in:
V  x1 x2 (a  bx1  cx )
E
2
1
V  x1 x2 (a  bx1  cx )
E
2
1
Numerical problem: find the values of a, b, and c that best fit
to the given set of data
a = 3448; b = -3202; c = 244.62
• (b) Determine the maximum value of VE and
the value of x1 at which this occurs.
dVE/dx1 = 0 and solve for x1
VEmax = 536.29 cm3/mol ; x1 =0.353
• c) From the results of part (a) find expressions
for V E and V E
1
2
• Prepare a plot and discuss its features
V  x1 x2 (a  bx1  cx )
E
2
1
E
dV
3
2
 4cx1  3(c  b) x1  2(b  a ) x1  a
dx1
V1  x (a  2bx1  3cx )
E
2
2
2
1
V2  x (a  b  2(b  c) x1  3cx )
E
2
1
2
1
4000
3000
VE
partial V1E
partial V2E
2000
1000
0
0
0.2
0.4
0.6
x1
-1000
-2000
0.8
1
Temperature dependence of
excess properties
example
• If CPE is a constant, independent of T, find
expressions for GE, SE, HE for an
equimolar solution of benzene(1)/nhexane(2) at 323.15K, given the following
excess-property values for an equimolar
solution at 298.15 K
• CPE = -2.86 J/mol K; HE = 897.9 J/mol;
• GE = 384.5 J/mol
2 E


G
E
C P  T 
2

T

From
 G E

 T

  a
 P,x

  a ln T  b
 P,x
G  a (T ln T  T )  bT  c
E
Also,
E

G
E
S  
 T

  a ln T  b
 P, x
H  G  TS  aT  c
E
E
E
using the values at 298.15K
• We already know the value of a = -2.86
(equal to CPE)
• From HE obtain c =1,750.6
• From GE obtain b = -18.0171
• Now calculate GE, SE, HE at 323.15 K
problem
• Given the following data for equimolar
mixtures of organic liquids. Use all the
data to estimate values of GE, HE, and TSE
for the equimolar mixture at 25oC
– At T = 10oC, GE =544 and HE =932.1
– At T = 30oC, GE = 513.2, HE =893.4
– At T = 50oC, GE = 494.2, HE = 845.9
– Energies are in J/mol
Assume CpE is constant (a)
• Then HE = aT + c
• Use the three sets of
HE data and get the
best values of a and
c; a= -2.155; c = 1544
• With a and c get b for
each set of GE data;
then take b average.
2
E


G
C PE  T 
2

T


  a
 P,x
G E   a (T ln T  T )  bT  c
H E  aT  c
S E  a ln T  b
Finally calculate GE, HE, TSE at 25oC using the a, b, and
c parameters