Ideal solutions and excess functions

Part V

problem

• The data in Table 11.2 are experimental values of VE for binary liquid mixtures of 1,3-dioxolane(1) and isooctane(2) at 298.15K and 1 atm.

• (a) Determine from the data numerical values for a, b, and c in:

)( 21121 cxbxaxxV E

)( 21121 cxbxaxxV E

a = 3448; b = -3202; c = 244.62

Numerical problem: find the values of a, b, and c that best fit to the given set of data

• (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

• Prepare a plot and discuss its features

EE V andV 21

)3)(2(

)32(

)(2)(34

)(

211

212

211

221

121

31

1

21121

cxxcbbaxV

cxbxaxV

axabxbccxdx

dV

cxbxaxxV

E

E

E

E

-2000

-1000

0

1000

2000

3000

4000

0 0.2 0.4 0.6 0.8 1

x1

VE

partial V1E

partial V2E

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)/n-hexane(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

From

cbTTTTaG

bTaT

G

aT

GTC

E

xP

E

xP

EEP

)ln(

ln,

,

2

2

caTTSGH

bTaT

GS

EEE

xP

EE

ln,

Also,

using the values at 298.15K

• We already know the value of a = -2.86 (equal to CP

E)

• 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.

bTaS

caTH

cbTTTTaG

aT

GTC

E

E

E

xP

EEP

ln

)ln(

,

2

2

Finally calculate GE, HE, TSE at 25oC using the a, b, and c parameters