Ideal solutions and excess functions Part V. problem The data in Table 11.2 are experimental values...
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Transcript of Ideal solutions and excess functions Part V. problem The data in Table 11.2 are experimental values...
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