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MSE 303 Note7_Solution Theory
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1
MSE 303
Thermodynamics & Equilibrium Processes
Solution Theory
(Gaskell Chapter 9)
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Terminology
3
Liquid A
T
Vapor Pressure
Pressure
gauge
PA0
4
At equilibrium
Rates of Evaporation & Condensation for
Single Component
0
)( AAc kpr
0
)( AAe kpr
Eqn 9.1
5
Rates of Evaporation & Condensation for
Single Component
)()( BcBe rr
0'
)()( BBcBe pkrr
E
6
Rates of evaporation & condensation
for a solution
• If the mole fraction of A in the solution is XA and the atomic
diameters of A and B are similar, then assuming that the
composition of the surface of the liquid is the same as that
of the bulk liquid, the fraction of the surface sites occupied
by A atoms is XA.
Liquid A B
PA
PB
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• As A can only evaporate from surface sites occupied by A atoms, the
rate of evaporation of A scales by a factor XA
• Also, since at equilibrium, the rates of evaporation and condensation are
equal to one another, the equilibrium vapor pressure of A exerted by the
A-B solution is decreased from pA0 to pA.
AAAe kpXr )(
BBBe pkXr '
)(
Eqn 9.3
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Eq 9.5
Eq 9.6
Raoult’s Law
0
AAA pXp
0
BBB pXp
9
So far we made the assumption that
Deviation from Raoult’s Law
are independent
A B
B
A A B
B
B A B
B
10
'
)( Aer
AAAe kpXr '
)( Eqn. 9.7
(See Eqn 9.3)
B
B A B
B
11
0
)( AAe kpr AAAe kpXr '
)(
'
)( Aer
AAA Xkp Henry’s Law:
,
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Liquid A B
PA
PB
'
)( Aer
13
)(
'
)( AeAe rr
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Activity
0 ofactivity
i
ii
f
fai
.
the fugacity of a real gas is an effective pressure which replaces the true mechanical
pressure in accurate chemical equilibrium calculations.
At constant T,
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Activity for ideal solutions
0
i
ii
p
pa Eqn 9.12
ii Xa
which is an alternative
expression of Raoult’s law
16
Basically, the introduction of activity normalizes the vapor pressure-composition relationship with
respect to the saturated vapor pressure exerted in the standard state
0
i
ii
p
pa
iii Xkp iii Xka
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Gibbs-Duhem Equation
Let Q’ be a thermodynamic properties
At constant T and P, the variation in Q’ with the composition of the solution
Define:
Then:
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kth component entire solution
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How is this useful?
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Gibbs Free Energy formation of a Solution AAG
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Example:
The vapor pressures of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively.
An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of
methanol. Calculate the total vapor pressure for solution and the mole fraction of methanol
in the vapor.
Mol. mass of ethyl alcohol = C2H2OH = 46
No. of moles of ethyl alcohol = 60/46 = 1.304
Mol. mass of methyl alcohol = CH3OH = 32
No. of moles of methyl alcohol = 40/32 = 1.25
Then,
‘XA’, mole fraction of ethyl alcohol = 1.304/(1.304+1.25) = 0.5107
‘XB’, mole fraction of methyl alcohol = 1.25/(1.304+1.25) = 0.4893
Partial pressure of ethyl alcohol = XA. pA0 = 0.5107 × 44.5 = 22.73 mm Hg
Partial pressure of methyl alcohol = XB.pA0 =0.4893 × 88.7 = 43.73 m Hg
Total vapour pressure of solution = 22.73 + 43.40 = 66.13 mm Hg
Mole fraction of methyl alcohol in the vapour
= Partial pressure of CH3OH/Total vapour pressure = 43.40/66.13 = 0.6563
Solution:
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Change in Gibb’s Free Energy Due to
the Formation of a Solution
dpP
RTdG
i
i
p
pRTG
0
ln
And Recall 0
i
ii
p
pa
iiii aRTpureGsolutioninGG ln)()(
The difference between the two G’s (solution vs pure) is the change in the Gibbs free
energy accompanying the introduction of 1 mole of component i into the solution
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Is there any change of volume in mixing?
For binary A-B solution,
Hence,
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The change in the volume in the formation of an ideal solution is zero
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Heat of formation of ideal solutions
Heat of formation of ideal solutions i
M
i XRTG lnFor an ideal solution
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Entropy of formation of ideal solutions
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With Sterling’s approximation
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Substitute for nA and nB
Term inside brackets is always negative
So Sconf is always positive during the formation of a solution
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Non-ideal Solutions
Non-ideal Solutions
0
i
ii
p
pa
32
Statistical Model for Regular Solutions
Z
Then,
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Statistical Model for Regular Solutions
34
35
Statistical Model for Regular Solutions
=
36
Then,
37
Of course
then
38 This is also equal to GXS, the excess Gibbs free energy of the solution
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From the Gibbs-Duhem equation,
Because
40
On the other hand, according to the definition of activity,
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Example
Solution
So,
similarly
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From Table A-4, Gaskell