thermodynamic activity coefficient

Advanced Chemical Engineering Thermodynamics

Class 26

Theory of Liquid MixturesChapter 7, Prausnitz

Theoretical Derivation of van Laar’s Equations

Let’s focus on a mixture of 2 liquids

x1 moles of liquid 1x2 moles of liquid 2

The two liquids mix at constant T and P




We assume that during mixing

1. There is no volume change

2. The entropy is that of an ideal solution





We assume that during mixing

1. There is no volume change

2. The entropy is that of an ideal solution

vE = 0

sE = 0

Then





We remember that

gE = uE + P vE - T sE





vE = 0

sE = 0If then

We remember that

gE = uE + P vE - T sE

gE = uE





This is the basic assumption behind van Laar’s theory

gE = uE



Let’s use it within the thermodynamic cycle

gE = uE

Pure liquidsP

Liquid MixtureP

Pure ideal gasesVery low P

Ideal gas mixtureVery low P

T = const

Pure liquidsP

Liquid MixtureP



T = constI

II

III

Step I: The liquids are vaporized isothermally

(∂u/∂v)T = T (∂P/∂T)v - P

Because, we need to find the variation in u

Rigorously,

gE = uE

Pure liquidsP

Liquid MixtureP



T = constI

II

III


(∂u/∂v)T = a/v2

Because, we need to find the variation in u

However, van Laar used the VdW EOS, and then

gE = uE

a is the VdW parameter

Pure liquidsP

Liquid MixtureP



T = constI

II

III


a1 x1 / v1L

By considering the expansion of both liquids, integrating we get

ai is the VdW parameter for component i

a2 x2 / v2L

Pure liquidsP

Liquid MixtureP



T = constI

II

III


ΔuI = a1 x1 / b1 + a2 x2 / b2

If we substitute the molar liquid volumes with the VdW parameter b (not a bad assumption), then we get

ai is the VdW parameter for component i

√

Pure liquidsP

Liquid MixtureP



T = constI

II

III

Step II: The two ideal gases are mixed isothermallyWe need to evaluate the variation in internal energy

√

Pure liquidsP

Liquid MixtureP



T = constI

II

III

Step II: The two ideal gases are mixed isothermallyWe need to evaluate the variation in internal energy

√

ΔuII = 0

√

Pure liquidsP

Liquid MixtureP



T = constI

II

III

Step III: The ideal mixture is compressed to the initial pressureWe need to evaluate the variation in internal energy

√

√

Pure liquidsP

Liquid MixtureP



T = constI

II

III

Step III: The ideal mixture is compressed to the initial pressure

√

√

Using, in reverse, the derivation used in step I, and remembering that the total mixture is composed by x1+x2=1 moles, we obtain

ΔuIII = - amix / bmix

Pure liquidsP

Liquid MixtureP



T = constI

II

III


√

√

To finish, we need to evaluate the a and b parameters for the mixture


Pure liquidsP

Liquid MixtureP



T = constI

II

III


√

√

Van Laar used:


bmix = x1 b1 + x2 b2

Pure liquidsP

Liquid MixtureP



T = constI

II

III


√

√

Van Laar used:


amix = x12 a1 + x2

2 a2 + 2 x1 x2 sqrt (a1 a2)

Pure liquidsP

Liquid MixtureP



T = constI

II

III


√

√

Van Laar used:

Note that only interactions between 2 molecules are important


amix = x12 a1 + x2

2 a2 + 2 x1 x2 sqrt (a1 a2)

√

Pure liquidsP

Liquid MixtureP



T = constI

II

III

Now we need to put things together

√

√


√

ΔuII = 0

ΔuI = a1 x1 / b1 + a2 x2 / b2


ΔuII = 0

ΔuI = a1 x1 / b1 + a2 x2 / b2


uE = ΔuI + ΔuII + ΔuIII

Substituting:uE = a1x1/b1 + a2x2/b2 + 0 - [x1

2a1 + x22a2 + 2x1x2 sqrt(a1a2) ] / [(x1b1 + x2b2) b1b2 ]

Playing around, we get

uE = x1x2b1b2 / [x1b1 + x2b2] { sqrt(a1)/b1 - sqrt(a2)/b2) }2


uE = x1x2b1b2 / [x1b1 + x2b2] { sqrt(a1)/b1 - sqrt(a2)/b2) }2

According to the initial assumption,

gE = uE

Thus, we obtain

gE = x1x2b1b2 / [x1b1 + x2b2] { sqrt(a1)/b1 - sqrt(a2)/b2) }2

This is an expression for the excess Gibbs free energy, which we can use to obtain the activity coefficients



Remembering that

giE = RT ln [ γi ]

GE = RT ∑i ni ln γi

We can calculate the fugacity coefficients of the two components



ln γ2 = B’ / [ 1 + B’/A’ x2/x1 ]2

ln γ1 = A’ / [ 1 + A’/B’ x1/x2 ]2

But now we have a relationship between molecular parameters and A’ B’:

A’ = b1 / RT [ sqrt(a1)/b1 - sqrt(a2)/b2 ]2

B’ = b2 / RT [ sqrt(a1)/b1 - sqrt(a2)/b2 ]2

Comments• According to the van Laar’s theory, the activity

coefficients are never less than unity

• In other words, the theory predicts always positive deviations from the Raoult’s law

• This is a consequence of the mixing rule,

• which implies

• This means that the force of attraction in the mixture are less than they would be if they were additive on a molar basis

amix = x12 a1 + x2

2 a2 + 2 x1 x2 sqrt (a1 a2)

amix < x1 a1 + x2 a2

Comments

• If we had used mixing rules that result in

• Like, for example,

• Then we would have obtained negative deviations from the Raoult law in all cases

• What if we had used

amix = x1 a1 + x2 a2 + 2 x1 x2 sqrt (a1 a2)

amix > x1 a1 + x2 a2

amix = x1 a1 + x2 a2 ?

Chapter 7The goal of this chapter is to see whether we can use our understanding (theory) of the structure and properties of the liquid mixture to calculate the activity coefficients of the various compounds

The first example was that of the van Laar’s theory

The main problem there was the use of the VdW EOS to calculate ΔuI

However, it was instructive because a simple vision of the solution yield a useful relation often used to fit activity coefficients

The main assumption was that excess entropy and excess volume were 0

The Scatchard-Hildebrand Theory

Stimulated by the results of van Laar’s, Hildebrand introduced the concept of ‘regular solutions’

Regular solutions are those in which the components mix isothermally with no excess entropy provided that there is no volume change upon mixing

Equivalently, the excess entropy is negligiblewhen T=const and V=const

The theory then builds similarly to the procedure van Laar adopted

The Scatchard-Hildebrand Cycle

Pure liquidsP

Liquid MixtureP



T = constI

II

III


Pure liquidsP

Liquid MixtureP



T = constI

II

III

The main problem for van Laar was the assumption he used in I


Pure liquidsP

Liquid MixtureP



T = constI

II

III

The improvement proposed by both Scatchard and Hildebrand was the introduction of a parameter ‘c’ defined as:

c ≡ Δvap u / vL

Δvap u is the energy of complete vaporization (liquid to ideal gas)vL is the molar volume of the liquidc is the cohesive energy density

The Scatchard-Hildebrand TheoryFollowing the same process as described in the case of van Laar’s equations, Scatchard and Hildebrand obtained:

Defining the volume fractions (vL is replaced by v)

φ1 = x1 v1 / [ x1 v1 + x2 v2 ]

φ2 = x2 v2 / [ x1 v1 + x2 v2 ]

Using the fact that

uEideal gases = 0

uE for the mixture is given by

uE = ( c11 + c22 - 2 c12 ) φ1 φ2 ( x1v1 + x2v2 )


uE = ( c11 + c22 - 2 c12 ) φ1 φ2 ( x1v1 + x2v2 )

To use this relation, we need an expression for c12

Here comes the most important assumption:

c12 = sqrt [ c11 c22 ]


uE = ( c11 + c22 - 2 c12 ) φ1 φ2 ( x1v1 + x2v2 )

To use this relation, we need an expression for c12

Here comes the most important assumption:

c12 = sqrt [ c11 c22 ]

Which, upon substitution, leads to

uE = ( x1v1 + x2v2 ) φ1 φ2 [ c11 + c22 - 2 sqrt (c11 c22 )]


The usefulness of this expression is related to the definition of the solubility parameters:

uE = ( x1v1 + x2v2 ) φ1 φ2 [ c11 + c22 - 2 sqrt (c11 c22 ) ]

uE = ( x1v1 + x2v2 ) φ1 φ2 [ sqrt(c11) - sqrt(c22) ]2

δ1 ≡ sqrt [ c11 ] = sqrt [ Δvap u / v ]1δ2 ≡ sqrt [ c22 ] = sqrt [ Δvap u / v ]2


The usefulness of this expression is related to the definition of the solubility parameters:

uE = ( x1v1 + x2v2 ) φ1 φ2 ( δ1 - δ2 )2

δ1 ≡ sqrt [ c11 ] = sqrt [ Δvap u / v ]1δ2 ≡ sqrt [ c22 ] = sqrt [ Δvap u / v ]2

Which allows us to write uE as:

uE = ( x1v1 + x2v2 ) φ1 φ2 [ c11 + c22 - 2 sqrt (c11 c22 ) ]

uE = ( x1v1 + x2v2 ) φ1 φ2 [ sqrt(c11) - sqrt(c22) ]2


uE = ( x1v1 + x2v2 ) φ1 φ2 ( δ1 - δ2 )2


uE = ( x1v1 + x2v2 ) φ1 φ2 ( δ1 - δ2 )2

If, at constant T and P, sE vanishes (notice that this is an additional assumption, although not too strong because we had assumed already that sE is 0 when the volume is constant), then

gE = uE

Thus we can write:

gE = ( x1v1 + x2v2 ) φ1 φ2 ( δ1 - δ2 )2


gE = ( x1v1 + x2v2 ) φ1 φ2 ( δ1 - δ2 )2

From

Remembering that

giE = RT ln [ γi ]

GE = RT ∑i ni ln γi

We can calculate the fugacity coefficients of the two components


gE = ( x1v1 + x2v2 ) φ1 φ2 ( δ1 - δ2 )2

From

We obtain the regular-solution equations:

RT ln γ1 = v1 φ22 [ δ1 - δ2 ]2

RT ln γ2 = v2 φ12 [ δ1 - δ2 ]2

Which relate the activity coefficients to the solubility parameters and the molar liquid volumes of the compounds

Comparison: Van Laar’s EquationsgE = x1x2b1b2 / [x1b1 + x2b2] { sqrt(a1)/b1 - sqrt(a2)/b2) }2

ln γ2 = B’ / [ 1 + B’/A’ x2/x1 ]2

ln γ1 = A’ / [ 1 + A’/B’ x1/x2 ]2

Which provide a relationship between molecular parameters a and b and A’ B’:

A’ = b1 / RT [ sqrt(a1)/b1 - sqrt(a2)/b2 ]2

B’ = b2 / RT [ sqrt(a1)/b1 - sqrt(a2)/b2 ]2


gE = ( x1v1 + x2v2 ) φ1 φ2 ( δ1 - δ2 )2

From

We obtain the regular-solution equations:

RT ln γ1 = v1 φ22 [ δ1 - δ2 ]2

RT ln γ2 = v2 φ12 [ δ1 - δ2 ]2

Which relate the activity coefficients to the solubility parameters and the molar liquid volumes of the compounds

These expressions are similar to van Laar’sHowever, we accept that they are MUCH IMPROVED

How can we say this?

CommentsThe regular solution equations always predict positive deviations from ideality, as van Laar’s equations do

This is a consequence of the assumptionc12 = sqrt [ c11 c22 ]


This is a consequence of the assumption

The solubility parameters δ1 and δ2 are in general a function of THowever, their difference

δ1 - δ2is almost independent of T

c12 = sqrt [ c11 c22 ]


This is a consequence of the assumption

The solubility parameters δ1 and δ2 are in general a function of THowever, their difference

δ1 - δ2is almost independent of T

The main result is that the difference in solubility parameters between the components in a mixture gives us an idea about the extent of the deviations from ideality

c12 = sqrt [ c11 c22 ]

Table 7.1

Compound vL @ 298K (cm3/mol) δ (J/cm3)1/2

Isopentane 117 13.9

n-pentane 116 14.5

n-octane 164 15.3

Toluene 107 18.2

Benzene 89 18.8

Table 7.1Compound vL @ 298K (cm3/mol) δ (J/cm3)1/2

Isopentane 117 13.9

n-pentane 116 14.5

n-octane 164 15.3

Toluene 107 18.2

Benzene 89 18.8

Regular-solution equations give a good semi-quantitative representation of activity coefficients for many solutions

containing non polar components

Example: Fig. 7.2

1 - CO2 - CH4

T = 90.7 K VLE

Example: Fig. 7.3

1 - C6H62 - n-C7H16

T = 70 C VLE

Example: Fig. 7.4

1 - neo-C5H122 - CCl4

T = 0 C VLE

CommentsRT ln γ1 = v1 φ2

2 [ δ1 - δ2 ]2

RT ln γ2 = v2 φ12 [ δ1 - δ2 ]2

Because of the mathematical formalism of the regular-solution equations, they can behave poorly when they are applied on systems of compounds very similarThe reason is that small errors in the geometric mean used for evaluating c12 have large impact in the calculation for gE when the solubility parameter for two liquids are similar

CommentsRT ln γ1 = v1 φ2

2 [ δ1 - δ2 ]2

RT ln γ2 = v2 φ12 [ δ1 - δ2 ]2

Because of the mathematical formalism of the regular-solution equations, they can behave poorly when they are applied on systems of compounds very similarThe reason is that small errors in the geometric mean used for evaluating c12 have large impact in the calculation for gE when the solubility parameter for two liquids are similarAs a consequence, they are most useful for non-polar mixtureshaving appreciable non idealityEven better, when the compounds have different solubility parameter

Multi-Component MixturesAnother advantage for the Scatchard-Hildebran equations is their simplicityIn fact, they can be easily extended to multi-component mixtures

It can be demonstrated that for a component j in the mixture

RT ln γj = vj [ δj - δ ]2

Whereδ = ∑ φi δi

i = 1

m

The sums are extended to all m compounds in the

mixtureAnd

φj = xj vj / [ ∑ xi vi ]i = 1

m

thermodynamic activity coefficient

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