computing liquid-vapor phase diagrams for non-ideal binary mixtures

8/18/2019 Computing Liquid-Vapor Phase Diagrams for Non-ideal Binary Mixtures

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Computing Liquid-Vapor Phase Diagrams for non-ideal binary mixtures

by

Franklin Chen

Department of Natural and Applied Science

University of Wisconsin-Green Bay

Green Bay, WI [email protected]

© Copyright 2005 by the Division of Chemical Education, Inc., American Chemical Society. All

rights reserved. For Classroom use by teachers, one copy per student in the class may be made

free of change. Write to JCE Online, [email protected], for permission to place a

document, free of charge on a class intranet.

Goal

The goal of this document is to introduce the mathematical models that are used to describe

vapor-liquid equilibrium of binary mixtures. This document focuses on the solubility parameter theory

approach to calculate the activity coefficients of non-ideal mixtures.

Prerequisites

1. Experience with concepts of the 2nd Law of thermodynamics, chemical potentials,

and equilibrium constants.

2. Moderate skill with Mathcad.

Performance Objectives

At the end of this exercise you will be able to:

relate the phase rule to liquid-phase and vapor-phase compositions of binary mixtures at1.equilibrium;

explain why azeotrope arises only in the non-ideal mixtures;2.

calculate activity coefficients using solubility parameter theory and the variation of the solubility3.

parameter approaches;

construct a phase diagram of a given non-ideal binary mixture.4.

For Instructors

Although most of the materials in this document are suitable for undergraduate physical chemistry

students, theories of non-ideal binary mixtures may be beyond the scope of introductory physical

chemistry. These theories are collected in the collapsible section embedded in the main

document. Beginning physical chemistry students are encouraged to skip this more advanced

theory section and move on to work on the algorithm for constructing the liquid vapor phase

diagram. Two flow charts were constructed to help students follow the algorithm to construct a

phase diagram. Guided by the flow charts, the algorithm can be understood more easily.

Author: Franklin Chen VapLiqEquilMC11.mcd Page 1


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Exercise 2: Can activity coefficient be greater than 1?

Eq. [5]111 X a γ=

Activity (a1) is related to mole fraction (X1) through activity coefficient, γ1 in Eq.[5]

Write the equilibrium constant for the equilibrium reaction shown in Eq.[2] when chemical potential is

written as Eq.[4] ?

Exercise 1

Throughout the remainder of this document the blue color indicates an

exercise or problem for students to solve.

where a1 is the activity of component 1.

Eq. [4]1

0

11 lna RT += µ

The general expression for a chemical potential is expressed in Equation 4.

Eq. [3]b B

a

A

d D

cC

X X

X X K

K RT G

=

−=∆ ln0

the equilibrium constant and the standard free energy change (in the ideal case) are related as:

Eq. [2]dDcC bBaA +=+

when the mixture is an ideal mixture. For a chemical equilibrium,

Eq. [1]1

0

11 ln X RT += µ

Chemical potential is a partial molar quantity. Raoult's Law allows us to write

Chemical potential, activity, and activity coefficient



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Exercise 3: The experimental data allowing us to calculate the activity coefficients of chlorobenzene

in equilibrium with a chlorobenzene/1-nitropropane solution at 75 o C are listed as follows:

xchlorobenzene

0.119

0.289

0.460

0.691

1

:= pchlorobenzene

19.0

41.9

62.4

86.4

119

:= units in torr

Navy blue color text explainshow to effectively use Mathcad.

i 0 1, 4..:= There are two kinds of subscripts. For asubscript which is part of a variable

name, we use period "." after the variable

name.

When we want an entity from a vector,

we use "[".

In Mathcad, the vector entity starts with

0. This is why we wrote i=0, 1 (;) 4 onthe left. The semicolon appears as ".."

in the document.

achlorobenzenei

pchlorobenzenei

119:=

achlorobenzene

0.16

0.352

0.524

0.726

1

=

Now that you know the activities you should be able to write a vector for the activity coefficients for chlorobenzene at various compositions in the mixture. At what mole fraction of chlorobenzene in the

mixture would its activity coefficient be equal to 1? Are the activity coefficients at other mixture

conditions greater than or less than zero?

Next plot real and ideal vapor pressures of chlorobenzene against mole fractions of chlorobenzenes

and explain the deviations of real vapor pressure from ideal vapor pressure.



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pidealchloroi

xchlorobenzene( )i

119⋅:= Remember Raoult's Law.....

0 0.25 0.5 0.75 110

37.5

65

92.5

120

Ideal

Real

Vapor pressure (p)

torr

Mole fraction of chlorobenzene (x)

Phase rule, Vapor and liquid phase compositions at equilibrium

You may recall that the Gibbs phase rule states that the degree of freedom, f , is related to the

number of components, c, and the number of phases, p, by the following relation:

f=c-p+2 Eq. [6]

Exercise 4 Consider the vapor-phase equilibrium of a binary mixture, how would you assign c and

p ? What is the degree of freedom when both phases are present at constant temperature and

pressure?

The text with maroon color indicates sections that

instructors can delete for copies of this template given

to students.

For a binary mixture, we have c=2. When both liquid and vapor are present, p=2. At constant p

and T, and when both vapor and liquid phases are present , f =0.



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We are now ready to construct vapor-liquid diagrams in which the y-axis is the boiling point of the

mixture while the x-axis is either mole fraction of the component in the liquid phase, or the mole

fraction of the component in the vapor phase.

Exercise 5 What is the definition of a boiling point? What is the total vapor pressure when the

mixture is at the boiling point?

Boiling point of a liquid is the temperature at which the total vapor pressure of the liquid is equal to

the external pressure.

Since boiling points vary with the compositions of mixtures, and the total vapor pressures of the

mixture are also functions of temperature, we need to construct vapor pressure versus temperature

data. Some data at selected temperatures are available in the CRC Chemistry Handbook. We

still, however, need to construct regression equations to interpolate between published points. For

this we will use the Clausius-Clapeyron equation, which you learned in General Chemistry.

−

∆−=

−

121

2 11lnT T R

H p p vap

Eq. [7]

where p1 and T 1 are the reference point vapor pressure and temperature, ∆vapH is heat of

vaporization of the liquid, and R is the gas constant. Data on ∆vapH and boiling points are available

in CRC Handbook.

Exercise 6: What are the assumptions used when Eq[7] is used to calculate p2 at T 2 ?

There are 2 assumptions: 1) Molar volume of a liquid is insignificant when compared to that of agas. 2) Heat of vaporization is a constant from T1 to T2.

We now will set up vapor pressure data for both toluene and benzene

from 350 K to 420 K at 0.7 K interval

R 8.314 joule⋅ K 1−

⋅ mole 1−

⋅:= bar 105

Pa⋅:= kJ 103

joule⋅:=

The following reference gives the triple point and normal boiling point data for toluene

(http://www.nist.gov/srd/PDFfiles/JPCRD3711.pdf) Goodwin R.D., J. Phy. Chem. Ref. Data, 18,

1565-1636.

Toluene, triple point: 178.15K, 4.362*10 -7 bar. boiling point: 383.764K, 1.01325 bar.

Heat of vaporization at 388 K is 33.11 kJ/mol.

Benzene, the reference is http://www.nist.gov/srd/PDFfiles/JPCRD350.pdf, Goodwin,R.D., J Phy.

Chem. Ref. Data, 17, 1541 (1988)

Benzene, triple point: 278.65K, 0.04785 bar. boiling point: 353.24K, 1.01325 bar.

Heat of vaporization at 300K is 33.737 kJ/mol



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Block #1Y j

p2tolu j

X j⋅

pext:=X j

pext p2benz j

−

p2tolu j

p2benz j

−:=

pext 1.013 bar ⋅:=

For the ideal case, the strategy of calculation is to set Tj as the boiling temperature. We will then use

both Raoult's law and Dalton's law to find both Xj (liquid mole fraction for component j) and Yj (vapor

mole fraction for component j) such that pbenzj + ptoulj =pext=1.013 bar

Exercise 8: Let us assume benzene and toluene form an ideal solution. Is this a good

assumption? Why or why not?

Ideal Case:

p2benz j

p1benz exp∆Hvapbenz

R

1

T1benz

1

T j−

⋅

⋅:=

p2tolu j

p1tolu exp ∆HvaptoluR

1T1tolu

1T j

−

⋅

⋅:=

T j 350 j 0.7⋅+( ) K⋅:=

j 0 1, 100..:=

∆Hvapbenz 33.737 kJ⋅ mole 1−

⋅:=∆Hvaptolu 33.11 kJ⋅ mole 1−

⋅:=

p1tolu 1.013 bar ⋅:=p1benz 1.013 bar ⋅:=

T1benz 353.24 K⋅:=T1tolu 383.76 K⋅:=

Exercise 7: We choose our reference points for toluene and benzene as their normal boiling

points. Explain why these fixed points are chosen as reference points?



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Liquid Vapor Equilibrium of Benzene Toluene Mixture

0 0.25 0.5 0.75 1340

352.5

365

377.5

390

Liquid

Vapor

b.p. (K)

Mole Fraction of Toluene

Exercise 9: This diagram correctly predicts the b.p. of pure benzene which is at 353 K,and

b.p. of toluene which is at 383.76K. Practice yourself using the triple point of the toluene as

the reference point. What do you see?

Now, in the ideal case plot 1-Yi vs (1-X

j) (benzene vapor phase composition against benzene

liquid phase composition), what do you see? Would this plot convince you that there will be

no azeotrope when (1-Yi)>(1-Xi) whether 1-X-->0 or 1-X-->1?

0 0.5 10

0.5

1

1 Y j

−

1 X j

−



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∆U and ∆H are related as ∆U=∆H-RT( ) 2/1/V U ∆=δ

Exercise C1: Why do we subtract RT from ∆vapH in Eq. [C4]? (This Exercise is the same

as Exercise 10 in the main document).

where V m is the molar volume of the mixture, d1 and d2 are the solubility parameters of

components 1, and 2 respectively, V is the molar volume. f 1 and f 2 are the component

volume fractions of components 1 and 2.

Eq [C4]( )[ ] 2/1/V RT H vap −∆=

Eq.[C3]( )2

2121 −=∆ mcontact

m V H

Different models propose different expressions for the excess chemical potential.

Hildebrand estimated the enthalpy of mixing, ∆Hmcontact using differences of solubility

parameters.

Eq [C2]( ) 1

Excess0

11 ln RT γ µ µ =−

Eq. [C1]( ) 1011 ln X RT ideal

=− µ

Non-ideal Case comes from enthalpy and entropy changes induced by interactions taking

place when solutions form. The effects of the interactions is treated in terms of a contact free

energy change,∆

Gcontact m .

This collapsible section covers theory for the non ideal case.

In this document, van Laar's theory is used to calculate activity coefficient of each

component in a binary mixture: Activity coefficients ( γ1 and γ2 ) are calculated based on

differences in solubility parameters.

Nonideal Solutions



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10/22

( )2212

122ln −= −

V RT Eq [C6]

If we define Van Laar Coefficients as:

( )

( )2212

21

221

112

−=

−=−

−

RT

V A

RT V A

Eq.[C7]

Equation [C7] is the same as Eq.[11] in the main document.

Then, we can show that

( )212121212112

A X A X

X X A A

RT

H g

contact

m

+=∆

= Eq [C8]

2

112

221

21

2

2

2

221

112

12

1

1

1

ln

1

ln

+

=∂∂

=

+

=∂∂

=

X A

X A

A

X

g

X A

X A

A

X

g

γ

γ This is because:

1

1

lnγ RT n

H contact

m =∂

∆∂Eq [C9]

Equation [C9] is the same as Eq.[12] in the main document.

Although Van Laar theory is not different from the solubility parameter theory in principle, it

does provide an algorithm to calculate the activity coefficients for a non-ideal mixture if

azeotrope data of the mixtures are available.

The tasks are to determine A12 and A21 , two Van Laar coefficients. Once these

two coefficients are found, Eq.[C9] allows us to calculate g1 and g2 at any compositions.

Looking at Eq.[C7], A12 and A21 can be calculated from solubility parameters and partial

molar volumes of the components. Solubility parameters can be calculated from heat of

vaporization, and molar volume. Partial molar volume can be calculated fromdensity-composition data. In principle, they can be calculated, although the calculations

may be quite tedious.

If we know the azeotrope point at one known external pressure, then Eq.[C8] and Eq[C9]

allow us to calculate g1 and g2 at any compositions. These data, in turn, allow us to

calculate azeotrope points at any other external pressures. We shall use water-propanol

binary system as the example.



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Before we discuss the calculations of g1 and g2 at the azeotrope conditions, we shall simplify the

expressions in the brackets of Eq[C9]

Exercise C3: Show that

221

112

11

22

ln

ln

x A

x A

x

x

⋅

⋅=

⋅

⋅

γ

γ Eq [C10]

Before we proceed to prove Eq[C10], we shall see the utility of Eq[C10]. This relation

suggests that A12 and A21 can be obtained from Eq[C9] using the relation of Eq[C10].

( )

( )( )( )

2

22

112

2

112

221221

2

11

22

1

2

221

112

112

ln

ln1)ln(1)ln(

ln

ln1)ln(1)ln(

⋅⋅

+⋅=

⋅⋅

+⋅=

⋅

⋅+⋅=

⋅

⋅+⋅=

x

x

x A

x A A

x

x

x A

x A A

γ

γ γ γ

γ

γ γ γ

Eq [C11]

Eq[C11] is the same as Eq.[13] in the main document. Eq.[C13] indicates that A 12 and A12

can be calculated for a given x1, x2, g1, and g2. Once A12 and A21 are determined, Eq[9] can

the be used to determine g1 and g2 at any compositions

We now shall proceed to prove Eq [C10]

In Eq[C9], multiplying Eq[C9 b] with x2 and Eq[C9 a] with x1 and take the ratio for these 2

expressions. We have:

221

112

2

1

2

12

2

1

2

12212112

2

2

2

21

112

2

2

2

21

2

1

2

12212112

2

2

2

21

221

112

2

221

112

2

112

221

221

11

22

x A

x A

x A

x A x x A A2 x A x A

x A

x A x x A A2 x A x A

x A

x A

x A1

x A

x A1

x A

ln x

ln x

⋅⋅=

⋅

⋅+⋅⋅⋅⋅+⋅⋅⋅

⋅

⋅+⋅⋅⋅⋅+⋅⋅⋅

=⋅

⋅⋅

+

⋅

⋅

⋅+

⋅=

⋅

⋅

γ

γ

You can go back to the original document to continue for constructing a vapor-liquid

diagram (VLE) for water-propanol binary system.



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Then, we can show that

Eq.[11]

( )

( )2212

21

2

21

1

12

−=

−=

−

−

RT

V A

RT

V A

Van Laar Coefficients are defined as:

Eq [10]( )2212

122ln −= −

V RT

Eq [9]( )2212

211ln −= −

V RT

Van Laar theory allows us to compute γ1 and γ2 using the differences in the solubility

parameters.

Van Laar Theory

When two solubility parameter numbers are really close, the mixture is close to an ideal

mixture. The enthalpy of interaction between toluene and benzene is approximately zero. The

mixture is close to ideal. This is shown in exercise C2 in the collapsed section.

∆U and ∆H are related as ∆U=∆H-RT( ) 2/1/V U ∆=δ

The definition of the solubility parameter is actually:

Exercise 10: Why do we subtract RT from ∆vapH in Eq. [8]?

where δ1 and δ2 are the solubility parameters of components 1, and 2 respectively, V is the molar

volume of the volatile liquid. φ1 and φ2 are the component volume fractions of components 1 and

2.

Eq [8]( )[ ] 2/1/V RT H vap −∆=

The solubility parameter, δ, is defined as follows:

Here we are just going to use several essential equations to allow students compute vapor-liquid

equilibrium in the non-ideal case.



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2

112

221

212

2

221

112

121

1

ln

1

ln

+

=

+

=

X A

X A

A

X A

X A

A

γ

γ

Eq [12]

( )( )

( )( )

2

22

112

2

112

221221

2

11

221

2

221

112112

ln

ln1)ln(1)ln(

ln

ln1)ln(1)ln(

⋅⋅

+⋅=

⋅⋅

+⋅=

⋅

⋅+⋅=

⋅

⋅+⋅=

x

x

x A

x A A

x

x

x A

x A A

γ

γ γ γ

γ

γ γ γ

Eq [13]

Eq[13] indicates that A12 and A12 can be calculated for a given x1, x2, γ1, and γ2. Once A12and A21 are determined, Eq[12] can they be used to determine γ1 and γ2 at any other

compositions.

Exercise 11 In a binary mixture, x1/x2 are liquid compositions and y1/y2 are vapor compositions of

component 1 and 2. At azeotropic composition, which one of the following is correct?

(a) x1=x2; y1=y2; (b) x1=y1, x2=y2



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The following flow chart illustrates the strategy to calculate activity coefficients γ1and γ2 at any composition.

Example : Water-propanol mixture

Azeotrope data for water-propanol

For the n-propanol/water binary system, xpropanol=ypropanol=0.42 for pext=1.013bar;

Water, Tb=373.15K, ∆vapH=44.016 kJ/mol at 298 K

Propanol, Tb=370.3 K ∆vapH=47.5 kJ/mol

Molar volume for water is 18mL/mole

Molar volume for n-propanol is 74.84 mL/mole

References: Majer,V., and Svoboda,V., [1985] Lee and Shen[2003]

Eq [14]



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These are the partial pressures that the pure substances would have at the azeotrope if

they acted as ideal gases. The activity coefficients at the azeotrope can be calculated as

follows:

Exercise 13: Why the sum of the partial pressures of propanol and water are greater the

total external pressure which is 1.013 bar?

p2wateraz p2propanolaz+ 1.306 bar =

p2propanolaz 0.679 bar =

p2propanolaz p1propanol exp∆Hvappropanol

R

1

T1propanol

1

Taz−

⋅

⋅:=

Eq [16]

p2wateraz 0.627 bar =p2wateraz p1water exp∆Hvapwater

R

1

T1water

1

Taz−

⋅

⋅:=

Taz 360.95 K⋅:=

T1water 373.15 K⋅:=p1water 1.013 bar ⋅:=∆Hvapwater 44.016 kJ⋅ mole

1−⋅:=

p1propanol 1.013 bar ⋅:=∆Hvappropanol 47.5 kJ⋅ mole 1−

⋅:=T1propanol 370.3 K⋅:=

pext 1 bar ⋅:=X1az 0.42:=Taz 360.95:=

Propanol Mole Fraction at the azeotrope (assertive) point

The difference is big enough to treat the case with nonlinear approach. That means we need to

calculate activity coefficients for water and propanol at many compositions.

δpropanol 24.527 joule0.5

m 1.5−

⋅ 103

⋅=δpropanol47.5 kJ⋅ mole

1−⋅ R 298⋅ K⋅−

74.84 mL⋅ mole 1−

⋅

0.5

:=

Eq [15]

δwater 48.038 joule0.5 m 1.5−⋅ 103⋅=δwater

44.016 kJ⋅ mole 1−

⋅ R 298⋅ K⋅−( )18 mL⋅ mole

1−⋅

0.5

:=

Exercise 12: Given the data in Eq[14], calculate solubility parameters for water and propanol.



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γ2 j 1+

exp A21

1

1 XN j 1+

−( ) A21⋅XN

j 1+( ) A12⋅+

2

:=γ1 j exp A12

1XN j A12⋅

1 XN j−( ) A21⋅+

2

:=

XNii

30:=

j 0 29..:=i 0 30..:=

Activity coefficients at other compositions--water-propanol mixture

A21 1.195=

A12 2.749=Eq [20]

A12 ln γ1az( ) 1X2az ln γ2az( )⋅

X1az ln γ1az( )⋅+

2

⋅:= A21 ln γ2az( ) 1

X1az ln γ1az( )⋅

X2az ln γ2az( )⋅+

2

⋅:=

X2az 1 X1az−:=Eq [19]

Using the van Laar equation (Eq[13]), one can calculate the van Laar constants for this system.

Van Laar constants at azeotrope for water-propanol

γ2az 1.595=γ1az 1.472=

Eq [18]

γ2azpext

p20az:=γ1az

pext

p10az:=

Eq [17]Recall that a=(pext y az )/p*=(pext x az )/p*; γ=a/x az; γ=pext /p*=pext /paz

p20az p2wateraz:=p10az p2propanolaz:=

Activity coefficients at azeotrope for water-propanol



17/22

γ130

1:=γ2

0 1:= Block #2

The algorithm of Block #2 is straightforward, except you need to pay attention the indices i and j

because you don't want XNj=1 in calculating γ1j, and XN j+1 =0 in calculating γ2 j+1.

To complete the phase diagram, we must find the temperatures at which a solution

of mole fraction X1 will boil.

This calculation is performed by using the temperature-dependent values of the

partial pressures from the Clausius Clapeyron equation and the activity coefficients from the

van Laar equation.

Phase diagram Construction: Water-Propanol

The following diagram illustrates the algorithm strategy of constructing water-propanol phase diagram:

MATHCAD uses a function in which an initial guess is made, and a Leavenberg-Marquardt

algorithm modifies the guess, minimizing the difference on either side of the = sign. You should

refer to Mathcad Resource Center " Solving a Nonlinear System Equations for help.

As an initial guess, try 350K., which is slightly below the boiling points of either

of the two species.



18/22

Eq [31]

YNi

γ1i XNi⋅ p1p exp ∆H1pR

1T1p

1Tni

−

⋅

⋅

⋅

γ1i XNi⋅ p1p exp∆H1p

R

1

T1p

1

Tni−

⋅

⋅

⋅ γ2i 1 XNi−( )⋅ p1w exp

∆H1w

R

1

T1w

1

Tni−

⋅

⋅

⋅+

:=

YN is a vector containing the vapor-phase mole fractions. It is found directly

from Dalton Law of Partial Pressures.

This is a vector of the temperatures corresponding to

the liquid mole fraction vector XN. Type Tn= in the space

to the right to see all of the components of the vector.

Tn Temp XN γ1, γ2,( )→

:=

Exercise 14: What is the physical meaning of Eq.[22] ?

End of the solve block #3.

Temp XN γ1, γ2,( ) Find Tn( ):=

Note here we use Ctr= for the

equal sign on the equation

above. XN, γ1, and γ2 arevectors.

Eq [22]p1p exp ∆H1pR

1T1p

1Tn

−

⋅

⋅ XN⋅ γ1⋅ p1w exp ∆H1wR

1T1w

1Tn

−

⋅

⋅ 1 XN−( )⋅ γ2⋅+ pext=

GivenBeginning of the solve block.

Tn 350 K⋅:=

Eq [21]T1w T1water :=

∆H1w ∆Hvapwater :=T1p T1propanol:=

p1w p1water := ∆H1p ∆Hvappropanol:=p1p p1propanol:=

Let us simplify the notation as follows:



19/22

Exercise 14: Explain the physical meanings of the numerator and the denominator of Eq.[31]

Exercise 15: Plot the non-ideal liquid-vapor phase diagram for water-propanol system.

NON-IDEAL LIQUID-VAPOR PHASE DIAGRAM

0 0.2 0.4 0.6 0.8350

355

360

365

370

375

380

liquid

vapor

Tni

Tni

XNi

YNi

,

Temp

Mole Fraction of propanol



20/22

Now plot activity coefficients as a function of mole fraction.

Activity Coefficients

0 0.2 0.4 0.6 0.80

5

10

15

20

γ1i

γ2i

XNi

References

Gentilcore,M., and Healthcare, T., (2004) Apply Solubility Theory for Process Improvement,

www. cepmagzine.org; 38-41

Lee,L.S., and Shen, H.C., Ind. Eng. Chem. Res; 42, 5905-5914(2003)

Majer,V., and Svoboda,V., Enthalpies of vaporization of organic compounds: a critical review and

data compilation.International Union of Pure and Applied Chemistry Chemical Data Series, 1985,

N 32, 300 pp

Young,S., (1996) Computing a Liquid-Vapor Phase Diagram ; Symbolic Mathematics Document

for Physical Chemistry http://bluehawk.monmouth.edu/~tzielins/mathcad/index.htm



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computing liquid-vapor phase diagrams for non-ideal binary mixtures

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