Chapter 11

c° 2011 Ismail Tosun

Chapter 11

Liquid-Liquid Equilibrium

When two liquids are mixed, depending on the temperature and/or composition, either theyare completely miscible in each other and form a single phase, or they are partially miscible(or totally immiscible) in each other and form two separate phases. In the case of completelymiscible, i.e., single phase, systems, the equations developed in Chapter 6 are applicable. In thischapter, we are interested in the case when a liquid is partially miscible (or totally immiscible)in another liquid. First, the reason for making liquids partially miscible in each other willbe investigated. Then the procedure to determine the compositions in two separate liquidphases that are in equilibrium with each other will be described. Such equilibrium data areneeded in various chemical and physical processes such as liquid-liquid extraction and tertiaryoil recovery.

11.1 STABILITY OF MULTICOMPONENT LIQUID MIXTURES

For thermodynamically allowable changes at constant temperature and pressure, the criterion

(dG)T,P ≤ 0 (11.1-1)

indicates that all irreversible changes taking place at constant temperature and pressure mustdecrease the total Gibbs energy of the system. When the system reaches equilibrium, no furtherchanges can occur and Gibbs energy reaches its minimum value.

Let us consider the mixing of two liquids, 1 and 2, at constant temperature and pressure.The Gibbs energy of this binary mixture is

Gmix = n1 eG1 + n2 eG2 +∆Gmix (11.1-2)

If the Gibbs energy of the mixture is lower than the summation of the Gibbs energies of pureliquids, then liquids 1 and 2 will be miscible (totally or partially) in each other. Otherwise,liquids 1 and 2 tend to remain unmixed like olive oil and water. The condition of miscibility ismathematically expressed as

Gmix < n1 eG1 + n2 eG2 (11.1-3)

or, in other words,∆Gmix < 0 (11.1-4)

which is the criterion for miscibility (partial or complete).The Gibbs energy of mixing is given by

∆ eGmix = ∆ eHmix| {z }X

− T ∆eSmix| {z }Y

(11.1-5)

Since mixing increases the degree of disorder within the system, the entropy change on mixing,∆eSmix, is always positive. Thus, the term Y in Eq. (11.1-5) is always positive. The heat of

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mixing, ∆ eHmix, is related to the energetic effects. In order to make a mixture of 1 and 2 frompure components, it is necessary to break 1-1 and 2-2 bonds and form 1-2 bonds1. Depending onthe magnitude of the interactions between like and unlike molecules, ∆ eHmix may take positiveor negative values. Therefore, it is necessary to consider the following two cases.

Case (i): Exothermic mixing (∆ eHmix < 0)

When ∆ eHmix < 0, interactions between unlike molecules are greater than those between likemolecules, i.e., negative deviation from Raoult’s law leading to γi < 1. In this case, ∆ eGmix isnegative for all compositions of components 1 and 2. Therefore, liquids are completely misciblein each other and form a stable single phase. This situation is illustrated in Figure 11.1.

0

mixG~Δ

mixH~Δ

mixS~TΔ−

1 0 1x

0

mixG~Δ

mixH~Δ

mixS~TΔ−

1 0 1x

Figure 11.1 Effects of ∆eSmix and ∆ eHmix on ∆ eGmix when ∆ eHmix < 0.

Case (ii): Endothermic mixing (∆ eHmix > 0)

When ∆ eHmix > 0, interactions between like molecules are stronger than those between unlikemolecules, i.e., positive deviation from Raoult’s law leading to γi > 1. In this case, examinationof Eq. (11.1-5) indicates that the terms X and Y compete with each other and the magnitudeof the temperature decides the dominant term and, hence, the sign of ∆ eGmix.

a) At low temperatures, ∆ eHmix À T ∆eSmix and, as shown in Figure 11.2, this indicates that∆ eGmix > 0. Therefore, liquids 1 and 2 tend to remain as two separate phases throughout thewhole composition range, i.e., liquids 1 and 2 are immiscible in each other.

mixS~TΔ−

0

mixH~Δ

mixG~Δ

1 0 1x

Figure 11.2 Effects of ∆eSmix and ∆ eHmix on ∆ eGmix when ∆ eHmix > 0 with∆ eHmix À T∆eSmix.

b) When temperature is high enough so that ∆ eGmix < 0, we may encounter two differentsituations. When T ∆eSmix À ∆ eHmix, liquid phases mix with each other throughout the wholecomposition range as shown in Figure 11.3.

1As stated in Section 6.3, while bond breaking is an endothermic process, bond formation is exothermic.

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0

mixG~Δ

mixH~Δ

mixS~TΔ−

1 0 1x

Figure 11.3 Effects of ∆eSmix and ∆ eHmix on ∆ eGmix when ∆ eHmix > 0 withT∆eSmix À ∆ eHmix.

An interesting case may arise when T ∆eSmix > ∆ eHmix. Under these circumstances, varia-tion in ∆ eGmix with respect to composition may result in more than one minimum, as shownin Figure 11.4. Examination of this case requires the following mathematical preliminaries.

1x 1

0

0

mixG~Δ

Figure 11.4 Variation in ∆ eGmix with composition when ∆ eHmix > 0 withT∆eSmix > ∆ eHmix.

A function f(x) is said to be a concave function in x if and only if the points on a chordconnecting two points on a function lie beneath the function. On the other hand, a functionf(x) is said to be a convex function in x if and only if the points on a chord connecting twopoints on a function lie above the function. Concave and convex functions are shown in Figure11.5.

x

)x(f

x

)x(fFunctionConcave FunctionConvex

Figure 11.5 Concave and convex functions.

Consider a function f that is dependent on a single variable x. Various possibilities existfor the variation of f(x) with x as shown in Figure 11.6. Note that the functions on the left(a and c) are convex, while the functions on the right (b and d) are concave.

383

x

)x(f 00 2

2>>

dxfd

dxdf

x

)x(f 00 2

2<>

dxfd

dxdf

x

)x(f 00 2

2><

dxfd

dxdf

x

)x(f 00 2

2<<

dxfd

dxdf

(a) (b)

(c) (d)

Figure 11.6 Variation of the first and second derivatives of convex and concave functions.

The first derivative of f(x), df(x)/dx, measures the rate of change of the function f(x).The second derivative of f(x), d2f(x)/dx2, measures the rate of change of the first derivative,df(x)/dx. The signs of the first and second derivatives of f(x) imply the following:

df(x)

dx

½> 0 Value of f(x) increases with increasing x< 0 Value of f(x) decreases with increasing x

(11.1-6)

d2f(x)

dx2

½> 0 Slope of f(x) vs x tends to increase< 0 Slope of f(x) vs x tends to decrease

(11.1-7)

Note that the second derivative can be used to determine whether a function is concave orconvex. The second derivatives of concave and convex functions are negative and positive,respectively.

Let us consider a concave function of a single variable, f(x), and draw a chord joining anytwo points on the function as shown in Figure 11.7. The chord intersects the function at xAand xB. Let xc be any value between xA and xB such that

xc = λxA + (1− λ)xB (11.1-8)

where 0 ≤ λ ≤ 1.

x

[ ]BA x)(x λλφ −+ 1

Cx BxAx

)x(f A

[ ]BA x)(xf λλ −+ 1

)x(f

)x(f B

Figure 11.7 A concave function.

384

The equation of the chord is

φ(x) =f(xB)− f(xA)

xB − xA(x− xA) + f(xA) (11.1-9)

From Figure 11.7 it is apparent that

φhλxA + (1− λ)xB

i< f

hλxA + (1− λ)xB

i(11.1-10)

When x = λxA + (1− λ)xB, from Eq. (11.1-9) we have

φhλxA + (1− λ)xB

i= λ f(xA) + (1− λ) f(xB) (11.1-11)

Substitution of Eq. (11.1-11) into Eq. (11.1-10) gives

λ f(xA) + (1− λ) f(xB) < f£λxA + (1− λ)xB

¤(11.1-12)

Now let us tackle the case shown in Figure 11.4 by drawing a common tangent line to theminima2, as shown in Figure 11.8. The so-called common tangent rule states that when thecomposition of the mixture is between xα1 and x

β1 , instead of having a homogeneous mixture, we

have two separate phases, α and β, that are in equilibrium with each other. The compositionsof these coexisting equilibrium phases lie at the points of co-tangency, i.e., xα1 and xβ1 . Thecoexisting compositions are called binodal points. The reason for the phase separation can beexpressed as follows.

Common tangent to the minima

α1x β

1x 1x 1

0

0

mixG~

Δ

Points of co-tangency (not minima)

Inflection points

021

2

=Δ

dxG~d mix

Figure 11.8 Common tangent rule.

Let x∗1 be the mole fraction of component 1 between xα1 and xβ1 , i.e.,

x∗1 = λxα1 + (1− λ)xβ1 (11.1-13)

where 0 ≤ λ ≤ 1. Note that the solution of Eq. (11.1-13) for λ gives

λ =xβ1 − x∗1

xβ1 − xα1

(11.1-14)

2Keep in mind that drawing a common tangent to the minima does not imply joining the minimum pointsby a straight line.

385

From the lever rulexβ1 − x∗1

xβ1 − xα1

=nα

nα + nβ(11.1-15)

Comparison of Eqs. (11.1-14) and (11.1-15) reveals that the term λ represents the mole fractionof the α-phase.

According to Eq. (11.1-12)

λ∆ eGmix(xα1 ) + (1− λ)∆ eGmix(x

β1 ) < ∆

eGmix

hλxα1 + (1− λ)xβ1

i(11.1-16)

which states that the combined Gibbs energies of the two separate liquid phases are lower thanthe Gibbs energy of a homogeneous mixture. Since the system tries to minimize its Gibbsenergy, over this composition range we have two separate phases of compositions xα1 and xβ1 ,and the common tangent acts as a "tie line".

When xα1 < x1 < xβ1 , there are two inflection points at which d2∆ eGmix/dx21 = 0. The

inflection points are called spinodal points. Between the two inflection points, ∆ eGmix is aconcave function with d2∆ eGmix/dx

21 < 0.

Figure 11.8 shows the variation of ∆ eGmix as a function of composition at a fixed tempera-ture, say T1. By using these data, it is possible to locate binodal and spinodal compositions ona temperature-composition (or solubility) diagram, as shown in Figure 11.9. If ∆ eGmix versusx1 data are known at various temperatures, then it is possible to draw binodal and spinodalcurves.

T curveBinodal

1 0

1T

1x

1x 1

0

0

mixG~Δ 1TT =

curveSpinodal

Figure 11.9 Liquid-liquid solubility diagram.

The binodal and spinodal curves coincide at the critical solution ( or consolute) tempera-ture, i.e., the temperature at which two partially miscible liquids become fully miscible. Inother words, when the temperature is above the critical solution temperature, the mixtureis completely miscible and forms a homogeneous phase. At temperatures below the critical

386

solution temperature, the mixture is partially miscible and the binodal curve is the boundarybetween the two-phase and one-phase regions. Within the two-phase region, the region betweenthe spinodal and binodal curves is metastable, and the time it takes for the phase separationis not definite. On the other hand, within the spinodal curve, the mixture is unstable andphase separation takes place immediately. Since the end points of the spinodal curve representinflection points, then the condition of instability of liquid mixtures can be stated as

d2∆ eGmix

dx21

< 0 Condition of instability (11.1-17)

where

∆ eGmix

RT=∆ eGIM

mix

RT+eGex

RT

=NXi=1

xi lnxi +eGex

RT(11.1-18)

For a binary system, substitution of Eq. (11.1-18) into Eq. (11.1-17) gives

1

x1x2+

d2( eGex/RT )

dx21

< 0 (11.1-19)

Therefore, the concentrations at the spinodal points are obtained from the solution of thefollowing equation:

1

x1x2+

d2( eGex/RT )

dx21

= 0 (11.1-20)

One should keep in mind that the criterion given by Eq. (11.1-4) is a necessary but notsufficient condition for complete miscibility of components. When ∆Gmix (or ∆ eGmix) is aconvex function of composition with only one minimum, components are miscible in each otherover the entire composition range and form a homogeneous phase. When∆Gmix (or∆ eGmix) is aconvex function of composition with more than one minimum, phase separation, i.e., formationof two (or more) phases, takes place.

Example 11.1 In a binary liquid mixture of 1 and 2 at constant temperature and pressure,the molar excess Gibbs energy is expressed as

eGex

RT= x1x2

h1.8 + 0.3 (x1 − x2)− 0.85 (x1 − x2)

2i

Determine the region of instability.

Solution

Substitution of eGex/RT into Eq. (11.1-20) and differentiation give

408x41 − 852x31 + 511x21 − 67x1 − 10 = 0 (1)

The solution of Eq. (1) by MATHCADR°gives x1 = − 8.54 × 10−2, x1 = 0.378, x1 = 0.682

and x1 = 1.114. Therefore, the limits of instability are given as

0.378 < x1 < 0.682

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In other words, these are the mole fractions at the spinodal points.

Comment: Equation (11.1-18) gives

∆ eGmix

RT= x1 lnx1 + x2 lnx2 + x1x2

h1.8 + 0.3 (x1 − x2)− 0.85 (x1 − x2)

2i

The variation of ∆ eGmix/RT versus x1 is given in the figure below.

0 0.2 0.4 0.6 0.8 1

0.3−

0.2−

0.1−

00.043−

0.35−

G x( )

0.990.01 x

The common tangent rule gives the binodal compositions as x1 = 0.26 and x1 = 0.80.

The simplest function to express molar excess Gibbs energy as a function of composition isthe one-constant Margules model, Eq. (8.4-1),

eGex

RT= Ax1x2 (11.1-21)

Substitution of Eq. (11.1-21) into Eq. (11.1-18) gives

∆ eGmix

RT= x1 lnx1 + x2 lnx2 +Ax1x2 (11.1-22)

Note thatd

dx1

Ã∆ eGmix

RT

!= ln

µx1x2

¶+A(x2 − x1) (11.1-23)

indicating that d(∆ eGmix/RT )/dx1 = 0 at x1 = x2 = 0.5. On the other hand, the use of Eq.(11.1-21) in Eq. (11.1-19) indicates that phase splitting takes place when

A >1

2

µ1

x1x2

¶(11.1-24)

The function 1/x1x2 becomes +∞ for pure components, i.e., either x1 = 1 or x2 = 1, and fallsto a minimum value of 4 at x1 = 0.5. Therefore, a binary liquid mixture becomes unstable andexists as two separate phases when A > 2. From Eq. (8.4-4)

A = ln γ∞i i = 1, 2 (11.1-25)

388

Therefore, phase separation takes place when ln γ∞i > 2 (or γ∞i > 7.4). A plot of Eq. (11.1-22)in the form of ∆ eGmix/RT versus x1 is presented in Figure 11.10 for various values of A, i.e.,0, 1, 2, 3, 4.

0 0.2 0.4 0.6 0.8 10.8−

0.6−

0.4−

0.2−

0

0.2

0.40.307

0.693−

G x1 1, ( )

G x1 0, ( )

G x1 2, ( )

G x1 3, ( )

G x1 4, ( )

0.9991 10 3−× x1

Figure 11.10 A plot of Eq. (11.1-22) in the form of ∆ eGmix/RT versus x1 with A being aparameter.

The parameter of the two-suffix Margules equation, A, is dependent on temperature. IfA decreases with increasing temperature as shown in Figure 11.11-a, then the temperaturecorresponding to A = 2 is the upper critical solution temperature (UCST). At temperaturesgreater than UCST, a binary mixture becomes homogeneous. When A increases with increas-ing temperature as shown in Figure 11.11-b, then the temperature corresponding to A = 2 isthe lower critical solution temperature (LCST). In this case, a binary mixture becomes homo-geneous at temperatures less than LCST. If the variation of A with respect to temperatureexhibits a maximum, then a binary mixture has both UCST and LCST as shown in Figure11.11-c. Finally, if the variation of A with respect to temperature exhibits a minimum, thenthe corresponding solubility diagram is as shown in Figure 11.11-d.

11.2 LIQUID-LIQUID PHASE EQUILIBRIUM CALCULATIONS

Consider a multicomponent mixture of k species distributed in two liquid phases, α and β.The condition of equilibrium states that the fugacities of each species in α- and β-phases mustbe equal to each other, i.e.,

bfαi (T, P, xαi ) = bfβi (T, P, xβi ) i = 1, 2, ..., k (11.2-1)

In terms of activity coefficients, Eq. (11.2-1) takes the form

γαi (T, P, xαi )x

αi fi(T, P ) = γβi (T, P, x

βi )x

βi fi(T, P ) (11.2-2)

orγαi (T, P, x

αi )x

αi = γβi (T, P, x

βi )x

βi i = 1, 2, ..., k (11.2-3)

In each phase, the mole fractions are related to each other by the following equations:

kXi=1

xαi = 1.0 andkXi=1

xβi = 1.0 (11.2-4)

389

Simultaneous solution of Eqs. (11.2-3) and (11.2-4) gives the coexistence curve for the two-phase system.

UCST

x1

x1

T

(b)

T

T* (LCST)

TWO PHASES

2

A

T

ONE PHASE

TWO PHASES

ONE PHASE T* (LCST)

x1

→

T

TWO PHASES

ONE PHASE

T* (UCST)

TWO PHASES

2

A

T

T* (UCST)

x1

ONE PHASE

→

(a)

T

LCST

LCST UCST

TWO PHASES

ONE PHASE

T

2

A

ONE PHASE →

TWO PHASES

UCST

ONE PHASE

UCSTLCST

TWO PHASES

ONE PHASE

T

2

A

ONE PHASE LCST

→ TWO PHASES

TWO PHASES

(c)

(d)

Figure 11.11 Four different types of solubility diagram.

For a binary system, Eqs. (11.2-3) and (11.2-4) become

ln

Ãγα1

γβ1

!= ln

⎛⎝xβ1

xα1

⎞⎠ (11.2-5)

ln

Ãγα2

γβ2

!= ln

⎛⎝1− xβ1

1− xα1

⎞⎠ (11.2-6)

390

In practice, one encounters two types of liquid-liquid equilibrium calculations:

• When compositions are known, Eqs. (11.2-5) and (11.2-6) can be solved for the parametersof an activity coefficient model.

• When parameters of an activity coefficient model are known, Eqs. (11.2-5) and (11.2-6) canbe solved for the compositions in α- and β-phases.

Example 11.2 Diethyl ether (1) and water (2) form two partially miscible liquid phases. At308K and atmospheric pressure, Villamanan et al. (1984) reported the following compositionsof the two phases:

xα1 = 0.01172 and xβ1 = 0.9500

If the system is represented by the three-suffix Margules equation, estimate the parameters Aand B.

Solution

Substitution of Eqs. (8.4-6) and (8.4-7) into Eqs. (11.2-5) and (11.2-6), respectively, gives

A

∙(1− xα1 )

2 −³1− xβ1

´2¸+B

h(1− xα1 )

2 (4xα1 − 1)− (1− xβ1 )2(4xβ1 − 1)

i= ln

⎛⎝xβ1

xα1

⎞⎠ (1)

A

∙(xα1 )

2 −³xβ1

´2¸+B

h(xα1 )

2 (4xα1 − 3)− (xβ1 )2(4xβ1 − 3)

i= ln

⎛⎝1− xβ1

1− xα1

⎞⎠ (2)

The use of matrix algebra gives the parameters A and B as

µAB

¶=

⎡⎣(1− xα1 )2 − (1− xβ1 )

2 (1− xα1 )2 (4xα1 − 1)− (1− xβ1 )

2(4xβ1 − 1)

(xα1 )2 − (xβ1 )2 (xα1 )

2 (4xα1 − 3)− (xβ1 )2(4xβ1 − 3)

⎤⎦−1

×

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣ln

⎛⎝xβ1

xα1

⎞⎠

ln

⎛⎝1− xβ1

1− xα1

⎞⎠

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦(3)

or

A =X (Λ−Θ)− 6Λ (xα1 + xβ1 − 1)

2 (xα1 − xβ1 )3

and B =(xα1 + xβ1 ) (Θ− Λ) + 2Λ

2 (xα1 − xβ1 )3

(4)

where

X = (xα1 + xβ1 )(4xα1 + 4x

β1 − 3)− 4xα1x

β1 Θ = ln

⎛⎝xβ1

xα1

⎞⎠ Λ = ln

⎛⎝1− xβ1

1− xα1

⎞⎠ (5)

Substitution of the numerical values into Eqs. (4) and (5) gives

A = 3.854 and B = − 0.683

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Comment: The use of Eq. (8.4-9) gives

γ∞1 = 93.4 and γ∞2 = 23.8

As stated in Section 8.3, activity coefficients are greater than unity when components repeleach other, i.e., unlike interactions are weaker than like interactions. When γi (or γ

∞i ) is very

much greater than unity, it is more likely that phase separation will take place.

Example 11.3 Estimate the compositions of the coexisting liquid phases in a mixture of n-pentane (1) and sulfolane (2) at 374.11K. The system is represented by the NRTL model andKo et al. (2007) reported the following parameters :

τ12 = 2.329 τ21 = 1.061 α = 0.3

Solution

From Eq. (8.4-27)

G12 = exp(−ατ12) = exph− (0.3)(2.329)

i= 0.497

G21 = exp(−ατ21) = exph− (0.3)(1.061)

i= 0.727

Therefore, activity coefficients defined by Eqs. (8.4-29) and (8.4-30) become

ln γ1 = (1− x1)2

"1.061

µ0.727

0.727 + 0.273x1

¶2+

1.158

(1− 0.503x1)2

#(1)

ln γ2 = x21

"2.329

µ0.497

1− 0.503x1

¶2+

0.771

(0.727 + 0.273x1)2

#(2)

Substitution of Eqs. (1) and (2) into Eqs. (11.2-5) and (11.2-6), respectively, results in twohighly nonlinear equations given by

(1− xα1 )2

⎡⎣1.061Ã 0.727

0.727 + 0.273xα1

!2+

1.158

(1− 0.503xα1 )2

⎤⎦−(1− xβ1 )

2

⎡⎣1.061Ã 0.727

0.727 + 0.273xβ1

!2+

1.158

(1− 0.503xβ1 )2

⎤⎦ = ln⎛⎝xβ1

xα1

⎞⎠ (3)

(xα1 )2

⎡⎣2.329Ã 0.497

1− 0.503xα1

!2+

0.771

(0.727 + 0.273xα1 )2

⎤⎦− (xβ1 )2

⎡⎣2.329Ã 0.497

1− 0.503xβ1

!2+

0.771

(0.727 + 0.273xβ1 )2

⎤⎦ = ln⎛⎝1− xβ1

1− xα1

⎞⎠ (4)

392

Simultaneous solution of Eqs. (3) and (4) requires a numerical technique3. Using MATH-

CADR°

xα1 = 0.234 and xβ1 = 0.920

It is much easier to determine compositions graphically by drawing a common tangent to theminima of ∆ eGmix/RT versus the x1 curve. The molar excess Gibbs energy for the NRTLmodel is given by Eq. (8.4-26). Thus, the molar Gibbs energy change on mixing is given by

∆ eGmix

RT= x1x2

µ0.771

x1 + 0.727x2+

1.158

x2 + 0.497x1

¶+ x1 lnx1 + x2 lnx2 (8)

The values of ∆ eGmix/RT as a function of x1 are given in the table below:

x1 ∆ eGmix/RT x1 ∆ eGmix/RT x1 ∆ eGmix/RT

0.00 0 0.35 − 0.114 0.70 − 0.0590.05 − 0.093 0.40 − 0.104 0.75 − 0.0590.10 − 0.123 0.45 − 0.093 0.80 − 0.0600.15 − 0.135 0.50 − 0.083 0.85 − 0.0620.20 − 0.137 0.55 − 0.074 0.90 − 0.0630.25 − 0.132 0.60 − 0.067 0.95 − 0.0560.30 − 0.124 0.65 − 0.062 1.00 0

The figure shown below shows the variation in ∆ eGmix/RT with composition. The equilibriumcompositions in the two-phase region can be determined by drawing a line tangent to the minimawith the result xα1 = 0.23 and xβ1 = 0.92.

0 0.2 0.4 0.6 0.8 10.15−

0.1−

0.05−

00.026−

0.137−

G x.1( )

0.990.01 x.1

Liquid-liquid equilibrium separations are analogous to the flash calculations mentioned inChapter 9. Consider a binary liquid mixture with a solubility diagram as shown in Figure11.12-a. Initially the temperature is at T1 and the overall mole fraction of component 1 is z1.This liquid mixture enters a separation tank with a molar flow rate of F as shown in Figure

3Newton’s method for solving nonlinear systems of equations is explained in Problem 11.6.

393

11.12-b. Phase separation is initiated4 within the tank by decreasing temperature to T2. Themolar flow rates of the resulting α- and β-phases are Lα and Lβ, respectively. Throughout theprocess, the pressure is kept constant at a value above the bubble point pressure of the mixtureso that no vapor phase is present.

α-Phase F1 z1

Lβ , β1x

Lα , α1x

β-Phase

0 1

T2

T1

F Lβ Lα

z1 β1x α

1x x1

T

(a) (b)

Figure 11.12 Liquid-liquid flash problem.

The overall and component material balances around the separation chamber are given by

F = Lα + Lβ (11.2-7)

F zi = Lα xαi + Lβ x

βi i = 1, 2 (11.2-8)

For component 1, combination of Eqs. (11.2-7) and (11.2-8) leads to

Lα

F=

xβ1 − z1

xβ1 − xα1

(11.2-9)

Using the solubility diagram given in Figure 11.12, Eq. (11.2-9) can also be obtained by theapplication of the lever rule.

From the equilibrium relation given by Eq. (11.2-3)

γα1 xα1 = γβ1 x

β1 ⇒ xβ1 =

Ãγα1

γβ1

!xα1 (11.2-10)

Substitution of Eq. (11.2-10) into Eq. (11.2-9) and rearrangement yield

xα1 =

z1

⎛⎝γβ1

γα1

⎞⎠1 +

Lα

F

⎛⎝γβ1

γα1

− 1

⎞⎠(11.2-11)

4Thermodynamics is not concerned with the question of "How long will it take for the phase separation totake place?"

394

A similar development for xα2 gives

xα2 =

z2

⎛⎝γβ2

γα2

⎞⎠1 +

Lα

F

⎛⎝γβ2

γα2

− 1

⎞⎠(11.2-12)

Addition of Eqs. (11.2-11) and (11.2-12) results in

z1

⎛⎝γβ1

γα1

⎞⎠1 +

Lα

F

⎛⎝γβ1

γα1

− 1

⎞⎠+

z2

⎛⎝γβ2

γα2

⎞⎠1 +

Lα

F

⎛⎝γβ2

γα2

− 1

⎞⎠= 1 (11.2-13)

which is analogous to the equations developed for flash calculations, i.e., Eqs. (9.3-19) and(9.3-20).

At a given z1 and F , calculation of Lα (or Lβ), xα1 , and xβ1 requires the following iterativeprocedure:

1. Assume xα1 and calculate xβ1 from Eq. (11.2-10). Note that the solution of Eq. (11.2-10) is

not straightforward since γβ1 may be a highly nonlinear function of xβ1 .

2. Calculate Lα/F from Eq. (11.2-9).3. Substitute the values into the left-hand side of Eq. (11.2-13) and check whether the sum-mation is equal to unity.

11.3 DISTRIBUTION OF A SOLUTE BETWEEN TWO IMMISCIBLE LIQUIDS

Consider two immiscible solvents forming two separate phases, α and β. Let component 1 bethe solute which is distributed between the two phases. Under equilibrium conditionsbfα1 = bfβ1 (11.3-1)

orxα1 γ

α1 = xβ1 γ

β1 (11.3-2)

The distribution of a solute between the α- and β-phases is quantified by the partition coefficientor distribution coefficient, Kαβ

1 , defined as

Kαβ1 =

xα1

xβ1(11.3-3)

The use of Eq. (11.3-2) in Eq. (11.3-3) results in

Kαβ1 =

γβ1γα1

(11.3-4)

On the other hand, the material balance for the solute is written as

n1 = nα xα1 + nβ x

β1

=³nαK

αβ1 + nβ

´xβ1 (11.3-5)

395

Example 11.4 An organic acid is to be extracted from a 20 mol % acid and 80% hexanemixture by liquid-liquid extraction using water at 298K. It is required to estimate the molesof water per mole of this mixture for removing 90% of the acid from the hexane. Assume thathexane and water are completely immiscible and that the phases leaving the extractor are inequilibrium.

The following data are given at 298K:

For organic acid (1) - hexane (2) mixture: eGex/RT = 0.2x1 x2

For organic acid (1) - water (3) mixture: eGex/RT = 0.012x1 x3

Solution

Let α and β be the hexane and water phases, respectively. Choosing 100mol of acid-hexanemixture as a basis, final amounts of acid and hexane in the α-phase are

nα2 = 80mol

nα1 = (20)(0.1) = 2mol

Thus,

xα1 =2

82= 0.02439

From the given data, the activity coefficients of acid in the α- and β-phases are

γα1 = exph0.2(1− xα1 )

2i

and γβ1 = exph0.012(1− xβ1 )

2i

The condition of equilibrium is expressed as

xα1 γα1 = xβ1 γ

β1

orxα1 exp

h0.2(1− xα1 )

2i= xβ1 exp

h0.012(1− xβ1 )

2i

Substitution of the numerical values gives

(0.02439) exph0.2(1− 0.02439)2

i= xβ1 exp

h0.012(1− xβ1 )

2i

The solution gives xβ1 = 0.02917. Therefore, the number of moles of the β-phase is

0.02917 =20− 2nβ

⇒ nβ = 617mol

The number of moles of water is

nβ3 = 617− 18 = 599mol

The desired molar ratio is

599

100= 5.99mol of water per mol of acid-hexane mixture

396

11.3.1 Octanol-Water Partition Coefficient

Octanol [CH3(CH2)7OH] and water are partially immiscible and the distribution of an organiccompound i between these two phases is known as the octanol-water partition coefficient, Kow

i ,i.e.,

Kowi =

coi

cwi(11.3-6)

Since Kowi values may range from 10−4 to 108 (encompassing 12 orders of magnitude), it is

usually reported as logKowi . Octanol-water partition coefficients of various substances are given

in Table 11.1.

Table 11.1 Octanol-water partition coefficients of various substances5.

Substance Chemical Formula logKow

Methanol CH3OH − 0.77Chloroform CHCl3 1.97Benzene C6H6 2.131,1,2,2-Tetrachloroethane C2H2Cl4 2.391,1,1-Trichloroethane CH3CCl3 2.49Naphthalene C10H8 3.29Hexachlorobenzene C6Cl6 6.18

Cells are mainly made of lipids and they are generally modeled as a lipid bilayer model,with a long hydrophobic (water disliking) chain and a polar hydrophilic (water liking) end. Thereason for choosing n-octanol is the fact that it exhibits both a hydrophobic and a hydrophiliccharacter, and its carbon/oxygen ratio is similar to that of lipids. In other words, n-octanolmimics the structure and properties of cells and organisms.

Since octanol-water partition coefficient quantifies how a substance distributes itself be-tween lipid and water, it is extensively used to describe lipophilic (lipid liking) and hydrophilicproperties of a particular substance. In that respect, it is one of the key physical/chemicalproperties, such as vapor pressure and solubility in water, used to assess the impact of agri-cultural and industrial chemicals on the environment. For example, polychlorinated biphenyls(PCBs) have low solubilities in water and high octanol-water partition coefficients. If PCBs areaccidentally released into the lake, then they are most probably found in higher concentrationsin the sediment layer.

11.4 STEAM DISTILLATION

If two immiscible liquids are placed in a tank, the one with the lower density floats on the toplayer and solely contributes to the pressure in the vapor phase. The liquid with the higherdensity has no contribution to the pressure in the vapor phase since it is located in the lowerlayer. Therefore, when we talk about the vapor-liquid-liquid equilibrium (VLLE) calculations,we are implicitly assuming that the liquid phases are continuously agitated in such a way thatthere will be droplets of both liquids on the surface, which is in contact with the vapor phase.

5Compiled from US National Library of Medicine, Hazardous Substances Data Bank (HSDB),http://toxnet.nlm.nih.gov. See also A Databank of Evaluated Octanol-Water Partition Coefficients (LOGKOW),http://logkow.cisti.nrc.ca/logkow/index.jsp.

397

A pure liquid starts to boil when its vapor pressure equals the surrounding pressure. Forexample, using the values given in Appendix C, the vapor pressures of toluene and water areexpressed as

lnP vaptoluene = 9.3935−

3096.52

T − 53.67

lnP vapwater = 11.6834−

3816.44

T − 46.13where P vap is in bar and T is in K. The normal boiling point temperatures of pure tolueneand pure water are

T sattoluene = 53.67 +

3096.52

9.3935− ln 1.01325 = 383.78K

T satwater = 46.13 +

3816.44

11.6834− ln 1.01325 = 373.15K

In the case of immiscible liquids, each component contributes to the vapor pressure of themixture. The total pressure is simply the sum of the vapor pressures of each component, i.e.,

P =kXi=1

P vapi Immiscible mixture (11.4-1)

Since toluene and water are essentially immiscible as liquids, Eq. (11.4-1) is expressed as

exp

µ9.3935− 3096.52

T − 53.67

¶+ exp

µ11.6834− 3816.44

T − 46.13

¶= 1.013

The solution of the above equation gives T = 357.48K. Thus, under atmospheric pressure, atoluene-water mixture boils at 357.48K, lower than the boiling points of pure toluene (383.78K)and water (373.15K).

High boiling point liquids at atmospheric pressure, i.e., essential oils6, waxes, and complexfats, may decompose at high temperatures and cannot be purified by distillation. Since oils areusually insoluble in water, then the mixture of oil and steam boils at a temperature well belowthe boiling point of pure oil. Therefore, to decrease the boiling point, steam is directly injectedinto the distillation column. When the resulting vapor is condensed, the two immiscible liquidphases are separated easily. Such a process is called steam distillation.

REFERENCES

Abedinzadegan, M. and A. Meisen, 1996, Fluid Phase Equilibria, 123, 259-270.

Abraham, M.H., G.S. Whiting, R. Fuchs and E.J. Chambers, 1990, J. Chem. Soc. PerkinTrans., 2, 291-300.

Baudot, A. and M. Marin, 1996, J. Membrane Science, 120, 207-220.

Furuya, T., T. Ishikawa, T. Funazukuri, Y. Takebayashi, S. Yoda, K. Otake and T. Saito, 2007,Fluid Phase Equilibria, 257, 147-150.

6Essential oils are the concentrated extracts of plants and herbs.

398

Ko, M., J. Im, J.Y. Sung and H. Kim, 2007, J. Chem. Eng. Data, 52, 1464-1467.

Lipinsky, C.A., F. Lombardo, B.W. Dominy and P.J. Feeney, 1996, Advanced Drug DeliveryReviews, 23, 3-25.

May, W.E., S.P. Wasik, M.M. Miller, Y. B. Schult, C.J., B.J. Neely, R.L. Robinson, K.A.M.Gasem and B.A. Todd, 2001, Fluid Phase Equilibria, 179, 117-129.

Van Ness, H.C. and M.M. Abbott, 1982, Classical Thermodynamics of Nonelectrolyte Solutions,McGraw-Hill, New York.

Villamanan, M.A., A.J. Allawl and H.C. Van Ness, 1984, J. Chem. Eng. Data, 29, 431-435.

PROBLEMS

Problems related to Section 11.1

11.1 Show that an ideal mixture always forms a homogeneous phase over the entire compo-sition range.

11.2 Start with Eq. (8.4-18) and show that

d2( eGex/RT )

dx21

=Λ212

x1 (x1 + Λ12x2)2 +

Λ221

x2 (x2 + Λ21x1)2 (1)

for the Wilson model. Since Eq. (1) is always positive (Why?), conclude that the Wilson modelcannot be used to predict phase separation.

11.3 For a binary liquid mixture of water (1) and diacetyl (2), Baudot and Marin (1996)reported the following activity coefficients at infinite dilution

γ∞1 = 3.3 γ∞2 = 13 303 < T < 323

Estimate the spinodal compositions if the system is represented by the three-suffix Margulesmodel.

(Answer: x1 = 0.531 and x1 = 0.799)

11.4 When compositions, i.e., xα1 and xβ1 , are known, the parameters of an activity coefficient

model can be estimated by solving Eqs. (11.2-5) and (11.2-6).

a) If the system is represented by the two-suffix Margules model, show that

A =ln(xβ1/x

α1 )

(1− xα1 )2 − (1− x

β1 )2=

lnh(1− xβ1 )/(1− xα1 )

i(xα1 − xβ1 )(x

α1 + x

β1 )

(1)

b) If the system is represented by the van Laar model, show that

A = − (X − Y )(X Θ+ Λ)2(Y Θ+ Λ)2£Θ (X + Y ) + 2Λ

¤ £Y Λ+X (2Y Θ+ Λ)

¤2 (2)

399

B =(X − Y )(X Θ+ Λ)2(Y Θ+ Λ)2£

Θ (X + Y ) + 2Λ¤2 £

Y Λ+X (2Y Θ+ Λ)¤ (3)

where

X =xα1

1− xα1Y =

xβ1

1− xβ1

Θ = ln

⎛⎝xβ1

xα1

⎞⎠ Λ = ln

⎛⎝1− xβ1

1− xα1

⎞⎠ (4)

c) Abedinzadegan and Meisen (1996) studied liquid-liquid equilibrium of diethanolamine (1)and octadecane (2) mixtures and reported the following solubility values at 492K and at-mospheric pressure:

xα1 = 0.0959 and xβ1 = 0.9924

If the system is represented by the van Laar model, estimate the parameters A and B.

(Answer: c) A = 2.607 B = 4.933)

11.5 A mixture of acetonitrile (1) and n-hexadecane (2) forms two partially miscible liquidphases. Liquid-liquid equilibrium data on such systems are needed for the design of the oxida-tive desulfurization process. Furuya et al. (2007) obtained the following NRTL parameters at333K:

τ12 = 4.852 τ21 = 0.2818 α = 0.2

a) If the composition of acetonitrile in the acetonitrile-rich phase is 0.994, estimate its compo-sition in the n-hexadecane-rich phase.

b) Are the like interactions stronger or weaker than the unlike interactions?(Answer: a) 0.168)

11.6 Using the Newton’s method as described in Section 9.4.1, rearrange Eqs. (3) and (4) ofExample 11.3 in the form

f1(x, y) = (1− x)2

"1.061

µ0.727

0.727 + 0.273x

¶2+

1.158

(1− 0.503x)2

#−

(1− y)2

"1.061

µ0.727

0.727 + 0.273 y

¶2+

1.158

(1− 0.503 y)2

#− ln

³yx

´= 0 (1)

f2(x, y) = x2

"2.329

µ0.497

1− 0.503x

¶2+

0.771

(0.727 + 0.273x)2

#

− y2

"2.329

µ0.497

1− 0.503 y

¶2+

0.771

(0.727 + 0.273 y)2

#− ln

µ1− y

1− x

¶= 0 (2)

wherexα1 = x and xβ1 = y (3)

a) Choose x(0) = 0.2, y(0) = 0.9, and = 1 × 10−5 to begin iterations for the solution of thissystem.b) For the first iteration, i.e., k = 1, show that Eq. (9.8-3) takes the form∙

1.834147 − 0.310829− 0.461743 2.953479

¸ ∙∆1∆2

¸= −

∙− 4.49971× 10−2− 5.73517× 10−2

¸(4)

400

c) Solve Eq. (4) to obtain

∆1 = 2.85810× 10−2 and ∆2 = 2.38867× 10−2

d) Since ∆max ' 0.03 > , continue iteration with

x(1) = x(0) +∆1 = 0.228581 y(1) = y(0) +∆2 = 0.923887

e) Show that when k = 5

x(5) = 0.233811 y(5) = 0.920410

Keep in mind that good initial estimates are extremely important in numerical solutions.

11.7 Infinite dilution activity coefficients of various liquids in benzene at 293K are given asfollows:

Component Acetone Acetonitrile Carbon tetrachloride n-Hexane

γ∞ 1.77 3.21 1.13 2.21

Arrange these substances in order of decreasing solubility in benzene.

Problems related to Section 11.3

11.8 Since the solubility of water in n-hexadecane is 0.0059 mole fraction and that of n-hexadecane in water is 0.0072 mole fraction at 298.15K, the n-hexadecane-water system isregarded as a system containing the two pure solvents. Consider a solute (1) that is partitionedbetween n-hexadecane (α-phase) and water (β-phase). The following data are provided:

Component ρ ( kg/m3 ) Molecular Weight

n-Hexadecane 770.20 226.44

Water 997.05 18.02

The condition of equilibrium states that

xα1 γα1 = xβ1 γ

β1 (1)

a) The hexadecane-water partition coefficient, Kαβ1 , is defined as the molar concentration of

the solute in the n-hexadecane phase to the molar concentration of the solute in the waterphase, i.e.,

Kαβ1 =

cα1

cβ1(2)

Combine Eqs. (1) and (2) to obtain

γβ1 = 16.3Kαβ1 γα1 (3)

If the solute is infinitely dilute in this system, note that Eq. (3) takes the form

γβ∞1 = 16.3Kαβ1 γα∞1 (4)

If the infinite-dilution activity coefficient of solute in n-hexadecane and the hexadecane-waterpartition coefficient are known, then Eq. (4) can be used to estimate the infinite-dilutionactivity coefficient of solute in water.

b) The infinite-dilution activity coefficient of toluene in n-hexadecane is given as a function oftemperature as follows (Schult et al., 2001):

401

Temperature (K) 323.2 343.2 353.2 373.2

γ∞ 0.941 0.870 0.846 0.808

The hexadecane-water partition coefficient of toluene is (Abraham et al., 1990)

K = 575.4

Express the infinite-dilution activity coefficient in the form

ln γ∞ = A+B

T

and estimate the infinite-dilution activity coefficient of toluene in water at 298K.

(Answer: 9660)

11.9 Consider two liquid phases, α and β, in which the α-phase is almost pure 1 and theβ-phase is almost pure 2.

a) Using the condition of equilibrium show that

γ∞1 =1

xβ1

and γ∞2 =1

xα2

(1)

Therefore, determination of solubility enables one to estimate the infinite-dilution activitycoefficient. This method, known as inverse solubility, is suitable for organics that are sparinglysoluble in water, i.e., xi < 10−3.

b) The solubility of benzene (2) in water (1) at 298.15K is reported by May et al. (1983) as

x2 = 0.4129× 10−3

Estimate the infinite-dilution activity coefficient of benzene in water.

(Answer: 2422)

402