Chapter 12

c° 2011 Ismail Tosun

Chapter 12

Phase Equilibria Involving Dissolved Solidsin Liquids

Crystallization is the formation of solid particles from within a homogeneous phase and it isone of the unit operations used to separate substances in pure solid form from a liquid mixture.Therefore, reliable data on the solubility of solids in liquid solvents are needed when designingcrystallizers. This chapter first provides equations to estimate solid solubility in solvents. Thenhow boiling and freezing points of liquids are affected by the presence of dissolved solids willbe discussed.

12.1 EQUILIBRIUM BETWEEN A PURE SOLID AND A LIQUID MIXTURE

Let substance i be present as a pure solid and also as a component in a liquid solution atconstant temperature and pressure. The condition of equilibrium states that

fSi (T, P ) =bfLi (T, P, xi)

= fLi (T, P ) γi(T, P, xi)xi (12.1-1)

Rearrangement of Eq. (12.1-1) gives

ln (γixi) = ln

"fSi (T, P )

fLi (T,P )

#(12.1-2)

The fugacities of pure solid and liquid are related to the molar Gibbs energies by Eq. (5.2-1),i.e.,

ln

"fSi (T, P )

P

#=eGSi (T, P )− eGIG

i (T, P )

RT(12.1-3)

ln

"fLi (T, P )

P

#=eGLi (T,P )− eGIG

i (T, P )

RT(12.1-4)

Thus,

ln

"fSi (T, P )

fLi (T, P )

#=eGSi (T, P )− eGL

i (T, P )

RT

= −∆ eGfus

i (T, P )

RT= −

∆ eHfusi (T,P )− T ∆eSfus

i (T, P )

RT(12.1-5)

The use of Eq. (12.1-5) in Eq. (12.1-2) leads to

403

ln (γixi) = −∆ eHfus

i (T, P )− T ∆eSfusi (T, P )

RT(12.1-6)

Since the solution temperature, T , is usually much lower than the triple point temperature,Tt, of pure solid, a pure substance does not exist in the liquid form at the solution temperature.Thus, extrapolation of the liquid-vapor pressure curve from the triple point temperature to thesolid region is necessary until the temperature reaches the solution temperature, as shown inFigure 12.1.

vapsolidP

vap

liquidsubcooledP

Pre

ssur

e

Temperature T tT

VAPOR

LIQUIDSOLID

Figure 12.1 Extrapolation of vapor-liquid pressure curve to the solution temperature, T .

Systematic extrapolation can be carried out by devising a following path as shown in Figure12.2.

Step 1: Solid is heated at constant pressure from solution temperature, T , to triple pointtemperature, Tt.

Step 2: Solid is melted to form a liquid (∆ eHfusi and ∆eSfusi at Tt are known).

Step 3: Liquid is subcooled without solidification from Tt to T .

Tem

pera

ture

tT

T

Solid Subcooled Liquid Phase

Figure 12.2 Calculation path for ∆ eHfusi (T ) and ∆eSfus

i (T ).

For steps 1 and 3, the changes in enthalpy and entropy can be calculated from Eqs. (3.3-1)and (3.4-2), respectively. Since pressure remains constant, the changes in enthalpy and entropyfor the overall process are expressed as

∆ eHfusi (T ) =

Z Tt

T

eCSPi dT +∆

eHfusi (Tt) +

Z T

Tt

eCLPi dT

=∆ eHfusi (Tt) +

Z T

Tt

∆ eCLSPi dT (12.1-7)

404

∆eSfusi (T ) =

Z Tt

T

eCSPi

TdT +∆eSfus

i (Tt) +

Z T

Tt

eCLPi

TdT

=∆eSfusi (Tt) +

Z T

Tt

∆ eCLSPi

TdT (12.1-8)

where∆ eCLS

Pi =eCLPi − eCS

Pi (12.1-9)

At the triple point, solid and liquid phases of pure component i are in equilibrium with eachother, i.e.,

∆ eGfusi (Tt) = ∆ eHfus

i (Tt)− Tt∆eSfusi (Tt) = 0 ⇒ ∆eSfus

i (Tt) =∆ eHfus

i (Tt)

Tt(12.1-10)

The use of Eq. (12.1-10) in Eq. (12.1-8) leads to

∆eSfusi (T ) =

∆ eHfusi (Tt)

Tt+

Z T

Tt

∆ eCLSPi

TdT (12.1-11)

Substitution of Eqs. (12.1-7) and (12.1-11) into Eq. (12.1-6) results in

ln (γixi) =∆ eHfus

i (Tt)

RTt

µ1− Tt

T

¶− 1

RT

Z T

Tt

∆ eCLSPi dT +

1

R

Z T

Tt

∆ eCLSPi

TdT (12.1-12)

For most substances, the melting (or fusion) curve has a very steep slope. As a result, thetriple point temperature is very close to the normal melting temperature, Tm. Moreover, it isplausible to assume that the enthalpy of fusion is essentially the same at these two temperatures.Under these circumstances, Eq. (12.1-12) becomes

ln (γixi) =∆ eHfus

i (Tm)

RTm

µ1−

Tm

T

¶− 1

RT

Z T

Tm

∆ eCLSPi

dT +1

R

Z T

Tm

∆ eCLSPi

TdT (12.1-13)

Equation (12.1-13) can be used twofold: When it is applied to a solid, it gives the solubilityof solid in a solvent. On the other hand, if it is applied to a solvent, it gives its freezing pointdepression.

12.1.1 Solubility of Solid in a Solvent

Let i be the solid (2) dissolved in a solvent (1). Then Eq. (12.1-13) gives the mole fractionsolubility, x2, of solute as

ln (γ2x2) =∆ eHfus

2

RTm

µ1−

Tm

T

¶− 1

RT

Z T

Tm

∆ eCLSP2

dT +1

R

Z T

Tm

∆ eCLSP2

TdT (12.1-14)

The term γ2 takes into account the intermolecular forces between the solid and the solvent.As a result, the solubility of solid is dependent on the type of solvent. Activity coefficient isestimated from either experimental data or a liquid solution model, i.e., regular mixture theorygiven in Section 8.5.

405

In the literature, solubility is also expressed as " g of solid per 100 g of solvent", S, whichis related to the mole fraction based solubility, x2, in the form

S = 100

µx2

1− x2

¶µM2

M1

¶(12.1-15)

where M1 and M2 are the molecular weights of solvent and solid, respectively.The simplification of Eq. (12.1-14) can be done for the following cases.

• Heat capacities are independent of temperatureWhen ∆ eCLS

P2is constant, Eq. (12.1-14) simplifies to

ln (γ2x2) =∆ eHfus

2

RTm

µ1− Tm

T

¶−∆ eCLS

P2

R

∙1− Tm

T+ ln

µTmT

¶¸(12.1-16)

• ∆ eCLSP2' 0 and/or ∆ eCLS

P2= constant with T 'Tm

When∆ eCLSP2is close to zero1, the second and third terms on the right-hand side of Eq. (12.1-14)

vanish, with the result

ln (γ2x2) =∆ eHfus

2

RTm

µ1− Tm

T

¶(12.1-17)

On the other hand, when ∆ eCLSP2is constant, and if T and Tm are close to each other, expansion

of the term ln(Tm/T ) in Taylor series yields

ln

µTmT

¶=

µTmT− 1¶− 12

µTmT− 1¶2+1

3

µTmT− 1¶3− ...| {z }

Negligible when T and Tm do not differ greatly.

(12.1-18)

The use of Eq. (12.1-18) in Eq. (12.1-16) also leads to Eq. (12.1-17). In the literature, Eq.(12.1-17) is known as the Schröder equation2 and it is applicable to compounds that do notionize or dissociate in solution, and when the solid phase is pure.

When the solubility is very low, i.e., x2 is of the order of 10−3 or less, it is plausible toreplace γ2 by γ

∞2 and Eq. (12.1-17) becomes

ln (γ∞2 x2) =∆ eHfus

2

RTm

µ1− Tm

T

¶x2 ≤ 10−3 (12.1-19)

The term ideal solubility, xideal2 , is used when the liquid mixture is ideal, i.e., γ2 = 1. Inthat case Eq. (12.1-17) reduces to

lnxideal2 =∆ eHfus

2

RTm

µ1− Tm

T

¶(12.1-20)

indicating that the ideal solubility is independent of the type of solvent. It follows from Eq.(12.1-20) that the solubility of a solid increases with an increase in temperature.

1For compounds with high melting temperatures, i.e., pharmaceuticals and organic pollutants, instead oftaking ∆ eCLS

P2 as zero, the approximation ∆ eCLSP2 = ∆eSfus2 (Tm) gives better results (Pappa et al., 2005).

2The Schröder equation can also be derived by combining Eq. (5) of Example 5.13 with Eq. (12.1-2).

406

It can be shown from Eqs. (12.1-17) and (12.1-20) that

x2 =xideal2

γ2(12.1-21)

indicating the higher the activity coefficient of solid in a solvent, the lower its solubility.

Example 12.1 Calculate the ideal solubility of naphthalene (2) in a solvent (1) at 313.15K.For naphthalene, Tm = 353.35K and ∆ eHfus

2 = 18, 803 J/mol.

Solution

The use of Eq. (12.1-20) gives

lnxideal2 =18, 803

(8.314)(353.35)

µ1− 353.35

313.15

¶⇒ xideal2 = 0.44

Comment: While the ideal solubility is independent of the type of solvent, solubilities ofnaphthalene (2) in various solvents at 313.15K are reported by Gmehling et al. (1978) asfollows:

Solvent x2 Solvent x2

Methanol 0.044 n-Hexane 0.222Ethanol 0.073 Cyclohexanol 0.225n-Propanol 0.094 Acetic acid 0.117Isopropanol 0.076 Acetone 0.378n-Butanol 0.116 Chloroform 0.473

Example 12.2 At 313.15K, solubility of naphthalene (2) in n-hexane (1) is 0.222. Estimatethe solubility of naphthalene in n-hexane at 325K if its activity coefficient is represented by thevan Laar model. The molar volumes of liquid naphthalene and n-hexane are given as 125.03and 131.61 cm3/mol, respectively.

Solution

A single set of activity coefficient data must be known to determine the parameters in the vanLaar model. The use of Eq. (12.1-21) gives the activity coefficient of naphthalene at x2 = 0.222as

γ2 =xideal2

x2=0.44

0.222= 1.982

For the van Laar model, Eq. (8.4-17) gives

B

A=eV L2eV L1

=125.03

131.61= 0.95 (1)

The use of Eq. (8.4-14) gives the parameter B as

B =

∙1 +

B

A

µx2

1− x2

¶¸2ln γ2 =

∙1 + 0.95

µ0.222

1− 0.222

¶¸2ln(1.982) = 1.105

407

Therefore, the activity coefficient of naphthalene is expressed as

ln γ2 =1.105∙

1 + 0.95

µx2

1− x2

¶¸2 (2)

Substitution of Eq. (2) into Eq. (12.1-17) leads to

lnx2 +1.105∙

1 + 0.95

µx2

1− x2

¶¸2 = 18, 803

(8.314)(353.35)

µ1− 353.35

T

¶(3)

When T = 325K, the solution of Eq. (3) gives x2 = 0.356.

Comment: IUPAC-NIST Solubility Database3 gives the solubility of naphthalene in n-hexaneas a function of temperature as

lnx2 = −3849.6

T+ 10.789 (4)

When T = 325K, Eq. (4) gives x2 = 0.348.

Example 12.3 Estimate the solubility of anthracene (2) in chlorobenzene (1) at 298K. Theliquid mixture obeys the regular solution theory and the following data are provided:

Chlorobenzene: eV L1 = 102 cm

3/mol, δ1 = 19.4( J/ cm3)1/2

Anthracene: eV L2 = 150 cm

3/mol, Tm = 490K, ∆ eHfus2 = 29, 372 J/mol

lnP sub2 = 19.9724− 11, 467

T

where P sub is in bar and T is in K.

Solution

To calculate the solubility of anthracene in chlorobenzene with the help of Eq. (12.1-17), theactivity coefficient of anthracene must be expressed as a function of composition by using Eqs.(8.5-3) and (8.5-4). For this purpose, it is first necessary to determine the solubility parameterof anthracene by using Eq. (8.5-5), i.e.,

δ2 =

vuut∆ eHvap2 −RTeV L2

(1)

The heat of vaporization can be indirectly calculated as

∆ eHvap2 = ∆ eHsub

2 −∆ eHfus2 (2)

The heat of sublimation of anthracene can be calculated from Eq. (5.6-30), i.e.,

d lnP sub2

dT=∆ eHsub

2

RT 2=11, 467

T 2

3http://srdata.nist.gov/solubility

408

or∆ eHsub

2 = (11, 467)(8.314) = 95, 337 J/mol

Therefore, the heat of vaporization and the solubility parameter are calculated as

∆ eHvap2 = 95, 337− 29, 372 = 65, 965 J/mol

and

δ2 =

r65, 965− (8.314)(298)

150= 20.6( J/ cm3)1/2

Equation (12.1-17) is expressed as

eV L2 (δ1 − δ2)

2

RT

Ãx1 eV L

1

x1 eV L1 + x2 eV L

2

!2+ lnx2 =

∆ eHfus2

RTm

µ1− Tm

T

¶(3)

or(150)(19.4− 20.6)2(83.14)(298)

∙102 (1− x2)

102 (1− x2) + 150x2

¸2+ lnx2 =

29, 372

(8.314)(490)

µ1− 490

298

¶(4)

The solution of Eq. (4) by MATHCADR°gives x2 = 9.53× 10−3.

As stated in Section 8.3.2, activity coefficients are generally estimated from vapor-liquidequilibrium (VLE) measurements. If the boiling points of the components are very high, then itis difficult to obtain VLE data. In this case, it is more convenient to obtain activity coefficientsfrom solid-liquid equilibrium measurements4.

12.1.2 Freezing Point Depression

In this case, Eq. (12.1-13) is applied to a solvent (1). Assuming the second and third terms onthe right-hand side of Eq. (12.1-13) to be negligible, the result is

ln(γ1x1) =∆ eHfus

1 (Tm)

RTm

µ1− Tm

T

¶(12.1-22)

When the freezing point depression is small, i.e., T ' Tm, then T Tm ' T 2m. Consequently, Eq.(12.1-22) takes the form

Tm − T = −RT 2m

∆ eHfus1

ln(γ1x1) (12.1-23)

If solute (2) is dilute enough that the liquid can be treated as an ideal mixture, i.e., γ1 ' 1,then

ln(γ1x1) = lnx1 = ln(1− x2) = −x2 (12.1-24)

Substitution of Eq. (12.1-24) into Eq. (12.1-23) gives

Tm − T =RT 2m

∆ eHfus1

x2 (12.1-25)

4See Problem 12.4.

409

indicating that the freezing point depression depends on the mole fraction and not on thechemical structure of solute5.

The mole fraction of solute, x2, is related to the molality of solute6, m2, as

x2 =m2

m2 +1000

M1

' m2

µM1

1000

¶(12.1-26)

The use of Eq. (12.1-26) in Eq. (12.1-25) leads to

Tm − T =

ÃRT 2m

1000∆ bHfus1

!m2 (12.1-27)

In the literature, Eq. (12.1-27) is usually expressed as

Tm − T = iKf m2 (12.1-28)

where i is called the van’t Hoff factor7 and Kf is the freezing point depression constant of asolvent defined by

Kf =RT 2m

1000∆ bHfus1

(12.1-29)

The van’t Hoff factor, i, is associated with the degree of dissociation of the solute in the solventand it represents the moles of particles in solution per mole of solute dissolved. For example,

i =

⎧⎨⎩1 Nonelectrolytes2 Electrolytes that ionize into two ions, i.e., NaCl3 Electrolytes that ionize into three ions, i.e., MgCl2

(12.1-30)

Freezing point depression constants for some solvents are given in Table 12.1.

Table 12.1 Freezing point depression constants for various solvents.

SolventFreezing Point at 1 atm

( ◦C)Kf

[ ◦C/(mol/ kg)]

Acetic acid 16.6 3.90Benzene 5.5 5.12CCl4 − 22.8 5.22Chloroform − 63.5 4.68Ethanol − 114.6 1.99Water 0 1.86

Note that either Eq. (12.1-25) or Eq. (12.1-28) can be used not only to estimate the freezingpoint depression but also to estimate the heat of fusion of a solvent, ∆ eHfus

1 , or the molecularweight of a solute.

5The freezing point depression is known as a colligative property. Colligative means depending on the numberof particles and not on the nature of particles. Other colligative properties will be discussed in Section 12.2.

6Molality of a solute is defined as the number of moles of solute per kg of solvent.7Jacobus Henricus van’t Hoff was a Dutch physical and organic chemist who lectured at various institutions

(Veterinary College in Utrecht, University of Amsterdam, University of Berlin). He was the recepient in 1901 ofthe first Nobel prize for chemistry .

410

Example 12.4 Calculate the freezing point of a solution prepared by dissolving 171 g ofAl2(SO4)3 in 750 g of water.

Solution

The molality of Al2(SO4)3 (molecular weight = 342 g/mol) is

m2 =171/342

0.75= 0.667mol/ kg

If Al2(SO4)3 completely dissociates, i.e.,

Al2(SO4)3 → 2Al+3 + 3SO−24

then i = 5. The use of Eq. (12.1-28) gives

0− T = (5)(1.86)(0.667) = 6.203 ◦C

Thus, the freezing point temperature is T = − 6.203 ◦C.

12.1.3 Solid-Liquid Equilibrium Phase Diagrams

Solid-liquid equilibrium (SLE) data are obtained experimentally by cooling a liquid mixture ofknown composition and recording the temperature continuously as a function of time8. A breakpoint in this curve indicates the formation of a solid phase. The temperature at this point isthe solid-liquid equilibrium temperature, and the composition of the liquid mixture gives thesolubility of solute in the solvent. SLE data are generally presented in the form of temperatureversus solubility phase diagrams under isobaric conditions. Binary SLE phase diagrams can beclassified as (i) systems exhibiting eutectic behavior, (ii) systems forming solid solutions.

12.1.3.1 Systems exhibiting eutectic behavior

Among the phase diagrams exhibiting eutectic behavior, the simplest one is shown in Figure12.3.

E

D

C

B

Liquid

Solid 1 + Solid 2

Liquid + Solid 2

Liquid + Solid 1

x2

1

0

T

A

P = constant

Figure 12.3 Solid-liquid phase diagram for a binary mixture.

8Besides this traditional technique of cooling curve and visual measurement, the differential scanning calorime-try (DSC) method is also used (Kousksou et al., 2007).

411

The temperatures at points A and C represent the freezing (or melting) temperatures ofpure 1 and pure 2, respectively. Only homogeneous liquid phase exists over the curve AEC. Thecurve AE represents the liquidus curve on which pure solid 1 is in equilibrium with the liquidmixture. Similarly, the curve CE is the liquidus curve on which pure solid 2 is in equilibriumwith the liquid mixture. In other words, while the curve AE can be regarded as the solubilitycurve of component 1 in solvent 2, the curve CE represents the solubility curve of component2 in solvent 1. Therefore, liquidus curves are determined from the solubility equation. Forexample, when ∆ eCLS

P ' 0 Eq. (12.1-17) gives

ln(γ2x2) =∆ eHfus

2

RTm2

µ1−

Tm2

T

¶Curve AE (12.1-31)

ln(γ1x1) =∆ eHfus

1

RTm1

µ1−

Tm1

T

¶Curve CE (12.1-32)

The curves AE and CE intersect at the point E, known as the eutectic point. At the eutecticpoint, the liquid phase is in equilibrium with the two solid phases. Since their compositionsare the same, no separation is possible at the eutectic point. The corresponding eutectic tem-perature represents the minimum freezing (or melting) temperature of the mixture. Note thatthe eutectic point is analogous to the azeotropic point in vapor-liquid equilibrium. The curveBED represents the solidus curve below which nonhomogeneous two-phase solid consisting ofpure 1 and pure 2 exists.

Consider a mixture at point F as shown in Figure 12.4. As this liquid mixture is cooled atconstant pressure, pure solid 1 forms when the temperature drops to T ∗ (point G) and therelationship between T ∗ and x∗2 is given by Eq. (12.1-31), i.e.,

lnhγ2(P, T

∗, x∗2)x∗2

i=∆ eHfus

2 (Tm)

RTm2

µ1− Tm2

T ∗

¶(12.1-33)

Thus, solid-liquid equilibrium data can be used to determine activity coefficients of the mixturecomponents.

*2x

T*

G

H L K

F

E

D

C

B

Liquid

Solid 1 + Solid 2

x2

1

0

T

A

P = constant

Figure 12.4 Solid-liquid phase diagram for a binary mixture.

As the temperature is further decreased, pure solid 1 continues to form and the liquidmixture becomes richer in component 2. For example, at point H, the amount of solid 1 andliquid mixture can be calculated by the application of the lever rule in the form

Amount of solid 1Amount of liquid mixture

=HL

HK(12.1-34)

412

Example 12.5 Prepare a solid-liquid phase diagram for naphthalene (1) and p-dichlorobenzene(2) if the mixture is represented by the regular solution theory. The following data are providedfor this system:

Naphthalene p-dichlorobenzene

Tm (K) 353.35 326.24

∆ eHfus( J/mol) 18, 803 18, 082eV L( cm3/mol) 125.03 101.8δ ( J/ cm3) 20.3 19.8

Solution

The prediction of the liquidus curve requires the determination of the solubility of solid i insolvent j. Assuming ∆ eCLS

P ' 0, Eq. (12.1-17) gives the solubility of solid i in solvent j in theform

ln γi + lnxi =∆ eHfus

i

RTmi

µ1−

Tmi

T

¶(1)

in which the activity coefficient of component i is given by Eqs. (8.5-3) and (8.5-4) as

ln γi =eV Li (δj − δi)

2

RT

Ãxj eV L

j

xj eV Lj + xieV L

i

!2(2)

Thus, the solubility of solid naphthalene in p-dichlorobenzene ( i = 1, j = 2) is

(125.03)(19.8− 20.3)28.314T

µ101.8x2

101.8x2 + 125.03x1

¶2+ lnx1 =

18, 803

(8.314)(353.35)

µ1− 353.35

T

¶(3)

On the other hand, the solubility of solid p-dichlorobenzene in naphthalene ( i = 2, j = 1) is

(101.8)(20.3− 19.8)28.314T

µ125.03x1

101.8x2 + 125.03x1

¶2+ lnx2 =

18, 082

(8.314)(326.24)

µ1− 326.24

T

¶(4)

The easiest way of preparing a solid-liquid equilibrium diagram is to give values to x1 (or x2)and to calculate the corresponding temperatures from Eqs. (3) and (4). The results presentedbelow are plotted in the following figure.

T (K)

x1 Solid naphthalene in p-dichlorobenzene Solid p-dichlorobenzene in naphthalene

0 − 326.240.1 260.2 321.20.2 282.6 315.70.3 297.6 309.70.4 309.3 303.10.5 318.9 295.60.6 327.3 287.00.7 334.7 276.50.8 341.5 263.10.9 347.6 242.81.0 353.35 −

413

0 0.2 0.4 0.6 0.8 1250

300

350360

250

T1

T

10 x1 x,

From the point of intersection of the liquidus curves, composition and temperature at the eutecticpoint are x1 = 0.36 and 305.7K, respectively. The experimental values are x1 = 0.39 and302.85K (Wei and Jin, 2009).

Comment: A similar problem can be rephrased as follows: "A liquid mixture consisting of 20mol % naphthalene (1) and 80% p-dichlorobenzene is cooled at constant pressure. Estimate thetemperature at which the solid phase first appears and identify the solid."

For this purpose it is necessary to calculate the freezing (or melting) temperatures of naphthaleneand p-dichlorobenzene at the given composition.

For naphthalene from Eq. (3)

(125.03)(19.8− 20.3)28.314T

∙(101.8)(0.8)

(101.8) (0.8) + (125.03) (0.2)

¸2+ ln(0.2) =

18, 803

(8.314)(353.35)

µ1− 353.35

T

¶=⇒ T = 282.6K

For p-dichlorobenzene from Eq. (4)

(101.8)(20.3− 19.8)28.314T

∙(125.03) (0.2)

(101.8) (0.8) + (125.03) (0.2)

¸2+ ln(0.8) =

18, 082

(8.314)(326.24)

µ1− 326.24

T

¶=⇒ T = 315.7K

Therefore, pure p-dichlorobenzene solidifies first at 315.7K. Further cooling will result in anincrease in the concentration of naphthalene in the solution.

12.1.3.2 Systems forming solid solution

The components of a solid solution completely dissolve in one another. As a result, the systemdoes not have a eutectic point and the solid-liquid phase diagram shown in Figure 12.5 resemblesthe temperature-composition diagram for a vapor-liquid mixture, i.e., Figure 9.3. The two-phase region in the middle is bounded by the liquidus and solidus curves. Contrary to theprevious case, components 1 and 2 are miscible in the solid phase.

414

Solidus curve

Liquidus curve

B

A

T

10 x2

Solid + Liquid

SOLID

P = constant

LIQUID

Figure 12.5 Solid-liquid phase diagram for a binary mixture (solid solution).

12.2 COLLIGATIVE PROPERTIES

Colligative means depending on the number of particles and not on the nature of particles.Colligative properties can only be applied to solutions and, as the name implies, they areindependent of the identity of the solute. In Section 12.1.2, we showed that the freezing pointdepression is a colligative property. Now, we’ll consider two other major colligative properties,namely boiling point elevation and osmosis.

12.2.1 Boiling Point Elevation

The boiling point of a pure liquid is the temperature at which the vapor pressure of the liquidis equal to the surrounding pressure. Consider a pure solvent (1) at a given temperature T .At the surface of a liquid there is competition between the kinetic energy of the molecules andthe intermolecular forces. If the kinetic energy of the molecules is high enough to overcomethe intermolecular forces, then the molecules will escape into the gas phase. In the meantime,gas molecules may return to the liquid phase when they collide with the surface. When thesolvent is in equilibrium with the gas phase above it, then the partial pressure of the solventin the gas phase is P vap

1 . In this case the number of molecules leaving the surface is equal tothose returning to the surface.

When a nonvolatile solute (2) is added to the solvent, some of the solvent molecules onthe surface of the liquid will be replaced by solute molecules. Since there are fewer solventmolecules on the surface to escape, then the vapor pressure goes down. Since x2 is small, x1 isclose to unity and we can use Raoult’s law for the solvent, i.e.,

Py1 = x1Pvap1 = (1− x2)P

vap1 (12.2-1)

indicating a decrease in the partial pressure of solvent.The decrease in vapor pressure implies an increase in the boiling temperature. To quantify

the elevation of boiling point, let us consider the condition of equilibrium between vapor andliquid phases. Since the vapor phase contains only solvent, then we can write

fV1 (T, P ) =bfL1 (T, P, x1)

= γ1x1fL1 (T, P ) (12.2-2)

415

or

ln (γ1x1) = ln

"fV1 (T, P )

fL1 (T, P )

#(12.2-3)

Fugacities of pure solvent in the vapor and liquid phases are related to molar Gibbs energiesby Eq. (5.2-1), i.e.,

ln

"fV1 (T, P )

P

#=eGV1 (T, P )− eGIG

1 (T, P )

RT(12.2-4)

ln

"fL1 (T, P )

P

#=eGL1 (T,P )− eGIG

1 (T, P )

RT(12.2-5)

Therefore,

ln

"fV1 (T, P )

fL1 (T,P )

#=eGV1 (T, P )− eGL

1 (T, P )

RT

=∆ eGvap

1 (T, P )

RT=∆ eHvap

1 (T, P )− T ∆eSvap1 (T, P )

RT(12.2-6)

The use of Eq. (12.2-6) in Eq. (12.2-3) leads to

ln (γ1x1) =∆ eHvap

1 (T, P )− T ∆eSvap1 (T, P )

RT(12.2-7)

The enthalpy and entropy of vaporization at the normal boiling point Tb are generally known.To find the enthalpy and entropy of vaporization at any given temperature, it is necessary todevise an alternative path as shown in Figure 12.6.

T

Tb

Liquid Vapor

Tem

pera

ture

Phase

Figure 12.6 Calculation path for ∆ eHvap1 (T ) and ∆eSvap

1 (T ).

The change in enthalpy is expressed as

∆ eHvap1 (T ) =

Z Tb

T

eCLP1 dT +∆

eHvap1 (Tb) +

Z T

Tb

eCVP1 dT (12.2-8)

or

∆ eHvap1 (T ) = ∆ eHvap

1 (Tb) +

Z T

Tb

∆ eCV LP1 dT (12.2-9)

where∆ eCV L

P1 = eCVP1 − eCL

P1 (12.2-10)

416

On the other hand, the entropy change is

∆eSvap1 (T ) =

Z Tb

T

eCLP1

TdT +∆eSvap

1 (Tb) +

Z T

Tb

eCVP1

TdT (12.2-11)

At a temperature of Tb, vapor and liquid phases of pure solvent are in equilibrium with eachother, i.e.,

∆ eGvap1 (Tb) = ∆ eHvap

1 (Tb)− Tb∆eSvap1 (Tb) = 0 ⇒ ∆eSvap

1 (Tb) =∆ eHvap

1 (Tb)

Tb(12.2-12)

The use of Eq. (12.2-12) in Eq. (12.2-11) leads to

∆eSvap1 (T ) =

∆ eHvap1 (Tb)

Tb+

Z T

Tb

∆ eCV LP1

TdT (12.2-13)

Substitution of Eqs. (12.2-9) and (12.2-13) into Eq. (12.2-7) gives

ln (γ1x1) =∆ eHvap

1 (Tb)

RT

µ1

T− 1

Tb

¶+

1

RT

Z T

Tb

∆ eCV LP1

dT − 1

R

Z T

Tb

∆ eCV LP1

TdT (12.2-14)

If heat capacities are independent of temperature, Eq. (12.2-14) simplifies to

ln (γ1x1) =∆ eHvap

1 (Tb)

R

µ1

T− 1

Tb

¶+∆ eCV L

P1

R

∙1− Tb

T+ ln

µTbT

¶¸(12.2-15)

The term ln(Tb/T ) is expanded in Taylor series as

ln

µTbT

¶=

µTbT− 1¶− 12

µTbT− 1¶2+1

3

µTbT− 1¶3− ...| {z }

Negligible when T and Tb do not differ greatly.

(12.2-16)

If solute (2) is dilute, the solvent (1) can be treated as an ideal solution, i.e., γ1 = 1, and theboiling point elevation is small, i.e., T ' Tb. Consequently, Eq. (12.2-15) reduces to

lnx1 = ln(1− x2) =∆ eHvap

1 (Tb)

R

µ1

T− 1

Tb

¶(12.2-17)

Further simplification of Eq. (12.2-17) with ln(1− x2) ' −x2 and T Tb ' T 2b leads to

T − Tb =RT 2b

∆ eHvap1

x2 (12.2-18)

indicating that the boiling point elevation is independent of the chemical nature of a solute.Equation (12.2-18) can be used not only for calculating the boiling point elevation, T − Tb,but also for determining the heat of vaporization (or latent heat) of a solvent, ∆ eHvap

1 , or themolecular weight of a solute.

The mole fraction of solute, x2, is related to the molality of solute, m2, by Eq. (12.1-26).Substitution of Eq. (12.1-26) into Eq. (12.2-18) gives

T − Tb =

ÃRT 2b

1000∆ bHvap1

!m2 (12.2-19)

417

In an analogous fashion to Eq. (12.1-28), Eq. (12.2-19) is usually expressed as

T − Tb = iKbm2 (12.2-20)

where i is the van’t Hoff factor taking into account the degree of dissociation of the solute inthe solvent, and Kb is the boiling point elevation constant of a solvent defined by

Kb =RT 2b

1000∆ bHvap1

(12.2-21)

Boiling point elevation constants for some solvents are given in Table 12.2.

Table 12.2 Boiling point elevation constants for various solvents.

SolventBoiling Point at 1 atm

( ◦C)Kb

[ ◦C/(mol/ kg)]

Acetic acid 118.1 3.07Benzene 80.1 2.63CCl4 76.8 5.22Chloroform 61.3 3.63Ethanol 78.4 1.23Water 100 0.513

Example 12.6 Calculate the boiling point of a solution prepared by dissolving 68.4 g of su-crose (molecular weight = 342 g/mol) in 1000 g of water.

Solution

The molality of sucrose is

m2 =68.4/342

1= 0.2mol/ kg

Sucrose is a nonelectrolyte, i.e., i = 1, and Kb = 0.513 for water. The use of Eqs. (12.2-20)gives

T − 100 = (1)(0.513)(0.2) = 0.103 ◦CTherefore, the boiling point is T = 100.103 ◦C.

Example 12.7 Calculate the boiling point of seawater if the concentration of NaCl is 3.5%by weight.

Solution

The molality of salt (molecular weight = 58.5 g/mol) is

m2 =3.5/58.5

(100− 3.5)/1000 = 0.62mol/ kg

If we assume that the NaCl completely dissociates into sodium and chloride9, i.e.,

NaCl→ Na+1 +Cl−1

then i = 2. The use of Eq. (12.2-20) gives

T − 100 = (2)(0.513)(0.62) = 0.636 ◦CTherefore, seawater boils at 100.636 ◦C.

9Experimental values indicate that i = 1.8 for NaCl.

418

12.2.2 Osmosis

Consider two compartments separated by a semipermeable membrane as shown in Figure 12.7.While the compartment on the left is filled with pure solvent (1), the other compartment con-tains a dilute amount of solute (2) in solvent (1). The membrane allows passage of solvent butprevents flow of the solute.

h

Membrane

Solvent Solution

Figure 12.7 Osmotic pressure at equilibrium.

Since the fugacity of pure solvent in the left compartment is greater than the fugacity ofsolvent in the solution, solvent molecules start passing from the left to the right compartmentuntil the fugacities (or partial molar Gibbs energies) become equal to each other. The transportof a pure solvent into a concentrated solution through a semipermeable membrane is calledosmosis10. In general, solvent moves from a solution with lower solute concentration into asolution with higher solute concentration.

When the movement of solvent stops, i.e., the system reaches equilibrium, the pressure ofthe solution, P soln, is greater than the pressure of the pure solvent, P . This extra pressure,i.e., P soln − P , is called the osmotic pressure, Π, i.e.,

Π = P soln − P = ρsolngh (12.2-22)

The flow direction of the solvent can only be reversed by applying a pressure in excess ofosmotic pressure to the compartment on the right-hand side. This process is called reverseosmosis.

At the membrane, the condition of equilibrium indicates that the fugacity of pure solventis equal to the fugacity of solvent in the solution, i.e.,

fL1 (T,P ) =bf soln1 (T,P soln, x1)

= γ1x1fL1 (T,P

soln) (12.2-23)

or

− ln(γ1x1) = ln"fL1 (T, P

soln)

fL1 (T, P )

#(12.2-24)

The use of Eq. (5.4-11) gives

fL1 (T, Psoln)

fL1 (T, P )

= exp

" eV1(P soln − P )

RT

#(12.2-25)

10The word "osmosis" comes from the Greek word for "push."

419

Substitution of Eq. (12.2-25) into Eq. (12.2-24) yields

− ln(γ1x1) =eV1(P soln − P )

RT(12.2-26)

If solute is dilute enough, the mixture forms an ideal solution, i.e., γ1 ' 1, and the left-handside of Eq. (12.2-26) is simplified as

ln(γ1x1) ' lnx1 = ln(1− x2) ' −x2 (12.2-27)

Therefore, Eq. (12.2-24) becomes

Π =RT x2eV1 (12.2-28)

indicating that the osmotic pressure of a solution depends on the number (or mole fraction)of solute in the solution and not on the chemical structure of the solute. In the literature, Eq.(12.2-28) is known as the van’t Hoff equation for osmotic pressure.

The mole fraction of solute is expressed as

x2 =n2

n1 + n2'

n2

n1(12.2-29)

On the other hand, molar volume of solvent is given by

eV1 = V1

n1(12.2-30)

Substitution of Eqs. (12.2-29) and (12.2-30) into Eq. (12.2-28) gives

Π =

µn2

V1

¶RT (12.2-31)

which is analogous to the ideal gas equation of state, i.e., P = (n/V )RT . Note that the termn2/V1 represents the molarity of solute.

Example 12.8 Calculate the osmotic pressure of 1 M solution of any nonelectrolyte, i.e.,glucose, dextrose, or sucrose, in water at 298K.

Solution

Taking R = 8.314× 10−2 bar.L/(mol.K), the use of Eq. (12.2-31) leads to

Π = (1)(8.314× 10−2)(298) = 24.8 bar

Comment: How fast this osmotic pressure is generated is not a subject in thermodynamics.

420

For ionized solutes, Eq. (12.2-31) is modified as

Π = i

µn2

V1

¶RT (12.2-32)

where i is the van’t Hoff factor defined by Eq. (12.1-27).

Example 12.9 List the osmotic pressures of the following solutions in decreasing order:0.1 M NaNO3, 0.1 M KNO3, 0.1 M NaCl, 0.1 M glucose, 0.1 M K2SO4.

Solution

Since molarities are equal, osmotic pressure will be dependent on the van’t Hoff factor, i. Thevan’t Hoff factors of the species are

Species NaNO3 KNO3 NaCl Glucose K2SO4

i 2 2 2 1 3

Therefore,Πglucose < ΠNaNO3 = ΠKNO3 = ΠNaCl < ΠK2SO4

Example 12.10 It is required to obtain fresh water from seawater by reverse osmosis. Sea-water contains about 3.5 wt % NaCl.

a) What is the minimum pressure required to carry out this process at 298K?

b) Estimate the minimum work required to obtain 10 L of fresh water.

Solution

a) Taking the density of water as 1 g/ cm3 and the molecular weight of NaCl as 58.5 g/mol,the use of Eq. (12.2-32) gives

Π = (2)

∙(3.5/58.5)

(96.5/1)× 10−3

¸| {z }

molL

(8.314× 10−2)| {z }bar.Lmol.K

(298)| {z }K

= 30.7 bar

The process of reverse osmosis requires an applied pressure just in excess of the osmotic pres-sure. The minimum applied pressure is then 30.7 bar. In practice, desalination of seawater iscarried out at 50-60 bar.

b) The minimum work required is

W = −Π∆V = −(30.7)(10× 10−3) = − 0.307 bar.m3 = − 30.7 kJ

The minus sign indicates that the work must be supplied by external means.

Comment: In the case of a batch process, the solution concentration increases as the water isremoved. Hence, the applied pressure must be continuously increased.

421

Note that Eq. (12.2-31) can also be expressed in the form

Π =

µm2/M2

V1

¶RT (12.2-33)

If c2 is the mass of solute per unit volume of solvent, i.e., m2/V1, then Eq. (12.2-33) reduces to

Π =RTc2M2

(12.2-34)

It should be kept in mind that Eq. (12.2-34) is only valid for very dilute solutions, i.e.,

limc2→0

Π

c2=

RT

M2(12.2-35)

For finite concentrations, expansion of the right-hand side of Eq. (12.2-34) by power series interms of c2 gives

Π

c2= RT

µ1

M2+B∗ c2 + C∗ c22 + ...

¶(12.2-36)

where B∗ and C∗ are the second and third virial coefficients, respectively. Once the osmoticpressure data are obtained as a function of solute concentration, a plot of reduced osmoticpressure, (Π/RT )/c2, versus c2 gives a straight line with slope B∗ and an intercept 1/M2. Anegative value of B∗ is indicative of a good solvent. A positive value of B∗ indicates the soluteis insoluble in the solvent.

Measurement of osmotic pressure is often used to determine the molecular weights of macro-molecules. The instrument for carrying out these measurements is called an osmometer. In asimple osmometer, shown in Figure 12.8, the pressure generated across a semipermeable mem-brane with solvent on one side and the solution in question on the other is measured in termsof the height of a solution. The molecular weight of the solute is then calculated from Eq.(12.2-34).

h

Solvent

Semipermeable membrane

Solution

Figure 12.8 A simple osmometer.

Example 12.11 To determine the molecular weight of insulin, 0.58 g of insulin is first dis-solved in 100mL of water. It is then placed in an osmometer operating at 291K. If the heightof the dilute solution rises 28mm above pure water, determine the molecular weight of insulin.

Solution

The osmotic pressure is

Π = ρgh = (1000)(9.8)(28× 10−3) = 274.4Pa

422

The mass concentration of the insulin is

c2 =0.58

100= 5.8× 10−3 g/ cm3 = 5.8× 103 g/m3

From Eq. (12.2-34)

M2 =RT c2Π

=(8.314)(291)(5.8× 103)

274.4= 51, 138 g/mol

12.2.2.1 Applications of osmosis

• Biology

Cells are composed of three parts: membrane, cytoplasm, and nucleus. The watery or aqueouscomponent of the cytoplasm is the cytosol, which includes ions and soluble macromolecules,for example enzymes. The insoluble constituents of the cytoplasm include the organelles andthe cytoskeleton, which gives to the cytoplasm a gel-like structure and consistency.

Solutions that contain the same concentration of water and solutes as the cell cytoplasmare called isotonic solutions. Cells placed in an isotonic solution will neither shrink nor swellsince there is no net gain or loss of water. Cell cytoplasm, body fluids (including blood, sweat,and tears), and 0.9 wt. % NaCl solution contain approximately the same solute concentration.Therefore, they are all isotonic solutions and blood freezes at − 0.52 ◦C lower than water11.

Solutions that contain a lower concentration of solute than the cytoplasm are said to behypotonic solutions. When a cell is placed in a hypotonic solution, the water diffuses into thecell as a result of osmosis, causing the cell to swell and possibly explode. This phenomenon iscalled lysis. On the other hand, solutions that contain a higher concentration of solute than thecell cytoplasm are called hypertonic solutions. When a cell is placed in a hypertonic solution,the water diffuses out of the cell as a result of osmosis, causing the cell to dehydrate andcollapse. For example, people drinking ocean water could die as a result of dehydration.

The three cases described above are schematically shown in Figure 12.9. Note that thebody loses water and minerals during physical activity. To make up for these losses, an isotonic

beverage, i.e., GatoradeR°, must be drunk.

Cell

Pure water

Cell

Very salty water

Cell

Saline solution (0.9% NaCl)

Isotonic Solution Hypotonic Solution Hypertonic Solution

Figure 12.9 Direction of water movement as a result of osmosis.

11Taking i = 1.8, T = −(1.8) ( 1.86)µ0.9/58.5

0.0991

¶= − 0.52 ◦C

423

Example 12.12 If a red blood cell is placed in a large volume of 0.25 M NaCl solution, whathappens to the volume of the cell?

Solution

The concentration of NaCl in the cell is approximately 0.9 wt. %. The corresponding molarityis

0.9/58.5

[(100− 0.1)/1]× 10−3= 0.155 M

Since the cell is exposed to a hypertonic solution, water moves out of the cell with a concomitantdecrease in cell volume. The cell shrinks until the inside solute concentration reaches 0.25 M.Since the total mass of solutes within the cell does not change, then

Vf

Vi=0.155

0.25= 0.62

where Vf and Vi represent the cell volume at the final and initial states.

When making pickles, the use of hypertonic salt solution is preferred since it causes thecucumbers to lose water as the lack of water prevents bacterial activities. This is the reasonwhy plain old cucumbers become infested with mold if they remain in the refrigerator over along period of time.

• Osmotic Power PlantThe world’s first osmotic power plant has been in operation in Norway since 2009. In theprototype built by Statkraft, seawater and fresh water, filtered to remove silt, are fed intopipes leading to membrane modules, made up of spiral coils to maximize surface area. In themembrane module, 80-90% of fresh water is transferred by osmosis across the membrane intothe seawater. As a result, the volumetric flow rate and pressure on the seawater side increase.The seawater leaving the membrane module is then sent to a hydropower turbine to generateelectricity. The details of the plant can be found on Statkraft’s web page12. Interested readersmay also refer to Skilhagen et al. (2008) and Gerstandt et al. (2008).

• Osmotic PumpThe most general form of an osmotic pump is shown in Figure 12.9. A collapsible partitionseparates the drug and osmotic driving agent chambers. When the osmotic pump is placed in anaqueous medium, water moves across the semipermeable membrane into the chamber containingosmotic driving agent, leading to an increase in pressure. As a result of the pressure exertedon the collapsible partition, the contents of the drug compartment are displaced through the

delivery orifice. Osmotic pumps manufactured by ALZETR°13 are extensively used for research

in mice, rats, and other laboratory animals.

Semipermeable membrane

Drug Osmotic driving

agent

Movable partition

Pump housing

Delivery orifice

Figure 12.10 A typical osmotic pump (Theeuwes and Yum, 1976).12http://www.statkraft.com13http://www.alzet.com

424

REFERENCES

Gentilcore, M.J., 2004, Chem. Eng. Prog., 100 (3), 38-41.

Gerstandt, K., K.V. Peinemann, S.E. Skilhagen, T. Thorsen and T. Holt, 2008, Desalina-tion, 224, 64-70.Gmehling, J.G., T.F. Anderson and J.M. Prausnitz, 1978, Ind. Eng. Chem.

Fundam., 17 (4), 269-273.

Goff, M.J., G.J. Suppes and M.A. Dasari, 2005, Fluid Phase Equilibria, 238, 149-156.

Hojjati, H. and S. Rohani, 2006, Organic Process Research & Development, 10 (6), 1110-1118.

Huang, C.C. and Y.P. Chen, 2000, Chem. Eng. Sci., 55, 3175-3185.

Kotula, I. and B. Marciniak, 2001, J. Chem. Eng. Data, 46, 783-787.

Kousksou, T., A. Jamil, Y. Zeraouli and J.P. Dumas, 2007, Chem. Eng. Sci., 62, 6516-6523.

Levenspiel, O. and N. de Nevers, 1974, Science, 183 (4121), 157-160.

Moon, Y.U., C.O. Anderson, H.W. Blanch and J.M. Prausnitz, 2000, Fluid Phase Equilibria,168, 229-239.

Nicola, D.G., G. Giuliana, P. Polonara, G. Santori and R. Stryjek, 2007, J. Chem. Eng. Data,52, 2451-2454.

Nti-Gyabaah, J., R. Chmielowski, V. Chan and Y.C. Chiew, 2008, Int. J. Pharm., 359, 111-117.

Nti-Gyabaah, J., V. Chan and Y.C. Chiew, 2009, Fluid Phase Equilibria, 280, 35-41.

Pappa, G.D., E.C. Voutsas, K. Magoulas and D.P. Tassios, 2005, Ind. Eng. Chem. Res., 44,3799-3806.

Skilhagen, S.E., J.E. Dugstad and R.J. Aaberg, 2008, Desalination, 220, 476-482.

Theeuwes, F. and S.I. Yum, 1976, Annals of Biomedical Eng., 4 (4), 343-353.

Wang Z. and L. Dang, 2009, Fluid Phase Equilibria, 276, 94-98.

Wei, D., L. Chen, J. Xu and F. Li, 2009, Fluid Phase Equilibria, 287, 39-42.

Wei, D. and K. Jin, 2009, J. Chem. Thermodynamics, 41, 145-149.

Verduzco, L.F.R, A.J. Aguilar, J.A.D.L. Reyes, J.A.M. Arroyo and F.M. Guevara, 2007, J.Chem. Eng. Data, 52, 2212-2219.

425

PROBLEMS

Problems related to Section 12.1

12.1 Using the following experimental data for the solubility of drug (2) in a solvent (1) as afunction of temperature, estimate the heat of fusion of the drug. The molecular weights of thesolvent and drug are 72.1 and 284.5 g/mol, respectively.

S ( g/100 g solvent) T ( ◦C) S ( g/100 g solvent) T ( ◦C) S ( g/100 g solvent) T ( ◦C)

0.725 8.5 2.274 18.5 5.190 26.11.547 15.0 2.770 20.7 7.441 29.5

(Answer: 77.7 kJ/mol)

12.2 For a binary system of stearic acid (2) and acetone (1), Goff et al. (2005) estimated thefollowing three-suffix Margules parameters from VLE measurements:

A = 0.045 B = 0.585

Estimate the solubility of stearic acid in acetone at 323K. The heat of fusion and the meltingpoint temperature of stearic acid are 61, 300 J/mol and 341K, respectively.

(Answer: x1 = 0.315)

12.3 Verduzco et al. (2007) studied solid-liquid equilibria of dibenzothiophene (2) in cyclo-hexane (1). The properties of dibenzothiophene are

∆ eCLSP = 31.4 J/mol.K Tm = 372K ∆ eHfus = 21, 000 J/mol

The activity coefficients of this binary liquid mixture are represented by the Wilson model with

λ12 − λ11 = 1969 J/mol λ21 − λ22 = 3709 J/mol

The molar volumes of liquid cyclohexane and dibenzothiophene are 119.8 and 163.6 cm3/mol,respectively. Calculate the solubility of dibenzothiophene in cyclohexane at 328.8K. Theexperimental value is reported as 0.1291.

(Answer: 0.1302)

12.4 When the constituents of a mixture have high boiling points, determination of activitycoefficients from vapor-liquid equilibrium experiments is difficult. In this case, the solid-liquidequilibrium data in the form of "solubility versus temperature" provide the means to estimateactivity coefficients as a function of composition. The following experimental data are providedby Verduzco et al. (2007) for the solubility of dibenzothiophene (2) in n-hexadecane (1) atatmospheric conditions:

x2 T (K) x2 T (K) x2 T (K)

0.0463 291.0 0.1008 315.0 0.3737 350.90.0475 292.2 0.1201 319.9 0.5021 356.90.0487 293.0 0.1783 331.5 0.5094 357.00.0504 293.7 0.2000 334.7 0.6047 360.10.0726 304.4 0.2565 341.5 0.6954 362.7

426

The properties of dibenzothiophene are reported as

∆ eCLSP = 31.4 J/mol.K Tm = 372K ∆ eHfus = 21, 000 J/mol

a) Determine the activity coefficients of dibenzothiophene as a function of composition.b) If the activity coefficients are represented by the Wilson model, determine the parametersusing the procedure explained in Problem 8.19.

12.5 When ∆ eCLSP 6= 0, calculation of ideal solubility from Eq. (12.1-20) is valid as long as T

and Tm do not differ greatly. Otherwise, the second and third terms on the right-hand side ofEq. (12.1-14) should be taken into consideration.

a) If ∆ eCLSP varies linearly with temperature, i.e.,

∆ eCLSP = α+ βT (1)

show that the ideal solubility is given by

lnxideal2 =∆ eHfus

2

RTm

µ1− Tm

T

¶+(βT − α)

R

µ1− Tm

T

¶− βT

2R

"1−

µTmT

¶2#− α

Rln

µTmT

¶(2)

b) Use Eq. (2) and calculate the ideal solubility of paracetamol at 298.2K. The following dataare provided by Hojjani and Rohani (2006) for paracetamol:

∆ eHfus = 28.1 kJ/mol Tm = 442.2K ∆ eCLSP ( J/mol.K) = 302.54− 0.4589T

Compare the calculated value with the one obtained from Eq. (12.1-20).

(Answer: b) 0.093 and 0.025)

12.6 Nti-Gyabaah et al. (2008) reported the solubility of lovastatin (2) in n-butanol (1) as

T (K) x2 × 103 T (K) x2 × 103

285.70 4.6046 304.10 7.6196288.70 4.9877 305.20 7.8606290.65 5.2554 306.70 8.2028296.20 6.1126 308.70 8.6850301.20 7.0227 310.10 9.0410

The heat of fusion and the melting temperature of lovastatin are 43, 136 J/mol and 445.5K,respectively. It is required to express the infinite dilution activity coefficients as a function oftemperature in the form

ln γ∞2 = A+B

T(1)

a) If ∆ eCLSP2= 255 J/mol.K, use Eq. (12.1-16) with γ2 = γ∞2 (Why?) and show that

A = − 3.898 and B = 1798

b) Assume ∆ eCLSP2= 0 and use Eq. (12.1-19) to obtain

A = 8.456 and B = − 2737

427

c) If ∆ eCLSP2

is approximated to be equal to the entropy of fusion at the melting temperature,show that Eq. (12.1-16) simplifies to

ln (γ2x2) = −∆ eHfus

2

RTmln

µTmT

¶(2)

Use Eq. (2) with γ2 = γ∞2 and show that

A = 3.765 and B = − 1015

12.7 For low solubility values, i.e., x2 ¿ 1, it is plausible to replace γ2 by γ∞2 in Eq. (12.1-16). The temperature dependence of the infinite dilution activity coefficient can be expressedas

ln γ∞2 = A+B

T(1)

a) Show that the substitution of Eq. (1) into Eq. (12.1-16) and rearrangement lead to

lnx2 = α1 +α2

T+ α3 lnT (2)

where

α1 =∆ eHfus

2

RTm−∆ eCLS

P2

R(1 + lnTm)−A (3)

α2 =∆ eCLS

P2Tm

R−∆ eHfus

2

R−B (4)

α3 =∆ eCLS

P2

R(5)

Expressing solubility in the form of Eq. (2) is known as the Apelblat model.

b)Wang and Dang (2009) reported the following data for the solubility of erythromycin (2) inacetone (1) as a function of temperature:

x2 × 104 83.50 130.5 195.2 339.9 444.2 654.0

T (K) 298 303 308 313 318 323

Fit the data to Eq. (2) by using the method of least squares as explained in Example 6.15 andshow that

α1 = − 98.6326 α2 = 4358.31 α3 = 14.7472

12.8 Irganox 1010 (C73H108O12) is a widely used phenolic antioxidant in plastic, rubber,synthetic fiber, and elastomer. Its solubility in various solvents is of interest for the properdesign of a crystallizer. Wei et al. (2009) reported that the binary system of n-hexane (1) andIrganox 1010 (2) follows the Wilson model with the following parameters:

λ12 − λ11 = 3289.60 J/mol and λ21 − λ22 = 4482.60 J/mol

Estimate the solubility of Irganox 1010 in n-hexane at 328.15K. The characteristic data forpure components are given by

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Component ∆ eHfus ( kJ/mol) Tm (K) eV L ( cm3/mol)

Irganox 1010 65.95 385.75 1125.7n-Hexane 134.9

The experimental value reported by Wei et al. (2009) is x2 = 3.703× 10−3.(Answer: x2 = 3.625× 10−3)

12.9 Nti-Gyabaah et al. (2009) reported the solubility of simvastatin (2), a lipid-loweringagent, in methyl acetate (1) at 285.3K as x2 = 18.86 × 10−3. Using the following data,estimate the solubility of simvastatin in butyl acetate at the same temperature:

DATA: Methyl acetate (1) - simvastatin (2) and butyl acetate (3) - simvastatin (2) mixturesare represented by the NRTL model with the following parameters:

g12 − g22 = 5516 J/mol g21 − g11 = − 369 J/mol

g32 − g22 = 6718 J/mol g23 − g33 = − 1466 J/mol

α = 0.4

(Answer: x2 = 24.48× 10−3)

12.10 Solubility of acenaphthene (2) in chloroform (1) at 305K is reported by Kotula andMarciniak (2001) as x2 = 0.2322. Estimate the solubility of acenaphthene in 1,1-dichloroethaneat the same temperature.

Chloroform-acenaphthene and 1,1-dichloroethane-acenaphthene systems are represented by theregular mixture theory with the following parameters:

Component eV L ( cm3/mol) δ ( J/ cm3)1/2

Chloroform 88.27 16.616Acenaphthene 149.8 18.931,1-dichloroethane 93.28 15.984

(Answer: x2 = 0.2066)

12.11 Prediction of solid-liquid equilibrium is of importance in the cryogenics industry. It isrequired to estimate the solubility of solid n-hexane (2) in liquid methane (1) at 150K.

a) Estimate the solubility by using the regular mixture theory with the following data:

Component ∆ eHfus ( kJ/mol) Tm (K) eV L ( cm3/mol) δ ( J/ cm3)1/2

Methane 52 11.6n-Hexane 13.08 177.8 134.9 14.9

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b) The activity coefficient of species i in a mixture is defined by Eq. (8.3-5) as

γi =bφLi (T,P, xi)φLi (T, P )

=bφLi (T, P, xi)

(P vapi /P ) φ

Vi (T, P

vapi ) exp

" eV Li (P − P vap

i )

RT

# (1)

If the liquid phase is represented by the Redlich-Kwong equation of state, show that thesubstitution of Eqs. (5.4-12) and (7.6-2) into Eq. (1) leads to

ln γi =Bi

Bmix

¡ZLmix − 1

¢−¡ZLi − 1

¢+ ln

ÃZLi −Bi

ZLmix −Bmix

!+

Aii

Biln

Ã1 +

Bi

ZLi

!

− Amix

Bmix

⎛⎜⎜⎜⎝2

kPj=1

xjAij

Amix− Bi

Bmix

⎞⎟⎟⎟⎠ lnÃ1 +

Bmix

ZLmix

!(2)

Estimate the solubility of solid n-hexane in liquid methane using the Redlich-Kwong equationof state. Mateo and Furata (1975) reported that k12 = 0.0567. In the calculation of com-pressibility factors, use the vapor pressure of pure methane, which is 10.1 bar at 150K, as thepressure.(Answer: a) x2 = 9.3× 10−2 b) x2 = 1.25× 10−2)

12.12 The "proof" of an alcoholic beverage is twice the volume percent of its alcoholic (ethylalcohol) content. Estimate the freezing point temperatures of whisky (80 proof), gin (70 proof),and vodka (65 proof). The density of ethyl alcohol is 0.79 g/ cm3.

(Answer: − 21.26 ◦C, − 17.17 ◦C, − 15.36 ◦C)

12.13 When 8 g of nonelectrolyte X is dissolved in 92 g of benzene, the observed freezingpoint is 2.8 ◦C. Estimate the molecular weight of X.

(Answer: 164.9 g/mol)

12.14 A liquid mixture contains 70 mol % benzoic acid (1) and 30% 1-naphthol (2). If thismixture is cooled at constant pressure, estimate the temperature at which the solid phase firstappears and identify the solid. The mixture is represented by the Wilson equation with thefollowing parameters (Huang and Chen, 2000):

Λ12 = 1.5484 and Λ21 = 0.6326

For pure components

Component ∆ eHfus ( J/mol) Tm (K)

Benzoic acid 16, 230 395.41-Naphthol 23, 220 369.0

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(Answer: 368.1K, Benzoic acid)

12.15 180 kg of a liquid mixture consisting of 70 weight % component 1 and 30% component2 is fed to a crystallizer. Within the crystallizer, the temperature of the feed solution is slowlyreduced from 50 ◦C to 30 ◦C and one-third of the liquid mixture is removed from the crystallizer.The contents of the crystallizer are then heated up to 40 ◦C. Using the phase diagram givenbelow, estimate the amount of component 1 crystals at the end of the overall process.

0 0.8 0.9 1.0 0.7 0.6 0.5 0.4 0.3 0.2 0.1 10

30

20

40

60

50

Tem

pera

ture

(o C

)

Weight fraction of component 1

(Answer: 24 kg)

Problems related to Section 12.2

12.16 When 38 g of polymer X is dissolved in 80 g of benzene, the normal boiling point ofbenzene increases by 1.014K. Calculate the molecular weight of X if the only available dataare the variation of vapor pressure of benzene as a function of temperature in the form

lnP vap = 9.2806− 2788.51

T − 52.36

where P is in bar and T is in K.

(Answer: 1151 g/mol)

12.17When 1.3 g of an unknown protein is dissolved in 125 cm3 of water at 298K, the osmoticpressure is measured as 0.005153 bar. Estimate the molecular weight of the protein.

(Answer: 50, 000 g/mol)

12.18 When you soak your hands in water for a long period of time, you notice that they gowrinkly. Explain.

12.19 Explain what happens to your eyesight when you swim

a) in a pool,b) in the ocean. Why?(Answer: a) short-sightedness b) long-sightedness)

12.20 Explain the following:

a) What is the purpose of using excessive sugar in making jams?b) What is the cause of diarrhea?

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12.21 Determine the molarity of a nonelectrolyte having an osmotic pressure of 70 bar at293K.

(Answer: 2.87)

12.22 Moon et al. (2000) reported the following osmotic pressure data for dilute concentra-tions of lysozyme (2) in ammonium sulfate solutions at 298.15K for an ionic strength of 1 Mat pH 8:

c2 ( g/L) Π (mmHg) c2 ( g/L) Π (mmHg)

4.06 4.40 14.26 14.556.13 6.50 16.31 16.308.18 8.65 18.38 18.3510.42 10.75 20.44 20.1512.26 12.55

Determine the molecular weight of lysozyme.

(Answer: 16, 910 g/mol)

12.23 Consider a long hollow pipe with a reverse osmosis membrane attached to its bottomend.

a) How deep should this pipe be lowered into the ocean so as to initiate reverse osmosis?Assume uniform temperature and salinity of the ocean and use the following data:

• Osmotic pressure of seawater = 26bar• Atmospheric pressure = 1bar• Density of seawater = 1025 kg/m3• Density of pure water = 1000 kg/m3• Acceleration of gravity = 9.8m/ s2

b) If the pipe is lowered to a depth of 500m, estimate the level of fresh water within the pipe.c) Calculate the length of the pipe required to bring fresh water just to the ocean surface. Isthis a feasible project?

For more details on this problem see Levenspiel and de Nevers (1971).

(Answer: a) 259m b) 253m below the sea level c) 10, 612m)

12.24 The primary function of red blood cells (erythrocytes) is to carry oxygen to the tissuesand to remove waste. The cytoplasm of red blood cells is congested with protein molecules,called hemoglobin, occupying a quarter of the cell volume. Red blood cells get their colorfrom hemoglobin, which is bright red. Consider a spherical red blood cell with a diameter of8.4× 10−6m. The average diameter of a hemoglobin molecule is about 6× 10−9m.

a) If a red blood cell is placed in a large volume of pure water at 298K, estimate the osmoticpressure experienced by the cell membrane.

Hint: 1 mol of hemoglobin contains 6.02× 1023 molecules, i.e., Avogadro’s number.b) The surface tension, σ, of a spherical droplet of radius R is given by

σ =R∆P

2

432

where ∆P represents the pressure differential across the droplet surface. If the cell membranecan withstand a maximum tension of 2× 10−4N/m, determine whether lysis of the red bloodcell takes place.

(Answer: a) 12.1 kPa)

12.25 Experimental determination of the freezing point depression is much easier than that ofthe osmotic pressure. For this reason, it is generally preferred to estimate the osmotic pressureof a solution from its freezing point depression. When the solvent is water at a temperature of273.15K, combine Eqs. (12.1-27) and (12.2-31) to show that

Π = 12.2 (Tm − T )

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434