Chapter 14-Part VIII Solid-vapor equilibrium (SVE) and Solid-liquid equilibrium (SLE)

Chapter 14-Part VIII

Solid-vapor equilibrium (SVE)and

Solid-liquid equilibrium (SLE)

A solid can vaporize at T < T triple point; pressures along the sublimation curve are called saturation pressures of the solid

Lets consider SVE of a pure solid (1) and a vapor mixture

Solid 1

Vapor mixtureof 1 and 2

Species 2 does not dissolve in the solid phase;

In the vapor phase usually 2 is the solvent; 1 is the solute

We want to calculate the solubility of 1 in the vapor phase as afunction of T and P

SVE for component 1:

vs ff 11ˆ

We model the solid phase with the same equation of the liquid(is a condensed phase)

RT

PPVPf

satsolsatsats )(

exp 11111

Pyf vv111ˆˆ

Solubility of 1 in the vapor phase

RT

PPV

P

Py

satsol

v

satsat )(exp

ˆ11

1

111

How this expression may reduce at low pressures?

Simplifications to the equationSolubilities of solids in fluids at high pressures important for separation processes

Examples: extraction of caffeine from coffee, separation of asphaltenes from heavy petroleum fractions

Usually P1sat is very small;

Saturated vapor can be considered ideal gasAlso if y1 is very small, 11

ˆˆ

RT

PV

P

Py

RT

PPV

P

Py

solsat

satsolsatsat

1

1

11

11

1

111

expˆ

)(exp

ˆ

Simplified equation

RT

PV

P

Py

solsat1

1

11 exp

ˆ

The fugacity coefficient at infinite dilution can be calculated from an EOS

Where aij is calculated as

aij =(1-lij)(aiaj)1/2

lij is a cross-coefficient for the mixture

Solubility of a solid in a gasEstimate the solubility of naphthalene in carbon dioxide at 1 bar and temperatures of 35 and 60.4 oC assuming that the solid is incompressible, and the solid and fluid phases may be considered ideal.

P

Py

RT

PV

P

Py

RT

PPV

P

Py

sat

solsat

satsolsatsat

11

1

1

11

11

1

111

expˆ

)(exp

ˆ

Solving at each temperature

T = 35 oC; Pnsat = 2.789 x10-4 bar; yN =0.00028

T = 60.4 oC; Pnsat = 2.401 x10-3 bar; yN =0.0024

Lets consider the effect of pressureEstimate the solubility of naphthalene in carbon dioxide at 1 bar and temperatures of 35 oC and pressures from 1 bar to 60 bar using the virial equation of state with the following values for the second virial coefficient

B(CO2-naphthalene) = -345 cc/molB (CO2-CO2) = -96.5 cc/molB(naphthalene-naphthalene) = -1850 cc/mol

Assume that CO2 is insoluble in solid naphthalene, and therefore only equate thefugacities of naphthalene in the solid and vapor phases.

RT

PPV

P

Py

satsolsatsat )(exp

ˆ11

1

111

Since the fugacity coefficient

vi is a function of y1

iterate

SLE: solid-liquid equilibria

Solid-liquid equilibria (SLE)

Solid-Liquid Equilibria (SLE)

i

s

i

s

i

l

i

l

i

si

s

i

s

ili

l

i

l

i

s

i

l

i

xx

fxfx

ff

ˆˆ

SLE-cont.

),(

),(

),(

),(

),(

),(

),(

),(

),(

),(

),(

),(

PTf

PTf

PTf

PTf

PTf

PTf

PTf

PTf

PTf

PTf

PTf

PTf

li

mli

msi

si

li

mli

mli

imsi

msi

si

li

si

i

i

i

i

ii

T-effect on fugacity

SLE-cont.

T

T

si

li

i

T

T

Ri

mi

i

Ri

P

i

im

imi

dTRT

HH

dTRT

H

PTf

PTf

RT

H

T

f

2

2

2

exp

exp),(

),(

ln

The enthalpies are functions of T (through the Cp dependence on T)

SLE (cont.)

IT

TT

T

T

R

C

T

TT

RT

H

dTRT

HH

i

i

i

i

i

im

m

m

P

m

m

sli

T

T

si

li

i

ln

exp

exp2

dTdTdTT

CC

RTI

T

T

T

T

T

sP

lPT

Tim im

ii

im

)(12

Simplifications

• Triple integral I is usually neglected• Heat capacity change of melting usually not

available• Therefore

T

TT

RT

Hi

i

m

m

sli

i exp

i

s

i

s

i

l

i

l

i xx For all components

SLE-Typical cases

A) Assume ideal-solution behavior for both phases

for all T and compositions

1;1 11 sl

222

111

sl

sl

xx

xx

T

TT

RT

Hi

i

m

m

sli

i expGives a normalT-x phase diagram

SLE-typical casesB) Assume ideal behavior

for the liquid phase

and complete immiscibility for all species in the solid state=>

11 l

1sis

ix

22

11

l

l

x

x

For case B:11 lx

Both equations apply at the eutectic point

Liquid + solid 1Liquid + solid 2

22 lx

Example: SLEEstimate the solubility of solid naphthalene in liquid n-hexane at 20 oC.

DataNaphthalene MW = 128.19Melting point: 80.2 oCHeat of fusion: 18.804 kJ/molDensity of the solid: 1.0253 g/cc at 20oCDensity of the liquid: 0.9625 g/cc at 100 oCVapor pressure of the solid: log P (bar) = 8.722 -3783/T (T in K)The heat capacities of liquid and solid naphthalene may be assumed to be equal

If Cp =0,

1

1ln 11

m

fusion

T

T

RT

Hx

The result is x1 = 0.269, the experimental result is x1 = 0.09How can we correct the answer?

Same example using UNIFAC to estimate the naphthalene activity coefficient

Naphthalene has 8 aromatic CH (subgroup 10) and 2 aromatic C (subgroup 11)

N-hexane has 2 CH3 (subgroup 1) and 4 CH2 (subgroup 2) groups.

11

1

11

/)314.1exp(

1exp

0 since

ln11exp

m

fus

P

mmP

m

fus

TT

RT

H

x

C

T

T

T

T

RC

TT

RT

H

x

Since 1 is a function of x1, we need to iterate in x1

Solving for x1 yields x1 =0.124, 38% larger than the experimental value

Calculation of activity coefficient from solubility data

The following data has been reported for benzo-pyrene and its solubility in water at 25 oC:

Melting point: 178.1 oCHeat of fusion: 15.1 kJ/molSolubility in water: xBP =3.37x10-10

Estimate the activity coefficient of benzo-pyrene in water at 25 oC

Calculation is similar to previous example

8

11 1076.3

1exp

0 if

xx

TT

RT

H

C

m

fus

P

Note that in this case the correction given by the activity coefficient is HUGE !!!!

Important in applications when we are interested in the distribution of a chemical speciesbetween air, water, soil. Since the concentration of benzopyrene is small, thecalculated activity coefficient is the infinite diluted activity coefficient.