CHEE 311 J.S. Parent 1

8. Solute (1) / Solvent (2) Systems 12.7 SVNA

Until now, all the components we have considered in VLE calculations have been below their critical temperature. Their pure component liquid fugacity is calculated using:

Our VLE equation that describes the distribution of each component between liquid and vapour has the form:

How do we deal with components that, at the temperature of interest, are above Tc and no longer have a Pi

sat?

RT

)PP(VexpPf

sati

liqisat

isati

liqi

RT

)PP(VexpPx

fxPˆysati

lsati

satiii

liii

vii

i

CHEE 311 J.S. Parent 2

VLE Above the Critical Point of Pure Components

CHEE 311 J.S. Parent 3

Pure Species Fugacity of a Solute

The difficulty in handling a component that is above its critical temperature or simply unstable as a pure liquid is to define a pure component fugacity for the purpose of VLE calculations.

While this component must have a liquid solution fugacity, f1

l, it does not havea pure liquid fugacity, f1

l atx1 = 1.

The tangent line at x1=0 isthe Henry’s constant, k1.It is useful for predicting the mixture fugacity of a dilutecomponent, but it cannotbe extrapolated to x1=1with any degree of accuracy.

CHEE 311 J.S. Parent 4

Pure Component Fugacity of a Solute

The pure component fugacity of a solute is calculated from a combination of Henry’s Law and an activity coefficient model.

Recall that Henry’s Law may be used to represent the mixture fugacity of a minor (xi<0.02) component in a liquid.

defines the Henry’sconstant

andis accurate as longas x1 < 0.02

Unfortunately, we cannot extrapolate the above equation to x1 = 1 to give us the pure component f1.

An activity coefficient model can refine this approach

1

l1

0x1 x

f̂limk1

11l1 xkf̂

CHEE 311 J.S. Parent 5

Pure Component Fugacity of a Solute

Recall that the activity coefficient is the ratio of the mixture fugacity of a component to its ideal solution fugacity:

At infinite dilution (x10), the activity coefficient becomes:

Since the pure component fugacity is a constant at a given T, we can write this expression as:

Using the definition of the Henry’s Constant, ki, we have:

or 12.34

l11

l1

0x1

0x1

fx

f̂limlim11

1

l1

0xl1

1 xf̂

limf

1

1

l1

11

f

k

1

1l1

kf

l11

l1

1 fx

f̂

CHEE 311 J.S. Parent 6

Pure Component Fugacity of a Solute

Equation 12.34 is a rigorous thermodynamic equation,

12.34

for the fugacity of a “pure” solute. However, it is evaluated at P2sat

(where x1 = 0) and its use requires us to assume that pressure has an insignificant influence on the solute’s fugacity.

To apply 12.34, we require a Henry’s constant for the system at the temperature of interest, ki(T), and an excess Gibbs energy model for the system, also at the T of interest.

1

1l1

kf

CHEE 311 J.S. Parent 7

VLE Relationship for a Supercritical Component

Consider a system where one component is above Tc (species 1) and the other component is below Tc (species 2).

The equilibrium relationship for component 2 is unchanged:

or

However, component 1 is handled differently, using a Henry’s constant (k1) and the infinite dilution activity coefficient (1

). Both are properties specific to this mixture.

12.36

RT

)PP(VexpPxPˆy

sat2

lsat2

sat222

v22

2

1

111

v11

kxPˆy

sat22222 PxPy

CHEE 311 J.S. Parent 8

Solute (1) / Solvent (2) Systems: Example

Carbon Dioxide(1) / n-butane(2) at 71C

0

10

20

30

40

50

60

70

80

90

0.000 0.200 0.400 0.600 0.800 1.000

x1, y1

Pre

ss

ure

(b

ar)

P-x1

P-y1

CHEE 311 J.S. Parent 9

9. Phase Stability and Liquid-Liquid Equilibria

Throughout the course we have developed methods of calculating the thermodynamic properties of different systems:

Gibbs energy of pure vapours and liquids Gibbs energy of ideal and real mixtures Definition of vapour liquid equilibrium conditions

As we apply these methods, we assume that the phases are stable.

Recall our calculation of the Gibbs energy of a hypothetical liquid while developing Raoult’s law.

In our flash calculations that we calculated Pdew and Pbubble before assuming that two phases exist

A slight extension of the thermodynamic theory covered in CHEE 311 provides us with a means of assessing the stability of a phase.

Answers the question: “Will the system actually exist in the state I have chosen?”

CHEE 311 J.S. Parent 10

Phase Stability

A system at equilibrium has minimized the total Gibbs energy. Under some conditions (relatively low P, high T) it assumes a

vapour state Under others (relatively high P, low T) the system exists as a

liquid Mixtures at specific temperatures and pressures exist as a

liquid and vapour in equilibrium

Consider the mixing of two, pure liquids. We can observe two behaviours:

complete miscibility which creates a single liquid phase partial miscibility which creates two liquid phases

» in the extreme case, these phases may be considered completely immiscible.

CHEE 311 J.S. Parent 11

Stability and the Gibbs Energy of Mixing

We have already discussed the property changes of mixing, in particular the Gibbs energy of mixing.

Before After

GA GB G

nA moles nB moles nA + nB moles liquid A liquid B of mixture

The Gibbs energy of mixing is defined as:

which in terms of mole fractions becomes:

+

BBAABAmixBA GnGnG)nn(G)nn(

BBAAmix GxGxGG

CHEE 311 J.S. Parent 12

Stability and the Gibbs Energy of Mixing

The mixing of liquids changes the Gibbs energy of the system by:

Clearly, this quantity must be negative if mixing is to occur, meaning that the mixed state is lower in Gibbs energy than the unmixed state.

i

iiimix xlnxRTG

CHEE 311 J.S. Parent 13

Stability Criterion Based on Gmix

If the system can lower its Gibbs energy by splitting a single liquid phase into two liquids, it will proceed towards this multiphase state.

A criterion for single phase stability can be derived from a knowledge of the composition dependence of Gmix.

For a single phase to be stable at a given temperature, pressure and composition:

Gmix and its first and second derivatives must be continuous functions of x1

The second derivative of Gmix must satisfy:

14.50

dx

)RT/G(d21

mix2

CHEE 311 J.S. Parent 14

Phase Stability Example: Phenol-WaterTemp 298 K Composition: Phase Property Data:

xphenol xwater phenol water Gmix (J)NRTL Equation Coefficients: 0.00 1.00 30.781 1.000 0

b12 -563.6 K 0.05 0.95 7.786 1.033 -161b21 1843.9 K 0.10 0.90 3.291 1.106 -285

0.2 0.15 0.85 1.909 1.194 -4330.20 0.80 1.354 1.283 -595

-1.8912 =b12/T 0.25 0.75 1.095 1.364 -760 6.1876 =b21/T 0.30 0.70 0.964 1.431 -919

0.35 0.65 0.899 1.480 -1066G12 1.4597 =exp(- 0.40 0.60 0.869 1.509 -1195G21 0.2901 =exp(- 0.45 0.55 0.861 1.519 -1302

0.50 0.50 0.866 1.511 -13840.55 0.45 0.879 1.486 -14390.60 0.40 0.897 1.446 -14630.65 0.35 0.916 1.396 -14560.70 0.30 0.936 1.336 -14130.75 0.25 0.954 1.270 -13330.80 0.20 0.970 1.199 -12110.85 0.15 0.983 1.127 -10400.90 0.10 0.992 1.053 -8100.95 0.05 0.998 0.980 -4991.00 0.00 1.000 0.908 0

CHEE 311 J.S. Parent 15

Phase Stability Example: Phenol-Water (25C)

-1600

-1400

-1200

-1000

-800

-600

-400

-200

0

0.00 0.20 0.40 0.60 0.80 1.00xphenol

Gm

ix:

J/m

ole

CHEE 311 J.S. Parent 16

Liquid-Liquid Equilibrium: Phenol-Water

CHEE 311 J.S. Parent 17

9. Liquid Stability 14.1 SVNA

Whether a multi-component liquid system exists as a single liquid or two liquid phases is determined by the stability criterion:

14.5

If this condition holds, the liquid is stable. If not, it will split into two (or more) phases.

Substituting Gmix for a binary system,

(A)

we derive an alternate stability criteria based on component 1:

(A) into 14.5

This quantity must be positive for a liquid to be stable.

0dx

RTGd

P,T21

mix2

222111mix xlnxxlnxRT/G

0dx

)x(lnd

P,T1

11

CHEE 311 J.S. Parent 18

Wilson’s Equation and Liquid Stability

Given our phase stability criterion:

what we require to gauge liquid stability is an activity coefficient model. Wilson’s equation for component 1 of a binary system gives:

11.17

Applying our stability criterion to 11.17 yields:

Given that all Wilson’s coefficients ij are positive, all quantities on the right hand side are greater than zero for all compositions

Wilson’s equation cannot predict liquid instability, and cannot be used for LLE modeling.

0dx

)x(lnd

P,Ti

ii

2112

21

1221

12212211 xxxx

x)xxln(ln

22112

221

212211

2122

1

11

)xx(xxx

xdx

)x(lnd

CHEE 311 J.S. Parent 19

Liquid-Liquid Equilibrium (LLE) SVNA 14.2

CHEE 311 J.S. Parent 20

Liquid-Liquid Equilibrium: Phenol-Water

CHEE 311 J.S. Parent 21

Liquid-Liquid Equilibrium Relationships

Two liquid phases () at equilibrium must have equivalent component mixture fugacities:

In terms of activity coefficients:

If each component can exist as a liquid at the given T,P, the pure component fugacities cancel, leaving us with:

14.10

Note that the same activity coefficient expression applies to each phase.

iiiiii fxfx

ii f̂f̂

iiii xx

)P,T,x,...,x,x( n21ii

)P,T,x,...,x,x( n21ii

CHEE 311 J.S. Parent 22

Liquid-Liquid Equilibrium-NRTL

Consider a binary liquid-liquid system described by the NRTL excess Gibbs energy model.

For phase , we have:

11.20

11.21 For phase , we have:

11.20

11.21

The activity coefficients of the two liquids are distinguished solely by the mole fractions of the phases to which they apply.

2

2121

2121

2

1212

121222

Gxx

G

Gxx

Gxln

2

1212

1212

2

2121

211211

Gxx

G

Gxx

Gxln

2

2121

2121

2

1212

121222

Gxx

G

Gxx

Gxln

2

1212

1212

2

2121

211211

Gxx

G

Gxx

Gxln

CHEE 311 J.S. Parent 23

Liquid-Liquid Equilibrium Calculations

In CHEE 311, we will consider only binary liquid-liquid systems at conditions where the excess Gibbs energy is not influenced by pressure.

The phase rule tells us F=2-+C=2-2+2 = 2 degrees of freedom

If T and P are specified, all intensive variables are fixed

For this two-component system we can write the following equilibrium relationships:

and 14.10

The latter can be stated in terms of component 1, to yield:

and 14.12

The activity coefficients i and i

are functions of xi and xi

.

1111 xx 2222 xx

1111 xx 2121 )x1()x1(

CHEE 311 J.S. Parent 24

Liquid-Liquid Equilibrium Calculations

Once we establish that a LLE condition exists, we are interested to know the composition of the two phases.

Given T (and P), find the composition of the two liquids.

Start with our equilibrium relationships.Component 1:

Component 2:

The natural logarithm is usually simpler to solve:

We have two equations, and twounknowns xi

and xi.

1111 xx

2121 )x1()x1(

1

1

1

1

x

xlnln

)x1(

)x1(lnln

1

1

2

2