chemthermo-chap3

CAB 2023 : CHEMICAL THERMODYNAMICS July 2007

1

3. PHASE EQUILIBRIUM - FUNDAMENTALS3. PHASE EQUILIBRIUM - FUNDAMENTALS

Learning Objectives

Student should be able to ;

i. Explain the use of phase equilibrium in chemical engineering processesii. Define and explain the fundamental criteria of phase equilibrium particularly using

Gibbs Free Energy.iii. Define the concept of chemical potential and able to explain its role in defining phase

equilibria.iv. Define the concept of fugacity and the reason for its introduction as a means for

performing calculation related to phase equilibria.v. Define and explain partial molar property concept and able to calculate property

changes of a system as a result of mixing.vi. Derive the Maxwell relations for mixtures.vii. Perform calculation related to phase equilibria and establish the vapour liquid

equilibrium plot (P-x,y or T-x,y) for ideal mixtures.


2

vapour

liquid

phase changes

Phase Equilibrium is commonly encountered in Chemical Processes, eg. distillation

What happen in the system ?

- equilibrium between liquid and vapour- vaporisation/condensation taking place i.e., phase changes

To make matter more complex, system containsmulti-components.

Recall on what we have learnt before ?

Thermodynamic relationships involving Free Energy function i.e. Gibbs Free Energy in particular - homogenous fluid of pure component/const. composition.

Thermodynamic calculations (enthalpy & saturation vapour pressure) involving phase changes.

Network of Thermodynamic Equations - Maxwell Equations

Now, these thermodynamic property relationships will be extended to systems undergoing phase & composition changes

Application of Phase Equilibrium


3

Equilibrium Criteria

Let us review the concept of thermal and mechanical equilibrium….

Thermal Equilibrium Mechanical Equilibrium

Under phase equilibria of pure component,

Phase boundary

Phase

Phase

T = T

Driving Force T

P = P

Driving Force P


4


Now consider a system of multi component…..

Thermal Equilibrium Mechanical Equilibrium

Phase boundary

Phase

Phase

T = T

Driving Force T

P = P

Driving Force P

Another equilibrium criteria involvingcomposition in the two phase

? = ?

Driving Force

The tendency would be to say composition difference….


5


Now consider a system of multi component…..

Phase boundary

Phase

Phase

If composition, then the criteria at equilibrium in addition to the pressure and temperature should be x = x . But is it true ?

Air-Water system containing small amount of methanol

Suppose the composition of methanol in water is 0.01

Will the composition of methanol in air is 0.01 too at equilibria?

The answer is no. Therefore, composition is not necessary the driving force…..

We will look at this matter now …


6


Consider a closed system in terms of changes in matter but not isolated from surroundings in terms of heat and work flow. Let us start with a pure system first….

1st lawdU = dQ + dW = dQ - Pb dV

dS = dQ/Tb + Sp 2nd law

SYSTEM

QW

Sp - entropy production due to irreversibilitysubscipt 'b' - boundary of system

dU = Tb dS - TbSp - Pb dV

combining the two eqn. will yield ;

Since, Sp > 0,

dU < T dS - P dV or dU + P dV - T dS < 0

suppose there is an imbalance of T or P,process leading to equilibrium will continue until T and P for the system boundary equal to surrounding by means of rejecting/accepting heator doing or receiving work.

for the above equation to be true.


7


Using the equation as the basis,

At const. T and P, the eqn. becomes;

Taking the total differential from the equation, we can derive

dG = dU + P dV + V dP - T dS - S dT

Introduce the Gibbs Free EnergyG = H - TS = U + PV - TS We know that H = U + PV

dG T,P = dU + P dV - T dS



8


Referring back to the equation,

Hence, at equilibrium we could use this basic criteria

Implication : Any real process occuring spontaneously when system is at const. T and P incurs a decrease in Gibbs Free Energy. And the system will attain its thermodynamic equilibrium when the Gibbs Free Energy value has attained the lowest (minimum) value.

dGT,P = 0

We can derive the criteria below,


dGT,P < 0

and comparing it to



9

For mixtures, we could write the equation in the form below,

Gi = ni gi

+ ni gi

+ ni gi

+ ……. at constant T, P

, and are the respective different phases and ni is the no. of moles of component i.

dGi = d(ni gi

) + d(ni gi

) + d(ni gi

) + ……. < 0

dGi = (ni dgi

+ gi dni

) + (ni dgi

+ gi dni

(ni dgi

+ gi dni

) + ..

Taking its differential,.

At equilibrium (const. T and P), dgi = 0. Therefore, the equation becomes..

dGi = gi dni

+ gi dni

gi dni

+ ……. < 0In a closed system and considering two phases ( and ), any mass loss in one phase will go to the other phase. Therefore,

dni = - dni

Substituting in the above equation

(gi - gi

) dni < 0

Since at equilibrium dGT,P = 0, therefore….

(gi - gi

) = 0 and gi = gi


at constant T, P


10

dGT,P = 0


Also from the same equation shown earlier,

other equilibrium criteria subject to different constraint could be derived.

The other equilibrium criteria consist of ;

dSU,V = 0dAT,V = 0dUS,V = 0dSH,P = 0dHS,P = 0

but the earlier relation involving Gibbs Free Energy is the most popularly used !



11

The equilibrium criteria established so far seems to be very general. Question is how can we apply such criteria in the calculation for phase equilibria ?

Need to introduce further concepts !

For a constant mass system at equilibrium, we can write the following total differentialeqn. for the Gibbs Free Energy (G)

dG = (G/T)P dT + (G/P)T dP = - S dT + V dP

Recall the total differential

dG = dU + P dV + V dP - T dS - S dT

(G/T)P = -S

(G/P)T = V

Now if we consider a system with variable mass, clearly there should be an additional term to account for this.

dG = (G/T)P dT + (G/P)T dP + (G/n)T,P dn

= - S dT + V dP + dnaccount for changes in G with addition/removal of material = (G/n)T,P where

We now introduce Chemical Potential () which is defined as the rate of changes in G with changes in material amount in the system.

Chemical Potential

Consider the derivation below for pure component system!


12

Following from the earlier relation for vapour-liquid equilibrium, we can then write ;

dG + dG = + = 0 =

at const T,P

for pure component system

Chemical Potential

Gi = ni gi

+ ni gi

+ ni gi

+ …….

dG = dG + dG + dG + …… = 0

at equilibrium

Thus, for 2 phase system,


13

Chemical potential could also be defined using other functions,

Chemical Potential

= (G/n)T,P Shown earlier

Using the internal energy, U function, it can be defined as;

= (U/n)S,V,n

The eqn. is derived from the total differential eqn. for U.

dU = T dS - P dV + dn

Now, from the definition H = U + PV and dH = dU + P dV + V dP

dH = T dS + V dP + dn

recall the eqn. by combining 1st & 2nd law

dU = TdS - PdV

We could write ;

dH = (H/S)P,n dS + (H/P)S,n dP + (H/n)P,S dn

and by comparing with the equation below;

can equate the two variables

Therefore we could developed another definition for Chemical Potential (H/n)P,S =

j

You can also derive similar relation involving A (Helmholtz Free Energy) at constant T,V, n

dU = (U/S)V,n dS + (U/V)S,n dV + (U/n)S,V dn

Let us attempt another one,


14

Equilibrium of Mixtures

Recall the earlier example on a system.

Air-Water system containing small amount of methanol.

Suppose the composition of methanol in water is 0.01

Will the composition of methanol in air is 0.01 too at equilibria?

The answer is no. Therefore, composition is not necessary the driving force…..

If we look carefully at the system, it contains more than one component I.e., exist in the form of mixture.

In separation processes which forms part of a whole chemical process, often mixtures of various components have to be separated either physically or chemically. For a physical based separation, often the understanding of phase equilibria for multicomponent system is important for the design and operation of the process. Distillation, absorption, adsorption etc. are some of the examples of the separation processes.

Earlier, the understanding of phase equilibria was developed around pure component. Using the fundamental developed, we will now explore the multi component system (mixtures).


15

Extending the same development to mixtures of components ;

1 dn1 + 2 dn2 + 1 dn1 + 2 dn2 = 0

dn1 = - dn1

dn2 = - dn2

1 = 1

2 = 2

In general terms,

Basis for multicomponent VLE

calculation

( i - i ) dni = 0

i = i

i

dG + dG = + = 0

For 2For 2

componentcomponent


From the earlier equation for phase equilibria for pure component system.

(gi - gi

) = 0 and gi = gi

at constant T, P

This is comparable to the equation developed earlier,,

at constant T, P


16


Chemical potential is an abstract concept. However, it is useful since it provides a simple criterion for equilibria measures. Unfortunately, in application, it turns out that chemical potential has some inconvenient mathematical behaviours. This has led to the introduction of fugacity.

= + RT ln f0

- chemical potential value at f = unity- function of only T

Fugacity (f) - a measure of volatility of a component in solution

The above relation is derived from ;

d = dg = vdP - sdTd = dg = vdP = RT dln P

T Tin the case of real gas, P is substituted by f

integrate !

f (fugacity) has the unit of pressure (atm, kPa, bar etc)

Extending the relation to mixtures ;

= + RT ln f0

i i i

For ideal gas, the fugacity could be simplified to be equal to the partial pressure exerted by the component in the mixture

f = y P = pi i i

at const. T

The relation between chemical potential and fugacity is described by the equation below;

Fugacity.


17


Writing the eqn. for 2 different phase ( and )

= + RT ln f0

i i i = + RT ln f

0i i i

from equilibrium criteria ;

i = i

it can easily be seen that the following criteria also holds at equilibrium ;

fi = fi

The fugacity of the individual component is equal for the 2 phases.Also basis for

multicomponent VLE calculation

Now if we are dealing with vapour liquid equilibria, clearly the two phases will refer to liquid and vapour !

fi = fivl

Fugacity is something easier to handle.Later we will be looking at how to calculate fugacity


18

Partial Molar Properties.

When a substance becomes part of a mixture, it losses its identity. However, it still contributes to the mixture properties. For mixtures that do not behave ideally, the mixture properties are not simply the molar-weighted sum of the pure component.

as what you have seen in MEB : Calculation of properties based on Pseudo-Component Approach

Introducing the Partial Molar Properties which has similar appearance in the equation as the chemical potential.

Writing it in a total differential form of M ;

dM = M /P) dP + M/T) dT + M/n ) dnT,n P,n ii

iP,T,n = ij

T,n P,n ii

isum of partial molar properties

dM = M /P) dP + M/T) dT + M dn

The total solution properties depend on the amount present of each species and its resultant interactions. Therefore, need to develop method to enable calculation of the properties in mixture form.

M = ( M/ni ) T,P,nj = 1

Property Changes for Mixing Process


19

In respect to the various thermodynamic properties, the partial molar properties could be written as:

dV = (V/T)P,ni + (V/P)T,ni

+ (V/ni)T,P,nj = i dnii = 1

k

dH = (H/T)P,ni + (H/P)T,ni

+ (H/ni)T,P,nj = i dnii = 1

k

For partial volume,

at constant T,P

dV = Vi dnii = 1

k

at constant T,P

dH = Hi dnii = 1

k

Integrate V = Vi ni

k

i = 1

H = Hi ni

k

i = 1Integrate

For partial enthalpy,

or v = vi xi

k

i = 1

(molar volume)

or h = hi xi

k

i = 1

(molar enthalpy)



20

The significance of Partial Molar Properties

UU

consider adding a small portionsof identical mixture to the larger one. What happen is ;

Utot = U + U

By knowing the partial molar properties, we can determine the actual changes in the properties as a result of the addition of material.

Utot = U + U

supposedly from pure molar property

but the actual one

The phenomena could be seen when mixing two species of different molecular sizeeg. benzene with water.



21

The Gibbs Duhem EquationThe Gibbs-Duhem equation provides a very useful relationship between the partial molar properties of different species in a mixture. It provides constraints between the partial molar properties of the different species i.e. if the partial molar property of one component is known in a 2-component mixture, the partial molar property of the other component could be calculated.

Integration of the above eqn. at const T and P ;

differentiating the above ;

equating withthe above eqn.

For the condition of constant T and P (under which equilibrium normaly established),

n dM = 0

i i

Gibbs Duhem Eqn.

i = 1

i = 1 i = 1

M = Mi nii = 1

k

M = Mi nii = 1

k

k

dM = Mi dni + ni dMi k k

we can show that ;

M/P) dP + M/T) dT = n dMT,n P,n i ii = 1

k

dM = M /P) dP + M/T) dT + M dn

T,n P,n i i

Consider again the general eqn. for partial molar properties ;

i = 1

k

at constant T and P



22

Alternatively, we start from the eqn. .

T,n P,n ii

idM = M /P) dP + M/T) dT + M dn

At constant T and P

dM = Mi dnii = 1

k

Shown earlier !

Based on definition of partial molar property,

M = Mi ni

We differentiate the equation,

i = 1

k

dM = Mi dni + ni dMi k

i = 1

k

i = 1

For both equation to be true, ni dMi must be equal to 0i = 1

k

n dM = 0

i iGibbs Duhem Eqn.

i = 1

k

at constant T and P



23

Now let us apply the derivation using Gibbs Free Energy as the basis.

G = i ni which on a molar basis is written as (divide by n),

G = Gi ni

Using Gibbs Free Energy

by differentiation,

ni di = 0This is the Gibbs-Duhem Eqn. using Chemical Potential as the Extensive Property.

G/n = i yi = Gi yi = g

we can prove that at const T and P,

i = 1

k

i = 1 i = 1

k k

By definition of partial molar property

i = 1

k

dG = Gi dni + ni dGi = i dni + ni dii = 1 i = 1 i = 1 i = 1

k k k k

through comparison to eqn :

- S dT + V dP - i dni = 0 at equilibrium dG = i = 1

k


i = 1

k


24

Experimental Determination of Partial Molar PropertiesExperimental Determination of Partial Molar Properties

We know now that all thermodynamic properties must be either measured directlyor calculated from experimentally determined data. We will now focussed on determining property changes on mixing and partial molar properties.

For the property changes, we can write a general eqn ;

M = M - nimi

property of the pure component

mixture property

property changes

M = Mi ni

Using the eqn.,

Lead to M = niMi - nimi = niMi

partial molar property change

partial molar property

G = H - TS = i ni

For Gibbs Free Energy changes on forming a liquid solution, we can write,

G = niGi - nigi = niGi

alternatively from eqn. we can write,

G = nii - nii = nii

Chem & Process Therm. by B.G. Kyle (pg 378-379)

i ii

i i i

Additional Note !

i = 1

k

i = 1

k

i = 1

k

i = 1

k

i = 1

k

i

To simplify writing the eqn., let us assume,

= i = 1

k

i



25

Consider the equation again ;

M = niMi - nimi = niMi

After conducting few experiments, we can have a relation of V against some combinations of compositions.

If a plot is to be generated for a binary mixture ;

V

x1 1

i i i

let say we do it for volume (V)

V = niVi - nivi = niVi i i i

0

Positive deviation !



26

From the eqn.,

we can write ;

divide by n1 + n2

V / (n1 + n2) = v = x1V1 + x2V2

differentiate v wrt x1 ;

In accordance with eqn. above,the last 2 terms can be written as

x1dV1/dx1 - x1dv1/dx1 + x2 dV2/dx1 - x2dv2/dx1

as v1 and v2 are property of pure components, their derivatives are 0. And from Gibbs Duhem eqn.,the terms x1dV1/dx1 + x2dV2/dx2 is also equal to 0.

Thus, the remaining of the eqn. becomes ;

dv/dx1 = V1 - V2

eliminating V2 from the eqn.,

V1 = v + (1-x1)dv/dx1 (continue next page)

i i i

ni dM i = 0

Gibbs Duhem eqn.

i

V = n1V1 + n2V2

V = niVi - niVi = niVi

dv/dx1 = V1 - V2 + x1 dV1/dx1 + x2 dV2/dx1

(terms appear in mol fraction)



27

V

x1

A

G

E

F D

C

B

K

V1 = v + (1-x1)dv/dx1The eqn.

can be viewed graphically using the figure below. Using point A as the reference ;

v = AG or BD, (1-x1) = GD or AB, and the slope dm/dx1 = CB/AB or KC

V1 (at point A) = BD + AB(CB/AB) = BD + CB = CD

and also we can write the eqn.;

V2 = V1 - dv/dx1 = CD - KC/1 = KD = EF

slope

Thus from experimental data, we could find the respective value for V1 and V2 which shows the differences between partial molar volume to pure species volume with the respective compositions.



28

To demonstrate the applicability of the equations, lets suppose we have experimental data representing volume changes with respect to composition for a binary system.

Experimental data of the molar volume for a binary system of components 1 and 2.

2000

1750

1500

1250

1000

0 10.2 0.4 0.6 0.8

v = x1 V1 + x2 V2 Using the equation :

v = (1-x2) V1 + x2 V2



29

Differentiating the eqn wrt x2 and applying the Gibbs Duhem Eqn.,

v = (1-x2) V1 + x2 V2

x2 (dv/dx2)= -x2 V1 + x2 V2

x2 (dv/dx2)= - V1 + ( x1V1 + x2 V2 )

= vv = V1 + x2 (dv/dx2)

2000

1750

1500

1250

1000

0 10.2 0.4 0.6 0.8

V2

V1

v1

v2

X2 mix = 0.7

vmix

At x2 = 0.7



30

Let’s try with another example involving molar enthalpy

Develop an expression for the partial molar enthalpy of sulphuric acid in water at 21 oC using equation;

HH2SO4 = h - xH20 (dh/dxh20)

Given the enthalpy expression

h = 1.596 xH2SO4 + 1.591xH20 - 74.40 xH2SO4

xH20 ( 1 - 0.561 xH2SO4 ) kJ/kmol

Differentiate the eqn.,

dh/dxH20 = 1.591 – 74.4 [ (xH2SO4 - xH20) (1 - 0.561 xH2SO4

) + 0.561 xH2SO4 xH20 ] kJ/kmol

Substitute the two eqn. above into the top equation,

HH2SO4 = 1.596 – 74.40 xH20

2 + 83.48 xH2SO4 xH20

2 kJ/kmol

Applying Gibbs Duhem Equation



31

Analytical Determination of Partial Molar PropertiesAnalytical Determination of Partial Molar Properties

Often an analytical expression for total solution property is known as a function of composition. An example would be the expression developed using virial equation for mixtures (will look at the expression in detail later!).

Additional Note !

v = (RT/P) [ 1 + (Bmix) / RT)] = (RT/P) + y12 B11 + 2y1y2 B12 + y2

2 B22

for 2 component

V = ( n1 + n2) v = ( n1 + n2) (RT/P) + [ ( n12 B11 + 2n1n2 B12 + n2

2 B22 ) / (n1 + n2) ]

Molar volume

System volume

Taking the differentiation,

V1 = (V/n1)T,P,n2 = RT + 2n1

B11 + 2n2 B22 - n1

2 B11 + 2n1n2 B12 + n22 B22

P (n1 + n2) (n1 + n2) 2

Simplifying the eqn,

Partial molar volume of species 1

V1 = RT + ( y12 + 2y1y2

) B11 + 2 y22 B12 - y2

2 B22

P



32

Additional Note !

Similarly, the expression for species 2 could be derived.

The molar volume for pure species 1 is obtained by setting y2 = 0

V2 = RT - y12 B11 + 2y1

2 B12 + ( y22 + 2 y1y2 ) B22

P

v1 = RT/P + B11

The volume change for mixing is given by,

vmix = v - ( y1v1 + y2v2 )

Substituting for v, v1 and v2 and simplifying the eqn. will yield the expression ;

vmix = 2 y1y2 [ B12 - ( B11 + B22 ) / 2 ]



33

A property changes of mixing describes the extent of property changes as a result of mixing. It is defined as the difference between the total solution property in the mixture and the sum of the pure species properties of its constituents, each in proportion to how much is present in the mixture.


Mmix = M - ni Mii = 1

k

Hmix = H - ni Hii = 1

k

Writing it for the respective properties,

Vmix = V - ni Vii = 1

k

Smix = S - ni Sii = 1

k

The properties of the pure component are determined at the set T and P.

For cases where partial molar properties could be used to determine the total mixture property eg. volume,

Vmix = niVi - ni Vii = 1

k

OVERALL VIEW


34


When involving partial molar properties, the equations can also be written in the form,


k

i = 1

k

Vmix = ni ( Vi - Vi )i = 1

k

Hmix = niHi - ni Hii = 1

k

i = 1

k

Hmix = ni ( Hi - Hi )i = 1

k

and

For volume

For enthalpy


35

At a molecular level, the property change of mixing reflects how the interactions between unlike species in the mixture compare to the like interactions of the pure species they replaced.


For the case of volume changes as a result of mixing between different species,


k

The volume change as a result of mixing reflects the differences in how closely species in a mixture can pack together in comparison to how they pack as pure species. A negative Vmix when the unlike interaction “pull” the species in the mixture closer together while a positive Vmix when the unlike interaction “push” the species apart when they are mixed. However, if the Vmix is zero, then the different species behave identically in the mixture as the pure species.

For the case of enthalpy changes as a result of mixing between different species,

Hmix = niHi - ni Hii = 1

k

The enthalpy change as a result of mixing reflects the energetic interactions between the different species. A negative Vmix reflects more stable presence of the species in the mixture as compared to at their pure state while a positive Vmix reflects less stable presence of the species in the mixture as compared to at their pure state . However, if the Vmix is zero, there is no changes in the stability of their presence whether in mixture or pure state.


36

The earlier shown example demonstrate the applicability of the equation below for volume changes as a result of mixing;


Vmix = niVi - ni Vi

k

2000

1750

1500

1250

1000

0 10.2 0.4 0.6 0.8V1

V2

v1

v2

X2 mix = 0.7

vmix

At x2 = 0.7

Experimental data of the molar volume for a binary system of components 1 and 2.

x2


37

Recall on the development of Maxwell eqn. for single component (pure) system .

dG = V dP - S dT

(G/T)P = -S(G/P)T = V

(G/PT)P = (V/T)P2 (G/TP)T = -(S/P)T

2

Equal

eg.

The same procedure could be applied to equations representing a mixture.

dG = - S dT + V dP + dn

From the earlier derivation involving pure system, we could show that the following applies to mixture system.

= (G/n)T,P where

(G/T)P,n = -S (G/P)T,n = V

( G/TP)n = - (S/P)T,n

2 ( G/PT)n = (V/T)P,n

2

Equal

Maxwell Equations for Mixtures.


38

Following the same procedure but using other properties (H, U and A) will lead to the following Maxwell Equations.

- (S/P)V,n = (V/T)S,n

(S/V)P,n = (P/T)S,n

(S/V)T,n = (P/T)V,n

Now if n is not being held constant, we could derive

dG = - S dT + V dP + dn = (G/n)T,P where

(G/T)P,n = -S (G/n)T,P =

( G/Tn)P = - (S/n)T,P

( G/nT)P = (/T)P,n

22

Equal

More Maxwell equations could be derived using other properties (H,U and A)

Maxwell Equations for Mixtures.


39

Phase Rule – A Review

Consider a system at phase equilibrium

Vapour & Liquid in equilibrium

Conditions at equilibrium ;

Tl = Tv

l = v

Pl = Pv

l = f(Tl, Pl, xi)

v = f(Tv, Pv, yi)2 phases

From phase rule ;

F = C - P + 2F - degree of freedomC - No. of componentP - No. of phases

For the above case, clearly F =1 for a single component

As we add-in additional component, we will have 3 additional eqns and 4 variables !

l = v

l = f(Tl, Pl, xi)

v = f(Tv, Pv, yi)

l , v

xi , yi

1 extra degrees of freedomi.e., 1 extra spec. on system variable

No of spec. on system variables to be made in orderto fully specified the system.


40

Ideal Mixtures

There are 2 classes of extensive properties

1. They are functions of entropy (S,A,G). Their partial molar properties is not equal to the pure molar properties even for ideal mixtures.

2 They are not function of entropy (V,U,H). Their partial molar properties is equal to the pure molar properties for ideal mixtures.

The free energy function i.e. Gibbs (or Helmholtz) free energies, belong to the first group.

G = ini = Gi ni ( = Gi ni )

partial molarGibbs Free Energy

pure molarGibbs Free Energy

Total Gibbs Free Energyof a mixture system

iii

M = xi mii = 1

k

mi – pure molar properties

To simplify writing the eqn., let us assume,

= i = 1

k

i


41

G = i ni + RT ni ln fi

recall the formula ;

The Gibbs free energy equation can be written in terms of fugacities using the eqn. below ;

For an ideal mixtures, we can write the equation (using partial pressures) ;

G = i ni + RT ni ln pi

For a number of ideal gas mixed at temperature T and initial partial pressure pio we can write ;

G = i ni + RT ni ln pio sum of the ideal gas components G prior to mixing

0

0

0

mixed all the ideal gas components together at const. T and P. The changes in the free energy resulted from mixing is ;

Gmix = RT ni ln ( pi / pio )

i.e. the difference between the two eqn. above.

i

i

i

i

i

i

i = + RT ln f0

i i i

Ideal Mixtures


42

Consider the mixing of pure gases at same initial pressure (p io) together to form mixture which will also be at the same pressure.

pio = P pi = yi P

Therefore

initial pressure = final pressure partial pressure (final)

Using the 2 above relation and yi = ni / n into the equation,

Gmix = RT ni ln ( pi / pio )

Gmix = RT ni ln yiwhich can be derived into the form

Note that the answer will always be negative. Why ?

Remember when a process took place spontaneously, the Gibbs Free Energy will always seek for a minimum value shown earlier.

Gases will always tend to mix and not unmix unless being forced to be separated. And this is the natural direction of the process which is towards the minimum value for the Gibbs Free Energy

i

Ideal Mixtures

i = 1

k


43

The entropy changes for the mixing proces could be derived from the definition of G = H - TS

Gmix = Hmix - T Smix

Smix = - Gmix / T

For ideal mixture, there are no changes in enthalpy due to mixing (Hmix = 0) as H is not a function of entropy.Therefore, their partial molar properties is equal to the pure molar properties for ideal mixture.

Hence,

Smix = - R ni ln yiWill always be

positive

The logical explanation is that for any spontaneous process, the entropy for the system will always increase. Principles of increasing entropy

This also help to explain that for entropy dependent partial molar properties (S, A, G), they have different values from the pure molar properties (S, A, G) even for ideal mixtures

On the contrary to the Gibbs Free Energy, the answer for entropy changes due to mixing will always be positive. Why ?

Ideal Mixtures

i = 1

k


44

Using similar method of derivation but applying it to ideal liquid solution mixtures will lead to the following results :-

Smix = - R ni ln xi

Gmix = RT ni ln xi

For Gibbs Free Energy of mixing ideal solutions

For entropy changes from the mixing process of ideal solution

Ideal Mixtures

i = 1

k

i = 1

k


45

Ideal Mixtures

Example : Find the ideal change in G and S when 3 mol of Argon (1), initially at 1 atm, 298 K are allowed to mix with 2 mol of Xenon (2) at the same initial condition.

The total moles are 3 + 2 = 5

Calculating for GmixGmix = RT ni ln xi

i = 1

k

= 8.314 . 298 [ 3 ln (3/5) + 2 ln (2/5) ]= - 8339 J

Calculating for Smix Smix = - R ni ln xii = 1

k

= - 8.314 [3 ln (3/5) + 2 ln (2/5) ]= + 28 J/K


46

Vapour Liquid Equilibria for Ideal Mixtures

Recall the concept of fugacity of each component in a mixture.

= + RT ln f0

i i i

d = dg = v dP – s dT

dT = dgT = v dP = RT dln P

The relation was derived from

= + RT ln f / f

Introduce fugacity and substitute it into the equation,

0 0

Taking the f 0 at atmospheric pressure as the reference f 0 = P = 1and writing the equation for components in mixtures ;

= + RT ln f0

i i i

Reviewing the derivation for fugacity ….


47


Now, the fugacity of each component in a simplest liquid mixture could be calculated by ;

fi = fi xi0

fi - fugacity of pure liquid at T & P of mixture0

At low pressure (under ideal gas condition), fi0 = Pi

0

where Pi0 is the partial pressure of the pure liquid at T & P of the mixture

and under the condition where the vapour phase is an ideal gas mixture, we can write

fi = P yi

Following the equilibrium criteria where

fi = fil v

we can derive that ;

yi P = Pi xi0

This is known as Raoult's Law

fi = fi

The fugacity of the individual component is equal for the 2 phases.

Recall the basis for equilibrium

Pi0 = saturation vapour pressure for component I at specified T,P


48

The Raoults law (equation) enable a vapour liquid equilibria relation be established for an ideal liquid/gas system mixture.

For ideal mixture, can be calculated from : Raoult's Law

yi P = xi Pi

o

only need information on,- Total Pressure- Vapour Pressure Pi

can determine x-y diagram

x

yWe can also go

directly from left to right ?

P

x,y

liquid

vapour

IsothermalVLE diagram

P1

o

P2

o

o



49

T

x,yFor ideal mixture, can be calculated from :

Raoult's Law

x

y

liquid

vapourIsobaricVLE diagram

T1

o

T2

o

yi = xi Pi / Po

- have to calculate Pi

for T value in between boiling point.

o

Use Antoinne

Eqn.

Can we go directly from left to right ?

can determine x-y diagram



50


Example : Find the vapour composition above a liquid mixture of 50 mol% benzene(1) and hexane(2) at 25 C if it is assumed that both the liquid and vapour form ideal solutions. The vapour pressure of benzene and hexane at 25 C are 93.9 and 149.2 mmHg respectively.

yi P = Pi0 xi

Using equation,

Calculate for y1 and y2.

y1 = 0.5 . 93.9 / P y2 = 0.5 . 149.2 / P

P = y1 P + y2 P = x1 P10 + x2 P2

0 = 121.5 mmHg

The composition in vapour could then be calculated

y1 = 0.386 y2 = 0.614


51


Example : Develop a T-x,y diagram for a system containing n-pentane(1) and n-heptane(2) at pressure 101.3 kPa. Given the Antoinne eqn. for saturation vapour pressure for the 2 components as below;

Component 1 : ln P10 = 13.818 – 2477.07 / (T + 233.21)

Component 2 : ln P20 = 13.8587 – 2911.32 / (T + 216.64)

First, find the two temperature limit I.e., the boiling points at specified pressure.

Component 1 : T = 36.04 C

Component 2 : T = 98.42 C

T

x-y

36.04

98.42


52


To solve for the composition at the intermittent temperature between the boiling temperatures, use equation;

y1 = x1 P10 / P

x2 = 1 – x1

P = y1 P + y2 P = x1 P10 + x2 P2

0

Solve for x1

Then

T

x-y

36.04

98.42

T x1 y1

40 0.86 0.98370 0.250 0.70080 0.142 0.517


53

Solubility in an Ideal Solution

Gas solubility in a liquid solvent is determined by equating vapour and liquid fugacities of the species.

fi = pi = yi P = xi Pi0

this is only for ideal mixture !

For solid solubility, the equation used for determining solid solubility in ideal solution is :

( ln xi / T)P = hm/RT2

enthalpy of melting

Note : The equation is independent of solvent properties (In actual fact, it is not true for actual solution)

If we integrate with assumption that the enthalpy remain constant, the result is

ln xT = (hm/R) (dT/T ) = (hm/R) [ 1/Tm - 1/T ] T

Tm

2

Tm - melting temperature of the soluteT - temperature of solvent

The principle of equilibrium

Now, we got ourself an equation to determine solid solubility in ideal solution.


54

Freezing point depressionFreezing point depression

The extend of lowering in the melting temperature. Example : Addition of salt to water

The effect is to lower the activity of the solvent in the liquid phase primarily by loweringits mole fraction from unity to some lower value.

The equation used to estimates the lowering in melting temperature

T / ln x2 = T / ln (1-x1) = RT / hm2

solute composition in solvent solvent composition after mixing with solute

Integrating the above equation

ln (1-x1) = hm/T) . [ 1/Tm - 1/T ]

This is almost equal to -x1

as its value is very small

rearrange to give : T = x1 RTm / hm 2

refer to solvent

refer to solvent

Thus, we can predict the loweringof freezing point for a solution dueto addition of solute.



55

Boiling Point ElevationBoiling Point Elevation

A similar form of analysis can be done to determine the extent of boiling point elevation

eg : salt dissolved in water will result in higher boiling temp.

A similar form of equation could be derived

ln (1-x1) = hvap/T) . [ 1/T - 1/Tvap ]

solute composition in solvent x1 << 1

T = x1 RTvap / hvap

After rearranging,

2

refer to solvent

This is almost equal to -x1

as its value is very small

Thus, we can predict the increasein boiling point for a solution dueto addition of solute.



56

Conclusions and Review

The importance of phase equilibria in Chemical Eng. Separation Processes

Fundamental of Phase Equilibria

Fundamental Relation using Gibbs Free Energy


dGT,P = 0

Deriving fundamental criteria at equilibrium

extended to mixture…

Introduce chemical potential to provide more sense…

ii And relate them through the equation involving fugacity

fRTii ln0 Thus allowing for the criteria to be expressed in fugacity

ii ff v

ili ff

Easier to perform calculation !


57


Property variation in mixture as composition changes

Chemical engineers have to deal mainly with mixtures

Introduce partial molar property

1,,

jnPTin

MM To account for property changes as a result of

composition change

Equation used for computing the changes (for volume and enthalpy) ;

i

k

iimix VnVV

1

i

k

iimix HnHH

1

i

k

iimix vxvv

1

i

k

iimix hxhh

1

Molar property (per mol basis)

Could be determine from pure molar property using the eqn. derived from Gibbs Duhem relation.

221dx

dMxMM for binary mixture

Has also established for Gmix and smix.


58


Demonstration on the application of the phase equilibria

Consider ideal mixture, where Raoult’s law is used to simplify the phase equilibria complexity

0iii PxPy

Fixing the system pressure and knowing the saturation vapour pressure at respective temperature, would allow for VLE composition to be computed with respect to the temperature range between the two boiling points.

T

x,y

yx

chemthermo-chap3

Documents

equilibrium criteria

use of phase equilibrium

t ds p dv

p dv t ds

thermodynamic equilibrium

phase te

imbalance of t

constant t