9. lle calculations

CHEE 311 J.S. Parent 1

9. LLE Calculations

For two liquid phases at equilibriumthe fugacity of each component inthe phases must be equal.

For the binary case shown:

are the two relationships thatgovern the partitioning of species1 and 2 between the two phases.

1111 xx

2121 )x1()x1(


Binary LLE Separations

The equivalent of a VLE flash calculation can be carried out on liquid-liquid systems.

Given: T, P and the overall composition of the systemF, z1, z2

Find: L, x1, x2

L, x1, x2

Feed @ T, P L, x1, x2

z1, z2 L, x1, x2


Binary LLE Separations - Governing Eqn

Solving these problems requires a series of material balances:Using a unit feed as our basis, an overall material balance yields:

(A)

A material balance on component 1 give us:(B)

Substituting for L from A into equation B:(C)

An analogous material balance on component 2, yields:(D)

We have two equations (C,D) and three unknowns (L, x1 and x1

). We need an equilibrium relationship between xi

and xi

LL1F

LxLx)1(z 111

)L1(xLxz 111

)L1(xLxz 222


Binary LLE Separations - Governing Eqn

Our LLE expression is:(14.10)

orand (E)

The governing equation we require to solve the problem is generated from a final material balance on one of the liquid phases:

(F)

Substituting equations C, D, E into the material balance F gives us the final equation:

iiii xx

1

111

xx

2

222

xx

1xx 21

1

1L1

z

1L1

z

2

2

2

1

1

1


Solving Binary LLE Separation Problems

Given: T, P,F, z1, z2 Find: L, x1, x2

L, x1, x2

The solution procedure follows that of binary VLE flash calculations very closely.

You can immediately solve for x1 and x1

using the LLE relationships

Or You can solve the governing equation by iteration, starting

with estimates of x1 and x1

to determine activity coefficients, and refining these estimates and L by successive substitution.

1

1L1

z

1L1

z

2

2

2

1

1

1


Vapour-Liquid-Liquid Equilibrium (VLLE)

In some cases we observeVLLE, in which threephases exist at equilibrium.

F = 2 - + C = 2 - 3 + 2 = 1

Therefore, at a given P,all intensive variablesare fixed, and we havea single point on a binaryTx,x,y diagram


Vapour-Liquid-Liquid Equilibrium (VLLE)

At a given T, we cancreate Px,x,y diagramsif we have a goodactivity coefficientmodel.

Note the weakdependence of theliquid phasecompositions on thesystem pressure.


10. Chemical Reaction Equilibrium SVNA 15

If sufficient data exists, we can describe the equilibrium state of a reacting system.

If the system is able to lower its Gibbs energy through a change in its composition, this reaction is favourable.

However, this does not imply that the reaction will occur in a finite period of time. This is a question of reaction kinetics.

There are several industrially important reactions that are both rapid and “equilibrium limited”.

Synthesis gas reaction

production of methyl-t-butyl ether (MBTE)

In these processes, it is necessary to know the thermodynamic limit of the reaction extent under given conditions.

3332323 )CH(OCCH)CH(CCHOHCH

224 H3COOHCH


Reaction Extent

Given a feed composition for a reactive system, we are most interested in the degree of conversion of reactants into products.

A concise measure is the reaction extent, .

Consider the following reaction:

In terms of stoichiometric coefficients:

where, CH4 = -1, H20 = -1, CO = 1, H2 = 3

For any change in composition due to this reaction,

15.2

where d is called the differential extent of reaction.

224 H3COOHCH

2HCO2OH4CH HCOOHCH224

ddndndndn

2

2

2

2

4

4

H

H

CO

CO

OH

OH

CH

CH


Reaction Extent

Another form of the reaction extent is:(i=1,2,…,N) 15.3

The second part of our definition of reaction extent is that it equals zero prior to the reaction.

Given that we are interested in the reaction extent, and not its differential, we integrate 15.3 from the initial, unreacted state to any reacted state of interest:

or

15.4

ioi nnat0

ddn ii

oi

n

ni ddn

i

io

iioi nn


Reaction Extent and Mole Fractions

Translating the reaction extent into mole fractions is accomplished by calculating the total number of moles in the system at the given state.

Where,

Mole fractions for all species are derived from:

15.5

o

iioi

n

)n(nn

iiooi nnnn

o

iioii n

nnn

y


Multiple Reactions and the Reaction Extent

The reaction extent approach can be generalized to accommodate two or more independent, simultaneous reactions.

For j reactions of N components:(i=1,2,…,N)

and the number of moles of each component for given reaction extents is:

15.6

and the total number of moles in the system becomes:

where we can write:

j

jj,ii ddn

j

jj,iioi nn

j i

jj,io )(nn

j,ijiooi nnnn

j

jjonn


Chemical Reaction Equilibrium Criteria

To determine the state of a reactive system at equilibrium,we need to relate the reactionextent to the total Gibbs energy, GT.

We have seen that GT of aclosed system at T,Preaches a minimum atan equilibrium state:

Eq. 14.4

0dG P,TT


Reaction Extent and Gibbs Energy

For the time being, consider a single phase system in which chemical reactions are possible.

The changes in Gibbs energy resulting from shifts in temperature, pressure and composition are described by the fundamental equation:

At constant temperature and pressure, this reduces to:

and the only means the system has to lower the Gibbsenergy is to alter the number of moles of individualcomponents.

What remains is to translate changes in moles to the reaction extent.

i

iidndT)nS(dP)nV()nG(d

i

iiP,T dn)nG(d


Criterion for Chemical Equilibrium

For a single chemical reaction, we can apply equation 15.3 which relates the reaction extent to the changes in the number of moles:

15.3

Substituting for dni in the fundamental equation yields:

At equilibrium, we know that d(nG)T,P, = 0. Therefore, for the above equation to hold at any reaction extent, we require that

15.8

ddn ii

i

iiP,T d)nG(d

0i

ii


Reaction Equilibrium and Chemical Potential

We have developed a criterion for chemical equilibrium in terms the chemical potentials of components.

15.8

While this criterion is complete, it is not in a useable form. Recall our definition of fugacity which applies to any species

in any phase (vapour, liquid, solid)

In dealing with reaction equilibria, we need to pay particular attention to the reference state, i(T). We can assign a standard state, Gi

o, as:

0)nG(

P,Tiii

iii flnRT)T(

oii

oi flnRT)T(G


Standard States 4.4 SVNA

For our purposes, the Gibbs energy at standard conditions is of greatest interest.

This is the molar Gibbs energy of: pure component i at the reaction temperature in a user-defined phase at a user-defined pressure (often 1 bar)

A great deal of thermodynamic data are published as the standard properties of formation at STP (Table C.4 of the text)

Gfo is standard Gibbs energy of formation per mole of the

compound when formed from its elements in its standard state at 25oC.

»Gases: pure, ideal gas at 1 bar»Liquids: pure substance at 1 bar

oii

oi flnRT)T(G


Chemical Potential and Activity

Substituting our standard Gibbs energy (Gio) in the place of i(T),

the chemical potential of component i in our system becomes:

15.9

We define a new parameter, activity, to simplify this expression:15.11

where,

The activity of a component is the ratio of its mixture fugacity to its pure component fugacity at the standard state.

oi

ioii

f

flnRTG

ioii alnRTG

oiii ffa


Reaction Equilibrium and Activity

When a reactive system reaches an equilibrium state, we know that the equilibrium criterion is satisfied. Recall that chemical reaction equilibrium requires:

where i is the stoichiometric coefficient of component i and i is the chemical potential of component i at the given P,T, and composition.

Substituting our expression for chemical equilibrium into the above equation gives us :

Or,

i

ioiii

ii 0alnRTG

i

oii i

ii RTG

aln

0ii

i


The Equilibrium Constant

Our equilibrium expression for reactive systems can be expressed concisely in the form:

15.12

where signifies the product over all species.

The right hand side of equation 15.12 is a function of pure component properties alone, and is therefore constant at a given temperature.

The equilibrium constant, K, for the reaction is defined as:

15.13

K is calculated from the standard Gibbs energies of the pure components and the stoichiometric coefficients of the reaction.

RTG

alnoii i

ii

i

ii

oii i ia

RTG

expK


Standard Gibbs Energy Change of Reaction

The conventional means of representing the equilibrium constant uses Go, the standard Gibbs energy change the reaction.

Using this notation, our equilibrium constant assumes the familiar form:

15.14

When calculating an equilibrium constant (or interpreting a literature value), pay attention to standard state conditions.

Each Gio must represent the pure component at the

temperature of interest

The state of the component and the pressure are arbitrary, but they must correspond with fi

o used to calculate the activity of the component in the mixture.

RTG

expKo

ioii

o GG


Temp. Dependence of Reaction Equilibrium

Defined by the following relationship,

the equilibrium constant is a function of temperature. Recall that Go represents the standard Gibbs energy of

reaction at the specified temperature.

We know that: 15.15

From which we can derive the temperature dependence of K:

15.16

If we assume that Ho is independent of temperature, we can integrate 15.16 directly to yield:

15.17

KlnRTGo

dT)RT/G(d

RTHo

2o

2

o

RT

HdT

Klnd

1

o

1 T1

T1

RH

KK

ln


K vs Temperature

Equation 15.17 predicts that ln Kversus 1/T is linear. This is based onthe assumption that Ho is a weakfunction of temperature over therange of interest.

This is true for a number ofreactions, including thosedepicted by Figure15.2.

A rigorous development of temperature dependenceof K may be found in the text(Equation 15.20)


Equilibrium State of a Reactive System

Given that an equilibrium constant for a reaction can be derived from the standard state Gibbs energies of the pure components, we can define the composition of the system at equilibrium.

15.13

Consider the gas phase reaction:

The equilibrium constant gives us:

Or

ii

oii i iaK

RTG

exp

224 H3COOHCH

OHCH

3HCO

24

2

aa

aaK

)f/f)(f/f(

)f/f)(f/f(K

oOHOH

oCHCH

3oHH

oCOCO

2244

22

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