Thermodynamics

I. Thermodynamics of mixtures

I. 1. IntroductionThe chemical industry typically deals with mixtures in the process of making pure components or even mixtures as a product. Dealing with pure substances is the exception, and in some cases streams can be approximated by assuming that they are pure substances. It is then important to know when this assumption will fail. It is often the task of a chemical engineer to ensure the separation of a relatively pure target product from a mixture.

The task of a chemical engineer is to design, optimize, and operate chemical processes. The starting point for these activities are the mass and energy balance around a system. On a plant, we typically measure volumetric flow rates (e.g. via an orifice plate, Venturi meter, etc.), although more mass flow meters are coming into practice (even taken the ‘old’ bucket and stopwatch method).

SYSTEM

Vin,1

Tin,1,pin,1,xin,1

State of aggregation

Vin,2

Tin,2,pin,2,xin,2


Vout

Tout,pout,xout


.

..

Hence, to understand and being able to model an existing process, the volumetric flow rate must be converted into a mass flow rate. This can be done using the density of the particular mixture (mixture):

1ρmixture

=w1ρ1

+w2ρ2

+w3ρ3

+⋯

with wi: the mass fraction of component i in the mixturei: the density of the pure component i

The underlying assumption here is, that the volume of the various components added to the mixture is additive. This assumption needs to be investigated.

Knowing the temperature and pressure (easily measured quantities) and the composition (not that easily determined), the system is completely described. This means that all other thermodynamic quantities are now fixed. For the energy balance around a system, it is important to know what the enthalpy of the streams are (in relation to a pre-defined reference state), so that the heat required to be added or removed can be predicted. We need therefore a tool to calculate the enthalpy of the streams relative to the reference state knowing the temperature, pressure and composition of a stream.

For a chemical engineer it is not only important to know how much of a product can be produced (as given by the mass balance, thermodynamic constraints, and the size of the equipment), but also whether the production is as efficient as possible. Thermodynamics can set targets with respect to the amount of product formed (chemical equilibrium considerations), but also with respect to the efficiency of the process. The thermodynamic efficiency of a process can be defined as

ηthermodynamic=∆Gsystem

∆H system

1

(for inlet and outlet streams at 298 K). Knowing the maximum theoretical efficiency of a process sets a target for the optimization of the process and can be used to analyze the part of the process with the largest thermodynamic losses.

It is thus important to understand and possibly predict the behavior of mixtures in order to be able to determine the volumetric, enthalpic and entropic properties. Mixtures differ from pure substances. Mixtures differ from pure components upon phase change. Pure substances will melt/evaporate at a particular temperature (for a given pressure) or at a particular pressure (for a given temperature). This implies that there is only one single state of aggregation possible for a pure substance (i.e. either gas, liquid or solid). In general, mixtures of a particular composition do not melt/evaporate at a single temperature (for a given pressure) or at a single pressure (for a given temperature). They typically show a range where two or more phases can be present (depending on the number of components present in the system).

Mixing of pure components does yield some interesting properties, such as change in volume and change in temperature upon adiabatic mixing of streams.

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Mixing equal amount of water and ethanol Mixing 50 ml of water with 50 ml of ethanol yields a mixture of containing 50 vol.-% ethanol. This process is associated with a volume contraction (see Fig. 1.1). What is the volume of this mixture?Data: H2O = 0.99708 g/cm3; MH2O = 18.02 g/mol; ethanol = 0.78506 g/cm3; Methanol = 46.07 g/mol

-1.2

-0.9

-0.6

-0.3

0

0 0.2 0.4 0.6 0.8 1

Ch

an

ge

in

vo

lum

e u

po

n

mix

ing

, c

m3/m

ol

of

mix

ture

Mole fraction of ethanol in mixture, xEtOH

DV (xEtOH = 0.2364) =-0.9552 cm3/mol

Figure I.1.1: Change in the volume of the mixture per mole of mixture as a function of the mole fraction of ethanol in the mixture (redrawn from J.-P.E. Grolier and E. Wilhelm, Fluid Phase Equilibria 6 (1981), 283-287)

50 ml of ethanol contains 39.465 g ethanol 0.8567 mol ethanol

50 ml of water contains 49.850 g water 2.7664 mol water

A mixture of 50 ml ethanol and 50 ml of water contains the same number of moles of water and ethanol as for the pure substances (mole balance!). Hence, the mole fraction of ethanol in the mixture is given by:

xethanol∈mixture=0.8567

0.8567+2.7664=0.2364

Hence, the change in the volume upon mixing is -0.9552 cm3/mol of mixture (see Fig. 1.1). This mixture contains 3.6230 moles (0.8567 mol of ethanol and 2.7664 mol of water). Thus, the volume reduction is now

2

∆V mixture=3.6230 ∙mol ∙(−0.9552 ∙ cm3

mol )=−3.461 ∙ cm3

And the total volume of the mixture isV mixture=V ethanol+V water+∆V mixture=50+50−3.461=96.53 ∙ cm

3

(AND NOT 100 ml!).The reduction in the volume is due to the molecular interaction between ethanol and water. The resulting interaction (due to H-bonding) will result in a closer packing of water and ethanol molecules resulting in a volume reduction in comparison to the pure liquids.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••The molar volume of the mixture can thus be described in terms of the molar volume of the pure components and the change in the volume upon mixing to form 1 mole of mixture:

V mixture=∑i=1

C

x i ∙V i+∆mixV

The change in the volume upon mixing to form 1 mole of mixture is dependent on temperature, pressure and the composition of the mixture: ∆mixV=∆mixV (T , p , x ) (V mixture=V mixture (T , p , x ); (V i=V i (T , p ))(the dependency on pressure is typically neglected when dealing with liquids, since liquids can be considered to be incompressible).

Furthermore, the mixing process may involve heat effects. The mixing process can be exothermic, upon adding two pure components together, i.e. the mixture has a higher temperature upon adiabatic mixing than the original components or otherwise stated heat has to be removed to keep the temperature constant (e.g. in the case of mixing sulfuric acid with water or to a lesser extent mixing water and ethanol), or endothermic, i.e. the mixture has a lower temperature upon adiabatic mixing than the original components or otherwise stated heat has to be added to keep the temperature constant (e.g. in the case of mixing ethanol with benzene). It may even have a more complex dependency as given in the example below.

The molar enthalpy of a mixture can be defined in a similar manner as the molar volume of a mixture, i.e. as a function of the molar enthalpy of the pure components and the change in the enthalpy upon mixing to form 1 mole of a mixture:

Hmixture=∑i=1

C

x i ∙ H i+∆mix H

The change in the volume upon mixing to form 1 mole of mixture is dependent on temperature, pressure and the composition of the mixture: ∆mix H=∆mix H (T , p , x ) (Hmixture=Hmixture (T , p , x ); (H i=H i (T , p ))(the dependency on pressure is typically neglected when dealing with liquids, since molar volume of liquids is rather small).

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Isothermal mixing of water and acetone For a process it is necessary to separate isothermally at 25oC a liquid stream (2 m3/s) containing 20 mol-% acetone in water into its pure components. How much heat has to be added/removed for the separation process? What is the volumetric flow rate of each of the pure component streams?Data: H2O = 0.9970 g/cm3; MH2O = 18.02 g/mol; acetone = 0.79042 g/cm3; Macetone = 58.08 g/mol

3

-1.6

-1.2

-0.8

-0.4

0

0 0.2 0.4 0.6 0.8 1

Ch

an

ge

in v

olu

me

up

on

m

ixin

g,

cm3/m

ol o

f m

ixtu

re

Mole fraction of acetone in mixture, xacetone

DV (xacetone= 0.2) = -1.1929 cm3/mol

-800

-400

0

400

0 0.2 0.4 0.6 0.8 1

Ch

an

ge

in e

nth

aply

up

on

m

ixin

g,

J/m

ol

of

mix

ture

Mole fraction of acetone in mixture, xacetone

DH (xacetone= 0.2) = -651.768 J/mol

Figure I.1.2: Change in the volume (left) and change in the enthalpy (right) for mixtures of acetone and water at 298.15 K (data: change in volume – H.K. Bae and H.-C. Song, Korean Journal of Chemical Engineering 15(6) (1998), 615-618; change in enthalpy – B. Löwen and S. Schulz, Thermochimica Acta 262 (1995). 69-82)

Sepa

ratio

n pr

oces

sMixture (2 m3/s)xacetone = 0.2xwater = 0.8T = 298.15 K

Acetone (? m3/s)xacetone = 1.0xwater = 0.0T = 298.15 K

water (? m3/s)xacetone = 0.0xwater = 1.0T = 298.15 K

A mole balance on the process:Acetone xacetone ∙nmixture=nacetone leaving

Water xwater ∙ nmixture= nwater leaving

The main problem is now to find the molar flow rate of the mixture entering the process. Taking a basis of 1 mol of a mixture containing 0.8 mol of water and 0.2 mol of acetone, the mass of water and acetone can be determined: mwater=xwater ∙Mwater=0.8 ∙18.02=14.416 ∙ g macetone= xacetone ∙ M acetone=0.2 ∙58.08=11.616 ∙ g

The change in the volume for a mixture containing 20 mol-% acetone is -1.1929 cm3/mol.The volume of 1 mole of the mixture is given by:

V mixture=V water+V acetone+∆V mixture=mwater

ρwater

+macetone

ρacetone

+∆V mixture

V mixture=14.4160.9970

+ 11.6160.76042

−1.1929=28.452∙ cm3

4

Thus the molar volume of the mixture is:

V mixture=28.452∙cm3

mol

Hence, a flow of the mixture of 2m3/s corresponds to a flow of 70.07 kmol/s (i.e. 56.06 kmol/s of water and 14.01 kmol/s of acetone). The flow rates of the pure streams leaving the process are thus given by

V water leaving=mwater leaving

ρwater

=nwater leaving ∙ Mwater

ρwater

=nwater entering∙ Mwater

ρwater

=56.06 ∙

kmols

∙18.02∙kg

kmol∙103 ∙

gkg

0.9970 ∙g

cm3 ∙106 ∙

cm3

m3

V water leaving=1.013 ∙m3

sSimilarly for acetone

V acetone leaving=nacetone entering ∙ M acetone

ρacetone

=14.01 ∙

kmols

∙58.08∙kg

kmol∙103∙

gkg

0.79042∙g

cm3 ∙106 ∙cm3

m3

=1.070∙m3

s

The heat added/removed can now be determined using an energy balance. For this, the definition of the change in the enthalpy as given in Fig. 1.2 needs to be defined

∆ H (T )=Hmixture (T )−xwater ∙H water (T )−xacetone ∙ H acetone (T )The enthalpy is a function of temperature. It is important to realize that all enthalpies must be taken all at the same temperature (298.15 K). The energy balance on the process:

Hmixture+Q=Hwater leaving+ H acetone leaving

with Hmixture=nmixture ∙ Hmixture

Hmixture=nmixture ∙(∆ H (T inlet )+ xwater ∙ Hwater (T inlet )+xacetone ∙H acetone (T inlet )) Hmixture=nmixture ∙∆ H (T inlet )+nwater entering ∙H water (T inlet )+ nacetone entering ∙H acetone (T inlet )

and Hwater leaving= nwater leaving ∙ Hwater (T outlet ) H acetone leaving=nacetone leaving ∙H acetone (T outlet )

Substituting the mole balance and realizing that Tinlet=Toulet: Hwater leaving= nwater entering ∙ Hwater (T inlet ) H acetone leaving=nacetone entering∙ H acetone (T inlet )

Substituting into the energy balancenmixture∙∆ H (T inlet )+ nwater entering∙ Hwater (T inlet )+ nacetone entering ∙ H acetone (T inlet )+Q=nwater entering∙ Hwater (T inlet )+nacetone entering∙ H acetone (T inlet )

nmixture∙∆ H (T inlet )+Q=0

Q=− nmixture ∙∆ H (T inlet )=−70.07 ∙ kmols

∙(−651.768 ∙ Jmol )=45.67 ∙MW

5

Thus, the heat to be added to the system amounts to 45.67 MW.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

The change in the volumetric properties and enthalpy upon mixing can be understood in terms of molecular interactions between the various molecules present in the mixture. The molecular interactions are associated with energetic effects (internal energy, enthalpy). Furthermore, mixing results in an increase in the entropy of the mixture.

For pure substances, the interaction between the molecules was taken into account using an appropriate equation of state (and it was shown that cubic equations of state were useful to predict properties around the vapour-liquid equilibrium). The thermodynamic functions were then developed around the idea that the property can be described as a deviation of the thermodynamic function from its value, when the compound at a particular temperature and pressure could be considered to be an ideal gas.Enthalpy

Temperature change H IG (T2 , p=0 )−H IG (T1 , p=0 )=∫T1

T2

c p ∙ dT

pressure change H (T 1 , p )−H IG (T 1 , p=0 )= ∫p=0

p=p

(V −T ∙( ∂V∂T )p)∙ dp

Entropy

Temperature change S IG (T 2 , p=0 )−S IG (T 1 , p=0 )=∫T1

T2 c p

T∙dT

pressure change S (T 1 , p )−S IG (T 1 , p=p )=− ∫p=0

p=p

(( ∂V∂T )p

−Rp ) ∙ dp

Similarly, the departure of the Gibbs free energy from ideality was formulated in terms of the fugacity:

f=p ∙ eG (T1 , p )−GIG (T1 , p=0)

RT =p ∙ e

∫T 1

T 2

(V −RTp )∙dp

RT

Hence, the appropriate equation of state relating the molar volume to temperature and pressure yields the information on the thermodynamic functions at a given temperature and pressure for a pure substance. The specific molecular interactions between the molecules were introduced through the critical properties of the pure components.

6

I.2. Ideal gas mixtures – non-interacting mixturesMixtures differ from pure components due to the interaction of the different components in the mixture with each other. An ideal gas is a state, in which molecules are point-like objects, which do not interact with each other, and an ideal gas mixture is thus a mixture where the molecules do not interact with each other. The difference between an ideal gas and an ideal gas mixture lies in the fact that the ideal gas contains indistinguishable molecules, whereas the ideal gas mixture contains molecules which are distinguishable.

In an ideal gas mixture (IGM), the pressure is given by the ideal gas law:

p=N mixture∙RT

V IGM

The partial pressure of each component in the mixture is defined in terms of the total pressure of the system, and the mole fraction of each component:

pi=x i ∙ p=N i

Nmixture

∙ p

or otherwise stated the partial pressure of a component i in an ideal gas mixture (piIGM)

piIGM=x i ∙ p=N i ∙

RT

V IGM

Thus, the partial pressure of a component in an ideal gas mixture is the pressure exerted by the same number of molecules as present in the mixture, at the same temperature of the mixture (and thus average velocity), and in the same volume as the volume of the mixture.

It should be noted that the volume of the system does not change upon mixing to form an ideal gas mixture. In an ideal gas mixture molecules do not interact. The volume change, which is often observed upon mixing, is due to the interactions between molecules. Hence, in an ideal gas mixture there will be no volume change upon mixing.

In an ideal gas mixture, molecules do not interact. Hence, the internal energy associated with each component is not changed due to the presence of other molecules. Hence, interactions, such as van der Waals interactions (dipole-dipole, dipole-induced dipole, induced dipole-induced dipole interactions), can be neglected. The only contributing factor to the internal energy is the energy associated with a single molecule. Hence, the internal energy of an ideal gas mixture can be expressed in terms of the internal energy of the individual components present in the mixture:

U IGM (T ,N )=∑i

N i ∙U iIG (T )

(Note: d U=cv ∙ dT+(T ∙( pT )V

−p)∙ d V d U IG=cv ∙ dT )

This immediately leads to the conclusion that the change in the internal energy upon mixing ideal gases to form an ideal gas mixture (mixUIGM) equals zero:

∆mixUIGM=U IGM−∑

i

N i ∙U iIG (T )=0

The enthalpy is defined as: H ≡U+ p ∙V

Hence, the change in enthalpy upon mixing two ideal gases forming an ideal gas mixture:∆mix H

IGM=∆mixUIGM+∆mix (p ∙V ) IGM=0

7

8

The entropy will change due to mixing. A mixing process at constant temperature and pressure (p total) is equivalent to reducing the pressure of each component from the total pressure p total to its partial pressure, pi, in the mixture. The change in entropy upon isothermal reduction of the pressure for an ideal gas is given by

( ∂S∂ p )T

=−( ∂V∂T )p

∙ dp=−Rp

∙dp

S IG (T , pi )−S IG (T , psystem)=−R ∙ ln( pi

psystem)

Thus, the entropy of a component in an ideal gas mixture can be deduced from the entropy of the pure component as an ideal gas at the same temperature and pressure of the system taking into account the reduction of the pressure from psystem to its partial pressure pi in the ideal gas mixture.

SiIGM (T , psystem )=Si

IG (T , psystem )−R ∙ ln ( pi

psystem)

Hence the change in entropy upon mixing ideal gases to yield an ideal gas mixture is given by

∆mixSIGM=S IGM−∑

i

N i ∙ SiIG (T , psystem )

∆mixSIGM=∑

i

N i ∙ SiIG (T , psystem )−∑

i

N i ∙ R ∙ ln( pi

psystem)−∑

i

N i ∙ S iIG (T , psystem )

∆mixSIGM=−∑

i

N i ∙R ∙ ln( p i

psystem)

or in terms of the mole fractions in the ideal gas mixture

∆mixSIGM=−R ∙∑

i

N i ∙ ln (x i )

The mole fraction of each component is less than 1, and hence ln(x i) is less than zero. This means that the entropy of the system increases upon mixing.

The Gibbs free energy is defined as:G≡H−T ∙S

Hence, the change in the Gibbs free energy upon mixing ideal gases to form an ideal gas mixture is given by:

∆mixGIGM=∆mix H

IGM−T ∙ ∆mixSIGM

∆mixGIGM=RT ∙∑

i

N i ∙ ln (xi )Hence, the Gibbs free energy of the ideal gas mixture will be lower than the Gibbs free energy of its contributing pure components.

The consideration of an ideal gas mixture leads to the conclusion that the change in the internal energy and the enthalpy equals zero since molecules do not interact, i.e. no energy associated with the interaction. However, the change in the entropy and the Gibbs free energy is not equal to zero despite the absence of any interaction between the molecules. This originates from an effective reduction of the pressure of each component from the system pressure to its partial pressure in the mixture (similarly, it can be shown that the Helmholtz free energy changes upon mixing, since it is defined as the difference between the internal energy and T.S).

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Separation of an ideal gas mixture Air at 298.15K, which at low pressure can be considered to be an ideal gas mixture, is to be separated into its components oxygen and nitrogen in a continuous, reversible, isothermal and isobaric process.

9

Determine the amount of work required per mole of air fed to the process, and the amount of heat released/required.

Mole balance nair,∈¿=nO2,out +nN

2,out ¿

10

Energy balance H air ,∈¿−HO2, out− HN2 ,out

+Q+W=0¿

nair,∈¿ ∙ H air,∈¿ ( T¿ ,p¿ )−nO2,out ∙H O

2,out (T out ,pout ) −nN

2,out ∙H N

2,out (T out ,pout ) +Q+W =0¿ ¿

For air as an ideal gas mixturenair,∈¿ ∙ H air,∈¿ ( T , p)=nO

2,out ∙H O

2,out

(T , p)+ nN2,out ∙H N

2,out

( T , p) ¿¿

Hence, for the isothermal and adiabatic separation of air:Q+W=0

Entropy balance Sair ,∈¿−SO2 ,out−SN 2,out+

QT

+S gen=0¿

For a reversible process entropy generation is zeroSair ,∈¿−SO2 ,out−SN 2,out+

QT

=0¿

S IGM (T , psystem )=∑i=1

C

x i ∙ SiIG (T , psystem )−R ∙∑

i=1

C

x i ∙ ln (x i )

nO 2 ,out∙ SO 2

IG (T ¿ , p¿ )+nN2 ,out∙ SN 2

IG (T ¿ , p¿)−R ∙ (xO2∙ ln (xO2 )+xN 2

∙ ln (xN2 ))−nO 2 ,out∙ SO 2

IG (T out , pout )−nN2 , out∙ S N2

IG (T out , pout )+QT

=0

Hence, for the isothermal and isobaric separation process:

−R ∙( xO2∙ ln (xO2 )+ xN2

∙ ln ( xN 2) )+QT

=0

Q=−RT ∙ ( xO 2∙ ln (xO2 )+x N2

∙ ln (x N2 ))=−1.274 ∙ kJmol

W=−Q=1.274 ∙ kJmol

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

11

I.3 Describing mixtures with interactions – partial molar propertiesMolecules can be polarized resulting in a charge distribution within the molecule (possibly leading to dipole, quadrupole or multipoles) resulting an electro-static interaction between the molecules. These interactions will be minimal, when the molecules are far from each other, e.g. in a gas/vapour at low pressure, but will affect the energy of the system severely in dense systems, such as liquids or even gases at high pressure. In systems containing pure compounds, this was taken into account via the critical properties of the pure component involved in the equation of state.

In mixtures, different types of molecules are interacting with each other. The interaction between the various molecules manifests itself in various forms. One of the forms is the molar volume of the mixture. The molar volume of a mixture cannot be represented by a linear relationship between the molar volume of the components making up the mixture (see Figure I.3.1).

V mixture≠∑i

C

x i ∙V i

0

10

20

30

40

0 0.2 0.4 0.6 0.8 1

Mo

lar

vo

lum

e o

f m

ixtu

re,

cm

3/m

ol o

f m

ixtu

re

Mole fraction of water, xw

Figure I.3.1: Molar volume of methanol-water mixture (in cm3 per mol of mixture) at 293.15 K as a function of mole fraction of water (dashed line represents the linear relationship between the molar volume of the mixture and the molar volumes of pure methanol and water; Data from Handbook of Physics and Chemistry, D-238, 67 th ed. (R.C. Weast, M.J. Astle, W.H. Beyer, Eds.), CRC Press, Boca Raton, 1986)

A small difference exists between the molar volume of the mixture and the molar volume of the pure components. The difference originates from the structure of the liquid, which is a consequence of the inter-molecular interactions. This small difference is better illustrated by looking at the change in the molar volume upon mixing:

∆mixV=V mixture−∑i

C

x i ∙V i

or in the case of the methanol-water mixture: ∆mixV=V methanol−watermixture−xwater ∙V water−xmethanol ∙V methanol

which shows a contraction in the volume upon mixing the two components together (see Fig. 3.2). The highest contraction in the volume (1 cm3/mol of mixture) is obtained at a mole fraction of water of 0.52. The change in volume upon mixing is thus not symmetric with respect to the mole fraction of its components!

12

-1.5

-1

-0.5

0

0 0.2 0.4 0.6 0.8 1

Dm

ixV

, cm

3/m

ol

of

mix

ture


Figure I.3.2: Change in the molar volume upon mixing pure methanol and pure water to form a mixture as a function of the mole fraction at 293.15 K (solid curve represents fit to the Redlich-Kister equation - see text; Data of mixture from Handbook of Physics and Chemistry, D-238, 67th ed. (R.C. Weast, M.J. Astle, W.H. Beyer, Eds.), CRC Press, Boca Raton, 1986)

The change in the molar volume as a function of the mole fraction of water can be represented by the empirical Redlich-Kister equation:

∆mixV=x1 ∙ x2∙∑i=0

i=i

ai ∙ (x1−x2 )i in its general form

The change in the molar volume upon mixing water and methanol can be reasonably fitted with a 3 rd

order Redlich-Kister equation (i.e. i goes up to 3)

∆mixV=xwater ∙ xmethanol ∙∑i=0

i=3

ai ∙ ( xwater−xmethanol )i

∆mixV=xwater ∙ xmethanol ∙ (a0+a1 ∙ (xwater−xmethanol )+a2 ∙ (xwater−xmethanol )

2+a3∙ (xwater−xmethanol )3 )

(with the coefficients a0 = -4.0012 cm3/mol; a1 = -0.3961 cm3/mol, a2 = 0.3827 cm3/mol; a3 = 1.2836 cm3/mol)

The molecular interaction between the different molecules in the mixture results in a change in the molar volume upon mixing two pure components, as well as other thermodynamic properties (such as mixU, mixH, mixS, mixG). This implies that the molar properties of the mixture cannot be represented in terms of the molar properties of the pure components (Vi, Ui, Si,. Hi, Gi). Mixtures are described in terms of the partial molar properties.

A partial molar property is the change in the property of the mixture upon the addition of a small amount of compound i to the mixture keeping all other variables (T,p and the number of moles of all other components in the mixture constant)

θi (T , p , x )=( ∂ (Nmixture ∙ θmixture )∂N i

)T , p , N j≠ i

in its general form

partial molar volume V i (T , p , x )=( ∂ (Nmixture ∙V mixture )∂N i

)T , p , N j≠ i

13

partial molar enthalpy H i (T , p , x )=( ∂ (Nmixture ∙Hmixture )∂ N i

)T , p , N j≠ i

The partial molar property is compound specific and is a function of temperature, pressure and the molar composition of the mixture. It describes the change in the molar property of the mixture upon addition of an infinite small number of moles of component i to the mixture.

The thermodynamic property of the mixture can now be expressed in terms of the partial molar properties of the components in the mixture:

Θmixture (T , p , x )=∑i

C

x i ∙ θi in its general form

Molar volume of mixture V mixture (T , p , x )=∑i

C

x i ∙V i

Molar enthalpy of mixture Hmixture (T , p , x )=∑i

C

x i ∙ H i

The partial molar property can be obtained from the experimentally observed change in the property upon mixing. For instance, the change in the molar volume upon mixing methanol and water was modelled as:

∆mixV=xwater ∙ xmethanol ∙∑i=0

i=3

ai ∙ ( xwater−xmethanol )i=f (xw)

The change in the molar volume upon mixing is defined as: ∆mixV (T , p , x )=V mixture (T , p , x )−xwater ∙V water−xmethanol ∙V methanol

V mixture (T , p , x )=∆mixV (T , p , x )+ xwater ∙V water+xmethanol ∙V methanol

V mixture (T , p , x )=f (xB)+xwater ∙V water+ xmethanol ∙V methanol

The partial molar volume of water is defined as:

V water (T , p , x )=( ∂ (N mixture∙V mixture )∂ Nwater

)T , p , NMeOH

V water (T , p , x )=( ∂ (N mixture ∙ f (xw )+Nwater ∙V water+Nmethanol ∙V methanol )∂N water

)T , p , N MeOH

V water (T , p , x )=( ∂ (N mixture ∙ f (xw ))∂N water

)T , p , N MeOH

+( ∂ (Nwater ∙V water )∂ Nwater

)T , p , NMeOH

+( ∂ (Nmethanol ∙V methanol )∂N water

)T , p ,N MeOH

Evaluating each of the three terms in the equation:

( ∂ (Nmethanol ∙V methanol )∂N water

)T , p , NMeOH

=Nmethanol ∙( ∂V methanol

∂N water)T , p , N MeOH

+V methanol ∙( ∂N methanol

∂N water)T , p , NMeOH

( ∂ Nmethanol


=0, since the number of methanol is kept constant

( ∂V methanol

∂ Nwater)T , p , N MeOH

=0, since the molar volume of pure methanol is only a function of

temperature and pressure

14

( ∂ (Nwater ∙V water )∂ Nwater

)T , p , NMeOH

=Nwater ∙( ∂V water

∂ Nwater)T , p ,N MeOH

+V water ∙( ∂ Nwater

∂ Nwater)T , p , NMeOH

Nwater ∙( ∂V water


=0, since the molar volume of pure water is only a function

temperature and pressure

V water ∙( ∂ Nwater


=V water , since ( dXdX )=1

( ∂ (Nmixture ∙ f (xw ))∂ Nwater

)T , p , NMeOH

=Nmixture ∙( f (xw )∂N water

)T , p , N MeOH

+f (xw ) ∙( ∂ Nmixture

∂ Nwater)T , p , N MeOH

f (xw ) ∙( ∂ Nmixture


=f (xw )∙( ∂ (N water+Nmethanol )∂N water

)T , p , N MeOH

=f (xw )

Nmixture ∙( f (xw )∂ Nwater

)T , p ,N MeOH

=N mixture ∙( f (xw )∂ xw

)T , p , NMeOH

∙( ∂ xw


( ∂ xw

∂ Nwater)T , p ,N MeOH

=( ∂ N water

N water+Nmethanol

∂N water)T , p , NMeOH

=N methanol

(N water+Nmethanol )2


)T , p ,N MeOH

=N mixture ∙( f (xw )∂ xw

)T , p , NMeOH

∙N methanol

(N water+Nmethanol )2


)T , p ,N MeOH

= xmethanol ∙( f (xw )∂ xw

)T , p , N MeOH

Substituting it all back:

V water (T , p , x )=f (xw)+xmethanol ∙( f (xw )∂xw

)T , p , N MeOH

+V water

Knowing the fitting function f(xw), the partial molar volume of water can be found:Realize:

f (xw )=xwater ∙ xmethanol ∙∑i=0

i=3

ai ∙ (xwater−xmethanol )i

f (xw )=xwater ∙ (1−xwater ) ∙∑i=0

i=3

ai ∙ (2xwater−1 )i

V water (T , p , x )−V water=xwater ∙ (1−xwater ) ∙∑i=0

i=3

ai ∙ (2xwater−1 )i+xmethanol ∙( xwater ∙ (1−xwater )∙∑i=0

i=3

ai ∙ (2 xwater−1 ) i

∂ xwater)T , p ,N MeOH

15

V water (T , p , x )−V water=xwater ∙ (1−xwater ) ∙∑i=0

i=3

ai ∙ (2xwater−1 )i+(1−xwater ) ∙((1−xwater ) ∙∑i=0

i=3

ai ∙ (2 xwater−1 )i−xwater ∙∑i=0

i=3

ai ∙ (2 xwater−1 )i+xwater ∙ (1−xwater ) ∙∑i=1

i=3

2∙ ai ∙i ∙ (2xwater−1 )i−1)

V water (T , p , x )−V water=(1−xwater )2∙∑i=0

i=3

ai ∙ (2 xwater−1 ) i+(1−xwater ) ∙ xwater ∙ (1−xwater ) ∙∑i=1

i=3

2∙ ai ∙i ∙ (2xwater−1 )i−1

The partial molar volume can be determined without previous knowledge on the fit of the change in the volume upon mixing as a function of the mole fraction in the mixture. For instance, the change in the molar volume upon mixing methanol and water is defined as: ∆mixV (T , p , x )=V methanol−watermixture (T , p , x )−xwater ∙V water−xmethanol ∙V methanol

∆mixV (T , p , x )=xwater ∙V water (T , p , x )+xmethanol ∙V methanol (T , p , x )−xwater ∙V water−xmethanol ∙V methanol

∆mixV (T , p , x )=xwater ∙ (V water−V water )+xmethanol ∙ (V methanol−V methanol )

This equation has two variables which are a function of the mole fraction, viz. V water ,V methanol, and we need therefore another equation to obtain the partial molar volumes of the individual compounds. This can be obtained by differentiating the change in molar volume upon mixing with respect to the mole fraction of water (keeping the temperature and pressure constant):

( ∂ (∆mixV (T , p , x ) )∂ xwater

)T , p

= (V water−V water ) ∙( ∂ xwater

∂ xwater)T , p

+xwater ∙( ∂ (V water−V water )∂ xwater

)T , p

+(V methanol−V methanol )∙( ∂xmethanol

∂ xwater)T , p

+xmethanol ∙( ∂ (V methanol−V methanol )∂ xwater

)T , p

For a binary mixture of methanol-water:

xmethanol=1− xwater and ( ∂ xmethanol

∂x water)T , p

=−1

The molar volume of a pure component is independent of the composition in the mixture, i.e.

( ∂V i

∂ xwater)T , p

=0

Thus, the differential equation simplifies to:

( ∂ (∆mixV (T , p , x ) )∂ xwater

)T , p

= (V water−V water )+x water ∙( ∂V water

∂ xwater)T , p

−¿

(V methanol−V methanol )+xmethanol ∙( ∂ (V methanol)∂ xwater

)T , p

The right hand side of the equation still contains two partial differentials. We can show that the Gibbs-Duhem equation shows that

xwater ∙( ∂V water

∂xwater)T , p

+xmethanol ∙( ∂V methanol

∂ xwater)T , p

=0

16

Hence, the two equations required to obtain the partial molar volume of methanol and the partial molar volume of water as a function of the mole fraction of water are: ∆mixV (T , p , x )=xwater ∙ (V water−V water )+xmethanol ∙ (V methanol−V methanol )and

( ∂ (∆mixV (T , p , x ) )∂ xwater

)T , p

= (V water−V water )−(V methanol−V methanol )

Now working out the two unknowns:

(V water−V water )=∆mixV (T , p , x )+xmethanol ∙( ∂ (∆mixV (T , p , x ))∂xwater

)T , p

(V methanol−V methanol )=∆mixV (T , p , x )+xwater ∙( ∂ (∆mixV (T , p , x ) )∂ xmethanol

)T , p

Now, the actual experimental data or the fitted Redlich-Kister equation can be used to find the partial molar volume of methanol and water in a water-methanol mixture.

The so deduced partial molar volume of water in a water-methanol mixture shows a monotonic increase with increasing mole fraction of water (see Fig. 3.3). The monotonic increase can be interpreted as a steady decrease in the number of molecules of methanol surrounding a water molecule with increasing mole fraction of water. The partial molar volume of methanol as a function of the mole fraction of water shows a peculiar behavior with a minimum for the mole fraction of water of ca. 0.62. This behavior can be explained in terms of a preferential surrounding of methanol with water due to the formation of H-bridges.

39

41

43

45

47

12

14

16

18

20

0 0.2 0.4 0.6 0.8 1

Pa

rtial m

ola

r vo

lum

e o

f m

eth

an

ol, V

me

tha

no

l , cm

3/mo

lP

art

ial m

ola

r v

olu

me

of

wa

ter,

Vw

ate

r, c

m3/m

ol


Figure I.3.3: Partial molar volume of water (left axis) and methanol (right axis) in a water-methanol mixture at 293.15 K as a function of the mole fraction (Data of mixture from Handbook of Physics and Chemistry, D-238, 67th ed. (R.C. Weast, M.J. Astle, W.H. Beyer, Eds.), CRC Press, Boca Raton, 1986)

The partial molar enthalpy can be easily determined from calorimetric measurements on the heat release/removal upon the addition of a compound to another compound. From here the partial molar specific heat can be determined. The specific heat of a mixture is defined as:

17

C p ,mixture=( ∂ Hmixture

∂T )p ,N j

The partial molar heat capacity is thus defined as:

C p ,i=( ∂N mixture∙CP,mixture

∂N i)T , p , N j≠ i

=( ∂∂N i

∙( ∂N mixture ∙Hmixture

∂T )p , N i

)T , p , N j ≠i

C p ,i=( ∂∂T

∙( ∂ Nmixture ∙ Hmixture

∂N i)T , p , N j ≠i

)p , N i

=( ∂H i

∂T )p ,N i

Hence, the determination of the partial molar enthalpy at a variety of temperatures will yield the partial molar heat capacity.

How to determine the partial molar Gibbs free energy?A useful relationship between the Gibbs free energy is given by considering the change in the Gibbs free energy relative to temperature with respect to temperature:

( ∂(GT )∂T )

p

=1T∙( ∂G∂T )

p

−G

T 2

Knowing that ( ∂G∂T )p

=−S and G=H−T ∙S

( ∂(GT )∂T )

p

=−ST

−H−T ∙S

T2

( ∂(GT )∂T )

p

=−H

T2

And thus also for the partial molar properties:

( ∂(Gi

T )∂T

)p

=−H i

T 2

The enthalpy is in principle a function of temperature

H i=H i . T=0+ ∫T=0

T=T

C p ,i ∙dT

Integration between T=0 and T=T yields

(GT )T=T

−(GT )T=0

=H i ,T=0

T−

H i , T=0

T=0+ ∫

T=0

T=T C p ,i

T 2∙ dT

At 0K the entropy goes to zero (3rd law of thermodynamics), and the Gibbs free energy at 0K equals the enthalpy at 0 K:

(Gi

T )T=T

=H i ,T=0

T+ ∫

T=0

T=T Cp , i

T 2∙ dT

More often however, the partial molar Gibbs free energy is determined from VLE-data using the excess Gibbs free energy (vide verde).

The partial molar entropy can then be determined from:

18

Si=H i−Gi

T

19

I.4 Gibbs-Duhem equationAny thermodynamic property of a mixture can be represented by:

θmixture (T , p , x )=∑i

C

x i ∙ θi

or in terms of extensive property mixture:

θmixture (T , p , N i ,N j ,N k , N l ,⋯ )=N mixture∙ θmixture (T , p , x )=∑i

C

N i ∙θ i (3.1.1)

The thermodynamic property is a complete function of T,p and the number of moles of the various components in the mixture. The complete differential is thus given by:

dθmixture (T , p ,N i , N j , N k ,N l ,⋯ )=( ∂ (Nmixture ∙ θmixture )∂T )

p ,N i

∙ dT+( ∂ (Nmixture ∙ θmixture )∂ p )

T , N i

∙dp+∑i

C ( ∂ (Nmixture ∙ θmixture )∂ N i

)T , p , N j≠ i

∙ d N i

Substituting the definition for the partial molar property:

dθmixture (T , p ,N i , N j , N k ,N l ,⋯ )=( ∂ (Nmixture ∙ θmixture )∂T )

p ,N i


T , N i

∙dp+∑i

C

θi ∙ d N i

(3.1.2)

However, the complete differential is also equal to

d θmixture (T , p , N i ,N j ,N k , N l ,⋯ )=∑i

C

N i ∙ dθi+∑i

C

θ i ∙ dN i (3.1.3)

Equating (3.1.3) and (3.1.2):

( ∂ (Nmixture ∙ θmixture )∂T )

p ,N i


T , N i

∙dp+∑i

C

θi ∙ d N i=∑i

C

N i ∙ d θi+∑i

C

θi ∙ dN i

or

( ∂ (Nmixture ∙ θmixture )∂T )

p ,N i


T , N i

∙dp−∑i

C

N i ∙ dθ i=0

Thus at constant temperature and pressure:

∑i

C

N i ∙ dθ i=0

This can be expressed in terms of the mole fraction (by dividing both left and right hand side of the equation with the total number of moles in the mixture)

∑i

C

x i ∙ dθi=0

This can be transformed into a partial differential equation by dividing both sides of the equation with the same differential (e.g. dxA)

∑i

C

x i ∙dθi

d xA

=0

20

Thus for the mixing process of methanol and water (with i = water, methanol,; and A is water), this equation becomes:

xwater ∙( ∂V water

∂xwater)T , p

+xmethanol ∙( ∂V methanol

∂ xwater)T , p

=0

A consequence of the Gibbs-Duhem equation is that for a multi-component system with C components (and thus the variables T,p, N1, N2, …, NC) there are only C+1 independent variables!

I.4.1 Partial molar Gibbs free energy and the Gibbs-Duhem equationThe Gibbs free energy is for chemical engineers an important property, since this function is at its minimum at equilibrium for systems at constant temperature and pressure. The Gibbs free energy of a multi-component mixture is a complete function of temperature, pressure and the number of moles present of each species. G=G (T , p , N i )

d G=( ∂G∂T )p , N i

∙ dT +( ∂G∂ p )T , N i

∙ dp+∑i

C

( ∂G∂ N i)T , p , N j≠ i

∙ d N i

d G=−S ∙dT+V ∙dp+∑i

C

Gi ∙ d N i

(the partial molar Gibbs free energy has classically been termed the chemical potential)

We can do the same with other thermodynamic functions. For instance, the enthalpy of a multi-component mixture is a function of pressure, entropy, and the number of components H=H (S , p ,N i )

d H=( ∂H∂S )p , N i

∙ dS+( ∂ H∂ p )

S , N i

∙ dp+∑i

C

( ∂ H∂ N i

)S , p , N j≠ i

∙d N i

d H=T ∙dS+V ∙dp+∑i

C

( ∂ H∂ N i

)S , p , N j ≠ i

∙ d N i

NOTE: ( ∂ H∂ N i

)S , p , N j≠ i

≠( ∂H∂N i)T , p ,N j≠ i

=H i

The partial differential of the enthalpy with respect to the number of moles of i keeping the entropy, pressure and the number of moles of all other components constant is most conveniently evaluated using the definition of the Gibbs free energy H ≡G+T ∙ S

d H=dG+T ∙dS+S ∙dT=−S ∙dT+V ∙dp+∑i

C

Gi ∙ d N i+T ∙dS+S ∙dT


C

Gi ∙ d N i

Hence,

∑i

C

Gi ∙ d N i=∑i

C

( ∂ H∂ N i

)S , p , N j ≠i

∙ d N i

Thus, the change in enthalpy is given by

21


C

Gi ∙ d N i

Similarly

d U=T ∙dS−p ∙dV +∑i

C

Gi ∙ d N i

d A=−S ∙dT−p ∙dV +∑i

C

Gi ∙d N i

The partial molar Gibbs free energy is related to the fugacity of a species in the mixture. The fugacity of a species in a mixture is a departure function of the Gibbs free energy for a species from its value in an ideal gas mixture:

f i=x i ∙ P∙ eGi (T , p , x )−Gi

IGM (T , p , x )RT

The partial molar Gibbs free energy of a compound in a mixture can be obtained from:

∆mixGIGM=RT ∙∑

k

N k ∙ ln (xk )=N ∙GIGM−∑k

N k ∙GkIG

GiIGM (T , p , x )=( ∂ N ∙G IGM

∂ N i)T . p , N j≠ i

=( ∂(∑k N k ∙G kIG+RT ∙∑

k

N k ∙ ln ( xk ))∂N i

)T . p , N j≠ i

22

GiIGM (T , p , x )=Gi

IG+( ∂ (RT ∙∑k

N k ∙ ln ( xk ))∂N i

)T . p , N j≠ i

( ∂(∑k N k ∙ ln (xk ))∂ N i

)T . p , N j ≠i

=ln (x i )+∑k

N k ∙( ∂ ln ( xk )∂ N k

)T . p , N j≠ i

N k ∙( ∂ ln (xk )∂N i

)=N k ∙( ∂ ln (xk )∂ x i

)( ∂x i

∂N i)

For k=iN i ∙( ∂ ln (xi )

∂ N i)=N i ∙( ∂ ln (x i)

∂x i)( ∂ x i

∂N i)=N i ∙

1x i

∙∑k ≠i

C

N k

Nmixture2 =∑

k ≠i

C

xk

For k=i

N k ∙( ∂ ln (xk )∂N i

)=N k ∙( ∂ ln (xk )∂ x i

)( ∂x i

∂N i)=−N k ∙

1xk

∙∑k ≠i

C

N k

Nmixture2 =−∑

k ≠i

C

xk

GiIGM (T , p , x )=Gi

IG+RT ∙ ln (xi )

Thus, the fugacity of a species in the mixture is given by


IG−RT ∙ ln (xi )RT

f i=P ∙eGi (T , p , x )−Gi

IG

RT

23

I.5 Ideal mixture and excess mixture propertiesAn ideal mixture is defined as a mixture, in which the partial molar volume and enthalpy of each component in the mixture equals the molar volume/enthalpy of the pure component for all temperatures/pressures and compositions : H i

ℑ (T , p , x )=H i (T , p ) V i

ℑ (T , p , x )=V i (T , p )

The fugacity of a species in an ideal mixture is given by:


ℑ (T , p ,x )RT =x i ∙P ∙ e

1RT

∙ ∫p=0

p= p

(V i−V iIG) dp

f i=x i ∙ f i

The partial molar Gibbs free energy in an ideal mixture is given by: Gi

ℑ (T , p , x )=Gi (T , p )+RT ∙ ln (x i )

Other functions of ideal mixtures (similar to ideal gas mixtures): Si

ℑ (T , p , x )=S i (T , p )−R ∙ ln (x i ) U i

ℑ (T , p , x )=U i (T , p )

Excess property is defined as the property of a component relative to the property in an ideal mixture: θex=∆mixθ (T , p , x )−∆mixθ

ℑ (T , p , x ) θex=∑

i

x i ∙ θi−∑i

x i ∙θ iℑ

or otherwise stated:θmixture=∑

i

x i ∙ θi=∑i

xi ∙ θiℑ+θex

e.g. Gmixture=∑i

x ∙Gi=∑i

x ∙Gi+RT ∙∑i

xi ∙ ln (x i )+Gex

The excess properties are the basis of the definition of activity coefficients of compounds in a solution. The activity coefficient of a component (i) in a solution is defined in terms of the fugacity of the pure component and the fugacity of a component in a mixture: f i

L=x i ∙ γ i ∙ f iL

γ i=f iL

x i ∙ f iL

RTln (γ i )=RTln( f iL

x i ∙ P )−RTln( f iLP ) RTln (γ i )=Gi (T , p , x )−Gi

ℑ (T , p , x )

RTln (γ i )=Giex (T , p , x )=( ∂ (N ∙G ex)

∂ N i)T , p , N j≠ i

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

24

Example: Is a binary mixture for which the molar Gibbs free energy is given by

Gmix (T , p . x )=∑i=1

2

x i ∙Gi (T , p )+RT ∙∑i=1

2

x i ∙ ln (x i )+a ∙ x1 ∙ x2 an ideal mixture?

With a a constant independent of T,p,x.

Condition for ideal mixture: V i

ℑ (T , p , x )=V i (T , p ) H iℑ (T , p , x )=H i (T , p )

V mix (T , p . x )=( ∂Gmix

∂ p )T , N

=∑i=1

2

x i ∙( ∂Gi

∂ p )T , N

=∑i=1

2

x i ∙V i

(hence the first condition is fulfilled)

Hmix (T , p . x )=−T 2∙( ∂(Gmix

T )∂T

)T , N

=−T 2 ∙( ∂(∑i=12

x i ∙Gi (T , p )

T+R ∙∑

i=1

2

x i ∙ ln (x i )+aT∙ x1 ∙ x2)

∂T)T , N

Hmix (T , p . x )=−T 2∙(∑i=12

x i ∙(−H i (T , p )T 2 )− a

T 2∙ x1 ∙ x2)

Hmix (T , p . x )=∑i=1

2

x i ∙ H i (T , p )+a ∙ x1∙ x2

For an ideal mixture Hmix (T , p . x )=∑i=1

2

x i ∙ H i (T , p ); hence this mixture is not ideal.

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

25

I.6 Properties of activity coefficient

The activity coefficient of a particular component in a liquid mixture can be derived from the excess molar Gibbs free energy of the mixture, since

RTln (γ i )=Giex (T , p , x )=( ∂ (N ∙G ex)

∂ N i)T , p , N j≠ i

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Developing an expression of the activity coefficient Develop an expression for the activity coefficients of component 1 and component 2 in a binary mixture for which the molar Gibbs free energy is given by

Gmix (T , p . x )=∑i=1

2

x i ∙Gi (T , p )+RT ∙∑i=1

2

x i ∙ ln (x i )+a ∙ x1 ∙ x2

The excess Gibbs free energy is defined as: Gex=∆mixG (T , p , x )−∆mixG

ℑ (T , p , x )

∆mixGℑ (T , p , x )=∑

i=1

2

x i ∙Gi (T , p )+RT ∙∑i=1

2

x i ∙ ln ( xi ) Gex=a ∙ x1 ∙ x2=a ∙ x1 ∙ (1−x1 )=a∙ (x1−x1

2 )

G1ex=( ∂ N ∙Gex

∂N 1)T , p , N2

=( ∂ N ∙a ∙ (x1−x12 )

∂N 1)T , p , N 2

=a ∙ (x1−x12 )+N ∙( ∂a ∙( x1− x1

2 )∂ x1

∙( ∂ x1∂N 1

))T , p , N2

( ∂a ∙( x1−x12)

∂x1 )T , p , N2

=a ∙ (1−2∙ x1 )

N ∙( ∂ x1

∂N 1)T , p ,N 2

=N ∙( ∂ N 1

N1+N1

∂N 1)T , p , N2

=N ∙N2

(N1+N1 )2=N ∙

N2

N2=x2

G1ex=a ∙ x1 ∙ (1− x1 )+a∙ (1−2∙ x1) ∙ x2

G1ex=a ∙ x1 ∙ (1− x1 )+a∙ (1−2∙ x1) ∙ (1−x1 )

G1ex=a ∙ (1−x1)

2

G2ex=( ∂ N ∙Gex

∂N 2)T , p , N1

=( ∂ N ∙a ∙ (x1−x12 )

∂N 2)T , p , N 1

=a ∙( x1−x12)+N ∙( ∂a ∙(x1−x1

2 )∂ x1

∙( ∂ x1∂N 2

))T , p , N1

N ∙( ∂ x1

∂N 2)T , p ,N 1

=N ∙( ∂ N 1

N1+N1

∂N 2)T , p , N1

=−N ∙N1

(N1+N1 )2=−N ∙

N2

N2=−x1

G2ex=a ∙ x1 ∙ (1− x1 )−a ∙ (1−2 ∙ x1 ) ∙ x1

G2ex=a ∙ x1

2

The activity coefficient is defined as:

26

γ i=eGi

ex

RT γ 1=ea ∙ x2

2

RT γ 2=ea ∙ x1

2

RT

The fugacity of a component in the mixture is given by: f iL=γi ∙ x i ∙ f i

L

f 1L=e

a ∙ x22

RT ∙ x1 ∙ f 1L f 2

L=ea ∙ x1

2

RT ∙ x2 ∙ f 2L

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

27

Thermodynamic consistency of activity coefficientsBoth the real mixture and the ideal mixture satisfy the Gibbs-Duhem equation:

( ∂ (Nmixture ∙Gmixture )∂T )

p , N i

∙ dT+( ∂ (Nmixture ∙Gmixture )∂ p )

T , N i

∙ dp−∑i

C

N i ∙ d Gi ,mixture=0

−Smixture ∙ dT+V mixture ∙dp−∑i

C

x i ∙d Gi ,mixture=0

−Sℑ ∙ dT+V ℑ ∙ dp−∑i

C

x i ∙ dGiℑ=0

Subtracting these two equations yields:

−(Smixture−Sℑ) ∙ dT+ (V mixture−V ℑ )∙ dp−∑i

C

x i ∙ (dGi , mixture−d Giℑ )=0

−Sex ∙ dT+V ex ∙ dp−∑i

C

x i ∙ d Giex=0

Thus at constant temperature and pressure

∑i

C

x i ∙ dGiex=0

∑i

C

x i ∙ d (RTln ( γi ))=0

Thus, for a binary mixture: x1 ∙ d (ln ( γ1 ))+x2 ∙ d ( ln (γ 2 ))=0Hence, for a variation with respect to the mole fraction of component 1 in an isothermal and isobaric system:

x1 ∙( ∂ ln ( γ1 )∂x1 )

T , p

+x2 ∙( ∂ ln ( γ2 )∂x1 )

T , p

=0

Temperature dependency of the activity coefficient

The activity coefficient is defined as: ln ( γi )=Gi

ex

RT

and the temperature dependency is given by:

( ∂ ln (γ i )∂T )

p , x

=( ∂∂T (Gi

ex

RT ))p , x

=−H i

ex

RT2

For small temperature variations, we may assume that the partial molar enthalpy does not vary significantly, and hence

γ i (T2 , p , x )=γi (T 1 , p , x ) ∙ eH i

ex

R∙( 1T2−

1T1 )

Pressure dependency of the activity coefficient

The activity coefficient is defined as: ln ( γi )=Gi

ex

RT

28

and the pressure dependency is given by:

( ∂ ln (γ i )∂ p )

T , x

=( ∂∂ p (Gi

ex

RT ))T , x

=V i

ex

RT

For liquids, the pressure dependency on the excess volume is small (liquids can be considered to be incompressible fluids), and hence

γ i (T1 , p2 , x )=γi (T1 , p1 , x ) ∙ eV i

ex

R∙ (p2− p1 )

29

I.7 Fugacities in mixtures

The fugacity of a compound in a mixture is defined as:


IGM (T , p , x )RT

or lnf i

x i ∙P=Gi (T , p , x )−G i

IGM (T , p , x )RT

lnf i

x i ∙P= 1

RT∙∫0

p

(V i−V iIG) ∙ dp

Equations of state are typically functions of the volume of the system rather than the pressure of the system. Hence, we are changing the integration variable from pressure to volume. The integral can be split into two parts:

lnf i

x i ∙P= 1

RT∙(∫0

p

V i ∙ dp−∫0

p

V iIG ∙dp)

The first integral is over the partial molar volume of compound i. From the triple product rule:

( ∂V∂ N i)T , p , N j≠ i

∙( ∂ p∂V )

T , N j

∙( ∂N i

∂ p )T ,V ,N j≠ i

=−1

or V i ∙dp=( ∂V∂N i)T , p , N j≠ i

∙ dp=−( ∂ p∂ N i

)T ,V , N j ≠i

∙ dV=−N ∙( ∂ p∂ N i

)T , V , N j ≠i

∙ dV

The second integral can be evaluated realizing that:d ( p∙V )=p ∙d V +V ∙dp

dp= 1V

∙d ( p ∙V )− pV

∙dV

Furthermore p ∙V =Z ∙RT and at constant temperatured ( p∙V )=RT ∙dZ

Hence, ∫0

p

V iIG ∙ dp=∫

Z=1

Z V iIG

VRT ∙dZ− ∫

V=∞

V=ZRTp

V iIG ∙

pV

∙dV

(Note that the integration boundaries change with the change in integration variable. Each time we integrate from the ideal gas state - Z=1, infinite large volume, zero pressure – to real gas state)

Realizing that V iIG= RT

p

∫0

p

V iIG ∙ dp=∫

Z=1

ZRTp ∙V

RT ∙dZ− ∫V =∞

V=ZRTp

RTp

∙pV

∙dV

∫0

p

V iIG ∙ dp=∫

Z=1

ZRTZ

∙dZ− ∫V =∞

V=ZRTp

RTV

∙d V

Substituting the two evaluated integrals back into the expression for the fugacity of the mixture yields:

lnf i

x i ∙P= 1

RT∙( ∫

V =∞

V=ZRTp

(RTV −N ∙( ∂ p∂ N i

)T ,V , N j≠ i) ∙ dV−∫

Z=1

ZRTZ

∙dZ)30

lnf i

x i ∙P= 1

RT∙( ∫

V =∞

V=ZRTp

(RTV −N ∙( ∂ p∂ N i

)T ,V , N j≠ i) ∙ dV )−ln (Z )

Hence, knowing the equation of state for the mixture (in the appropriate state of aggregation) should yield the fugacity of the specific component in a mixture.

31

Gaseous mixtures

Some gas mixtures follow Amagat’s law, i.e. V mixture (T , p , N i )=∑1=1

C

y i ∙V i (T . p ) (yi is the mole

fraction of component i in the gas phase). This also means that the excess volume is zero. A consequence of this is that partial molar volume of a component i in this mixture equals molar volume of the pure component. Hence, the expression for the fugacity of a component i in the mixture becomes

lnf i

y i ∙ P= 1

RT∙∫0

p

(V i−V iIG) ∙ dp= 1

RT∙∫0

p

(V i−V iIG) ∙ dp=ln( f ip )

and the fugacity of the species in the mixture is given by the fugacity of the pure component times its mole fraction

f i= y i ∙ f i Lewis-Randall ruleThis expression is valid as long as Amagat’s law is valid. This is typically for gas mixtures at low pressures and at extreme high pressures.

For intermediate pressures an equation of state has to be used. For instance, the truncated form of the virial equation (neglecting the higher order terms):

Zmixture=pVRT

=1+Bmix (T , x )

Vwith Bmix (T , x )=∑

i∑

j

yi ∙ y j ∙Bij (T )

Thus

p= RTV

∙(1+Bmix (T ,x )V )=N ∙RT

V∙(1+ N ∙∑

i∑

j


V )= N ∙RTV

∙(1+∑i∑j

N i ∙ y j ∙B ij (T )

V )

( ∂ p∂ N i

)T , V , N j ≠i

=RTV

∙(1+N ∙∑i∑

j

y i ∙ y j ∙B ij (T )

V )+ N ∙RT

V 2 ∙( ∂(N ∙∑i∑

j

y i ∙ y j ∙Bij (T ))∂ N i

)T , V , N j≠ i

This can now be further developed into an expression for the fugacity of a species of a compound i according to the truncated form of the virial equation of state to:

N ∙( ∂ p∂N i

)T , V , N j≠ i

=RTV

+2 ∙RT

V 2 ∙∑i

y i ∙ Bij (T )

lnf i

y i ∙ P= 1

RT∙( ∫

V =∞

V =ZRTp

(RTV −N ∙( ∂ p∂N i

)T ,V , N j≠ i) ∙ dV )−ln (Z )

lnf i

y i ∙ P= 1

RT∙( ∫

V =∞

V =ZRTp

(−2∙ RTV 2 ∙∑j

y j ∙ Bij (T )) ∙ dV )−ln (Z )

lnf i

y i ∙ P= 2V

∙∑j

y j ∙Bij (T )−ln (Z )

32

lnf i

y i ∙ P= 2 ∙ p

Z ∙ RT∙∑

j

y j ∙ Bij (T )−ln (Z )

(with Z expressed as a function of pressure Z=12 (1+√1+ 4 ∙Bmix ∙ p

RT ) )Other equations of state (such as the Peng-Robinson equation of state) can be used as well. The Peng-Robinson equation of state is given as:

p= RTV−b

−a (T )

V ∙ (V +b )+b ∙ (V −b )The parameters a and b now refer to the parameters for the mixture. Mixing rules have been identified based on statistical mechanics:

amix=∑i=1

C

∑j=1

C

y i ∙ y j ∙ aij with a ij=√aii ∙ a jj ∙ (1−k ij )

(here aii represents the pure component parameter a)

bmix=∑i=1

C

yi ∙ bi with bi the pure component parameter b

These mixing rules introduce a new parameter, k ij, the binary interaction parameter. This parameter is obtained by fitting experimental data to Peng-Robinson equation of state. Tabe 9.4-1 (p. 424) gives a selection of binary interaction parameters for binary hydrocarbon mixtures. Others are given in specialist journals (e.g. H. Nishiumi, T. Arai, K. Takeuchi, Fluid Phase Equilibria 42 (1988), 43-62). This approach only takes into consideration binary interactions. Mixtures containing more than two components can be built up by considering the various binary pairs present in the mixture (D.S.H. Wong, S.I. Sandler, AIChE J. 38 (1992), 671-680).

The fugacity obtained from the Peng-Robinson equation of state is now given by:

lnf i

y i ∙ P=

Bi

Bmix

∙ (Z−1 )−ln (Z−1 )−Amix

2√2Bmix

∙( 2∑i

y i ∙ A ij

Amix

−Bi

Bmix)∙ ln( Z+(1+√2 )Bmix

Z+ (1−√2 )Bmix)

with Z= p ∙VRT

The compressibility is obtained by solving the cubic equation of state:Z3+∙ (Bmix−1 )∙ Z2+(Amix−3Bmix

2 −2Bmix )+(−Amix ∙Bmix+Bmix2 +Bmix

3 )=0

Aij=aij ∙ p

(RT )2Amix=

amix ∙ p

(RT )2

Bi=bi ∙ p

RTBmix=

bmix ∙ p

RT

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Fugacity of ethane in an ethane-butane mixture Calculate the fugacity of ethane (Et) in an ethane (Et) – n-butane (Bu) mixture at 373.15 K and 1 bar, 10 bar, 50 bar, 100 bar, 500 bar as a function of the mole fraction of ethane in the mixture using the Lewis-Randall rule, the truncated virial equation of state, and the Peng-Robinson equation of state.

Data for the virial equation of state:BEt-Et = -1.15.10-4 m3/mol; BEt-Bu = -2.15.10-4 m3/mol; ; BBu-Bu = -4.22.10-4 m3/mol

33

Lewis-Randall rule: f Et= yEt ∙ f Et

The fugacity of ethane at 373.15 K can be calculated using the Sandler program “Peng-Robinson equation of state”. The calculated fugacity of pure ethane at 373.15K is

p, bar 1 10 50 100 500fEt, bar 0.99571 9.5784 40.2516 65.1960 208.8461Et=fEt/p 0.99571 0.95784 0.80503 0.65196 0.41769

Truncated virial equation of state lnf i

y i ∙ P= 2 ∙ p

Z ∙ RT∙∑

j

y j ∙ Bij (T )−ln (Z )

∑j

y j ∙ Bij (T )= y Et ∙BEt−Et+ yBu ∙ BEt−Bu=BEt−Bu+ yEt ∙ (BEt−Et−BEt−Bu )

∑j

y j ∙ Bij (T )=−2.15 ∙10−4+1.0∙10−4 ∙ y Et

Z=12 (1+√1+ 4 ∙Bmix ∙ p

RT ) Bmix (T , x )=∑

i∑

j


Bmix (T , x )= y Et2 ∙BEt−Et+2∙ yEt ∙ yBu ∙BEt−Bu+ yBu

2 ∙BBu−Bu

Bmix (T , x )=−4.22∙10−4+4.14 ∙10−4 ∙ y Et−1.07 ∙10−4 ∙ yEt

2

It should be noted that this equation can only be used for pressures up to ca. 15 bar; higher

pressures will result in 1+4 ∙Bmix ∙ p

RT<0 and hence imaginary compressibilities. This is an indication

that at the higher pressures the higher order terms in the virial equation of state need to be taken into account.

Peng-Robinson equation of state:

lnf i

y i ∙ P=

Bi

Bmix

∙ (Z−1 )−ln (Z−1 )−Amix

2√2Bmix

∙( 2∑i

y i ∙ A ij

Amix

−Bi


Z+ (1−√2 )Bmix)

Hence, the fugacity of ethane in the mixture can be obtained using Sandler’s program “Peng-Robinson equation of state for mixtures”. The binary interaction parameter for the system ethane – n-butane is 0.010 (see p. 424)

Figure I.7.1 shows the fugacity coefficient for ethane in an ethane – n-butane mixture as a function of the mole fraction of ethane in the mixture. At low pressure the fugacity coefficient does not deviate significantly from 1, and the choice of the model to describe the fugacity of ethane in the mixture is immaterial. Even at 10 bar, the fugacity coefficient is still close to 1, and any one of the models can be used. At pressures in the range between 10 and 100 bar, significant deviations between the prediction using Lewis-Randall rule and the prediction from the Peng-Robinson equation of state are observed. However, at very high pressures the predicted values for the fugacity coefficient of ethane become similar. The question is always, which model is better. We cannot decide on the quality of the models without comparison to experiment. For this the experimental compressibility of the mixture at various compositions must be measured and compared with the model prediction (see Fig. 9.4-1).

34

0.95

0.96

0.97

0.98

0.99

1

1.01

1.02

0 0.2 0.4 0.6 0.8 1

Fu

ga

cit

y c

oe

ffic

ien

t,

f Et=

f Et/(

y Et. P

)

Mol fraction ethane, yEt

1 bar

10 bar

0.4

0.7

1

1.3

1.6

0 0.2 0.4 0.6 0.8 1

Fu

ga

cit

y c

oe

ffic

ien

t,

f Et=

f Et/(

y Et. P

)

Mol fraction ethane, yEt

50 bar

100 bar

500 bar

Figure I. 7.1:Fugacity coefficient of ethane in an ethane/n-butane mixture as predicted using the Lewis-Randall relationship (dashed lines), the truncated virial equation of state (dotted curves), and the Peng-Robinson equation of state (solid curves).

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Liquid mixturesThe fugacity of compounds in a liquid mixture can be obtained from an appropriate equation of state (e.g. Peng-Robinson equation of state), if the binary interaction parameter is known:

lnf i

y i ∙ P=

Bi

Bmix

∙ (Z−1 )−ln (Z−1 )−Amix

2√2Bmix

∙( 2∑i

y i ∙ A ij

Amix

−Bi


Z+ (1−√2 )Bmix)

with Z the compressibility of the liquids phase (i.e. the smallest root of the cubic equation).

For mixtures in which one or more components cannot be described by an equation of state, the fugacity of a component is described in terms of the activity coefficient. f i

L=x i ∙ γ i ∙ f iL

This then requires an appropriate model for the activity coefficient, or better for the excess Gibbs free energy.

Solid mixturesIn most cases, solids do not form solid mixtures, but separate domains with pure crystalline phases. Hence, a solid containing more than one component can be viewed as a collection of solid phases, each containing a pure compound. Hence, the fugacity of the solid component in this solid phase is the fugacity of the pure solid: f i

s=f is

It should however be noted that this is not true for all solids. Some solids can form mixtures, e.g. in alloys. In that case, the fugacity of a compound in this mixture (alloy) can be formulated as: f i

S=xi ∙ γi ∙ f is

This then requires an appropriate model for the excess Gibbs free energy to obtain the activity coefficient for the particular component in the mixture.

35

I.8 Activity coefficient models

I.8.1 Random mixturesThe excess Gibbs free energy for a binary mixture can be modeled using the Redlich-Kister equation:

Gex=x1 ∙ x2 ∙∑i=0

i=i

ai ∙ (x1−x2 )i

The excess Gibbs free energy is zero (and thus we are then dealing with an ideal solution), when a i is equal to 0 for all values of i. The use of the Redlich-Kister equation has as an underlying assumption that the microscopic structure of the mixture is identical to the macroscopic structure. This means that the mole fraction for compound j is a measure for the likelihood (probability) that the molecule i is surrounded by molecules j.

A simple mixture might be modeled with a0 not equal to zero, but all other constants a i equal to zero for I larger than 0: Gex=a0 ∙ x1 ∙ x2

The activity coefficient for component 1 is now given by:

RT ∙ ln (γ 1 )=G1ex=( ∂ N ∙Gex

∂N 1)T , p , N 2

=( ∂ (a0 ∙ x1 ∙ N2 )∂ N1

)T , p , N2

=a0 ∙ x22

and for component 2


∂N 2)T , p , N 1

=( ∂ (a0 ∙ N1 ∙ x2 )∂ N2

)T , p , N2

=a0 ∙ x12

These equations are called the one-constant Margules equations.

A more complex mixture might be modeled with a0 and a1 unequal zero and ai equal to zero for i larger than 1: Gex=x1 ∙ x2 ∙ (a0+a1 ∙ (x1−x2) ) Gex=x1 ∙ (1−x1) ∙ (a0+a1 ∙ (2 ∙ x1−1 ))


∂N 1)T , p , N 2

=( ∂ ( x1 ∙N2 ∙(a0+a1 ∙ (2 ∙ x1−1 ) ))∂ N1

)T , p , N2

RT ∙ ln (γ 1 )=( ∂ ( x1 ∙ N2 ∙ (a0+a1 ∙ (2∙ x1−1 )))∂x1

∙∂ x1∂ N1

)T , p , N2

( ∂ (x1 ∙N2 ∙(a0+a1∙ (2 ∙ x1−1 )) )∂ x1 )

T , p , N 2

=N2 ∙ (a0+a1 ∙ (2∙ x1−1 ))+2∙ a1 ∙ x1 ∙N2

( ∂ x1∂ N1

)T , p , N2

=N2

(N1+N2 )2

RT ∙ ln (γ 1 )=x22 ∙ (a0+a1 ∙ (2∙ x1−1 ))+2 ∙ a1 ∙ x1 ∙ x22

RT ∙ ln (γ 1 )=(a0 ∙ x22+a1 ∙ x22 ∙ ( x1−x2 ))+2∙ a1 ∙ (1−x2 ) ∙ x22

RT ∙ ln (γ 1 )=(a0+3 ∙ a1 )∙ x22−4 ∙ a1 ∙ x23

36


∂N 2)T , p , N 1

=( ∂ (N 1∙ x2 ∙(a0+a1 ∙ (1−2∙ x2) ))∂ N2

)T , p , N1

RT ∙ ln (γ 1 )=(a0−3 ∙ a1 ) ∙ x22+4 ∙ a1 ∙ x13

Van Laar equationThe van Laar equation has its origins in the van der Waals equation and takes into account the volume occupied by each molecule. It is assumed that binary mixtures are composed of species of similar size and similar energies of interaction, and that the van der Waals equation of state is applicable to both the mixture and the pure fluids. The implicit assumption here is that the molecules of each species are uniform distributed throughout the mixture (i.e. random mixture).

Since the molecules are of similar size, it can be assumed that ∆mixV=0 or V ex=0and since the molecules are randomly mixed:

∆mixS=−R ∙ (x1∙ ln (x1 )+x2 ∙ ln (x2 )) or Sex=0

Thus, the molar excess Gibbs free energy is given by: Gex=U ex−T ∙ Sex+ p ∙V ex=U ex=∆mixU

In order to evaluate the excess molar internal energy, we start with a cycle:Step 1: Start with two pure liquids, x1 moles of liquid 1 and x2 moles of liquid 2, and decrease the

pressure to evaporate each liquid to an ideal gasStep 2: Mix the two ideal gases to form an ideal gas mixtureStep 3: Compress the ideal gas mixture to form a liquid at the starting pressure P

The change in the internal energy associated with step 1:

∆U step1=x1 ∙∫V 1

∞

( ∂U∂V )T

∙ dV +x2 ∙∫V 2

∞

( ∂U∂V )T

∙ dV


∞

(T ∙( ∂ p∂T )

V

−p) ∙ dV +x2 ∙∫V 2

∞

(T ∙( ∂ p∂T )

V

−p)∙ d V


∞

( a1V 2 )∙ d V +x2 ∙∫V 2

∞

( a2V 2 ) ∙d V=x1 ∙a1V 1

+x2 ∙a2V 2

The change in internal energy associated with step 2 ∆U step 2=0since an ideal gas mixture is formed!

The change in internal energy associated with step 3 can be obtained in the same manner as for step 1:

∆U step3=−amix

V mixture

Hence, the excess molar Gibbs free energy is then given by:

37

Gex=U ex=x1 ∙a1V 1

+x2 ∙a2V 2

−amix

V mixture

In liquids, molecules are closely packed and the molar volume is (approximately) the volume taken in by the molecule, which is represented by the factor b in the van der Waals equation of state:

Gex=x1 ∙a1b1

+x2 ∙a2b2

−amix

bmixture

Based on statistical mechanics:amix=2 ∙ x1 ∙ x2 ∙√a1 ∙ a2 with k12 =0bmix=x1∙ b1+x2 ∙b2

Gex=x1 ∙a1b1

+x2 ∙a2b2

−2 ∙ x1 ∙ x2∙√a1 ∙ a2x1 ∙ b1+x2∙ b2

=2 ∙ x1 ∙ x2 ∙b1 ∙ b2x1 ∙ b1+x2 ∙ b2

∙(√a1b1−

√a2b2 )

2


RT ∙ ln (γ 1 )=G1

ex=( ∂ N ∙Gex

∂N 1)T , p , N 2

=( ∂(N ∙2∙ x1 ∙ x2 ∙ b1 ∙ b2x1∙ b1+x2 ∙b2

∙(√a1b1−√a2

b2 )2

)∂N 1

)T , p , N2

RT ∙ ln (γ 1 )=( ∂( 2∙ x1 ∙N 2∙ b1 ∙b2

x1 ∙b1+x2 ∙ b2∙(√a1b1

−√a2b2 )

2

)∂ N1

)T , p , N2

RT ∙ ln (γ 1 )=2 ∙N 2 ∙ b1 ∙ b2 ∙(√a1b1

−√a2b2 )

2

∙( ∂( x1 ∙ b1x1 ∙ b1+x2 ∙b2 )

∂ x1∙∂ x1∂ N1

)T , p , N2

RT ∙ ln (γ 1 )=2 ∙N 2 ∙ b1 ∙ b2 ∙(√a1b1−

√a2b2 )

2

∙b1 ∙b2

(x1 ∙ b1+x2∙ b2 )2∙

N 2

(N 1+N 2)2

RT ∙ ln (γ 1 )=2 ∙ b1 ∙ b2 ∙(√a1b1−

√a2b2 )

2

∙b1 ∙ b2 ∙ x2

2

(x1 ∙b1+ x2 ∙ b2 )2

RT ∙ ln (γ 1 )=2 ∙(√a1b1−

√a2b2 )

2

∙(b1 ∙b2 ∙ x2)

2

(x1 ∙b1+x2 ∙b2 )2

RT ∙ ln (γ 1 )=

2RT

∙(√a1b1−

√a2b2 )

2

∙ b12

(1+ b1b2

∙x1x2 )

2

In its general form the van Laar equation looks like:

38

ln ( γ1 )= α

(1+ αβ∙x1x2 )

2 ln ( γ2 )= β

(1+ βα∙x1x2 )

2

The parameters, a and b, can in principal be estimated from the van der Waals equation of state. However, few fluids can be accurately predicted using this equation of state, and hence these parameters are often estimated empirically. The coefficients and can be obtained experimentally from a measured activity coefficient for each component at a particular composition

α=(1+ x2 ∙ ln ( γ2 )x1 ∙ ln ( γ1 ) )

2

∙ ln (γ 1 ) β=(1+ x1 ∙ ln ( γ1 )x2 ∙ ln ( γ2 ) )

2

∙ ln (γ 2 )

Regular solution theoryNot many liquids and liquid mixtures obey the van der Waals equation of state. The basic assumptions of the van Laar model (i.e. Vex and Sex are zero) is however a good approximation for many mixtures. It was suggested to use the experimental change in the internal energy upon evaporation rather than an equation of state (G. Scatchard, Chem. Rev. 8 (1931), 321). Hence, the molar excess Gibbs free energy is given by:

Gex=U ex=x1 ∙∆vapU 1+ x2 ∙∆vapU 2−∆vapU mixture

(the change in the internal energy upon evaporation can easily be obtained from the enthalpy change upon evaporation, since U=H−p ∙V ≈ H−RT (for the evaporation yielding a gas at low pressure; note that the enthalpy upon evaporation at the temperature of interest must be used!). The change of the internal energy of the system upon evaporation can be approximated by:

∆vapUmixture=(x1 ∙√V 1 ∙∆vapU 1

V mixture

+x2∙√V 2 ∙∆vapU 2

V mixture)2

Gex=U ex=x1 ∙∆vapU 1+ x2 ∙∆vapU 2−(x1 ∙√V 1 ∙∆vapU 1

V mixture

+x2 ∙√V 2 ∙ ∆vapU 2

V mixture)2

Gex=x1 ∙∆vapU 1+x2 ∙∆vapU 2−x12 ∙V 1 ∙∆vapU 1

V mixture

−2∙ x1∙ x2V mixture

∙√V 1 ∙ ∆vapU 1V 2 ∙∆vapU 2−x22 ∙V 2 ∙∆vapU 2

V mixture

Gex=∆vapU 1 ∙( x1 ∙ x2 ∙V 2

V mixture)+∆vapU 2 ∙( x2 ∙ x1 ∙V 1

V mixture)−2 ∙ x1 ∙ x2V mixture

∙√V 1 ∙∆vapU 1 ∙V 2 ∙∆vapU 2

Gex=( x1 ∙ x2 ∙V 1 ∙V 2

V mixture)∙(∆vapU 1

V 1

+∆vapU 2

V 2

−2∙√ ∆vapU 1

V 1

∙∆vapU 2

V 2)

Gex=( x1 ∙ x2 ∙V 1 ∙V 2

V mixture)∙(√∆vapU 1

V 1

−√∆vapU 1

V 1)2

Defining the volume fraction of component 1 and 2 in the solution:

Φ1=x1∙V 1

V mixture

Φ2=x2∙V 2

V mixture

and Hildebrandt solubility parameter is defined as

39

δ 1=√ ∆vapU 1

V 1

δ 2=√ ∆vapU 2

V 2

Gex= (x1 ∙V 1+x2 ∙V 2 ) ∙Φ1 ∙Φ2 ∙ (δ 1−δ 2)2



∂N 1)T , p , N 2

=( ∂ (N ∙ (x1 ∙V 1+ x2 ∙V 2) ∙Φ1 ∙Φ2∙ (δ 1−δ 2 )2 )∂ N1

)T , p , N 2

RT ∙ ln (γ 1 )=(δ1−δ2 )2 ∙( ∂ ( (N1∙V 1+N2 ∙V 2 ) ∙Φ1 ∙Φ2 )∂ N1

)T , p , N2

RT ∙ ln (γ 1 )=(δ1−δ2 )2 ∙(Φ1 ∙Φ2 ∙∂ ( (N1 ∙V 1+N2 ∙V 2 ))

∂N 1

+(N1 ∙V 1+N 2 ∙V 2 )∙∂ (Φ1 ∙Φ2 )

∂ N1)T , p , N2

( ∂ ((N1 ∙V 1+N2 ∙V 2) )∂ N1

)T , p , N2

=V 1

( ∂ (Φ1 ∙Φ2 )∂ N1

)T , p , N2

=( ∂ (Φ1 ∙Φ2 )∂ x1

∙∂ x1∂N 1

)T , p ,N 2

( ∂ (Φ1 ∙Φ2 )∂ x1 )

T , p , N2

=Φ1 ∙( ∂Φ2

∂x1 )T , p , N2

+Φ2 ∙( ∂Φ1

∂ x1 )T , p , N2

( ∂Φ2

∂ x1 )T , p , N 2

=( ∂( (1−x1 )∙V 2

x1 ∙V 1+(1−x1 ) ∙V 2)

∂x1)T , p ,N 2

=−V 2 ∙ (x1 ∙V 1+(1−x1 ) ∙V 2)−(1−x1 ) ∙V 2∙ (V 1−V 2 )

(x1 ∙V 1+(1−x1) ∙V 2 )2

( ∂Φ2

∂ x1 )T , p , N 2

=−V 1∙V 2

( x1 ∙V 1+(1−x1 )∙V 2 )2

( ∂Φ1

∂ x1 )T , p , N2

=( ∂( x1 ∙V 1

x1 ∙V 1+(1−x1 ) ∙V 2)

∂x1)T , p ,N 2

=V 1 ∙ (x1 ∙V 1+(1−x1 ) ∙V 2 )−x1 ∙V 1 ∙ (V 1−V 2 )

( x1 ∙V 1+(1−x1) ∙V 2 )2

( ∂Φ1

∂ x1 )T , p , N2

=V 1 ∙V 2

( x1 ∙V 1+(1−x1 )∙V 2 )2

( ∂ (Φ1 ∙Φ2 )∂ x1 )

T , p , N2

=V 1 ∙V 2

( x1 ∙V 1+(1−x1) ∙V 2 )2 ∙ (Φ2−Φ1 )

( ∂ (Φ1 ∙Φ2 )∂ N1

)T , p , N2

=V 1 ∙V 2

( x1 ∙V 1+(1−x1) ∙V 2 )2 ∙ (Φ2−Φ1 ) ∙

x2N1+N2

=V 1 ∙Φ2

( x1∙V 1+ (1−x1 )∙V 2 )∙

(Φ2−Φ1 )N 1+N 2

RT ∙ ln (γ 1 )=(δ1−δ2 )2 ∙ (Φ1∙Φ2 ∙V 1+V 1 ∙Φ2 ∙ (Φ2−Φ1 ))

40

RT ∙ ln (γ 1 )=(δ1−δ2 )2 ∙Φ22∙V 1

Similarly for compound 2: RT ∙ ln (γ 2 )=(δ1−δ2 )2 ∙Φ1

2∙V 2

Extension of regular solution theory to multi-component mixtures: RT ∙ ln (γ 1 )=(δ1−δ )2 ∙Φ2

2 ∙V 1

with δ=∑j

Φ j ∙ δ j (i.e. use a volume average solubility parameter with Φ j=x j ∙V j

∑k

xk ∙V k

Note: The heat of evaporation must be taken at the temperature at which the activity coefficient needs to be determined (which is usually not equal to the normal boiling temperature)

∆vap H (T )=∫T

T boil

c p , liquid ∙ dT+∆vapH (T boil )−∫Tboil

T

c p , vapour ∙ dT

I.8.2 Non-random mixturesThe local composition in a mixture is not necessarily the same as the macroscopic composition. For instance, a molecule might be highly solvated, i.e. preferentially surrounded by the solvent molecules, in which case the dissolved molecules will only experience the interaction with the solvent molecules (up to some limiting concentration). In general, non-random mixtures can be formed due to the energetics or interaction and/or difference in the size of the molecules involved.

Wilson equationThe Wilson equation is a two-parameter model, which is based on the following expression for the excess molar Gibbs free energy:

Gex

RT=−x1 ∙ ln (x1+ Λ12 ∙ x2)−x2 ∙ ln (x2+ Λ21 ∙ x1 )

From which the activity coefficient is

ln ( γ1 )=−ln (x1+ Λ12 ∙ x2)+x2 ∙( Λ12x1+x2 ∙ Λ12

−Λ21

x1 ∙ Λ21+x2 )ln ( γ2 )=−ln (x2+ Λ21 ∙ x1)−x1 ∙( Λ12

x1+ x2 ∙ Λ12−

Λ21x1∙ Λ21+ x2 )

This model can be fitted to experimental data to obtain the fitting parameters 12 and 21.

NRTL modelThe non-random two liquid (NRTL) model is based on a slightly different formulation of the molar excess Gibbs free energy, which involves three parameters:

Gex

RT=x1∙ x2 ∙( τ21 ∙G 21

x1+G21 ∙ x2+

τ12 ∙G12

x2+G 12 ∙ x1 ) with G21=e−α ∙ τ21 and G12=e−α ∙ τ12

This model requires 3 parameters to be fitted to experimental data

41

Flory-Huggins modelMixtures of polymers and their solvents (e.g. n-hexane) have molecules with vastly different sizes. Due to the size difference, the entropy change upon mixing is not approximately zero. The change in the entropy upon mixing is controlled rather by the volume fraction of the various species in the solution rather than the mole fraction:

∆mixS=−R ∙ (x1∙ ln (Φ1 )+x2 ∙ ln (Φ2 )) with Φ1=x1 ∙ v1

x1 ∙ v1+x2 ∙ v2=

x1x1+ x2 ∙m

with Φ2=x2 ∙ v2

x1 ∙ v1+x2 ∙ v2=

x2 ∙m

x1+ x2 ∙m

(the parameter m is here the volume ratio of species 2 relative to species 1).

Hence, the excess molar entropy is given by:

Sex=∆mix S−∆mixS

ℑ=−R ∙(x1 ∙ ln(Φ1

x1 )+x2∙ ln(Φ2

x2 ))The enthalpy of mixing can be expressed by a one-constant term in terms of the volume fraction of the species in solution:H ex=∆mix H= χ ∙RT ∙ (x1+m∙x2 ) ∙Φ1 ∙Φ2

Thus, the molar excess Gibbs free energy is given by:

Gex

RT=

H ex−T ∙Sex

RT= χ ∙ ( x1+m ∙x2 ) ∙Φ1 ∙Φ2+(x1∙ ln(Φ1

x1 )+ x2 ∙ ln(Φ2

x2 ))The activity coefficient is defined as:


∂N 1)T , p , N 2

=( ∂(N ∙ χ ∙ ( x1+m ∙x2 ) ∙Φ1∙Φ2+N ∙(x1 ∙ ln(Φ1

x1 )+x2 ∙ ln(Φ2

x2 )))∂N 1

)T , p , N2

( ∂ (N ∙ χ ∙ (x1+m∙ x2) ∙Φ1 ∙Φ2 )∂N 1

)T , p , N2

=( ∂ ( χ ∙ (N1+m∙ N2 ) ∙Φ1 ∙Φ2 )∂ N1

)T , p , N2

( ∂ ( χ ∙ (N1+m∙ N2 ) ∙Φ1 ∙Φ2 )∂ N1

)T , p ,N 2

= χ ∙Φ1∙Φ2+ χ ∙ (N1+m ∙N2 ) ∙( ∂ (Φ1 ∙Φ2 )∂ N1

)T , p , N2

( ∂ (Φ1 ∙Φ2 )∂ N1

)T , p , N2

=v1∙Φ2

(x1 ∙ v1+x2 ∙ v2 )∙

(Φ2−Φ1 )N1+N2

(see regular solution theory)

( ∂ (Φ1 ∙Φ2 )∂ N1

)T , p , N2

=Φ2

(x1+m∙ x2 )∙

(Φ2−Φ1 )N1+N2

( ∂ (N ∙ χ ∙ (x1+m∙ x2) ∙Φ1 ∙Φ2 )∂N 1

)T , p , N2

= χ ∙Φ1 ∙Φ2+ χ ∙ (x1+m∙x2 ) ∙Φ2

(x1+m∙ x2 )∙ (Φ2−Φ1 )

( ∂ (N ∙ χ ∙ (x1+m∙ x2) ∙Φ1 ∙Φ2 )∂N 1

)T , p , N2

= χ ∙Φ1 ∙Φ2+ χ ∙Φ2 ∙ (Φ2−Φ1 )

42

( ∂ (N ∙ χ ∙ (x1+m∙ x2) ∙Φ1 ∙Φ2 )∂N 1

)T , p , N2

= χ ∙Φ22

( ∂(N ∙(x1 ∙ ln(Φ1

x1 )+x2 ∙ ln(Φ2

x2 )))∂N 1

)T , p , N 2

=( ∂(N 1∙ ln(Φ1

x1 )+N 2∙ ln(Φ2

x2 ))∂ N1

)T , p , N 2

( ∂(N 1 ∙ ln(Φ1

x1 ))∂N 1

)T , p , N2

=ln(Φ1

x1 )+N 1 ∙( ∂ ln(Φ1

x1 )∂x1

∙∂ x1∂ N1

)T , p , N2

( ∂ ln(Φ1

x1 )∂x1

)T , p , N2

=( ∂ ln( 1x1+m∙x2 )∂ x1

)T , p , N 2

=−( ∂ ln (x1+m∙ x2 )∂ x1 )

T , p , N2

=−(1−m )x1+m∙ x2

( ∂(N 1 ∙ ln(Φ1

x1 ))∂N 1

)T , p , N2

=ln(Φ1

x1 )−x1 ∙ x2 ∙(1−m )x1+m∙ x2

( ∂(N 2 ∙ ln(Φ2

x2 ))∂N 1

)T , p , N2

=N2∙( ∂ ln(Φ2

x2 )∂ x1

∙∂ x1∂N1

)T , p , N 2

( ∂ ln(Φ2

x2 )∂x1

)T , p , N2

=( ∂ ln( mx1+m∙x2 )∂ x1

)T , p , N 2

=−( ∂ ln( x1m+ x2)∂x1

)T , p , N2

=−( 1m−1)x1m

+x2

( ∂(N 2 ∙ ln(Φ2

x2 ))∂N 1

)T , p , N2

=−x22 ∙

( 1m−1)x1m

+x2

( ∂(N ∙(x1 ∙ ln(Φ1

x1 )+x2 ∙ ln(Φ2

x2 )))∂N 1

)T , p , N 2

=ln(Φ1

x1 )+ x1 ∙ x2 ∙−(1−m )x1+m∙ x2

−x22∙

( 1m−1)x1m

+x2

43

( ∂(N ∙(x1 ∙ ln(Φ1

x1 )+x2 ∙ ln(Φ2

x2 )))∂N 1

)T , p , N 2

=ln(Φ1

x1 )−x1 ∙ x2 ∙(1−m )x1+m∙ x2

+x22∙

(1−m )x1+m ∙x2

( ∂(N ∙(x1 ∙ ln(Φ1

x1 )+x2 ∙ ln(Φ2

x2 )))∂N 1

)T , p , N 2

=ln(Φ1

x1 )−x2 ∙(1−m )

x1+m ∙x2∙ (x1+x2 )

( ∂(N ∙(x1 ∙ ln(Φ1

x1 )+x2 ∙ ln(Φ2

x2 )))∂N 1

)T , p , N 2

=ln(Φ1

x1 )+Φ2(1− 1m )

ln ( γ1 )= χ ∙Φ22+ ln(Φ1

x1 )+Φ2(1− 1m )

(with the first term describing the contribution of the excess enthalpy and the latter two terms describing the contribution of the excess molar entropy to the activity coefficient)

Similarly,

ln ( γ2 )=m∙ χ ∙Φ12+ ln(Φ2

x2 )−Φ1 (m−1 )

UNIQAC modelThe universal quasi-chemical (UNIQAC) model is based on statistical mechanics and allows a different local composition based on differences of size and energy between the molecules in the mixtures. The basic idea centers on the additive properties of energy. The contribution of the size difference and the difference in energetics to the molar excess Gibbs free energy is separated into a combinatorial term (mainly for size differences) and a residual term (mainly for the energy differences):Gex

RT=G ex (combinatorial )

RT+Gex (residual )

RT

The combinatorial part of the excess Gibbs free energy is expressed in terms of volume fraction:Gex ( combinatorial )

RT=∑

i

x i ∙ ln(Φ i

x i)+ z2∙∑

i

x i ∙ q i ∙ ln( θ i

Φ i)

The first term in this expression is familiar and originates from the entropic contribution to the Gibbs free energy (see Flory-Huggins model). The second term expresses the direct neighbor-neighbor interaction in the mixture.

The volume fraction in a molecule can be thought to originate from the various segments, which each contribute to the size of a molecule (Group Contribution Method). In a mixture, each species can be thought to be made up from various segments. The volume fraction of a species is then given by:

44

Φ i=x i ∙ ri

∑i

x i ∙ ri

(the contributing volume parameters ri are given in Table 9.5-2)

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Using the group contribution method to estimate the volume fraction Determine the volume fraction of benzene in a mixture of benzene and 2,2,4 trimethylpentane as a function of the mole fraction of benzene in the mixture.

The volume parameter can be computed by breaking each molecule into its constituents. Benzene is made up by 6 aromatic CH-units (ACH) each contributing 0.5313. Hence, the volume parameter for benzene isrbenzene=6 ∙ r ACH=6 ∙0.5313=3.1878

2,2,4 trimethylpentane constitutes of 5 CH3-groups, 1 CH2-group, 1 CH-group and 1 C-group. Hence, the volume parameter for 2,2,4 trimethylpentane is

r2,2,4 trimethylpentane=5 ∙ rCH 3+rC H 2

+rCH+rC=5 ∙0.9011+0.6744+0.4469+0.2195=5.8463Thus, the volume fraction of benzene in the mixture is given by:

Φbenzene=xbenzene ∙3.1878

xbenzene ∙3.1878+x2,2,4trimethylpentane ∙5.8463

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

The second term contributing to the combinatorial excess Gibbs free energy is determined by the number of neighboring molecules, z (a closed packing can have between 8 and 12 nearest species; z is typically taken to be 10) and the surface area parameter for a species. The term i is the area fraction of species i and is defined as:

θi=x i ∙ qi

∑i

x i ∙q i

The surface area parameter is also thought to be made up from the contributing factors of the segments within a molecule.

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Using the group contribution method to estimate the surface area fraction Determine the surface area fraction of benzene in a mixture of benzene and 2,2,4 trimethylpentane as a function of the mole fraction of benzene in the mixture.

The surface area parameter can be computed by breaking each molecule into its constituents. Benzene is made up by 6 aromatic CH-units (ACH) each contributing 0.5313. Hence, the surface area parameter for benzene isqbenzene=6 ∙ q ACH=6 ∙0.4=2.42,2,4 trimethylpentane constitutes of 5 CH3-groups, 1 CH2-group, 1 CH-group and 1 C-group. Hence, the surface area parameter for 2,2,4 trimethylpentane is

q2,2,4trimethylpentane=5 ∙qCH 3+qCH 2

+qCH+qC=5 ∙0.8480+0.5400+0.2280+0=5.0080Thus, the surface area fraction of benzene in the mixture is given by:

θbenzene=xbenzene ∙2.4

xbenzene ∙2.4+x2,2,4 trimethylpentane ∙5.0080••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

45

Thus the combinatorial contribution to the excess Gibbs free energy for the system benzene (B)-2,2,4 trimethylpentane (TMP) is given by:Gex ( combinatorial )

RT=xB ∙ ln( 3.1878

x B∙3.1878+xTMP ∙5.8463 )+xTMP ∙ ln( 5.8463xB ∙3.1878+xTMP ∙5.8463 )+5 ∙ xB∙2.4 ∙ ln( θB

ΦB)+ z2∙ xTMP ∙ qTMP ∙ ln( θTMP

ΦTMP)

Gex ( combinatorial )RT

=xB ∙ ln(ΦB

x B)+xTMP ∙ ln(ΦTMP

xTMP)+ z2∙ xB ∙ qB∙ ln ( xB ∙3.1878+xTMP ∙5.8463

xB ∙2.4+xTMP ∙5.0080∙2.43.1878 )+5 ∙ xTMP ∙5.0080 ∙ ln( xB ∙3.1878+ xTMP ∙5.8463

xB ∙2.4+xTMP ∙5.0080∙5.00805.8463 )

The residual contribution to the excess Gibbs free energy is a function of the surface area contribution and the interaction energy between the various species (uij) in the mixture:Gex (residual )

RT=∑

i

q i ∙ xi ∙ ln (∑j

θ j ∙ τ ji) withτ ji=−u ij−u jj

RT

A consequence of the splitting of the excess Gibbs free energy into two terms is that the activity coefficient can also be viewed as having two contributions:ln γi=ln γi (combinatorial )+ ln γi (residual )

ln γi (combinatorial )=ln(Φi

xi )− z2∙q i ∙ ln(Φ i

θ i)+li−Φi

x i

∙∑j

x j ∙l j

with li=z2∙ (ri−q i )−( r i−1 )

ln γi (residual )=qi ∙(1−ln (∑j

θi ∙ τ ji)−∑j

θi ∙ τ ji

∑k

θk ∙ τ kj )The UNIQAC model is a (very advanced), empirical model, since the parameters ij still need to be determined from experiment.

UNIFAC modelThe UNIFAC (UNIqac Functional group Activity Coefficient) model is based on the UNIQAC model. The combinatorial contribution to the activity coefficient for the UNIFAC model is slightly different from the UNIQAC model

ln γi (combinatorial )=ln(Φ'ixi )+1−Φ' i

x i

−z2∙q i ∙(1+ ln(Φi

θi)−Φi

θi )with Φ' i=( xi ∙ ri

0.75

∑i

x i ∙ ri0.75 )

The residual term is in terms of the interaction parameter (interaction energy), which is now also evaluated in terms of a group contribution method.

COSMO-RS modelThe COSMO-RS (Conductor-like Screening Model for Real Solvents’-) model uses quantum-chemical models to predict the electro-static interaction energy, hydrogen bonding and van der Waals interactions between components in the mixture (see e.g. A. Klamt, F. Eckert, Fluid Phase Equil. 172

46

(2000), 43). In principle COSMO-RS is analogous to UNIFAC, in that in both models an ensemble of pair-wise interacting surface segments is considered. The description of the interaction energy (important for the residual contribution to the activity coefficient) is different. COSMO-RS uses the actual determined screening charge density as a measure for the interaction energy, whereas in UNIFAC the interaction of between groups is determined as an average value.

47

I.9 Fugacity of species in a non-simple mixture

The fugacity of a species in a mixture is defined in terms of the fugacity of the same species in the same phase as a pure compound at the same temperature and pressure of the mixture.Gas phase f i

v (T , p , y )= yi ∙ f iv (T , p ) Lewis-Randall rule

Liquid phase f iL (T , p , x )=x i ∙ γ i (T ,x ) ∙ f i

L (T , p )

Vapour phase fugacity of pure componentA compound in a mixture at a specific temperature and pressure may be normally in the liquid phase. For instance, water at 298.15K and 1 bar is in its liquid form. The fugacity of water as a vapour can be obtained from

f iv (T , p )=p ∙( f ivp ) with ln ( f ivp )= 1

RT∫p=0

p

(V−V IG )∙ dp

At low pressure the fugacity of the pure component at the temperature and pressure of the mixture can be approximated by:

f iv (T , p )≈ p with p the pressure of the system

Liquid phase fugacity of pure component

Dissolution of a solid in a liquidThe normal state of aggregation of the pure compound may be the solid state. This is commonly known as the dissolution process. The ratio of the fugacity of a component as a liquid relative to the fugacity of a component as a solid is given by:

ln ( f iL (T , p )

f is (T , p ) )=Gi

L (T , p )−Gis (T , p )

RT=

∆ fusG (T , p )RT

Hence, the fugacity of the component as a liquid can be estimated knowing the fugacity of the pure component as a solid and the change in the Gibbs free energy of fusion at the temperature and pressure of the mixture. The fusion and melting process are hardly dependent on pressure, and only the temperature change will be considered. At its normal fusion/meltingpoint ∆ fusG (T melt , p )=0. A three step process is used to determine the change in the Gibbs free energy upon fusion at the temperature and pressure of the mixture.

1. Heating the pure component from the temperature of the mixture to the melting temperature

2. Melting the pure component at its melting temperature3. Cooling the liquid formed down to the temperature of the mixture

Enthalpy change for this three step process:

∆ fusH (T mixture )= ∫T mixture

Tmelt

cps ∙ dT+∆ fusH (T melt )+ ∫

T melt

Tmixture

c pL ∙ dT

∆ fus S (T mixture )= ∫T mixture

Tmelt c ps

T∙dT +∆ fusS (T melt )+ ∫

T melt

Tmixture c pL

T∙dT

Defining the difference between the specific heat of the pure compound as a liquid and the specific heat as a pure component as a solid as Cp:

∆ fusH (T mixture )=∆fusH (T melt )+ ∫Tmelt

T mixture

∆Cp ∙ dT

48

∆ fus S (T mixture )=∆fus S (T melt )+ ∫T melt

T mixture ∆C p

T∙dT

At the normal melting temperature 0=∆ fusG (Tmelt , p )=∆ fusH (T melt )−T ∙∆ fusS (T melt )

or ∆ fus S (T melt )=∆ fus H (T melt )

T melt

Substituting it all in

∆ fusG (Tmixture , p )=∆ fusH (Tmixture )−T ∙∆mixtureS (Tmelt )=∆ fusH (Tmelt ) ∙(1− TTmelt

)+ ∫T melt

Tmixture

∆C p∙ dT + ∫Tmelt

Tmixture ∆C p

T∙dT

For many substances, the specific heat of the solid is approximately the same as the specific heat of the liquid:

∆ fusG (T mixture , p )≈∆ fusH (T melt ) ∙(1−T mixture

T melt)

And the fugacity coefficient of the pure component as a liquid is thus given by:

f iL (T mixture , p )=f i

s (T mixture , p ) ∙ e∆fusH (Tmelt )

R∙( 1Tmixture

−1

T melt )

Dissolving a vapour in a liquidThe pure compound may exist at the temperature and pressure of the mixture as a vapour. The fugacity of the pure component as a liquid at this temperature and pressure, if the pressure is not too high, is then represented by the vapour pressure of the pure compound even if the pressure of the system is lower than the vapour pressure of the pure compound:

f iL (T mixture , p )=pi

vap (T mixture ) can be higher than p!

Dissolving a gas in a liquidGases can be sparingly soluble in liquids (e.g. nitrogen is to some extent soluble in water). The mole fraction of the dissolved gas is then typically low. It has been experimentally observed that the fugacity of the dissolved gas in the mixture depends linearly on the mole fraction of that component in the mixture:

f iL (T mixture , p , x )= xi ∙H i (T mixture , p )

The proportionality factor is called the Henry coefficient (with the units bar/mole-fraction). The Henry coefficient can be interpreted as the fugacity of the pure hypothetical liquid.At higher mole fractions of the dissolved gas, a deviation from the linear behavior is typically observed, which can be accounted for by the introduction of a new (and differently defined!) activity coefficient: f i

L (T mixture , p , x )= xi ∙ γi¿ (T mixture , p , x ) ∙H i (T mixture , p )

The difference becomes clear by examining the limiting cases, i.e. this new activity coefficient i* goes to 1 for xi 0, whereas the usual activity coefficient i goes to 1 for xi 1. They are however linked sincef i

L (T mixture , p , x )= xi ∙ γi¿ (T mixture , p , x ) ∙H i (T mixture , p )=x i ∙ γi (Tmixture , p , x ) ∙ f i L (T mixture , p )

or γ i

¿ (T mixture , p , x )∙ H i (T mixture , p )=γ i (T mixture , p , x ) ∙ f i L (Tmixture , p )

49

The Henry coefficient can be found by taking xi 0, for which i* becomes 1:H i (Tmixture , p )=γ i (T mixture , p , x i=0 )∙ f iL (T mixture , p )

Thus,

γ i¿ (T mixture , p , x )=

γi (Tmixture , p , x)γ i (T mixture , p , x i=0 )

This approach can also be used to describe the fugacity of other sparingly soluble species in solution such as proteins.

50

I.10 Electrolyte solutions

Solutions containing ions pose another problem, since the fugacity of the ion as a pure liquid does not exist. All solution containing ions are electrically neutral, i.e. the negative charges balance out with the positive charges. For instance, in a dilute aqueous solution sulphuric acid dissociates completely:

H2SO4 2 H+ + SO42-

The solution is electro-neutral, i.e. the charges balance:nH +¿=2 ∙nS O4

2−¿¿¿

At high concentration the dissociation of sulphuric acid, the equilibrium between the protons and sulphate ions on the one side and the hydrogen sulphate ion on the other side becomes noticeable, i.e.

H2SO4 HSO4- + H+ 2 H+ + SO4

2-

The charge balance is given now by: nH +¿=2 ∙nS O4

2−¿+nHS O

4−¿¿¿

¿ (and the pH of the solution can be calculated knowing the concentration of sulphuric acid in solution, and the equilibrium constant for the dissociation of the hydrogen sulphate ion). In general, the charge balance of a solution containing cations with a charge z+ and anions with a charge z- is given by:z+¿ ∙ nA z +¿+z−¿ ∙n

Bz−¿=0¿

¿¿¿

The implication of the charge balance is that the anion and cation always go together. The Gibbs free energy of the mixture is in principle given in terms of the partial molar Gibbs free energy of its constituents (in an ionic solution, the solvent, S, the undissociated ionic compound, A +B-, the dissociated ions Az+ and Bz-):Gmixture=NS ∙G S+N A ν+ ¿B ν−¿ ∙G A

ν+¿

Bν−¿+ N

Az +¿

∙GAz

+¿+N

Bz−¿

∙GBz−¿

¿¿¿¿¿¿¿¿

Since we more mostly interested in electrolyte solutions with a low concentration, the activity coefficient of the ionic compounds and the ions can in principle be defined as:

Gi (T , p ,M 1 )=Gi∎ (T , p )+RT ∙ ln( γ i

∎∙ M i

M i=1molal )

Where Mi is the molality of the solution (units mole of solute per kg of solvent), Gi∎ (T , p ) is the

partial molar Gibbs free energy in an 1 molal ideal solution, and γ i∎ is the activity coefficient, which

approaches 1 when the molality goes to zero. According to this definition, the partial molar Gibbs free energy of the undissociated ionic compound and the ions is given by GA ν+ ¿B

ν−¿ ( T , p, M1)=G

Aν+¿B

ν−¿

∎ (T ,p ) +RT ∙ln ¿¿¿¿¿

GA z+¿

(T , p ,M 1 )=G A z +¿∎ (T , p)+RT ∙ ln¿ ¿¿

GB z−¿

(T , p ,M 1)=GB z−¿∎ (T , p )+RT ∙ ln ¿¿ ¿

However, the partial molar Gibbs free energy of the cation cannot be determined since the change in Gibbs free energy of the mixture upon changing the number of moles of the cation keeping the number of moles of the anion constant is physical nonsense, violating the principle of the electro-neutrality balance. Hence, the Gibbs free energy of the mixture is described in terms of the amount of dissociated ionic compound, NAB,D:Gmixture=NS ∙G S+N A ν+ ¿B ν−¿ ∙G A

ν+¿

Bν−¿+ N

AB, D∙G

AB ,D¿¿¿¿

Thus, N AB ,D ∙GAB ,D=NA z+ ¿

∙G A z +¿+N

Bz−¿

∙ GB z

−¿¿¿¿¿

51

GAB ,D=ν+¿ ∙ G

A z+¿+ν

−¿ ∙GBz−¿

¿¿

¿¿

Substituting the definition of the partial molar Gibbs free energy for the ions in terms of the activity coefficient:

GAB ,D=ν+¿ ∙ G

A z+¿∎ + ν

−¿ ∙GBz−¿

∎ +RT ∙ ln¿¿¿¿¿

Defining Mean ionic activity coefficient γ ±

¿¿ Mean ionic molality M ±

¿¿ Partial molar Gibbs free energy of the dissociated compound as an ideal 1 molal solution

GAB ,D∎=ν

+¿ ∙GA z+¿∎ + ν

−¿ ∙GBz−¿

∎ ¿¿¿¿

Thus,GAB ,D=GAB ,D

∎+RT ∙ ln ¿

The mean ionic coefficient for dilute solutions can be modeled using the Debye-Hückel limiting law, which is based on an electro-static model of ions in a solvent:

ln ( γ± )=−α ∙¿ Here I is the ionic strength of a solution containing cations with a charge of z + and anions with a charge of z-. The ionic strength is defined as:

I=12∙∑ions

zi2 ∙M i

The parameter a is solvent dependent and is for water given in Table 9.10-1. It should be noted that the mean ionic activity coefficient is less than 1 and decreases with increasing ionic strength.

The Debye-Huckel limiting law is exact for low concentrations, but significant deviations are observed at concentrations higher than ca. 0.01 molal. Empirical corrections have been proposed to the limiting Debye-Huckel law:

ln ( γ± )=−α ∙¿¿ with a the average radius of the hydration sphere of the ions (typically taken as 4Å) and the parameter for aqueous solutions given in Table 9.10-1.

For more concentrated solutions (ca. 0.5 molal and higher) a further empirical correction has been proposed to account for the increase in the mean ionic activity coefficient at high ionic strength

ln ( γ± )=−α ∙¿¿ with δ=0.1 ∙¿ It should however be noted that nowadays robust models for the description of electrolyte solutions at high ionic strength are available (K.S. Pitzer, J. Phys. Chem. 77 (1973), 268).

52

II Phase equilibria

II.1. IntroductionChemical processes produce typically mixtures, which have to be separated into streams of the required purity in order to obtain a useful product. Phase equilibria play an important role in these separation processes, which use transfer of a particular component into another phase as a method to separate mixtures. This often occurs in multiple steps (stages). Knowledge on the equilibrium position between the phases is a pre-requisite for the design of these processes.

In a phase equilibrium situation, each of the compounds in a mixture can be distributed over two (or more) phases, phase I and phase II. If the system is a closed, adiabatic, constant volume system: d N i=0 d N i

I=−d N iII

i.e. the decrease of number of moles of component i in phase I corresponds to an increase in the number of moles of component i in phase II dV=0 d V I=−d V II

dU=0 d U I=−d U II

In a closed, adiabatic, constant volume system, the entropy strives towards a maximum. This means that the change in the entropy is equal to zero dS=d S I+d S II=0

dS=d UT

+ pT∙dV−∑

i

C Gi

T∙d N i=0

dS=d S I+d S II=dU I

T I + d U II

T II + p I

T I ∙ d V I+ p I

T I ∙ d V I−∑i

C GiI

T I ∙ d N iI−∑

i

C GiII

T II ∙ d N iII=0

dS=( 1T I− 1

T II )∙ d U I+( pI

T I− p II

T II ) ∙ dV I−∑i

C (GiI

T I−G i

II

T II ) ∙ d N iI=0

Since, the differential of the entropy must be equal to zero with respect to all independent variables (i.e. dUI, dVI and dNi

I), we obtain 3 equilibrium conditions, viz. T I=T II temperature in each phase equal

p I=p II pressure in each phase equal Gi

I=GiII partial molar Gibbs free energy of each component in each phase equal

The same criteria can be derived for a system at constant temperature and pressure. The Gibbs free energy in such a system is at a minimum: dG=dG I+d GII=0

d G=−∑i

C (GiI

T I−

GiII

T II )∙ d N iI=0 (dT and dp are equal to zero)

T I=T II temperature in each phase equal (pre-condition of system)p I=p II pressure in each phase equal (pre-condition of system)

GiI=Gi

II partial molar Gibbs free energy of each component in each phase equal

The equilibrium condition can also be expressed in terms of the fugacity of the species in the mixture. The partial molar Gibbs free energy for each species must be equal between the phases, and hence

53

f iI=f i

II partial molar Gibbs free energy of each component in each phase equal

II.1.1 How many variables are needed to describe a multi-component mixture containing multiple phases? – Gibbs phase rule

A single phase system containing C components, has requires C+1 variables to describe it. The last variable is fixed due to the constraint of the Gibbs-Duhem equation. When the system contains P phases, would require P(C+1) number of variables (a full set of equations for each phase). However, at equilibrium

Temperature in each phase is the same instead of P temperatures we have only one temperature leading to a reduction in number of variables with P-1

Pressure in each phase is the same instead of P pressures we have only one pressure leading to a reduction in number of variables with P-1

Partial molar Gibbs free energy for each component in each phase is the same instead of CP partial molar Gibbs free energies we have only one partial molar Gibbs energy for each component leading to a reduction in number of variables with C(P-1)

Hence the total number of variables needed (or degree of freedom in the system, F) to describe a system at (phase) equilibrium: F=P ∙ (C+1 )−(P−1 )−(P−1 )−C ∙ (P−1 ) F=C−P+2

For instance, 1 variable is required to specify a system containing 1 component and 2 phases present. The transition of a liquid to a vapour for a pure component will occur at a certain temperature at a given pressure (the temperature cannot be freely chosen, when pressure has been chosen). Two variables are required to specify a system containing 2 components and 2 phases present. The description of the transition of a liquid to a vapour for a binary mixture requires two parameters to be specified (e.g. temperature/pressure or temperature/composition).

54

II.2. Vapour-liquid equilibrium (VLE) of ideal binary mixtures at low pressureThe conditions for a vapour-liquid equilibrium of a binary mixture are in addition to the:

f 1v=f 1

L f 2v=f 2

L and the temperature and pressure in each phase are equal.

At low pressure, vapours obey the Lewis-Randall rule. Hence the fugacity of a component in the vapour phase mixture is given by:

f 1v= y1 ∙ f 1

v f 2v= y2 ∙ f 2

v

Furthermore, at low pressure the fugacity of the pure component is equal to the system pressure:f 1v= y1 ∙ p f 2

v= y2 ∙ p

The fugacity of a component in the liquid phase is given by:f 1L=x1∙ γ 1∙ f 1

L f 2v=x2 ∙ γ2 ∙ f 2

L

The activity coefficient of in an ideal liquid mixture is equal to 1.f 1L=x1∙ f 1

L f 2v=x2 ∙ f 2

L

The fugacity of the pure, liquid component at low pressure is equal to the vapour pressure of the component:

f 1L=x1∙ p1

vap f 2v=x2 ∙ p2

vap

The equilibrium conditions for an ideal binary mixture at vapour-liquid equilibrium are thus:y1∙ p=x1 ∙ p1

vap y2 ∙ p=x2 ∙ p2vap

or p=x1 ∙ p1vap+ (1−x1 ) ∙ p2vap

y1=x1∙ p1

vap

x1 ∙ p1vap+(1− x1 ) ∙ p2

vap

Thus, for a given mole fraction in the liquid phase in a mixture at a given temperature, the pressure and the composition of the gas phase can be calculated. It should be noted that the pressure of the system depends linearly on the mole fraction of compound 1 in the liquid.

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example Low pressure p-x-y VLE-diagrams for ideal mixtures The vapour pressure of n-hexane (Hx) at 60oC is 0.7536 bar and the vapour pressure of n-heptane (Hp) at 60oC is 0.2766 bar. Calculate the equilibrium pressure for liquid mixtures of n-hexane (H) and n-heptane (Hp) at 60oC and the corresponding gas phase composition assuming that n-hexane (H) and n-heptane (Hp) form an ideal mixture.

From the equilibrium condition f iv=f i

L for an ideal mixture it is derived that

p=xHx ∙ pHxvap+(1−xHx) ∙ pHp

vap=pHpvap+xHx ∙ ( pHx

vap−pHpvap )=0.2766+0.477 ∙ xHx

yHx=xHx ∙ pHx

vap

p=

0.7536 ∙ xHx

0.2766+0.477 ∙ xHx

55

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

Mo

le f

rac

tio

n n

-he

xa

ne

in

va

po

ur

ph

ae

, yH

x

Mole fraction of n-hexane in liquid phase, xHx

T = 60oC

0.2

0.4

0.6

0.8

0 0.2 0.4 0.6 0.8 1

Pre

ss

ure

, ba

r

Mole fraction of n-hexane

Pressure as a function of mole fraction n-hexane

in liquid phase, xHx

Pressure as a function of mole fraction n-hexane

in vapour phase, xHx

T = 60oC

Liquid mixture

vapour mixture

Figure II.2.1: x-y diagram (left) and P-x-y diagram (right) for the system n-hexane/n-heptane at 60oC at vapour liquid equilibrium

In chemical processes, we often keep the pressure (approximately) constant and not the temperature. Hence, we are often interested in the position of the vapour-liquid equilibrium (VLE) at a constant pressure. The vapour pressure of the compounds is dependent on temperature, and is typically given by the Antoine equation. Hence, the temperature of the mixture in a vapour-liquid equilibrium at a given pressure must be solved from:

pivap=10

Ai−Bi

T+C i Antoine equation

p=x1 ∙10A 1−

B1T+C1+(1−x1) ∙10

A2−B2

T+C2

y1=x1 ∙10

A1−B1

T +C1

p with p the system pressure

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example Low pressure T-x-y VLE-diagrams for ideal mixtures

The vapour pressure of n-hexane (Hx) is given by pHxvap (mmHg )=10

6.88555− 1175.817T (℃ )+224.867 and the vapour

pressure of n-heptane is given by pHpvap (mmHg )=10

6.90253− 1267.828T (℃ )+216.823 . Calculate the equilibrium

temperature for liquid mixtures of n-hexane (Hx) and n-heptane (Hp) at a pressure of 0.5 bar and the corresponding gas phase composition assuming that n-hexane (Hx) and n-heptane (Hp) form an ideal mixture.

The procedure is the same as for the calculations of the P-x-y diagram, although the calculations now require solving a non-linear equation by iteration. From the equilibrium condition f i

v=f iL for an

ideal mixture it is derived thatp=xHx ∙ pHx

vap+(1−xHx) ∙ pHpvap

p (mmHg )=xHx ∙106.88555− 1175.817

T (℃ )+224.867+(1−xHx ) ∙106.90253− 1267.828

T (℃ )+216.823

0.5 ∙760=xHx ∙106.88555− 1175.817

T (℃ )+224.867+ (1−xHx ) ∙106.90253− 1267.828

T (℃ )+216.823

and yHx=xHx ∙ pHx

vap

p=

xHx ∙106.88555− 1175.817

T (℃ )+224.867

0.5 ∙760

56

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

Mo

le f

rac

tio

n n

-he

xa

ne

in

va

po

ur

ph

ae

, yH

x

Mole fraction of n-hexane in liquid phase, xHx

p = 0.5 bar

40

50

60

70

80

0 0.2 0.4 0.6 0.8 1

Tem

pe

ratu

re, o

C


Temperature as a function of mole fraction n-hexane

in liquid phase, xHx

Temperature as a function of mole fraction n-hexane

in vapour phase, xHx

p = 0.5 bar

Liquid mixture

vapour mixture

Figure II.2.2: x-y diagram (left) and T-x-y diagram (right) for the system n-hexane/n-heptane at 0.5 bar at vapour-liquid equilibrium (VLE)

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Bubble point calculationThe bubble point is defined as the pressure (in the case of bubble point pressure) or temperature (in the case of bubble point temperature), at which in a particular liquid mixture with a given composition the first vapour bubble is formed upon decreasing the pressure/increasing the temperature (see Figure 2.3). This means that the composition of the liquid phase is known. A minute amount of the liquid has evaporated, and thus the composition of the liquid phase is the same as the composition of the starting liquid mixture. At this point a large amount of liquid is in equilibrium with a very small amount of vapour

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example Calculating bubble point pressure/temperature for ideal mixtures Determine the bubble point pressure for a mixture containing 50 mole-% n-hexane and 50 mole-% n-heptane at 60oC, and the bubble point temperature for the same mixture at a constant pressure of 0.5 bar.

Vapour pressure of n-hexane (Hx): pHxvap (mmHg )=10

6.88555− 1175.817T (℃ )+224.867

Vapour pressure of n-heptane: pHpvap (mmHg )=10

6.90253− 1267.828T (℃ )+216.823 .

The bubble point pressure is given by the equilibrium relationship. Operating at a constant temperature of 60oC, the

pbubble=xHx ∙ pHxvap+(1−xHx ) ∙ pHp

vap=0.2766+0.477 ∙ xHx

pbubble=0.2766+0.477 ∙0.5=0.515 ∙ ¿The bubble point temperature is given by the same equilibrium relationship. Operating at a constant pressure of 0.5 bar, the bubble temperature is given by

p (mmHg )=xHx ∙106.88555− 1175.817

T (℃ )+224.867+(1−xHx ) ∙106.90253− 1267.828

T (℃ )+216.823

0.5 ∙760=0.5 ∙106.88555− 1175.817

T bubble(℃)+224.867+0.5 ∙106.90253− 1267.828

Tbubble (℃ )+216.823

T bubble=59.14℃(quite similar answers as could be expected after calculating the bubble pressure at a temperature of 60oC)

57

0.2

0.4

0.6

0.8

0 0.2 0.4 0.6 0.8 1

Pre

ss

ure

, ba

r


T = 60oCLiquid

vapour

pbubble

yvapour

40

50

60

70

80

0 0.2 0.4 0.6 0.8 1

Tem

pe

ratu

re, o

C


p = 0.5 bar

Liquid

vapour

Tbubble

yvapour

Figure II.2.3: Bubble point pressure at 60oC (left) and bubble point temperature at 0.5 bar (right) for the ideal mixture n-hexane/n-heptane containing 50 mole-% n-hexane

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Dew point calculationThe dew point is defined as the pressure (in the case of dew point pressure) or temperature (in the case of dew point temperature), at which in a particular vapour mixture with a given composition the first liquid droplet is formed upon increasing the pressure/decreasing the temperature (see Figure 2.4). This means that the composition of the vapour phase is known. A minute amount of the vapour has condensed out, and thus the composition of the vapour phase is still the same as the composition of the starting vapour mixture. At this point a large amount of vapour is in equilibrium with a very small amount of liquid.

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example Calculating dew point pressure/temperature for ideal mixtures Determine the dew point pressure for a mixture containing 50 mole-% n-hexane and 50 mole-% n-heptane at 60oC, and the dew point temperature for the same mixture at a constant pressure of 0.5 bar.


6.88555− 1175.817T (℃ )+224.867


6.90253− 1267.828T (℃ )+216.823 .

The starting point for the dew point pressure/temperature calculation is knowing that the mole fraction in the vapour phase is known, but the mole fraction in the liquid phase is unknown. From the equilibrium relationship, it is deduced that

yHx=xHx ∙ pHx

vap

p and yHp=

xHp ∙ pHpvap

p

or xHx=yHx ∙ p

pHxvap and xHp=

yHp ∙ p

pHpvap

The sum of the mole fractions of n-heptane and n-hexane equal 1, thus

or 1=xHx+xHp=yHx ∙ p

pHxvap +

yHp ∙ p

pHpvap

andp= 1

yHx

pHxvap+

yHp

pHpvap

Thus, the dew point pressure is given for a mixture containing 50 mole-% n-hexane at 60oC by

58

pdew=1

0.50.2766

+0.5

0.7536

=0.405∙ ¿

The dew point temperature for the mixture containing 50 mole-% n-hexane at 0.5 bar can be estimated from

p= 1yHx

pHxvap+

yHp

pHpvap

p= 1yHx

106.88555−

1175.817T dew (℃ )+224.867

+yHp

106.90253−

1267.828T dew (℃ )+216.823

=0.5 ∙760

T dew=65.87℃

The corresponding composition of the liquid phase can be estimated by back substitution into

xHx=yHx ∙ p

pHxvap

(Note the large difference between the bubble point and the dew point!)

0.2

0.4

0.6

0.8

0 0.2 0.4 0.6 0.8 1

Pre

ss

ure

, ba

r


T = 60oCLiquid

vapour

pdew

xliquid

40

50

60

70

80

0 0.2 0.4 0.6 0.8 1

Tem

pe

ratu

re, o

C


p = 0.5 bar

Liquid

vapour

Tdew

xliquid

Figure II.2.4: Dew point pressure at 60oC (left) and the dew point temperature at 0.5 bar (right) for the ideal mixture n-hexane/n-heptane containing 50 mole-% n-hexane

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Flash evaporationThe sudden decrease in pressure (or, but less common, the sudden increase in temperature) may result in the splitting of a liquid mixture into a liquid stream and a vapour stream. The liquid and vapour stream can be considered to be at equilibrium, and the composition of the liquid and the vapour is then given by the temperature and pressure at the exit point of the flash evaporator (Please note that flash evaporation is typically associated with a temperature decrease of the mixture, if this process is carried out adiabatically, since the high enthalpy of the vapour stream – heat of evaporation - has to be taken into account). The amount of liquid and the amount of vapour leaving the flash evaporator can be determined using a simple mass balance, since the composition of the vapour phase and liquid phase are determined by equilibrium. For a binary mixture at a given temperature and pressure, the composition of the liquid and vapour phase in equilibrium with each other are given, since the degree of freedom equals zero.

59

If 1 mole of a mixture, containing xA,mix moles of A and (1-xA,mix) moles of B, splits into a L moles of liquid, with a composition of xA, and V moles of vapour, with a composition of yA, in a flash evaporation, the mole balances yield:Overall mole balance 1=L+V Mole balance for A x A,mix=L ∙x A+V ∙ y A

Substitute the overall mole balance into the mole balance for A:Mole balance for A x A,mix=L ∙x A+ (1−L ) ∙ y A= y A+L ∙ ( xA− y A )

Thus, the fraction of the mixture, which appears as a liquid is given by:

L=x A,mix− y A

x A− y A

and the mole fraction appearing as a vapour is given by:

V=1−L=1−x A,mix− yA

x A− y A

V=x A−xA ,mix

x A− y A

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example Flash calculation for ideal mixtures The pressure of a mixture containing 50 mole-% n-hexane and 50 mole-% n-heptane initially at 5 bar and 85oC is suddenly reduced to 1 bar. How much vapour and how much liquid are being formed, if the decompression takes place isothermally?


6.88555− 1175.817T (℃ )+224.867


6.90253− 1267.828T (℃ )+216.823 .

At 90oC, the vapour pressure of the compounds is:


6.88555− 1175.81785 (℃ )+224.867=1233 ∙mmHg


6.90253− 1267.82885 (℃ )+216.823=503 ∙mmHg.

From the vapour pressure of the pure compounds, it is possible that we are in the VLE-region. To be certain, we estimate the bubble point pressure and dew point pressure at 85oC:

pbubble=xHx ∙ pHxvap+(1−xHx ) ∙ pHp

vap=616.5+251.5=868 ∙mmHg

pdew=1

yHx

pHxvap+

yHp

pHpvap

= 10.51233

+0.5503

=715 ∙mmHg

The pressure is between the bubble and dew point we are in the VLE-region

At the given temperature and pressure the composition of the liquid phase is given by the equilibrium relationship:

p=xHx ∙ pHxvap+(1−xHx) ∙ pHp

vap

760 ∙mmHg=xHx ∙1233+(1−xHx) ∙503xHx=0.3516

60

yHx=xHx ∙ pHx

vap

p=0.3516 ∙1233

760=0.5705

Hence, the fraction of the mixture that remains as a liquid is:

L=x A,mix− y A

x A− y A

= 0.5−0.57050.3516−0.5705

=0.322

V=x A−xA ,mix

x A− y A

= 0.3516−0.50.3516−0.5705

=0.678

In reality this type of processes will be occur rather under adiabatic than isothermal conditions, implying that the transformation of a liquid mixture into a vapour-liquid mixture will be associated with a decrease in temperature due to the heat of evaporation. The heat of evaporation of n-hexane is 30.1 kJ/mol and the heat of evaporation of n-heptane is 34.2 kJ/mol. Furthermore, the specific heat of n-hexane is 26.3 J/(mol.K) and of n-heptane is 22.4 J/(mol.K)

The final temperature of the streams leaving the flash evaporator, can be obtained from an energy balance for this adiabatic system:

Hmixture ,∈¿ ¿¿

(Remember that the resulting vapour and liquid phase are in equilibrium. Hence, the temperature of the vapour and liquid phase leaving are equal.)

nmixture∙ Hmixture ,∈¿ ¿¿

Hence, per mole of mixture coming into the system:Hmixture ,∈¿ ¿¿

We are dealing with ideal mixtures (thus, the change in the enthalpy upon mixing equals zero!). Thus, the enthalpy of the mixtures can be split into the enthalpy of the individual components with the excess enthalpy being equal to zero:

xHx ,mix ∙ H n−hexane ,∈¿¿¿

The enthalpy of a component in the gas phase equals the enthalpy of the same component in the liquid phase plus the enthalpy change upon evaporation:

H n−hexane (v ) (T )=H n−hexane (l) (T )+∆vap H n−hexane (T )H n−heptane (v ) (T )=H n−heptane (l ) (T )+∆vapH n−heptane (T )

xHx ,mix ∙ H n−hexane ,∈¿¿¿

The mole balance for n-hexane showsxHx ,mix=V ∙ yHx+L∙ xHx

xHx ,mix ∙¿

The change in the enthalpy of the liquid is mainly due to the change in temperature, since the enthalpy of the liquid phase is hardly dependent on temperature (incompressible fluid assumption):

xHx ,mix ∙ cP ,n−hexane ( l) ∙ (85−T (℃ ) )+(1−xHx,mix ) ∙ cP, n−heptane ( l) ∙ (85−T (℃ ) )=V ∙ yHx ∙∆vapH n−hexane (T )+V ∙ (1− yHx) ∙∆vapH n−heptane (T )

Hence, we have a set of equations which needs to be solved simultaneously:Equilibrium condition 1: p=xHx ∙ pHx

vap (T )+(1−xHx )∙ pHpvap (T )=760 ∙mmHg

61

xHx=760 ∙mmHg−pHp

vap (T )pHx

vap (T )−pHpvap (T )

Equilibrium condition 2: yHx=xHx ∙ pHx

vap (T )p

Mole balance V=xHx−xHx,mix

xHx− yHx

=xHx−0.5xHx− yHx

Energy balance: (0.5 ∙26.3+0.5 ∙22.4 ) ∙ (85−T (℃ ) )+(1−xHx,mix ) ∙ cP, n−heptane ( l) ∙ (85−T (℃ ) )=V ∙ yHx ∙30.1∙10

3+V ∙ (1− yHx) ∙34.2∙103

Solving these equations simultaneously: xHx=0.4993 yHx=0.7130 V=0.003 T=80.5℃Thus, under these conditions hardly any vapour is formed.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••II.3. Vapour-liquid equilibrium (VLE) of non-ideal binary mixtures at low pressureThe conditions for a vapour-liquid equilibrium of a binary mixture are in addition to the :

f 1v=f 1

L f 2v=f 2

L and the temperature and pressure in each phase are equal.

At low pressure, vapours obey the Lewis-Randall rule. Hence the fugacity of a component in the vapour phase mixture is given by:

f 1v= y1 ∙ f 1

v f 2v= y2 ∙ f 2

v

Furthermore, at low pressure the fugacity of the pure component is equal to the system pressure:f 1v= y1 ∙ p f 2

v= y2 ∙ p

The fugacity of a component in the liquid phase is given by:f 1L=x1∙ γ 1∙ f 1

L f 2v=x2 ∙ γ2 ∙ f 2

L

Now the activity coefficient for the liquid phase is no longer equal to 1, but given by an activity coefficient model. The fugacity of a component in the mixture is now given by:

f 1L=x1∙ γ 1∙ p1

vap f 2v=x2 ∙ γ2 ∙ p2

vap

Thus, the equilibrium relationships are:y1∙ p=x1 ∙ γ1 ∙ p1

vap y2 ∙ p=x2 ∙ γ2 ∙ p2vap

and hence the equilibrium pressure for a binary mixture is now given byp=x1 ∙ γ1 ∙ p1

vap+ x2 ∙ γ2 ∙ p2vap

If the activity coefficients of at least one of the species in the mixture is larger than 1p>∑

i

x i ∙ γi ∙ pivap

(and similar for mixtures for which at least one of the species’ activity coefficient is less than 1, p<∑

i

x i ∙ γi ∙ pivap

Of special interest are mixtures where the equilibrium pressure of the system passes a maximum or a minimum as a function of the composition:

62

( ∂ p∂ x1 )T=0

( ∂ p∂ x1 )T=γ1 ∙ p1

vap+ x1 ∙ p1vap∙ ( ∂γ 1∂x1 )T−γ 2 ∙ p2

vap+(1−x1) ∙ p2vap ∙( ∂γ 2∂x1 )T

( ∂ p∂ x1 )T=γ1 ∙ p1

vap∙(1+x1 ∙( ∂ ln (γ 1 )∂ x1 )

T)−γ2 ∙ p2vap∙(1−x2 ∙( ∂ ln ( γ2 )

∂x1 )T)

According to the Gibbs-Duhem equation:

x1 ∙( ∂ ln ( γ1 )∂x1 )

T

+x2∙( ∂ ln ( γ2 )∂ x1 )

T

=0

x2 ∙( ∂ ln ( γ2 )∂x1 )

T

=−x1 ∙( ∂ ln ( γ 1)∂ x1 )

T

Substitute it back:

( ∂ p∂ x1 )T=γ1 ∙ p1

vap∙(1+x1 ∙( ∂ ln (γ 1 )∂ x1 )

T)−γ2 ∙ p2vap∙(1+x1 ∙( ∂ ln (γ 1 )

∂ x1 )T)

( ∂ p∂ x1 )T=(γ 1∙ p1vap−γ2 ∙ p2

vap ) ∙(1+x1 ∙( ∂ ln ( γ1 )∂x1 )

T)=0

Since, there is no reason why the second term should be equal to zero (i.e. (1+x1∙( ∂ ln ( γ1 )∂x1 )

T)≠0),

it means that at the extreme point:

(γ1 ∙ p1vap−γ2 ∙ p2vap )=0 γ 1∙ p1

vap=γ 2 ∙ p2vap

From the equilibrium relationships:y1∙ p=x1 ∙ γ1 ∙ p1


γ 1∙ p1vap=

y1 ∙ p

x1 γ 2∙ p2

vap=y2 ∙ p

x2

Thus, at the extreme pointy1∙ p

x1=

y2 ∙ p

x2

y1x1

=y2x2

Furthermore, the sum of the mole fractions equals 1:y1x1

=1− y11−x1

y1∙ (1−x1 )=x1 ∙ (1− y1 ) y1− y1 ∙ x1=x1− y1 ∙ x1

y1=x1 And thus also y2=x2

63

Thus, at the extreme point, the composition of the liquid phase is identical to the composition of the vapour phase (for each component!). Thus at a fixed temperature, the extreme pressure (either maximum or minimum) results in the composition of the vapour phase and the liquid phase to be identical. This is called an azeotropic mixture. A similar phenomenon takes place for the vapour-liquid equilibrium of non-ideal mixtures at constant pressure. If the temperature is mixture at the azeotropic point is larger than the boiling temperature of each component at the given pressure, it is called a maximum boiling azeotrope. A similarly, it is called a minimum boiling azeotrope, if the equilibrium temperature is lower than the boiling temperature of each component at the given pressure.

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Constructing p-x-y and T-x-y diagrams for VLE involving non-ideal liquid mixtures Ethyl acetate (EA) and benzene (B) form a non-ideal mixture. The activity coefficient for this system can be described by the van Laar model with = 1.15 and = 0.92. The vapour pressure ofEthyl acetate ln ¿ Benzene ln ¿ Determine the p-x-y diagram at 75oC and the T-x-y diagram at 1 bar.

The equilibrium condition for vapour-liquid-equilibrium (VLE) is:f EAv =f EA

L f Bv =f B

L At low pressure, this translates into:

y EA ∙ p=x EA ∙ γEA ∙ pEAvap yB ∙ p=xB ∙ γB ∙ p2

vap

With ln ( γEA )= 1.15

(1+1.150.92∙xEA

xB)2

ln ( γB )= 0.92

(1+ 0.921.15∙xB

xEA)2

orp=x EA ∙ γEA ∙ pEA

vap+ xB∙ γB ∙ p2vap

p¿

At 75oC, the vapour pressure of ethyl acetate (EA) is 0.946 bar and of benzene is 0.862 bar. p¿p¿

The corresponding composition of the vapour phase is given by:y EA ∙ p=x EA ∙ γEA ∙ pEA

vap

y EA=xEA ∙ γEA ∙ pEA

vap

p=

x EA ∙ e

1.15

(1+1.150.92∙xEA

xB )2

∙0.946p

64

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1mo

le f

rac

tio

n o

f e

thyl

ae

tate

in

va

po

ur

ph

as

e, y

EA

Mole fraction of ethyl acetate in liquid phase, xEA

T = 75oC

azeotropic point

0.8

0.9

1

1.1

1.2

0 0.2 0.4 0.6 0.8 1

pre

ss

ure

, ba

r

Mole fraction of ethyl acetate, xEA/yEA

T = 75oC

Liquid

Vapour

azeotropic point

Figure II.3.1: x-y diagram (left) and P-x-y diagram for the non-ideal mixture ethyl acetate/benzene at 75oC

The position of the azeotrope can easily be defined, since this occurs at the point where the equilibrium pressure of the system is maximum and

γ EA ∙ pEAvap=γB ∙ pB

vap

e

1.15

(1+1.150.92∙

xEA

1−xEA )2

∙0.946=e

0.92

(1+0.921.15∙1−xEA

xEA )2

∙0.862

Solving this equation results in the mole fraction of ethyl acetate in the liquid phase (and thus the mole fraction of ethyl acetate in the vapour phase) at the azeotropic point of 0.5180, and the corresponding pressure of 1.166 bar.

The construction of the T-x-y diagram is analogue to the construction of the P-x-y diagram, but now the pressure is kept constant at 1 bar:p¿

1=x EA ∙e

1.15

(1+1.150.92∙xEA

xB )2

∙ e9.6830− 2842.2

T (K )−56.3209+xB ∙ e

0.92

(1+0.921.15∙xBxEA)

2

∙ e9.3171− 2810.5

T (K )−51.2586

(It is assumed here that the coefficients in the van Laar equation do not change significantly over the temperature range considered. This is typically a good approximation, since the activity coefficient is not strongly dependent on temperature.)

This equation can be solved by assuming a mole fraction of ethyl acetate (and thus the mole fraction of benzene is known), and estimate the temperature for which the total pressure corresponds to 1 bar. This can be done using e.g. Goalseek in Excel®.

The corresponding composition of the vapour phase is given by:y EA ∙ p=x EA ∙ γEA ∙ pEA

vap

y EA=xEA ∙ γEA ∙ pEA

vap

p=xEA ∙ e

1.15

(1+1.150.92∙xEA

xB )2

∙ e9.6830− 2842.2

T (K )−56.3209

The azeotropic point is given by: γ EA ∙ pEA

vap=γB ∙ pBvap

e

1.15

(1+1.150.92∙

xEA

1−xEA )2

∙ e9.6830− 2842.2

T (K )−56.3209=e

0.92

(1+0.921.15∙1−xEA

xEA )2

∙ e9.3171− 2810.5

T (K )−51.2586

65

This is one equation with two unknowns. However, there is a constraint on the system, viz. the total pressure in the system must be equal to 1 bar:p¿

Hence, the composition at the azeotropic point and the temperature of the azotropic point can now be estimated (e.g. by using Solver in Excel®). I typically solve for the ratio of the mole fraction in the vapour phase relative to the mole fraction in the liquid phase (which should be 1) ensuring that the pressure is the set pressure and varying the temperature and the mole fraction in the liquid phase (i.e. for a given T and xEA, the activity coefficients and the vapour pressure of the pure components can be calculated and thus the total pressure; knowing the total pressure in conjunction with the mole fraction in the liquid phase, the activity coefficient and the vapour pressure of a component, the mole fraction in the vapour phase can be calculated).

T, K343.54 0.5146 1.2371 1.3482 0.8084 0.7417 1 0.5146 1

xEA gEA gB pvapEA pvap

B pcalc yEA yEA/xEA

Variables, varied during the optimization

Constrained during optimization

Target cell, containing the parameter to be solved for

The azeotrope for the constant pressure system is at a mole fraction of ethyl acetate of 0.515 and a temperature of 70.38oC. The position of the azeotrope depends on the applied conditions!

The system ethyl acetate/benzene shows a minimum boiling azeotrope, since this is the lowest temperature at which any mixture of ethyl acetate and benzene will evaporate.

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1mo

le f

rac

tio

n o

f e

thyl

ae

tate

in

va

po

ur

ph

as

e, y

EA

Mole fraction of ethyl acetate in liquid phase, xEA

p = 1 bar

azeotropic point

70

72

74

76

78

80

0 0.2 0.4 0.6 0.8 1

pre

ss

ure

, ba

r

Mole fraction of ethyl acetate, xEA/yEA

p = 1 bar

Liquid

Vapour

azeotropic point

Figure II.3.2: x-y diagram (left) and T-x-y diagram for the non-ideal mixture ethyl acetate/benzene at 1 bar

The presence of azeotropes causes difficulty in separation processes, since an azeotropic point cannot be passed, i.e. the end point of a mixture with an azeotrope will be the azeotropic mixture and one of the pure components. However, the position of the azeotrope is dependent on the

66

applied pressure for a system operating at a constant pressure or on the applied temperature for a system operating at a constant temperature. The condition for the azeotropic point is given by:

(γ1 ∙ p1vap−γ2 ∙ p2vap )=0 γ 1∙ p1

vap=γ 2 ∙ p2vap

and in the case of the system ethyl acetate (EA) and benzene (B)γEA

γB

=pBvap

pEAvap

or ln ( γEA )−ln (γB )=ln ( pBvap )−ln ( pEA

vap )

1.15

(1+ 1.150.92∙

xEA

1−xEA)2− 0.92

(1+ 0.921.15∙1−xEA

x EA)2=9.3171− 2810.5

T (K )−51.2586−9.6830+ 2842.2

T (K )−56.3209

Figure II.3.3 shows the shift in the position of the azeotrope as a function of the (constant) temperature. As a consequence, the pressure in the system has to change as well. The extent of the shift in the azeotropic point depends on the relative heat of evaporation of the two components (and the excess enthalpy for the mixture)

0.4

0.5

0.6

0 50 100 150 200 250

Mo

le fr

acti

on

eth

yl

acet

ate

at a

zeo

tro

pe,

xE

A

Temperature, oC

0.01

0.1

1

10

100

0 50 100 150 200 250

pre

ssu

re, b

ar

Temperature, oC

Figure II.3.3: Shift in the azeotropic point in the mixture ethyl acetate (EA) and benzene (B) as a function of temperature (left) and the associated shift in the equilibrium pressure (right)

The mode of operation for the separation of ethyl acetate and benzene into its pure components depends now on the initial composition. A mixture, which is rich in ethyl acetate (i.e. the mole fraction of ethyl acetate is higher than the mole fraction of ethyl acetate at the azeotropic point), will in a first distillation column operating at 1 bar separate into the azeotropic mixture corresponding to this pressure and pure ethyl acetate. The azeotropic mixture is fed into a column operating at a higher pressure (e.g. ca. 10 bar), where it will be separated into pure benzene and an azeotropic mixture corresponding to an equilibrium pressure of 10 bar (which contains more ethyl acetate than the mixture fed to this column). This azeotropic mixture is then fed back to the first column operating at 1 bar.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Determining activity coefficients from VLEMeasuring the equilibrium pressure at a given temperature for a mixture at its vapour-liquid equilibrium point allows the determination of the activity coefficient of the components at each point, if the composition of the gas phase and the composition of the liquid phase are measured as well.

67

For a mixture at its vapour-liquid equilibrium point (at low pressure):y1∙ p=x1 ∙ γ1 ∙ p1


γ 1=y1 ∙ p

x1 ∙ p1vap γ 2=

y2 ∙ p

x2 ∙ p2vap

The partial molar Gibbs free energy for each component in the mixture can now be determined from the activity coefficient as a function of the mole fraction in the liquid phase

G1ex=RT ∙ ln ( γ1 ) G2

ex=RT ∙ ln ( γ2 )and thus

Gex=x1 ∙G1ex+x2 ∙G2

ex Measuring the vapour-liquid equilibrium position at various temperatures lead to the excess enthalpy, since

H ex=−T2 ∙( ∂ Gex

T∂T )

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Determining excess Gibbs free energy in the system n-pentane/propionaldehyde The equilibrium pressure and the composition of the liquid and the vapour in equilibrium have been measured for the system n-pentane/propionaldehyde at 40oC. Model the molar excess Gibbs free energy as a function of the mole fraction

xn-pentane yn-pentane P (bar) xn-pentane yn-pentane P (bar)0 0 0.7609 0.4463 0.5877 1.3354

0.0503 0.2121 0.9398 0.5031 0.6146 1.34940.1014 0.3452 1.0643 0.561 0.6311 1.35680.1647 0.4288 1.1622 0.6812 0.6827 1.36360.2212 0.4685 1.2173 0.7597 0.7293 1.35670.3019 0.5281 1.2756 0.8333 0.7669 1.33530.3476 0.5539 1.2949 0.918 0.8452 1.28140.4082 0.5686 1.3197 1 1 1.1541

The activity coefficient as a function of the mole fraction can be determined from:

γ n−pentane=yn−pentane ∙ p

xn−pentane ∙ pn−pentanevap =

yn−pentane ∙ p

xn−pentane ∙1.1541

γ propionaldehyde=y propionaldehyde ∙ p

x propionaldehyde ∙ p propionaldehydevap =

(1− yn−pentane ) ∙ p(1−xn−entane ) ∙0.7609

Knowing the activity coefficient, the partial molar excess Gibbs free energy for each component can be calculated:

Giex=RT ∙ ln ( γi )=8.314 ∙ (273.15+40 ) ∙ ln ( γi )

and thusGex=x1 ∙G1

ex+x2 ∙G2ex

xn-pentane n-pentane propionaldehyde Gexn-pentane Gex

propionaldehyde Gex, J/mol0.0000 1.0000 0.0 0.00.0503 3.4337 1.0247 3211.8 63.5 221.90.1014 3.1394 1.0192 2978.6 49.6 346.6

68

0.1647 2.6218 1.0445 2509.4 113.3 507.90.2212 2.2340 1.0918 2092.7 228.7 641.00.3019 1.9334 1.1332 1716.5 325.6 745.50.3476 1.7879 1.1637 1512.8 394.6 783.30.4082 1.5928 1.2643 1212.0 610.6 856.10.4463 1.5237 1.3068 1096.4 696.7 875.10.5031 1.4284 1.3755 928.2 830.0 879.40.5610 1.3225 1.4984 727.8 1052.9 870.50.6812 1.1841 1.7837 440.0 1506.6 780.00.7597 1.1285 2.0086 314.8 1815.8 675.50.8333 1.0648 2.4539 163.5 2337.1 525.80.9180 1.0223 3.1792 57.3 3011.3 299.51.0000 1.0000 0.0 0.0

The molar excess Gibbs free energy can be modeled using a Redlich-Kister expression (see Figure 3.4) with a0 =3543.7 J/mol (taking more terms into account would reduce the error in the fit only marginally).

0

200

400

600

800

1000

0 0.2 0.4 0.6 0.8 1

Ex

ce

ss

mo

lar

Gib

bs

fre

e

en

erg

y, G

ex,

J/m

ol

Mole fraction n-pentane in liquid phase, xn-pentane

Figure II.3.4: Modeling the excess Gibbs free energy of the system n-pentane/propionaldehyde at 40oC with a one-constant Redlich-Kister equation:Gex=a0 ∙ xn−pentane ∙ x propionaldehyde ∙ (xn−pentane− xpropionaldehyde )

The activity coefficient model belonging to a one-constant Redlich-Kister expression is the one-constant Margules expression. Thus, the activity coefficients are given by:

RT ∙ ln (γ n−pentane)=a0 ∙ xpropionaldehyde2 =3543.7 ∙ x propionaldehyde

2

RT ∙ ln (γ propionaldehyde )=a0 ∙ xn−pentane2

γ n−pentane=ea0 ∙ xpropionaldehyde

2

RT =e1.3611 ∙ xpropionaldehyde2

γ propionaldehyde=ea0 ∙ xn−pentane

2

RT =e1.3611 ∙ xn−pentane2

69

With the model, the position of the azeotropic point at 40oC can be estimated. At the azeotropic point:

γ n−pentane ∙ pn−pentanevap =γ propionaldehyde ∙ p propionaldehyde

vap

ln ( γn−pentane )−ln (γ propionaldehyde)=ln ( ppropionaldehydevap

pn−pentanevap )

1.3611 ∙ (x propionaldehyde2 −xn−pentane

2 )=ln( ppropionaldehydevap

pn−pentanevap )

(1−xn−pentane )2−xn−pentane2 =

11.3611

∙ ln( ppropionaldehydevap

pn−pentanevap )

1−2 ∙ xn−pentane=1

1.3611∙ ln( p propionaldehyde

vap

pn−pentanevap )

xn− pentane=0.5−1

2 ∙1.3611∙ ln( ppropionaldehyde

vap

pn−pentanevap )=0.6530

At the azeotropic point yn−pentane=xn−pentane=0.6530 and the equilibrium pressure is given by:

p=xn−pentane ∙ pn−pentane

vap ∙ γ n−pentane

yn−pentane

p=pn−pentanevap ∙ γn−pentane

p=1.1541 ∙ e1.3611 ∙ (1−0.6530 )2=1.3596 ∙ ¿

0.7

0.9

1.1

1.3

0 0.2 0.4 0.6 0.8 1

Eq

uili

bri

um

pre

ss

ure

, ba

r

Mole fraction n-pentane in liquid phase or vapour phase, xn-pentane/yn-pentane

Figure II.3.5: Measured P-x-y diagram for the system n-pentane/propionaldehyde at 40oC and model prediction using one-constant Margules expression (solid line)

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

II.4. High pressure vapour-liquid equilibrium (VLE)

70

The phase equilibrium conditions for a binary mixture at its point of vapour-liquid equilibrium are:T V=T L pV=pL f 1

v=f 1L f 2

v=f 2L

At high pressure, the Lewis-Randall rule is no longer valid (unless we consider very high pressures) and the fugacity in the vapour phase cannot be expressed in terms of the total pressure and its mole fraction. Similarly, the fugacity of the pure component in the liquid phase cannot be set equal to the vapour pressure of that component. Now, an equation of state (e.g. the Peng-Robinson equation of state) can be used to determine the fugacity of each species in the respective phases. The fugacity obtained from the Peng-Robinson equation of state is now given by:

lnf iV

y i ∙ P=

Bi

Bmix

∙ (ZV−1 )−ln (ZV−1 )−Amix

2√2Bmix

∙( 2∑i

y i ∙ Aij

Amix

−Bi

Bmix)∙ ln( ZV+(1+√2 ) Bmix

ZV+(1−√2 ) Bmix)

lnf iL

x i ∙P=

Bi

Bmix

∙ (ZL−1 )−ln (ZL−1 )−Amix

2√2 Bmix

∙( 2∑i y i ∙ Aij

Amix

−Bi

Bmix) ∙ ln( Z L+(1+√2 )Bmix

Z L+(1−√2 )Bmix)

with Z= p ∙VRT

The compressibility is obtained by solving the cubic equation of state:Z3+∙ (Bmix−1 )∙ Z2+(Amix−3Bmix

2 −2Bmix )+(−Amix ∙Bmix+Bmix2 +Bmix

3 )=0

Aij=aij ∙ p

(RT )2Amix=

amix ∙ p

(RT )2

Bi=bi ∙ p

RTBmix=

bmix ∙ p

RT

amix=∑i=1

C

∑j=1

C

y i ∙ y j ∙ aij with a ij=√aii ∙ a jj ∙ (1−k ij )

(here aii represents the pure component parameter a)

bmix=∑i=1

C

yi ∙ bi with bi the pure component parameter b

The corresponding phase diagram can be computed (if the binary interaction parameter is known) by calculating the bubble point pressure for mixtures of a known composition (bubble line calculation; see Figure 10.3-5), and the dew point pressure for mixtures of a known composition (dew line calculation; see Figure 10.3-6). The critical point of the mixtures is defined as the point where the dew point line and the bubble point line meet. The calculation near the critical point is not easy to obtain convergence.

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Example: Determine the P-x-y diagram for the mixture ethane/propane at 298.15 K and 350 K. At 298.15 K both propane (Tc = 369.8 K) and ethane (Tc = 305.4 K) are below their respective critical temperature. However 350K is higher than the critical temperature for ethane. Hence, it is expected that at 298.15 K, the P-x-y diagram corresponds to a normal (quite ideal) P-x-y-diagram. At 350K no condensation of pure ethane is possible.

71

0

20

40

60

0 0.2 0.4 0.6 0.8 1

Eq

uili

bri

um

pre

ss

ure

, ba

r

Mole fraction propane

T = 298.15 K

T = 350 K

Figure II.4.1: P-x-y diagram for the system propane/ethane at 298.15 K and 350 K.

Interesting behavior can be observed close to the critical point. Figure II.4.2 shows the p-T-diagram for a mixture containing 40 mole-% n-heptane and 60 mole-% ethane. The critical point is not given by the maximum pressure or by the maximum temperature, but by the point at which the dew line and the bubble line meet. Interesting behavior can be observed when compressing this mixture from 10 bar to 80 bar at 478 K. Initially, the mixture is in the vapour phase. Increasing the pressure to 39 bar, and the first liquid start forming, since the dew line is crossed. The amount of liquid increases upon increasing the pressure further, but at some point the amount of liquid starts to decrease. At a pressure of ca. 68 bar the liquid disappears and the mixture is again in the vapour phase (since we passed the dew line again). This double crossing of the dew line at different conditions means that the mixture is initially present as a vapour, then as a vapour-liquid mixture, and then again as a vapour. This behavior is called retrograde condensation of the 1 st kind .

0

20

40

60

80

100

300 340 380 420 460 500

Eq

uili

bri

um

pre

ss

ure

, ba

r

Temperature, K

Mixture, 40% n-heptane/60% ethane

critical point

pmax>pc,mix

Tmax>Tc,mix

0

20

40

60

80

100

300 340 380 420 460 500

Eq

uili

bri

um

pre

ss

ure

, ba

r

Temperature, K

Mixture, 40% n-heptane/60% ethane

Figure II.4.2: P-T diagram for a mixture containing 40 mole-% n-heptane and 60 mole-% ethane.Left: Critical point determination for the mixture (i.e. the meeting point of the bubble

line (red) and the dew line (blue)Right: Retrograde condensation

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

72

II.5. Solubility of a gas in a liquidThe solubility of a gas in a liquid is given by the equilibrium conditions:

T V=T L pV=pL f 1v=f 1

L f 2v=f 2

L The fugacity of the dissolved gas in the liquid can be described using Henry’s law, which is based on the experimentally observed linear increase in the mole fraction of the dissolved gas with increasing pressure:

f iL (T mixture , p , x )= xi ∙ γi

¿ (T mixture , p , x ) ∙H i (T mixture , p ) (with i* describing the deviation from the linear increase in the mole fraction with increasing pressure). Hence the equilibrium condition for the dissolved gas is given by:

f iL (T mixture , p , x )= xi ∙ γi

¿ (T mixture , p , x ) ∙H i (T mixture , p )= y i ∙ p ∙( fp )i

=f iv

The activity coefficient i* can be related to the usual activity coefficients via:

γ i¿ (T mixture , p , x )=

γi (Tmixture , p , x)γ i (T mixture , p , x i=0 )

The activity coefficients of gases dissolved in a liquid can typically be estimated using e.g. the regular solution theory RT ∙ ln (γ 1 )=(δ1−δ2 )2 ∙Φ2

2∙V 1 or RT ∙ ln (γ 1 )=(δ1−δ )2 ∙Φ22 ∙V 1

(dissolution in pure solvent) (dissolution in a mixture)

which require pre-knowledge about the molar volume of the dissolved gas as a pure liquid and the solubility parameters. Table 11.1-1 lists these values for some common gases. It should be noted that these values represent typically a hypothetical liquid since these substances are typically non-condensable at the temperatures of interest.

For very dilute systems (as typically given for gases dissolved in liquids), the activity coefficient i* approaches 1 (it is typically larger than 0.98, but less than 1, if the volume fraction of the dissolved gas is less than 0.01). Hence, the Henry coefficient can typically be estimated from the measured mole fraction at atmospheric pressure (see Table 11.1-2 for the solubility of gases in water):

x i ∙ γ i¿ (T mixture , p , x )∙ H i (Tmixture , p )=p i

H i (Tmixture , p )≈p i

x i

If p is 1 atm (as in Table 11.1-2):

H i (Tmixture , p )≈ 1x i , measured

The Henry coefficient is dependent on pressure. The Henry coefficient is defined as:H i (Tmixture , p )=γ i (T mixture , p , x i=0 )∙ f iL (T mixture , p )

H i (Tmixture , p )=γ i (T mixture , p0 , x i=0 )∙ f iL (T mixture , p0 ) ∙e∫p0

p V (x i=0 )RT

∙ dp

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Determine the solubility of oxygen in water The equilibrium mole fraction of oxygen dissolved in water in equilibrium with oxygen at 1.013 bar at 298.15 K is 2.3.10-5 (see Table 11.1-2). Estimate the mole fraction of oxygen in water in equilibrium with oxygen at 10 bar.

73

Data: δH 2O=9.4 ∙√ cal

mol Handbook of Chemistry and Physics

δO2=4.0 ∙√ cal

mol Table 11.1-1

V H 2O=19 ∙ cm

3

mol Handbook of Chemistry and Physics

V O2=33.0∙ cm

3

mol Table 11.1-1

The Henry coefficient for oxygen in water at 298.15 K and 1.013 bar is:

HO2(T mixture , p )=

pO2

xO 2∙ γO2

¿ (T mixture , p , x )

The activity coefficient of oxygen in water, *, is given by:

γO2

¿ (T mixture , p , x )=γO2

(T mixture , p , x )γO2

(T mixture , p , x i=0 )

Suppose that the activity coefficient can be described using the regular solution theory: RT ∙ ln (γO2 )=(δO2

−δH2O )2 ∙ΦH 2O2 ∙V O2

Thus,

RT ∙ ln (γO2 (xO2=0 ))=(δO2

−δH 2O )2∙V O2= (4.0−9.4 )2 ∙33=962.3 ∙ cal

mol

γO2 (xO2=0 )=5.075

ΦH 2O ( xO2=2.3 ∙10−5 )=

xH2O∙V H 2O

xO 2∙V O2

+xH 2O∙V H 2O

=(1−2.3 ∙10−5 ) ∙18

2.3 ∙10−5 ∙33+ (1−2.3 ∙10−5 ) ∙18=0.99996

RT ∙ ln ( γO 2 (xO 2=2.3 ∙10−5 ))=(δO2

−δH 2O )2 ∙ΦH 2O2 ∙V O2

=(4.0−9.4 )2 ∙33=962.2∙ calmol

γO2(xO2

=2.3 ∙10−5 )=5.074

HO2¿

HO2¿

HO2¿

xO2=

pO2

HO2∙ γO2

¿ (T mixture , p , x )= 104.456 ∙104 ∙1

=2.244 ∙10−4

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

For some gas-solvent systems the Henry coefficient is not available. The solubility can be estimated from:

f iL (T mixture , p , x )= xi ∙ γi ∙ f i

L (Tmixture , p )= y i ∙ p∙ ( fp )i

=f iv

74

Now the fugacity of the dissolved gas as a pure liquid at the temperature and pressure of the mixture needs to be estimated. This can be obtained from the Prausnitz-Shair correlation (see Fig. 11.1-1), which relates the fugacity of a hypothetical liquid at atmospheric pressure relative to the critical pressure as a function of the relative temperature (i.e. the temperature relative to the critical temperature). Knowing the fugacity of a component as a liquid at 1.013 bar, the fugacity of this compound can be estimated at any pressure since


L¿ Since liquids are (relatively) incompressible

f iL (T mixture , p )≈ f i

L¿ II.5.1 Solubility of gases as a function of temperatureThe solubility of a gas in a liquid is given by:

f iL (T mixture , p , x )= xi ∙ γi ∙ f i

L (Tmixture , p )= y i ∙ p∙ ( fp )i

=f iv

In order to evaluate the temperature dependency of the solubility (given by the mole fraction xi), we will consider the derivative of this condition at constant pressure and constant gas phase composition:

( ∂ (x i ∙ γi ∙ f iL (T mixture , p ))∂T )

p , y

=( ∂( y i ∙ p ∙( fp )i)

∂T)p , y

=0

γ i ∙ f iL ∙( ∂ x i

∂T )p , y

+x i ∙ f iL ∙( ∂γ i

∂T )p , y

+x i ∙ γi ∙( ∂ f i L∂T )p , y

=0 divide by x i ∙ γ i ∙ f iL

1xi∙( ∂x i

∂T )p , y

+ 1γi∙( ∂γ i

∂T )p , y

+ 1f i

L ∙( ∂ f i L∂T )p , y

=0

( ∂ ln x i

∂T )p , y

+( ∂ ln γi∂T )p , y

+( ∂ ln f iL∂T )p , y

=0

( ∂ ln x i

∂T )p , y

=−( ∂ ln γ i

∂T )p , y

−( ∂ ln f iL∂T )p , y

The temperature dependency of the activity coefficient is given by:

( ∂ ln γ i

∂T )p , y

=−H i

ex (T , p , x )RT2

The temperature dependency of the fugacity of the gas as a pure liquid is given by:

( ∂ ln f iL

∂T )p , y

=( ∂ ln p ivap

∂T )p , y

=∆vapH i

RT 2

Thus, the temperature dependency of the solubility of a gas at a given pressure and a given gas phase composition is given by:

( ∂ ln x i

∂T )p , y

=H i

ex (T , p , x )−∆vapH i

RT 2

(Note, ∆vap H i−H iex (T , p , x )=(H 1

V−H 1L ) can be interpreted as the heat of evaporation from a

mixture).The heat of evaporation is typically much larger than the partial molar excess enthalpy. Hence,

( ∂ ln x i

∂T )p , y

≈−∆vap H i

RT2

75

Hence, the solubility of the gas will decrease if ∆vap H i>0. It should be noted that beyond the critical point the (extrapolated) heat of evaporation becomes less than zero. Thus, the solubility of gases can increase with increasing temperature if the temperature is much larger than the critical temperature of the gas.

76

II.6. Liquid-liquid equilibriumThe appearance of two or more liquid phases in equilibrium with each other implies that a single liquid phase is not stable. The equilibrium condition for a system at equilibrium (at constant temperature and pressure) is that the Gibbs free energy is at a minimum. This yields two conditions, i.e.

dG=0 and ∂2G>0 at constant N,T,pThe total Gibbs free energy for a binary liquid mixture is given by:

G=N 1 ∙G1+N 2 ∙G2+RT ∙ (N1 ∙ ln (x1 )+N 2 ∙ ln (x2 ))+N ∙G ex

or G=x1 ∙G1+x2 ∙G2+RT ∙( x1 ∙ ln (x1 )+x2 ∙ ln (x2 ))+Gex

G=x1 ∙G1+x2 ∙G2+RT ∙( x1 ∙ ln (x1 )+x2 ∙ ln (x2 ))+ x1 ∙G1ex+x2 ∙G2

ex

G=x1 ∙G1+x2 ∙G2+RT ∙( x1 ∙ ln (x1 )+x2 ∙ ln (x2 ))+ x1 ∙ ln (γ 1 )+ x2 ∙ ln (γ 2 )

For an ideal mixture (Gex=0), the Gibbs free energy of the mixture is always less than the sum of the Gibbs free energy of the pure components since ln(xi) is less than zero. Hence, the addition of two pure compounds forming an ideal mixture will always result in a lowering of the Gibbs free energy and hence yielding a mixture. However, the excess molar Gibbs free energy can be larger than zero resulting in a Gibbs free energy of the mixture larger than the Gibbs free energy of the pure components.

Suppose the excess Gibbs free energy s given by the one-constant Margules equation:Gex=A ∙ x1 ∙ x2

Then depending on the value of A, the Gibbs free energy of the mixture can be larger than the Gibbs free energy of the pure components (see Figure II.6.1).

-0.8

-0.6

-0.4

-0.2

0

0.2

0 0.2 0.4 0.6 0.8 1

(Gm

ixtu

re-x

1. G

1-x

2. G

2)/

RT

Mole fraction x1

A/RT = 0

A/RT = 1

A/RT = 2

A/RT = 3

Figure II.6.1: Difference between the molar Gibbs free energy of the mixture and the molar Gibbs free energy of the pure components as a function of the mole fraction for binary mixtures for which the excess Gibbs free energy can be described by the Margules expression (A/RT = 0 represents an ideal mixture)

Two liquid phases can form, if the difference between the Gibbs free energy of the mixture and the Gibbs free energy of its components as a function of the composition passes a minimum:

∂ (G−x1 ∙G1+x2 ∙G2)∂ x1

=0

77

since in that situation the formation of two liquid phases will result in a lower Gibbs free energy than the formation of a single liquid phase. The composition of the phases cannot be determined from this consideration. The equilibrium conditions have to be considered. For a binary system containing two liquids to be at equilibrium, the following equilibrium conditions must be valid:

T L I=T L II pLI=pLII f 1L I=f 1

L II f 2L I=f 2

L II The fugacity of each species in each of the liquid phases must be equal, and thus

(x i ∙ γi ∙ pivap )LI=( xi ∙ γi ∙ p i

vap )LII The vapour pressure of the pure component is independent of the phase:

(x i ∙ γi )LI=(x i ∙ γi )

LII Hence, the composition of the two liquid phases in equilibrium can be found using:

(x1 ∙ γ 1 )L I=(x1 ∙ γ 1 )L II (x2 ∙ γ 2 )LI=(x2 ∙ γ 2 )LII

With the boundary conditions (x1+x2)L I=1 and (x1+x2)

L II=1

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Phase splitting in the liquid furfural/iso-butane mixture The mixture furfural/iso-butane splits into two phases at 37.8oC and 5 bar. Determine the composition of the phases, if the activity coefficients in the mixtures can be represented by the van Laar expression:

ln ( γiso−butane )= 2.62

(1+2.623.02∙x iso−butane

x furfural)2

ln ( γ furfural )=302

(1+ 3.022.62∙

x furfural

x iso−butane)2

The equilibrium conditions are (besides constant T and p):

x iso−butaneL I ∙ e

2.62

(1+2.623.02∙x iso−butaneL I

x furfural

LI )

2

=x iso−butaneLII ∙ e

2.62

(1+2.623.02∙xiso−butaneLII

x furfural

LII )

2

x furfuralL I ∙ e

3.02

(1+3.022.62∙

xfurfuralLI

x iso−butane

LI )

2

=x furfuralL II ∙ e

3.02

(1+3.022.62∙

xfurfuralLII

x iso−butane

LII )

2

With the boundary conditions (x iso−butane+x furfural )LI=1 and (x iso−butane+x furfural )

LII=1

These 4 equations have to be solved simultaneously:

x iso−butaneL I ∙ e

2.62

(1+2.623.02∙

xiso−butaneLI

1−x iso−butane

LI )

2

=x iso−butaneLII ∙ e

2.62

(1+2.623.02∙

x iso−butaneLII

1−x iso−butane

LII )

2

x furfuralL I ∙ e

3.02

(1+ 3.022.62∙

x furfuralLI

1− xiso−butane

LI )

2

=x furfuralL II ∙ e

3.02

(1+3.022.62∙

x furfuralLII

1− xiso−butane

LII )

2

Solving these equations (e.g. by trial and error) yields:x iso−butaneL I =0.1128 and x iso−butane

L II =0.9284••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

As shown in Figure II.6.1 the likelihood for the formation of two liquid phases in equilibrium with each other diminishes with increasing temperature (with increasing temperature the term A/RT will

78

decrease). The highest temperature at which a mixture will split into two liquid mixtures is called the upper consolute temperature. The upper consolute temperature can be determined from the stability criterion of a single liquid. A phase is stable if:∂2G>0 at constant N,T,pThe change in the stability occurs, when

( ∂2G∂x1

2 )T , p

=0

since if ( ∂2G∂x1

2 )T , p

>0 a single phase will be stable and if ( ∂2G∂x1

2 )T , p

<0 a single phase will be

unstable. If the mixture can be described with a one constant Margules expression andG=x1 ∙G1+x2 ∙G2+RT ∙( x1 ∙ ln (x1 )+x2 ∙ ln (x2 ))+A ∙x1 ∙ x2 The 1st derivative is given by:

( ∂G∂ x1 )T , p

=G1−G2+RT ∙(ln (x1 )+x1x1

−ln (x2 )−x2x2 )+A ∙x2−A ∙ x2

( ∂G∂ x1 )T , p

=G1−G2+RT ∙( ln (x1 )−ln (x2 ))+A ∙ x2−A ∙x1

And the 2nd derivative is thus given by:

∂( ∂G∂ x1 )∂ x1 T , p

=RT ∙( 1x1+ 1x2 )−A−A

Thus, the highest temperature at which two liquid phases co-exist for a mixture of a given composition is given by:

RTmax ∙( 1x1+ 1x2 )−A−A=0

RT max ∙(1−x1+x1x1 ∙ x2 )−2∙ A=0

T max=2 ∙ A ∙ x1 ∙ x2

R

At temperatures below this maximum temperature, two liquid phases will co-exist and above the upper consolute temperature, a single liquid phase will exist for a Margules mixture. The highest possible temperature at which two liquid phases co-exists is given when the product of the mole fractions is maximum. This occurs for a Margules mixture at x1 = x2 = 0.5. This temperature is the upper consolute temperature, which for a Margules mixture is given by:

T upper consolute=2∙ A ∙ x1∙ x2

R=2 ∙ A ∙0.5 ∙0.5

R= A2∙ R

II.6.1 Distribution of compounds over two immiscible liquidsThe distribution of species over two in essential immiscible liquids is used industrially for the extraction of material from a process stream. For instance methyl bromide dissolved in water can be extracted by using a hydrocarbon, e.g. n-decane. The hydrocarbon and water are essentially immiscible. The distribution of the compound, which is to be extracted, over the two phases is given by the equilibrium relationship:

T L I=T L II pLI=pLII f iL I=f i

L II

79

(only the compound to be extracted needs to be considered, since the other components in this system remain in essence in the same phase). The fugacity of the compound to be extracted in each of the liquid phases must be equal, and thus

(x i ∙ γi ∙ pivap )LI=( xi ∙ γi ∙ p i

vap )LII


LII The distribution of the compound over the two phases is thus given by:

xiL I

xiL II

=γiL II

γiL I=K=e

Gi ,L II

ex −Gi ,L I

ex

RT

In literature, you will find often a concentration partition or distribution coefficient, which is defined as

K i , L I−LII

C =Ci

LI

CiL II

The partition coefficient is related to the thermodynamic distribution coefficient. If the mole fraction of the species involved in the solution is small, then

x iL I=

n iL I

niL I+nsolvent

L I

≈niL I

nsolventLI

Hence, the concentration of this species is given by:

x iL I≈

niLI

V solvent , L I

∙V solvent , LI

nsolventLI

=CiLI ∙V solvent , L I

K=x iL I

x iL II≈C i

L I ∙V solvent , L I

C iL II ∙V solvent , L II

=K i , LI−L II

C ∙V solvent ,L I

V solvent ,L II

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Distribution of bromine between an organic and aqueous phase The concentration distribution coefficient of bromine (Br2) between carbon tetrachloride (CCl4) and water at 25oC is

[Br2]in CCl4, kmol/m3 0.04 0.1 0.5 1.0 1.5 2.0 2.5

K c=CBr2

CCl 4

CBr2

water 26.8 27.2 29.0 30.4 31.4 33.4 35.0

Calculate the ratio of the activity coefficient of Br2 in water to that in carbon tetrachloride.

DataSpecies M, g/mol , kg/m3

CCl4 153.84 1.595.103

Water 18.0 1.0.103

Br2 159.83 3.119.103

The concentration in the water phase can now easily be calculated since

CBr2

water=CBr2

CCl4

K c

[Br2]in CCl4, kmol/m3 0.04 0.1 0.5 1.0 1.5 2.0 2.5

[Br2]in water, kmol/m3 1.5.10-3 3.7.10-3 1.7.10-2 3.3.10-2 4.8.10-2 6.0.10-2 7.1.10-2

80

The concentration can be re-calculated in terms of the mole fraction. The molar volume of the mixture containing the solvent and bromine assuming that the excess volume can be neglected is given by: V mixture=xsolvent ∙V solvent+xBr2

∙V Br2

V mixture=(1−x Br2 )∙M solvent

ρ solvent

+xBr 2∙MBr 2

ρBr2

The number of moles of bromine in 1 mole of the mixture:

nBr2=[Br2 ] ∙V mixture= [Br2 ] ∙( (1−xBr 2) ∙

M solvent

ρsolvent

+xBr2∙MBr2

ρBr2)=xBr2

Thus, the mole fraction of bromine is given by

xBr2=

[Br2 ] ∙M solvent

ρsolvent

1+[Br2 ] ∙( M solvent

ρ solvent

−M Br2

ρBr2)

(Be careful with the units!)

xBr2 in CCl4 3.85.10-3 9.60.10-3 4.72.10-2 9.23.10-1 1.35.10-1 1.77.10-1 2.17.10-1

xBr2 in water 2.69.10-5 6.62.10-5 3.11.10-4 5.93.10-4 8.61.10-4 1.08.10-3 1.29.10-3

The relative activity coefficient can now be calculated from the equilibrium condition:

(xBr 2∙ γBr2 )

CCl 4=(x Br2∙ γBr2 )

water

γBr2

water

γBr2

CCl4=

xBr2

CCl 4

xBr2

water

[Br2]in CCl4, kmol/m3 0.04 0.1 0.5 1.0 1.5 2.0 2.5

γBr2

water

γBr2

CCl4143.3 145.1 151.9 155.7 157.3 163.8 168.1

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

II.6.2 Ternary phase diagramsA system with three components (ternary systems) have F = C-P+2 = 5-P degrees of freedom. If systems at a given temperature and pressure are considered (i.e. taking away 2 degrees of freedom), the degrees of freedom is F= 3-P. Hence, 2 variables are required to describe a single phase ternary system (i.e. x1 and x2, with x3 = 1-x1-x2). To depict the dependency of the variables, ternary diagrams are being used (see Figure 6.1). The corners in this triangle graph represent the pure components. The lines opposite the corner representing the pure component are lines of constant composition. For instance, the top corner in Figure II.6.2 represents the pure component A. The horizontal lines represent mixtures with a constant mole fraction of A. A mixture containing 40% A and 30% B can be found from the intersection between the line representing mixtures containing 40% A and the line representing mixtures containing 30% B. The shape of the ternary diagram enforces the sum of the mole fractions being equal to 1.

81

0 20 40 60 80 100

A

C BxB

100

80

60

40

20

0

xc

100

80

60

40

20

0

xA

Figure II.6.2: Depicting mixtures in a ternary diagram

The ternary phase diagram is useful to assess whether the mixture will split into two or more liquid phases at the temperature considered. Figure II.6.3 shows the ternary diagram for the system water-methanol-acrylonitrile at 288.15K and atmospheric pressure. It is clear that in the methanol-poor region between mole fractions for acetonitrile of 2% to 91% the appearance of two liquid phases can be expected. It should be noted that a ternary system at a constant pressure and temperature may contain up to 3 different liquid phases (see Figure11.2-11e in Sandler, p. 620 and further examples on p. 621).

The presence of two liquid phases does not yield any information on the composition of these phases in equilibrium with each other. The phases in equilibrium with each other are connected with the tie-line (see Figure II.6.4).

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Liquid-liquid equilibrium in the system water/methanol/acetonitrile A stream contains 40 mole-% acetonitrile and 50 mole-% water with the balance methanol at 288.15 K and atmospheric pressure. Determine the phase composition of each of the liquid phases and the relative amount of each phase present.

The composition of the two liquid phases can be read off from the phase diagram. After locating the point representing the composition of the mixture on the ternary diagram, draw a tie line (almost parallel to the nearest given tie-line). At the crossing point of the tie-line with the LLE-curve, the composition of the two liquid phases can be read off.

82

0 20 40 60 80 100

methanol

water acetonitrilexacetonitrile

100

80

60

40

20

0

xwater

100

80

60

40

20

0

xmethanol

Two liquid region

single liquid region

Figure II.6.3: Ternary phase diagram for the system water-methanol-acrylonitrile at 288.15 K and 1 atm showing LLE-region, i.e. region where two liquids are in equilibrium with each other (data: X. Zhang and C. Jian, J. Chem. Eng. Data 52 (2012), 142)

0 20 40 60 80 100

methanol


100

80

60

40

20

0

xwater

100

80

60

40

20

0

xmethanol

tie-lines

Figure II.6.4: Ternary phase diagram for the system water-methanol-acrylonitrile at 288.15 K and 1 atm showing the composition of the phases in equilibrium with each other (data: X. Zhang and C. Jian, J. Chem. Eng. Data 52 (2012), 142)

83

0 20 40 60 80 100

methanol


100

80

60

40

20

0

xwater

100

80

60

40

20

0

xmethanol

tie-lines

Figure II.6.5: Ternary phase diagram for the system water-methanol-acrylonitrile at 288.15 K and 1 atm showing the splitting of a mixture containing 40 mole-% acetonitrile and 50 mole-% water into two liquid phases (data: X. Zhang and C. Jian, J. Chem. Eng. Data 52 (2012), 142)

Composition of LI: 85 mole-% water, 5 mole-% acetonitrile 10 mole-% methanolComposition of LII: 10 mole-% methanol, 77 mole-% acetonitrile 13 mole-% water

The relative amounts of LI and LII can be obtained from a mole balance:

Mole balance for water: (xwater ∙ nmix )¿=( xwater ∙nL I )L I

+(xwater ∙ nL II )LII

Mole balance for ACN (acetonitrile): (x ACN ∙ nmix )¿=( xACN ∙nL I )L I

+(x ACN ∙ nL II )LII

Define LI=nL I

nmix

and LII=nL II

nmix

Or xwater ,∈¿=xwater

L I ∙ L I+ xwaterLII ∙ LII ¿

xACN ,∈¿=x ACN

L I ∙ L I+x ACNL II ∙ L II¿

And substitute the obtained mole fractions:0.5=0.85 ∙ L I+0.13 ∙ L II

0.4=0.05 ∙ LI+0.77 ∙ LII

LI−0.486LII=0.514

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

84

II.6.3 Vapour-liquid-liquid equilibriaThe composition of co-existing liquids in a liquid-liquid equilibrium is (almost) independent of pressure, although dependent on temperature (see the section on the upper consolute temperature). Hence, decreasing the pressure of a system with co-existing liquids will yield a system where vapour-liquid-liquid equilibrium can be expected.

Determining the bubble point pressure for a system with co-existing liquidsAt the bubble point the overall composition of the liquid mixture is known, and the composition of the co-existing liquids is given by the condition(s):


LII for each component in the mixtureFurthermore, the co-existing liquids are in equilibrium with the vapour phase. For a system at low pressure, the condition is thus given by:

(x i ∙ γi )LI ∙ pi

vap=(x i ∙ γ i )L II ∙ p i

vap= y i ∙P for each component in the mixtureThe bubble point pressure of the mixture containing co-existing liquids is given by:

p=∑i

(x i ∙ γ i)L I ∙ p i

vap

The consequence is that the bubble point pressure is constant in the range of the co-existing liquids (as represented by the horizontal line in Figure 6.6), i.e. it does not vary with the overall composition of the mixture. In the region outside the liquid-liquid equilibrium, the bubble point pressure varies with the composition and is given by:

p=∑i

xi ∙ γi ∙ p ivap

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Bubble point determination in the presence of LLE The binary liquid mixture of nitromethane (NM) and 2,2,5 trimethylhexane (TMH) exhibits liquid-liquid equilibrium. The activity coefficient of nitromethane (NM) and 2,2,5 trimethylhexane (TMH) at 100oC can be described with the following van Laar equations:

ln ( γNM )= 1.5854

(1+1.58542.9428∙xNM

xTMH)2

ln ( γTMH )= 2.9428

(1+ 2.94281.5854∙xTMH

xNM)2

The vapor pressure of the pure components at 100oC are 1.0 bar and 0.5 bar for nitromethane and 2,2,5 trimethyl hexane, respectively. Determine the bubble point pressure as a function of the overall composition in the mixture.

The composition of the co-existing liquids at 100oC can be determined from: (xNM ∙ γNM )L I=(xNM ∙ γNM )L II (xTMH ∙ γTMH )L I=(xTMH ∙ γ TMH )LII

xNML I ∙ e

1.5854

(1+1.58542.9428∙xNMLI

xTMH

LI )

2

=xNML II ∙ e

1.5854

(1+1.58542.9428∙xNML II

xTMH

LII )

2

xTMHL I ∙e

2.9428

(1+2.94281.5854∙xTMHLI

xNM

LI )

2

=xTMHL II ∙ e

2.9428

(1+2.94281.5854∙xTMHL II

xNM

LII )

2

This can be developed into iteration formulae taking into account the sum of the mole fractions in a phase adds up to 1:

85

xNM ,newL I =(1−xTMH , old

L II ) ∙ e

1.5854

(1+1.58542.9428∙

(1− xTMH , oldLII )

xTMH ,oldL II )

2

e

1.5854

(1+1.58542.9428∙

xNM ,oldLI

(1−xNM ,old

LI ) )

2

xTMH ,newL II =(1−xNM , old

LI ) ∙ e

2.9428

(1+2.94281.5854∙1−xNM ,old

L I

xNM ,old

LI )

2

e

2.9428

(1+ 2.94281.5854∙

xTMH ,oldL

II

1− xTMH ,oldLII )

2

Starting values for the iterations would bexNML I =0.1 xTMH

L II =0.1 with the corresponding activity coefficients(note that you cannot take the pure substances as the starting point)

0.1 0.1 4.101585 1.009431 7.557284 1.0474390.22984 0.120214 3.24346 1.058078 6.50457 1.0670450.28944 0.12528 2.904149 1.099988 6.274701 1.072430.32301 0.124566 2.727289 1.130943 6.306373 1.0716590.34399 0.121407 2.621856 1.153482 6.449332 1.0682960.35799 0.117329 2.553647 1.170046 6.641081 1.0640620.36779 0.113111 2.506873 1.182432 6.848414 1.0598110.37494 0.109155 2.473285 1.191892 7.051674 1.0559420.38034 0.105649 2.44823 1.199278 7.239317 1.0526110.38452 0.102655 2.428946 1.205166 7.405377 1.0498390.38785 0.100164 2.413734 1.209939 7.547784 1.0475850.39054 0.09813 2.401511 1.213859 7.667032 1.0457790.39273 0.096491 2.391564 1.217107 7.765113 1.0443470.39454 0.095183 2.383404 1.21981 7.844721 1.0432190.39604 0.094145 2.376678 1.222066 7.908722 1.0423330.39728 0.093325 2.371121 1.223948 7.959839 1.0416390.39830 0.092678 2.366528 1.225517 8.000493 1.0410950.39915 0.092168 2.362733 1.226822 8.032747 1.0406680.39986 0.091766 2.3596 1.227905 8.058308 1.0403330.40043 0.091449 2.357018 1.228802 8.078563 1.040070.40091 0.091198 2.354894 1.229543 8.094624 1.039862

: Starting values

Hence, at 100oC mixtures with a mole fraction of nitro-methane between 0.4 and 0.91 are expected to exhibit phase splitting, i.e. co-existence of two liquids.

The bubble point pressure for mixtures with a mole fraction of nitro-methane between 0.4 and 0.9 is given by:

p=( xNM ∙ γNM )L I , xNM=0.4 ∙ pNMvap+ (xTMH ∙ γ TMH )LI , xNM=0.4 ∙ pTMH

vap p= (0.4 ∙2.35 )L I , xNM=0.4∙1+(0.6 ∙1.23 )LI , xNM=0.4 ∙0.5=1.31 ∙ ¿

For mixtures with a mole fraction of nitro-methane of less than 0.4 and more than 0.91, the bubble pressure is given by :

p=xNM ∙ γNM ∙ pNMvap+xTMH ∙ γTMH ∙ pTMH

vap

The corresponding dew line can be calculated from:

86

y i=x i ∙ γ i ∙ pi

vap

p

At a given composition of the liquid, the activity coefficient and the equilibrium pressure can be calculated. Then, the corresponding mole fraction in the vapour phase can be determined.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Determining the position of the azeotrope The systems exhibiting vapour-liquid-liquid equilibria are highly non-ideal and may thus exhibit an azeotrope. At the azeotrope:

y i ∙ p=xi ∙ γi ∙ p ivap p=γi ∙ p i

vap

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Azeotrope and liquid-liquid equilibrium in the system nitro-methane/2,2,5 trimethyl

hexaneThe binary liquid mixture of nitro-methane (NM) and 2,2,5 trimethyl hexane (TMH) exhibits liquid-liquid equilibrium. The activity coefficient of nitro-methane (NM) and 2,2,5 trimethyl hexane (TMH) at 100oC can be described with the following van Laar equations:

ln ( γNM )= 1.5854

(1+1.58542.9428∙xNM

xTMH)2

ln ( γTMH )= 2.9428

(1+ 2.94281.5854∙xTMH

xNM)2

The vapor pressure of the pure components at 100oC are 1.0 bar and 0.5 bar for nitro-methane and 2,2,5 trimethyl hexane, respectively. Determine the azeotropic point for this mixture at 100oC.

At the azeotropic point, the mole fraction in the vapour and liquid phase are identical, and thusy i ∙ p=xi ∙ γi ∙ p i

vap p=γi ∙ p ivap

The composition of the azeotropic point is given by:p=γNM ∙ pNM

vap=γ TMH ∙ pTMHvap

ln ( γNM )−ln (γ TMH )=ln( pTMHvap

pNMvap )

1.5854

(1+ 1.58542.9428∙x NM

xTMH)2− 2.9428

(1+ 2.94281.5854∙xTMH

xNM)2=−0.693

Solving this equations, yields a mole fraction of nitro-methane of 0.716, which lies within the region where the system exhibits a splitting of the liquid phase into two liquid phases. Hence, the pressure at the azeotropic point is given by

p= (0.4 ∙2.35 )L I , xNM=0.4∙1+(0.6 ∙1.23 )LI , xNM=0.4 ∙0.5=1.31 ∙ ¿ The overall phase diagram for the system nitro-methane/2,2,5 trimethyl hexane at 100oC is given in Figure 6.6. The vapour is either in equilibrium with the liquid L I or with the liquid LII. Only mixtures with the overall composition equal to the azeotropic composition show three phases in equilibrium with each other. This could have been expected, since for a two component system, the degrees of freedom is given by F = C-P+2=4-P. The choice of temperature reduces the degrees of freedom with 1. Hence, F=3-P. When 3 phases are in equilibrium (at a given temperature), the system is fully defined and hence the position at which three phases are in equilibrium with each other are defined by a point in the phase diagram.

87

0

0.5

1

1.5

2

0 0.2 0.4 0.6 0.8 1

Eq

uili

bri

um

pre

ss

ure

, ba

r

Mole fraction of nitromethane

V

LI LIILI+LII

VI+LII

VI+LI heterogeneousazeotrope

bubble line

dew line

Figure II.6.6: Phase diagram for the system nitro-methane/2,2,5 trimethyl hexane at 100oC showing the point of the heterogeneous azeotrope

88

II.7 Equilibrium involving solidsThe discussion on the equilibrium of solids is here limited to solids, which do not form solid mixtures and thus the fugacity of the solid species is that of the pure component

f is=f i

s Furthermore, only phase changes, in which the molecular structure of the component remains unaltered will be considered. Hence, the dissolution of ionic compounds, which (partially) dissociate upon dissolution, will not be considered her.

II.7.1 Dissolution of a solid component in a liquidThe equilibrium solubility of a solid is given by the mole fraction of that component in the liquid phase in equilibrium with the solid phase:

f iL (T mixture , pmixture , xmixture )=f i

s (T mixture , pmixture )

x i ∙ γ i ∙ f iL (T mixture , pmixture )=f i

s (T mixture , pmixture )

The product of the solubility (given by its mole fraction in the liquid phase) and the activity coefficient is thus given by the ratio of the fugacity of the pure component as a solid relative to the fugacity of the same component as a liquid.

x i ∙ γ i=f is (T mixture , pmixture )

f iL (T mixture , pmixture )

=eGi

s (Tmixture , pmixture )−GiL (T mixture , pmixture )

RT

x i ∙ γ i=e−∆ fusG (T mixture , pmixture )

RT

Hence, the product of the solubility and the activity coefficient can be found, when the Gibbs free of fusion at the temperature and pressure of the mixture is known. At the normal melting point, the Gibbs free energy of fusion equals zero (∆ fusG (T melt , p )=0). The fusion and melting process are hardly dependent on pressure, and only the temperature change will be considered. A three step process is used to determine the change in the Gibbs free energy upon fusion at the temperature and pressure of the mixture.

4. Heating the pure component from the temperature of the mixture to the melting temperature

5. Melting the pure component at its melting temperature6. Cooling the liquid formed down to the temperature of the mixture

Enthalpy change for this three step process:

∆ fusH (T mixture )= ∫T mixture

Tmelt

cps ∙ dT+∆ fusH (T melt )+ ∫

T melt

Tmixture

c pL ∙ dT

Entropy change for this three step process:

∆ fus S (T mixture )= ∫T mixture

Tmelt c ps

T∙dT +∆ fusS (T melt )+ ∫

T melt

Tmixture c pL

T∙dT

Defining the difference between the specific heat of the pure compound as a liquid and the specific heat as a pure component as a solid as Cp:

∆ fusH (T mixture )=∆fus H (T melt )+ ∫Tmelt

T mixture

∆Cp ∙ dT

∆ fus S (T mixture )=∆fus S (T melt )+ ∫T melt

T mixture ∆C p

T∙dT

89

At the normal melting temperature 0=∆ fusG (Tmelt , p )=∆ fusH (T melt )−T ∙∆ fusS (T melt )

or ∆ fus S (T melt )=∆ fusH (T melt )

T melt

Substituting it all in

∆ fusG (T mixture , p )=∆ fusH (Tmixture )−T ∙∆mixtureS (T melt )=∆ fusH (T melt ) ∙(1− TTmelt

)+ ∫T melt

Tmixture

∆C p ∙ dT + ∫Tmelt

Tmixture ∆C p

T∙dT

For many substances, the specific heat of the solid is approximately the same as the specific heat of the liquid:

∆ fusG (T mixture , p )≈∆ fusH (T melt ) ∙(1−T mixture

T melt)

And the fugacity coefficient of the pure component as a liquid is thus given by:


s (T mixture , p ) ∙ e∆fusH (Tmelt )

R∙( 1Tmixture

−1

T melt )

Hence, the solubility of a solid in a liquid is given by:

x i ∙ γ i=e

−∆ fusH (T melt )R

∙( 1T mixture

− 1Tmelt

)−∫T melt

T mixture

∆C p ∙dT + ∫T melt

T mixture ∆Cp

T∙ dT

RTmixture

x i ∙ γ i≈ e

−∆fusH (T melt )R

∙( 1T mixture

−1

T melt ) The activity coefficient itself is dependent on the composition and thus an iterative process is needed to determine the solubility of the solid in a liquid. A good starting point is often by using the ideal solubility (i.e. i = 1).

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Solubility of naphthalene in benzene Determine the solubility of naphthalene in benzene at 25oC for an ideal solution, and when the activity coefficient can be described using the regular solution theory. The difference in the specific heat between liquid naphthalene and solid naphthalene can be neglected.Data for naphthalene 1 :

M (g/mol) (g/cm3) Tmelt (K) fusH (kJ/mol) vapH (kJ/mol)128.17 1.0253 353.4 19.01 43.2

1 http://webbook.nist.gov/chemistry/

The Gibbs free energy of fusion at 298.15 K is given by:

∆ fusG (298.15K , p )≈18.96 ∙(1−298.15353,4 )=2.964 ∙ kJmol

Hence, the product of the solubility and the activity coefficient is given by:

xnaphthalene ∙ γ napthalene=e−2964

8.314 ∙298.15=0.302

In an ideal solution, the activity coefficient for naphthalene is equal to 1. Thus, the solubility in an ideal solution is 0.302

90

For a nono-ideal solution the activity coefficient must be considered. According to the regular solution theory, the activity coefficient of naphthalene in benzene is given by: RT ∙ ln (γ napthalene )=(δnaphthalene−δbenzene )2 ∙Φbenzene

2 ∙V naphthalene

The solubility parameter of benzene is given in Table 9.6-1 as 9.2 (cal/cm3)0.5 (or 18.8 (J/cm3)0.5) and the molar volume of 89 cm3/mol. The molar volume for naphthalene can be estimated from its density and molar mass:

V naphthalene=M naphthalene

ρnaphthalene

=128.171.0253

=125.0 ∙ cm3

molThe solubility parameter for naphthalene can then be estimated from the change in the enthalpy upon evaporation assuming that the change in enthalpy upon evaporation is not dependent on temperature (it should actually be evaluated at the temperature of the mixture, i.e. 298.15 K!):

δ naphthalene=√ ∆vapUnaphthalene

V naphtalene

=√ ∆vap H naphthalene−RTV naphtalene

=√ 43200−8.314 ∙298.15125.0=18.047 ∙( J

cm3 )0.5

Thus, the activity coefficient of naphthalene in the naphthalene/benzene mixture is given by:

ln ( γnapthalene )=(18.04−18.8 )2

8.314 ∙298,15∙ ( (1−xnaphthalene ) ∙89

(1−xnaphthalene) ∙89+xnaphthalene ∙125.0 )2

∙125.0

γ napthalene=e

0.03 ∙( (1−xnaphthalene ) ∙89(1−xnaphthalene ) ∙89+ xnaphthalene ∙125.0 )

2

and the solubility of naphthalene can be obtained from:

xnaphthalene=0.302γ napthalene

xnaphthalene ,new=0.302

e0.03 ∙( (1−xnaphthalene ,old ) ∙89

(1− xnaphthalene , old ) ∙89+ xnaphthalene ,old ∙ 125.0)2

Iteration:xnaphthalene benzene naphthalene

0.302461

0.621451

1.011655

0.298977

0.625342

1.011803

0.298933 0.62539

1.011804

0.298933

0.625391

1.011804

The solution of naphthalene in benzene is rather ideal resulting in a high solubility of this compound in benzene as a solvent (note: the regular solution theory overpredicts the solubility of naphthalene in benzene by ca. 5%.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

II.7.2 Freezing point depression

91

The melting point of a component is depressed in the presence of another compound. The point at which a solid starts forming is given by the equilibrium relationship:

f iL (T mixture , pmixture , xmixture )=f i

s (T mixture , pmixture ) x i ∙ γ i ∙ f i

L (T mixture , pmixture )=f is (T mixture , pmixture )

x i ∙ γ i≈ e

−∆fusH (T melt )R

∙( 1T mixture

−1

T melt )

ln (x i ∙ γi )≈−∆fusH (T melt )

R∙( 1T mixture

− 1T melt

) T mixture=

1

1T melt

−R ∙ ln (x i ∙ γ i )∆ fusH (T melt )

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Freezing point depression in the system naphthalene/benzene Pure naphthalene melts at 353.4 K. Determine the melting point in a mixture containing 10% benzene if the solution is ideal and when the activity coefficient can be described using the regular solution theory. The difference in the specific heat between liquid naphthalene and solid naphthalene can be neglected.Data for naphthalene 1,2 : M (g/mol) (g/cm3) Tmelt (K) fusH (kJ/mol) vapH (kJ/mol) (J/cm3)0.5 V (cm3/mol)

128.17 1.0253 353.4 19.01 43.2 18.05 1251 http://webbook.nist.gov/chemistry/2 determined in example in section 7.1

The melting point is given by:

T mixture=1

1T melt


At the melting point, the composition of the mixture is 90% naphthalene and 10% benzene.

γ napthalene=e

0.03 ∙( (1− xnaphthalene )∙ 89(1−xnaphthalene ) ∙89+ xnaphthalene ∙111.9)

2

γ napthalene=e0.03 ∙( 0.1 ∙ 89

0.1 ∙89+0.9 ∙125.0 )2

=1.0002Thus, the mixture is rather ideal!!

For an ideal mixture:

T mixture=1

1T melt

−R ∙ ln (x i )

∆ fusH (T melt )

= 11

353.4−8.314 ∙ ln (0.9 )19010

=347.7K

Thus, a mixture containing 90 mole-% naphthalene melts at a temperature of 6.7 K lower than pure naphthalene.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

II.7.3 Solid-liquid equilibrium phase diagramThe extent of the freezing point depression increases with increasing content of the second component in the mixture. However, also the freezing point of the second component decreases

92

http://webbook.nist.gov/chemistry/

with increasing content of the first component in the mixture. The crossing point of these curves represents the lowest temperature at which this mixture can exist as a liquid. This point is also called the eutectic point.

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Melting point for mixtures of naphthalene and benzene Pure naphthalene melts at 353.4 K and pure benzene at 278.57 K. Determine the melting point of mixtures of benzene/naphthalene as a function of the mole fraction of naphthalene. The heat of fusion of naphthalene is 19.01 kJ/mol and the heat of fusion of benzene is 9.9 kJ/mol. The differences in the specific heat of the solid and the liquid can be neglected and thus the heat of fusion can be considered to be independent of temperature.The napthlene/benzene mixture can be assumed to be ideal

The freezing point of naphthalene can be described by:

T melt ¿mixture ¿napthalene= 1

1T melt

−R ∙ ln ( xi ∙ γi )∆ fusH (Tmelt )

= 1

1353.4

−8.314 ∙ ln (xnaphthalene)

19010

Similarly, the freezing point of benzene can be described by:

T melt ¿mixture ¿benzene= 1

1Tmelt


= 1

1278.57

−8.314 ∙ ln (1−xnaphthalene )

9900

Hence, the curves for the freezing point of naphthalene and benzene from an indeal naphthalene/benzene mixture is given in Figure II.7.1. The lowest temperature for a mixture of naphthalene and benzene to be present as a liquid is at 269.5 K for a mixture containing 13.4 mole-% naphthalene.

200

250

300

350

0 0.2 0.4 0.6 0.8 1

T, K

Mole fraction naphthalene

Figure II.7.1: Freezing point of naphthalene and benzene in an ideal naphthalene/benzene mixture

Now the phase diagram for the solid-liquid transition in the system naphthalene/benzene can be drawn. A decrease in temperature for mixtures rich in naphthalene will precipitate out pure naphthalene, but solid benzene will precipitate out first from mixtures with a content of less than 13.4 mole-% naphthalene (the composition of the eutectic point). The separation of the compounds in a liquid and a pure solid can be used for purification of compounds (just like distillation, but is here called fractionated crystallization).

93

250

275

300

325

350

0 0.2 0.4 0.6 0.8 1

T, K

Mole fraction naphthalene

L

Sbenzene+Snaphthalene

L+SnaphthaleneL+Sbenzene

eutectic point

Figure II.7.2: Phase diagram for an ideal naphthalene/benzene mixture

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

94

III Chemical equilibria

III.1. IntroductionThe principle of chemical equilibria at a given temperature and pressure is the minimization of the Gibbs free energy of system due to the chemical reaction. The Gibbs free energy of a system is given by the sum of the number of moles of each component present in the system time the partial molar Gibbs free energy of these components

Gsystem=∑i=1

i=C

N i ∙Gi

The number of moles of a component is changing due to the reaction and can be can be expressed in terms of the extent of reaction or conversion (see notes on Reactor Design I). For instance, for the reaction

A + b B c C + d Dthe number of moles can be expressed as:

N A=N A, 0−ξN B=N B,0−b ∙ξNC=N C ,0+c ∙ξN D=N D, 0+d ∙ ξ

or in general N i=N i, 0+ν i ∙ ξ

Thus, the Gibbs free energy of the system is given by:

Gsystem=∑i=1

i=C

( N i ,0+ν i ∙ ξ )∙Gi=∑i=1

i=C

N i , 0 ∙Gi+∑i=1

i=C

νi ∙ ξ ∙Gi

The condition for the system to be at equilibrium requires that the change in the Gibbs free energy with changing extent of reaction is zero:

( ∂Gsystem

∂ξ )T , p

=∑i=1

i=C

ν i ∙Gi+∑i=1

i=C

(N i , 0+ν i ∙ ξ ) ∙ ξ ∙( ∂Gi

∂ξ )T , p

=0

It can be derived from the Gibbs-Duhem relationship, that

∑i=1

i=C

(N i ,0+νi ∙ ξ )∙( ∂Gi

∂ξ )T , p

=∑i=1

i=C

N i ∙( ∂Gi

∂ξ )T , p

=0

Hence, the equilibrium condition for a chemical reaction is:

( ∂Gsystem

∂ξ )T , p

=∑i=1

i=C

ν i ∙Gi=0

This can be easily extended to systems with multiple reactions occurring (note: only independent reactions need to be taken into consideration!)

The analysis of the equilibrium position requires therefore knowledge on the partial molar Gibbs free energy of the components in the system at the equilibrium pressure, temperature and composition. A standard reference state can be defined of the species at a defined composition (x i

0) and 1 bar. Hence,

Gi (T , p , x )=Gio (T , p=1 , x i

o )+(Gio (T , p , x )−Gi

o (T , p=1 , x io ))

Gi (T , p , x )=Gio (T , p=1 , x i

o )+RT ∙ ln( f i (T , p , x )

f i (T , p=1 , x io ) )

95

The ratio of the fugacity of the species in a mixture relative to the fugacity of the species at the standard state is called the activity, ai

a i=f i (T , p , x )

f i (T , p=1 , x io )

Thus, Gi (T , p , x )=Gio (T , p=1 , x i

o )+RT ∙ ln (ai )

The position of the equilibrium is thus given by:

( ∂Gsystem

∂ξ )T , p

=∑i=1

i=C

ν i ∙Gi=∑i=1

i=C

ν i ∙Gio (T , p=1 , x i

o )+ν i ∙ RT ∙ ln (ai )=0

∑i=1

i=C

ν i ∙ RT ∙ ln (ai )=−∑i=1

i=C

νi ∙G io (T , p=1 , xi

o )=−∆rxnG

∏i=1

i=C

aiνi=e

−∆rxnGRT

Hence, the thermodynamic equilibrium constant (Ka) is defined as:

Ka=∏i=1

i=C

aiν i=e

−∆rxnGRT

It should be noted that the definition of the activity of a component is strictly linked to the definition of the fugacity of the species at the reference state, and thus the Gibbs free energy of reaction!

96

III.2. Influence of pressure and temperature on the activity based equilibrium constantThe activity based equilibrium constant is defined as:

ln (K a )=−∆ rxnG

RT

Hence, the pressure dependency of the equilibrium constant is given by:

( ∂ ( ln (K a ))∂ p )

T

=( ∂(−∆rxnGRT )∂ p

)T

=−1RT

∙( ∂ (∆rxnG )∂ p )

T

( ∂ ( ln (K a ))∂ p )

T

=−1RT

∙( ∂ (∆rxnG )∂ p )

T

=−1RT

∙( ∂∑ ( ν i ∙Gio (T , p=1 , x i

o ))∂ p )

T

Since the partial molar Gibbs free energy of the components is defined at 1 bar, it is therefore not dependent on pressure:

−1RT

∙( ∂∑ (ν i ∙Gio (T , p=1 , x i

o ))∂ p )

T

=0

And hence the activity based equilibrium constant does not change with changing pressure at constant temperature.

The temperature dependency of the equilibrium constant is given by:

( ∂ ( ln (K a ))∂T )

p

=( ∂(−∆rxnGRT )∂T

)p

=∆ rxnH

RT 2

Hence, the change in the activity based equilibrium constant with changing temperature is given by

ln (K a (T2 ))−ln (K a (T 1 ))=∫T 1

T 2 ∆rxnH

RT2∙ dT

ln (K a (T 2 )K a (T 1 ) )=∫T 1

T 2 ∆rxnH

RT 2∙ dT

It should be noted that the enthalpy (and thus also the change in the enthalpy upon reaction) is dependent on temperature:

( ∂ H∂T )

p

=cP or H (T , p )=H (T ref , p )+∫T ref

T

c P∙ dT

The ideal gas specific heat can often be described as a polynomial as a function temperaturec P=a+b ∙T +c ∙T 2+d ∙T 3

Defining the change in the specific heat upon reaction as:∆ rxn cP=ν i ∙ cP, i=∆ a+∆b ∙T+∆c ∙T 2+∆d ∙T 3

And hence the enthalpy change upon reaction as a function of temperature is given by:

∆ rxnH (T , p )=∆rxn H (T ref , p )+∆ a∙ (T−T ref )+∆b2

∙ (T−T ref )2+∆c3

∙ (T−T ref )3+∆ d4

∙ (T−T ref )4

∆ rxnH (T , p )=(∆rxnH (T ref , p )−∆ a∙T ref−∆b2

∙T ref2 −∆c

3∙ T ref

3 −∆d4

∙ Tref4 )+∆a ∙T+ ∆b

2∙ T2+∆c

3∙ T 3+∆ d

4∙T 4

97

Thus, the temperature dependency of the activity based equilibrium constant is given by:

ln (K a (T 2 )K a (T 1 ) )=∫T 1

T 2 ∆rxnH

RT 2∙ dT

ln (K a (T 2 )K a (T 1 ) )=∫T 1

T 2 ((∆rxn H (T ref , p )−∆a ∙T ref−∆b2

∙T ref2 −

∆ c3

∙ T ref3 −

∆d4

∙ T ref4 )

RT2+

∆aR ∙T

+∆b2R

+∆c3R

∙T +∆d4 R

∙T 2) ∙ dT

ln (K a (T 2 )K a (T 1 ) )=

(−∆ rxnH (T ref , p )+∆a ∙T ref +∆b2

∙T ref2 +∆c

3∙T ref

3 +∆d4

∙ T ref4 )

R∙( 1T 2− 1

T 1 )+∆aR

∙ ln (T 2T 1 )+∆b2R

∙ (T 2−T 1 )+∆ c6R

∙ (T 22−T12 )+ ∆d12 R

∙ (T 23−T 13 )

98

III.3 Single gas phase reactionsFor gas phase reactions it is convenient to define the reference state as the pure components at 1 bar. Hence, the activity of the components in the mixture is given by:

a i=f i (T , p , x )

f iV ¿¿

The activity based equilibrium constant is thus related to the Kp, since

Ka=∏i=1

i=C

aiν i=∏

i=1

i=C

¿¿¿

The activity based equilibrium constant has the same numerical value as Kp (when the units bar are used!), if the Lewis-Randall rule is applicable and the fugacity coefficient is equal to 1, but does not have the same units. The activity based equilibrium constant is dimensionless, whereas Kp has the units of pressure.

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Single gas phase reaction with a constant number of moles – water gas shift reaction The water gas shift reaction H2O(g) + CO(g) CO2(g) + H2(g) is an important chemical reaction. This reaction is typically performed at temperatures of ca. 400oC and elevated pressure.The following data can be found for the various components:Compound fH (298.15K) fG (298.15K) c P=a+b ∙T +c ∙T 2+d ∙T 3, (J/mol.K)

kJ/mol kJ/mol a b c dH2O(g) -241.8 -228.6 32.218 0.192.10-2 1.055.10-5 -3.593.10-9

CO(g) -110.5 -137.2 28.142 0.167.10-2 0.537.10-5 -2.221.10-9

CO2 (g) -393.5 -394.4 22.243 5.977.10-2 -3.499.10-5 7.464.10-9

H2 (g) 0 0 29.088 -0.192.10-2 0.400.10-5 -0.870.10-9

1. Determine the change in the Gibbs free energy upon reaction at 298.15 K.2. Determine the change in the Gibbs free energy upon reaction at 673.15 K.3. Determine the change in the Gibbs free energy of reaction relative to the Gibbs free energy of

the starting mixture as a function of the conversion of CO for an equi-molar feed of water and CO at a temperature of 673.15 K and 1 bar

4. Determine the equilibrium conversion of CO for an equi-molar feed at 673.15 K and 1 bar5. Determine the equilibrium conversion of CO for an equi-molar feed at 673.15 K and 10 bar6. Determine the equilibrium conversion of CO for an equi-molar feed at 673.15 K and 10 bar

for a feed containing 20 mole-% CO and the balance water.

1 Determine the change in the Gibbs free energy upon reaction at 298.15 K.The change in the Gibbs energy upon reaction is given by:

∆ rxnG=∑i=1

i=C

ν i ∙Gio (T , p=1 , x i

o )

∆ rxnG=−GCOo (T , p=1 , x i

o )−GH2Oo (T , p=1 , x i

o )+GCO2

o (T , p=1 , x io )+GH 2

o (T , p=1 , x io )

Taking the reference states as the pure components at a pressure of 1 bar: ∆ rxnG=−GCO

o ¿

We can relate the Gibbs free energy of the various components (although the Gibbs free energy of the pure components is not known), by using the elements in their natural state at 298.15 K and 1 bar as the reference state. ∆ rxnG (298.15K )=−∆f GCO

o ¿

∆ rxnG (298.15K )=− (−137.2 )−(−228.6 )+(−394.4 )+(0 )=−28.6 ∙ kJmol

99

2 Determine the change in the Gibbs free energy upon reaction at 673.15 K.The change in the Gibbs free energy with temperature (at a constant pressure is given by):

( ∂(∆ rxnGT )

∂T)p

=−∆ rxnH

T2

This differential equation can be solved by separation of variables:

∫T 1

T 2

d (∆rxnGT )=∫

T1

T2 −∆ rxnH

T2∙ dT

(∆ rxnGT )

T 2

−( ∆rxnGT )

T1

=∫T1

T2 −∆ rxnH

T 2∙ dT

In order to evaluate the change in the Gibbs free energy upon reaction, the function of the reaction enthalpy as a function of temperature must be known.

∆ rxnH (T )=∆ rxnH (T ref , p )+∫T ref

T

∆ rxn cP∙ dT

∆ rxnH (298.15K )=− (−110.5)− (−241.8 )+(−393.5 )+(0 )=−41.2 ∙ kJmol

The change in enthalpy upon reaction at 298.15 K is given by: ∆ rxnH=−∆ f HCO

o ¿

∆ rxnH (298.15K )=− (−137.2 )−(−228.6 )+ (−394.4 )+(0 )=−41.2 ∙ kJmol

∆ rxn cP=∆rxna+∆ rxnb ∙T+∆rxn c ∙T2+∆rxn c ∙T

3

with ∆ rxna=−aCO−aH 2O+aCO2

+aH 2

∆ rxna=−28.142−32.218+22.243+29.088

1000=−9.029 ∙10−3 ∙ kJ

mol ∙ K

∆ rxnb=5.43 ∙10−5∙

kJ

mol ∙ K2

∆ rxn c=−5.09∙10−8 ∙kJ

mol ∙K3

∆ rxnd=1.24 ∙10−11 ∙

kJ

mol ∙ K3


T

(∆rxna+∆rxnb ∙T +∆ rxnc ∙T2+∆ rxnc ∙T

3 ) ∙ dT

∆ rxnH (T )=∆ rxnH (T ref , p )−∆ rxna ∙T ref −∆rxnb

2∙ T ref

2 −∆rxn c

3∙T ref

3 −∆ rxnd

4∙T ref

4 +∆ rxna ∙T+∆ rxnb

2∙T 2+

∆rxn c

3∙ T3+

∆ rxnd

4∙ T 4

∆ rxnH (T )=−40.4944+∆rxna∙T +∆rxnb

2∙ T 2+

∆ rxnc

3∙ T 3+

∆rxnd

4∙T 4

100

∫T 1

T 2 −∆ rxnH

T 2∙dT=∫

T 1

T 2 −−40.4944+∆rxna ∙T +∆rxnb

2∙ T 2+

∆ rxn c

3∙ T 3+

∆rxnd

4∙T 4

T 2∙ dT

∫T 1

T 2 −∆ rxnH

T 2∙dT=∫

T 1

T 2

40.4944

T2−

∆rxnaT

−∆ rxnb2

−∆rxn c3

∙T−∆ rxnd4

∙ T 2 ∙ dT

∫298.15

673.15 −∆rxn H

T 2∙ dT=−40.4944 ∙( 1

673.15− 1298.15 )−−9.029 ∙10−3 ∙ ln( 673.15298.15 )−

∆rxnb

2∙ (673.15−298.15 )−

∆rxn c

6∙ (673.152−298.152 )−

∆rxnd

12∙ (673.153−298.153 )

∫298.15

673.15 −∆rxn H

T 2∙ dT=0.0755

Thus, the change in the Gibbs free energy upon reaction at 673.15 K is given by:

(∆ rxnG

T )T=673.15K

−(∆rxnG

T )T=298.15K

= ∫T=298.15K

T=673.15K −∆ rxnH

T2∙ dT=0.0755

(∆ rxnG

T )T=673.15K

=(∆rxnG

T )T=298.15K

+0.0755=( −28.6298.15 )T=298.15K

+0.0755=−0.02042

∆ rxnG (673.15K )=673.15 ∙ (−0.02042 )=−13.75∙ kJmol

3 Determine the change in the Gibbs free energy of reaction relative to the Gibbs free energy of the starting mixture as a function of the conversion of CO for an equi-molar feed of water and CO at a temperature of 673.15 K and 1 bar

The number of moles in the system as function of the conversion of CO can be found from the stoichiometric table. Define XCO as the conversion of CO:

Compound Initial Final yi

H2O(g) N H 2O ,0=N CO, 0 NCO ,0 ∙ (1−X ) 0.5 ∙ (1−X )CO(g) NCO ,0 NCO ,0 ∙ (1−X ) 0.5 ∙ (1−X )CO2 (g) 0 NCO ,0 ∙ X 0.5 ∙ XH2 (g) 0 NCO ,0 ∙ X 0.5 ∙ X

∑ ¿2 ∙ NCO, 0

Thus the Gibbs free energy of the system can now be defined in terms of a single variable X Gsystem=NCO , 0 ∙ (1−X ) ∙GH 2O

+NCO ,0 ∙ (1−X ) ∙GCO+NCO ,0 ∙ X ∙GCO2+N CO, 0 ∙ X ∙GH 2

or per mole of CO entering the system:

Gsystem

N CO, 0

= (1−X ) ∙GH 2O+(1−X ) ∙GCO+X ∙GCO2

+X ∙GH 2

The partial molar Gibbs free energy can be related to the Gibbs free energy of the pure component at 1 bar: GCO=GCO¿

Thus, the Gibbs free energy of the system

101

Gsystem

N CO, 0

=GCO+GH 2O+X ∙ (GCO2

+GH2−GCO−GH 2O )+RT ∙ ( (1−X ) ∙ ln ( yH 2O )+ (1−X ) ∙ ln ( yCO )+X ∙ ln ( yCO2 )++X ∙ ln ( yH 2 ))

and the change in the Gibbs free energy of the system as a function of conversion is given by: Gsystem

N CO, 0

−GCO−GH 2O=X ∙∆ rxnG+RT ∙((1−X ) ∙ ln( 1−X

2 )+ (1−X ) ∙ ln( 1−X2 )+X ∙ ln( X2 )++X ∙ ln( X2 ))

The change in the Gibbs free energy of the system contains a linear term in terms of the conversion (proportional to the change in the Gibbs free energy upon reaction, rxnG) and a non-linear term, due to the mixing of the reactants and the products (see Fig. 3.1). The Gibbs free energy of the system is minimal at full conversion, if the products and reactants do not mix. A resulting partial equilibrium conversion is caused by the contribution of the mixing term to the Gibbs free energy of the system.

-20

-15

-10

-5

0

0 0.2 0.4 0.6 0.8 1

Gs

yste

m/N

CO

,0-G

CO-G

H2O

-2R

T.ln

(2)

CO-conversion, XCO

DmixG(mixing of reactants and poducts)

X.DrxnG

Figure III.3.1: The change in the Gibbs free energy of the system relative to the Gibbs free energy of the starting mixture at 673.15 K for an equi-molar feed of CO and water at 1 bar.

4 Determine the equilibrium conversion of CO for a an equi-molar feed at 673.15 K and 1 barThe equilibrium constant for the water-gas shift reaction at 673.15 K is given by:

Ka=e−∆rxnG (T=673.15K )

RT =e−(−13.75) ∙10008.314 ∙673.15 =11.67

The equilibrium composition in the gas in the gas phase is related to the equilibrium constant Ka

Ka=aH 2

∙ aCO2

aCO ∙ aH 2O

The activity of hydrogen is given as:

aH 2=

f H 2(T , p , x )

f H 2

V ¿¿ at low pressure

Thus, the activity based equilibrium constant:

Ka=yH 2

∙P

1 ¿ ∙

yCO2∙ P

1 ¿yCO ∙ P

1 ¿ ∙yH 2O

∙P

1 ¿=

yH 2∙ yCO2

yCO ∙ yH2O

¿¿¿

¿

102

The mole fractions are defined in terms of the conversion of CO:

Ka=yH 2

∙ yCO2

yCO ∙ yH 2O

= 0.5∙ X ∙0.5 ∙ X0.5 ∙ (1−X ) ∙0.5 ∙ (1−X )

= X2

(1−X )2=11.67

Solving for the conversion of CO XCO = 0.7735

5 Determine the equilibrium conversion of CO for a an equi-molar feed at 673.15 K and 10 barThe Gibbs free energy of reaction and thus the activity based equilibrium constant are not a function of pressure. Hence

Ka=e−∆rxnG (T=673.15K )

RT =e−(−13.75 ) ∙10008.314 ∙673.15 =11.67

The composition in the gas phase is also not dependent on the pressure. since

Ka=¿ X2

(1−X )2

both the left-had side of the equation and the right hand side of the equation are independent of pressure. Hence, the equilibrium conversion remains unchanged upon changing the pressure at 77.35%.

6 Determine the equilibrium conversion of CO for a an equi-molar feed at 673.15 K and 10 bar for a feed containing 20 mole-% CO and the balance water.

The equilibrium constant is not a function of the inlet composition. However, the resulting equilibrium composition of the gas phase is dependent on the starting composition of the mixture. Setting up the stoichiometric table for a feed containing 20 mole-% CO and the balance water:


H2O(g) N H 2O ,0=4 ∙ NCO, 0 NCO ,0 ∙ (4−X ) 0.2 ∙ (4−X )CO(g) NCO ,0 NCO ,0 ∙ (1−X ) 0.2 ∙ (1−X )CO2 (g) 0 NCO ,0 ∙ X 0.2 ∙ XH2 (g) 0 NCO ,0 ∙ X 0.2 ∙ X

∑ ¿5 ∙ NCO ,0

The gas phase composition is given by:

Ka=aH 2

∙ aCO2

aCO ∙ aH 2O

=yH 2

∙ P

1 ¿ ∙

yCO2∙P

1 ¿yCO ∙P

1 ¿ ∙yH 2O

∙P

1 ¿=

yH 2∙ yCO2

yCO ∙ yH 2O

=¿ 0.2∙ X ∙0.2∙ X0.2 ∙ (4−X ) ∙0.2 ∙ (1−X )

= X 2

(4−X ) ∙ (1−X )=11.67¿

¿¿

¿

Solving for the conversion of CO XCO = 0.9639

Hence, the equilibrium conversion can be improved by increasing the excess of water fed into the system (but not by changing the pressure).••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

103

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Single gas phase reaction with changing number of moles – ammonia synthesis The ammonia synthesis 3 H2(g) + N2(g) 2NH2(g) is an important chemical reaction. This reaction is typically performed at temperatures of ca. 450oC and pressures as high as 300 bar.The following data can be found for the various components:Compound fH (298.15K) fG (298.15K) c P=a+b ∙T +c ∙T 2+d ∙T 3, (J/mol.K)

kJ/mol kJ/mol a b c dH2 (g) 0 0 29.088 -0.192.10-2 0.400.10-5 -0.870.10-9

N2 (g) 0 0 28.883 -0.157.10-2 0.808.10-5 -2.871.10-9

NH3 (g) -46.1 -16.5 24.619 3.75.10-2 -0.138.10-5

1. Determine the change in the Gibbs free energy upon reaction at 723.15 K.2. Determine the equilibrium conversion of N2 for stoichiometric feed of H2 and N2 at 723.15 K

and 1 bar3. Determine the equilibrium conversion of N2 for stoichiometric feed of H2 and N2 at 723.15 K

and 10 bar4. Determine the equilibrium conversion of N2 for a feed containing 90% Ar, 7.5 % H2 and 2.5%

N2 at 723.15 K and 10 bar5. Determine the equilibrium conversion of N2 for stoichiometric feed of H2 and N2 at 723.15 K

and 100 bar

1 Determine the change in the Gibbs free energy upon reaction at 723.15 K.The change in the Gibbs energy upon reaction at 298.15K is given by: ∆ rxnG (298.15K )=−∆f GN2

o ¿(taking the reference states as the pure components at a pressure of 1 bar and relating the Gibbs free energy of the various components (although the Gibbs free energy of the pure components is not known), by using the elements in their natural state at 298.15 K and 1 bar as the reference state)

∆ rxnG (298.15K )=− (0 )−3 ∙ (0 )+2 ∙ (−16.5 )+ (0 )=−33.0 ∙ kJmol N2

The change in the Gibbs free energy with temperature (at a constant pressure of 1 bar is given by):

(∆ rxnGT )

T 2

−( ∆rxnGT )

T1

=∫T1

T2 −∆ rxnH

T 2∙ dT

In order to evaluate the change in the Gibbs free energy upon reaction, the function of the reaction enthalpy as a function of temperature must be known.


T

∆ rxn cP∙ dT

∆ rxnH (298.15K )=− (0 )−3 ∙ (0 )+2 ∙ (−46.1 )=−92.2∙ kJmol N2

∆ rxn cP=∆rxna+∆ rxnb ∙T+∆rxn c ∙T2+∆rxn c ∙T

3

with ∆ rxna=−aN 2−3 ∙ aH2

+2 ∙ aNH3

∆ rxna=−28.883−3∙29.088+2 ∙24.619

1000=−6.69 ∙10−2∙

kJ

(mol N2 )∙ K

∆ rxnb=7.92 ∙10−5∙

kJ

(mol N2 ) ∙K2

104

∆ rxn c=−2.28 ∙10−8 ∙kJ

(mol N2 ) ∙K3

∆ rxnd=5.48 ∙10−12 ∙

kJ

(mol N 2 )∙ K3


T

(∆rxna+∆rxnb ∙T +∆ rxnc ∙T2+∆ rxnc ∙T

3 ) ∙ dT

∆ rxnH (T )=∆ rxnH (T ref , p )−∆ rxna ∙T ref −∆rxnb

2∙ T ref

2 −∆rxn c

3∙T ref

3 −∆ rxnd

4∙T ref

4 +∆ rxna ∙T+∆ rxnb

2∙T 2+

∆rxn c

3∙ T3+

∆ rxnd

4∙ T 4

∆ rxnH (T )=−75.58−6.69∙10−2 ∙ T+ 7.92∙10−5

2∙ T 2+−2.28 ∙10−8

3∙ T 3+ 5.48 ∙10

−12

4∙T 4

∫298.15

723.15 −∆rxn H

T 2∙ dT=0.1930 ∙ kJ

(mol N 2) ∙ K

Thus, the change in the Gibbs free energy upon reaction at 723.15 K is given by:

(∆ rxnG

T )T=723.15K

−(∆rxnG

T )T=298.15K

= ∫T=298.15K

T=723.15K −∆rxn H

T2∙ dT=0.1930

∆ rxnG (723.15K )=59.48 ∙ kJmol

2 Determine the equilibrium conversion of N2 for stoichiometric feed of H2 and N2 at 723.15 K and 1 bar

The equilibrium composition in the gas phase is given by:

Ka=aN H 3

2

aN 2∙ aH 2

3

with the equilibrium constant given by Ka=e−∆rxnG (T=723.15K )

RT =e−(59.5 )∙ 10008.314 ∙ 673.15=5.05 ∙10−5

At low pressure, Ka=¿¿¿

The mole fractions of the individual components can be obtained from a stoichiometric table:


H2 (g) N H 2 ,0=3 ∙ NN 2 ,0 3 ∙N N2 , 0

∙ (1−X ) 3 ∙ (1−X )(4−2 ∙ X )

N2(g) N N2 ,0 N N2 ,0∙ (1−X )

(1−X )

(4−2 ∙ X )NH3 (g) 0 2 ∙N N2 , 0

∙ X 2∙ X

(4−2 ∙ X )

105

∑ ¿N N2 , 0∙ (4−2 ∙ X )

Thus, the equilibrium gas phase composition at 723.15 K and 1 bar is given by:

Ka=yN H 3

2

yN2∙ yH 2

3 ∙¿¿

Solving for the conversion of nitrogen yields X = 0.004597 (i.e. only 0.45% of nitrogen will be converted at atmospheric pressure).


The equilibrium constant is not dependent on pressure and thus

Ka=e−∆rxnG (T=723.15K )

RT =e−(59.5 )∙ 10008.314 ∙ 673.15=5.05 ∙10−5

However, the equilibrium gas phase composition is dependent on pressure. At low pressure, the equilibrium gas phase composition is given by:

Ka=¿¿¿

( 2 ∙ X(4−2 ∙ X ) )

2

( (1−X )(4−2 ∙ X ) ) ∙( 3 ∙ (1−X )

(4−2 ∙ X ) )3=Ka ∙¿¿

Thus, at 10 bar

( 2 ∙ X(4−2 ∙ X ) )

2

( (1−X )(4−2 ∙ X ) ) ∙( 3 ∙ (1−X )

(4−2 ∙ X ) )3=5.05 ∙10

−5 ∙100

Solving for the conversion of nitrogen X = 0.04413

Thus, increasing the pressure affects the equilibrium gas phase composition for reactions with a change in the number of moles.

4 Determine the equilibrium conversion of N2 for a feed containing 90% Ar, 7.5 % H2 and 2.5% N2

at 723.15 K and 10 barAt a first glance, the equilibrium conversion might be similar to one obtained for a total pressure of 1 bar, since the partial pressures of the reactants (N2 and H2) in this system add up to 1 bar. However, the partial pressure of the inert (Ar) does not remain at 9 bar during the reaction due to the contraction of the reaction. Setting up a stoichiometric table (including the inert component):


H2 (g) N H 2 ,0=3 ∙ NN 2 ,0 3 ∙N N2 , 0

∙ (1−X ) 3 ∙ (1−X )

(40−2∙ X )

106

N2(g) N N2 ,0 N N2 ,0∙ (1−X )

(1−X )

(40−2∙ X )NH3 (g) 0 2 ∙N N2 , 0

∙ X 2∙ X

(40−2∙ X )Ar (g) N H 2 ,0

=36 ∙ N N2 ,036 ∙N N 2 ,0

36(40−2∙ X )

∑ ¿N N2 , 0∙ (40−2 ∙ X )

The equilibrium conversion is given by:Ka=¿¿¿

( 2∙ X(40−2∙ X ) )

2

( (1−X )(40−2 ∙X ) )∙( 3 ∙ (1−X )

(40−2 ∙ X ) )3=Ka ∙¿¿

Thus, at 10 bar

( 2∙ X(40−2∙ X ) )

2

( (1−X )(40−2 ∙X ) )∙( 3 ∙ (1−X )

(40−2 ∙ X ) )3=5.05 ∙10

−5∙100

Solving for the conversion of nitrogen yields X = 0.004589 (i.e. only 0.45% of nitrogen will be converted at atmospheric pressure). The contraction of the chemical reaction (reduction in the number of moles due to the reaction) results in a (in this case slight) increase in the partial pressure of the inert component (Ar), resulting in a lower conversion than when the reaction pressure was 1 bar.


The equilibrium constant is independent of pressure:

Ka=e−∆rxnG (T=723.15K )

RT =e−(59.5 )∙ 10008.314 ∙ 673.15=5.05 ∙10−5

and the equilibrium gas phase composition is given through the equilibrium constant Ka:

Ka=aN H 3

2

aN 2∙ aH 2

3

The activities of the components are given as:

aN2=

f N2(T , p , x )

f N2

V ¿¿

aH 2=

f H 2(T , p , x )

f H 2

V ¿¿

aNH3=

f NH 3(T , p , x )

f H 2

V ¿¿

The mole fractions are given from the stoichiometric table.Ka=¿¿¿

107

Ka=(2 ∙ X )2 ( (4−2 ∙X ) )2

(1−X )4 ∙33∙¿¿

The problem here is that the fugacity coefficients ( f iy i ∙ P ) themselves depend on the composition in

the gas phase as well. Hence, we need an iterative procedure to compute the equilibrium

conversion. A reasonable starting value would be ( f iy i ∙ P )=1. The fugacity coefficients coefficient

can then be predicted using the Peng-Robinson equation of state (here the binary interaction coefficients have been set equal to zero; application of the binary interaction coefficient as defined by J.O. Valderama and E.A. Molina, Fluid Phase Equilibria 31 (1986), 209 does not seem to change the fugacity coefficients substantially.

XN2 yNH3 yN2 yH2 fH2/(yH2.p) fN2/(yN2

.p) fNH3/(yNH3.p)

0.279005 0.162118 0.20947 0.628411 1.0228 1.0356 0.994570.288888 0.168831 0.207792 0.623377 1.0229 1.0357 0.994190.288888 0.168831 0.207792 0.623377 1.0229 1.0357 0.99419

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

108

III.4 Multiple gas phase reactions taking place simultaneouslyOften, more than one reaction can take place in a system. In this case, various approaches are possible. The most robust approach is the use of Lagrange multipliers. The Gibbs free energy of the system must be minimal at equilibrium:

Gsystem=∑i=1

i=C

N i ∙Gi or Gsystem

RT=∑

i=1

i=C

N i ∙Gi

RT or

Furthermore, the elements are conserved in a system with chemical reactions occurring. This results in constraints for the system. The elemental balance for element C can be written as

gC=∑i=1

i=C

nC ∙N i−nC , feed=0

and for each element:

gk=∑i=1

i=C

nk ∙N i−nk , feed=0

A new function can be generated:

F=∑i=1

i=C

N i ∙Gi

RT− ∑

k=1

k=no . elements

λk ∙ gk

A new set of variables is introduced here (i, the Lagrange multipliers). This new function must be minimal at equilibrium as well. Hence, the variation of this of function with respect to N i and the Lagrange multipliers, i, must be equal to zero.

( ∂ F∂ N j

)T , p , N i≠ j, λ i

=G j

RT+∑

i=1

i=C N i

RT∙( ∂Gi

∂N j)T , p , N i≠ j , λi

+ ∑k=1

k=no . elements

λk ∙( ∂ gk

∂N j)T , p , N i ≠ j

=0

( ∂F∂ λ j)T , p , N i , λ j≠ i

=g j=0

Realizing that ∑i=1

i=C

N i ∙( ∂Gi

∂ N j)T , p , N i≠ j, λ i

=0 according to the Gibbs-Duhem equation, this set of C+k

equations simplifies to:

( ∂ F∂ N j

)T , p , N i≠ j, λ i

=G j

RT+ ∑

k=1

k=no. elements

λk ∙( ∂gk

∂ N j)T , p , N i≠ j

=0

( ∂F∂ λ j)T , p , N i , λ j≠ i

=g j=0

The partial molar Gibbs free energy of a compound in the mixture can be expressed in term of the Gibbs free energy of the pure compound and the activity of the compound

( ∂ F∂ N j

)T , p , N i≠ j, λ i

=G j0

RT+ ln (a j )+ ∑

k=1

k=no .elements

λk ∙( ∂ gk

∂ N j)T , p , N i≠ j

=0

( ∂F∂ λ j)T , p , N i , λ j≠ i

=g j=0

The activity of compound j can at low pressure be set equal to

a j=y j ∙ P

1 ¿=N j

∑ N i

∙P1 ¿ ¿

¿

(at higher pressure the fugacity coefficient has to be evaluated during the minimization process)

109

Thus,

( ∂ F∂ N j

)T , p , N i≠ j, λ i

=G j0

RT+ ln ¿

Hence, the following non-linear equations have to be solved simultaneously:G j0

RT+ ln ¿

g j=0

Alternatively, the Gibbs free energy of the system can be minimized subject to the constraints of the elemental balances:

Gsystem=∑i=1

i=C

N i ∙Gi=∑i=1

i=C

N i ∙¿¿

With the constraintsgk=0

Finding the minimum in the Gibbs free energy can be obtained using e.g. Solver® in Excel®.

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Multiple gas phase reactions – methanol/dimethylether synthesis The methanol synthesis from a feed containing 28% CO, 4% CO2 and 68% H2 at 250oC is equilibrium limited. The following reactions may occur in this system

CO + 2 H2 CH3OHCO2 + 3 H2 CH3OH +H2O

CO + H2O CO2 +H2

To obtain higher conversions per pass, a side reaction is allowed to occur, viz.2 CH3OH CH3OCH3 + H2O

What is the equilibrium conversion of CO with and without the formation of dimethylether for the reaction at 50 bar and 250oC?

The following data can be found for the various components:Compound fH (298.15K) fG (298.15K) c P=a+b ∙T +c ∙T 2+d ∙T 3, (J/mol.K)

kJ/mol kJ/mol a b c dH2O(g) -241.8 -228.6 32.218 0.192.10-2 1.055.10-5 -3.593.10-9

CO(g) -110.5 -137.2 28.142 0.167.10-2 0.537.10-5 -2.221.10-9

CO2 (g) -393.5 -394.4 22.243 5.977.10-2 -3.499.10-5 7.464.10-9

H2 (g) 0 0 29.088 -0.192.10-2 0.400.10-5 -0.870.10-9

CH3OH (g) -200.7 -162.0 19.038 9.146.10-2 -1.218.10-5 -8.034.10-9

CH3OCH3 (g)1 -184.1 -112.8 15.511 18.690.10-2 -6.379.10-5 -3.174.10-9

1: Data from Perry’s Chemical Engineers’ Handbook, 7th ed.

Step 1: Calculate the Gibbs free energy of the species at 250oC and 1 bar with respect to the elements at their natural state at 298.15 K and 1 bar

The Gibbs free energy of a species with respect to the elements at their natural state at 298.15 K is given by:

(Gi

T )T=523.15K

=( ∆f G

T )T=298.15K

+ ∫T=298,15

T=523,15 −H i

T 2∙ dT

with the enthalpy of a species with respect to the elements at their natural state at 298.15 K:

110

H i=∆f H (298.15K )+ ∫T=298.15K

T

c p ,i ∙ dT

H i=∆f H (298.15K )+ ∫T=298.15K

T

(a+b ∙T +c ∙T 2+d ∙T 3 )∙ dT

H i=∆f H (298.15K )−a∙298.15−b2∙298.152− c

3∙298.153−d

4∙298.154+a ∙T + b

2∙ T2+ c

3∙ T3+ d

4∙ T 4

∫T=298,15

T=523,15 −H i

T 2∙ dT=(∆ f H (298.15K )−a ∙298.15−b

2∙298.152− c

3∙298.153−d

4∙298.154) ∙( 1

523.15− 1298.15 )−a∙ ln (523.15298.15 )−b

2∙ (523.15−298.15 )− c

6∙ (523.152−298.152 )+ d

12∙ (523.153−298.153 )

Putting it all togetherCompound fH (298.15K) fG (298.15K) Gi (523.15K)1

kJ/mol kJ/mol kJ/molH2O(g) -241.8 -228.6 -210.4CO(g) -110.5 -137.2 -151.8CO2 (g) -393.5 -394.4 -378.8H2 (g) 0 0 -1.9CH3OH (g) -200.7 -162.0 -129.8CH3OCH3 (g) -184.1 -112.8 -61.1

1 Gibbs free energy of the compounds with respect to their elements at 298.15 K and 1 bar in their natural state

Step 2: Gibbs free energy of the systemThe Gibbs free energy of the system is given by

Gsystem=∑i=1

i=C

N i ∙Gi=∑i=1

i=C

N i ∙¿¿

Thus for the system excluding dimethyletherGsystem=N H 2O

∙¿ (in this equation the number of moles of H2O, CO, CO2, H2, and CH3OH are unknown)

Step 3: Elemental balancesThe following elements need to be considered: C, O and HC-balance gC=N CO+NCO2

+NC H 3OH+2 ∙N CH 3OCH 3−nC ,feed=0

O-balance gO=NH 2O+NCO+2 ∙NCO2

+NC H 3OH+NC H 3OCH 3−nO,feed=0

H-balance gH=2 ∙N H 2O+2∙ N H 2

+4 ∙ NCH 3OH+6 ∙NC H 3OCH 3−nH , feed=0

Taking a basis of 1 mol of CO fed to the system:

nC , feed=nCO , feed+nCO 2, feed=1+ 4

28=1.14

nO ,feed=nCO, feed+2 ∙ nCO2 , feed=1+2 ∙ 4

28=1.28

nH , feed=2 ∙ nH2 , feed=2 ∙ 68

28=4.86

Step 3: Setting up the spreadsheet

111

Finding the minimum Gibbs free energy of the system is a somewhat brute force method fraught with difficulties. Special attention is required for the initial values of the variable (i.e. the starting values for the number of moles of water, CO, CO2, H2, and methanol) and typically a scan of the possible initial values is required to obtain physically feasible solutions. Furthermore, this method of solution may result in the location of a local minimum rather than the global minimum on the Gibbs free energy surface.

Feed T 523.15 K28 1 1.142857

4 0.142857 1.28571468 2.428571 4.857143

Objective function (to be minimized)

Variables -143685

0.01 -2164.920751

0.4 -56708.25511

0.1 -37485.18161

1 12076.60939

0.5 -59403.04594

0

2.01 Constraints -0.142857143

-0.175714286

-0.837142857

Results

60.0%

30.0%

Gsystem

NH2O

NCO

NCO2

NH2

NCH3OH

NCH3OCH3

SNi gC

gO

gH

XCO

XCO2

After solving it using Solver:Results

XCO 62.4%

XCO2 4.3%

YCH3OH 55.2%

Step 4: Including non-idealitiesAt 50 bar, the fugacity coefficient is not expected to be equal to 1. The fugacity coefficients (as calculated using the Peng-Robinson equation of state using a binary interaction coefficient of zero for all components) can be taken into the calculation of the Gibbs free energy of the system:

Gsystem=∑i=1

i=C

N i ∙Gi=∑i=1

i=C

N i ∙¿¿

112

0.295134 methanol 0.2952 1.30E+01 8.80E-01

0 Compressibility 0.9865 of vapor phase

yCH3OH

yCH3OCH3

XCO 65.6%

XCO2 4.7%

YCH3OH 58.0%

The inclusion of the non-ideality through the fugacity coefficients results in an increase in the methanol yield and a decrease in the CO2-conversion. The non-ideality results in a virtual decrease in the mole fraction of methanol and thus shifting the equilibrium composition in the direction of methanol.

Step 5: Including the formation of dimethylether The spreadsheet for the minimization of the Gibbs free energy of the system is given on the following page. The inclusion of the formation of dimethylether into the system does result in an increase in the overall CO conversion:Results

XCO 97.1%

XCO2 -128.1%

YCH3OH 5.7% Carbon yield

YCH3OCH3 63.2% Carbon yield

The yield of methanol plus dimethylether on a carbon basis is ca. 10% higher than the methanol yield in the case when no dimethylether was formed. This is accompanied by a (strong) negative conversion of CO2 (i.e. CO2 is being formed in the system). The formation of dimethylether in the system yields water as a co-product, which may react with CO in the water-gas shift reaction yielding CO2. This is undesirable since the valuable CO is converted in the less reactive CO2. Hence, the strategy to increase the methanol yield per pass by co-producing dimethylether can only work, if a catalyst is found with a relative low activity for the water-gas shift reaction in comparison to its activity for the methanol synthesis and the dehydration of methanol yielding dimethylether. In that

113

case, the yield of methanol plus dimethylether on a carbon basis can be as high as 90.6 C-% (this was calculated by considering CO2 as an inert, i.e. the number of moles of CO2 was not optimized during the minimization of the Gibbs free energy of the system).

Step 6: Confirming the location of the global minimum using Lagrange multipliersTo find the minimum Gibbs free energy, we can also use Lagrange multipliers, and we need to solve the following set of non-linear equations:

G j0

RT+ ln ¿

g j=0

The first step is to find ( ∂gk

∂ N j)T , p , N i≠ j, λ i

for each component and each element

( ∂ gC

∂ NH 2O)T , p , N i≠CO

=0 ( ∂ gC

∂ NCO)T , p ,N i≠CO

=1 ( ∂gC

∂ NCO2)T , p , N i≠CO

=1

( ∂ gC

∂ NH 2)T , p , N i≠CO

=0 ( ∂ gC

∂ NCH 3OH )T , p , N i≠CO

=1 ( ∂gC

∂ NCH 3OCH 2)T , p , N i ≠CO

=2

( ∂ gO


=1 ( ∂ gO


=1 ( ∂ gO


=2

( ∂ gO

∂ NH 2)T , p , N i≠CO

=0 ( ∂gO


=1 ( ∂gO


=2

114

( ∂gH


=2 ( ∂gH


=0 ( ∂gH


=0

( ∂ gH

∂ NH 2)T , p , N i≠CO

=2 ( ∂ gH


=4 ( ∂gH


=6

Thus, the system in which the dimethylether formation is included, results in 9 non-linear equations:GH 2O0

RT+ ln ¿

GCO0

RT+ln ¿

GCO2

0

RT+ln( N CO2

∑ N i ∙∙ ∙

f CO2

yCO2∙P )+λC ∙1+ λO ∙2+ λH ∙0=0

GH 2

0

RT+ ln ¿

GCH3OH0

RT+ ln¿

GCH3OCH 3

0

RT+ ln ¿

gC=N CO+NCO2+NC H 3OH+2 ∙N CH 3OCH 3

−1.14=0gO=NH 2O

+NCO+2 ∙NCO2+NC H 3OH+NC H 3OCH 3

−1.28=0gH=2 ∙N H 2O

+2∙ N H 2+4 ∙ NCH 3OH+6 ∙NC H 3OCH 3

−4.86=0

This set of equation can be solved using a non-linear equation solver such as Polymath®. POLYMATH Results 04-24-2012, Rev5.1.225

NLES Solution

Variable Value f(x) Ini Guess gC -14.627517 5.107E-14 1 gO 49.871504 5.573E-14 1 gH -1.4416895 -6.217E-15 1 NH2O 0.1863727 -4.315E-12 0.1 NCO 0.0286006 -1.866E-09 0.05 NCO2 0.316985 -4.832E-12 0.25 NH2 1.1113426 -1.017E-13 1 NCH3OH 0.0676989 -2.277E-12 0.1 NCH3OCH3 0.3633578 1.643E-12 0.25

NLES Report (safenewt)

Nonlinear equations [1] f(gC) = NCO+NCO2 +NCH3OH+2*NCH3OCH3 -1.14 = 0 [2] f(gO) = NH2O+NCO+2*NCO2+NCH3OH+NCH3OCH3 -1.28 = 0 [3] f(gH) = NH2O+2*NH2+4*NCH3OH+6*NCH3OCH3 -4.86 = 0 [4] f(NH2O) = GH2O/RT+ln(NH2O/SUMNi*P*phiH2O)+0*gC+1*gO+2*gH = 0 [5] f(NCO) = GCO/RT+ln(NCO/SUMNi*P*phiCO)+1*gC+1*gO+0*gH = 0 [6] f(NCO2) = GCO2/RT+ln(NCO2/SUMNi*P)*phiCO2+1*gC+2*gO+0*gH = 0 [7] f(NH2) = GH2/RT+ln(NH2/SUMNi*P*phiH2)+0*gC+0*gO+2*gH = 0 [8] f(NCH3OH) = GCH3OH/RT+ln(NCH3OH/SUMNi*P*phiCH3OH)+1*gC+1*gO+4*gH = 0 [9] f(NCH3OCH3) = GCH3OCH3/RT+ln(NCH3OCH3/SUMNi*P*phiCH3OCH3)+2*gC+1*gO+6*gH = 0

Explicit equations [1] RT = 523.15*8.314/1000 [8] GCH3OH = -129.8 [2] P = 50 [9] GCH3OCH3 = -61.1 [3] SUMNi = NH2O+NCO+NCO2+NH2+NCH3OH+NCH3OCH3 [10] phiH2O = 8.8979e-01 [4] GH2O = -210.4 [11] phiCO = 1.0292e+00

115

[5] GCO = -151.8 [12] phiCO2 = 9.7154e-01 [6] GCO2 = -378.8 [13] phiH2 = 1.0328e+00 [7] GH2 = -1.9 [14] phiCH3OH = 8.8321e-01

[15] phiCH3OCH3 = 9.1517e-01

The resulting equilibrium composition of the gas phase is in agreement with those found from searching for the minimum in the Gibbs free energy of the system confirming the location of the minimum (and thus the conclusions made above).••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

116

III.5 Reactions involving solidsFor any reactive system, the Gibbs free energy of reaction strives towards a minimum:

Gsystem=∑i=1

i=C

N i ∙Gi is at a global minimum

The partial molar Gibbs free energy of a compound can be expressed in terms of the molar Gibbs free energy of the pure component at the same temperature and a pressure of 1 bar and the activity of that compound:

Gi (T , p , x )=Gio¿

The activity is defined as:

a i=f i (T , p , x )

f i ¿¿The fugacity of a component as a solid equals the fugacity of that pure component as a solid:

a i=f is (T , p , x )f is ¿¿

The fugacity of a solid is hardly dependent on pressure (unless extremely high pressures are applied):

a i=f is (T , p )f is ¿¿

and thus the partial molar Gibbs free energy is the molar Gibbs free energy of the pure component:Gi (T , p , x )=Gi

o¿

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Phase transition in a solid – transformation of red PbO into yellow PbO Lead oxide (PbO) can exist in two allotropic forms, also named red and yellow PbO. The following data can be found for the various components:

Compound fH (298.15K) fG (298.15K)kJ/mol kJ/mol

PbO-yellow (s) -217.3 -187.9PbO-red (s) -219.0 -188.9

Determine at which temperature red-PbO transforms into yellow-PbO.

For a solid-solid reaction, the Gibbs free energy of the system can be expressed as:

Gsystem=∑i=1

i=C

N i ∙Gi=N yellow−PbO ∙G yellow−PbO+N red−PbO ∙Gred−PbO

Gsystem=N yellow−PbO ∙G yellow−PbO+N red−PbO ∙Gred−PbO For a system containing N moles of Pb with xyellow the fraction of yellow-PbO and (1-xyellow) the fraction of red-PbO, the Gibbs free energy of the system is:

Gsystem

N Pb

=x yellow ∙G yellow−PbO+(1−x yellow ) ∙G red−PbO

Thus, the Gibbs free energy for a reaction involving only solids is minimal if the mole fraction of a single phase is 0 or 1. In the case of PbO, the Gibbs free energy is minimal at 298.15 K, if all PbO exists as red-PbO (see Figure 5.1). The difference between the behavior of reacting solids and reacting gases (and liquids) is their mixing behavior. In the case of gases and liquids, the mixing contribution towards the Gibbs free energy of the system results in a mixture being favored as the end product. Solids do (in general) not mix, and hence reactions involving only solids go to completion or do not proceed at all.

117

The Gibbs free energy of the system changes with temperature. The temperature at which the Gibbs free energy of yellow-PbO equals the Gibbs free energy of red-PbO, the phase transition will become thermodynamically possible. The change in the Gibbs free energy of the substances can be described with:

(∆ f GT )

T2

−(∆ f GT )

T1

=∫T 1

T 2 −∆f H

T 2∙ dT

-190

-189

-188

-187

0 0.2 0.4 0.6 0.8 1

Gs

yste

m(2

98

.15

K)/

nP

b,

kJ

/mo

l

Mole fraction of yellow PbO, xyellow

Figure III.5.1: Gibbs free energy of the system yellow-PbO/red-PbO at 298.15 K as a function of the mole fraction of yellow-PbO in the mixture

Assuming that the heat of reaction is independent of temperature:

(∆ f G

T )T2

−(∆ f G

T )T1

=∆ f H ∙( 1T 2− 1T 1 )

Thus, the transition temperature is given by:

(∆ f G yellow−PbO

T )=∆ f G yellow−PbO

298.15+∆f H yellow−PbO ∙( 1T− 1

298.15 )=¿

∆ f Gred−PbO

298.15+∆ f H red−PbO ∙( 1T − 1

298.15 )=(∆f Gred−PbO

T )or

∆ f H yellow−PbO−∆ f H red−PbO

T=

∆ f H yellow−PbO−∆f H red−PbO

298.15−

∆ f G yellow−PbO−∆f Gred−PbO

298.15

Substituting the values and solving for T T = 724 KThus, above 724 K red-PbO can be transformed into yellow-PbO.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example Formation of a gas by decomposition of a solid – transformation of CaCO 3

The decomposition of CaCO3 yields a solid, CaO, and a gas, CO2:CaCO3 (s) CaO(s) + CO2(g)

Hence, the products and the reactants do not mix either. Thus, this reaction may go to completion (or proceed not at all) depending on the pressure of CO2. The equilibrium pressure above which the solid decomposes is called the decomposition pressure. Determine the decomposition pressure of CaCO3 as a function of temperature in the range between 298.15 K and 1400 K.

118

The following data can be found for the various components:Compound fH (298.15K) fG (298.15K) c P=a+b ∙T +c ∙T 2+d ∙T 3, (J/mol.K)

kJ/mol kJ/mol a b c dCO2 (g) -393.5 -394.4 22.243 5.977.10-2 -3.499.10-5 7.464.10-9

CaCO3 (s) -1206.9 -1128.8CaO (s) -635.1 -604.0The heat capacity for CaCO3 and CaO (in J/(mol.K) are given as:

c p ,CaC O3=82.34+0.04975 ∙T−1287000

T 2

c p ,CaO=41.84+0.02025 ∙ T−451870T 2

The Gibbs free energy of the system can be expressed as:

Gsystem=∑i=1

i=C

N i ∙Gi=NCaCO3∙GCaCO3

+NCaO ∙GCaO+NCO2∙GCO2

The number of moles of CaCO3, CaO, and CO2 are related. Define Y as the number of moles of CaCO3

which decomposes:Moles of CaCO3 NCaCO3

=NCaCO3 , 0−Y

Moles of CaO NCaO=NCaO+YMoles of CO2 NCO2

=N CO2 ,0+Y

Gsystem=( NCaCO3 , 0−Y ) ∙GCaC O3

+ (NCaO, 0+Y ) ∙GCaO+ (NCO2+Y ) ∙GCO2

The Gibbs free energy of the system is minimum, if the variation of the Gibbs free energy of the system with respect to the extent of reaction (Y) is zero. Thus, taking into consideration the Gibbs-Duhem equation:

0=−GCaCO3+GCaO+GCO 2

The partial molar Gibbs free energyfor the solid compounds (CaCO3 and CaO) GCaCO3

=GCaCO3; GCaO=GCaO

for the gaseous compound CO2 GCO2=GCO 2

+RT ∙ ln (aCO2 )which at low pressure becomes GCO2

=GCO 2+RT ∙ ln ¿

Hence, the minimum of the Gibbs free energy of the system is given by: 0=−GCaCO3

+GCaO+¿ 0=∆ rxnG+RT ∙ ln ¿ with ∆ rxnG=GCaO+GCO2

−GCaC O3

Thus, the system strives to establish a partial pressure of CO2 equal to:pCO2

1 ¿=e−∆rxnG

RT ¿

by decomposing sufficient CaCO3 to yield the decomposition pressure, or if not sufficient CaCO3 is present by decomposing all CaCO3, if the partial pressure of CO2 is initially below the decomposition pressure, or by converting CaO into CaCO3 if the partial pressure of CO2 is higher than the decomposition pressure.

The change in the Gibbs free energy upon reaction can be obtained from:

119

(∆ rxnGT )

T 2

−( ∆rxnGT )

T1

=∫T1

T2 −∆ rxnH

T 2∙ dT

with ∆ rxnH=∆rxn H (298,15K )+ ∫298.15K

T

(−c p ,CaCO3+c p , CaO+c p ,CO 2) ∙ dT

∆ rxnH=178.3∙103+ ∫298.15K

T

(−18.257+0.03027 ∙T−3.499 ∙10−5 ∙T 2+7.464 ∙10−9 ∙T 3+ 835130T 2 ) ∙ dT

in J/mol

∆ rxnH=185.5∙103±18.257 ∙T + 0.030272

∙ T2−3.499 ∙10−5

3∙ T3+ 7.464 ∙10

−9

4∙ T 4−835130

T

∫298.15

T −∆rxn H

T 2∙ dT=185.5 ∙103 ∙( 1T− 1

298.15 )+18.257 ∙ ln( T298.15 )−0.030272

∙ (T−298.15 )+ 3.499 ∙10−5

6∙ (T2−298.152 )−7.464 ∙10

−9

12∙ (T 3−298.153 )−835130

2∙( 1T2− 1

198.15 )T DrxnG pCO2, bar Integral

298.15 130.4 1.42E-23 0300 130.1028 2.22E-23 -3.68774400 114.1008 1.26E-15 -152.112500 98.24608 5.44E-11 -240.872600 82.54286 6.51E-08 -299.792700 66.98582 1E-05 -341.67800 51.57117 0.000429 -372.9900 36.29808 0.007821 -397.033

1000 21.1683 0.078387 -416.1951100 6.185395 0.508475 -431.7411200 -8.64588 2.378807 -444.5691300 -23.3201 8.650645 -455.3021400 -37.8317 25.79707 -464.386

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Decomposition of a gas yielding a solid –formation of carbon black Carbon black is produced by the decomposition of of methane

CH4 (g) C (s) +2 H2(g)in a reactor maintained at 700oC and 1 bar. The equilibrium constant for this reaction at 700oC is 7.403 based on the pure component standard states (gaseous CH4 and H2 and solid C) Calculate the equilibrium gas phase composition in the reactor.

The activity based equilibrium constant is:

Ka=(aH 2)

2∙ aC

aCH 4

The activity of solid carbon is 1, and at low pressure the activity can be expressed in terms of the mole fraction and the total pressure (1 bar):

120

Ka=¿¿¿The fugacity coefficient will be equal to one at these low pressures. The mole fraction can be obtained from the stoichiometric table:


Gas phase Solid phaseCH4 (g) NCH 4 ,0 NC H 4 ,0

∙ (1−X ) (1−X )(1+X )

H2(g) 0 NCH 4 ,0∙2 ∙ X

2 ∙ X

(1+ ∙X )C (s) 0 NCH 4 ,0

∙ X

∑ ¿N CH4 ,0∙ (1+X )

Thus, the equilibrium gas phase composition is given by:

Ka=(2 X )2 ∙1

(1−X ) ∙ (1+X )=7.403

Solving for X X = 0.806

Thus, the equilibrium gas phase composition will contain yCH4 = 0.108 and yH2 =0.892. This means that in this case a partial conversion of methane can be obtained. The mixing contribution between the product H2 and the reactant CH4 makes this possible. Furthermore, the formation of carbon can continue in the reactor, if the flow of methane is maintained at such a level that the exit mole fraction of methane is larger than 0.108.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

121

III.6 Non-ionic reactions in the liquid phaseThe condition for equilibrium is that the Gibbs free energy for the system is minimal. Hence, the variation of the Gibbs free energy with respect to the extent of reaction must be equal to zero.

( ∂Gsystem

∂ξ )T , p

=∑i=1

i=C

ν i ∙Gi=0

With Gi (T , p , x )=Gio , L¿ if the reference state is chosen to be the pure compound as a liquid. The

activity of compound i is then at low pressure (fugacity coefficient equal to 1) given by

a i=f iL (T , p , x )f iL¿¿

Hence, the equilibrium position of the system is given by

Ka=∏i=1

i=C

aiν i=∏

i=1

i=C

(xi ∙ γ i)ν i=e

−∆rxnGRT

The activity based equilibrium constant can be regarded as two constants, one based on the mole fractions and one based on the activity coefficients:

Ka=∏i=1

i=C

(x i ∙ γi )νi=∏

i=1

i=C

(x i )ν i ∙∏

i=1

i=C

(γ i )ν i=K x ∙ K γ

In the older literature a concentration-based equilibrium constant is often given rather than the activity based equilibrium constant. The concentration of a component in the mixture and the mole fraction are linked, since:

x i=n i

∑ ni

; [ i ]=ni

V solution

=x i ∙∑ n i

V solution

=x i ∙∑ ni

V solution

=xi

V solution

K x=∏i=1

i=C

(x i )νi=∏

i=1

i=C

( [i ] ∙V solution )ν i=∏i=1

i=C

( [i ] )ν i ∙V solution∑ νi=K C ∙V solution

∑ ν i

Ka=K x ∙K γ=K C ∙ K γ ∙V solution∑ ν i=e

−∆ rxnGRT

Thus, for an ideal solution (K = 1), the concentration based equilibrium constant can be obtained from the activity based equilibrium constant from KC=K a∙V solution

−∑ νi

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Liquid phase esterification of methanol and phenol yielding anisole The esterification of phenol with methanol yielding anisole (methyl phenyl ether) may occur in the liquid phase.

CH3OH(l) +C6H5OH(l) C6H5OCH3(l) +H2O(l)A liquid mixture containing equi-molar amounts of methanol and phenol is allowed to react at 1 bar and 298.15 K. What is the expected conversion of methanol? The following data are available:

Compound fH (298.15K) fG (298.15K) Tmelt fusHkJ/mol kJ/mol K kJ/mol

CH3OH(l) -238.7 -166.3C6H5OH(s)1 -165.0 -50.9 314 11.514C6H5OCH3(l)2 -114.5 -0.1H2O(l) -285.8 -237.1

1 Phenol in the standard state at 298.15 K is a solid

122

2 Data from derived from Gibbs free energy as an ideal gas as given in Perry’s Chemical Engineers’ Handbook

The product mixture will contain four components, i.e. a quartenary system. However, starting from an equi-molar mixture of methanol and phenol, the mole fraction of methanol and phenol will always be the same and the mole fraction of water will always be identical to the mole fraction of methyl phenyl ether. The activity coefficients of the components can then be modeled using UNIFAC.The equilibrium liquid phase composition is given by:

Ka=e−∆rxnG

RT

123

The change in the Gibbs free energy upon reaction taking the compounds as a pure liquid state as the reference state, is given by:with ∆ rxnG=−∆ f GCH 3OH (l)−∆ fGC6 H 5OH (l)+∆ f GC6H 5OCH 3 (l )

+∆ f GH 2O (l )

We have the Gibbs free energy of formation of all compounds as a liquid with the exception of phenol. The Gibbs free energy of formation of phenol as a liquid can be calculated from the Gibbs free energy of phenol as a solid: ∆ f GC6H 5OH (l)−∆ f GC6H 5OH (s )=∆ fusGC6H 5OCH 3

(298.15K ) and thus the question reduces to finding the Gibbs free energy for fusion at 298.15 K. We have seen that neglecting the difference in the heat capacity between solid and liquid phenol

∆ fusG (T )=∆ fusH (Tmelt )∙(1− TT melt

)=11.514 ∙(1−298.15314 )=0.58 ∙ kJmol

Hence,

∆ f GC6H 5OH (l)−∆ f GC6H 5OH (s )=−50.9+0.58=−50.3 ∙ kJmol

∆ rxnG=−∆ f GCH 3OH (l)−∆ fGC6 H 5OH (l)+∆ f GC6H 5OCH 3 (l )+∆ f GH 2O (l )=−20.6 ∙ kJ

mol

Thus, the equilibrium constant at 298.15 Ka=e−∆rxnG

RT =4035

The composition of the liquid is given by

Ka=aH 2O (l ) ∙ aC6H 5OCH 3 (l )

aCH3OH (l) ∙ aC6H 5OH (l )=

xH 2O∙ xC6 H5OC H 3

xCH3OH ∙ aC6 H 5OH

∙γH 2O

∙ γC6 H 5OC H 3

γCH3OH ∙ γC6H 5OH

Setting up the stoichiometric table for a reaction mixture starting with an equi-molar mixture:Compound In Out xi

CH3OH(l) 1 1-X 0.5.(1-X)C6H5OH(s)1 1 1-X 0.5.(1-X)C6H5OCH3(l)2 0 X 0.5.XH2O(l) 0 X 0.5.X

Ka=aH 2O (l ) ∙ aC6H 5OCH 3 (l )

aCH3OH (l) ∙ aC6H 5OH (l )= X2

(1−X )2∙γ H2O

∙ γC6H 5OCH 3

γCH3OH ∙ γC6H 5OH

= X 2

(1−X )2∙ K γ

X=√ K a

K γ

1+√ Ka

K γ

The activity coefficients are dependent on the mixture composition. Start off the calculation assuming K = 1 and then iterate through.gMethanol gphenol gwater ganisole Kg Kx X xMethanol xphenol xwater xanisole

1 4035.254 0.984502 0.007749 0.007749 0.492251 0.4922511.6282 0.161 3.4295 1.9348 25.3124 159.4181 0.926611 0.036694 0.036694 0.463306 0.4633061.3866 0.1782 3.431 1.9381 26.91151 149.9453 0.924501 0.037749 0.037749 0.462251 0.462251

1.38 0.179 3.4315 1.9378 26.91912 149.9029 0.924491 0.037754 0.037754 0.462246 0.462246

124

Hence, a conversion of 92.4% can be achieved. This is lower than the one obtained for the ideal solution (which would result in a conversion of 98.4%), due to the non-ideality of the solution with K>1. ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example Combined chemical and phase equilibrium – esterification of phenol with methanol The esterification of phenol with methanol yielding anisole (methyl phenyl ether) may occur at 150 oC at a constant pressure of 1.5 bar. What is the equilibrium composition of the mixture starting from an equi-molar mixture of methanol and phenol? Data:

Compound pvap(373.15K) vapH(Tboil) Tboil

bar kJ/mol KCH3OH 3.303 35.21 337.8C6H5OH 0.055 45.91 455.0C6H5OCH3 0.186 38.97 426.8H2O 1.0135 40.65 373.15

At a temperature of 150oC and 1.5 bar, liquid phase and vapour phase may co-exist. Hence, both the chemical reaction and the vapour-liquid equilibrium need to be considered. The chemical and vapour-liquid split will ensure that the Gibbs free energy of the system strives towards a minimum:

Gsystem=∑i=1

i=C

N i ∙Gi is at a global minimum

The partial molar Gibbs free energy of a compound can be expressed in terms of the molar Gibbs free energy of the pure component at the same temperature and a pressure of 1 bar and the activity of that compound:

Gi (T , p , x )=Gio¿

Thus, the Gibbs free energy of the system can be expressed as:Gsystem=NC H 3OH

L ∙GCH 3OHL +N CH 3OH

V ∙GCH 3OHV +NC6H 5OH

L ∙GC6 H5OHL +NC6 H5OH

V ∙GC6 H 5OHV +NC6 H 5OCH 3

L ∙GC6H 5OC H3

L +NC6 H 5OCH 3

V ∙GC6 H5OC H3

V +N H 2OL ∙GH 2O

L +N H 2OV ∙GH 2O

V

withGi

L (T , p , x )=GiL, 0 ¿

GiV (T , p , x )=Gi

V , 0¿

The molar Gibbs free energy of the pure component in the liquid phase and in the vapour phase are linked:

GiV , 0¿

The change in the Gibbs free energy upon evaporation can be linked to the change in the heat of vaporization and the difference in the specific heat between the gas phase and the liquid phase: ∆vapGi¿(with ∆C p=C p ,i

V −Cp , iL )

This can be simplified to ∆vapGi¿, if the temperature range is rather small or the change in the specific heat is rather small. Thus,

GiV , 0¿

125

Step 1: Calculate the Gibbs free energy of the species in each phase at 150 oC and 1 bar with respect to the elements at their natural state at 298.15 K and 1 bar

Gibbs free energy of a species with respect to the elements at their natural state at 298.15 K is given by:

(Gi

T )T=373.15K

=( ∆f G

T )T=298.15K

+ ∫T=298,15

T=373,15 −H i

T 2∙ dT

126

with the enthalpy of a species with respect to the elements at their natural state at 298.15 K. The contribution of the integral of the specific heat to the enthalpy of the species will be rather small, since the temperature difference is not too large. Thus, the enthalpy of the species with respect to the elements at their natural state at 298.15 K can be approximated by the change in the enthalpy upon formation of the species, which is assumed to be constant:

(Gi

T )T=373.15K

=( ∆f G

T )T=298.15K

+ ∫T=298,15

T=373,15 −∆ f H

T2∙ dT=( ∆f G

T )T=298.15K

+∆ f H ∙( 1373.15

− 1273.15 )

Compound GiL (kJ/mol) Gi

V (kJ/mol)

kJ/mol kJ/molCH3OH -135.9 -144.8C6H5OH -7.1 -3.9C6H5OCH3 47.9 48.2H2O -216.7 -222.1

Step 2: Setting up the spreadsheetWe can now define the variables (the components in each phase – a total of 8 variables; in principle this number of variables can be reduced by considering that the total number of moles of phenol must be equal to the number of moles of methanol and similar for the moles of water and anisole), the function to be minimized (Gibbs free energy of the system), and the constraints (the elemental balances and the vapour-liquid equilibrium condition). The latter is best defined as

xi ∙ γi ∙ p i

vap

y i ∙ P−1=0

to take into account the large variation possible in the partial pressures of the components.

Step 3: Incorporating non-idealityThe non-ideality of the solution can be taken into account iteratively, i.e. first calculate the composition of the various phase using ideal solution, and then estimate the activity coefficient in the liquid phase using e.g. UNIFAC.

Compound Number of moles Mole fraction Activity coefficient Partial molar Gibbs free energyLiquid Gas Total Liquid Gas Liquid Liquid Vapour

Methanol 8.75312E-05 0.041575438 0.0417 0.0040 0.0425 1.2173 -154673 -154527 -13.5387 -6424.51Phenol 0.00472131 0.036941659 0.0417 0.2161 0.0378 0.6733 -13846.6 -13951.9 -65.374 -515.408Anisole 0.016320091 0.44201694 0.4583 0.7471 0.4519 0.9869 46790.33 46827.69 763.6225 20698.63Water 0.000715628 0.457621403 0.4583 0.0328 0.4678 4.5076 -223412 -223375 -159.88 -102221

0.0218 0.9782 1.0000

Temperature: 423.15 Deg K -87937.8-------------------------------No Name Mole Fraction Activity Coefficient

Elemental balances Vapour-liquid equilibrium -- ------------ ------------- --------------------C 0 methanol -4.92939E-14 1 methanol 0.0040 1.2173H 0 phenol -5.38458E-14 2 phenol 0.2160 0.6733O 0 anisole 1.64313E-14 3 yl phenyl ether 0.7472 0.9869

water 7.105427E-15 4 water 0.0328 4.5076

composition liquid phase Composition vapour phase

Iteration1 1 1 1 1 0.0046 0.1198 0.7221 0.1535 0.0400 0.0311 0.4426 0.48642 1.3414 0.547 1.0436 4.1894 0.0039 0.2599 0.7009 0.0354 0.0451 0.0369 0.4483 0.46973 1.1406 0.6993 0.9777 4.2526 0.0042 0.2064 0.7546 0.0348 0.0420 0.0375 0.4522 0.46844 1.2341 0.6662 0.9894 4.5483 0.0040 0.2186 0.7450 0.0325 0.0426 0.0378 0.4517 0.46785 1.2127 0.6751 0.9862 4.4956 0.0040 0.2155 0.7477 0.0328 0.0425 0.0377 0.4519 0.46796 1.2182 0.673 0.987 4.5105 0.0040 0.2162 0.7470 0.0327 0.0425 0.0378 0.4519 0.46787 1.2169 0.6735 0.9868 4.5065 0.0040 0.2160 0.7472 0.0328 0.0425 0.0378 0.4519 0.4678

G system

gMethanol gphenol gwater ganisole xMethanol xphenol xanisole xwater yMethanol yphenol yanisole ywater

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III.7 Reactions involving ionic speciesQuite a lot of systems of interest are aqueous systems containing ions. Chemical equilibria of interest in these systems are e.g. dissociation of acids, dissolution of salts, etc. The treatment of the chemical equilibrium condition (i.e. the Gibbs free energy of the system is minimal) is similar to that for gas- or other type of liquid-phase systems. The difference originates from the definition of the reference state for the Gibbs free energy. For ionic systems, it is common to take an ideal 1 molal solution (i.e. activity coefficient equal to 1) as the reference. The partial molar Gibbs free energy of an ion Z+ is then given in terms of the Gibbs free energy of formation of this ion from its elements as an ideal 1 molal ionic solution (see Table 13.1-4)GZ+¿ (T , p ,M )=∆f GZ ¿¿¿ ¿

The activity coefficient of the individual ions cannot be determined and a mean ionic activity coefficient has been defined. Thus, if the formation of ions Zz+ and Yy- in the stoichiometric amounts + and - are considered:ν+¿ ∙G Zz +¿ ( T , p, M) +ν−¿ ∙G

Yy−¿ (T, p,M )=ν+¿ ∙∆

fGZ¿ ¿¿

¿ ¿¿¿¿

with ln ( γ± )=−α ∙ ⌊ z+¿ ∙

z−¿ ⌋ ∙√ I

1+β ∙a ∙√ I+δ ∙ I ¿

¿ (extended Debye-Hückel model)

The minimization of the Gibbs free energy of the system, in which the dissociation of a salt Z +Y-

takes place, yields as an equilibrium condition:0=−GZν+¿Y ν−¿

(T ,p ,M )+ν+¿ ∙GZz +¿ (T, p, M) +ν−¿ ∙G

Yy−¿ (T ,p, M) ¿

¿ ¿¿¿¿

−∆disG=RTln¿

With−∆disG=−∆f GZν+¿Y ν−¿( T , p ,M=1molal )+ν+¿ ∙∆

fGZz+¿ (T ,p, M=1molal ) +ν−¿ ∙∆

fGYy−¿ (T , p,M) ¿

¿¿¿¿¿

or otherwise stated: Ka=¿¿¿ Ka≈ KC ∙ K ν

(the units for KC are molal to the power of ++--1). For very dilute solutions, the activity coefficients go to 1 (in this definition of the activity coefficients!) and KC

0 ≈ Ka ∙ (1molal )ν+¿+ν−¿−1¿ ¿

Thus,

KC=KC0

K ν

ln (KC )=ln (K C0 )−ln ( K ν )=ln (KC

0 )−¿

For dilute solution γ Zν+¿ Y ν−¿=1 ¿¿ and the limited Debye-Hückel approximation for the mean ionic activity

coefficient: ln (KC )=ln (K C

0 )+¿The ionic strength depends on the concentration of ions, and hence the concentration dependent equilibrium ‘constant’ is dependent on the concentration of ions in solution!

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Dissociation of acetic acid – determination of K a

MacInnes and Shedlovsky (J. Amer. Chem. Soc. 54 (1932), 1429) report the following data for the ionization of acetic acid in water at 25oC:

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Acetic acid added, CT, mol/liter [CH3COO-], mol/liter0.028.10-3 0.1511.10-4

0.1532.10-3 0.4405.10-4

1.0283.10-3 1.273.10-4

2.4140.10-3 2.001.10-4

5.9115.10-3 3.193.10-4

20.000.10-3 5.975.10-4

Calculate disG for this reaction.

The ionization of acetic acid in water can be seen as: C H 3COOH (aq)⇄C H 3COO−¿(aq )+H+¿(aq) ¿¿

Stoichiometric table (define I as the amount of acetic acid dissociated)Compound Initial FinalCH3COOH(aq) CT CT-ICH3COO-(aq) - IH+(aq) - I

Thus, the concentration based equilibrium ‘constant’ is given by:KC=¿¿

but the equilibrium constant depends also on the ionic strength of the solution. The ionic strength of the solution is given by:

I=12∙¿

We are dealing here with (extremely) diluted solutions. Thus, ln (KC )=ln (K C

0 )+¿

ln (KC )=ln (K C0 )+2 ∙1.175 ∙√I

Plotting the determined value for KC as a function of the square root of the ionic strength in a semi-logarithmic plot should yield a straight line.

Acetic acid added, CT, mol/liter

[CH3COO-], mol/liter

KC, mol/liter

2.80.10-5 1.51.10-4 1.77.10-5

1.53.10-4 4.41.10-4 1.78.10-5

1.03.10-3 1.27.10-4 1.8.10-5

2.41.10-3 2.00.10-4 1.81.10-5

5.91.10-3 3.19.10-4 1.82.10-5

2.00.10-2 5.98.10-4 1.84.10-5

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ln(KC) = -10.952+2.231.I0.5

R² = 0.997

-10.95

-10.94

-10.93

-10.92

-10.91

-10.9

0 0.01 0.02 0.03

ln(K

C/(

mo

l/lit

er)

)

I0.5, (mol/liter)0.5

Figure III.7.1: Dependency of the concentration-based equilibrium constant on the ionic strength of the solution

The semi-logarithmic plot of the concentration-based equilibrium constant as a function of the square root of the ionic strength is reasonable linear with a noticeable deviation at the high concentration level. A possible explanation is that the limited Debye-Hückel theory starts to fall off at this stage. In order to ascertain the origin of this deviation more data points would be required. It can be further noted that the slope of the line is not as expected equal to 2.35, but 2.231. This may indicate that the activity coefficient of acetic acid deviates from 1. However, the concentration at the high ionic strength represents still a mole fraction of acetic acid in water close to infinite dilution (i.e. xacetic acid less than 5.10-4) and hence the activity coefficient of acetic acid in this solution is not expected to change much and hence can be considered to be equal to 1 taking the ideal solution with a molality of 1 as a reference.

The intercept of the correlating line represents ln(KC0), which is equal to ln(Ka). Hence, Ka is

determined to be e-10.952 = 1.752.10-5. Hence, the Gibbs free energy for this dissociation reaction is given by:

−∆disG=RTln (Ka )=8.314 ∙298.15 ∙ (−10.952 )=−27.1 ∙ kJmol

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Influence of ionic strength on pH of aqueous acetic acid solutions The pH of aqueous acetic acid solutions is dependent on the ionic strength of the solution. Hence, the pH of the solution can be altered by adding NaCl to the solution. Calculate the pH of the solution as a function of the ionic strength of the solution for solutions with containing initially 10 mmol acetic acid per liter to which between 0 and 1 mol NaCl per liter is added.

In this ionic strength range, the extended Debye-Hückel needs to be taken. Hence, the concentration-based equilibrium constant can be determined using: ln (KC )=ln (K C

0 )−¿(the stoichiometric factors refer to the ions involved in the dissociation reaction)

The ionic strength of the solution is given by:

I=12∙∑ zi

2 ∙C i=12∙¿

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The concentrations can be obtained from a stoichiometric table (define A: initial concentration of acetic acid; N: initial amount of NaCl in the solution, which dissociates completely; X as the amount of acetic acid, which has dissociated; and assume :

Compound Initial FinalCH3COOH(aq) A A-XCH3COO-(aq) - XH+(aq) - XNaCl(aq) N -Cl-(aq) - NNa+(aq) - N

Thus, the ionic strength is given byI=X+N

The equilibrium constant ln (K C0 )=ln ( Ka )=−10.952

KC=¿¿

Putting it all together:

ln ( X2

A−X )=−10.952+2 ∙ 1.175 ∙√X+N1+0.3284 ∙4 ∙√X+N

−2 ∙0.1∙ (X+N )

and thus X can be calculated as a function of N (the initial concentration of NaCl added to the solution). X represents the concentration of H+ in the solution, and hence the pH = -log(X). The addition of NaCl to the solution decreases the pH (i.e. more acetic acid is dissociated; see Fig. 7.2). In this range of ionic strengths, the mean ionic activity coefficient decreases. The activity based equilibrium constant remains however constant. Thus, the ratio of the product of the dissociation products relative to the undissociated acetic acid must increase (with decreasing activity coefficient) in order to keep Ka constant.

3

3.1

3.2

3.3

3.4

3.5

0 0.5 1 1.5

pH

Amount of NaCl added, mol/liter

Figure III.7.2: Dependency of the pH of a 10 mmol/liter acetic acid solution on the amount of NaCl added to the solution due to change in the ionic strength

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Influence of ionic strength on pH of aqueous acetic acid solutions - common ion effect The pH of aqueous acetic acid solutions is dependent on the ionic strength of the solution and the type of ions present Hence, the pH of the solution can be altered by adding sodium acetate (Na(CH3COO)) to the solution. Calculate the pH of the solution as a function of the ionic strength of

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the solution for solutions with containing initially 10 mmol acetic acid per liter to which between 0 and 1 mol Na(CH3COO) per liter is added.

The concentrations can be obtained from a stoichiometric table (define A: initial concentration of acetic acid; N: initial amount of Na(CH3COO) in the solution, which dissociates completely; X as the amount of acetic acid, which has dissociated; and assume :

Compound Initial FinalCH3COOH(aq) A A-XCH3COO-(aq) - X+NH+(aq) - XNaCH3COOaq) N -Na+(aq) - N

The ionic strength of the solution is given by:

I=12∙¿

The equilibrium constant ln (K C0 )=ln ( Ka )=−10.952

KC=¿¿

Putting it all together:

ln ( ( X+N ) ∙ XA−X )=−10.952+2 ∙ 1.175 ∙√X+N

1+0.3284 ∙4 ∙√X+N−2 ∙0.1∙ (X+N )

The pH of the solution increases with increasing amount of sodium acetate added to the solution (see figure 7.3) indicating a decrease in the concentration of H+-ions, and thus indicating a decrease in the extent of dissociation of acetic acid. This is of course due to the presence of acetate ions in the solution originating from sodium acetate. Thus, at a constant value of KC the increase in the acetate concentration needs to be off-set by a proportional decrease in the H+-concentration. However, this decrease in the extent of dissociation of acetic acid is moderated by the change in the activity coefficient with increasing ionic strength (vide supra).

3

3.5

4

4.5

5

5.5

6

6.5

0 0.5 1 1.5

pH

Amount of NaCl added, mol/liter

Addition of NaCH3COO

Addition of NaCl

Figure III.7.3: Dependency of the pH of a 10 mmol/liter acetic acid solution on the amount of Na(CH3COO) or NaCl added to the solution – common ion effect opposed to the effect of ionic strength

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III.7.1 Solubility of saltsThe dissolution of salts is often given in terms of a solubility (g of salt dissolved per 100 cm3 of water) or in terms of the solubility product Ksp. The solubility in terms of g of salt per 100 cm3 of water can be easily recalculated in terms of the solubility product taking into account the molar mass of the salt and assuming that the change in the volume upon dissolving the salt is negligibly small. The solubility product is commonly defined as: K sp=¿¿

From a thermodynamic viewpoint, the dissolution of a salt can be represented as a chemical

reaction: Aν+¿Bν−¿(s)⇄ν

+¿ A

z +¿ (aq) +ν−¿Bz−¿ (aq ) ¿ ¿

¿

¿

¿¿

The activity based equilibrium constant is given by: Ka=¿¿¿but the activity of a solid (at not too a high a pressure) equals 1: Ka=¿¿

The molality of a solution is approximately the concentration in mol/liter in a solution (since the density of the solution is approximately 1 kg/liter): Ka≈ ¿¿

Hence, Ka≈ K sp ∙( γ ±

M=1molal )ν+¿+ν−¿ ¿¿

Similarly to the consideration on ionic solutions, the activity coefficient can be modeled using the Debye-Hückel theory.

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Calculation of solubility product from solubility data – influence of ionic strengthThe solubility of silver chloride (AgCl) in aqueous solutions of KNO3 has been determined at 25oC (S. Popoff, E.W. Neumann, J. Phys. Chem. 34 (1930), 1853; E.W. Neumann, J. Amer. Chem. Soc. 54 (1932), 2195).

[KNO3], mmol/liter

[AgCl]saturation, mol/liter

0.0 1.273.10-5

0.509 1.311.10-5

9.931 1.427.10-5

16.431 1.469.10-5

40.144 1.552.10-5

(a) Calculate the solubility product of silver chloride in each of these solutions(b) Calculate the theoretically predicted solubility product of AgCl from thermodynamic data

Data:fH (298.15 K), kJ/mol fG (298.15 K), kJ/mol

AgCl(s) -127.1 -109.8Ag+ (aq) 105.90 77.11Cl- (aq) -167.46 -131.17

(c) Compare the solubility product for AgCl in these solutions with the theoretically predicted solubility products

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The solubility product of AgCl is defined as: K sp=¿Assuming complete dissociation of AgCl [Ag+] = [Cl-] = [AgCl] K sp=[ AgCl ]2

[KNO3], mmol/liter

[AgCl]saturation, mol/liter

Ksp, (mol/liter)2

0.0 1.273.10-5 1.62E-100.509 1.311.10-5 1.72E-109.931 1.427.10-5 2.04E-10

16.431 1.469.10-5 2.16E-1040.144 1.552.10-5 2.41E-10

The theoretical estimate of the solubility product is given by:

K sp=K a

( γ± )2∙ (M=1molal )2

The activity based equilibrium constant is given in terms of the Gibbs free energy of reaction (dissolution)

Ka=e−∆dissolutionG

RT

with ∆dissolutionG=∆ f GAg+¿ (aq )+∆ fGCl−¿ (aq )−∆ f G AgCl( s )=55.74 ∙kJ

mol AgCl¿¿

Ka=1.715∙10−10

The activity coefficient can be modeled using the limited Debye-Hückel model seeing the low concentrations and low ionic strength:

K sp=1.715 ∙10−10

¿¿¿

with the ionic strength: I=12∙¿

I=[ AgCl ]+ [KNO3 ]

K sp=1.715 ∙10−10

¿¿¿[KNO3],

mmol/liter[AgCl]saturation,

mol/literKsp,

(mol/liter)2I,

mol/literKsp,theoretical,

(mol/liter)2

0 1.27E-05 1.62E-10 1.27E-05 1.73E-100.509 1.31E-05 1.72E-10 5.22E-04 1.81E-109.931 1.43E-05 2.04E-10 9.95E-03 2.17E-10

16.431 1.47E-05 2.16E-10 1.64E-02 2.32E-1040.144 1.55E-05 2.41E-10 4.02E-02 2.75E-10

The predicted solubility product is between 7 and 14% too large, and thus the estimated solubility is between 3 and 7% too larger. Hence, the theoretical approach can yield a good insight in the solubility of a compound, in the absence of experimental data.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

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••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Example: Common ion effect – solubility of AgCl in the presuence of TlCl The solubility of silver chloride (AgCl) in water at 25oC is 1.273.10-5 mol/liter, and that of thallium chloride (TlCl) is 0.144 mol/litre. Estimate the simultaneous solubility of AgCl and TlCl in water at 25oC.

Step 1: Calculate Ka for the dissolution of the pure salts The activity based equilibrium constant is defined as

Ka≈ K sp ∙( γ ±

M=1molal )2

For a pure AgCl-solution K sp=[ AgCl ]2=1.62 ∙10−10 ∙( molliter )

2

For a pure TlCl-solution K sp=[TlCl ]2=2.1 .144 ∙10−2 ∙( molliter )

2

Using the extended Debye-Hückel theory (with z+=1 and z-=-1)

ln ( γ± )=−α ∙ ⌊ z+¿ ∙

z−¿ ⌋ ∙√ I

1+β ∙a ∙√ I+0.1 ∙ ⌊ z+¿ ∙ z−¿ ⌋∙ I ¿¿

¿¿

ln ( γ± )=−1.175 ∙√I1+√ I

+0.1 ∙ I

For a pure AgCl-solution I=[ AgCl ]=1.273 ∙10−5 ∙molliter

For a pure TlCl-solution I=[TlCl ]=0.144 ∙ molliter

Hence, the activity coefficient according to the extended Debye-Hückel theory

for a pure AgCl-solution Ka , AgCl dissolution≈K sp ∙( γ±M=1molal )

2

=1.607 ∙10−10

for a pure TlCl-solution Ka , AgCl dissolution≈K sp ∙( γ±M=1molal )

2

=1.118 ∙10−2

For the simultaneous solubility of TlCl and AgCl, we need to set-up a stoichiometric table. Define A as the amount of AgCl dissolved and T as the amount of TlCl dissolved:

Compound Final concentrationAg+(aq) ACl-(aq) A+TTl+(aq) T

The variables A and T can be determined from the two equilibrium constants

Ka , AgCl dissolution≈K sp ∙( γ±M=1molal )

2

=1.607 ∙10−10

Ka , AgCl dissolution≈K sp ∙( γ±M=1molal )

2

=1.118 ∙10−2

which have to be solved simultaneously. Writing out in terms of the variables Ka , AgCl dissolution≈¿ Ka , AgCl dissolution≈¿

136

The ionic strength of the solution is now I=A+T . A first estimate is that the ionic strength is dominated by the concentration of thallium chloride and that the solubility of thallium chloride is unaffected by the presence of silver chloride.

γ ±=e−1.175∙ √I1+√I

+0.1 ∙ I=e

−1.175 ∙√0.1441+√0.144

+0.1 ∙ 0.144=0.7262

This results in a concentration of Ag+: ¿The assumption that the concentration of Ag+ is negligibly small seems justified, and the concentration of these ions will not contribute to the solubility of TlCl. Hence, the solubility of TlCl is not affected by the presence of AgCl, but the solubility of AgCl is strongly depressed. ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

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