MTK- Dilute Solution and Phase Rule

D E P A R T M E N T O F

Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)

Metallurgical Thermodynamics

Dilute Solution Henrys Law




Henrys Law




Henrys Law

The constant is equal to the slope of the curve at zero concentration of A, designated by coefficient of the solute A at infinite dilution).

Like Raoults law, Henrys law is valid within a concentration range where the extent varies from one system to another, but it is valid only at low concentration.

AAAAAAA

AA

AA

AA

xaorxconstaeixpk

pphavewepbydividing

kxpeixp

a

a

==

=

...

)1(..

000

activity(0Ag

AAA xa .0g=




Henrys Law

aA

xA

0A Henrys law constant




Henrys Law In concentratrated solution the standard state has been

defined as unit atmospheric pressure and unit activity i.e. pure substance at any temperature.

In dilute solutions relative standard states other than pure substance being used. Henrys law offers two such standard states, called alternative standard states.(1) Infinitely dilute, atom/mole fraction standard state.(2) Infinitely dilute, wt% (w/o or %) standard state.




Solubility of Gases

It is important to note that the validity of Henrys law depend upon the proper choice of solute species. For example, consider(a) solution of nitrogen in water(b) solution of nitrogen in liquid iron

In the first case nitrogen dissolves molecularly as N2

As solubility of N2 in water is low according to Henrys law we have

( ) )waterin dissolved(22 NgN 2

2

N

N

pa

K =

22 NNxka =

2

2

N

N

pkx

Kor =

222)lub( NNN pkpk

Kilitysox ==




Solubility of Gases

Thus the solubility of nitrogen in water is proportional to the partial pressure of nitrogen gas in equilibrium with water.

Solubility can be expressed as mole fraction, cc per 100 g of water or any other unit.

In the second case under consideration nitrogen dissolves atomically in solid or liquid metals:

)(2)(2 FeinNgN

22

2)(

2 )(

N

N

N

inFeN

pkx

paK ==

22')lub()( NNinFeN pkpk

Kilitysoxor ==




Solubility of Gases Sieverts law Since all the common diatomic gases N2, O2, H2

etc. dissolve atomically in metals, the general expression for solubility is given as:

This is known as Sieverts law and can be stated as solubility of diatomic gases in metals is directly proportional to the square root of partial pressure of the gas in equilibrium with the metal.

2NpkS =




Alternative Standard States

1. Infinitely dilute, atom fraction standard stateHenerian standard state is obtained from Henrys law which, strictly being limiting law obeyed by the solute in the dilute solution is expressed by

If the solute obeys Henrys law over a finite concentration range, then

tconslawsHenerytheisandstatedardsRoultiantrwAofactivitytheisawhere

xasxa

A

A

AAA

A

tan'tan..

0

0

0

g

g

AAA xa .0g




Alternative Standard States Henerian standard state is obtained by the extending the

Henrys law line to xA = 1. This state represents pure solute in the hypothetical,

nonphysical state in which it would exist as a pure component if it obeyed Henrys law over the entire composition range (i.e., as it does for a dilute solution)

Having defined the Henrian standard state, the activity of A in solution with respect to the Henrian standard state is given by:

activityheneriantheishandstatedardsheneriantheisThis

1=xat=1=h

A

A0AA

tan.

tcoefficienactivityheneriantheisfwherexfh

A

AAA .=





In the range of composition in which the solute obeys Henrys Law, fA =1 and solute exhibit the Henerian ideality

hA = xA

In the range of composition in which the solute obeys Henrys Law, fA =1 and A = A0

tconsxA

A

AA

AA

A

A

Affx

xha

tan..

=

==

gg

tconsxA

A

A

Aha

tan

0

=

= g





The free energy change accompanying the transfer of one mole of solute A from pure substance standard state (Raoultian standard state) to Henerian standard state, that is: A (in the Raoultian standard state) A( in the Henerian standard state)is given by

The partial molar free energy of the solute at constant concentration is independent of standard state. The value of GA0 remains unchanged if is added and is subtracted from the right hand side of the above equation

000)()(

)(RAHA

GGHRGA -=D

)(RAG )(HAG

( ) ( )0)(0)()()(

000

)()(

)()()(

HARA

RAHA

GGGG

GGGGHRG

HARA

HARAA

---=

-+-=D





AAA aTRGGBut ln0 =-

0

tan

0

ln

ln)(,

A

tconsxA

AA

RT

haRTHRGHence

A

g=

=D

=





2. Infinitely dilute, wt% standard state.

The use of this standard state eliminates the necessity of

converting weight percentages, obtained via chemical analysis, to

mole fractions for the purpose of thermodynamic calculations.

This standard state is particularly convenient to use in

metallurgical systems containing dilute solutes. This standard

state can formally be defined as:

( )0.%.%

0.%1.%

=

AwtA

A

Awtaor

AwtasAwt

a





If the concentration up to 1 weight-percent of solute A, then aA = 1 at wt%A =1 and this 1 weight-percent solution is then the standard state

W.r. t 1 weight-percent standard state, the activity of solute A is given by

Where fA(1wt%) is the 1 wt.% activity coefficient and in the range of composition in which A obeys the Henrys law fA(1wt%) = 1

Awtfh wtAwtA .%.%)1(%)1( =

Awth wtA .%%)1( =





We can also write

In the range of composition in which the solute obeys Henrys Law, fA(1wt%) =1 and A = A0 , therefore,

ncompositiotconswtA

AA

wtA

A

Awtfx

ha

tan%)1(%)1(%.

.

=

g

ncompositiotcons

AA

wtA

A

Awtx

ha

tan

0

%)1( %.

=g





We know that

Where MA and MB are the molecular weight of A and B. the first term in the denominator is small compared to the second and the relation may be simplified as

BA

AA

MAwt

MAwt

MAwt

x .%100.%

.%

-+

=

A

BA

A

B

B

AA

MM

AwtxTherefore

MMAwt

M

MAwt

x

.100.%,

.100..%

100

.%

=

=





The free energy change accompanying the transfer of one mole of solute A from pure substance standard state (Raoultian standard state) to 1 wt.% standard state, that is:

A (in the Raoultian standard state) A( in the 1 wt.% standard state)is given by

A

BA

constxWtA

AA

MMRTRT

haRTwtRG

A

.100lnln

ln%).1(

0

.%)1(

0

+=

=D

=

g





The free energy change accompanying the transfer of one mole of solute A from Henerian standard state to 1 wt.% standard state, that is:

A (in the Henerian standard state) A( in the 1 wt.% standard state)is given by

A

B

AA

BA

AAA

MMRT

RTM

MRTRT

HRGwtRGwtHG

.100ln

ln.100

lnln

)(%).1(%).1(

00

000

=

-

+=

D-D=D

gg




On differentiation,

In terms of logarithm




Inserting this into G-D equation, we get




Chemical Potential

The general equation for the free energy change of a system with temperature and pressure dG = VdP SdT, does not take into account any variation in free energy due to concentration changes.

We know From the fundamentals of partial differentiation we have

The coefficient called the chemical potential and is denoted by m hence

....),,,,( 21 nnPTGG =

......2....,,2

1....,,1,,

21

12

+

+

+

+

= dnnGdn

nGdP

PGdT

TGGd

nexceptnPT

nexceptnPTniTniP

i

nexceptnPTi

dnnG

+dPV+dTS=i

1

....,,

'''.

inexceptnTPin

G.......,, 1




Chemical Potential

m is an intensive variableThis gives a new sets of fundamental equations for the open

systems.

i

nExceptnPTi

i

nG m=

.....,, 1

++-= ii dndPVdTSGd m

+--= ii dnVPddTSAd m

++= ii dndPVSTdHd m

+-= ii dnVPdSTdUd m




Physical Meaning of Chemical potential Consider the change in free energy (dG/) of a system produced

by the addition of dnA mole of component A at constant pressure and temperature. The change in free energy of a system is given by is the partial molar free energy of component A in solution

Chemical potential of either 1 g mol or 1 g atom of a substance dissolved in a solution of definite concentration is the partial molar free energy. Thus

AAAA dnGdnGd == m

solutionofquantitylargeafor,,

A

BnTPA

A nGG m=

=

solutionof mole onefor,,

A

BnTPA

A nGG m=

=




Equality of chemical potentialamongst phases at equilibriumWe know:At constant T and P:Consider two phases(I and II) in the system. Then,

Consider moving an infinitesimal of quantity dn1 from phase

I to phase II. Then,

++-= ii dndPVdTSGd m ii dn=Gd




Equality of chemical potentialamongst phases at equilibriumTherefore, total free energy change of the system

is

For equilibrium at constant temp and pressure

Hence,

It can be generalized for all components at constant T and

P when phase I and II are at equilibrium as




Equality of chemical potentialamongst phases at equilibrium

Where P is the total no. of phases in the system




Phase Rule

Phase(P)A phase is defined as any homogeneous and physically distinct part of a system which is separated from other part of the system by a bounding surface. For example, at 273.15K, three phases ice, water and water vapour can exist in equilibrium. When ice exists in more than one crystalline form, each form will represent a separate phase because it is clearly distinguishable from each other.

Components(C)The number of components in a system at equilibrium is the smallest number of independently variable constituents by means of which the composition of each phase present can be expressed directly or in the form of a chemical equation.




Phase Rule

As an example let us consider decomposition of calcium carbonate :CaCO3 (s) = CaO(s) + CO2(g)

According to the above definition, at equilibrium this system will consist of two components since the third one is fixed by the equilibrium conditions.

Thus we have three phases two solids (CaCO3 and CaO) and a gas (CO2) and the system has only two components.

If CaO and CO2 are taken, the composition of calcium carbonate phase can be expressed as xCaO + xCO2 giving xCaCO3 (by the chemical reaction).

The composition of the three phases could be expressed equally by taking CaCO3 and CaO or CaCO3 and CO2 as the components.




Phase Rule

The dissociation of any carbonate, oxide or similar compounds involves two components; the same is true in the case of salt hydrate equilibria, for example : CuSO4.5H2O(s) = CuSO4.3H2O(s) + 2H2O(g) when the simplest components are

evidently CuSO4 and H2O.

In the slightly more complicated equilibrium : Fe(s) + H2O(g) = FeO(s) + H2(g) it is necessary to choose three components in order that the composition of each of the three phases can be expressed.

The composition of the two solid phases could be given in terms of Fe and O, but these alone are insufficient to define the gaseous phase which is a mixture of hydrogen and water vapour, a third component, viz., H2O is necessary.




Phase Rule

The water system for example consists of one component, viz., H2O each of the phases in equilibrium i.e. solid, liquid and vapour may be regarded as being made of this component only.

Degrees of freedom(F) The number of degrees of freedom is the number of variable

factors, such as temperature, pressure and concentration that need to be fixed in order that the condition of a system at equilibrium may be completely defined when referring to its equilibrium phase diagrams.




Derivation of the PhaseRule Equation




Phase rule in Reactive Components Consider a system consisting of N chemical species and there

are P number of phases.

In this case the number of components differ from number of species.

Let us consider there are three out of N species are chemically active and participate in the following reaction:

AB(s) = A(g) + B(g).

The number of total variables = P(N-1) + 2

Total number of constraints due to phase equilibrium= N(P-1).

There is another additional constraints: AB(s) = A(g) + B(g).




Phase rule in Reactive Components Additional, in the absence of A(g) and B(g) in the starting

reactant mixtures, stoichiometric consideration requires that PA= PB.

Some times, special constraints are placed on the system. For example, the system under consideration, the partial pressure of A has been fixed at 2 atm.

So this way total no. of constraints are = N(P-1)+1+1+1. F = [P(N-1)+2] [N(P-1)+1+1+1] = (N-2) P +1 = C P +1 Generalizing, for a system in which there are r independent

chemical equilibria, s stoichiometric relation and t special constraints we have

F = (N r s - t) P +2 = C- P + 2 t where C = N - r - s




Application of Phase rulein Reactive ComponentsProblem: A system is composed of a solid phase CaCO3, a solid phase CaO, and a gas phase CO2 . The following equilibrium occurs:

CaCO3(s) = CaO(s) + CO2(g)How many components are there and what are the degrees of freedom?Solution:Species: CaCO3(s) , CaO(s), CO2(g) : N =3, Phases : two solid and a gas phase

P = 3. No. of independent reaction equilibria r = 1. There is no stoichiometric or special constraints.

So s = 0 and t = 0C = N-r-s = 3-1-0= 2F = C-P+2-t = 2-3+2-0 = 1Either temperature or pressure must be specified.




Application of Phase rulein Reactive ComponentsProblem:A pure solid NH4Cl is introduced into an evacuated chamber. It is then allowed to decompose and equilibrium has been established by following reaction:

NH4Cl(s) = NH3(g) + HCl(g)Calculate the number of components and degrees of freedom.Solution:N = 3 (NH4Cl(s) , NH3(g) , HCl(g))P = 1 solid (NH4Cl(s) ) + 1 gases (NH3(g) + HCl(g ) = 2r = 1s = 1 as P NH3(g) = P HCl(g) t = 0C = N r s = 3 1 -1 = 1F = C P + 2 t = 1 -2 + 2 0 = 1




Application of Phase rulein Reactive ComponentsProblem:Show that the system in which the reaction Mn(s) + 2/3 AlCl3(g, 1atm) = MnCl2(l) + 2/3 Al (l)

is at equilibrium is invariant. Solution:N = 4 (Mn(s), AlCl3(g), MnCl2(l), Al (l))P = 1 solid (Mn) + 2 liquids(MnCl2(l)and Al (l)) + 1 gas (AlCl3(g)) = 4r = 1s = 0t = 1 (1 atm of AlCl3(g)) C = N r s = 4 1 0 = 3F = C P + 2 t = 3 4 + 2 1 = 0This is an invariant system




Application of Phase rulein Reactive ComponentsProblem:Consider reduction of FeO with CO under standard

conditions i.e. P = 1 atm. FeO(s) + CO(g) = Fe(s) + CO2(g).

Calculate the number of components and degrees of freedom.

Solution:In this system we have P = 3 (i.e. two solids FeO and Fe and a gaseous phase CO+CO2) and N = 4, r = 1, s = 0 and t = 1 (PCO + PCO2 =1 atm)C = N r s = 4 -1 -0 = 3F = C - P + 2-t = 3 3 + 2 -1 = 1Thus the above system has only one degree of freedom,

either temperature or pressure.

Documents

MTK- Dilute Solution and Phase Rule