D E P A R T M E N T O F Metallurgical Engineering Indian Institute of Technology( Banaras Hindu University) Metallurgical Thermodynamics Dilute Solution – Henry’s Law
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Dilute Solution Henrys Law
D E P A R T M E N T O F
Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Henrys Law
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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=
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Henrys Law
aA
xA
0A Henrys law constant
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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.
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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 ==
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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 ==
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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 =
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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
D E P A R T M E N T O F
Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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 .=
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Alternative Standard States
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
D E P A R T M E N T O F
Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Alternative Standard States
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
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Alternative Standard States
AAA aTRGGBut ln0 =-
0
tan
0
ln
ln)(,
A
tconsxA
AA
RT
haRTHRGHence
A
g=
=D
=
D E P A R T M E N T O F
Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Alternative Standard States
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
D E P A R T M E N T O F
Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Alternative Standard States
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( =
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Alternative Standard States
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
D E P A R T M E N T O F
Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Alternative Standard States
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
.%
=
=
D E P A R T M E N T O F
Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Alternative Standard States
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
D E P A R T M E N T O F
Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Alternative Standard States
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
D E P A R T M E N T O F
Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
On differentiation,
In terms of logarithm
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Inserting this into G-D equation, we get
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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
D E P A R T M E N T O F
Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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=
=
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Equality of chemical potentialamongst phases at equilibrium
Where P is the total no. of phases in the system
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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.
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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.
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Metallurgical Thermodynamics
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.
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Metallurgical Thermodynamics
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.
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Derivation of the PhaseRule Equation
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Derivation of the PhaseRule Equation
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
Derivation of the PhaseRule Equation
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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).
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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.
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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
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Metallurgical EngineeringIndian Institute of Technology( Banaras Hindu University)
Metallurgical Thermodynamics
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.