Equilibrium-Conditions-for-Solid-Solutions Notes | EduRev

: Equilibrium-Conditions-for-Solid-Solutions Notes | EduRev

 Page 1


MIT 3.00 Fall 2002 c ° W.C Carter 230
Lecture 33
Equilibrium Conditions for Solid Solutions
Last Time
Non-Ideal Solution Models
Regular Solutions
Spinodal Decomposition
Equilibria for Reactive Solids and Vapors (Oxidation)
Consider a reaction where a solid is reacting with a gaseous component to produce a solid
phase, such as the oxidation of aluminum metal to aluminum oxide:
4
3
Al
(solid)
+O
2(gaseous solution)
*
)
2
3
Al
2
O
3(solid)
(33-1)
or, for the case of pure silicon embedded to silica dissolved in alumina:
Si
(solid, pure)
+O
2(gaseous solution)
*
)SiO
2
(in solid sol. of Al
2
O
3
-SiO
2
)
(33-2)
In the most simple cases (e.g., Eq. 33-1), it is assumed that the reactions and products are
pure (i.e., the solubility of oxygen is neglected in the solid phases.).
Page 2


MIT 3.00 Fall 2002 c ° W.C Carter 230
Lecture 33
Equilibrium Conditions for Solid Solutions
Last Time
Non-Ideal Solution Models
Regular Solutions
Spinodal Decomposition
Equilibria for Reactive Solids and Vapors (Oxidation)
Consider a reaction where a solid is reacting with a gaseous component to produce a solid
phase, such as the oxidation of aluminum metal to aluminum oxide:
4
3
Al
(solid)
+O
2(gaseous solution)
*
)
2
3
Al
2
O
3(solid)
(33-1)
or, for the case of pure silicon embedded to silica dissolved in alumina:
Si
(solid, pure)
+O
2(gaseous solution)
*
)SiO
2
(in solid sol. of Al
2
O
3
-SiO
2
)
(33-2)
In the most simple cases (e.g., Eq. 33-1), it is assumed that the reactions and products are
pure (i.e., the solubility of oxygen is neglected in the solid phases.).
MIT 3.00 Fall 2002 c ° W.C Carter 231
In more realistic cases (e.g., Eq. 33-2), it should be clear that, as well as the free energies
of formation, considerations the free energy charges for forming a solution—such as those that
have been considered in previous lectures—must also be applied.
In any case, it simpli?es to divide a complex reaction into simpler steps.
For example Eq. 33-2, can be broken into two independent reactions:
reaction a: Si
(solid, pure)
+O
2(gaseous solution)
*
)SiO
2(solid, pure)
reaction b: SiO
2(solid, pure)
+Al
2
O
3(solid, pure)
*
)SiO
2(in solid solution with alumina
(33-3)
Inthisway,freeenergychangescanbecalculated(withrespecttosomestandardstate,usually
taken as a pure component at a particular temperature).
Reaction (a) in Eq. 33-3 will involve the molar free energies of formation with respect to
the pure components. These are usually tabulated in data books (e.g. the JANAF tables, the
book by O. Kubaschewski and C.B. Alcock)
Reaction (b) in Eq. 33-3 will involve the free energies of mixing were considered in the
construction of phase diagrams. Data is available for many practical systems and ThermoCalc
(software) is a method for extrapolating such data from known values and phase diagrams.
The Standard Approximation
Consider the general case:
aA+bB
*
)cC+dD (33-4)
Assuming the system is closed, the condition for equilibrium is just:
a¹
A
+b¹
B
=c¹
C
+d¹
D
(33-5)
which becomes, if reference is made to the pure states
a(¹
A
±
(T;P)+RT loga
A
)+b(¹
B
±
(T;P)+RT loga
B
)
=c(¹
C
±
(T;P)+RT loga
C
)+d(¹
D
±
(T;P)+RT loga
D
)
(33-6)
so the condition for equilibrium becomes
a
c
C
a
d
D
a
a
A
a
b
B
=e
¡
?G
Rx±
RT
(33-7)
where the activity of a component is a
A
=°
A
X
A
which is to be determined empirically.
However,therearestandardapproximationsinwhichthesolidphasesinthereactioncanbe
considered to be pure. In this approximation, and through use the additional thermodynamic
approximations for material behavior:
Page 3


MIT 3.00 Fall 2002 c ° W.C Carter 230
Lecture 33
Equilibrium Conditions for Solid Solutions
Last Time
Non-Ideal Solution Models
Regular Solutions
Spinodal Decomposition
Equilibria for Reactive Solids and Vapors (Oxidation)
Consider a reaction where a solid is reacting with a gaseous component to produce a solid
phase, such as the oxidation of aluminum metal to aluminum oxide:
4
3
Al
(solid)
+O
2(gaseous solution)
*
)
2
3
Al
2
O
3(solid)
(33-1)
or, for the case of pure silicon embedded to silica dissolved in alumina:
Si
(solid, pure)
+O
2(gaseous solution)
*
)SiO
2
(in solid sol. of Al
2
O
3
-SiO
2
)
(33-2)
In the most simple cases (e.g., Eq. 33-1), it is assumed that the reactions and products are
pure (i.e., the solubility of oxygen is neglected in the solid phases.).
MIT 3.00 Fall 2002 c ° W.C Carter 231
In more realistic cases (e.g., Eq. 33-2), it should be clear that, as well as the free energies
of formation, considerations the free energy charges for forming a solution—such as those that
have been considered in previous lectures—must also be applied.
In any case, it simpli?es to divide a complex reaction into simpler steps.
For example Eq. 33-2, can be broken into two independent reactions:
reaction a: Si
(solid, pure)
+O
2(gaseous solution)
*
)SiO
2(solid, pure)
reaction b: SiO
2(solid, pure)
+Al
2
O
3(solid, pure)
*
)SiO
2(in solid solution with alumina
(33-3)
Inthisway,freeenergychangescanbecalculated(withrespecttosomestandardstate,usually
taken as a pure component at a particular temperature).
Reaction (a) in Eq. 33-3 will involve the molar free energies of formation with respect to
the pure components. These are usually tabulated in data books (e.g. the JANAF tables, the
book by O. Kubaschewski and C.B. Alcock)
Reaction (b) in Eq. 33-3 will involve the free energies of mixing were considered in the
construction of phase diagrams. Data is available for many practical systems and ThermoCalc
(software) is a method for extrapolating such data from known values and phase diagrams.
The Standard Approximation
Consider the general case:
aA+bB
*
)cC+dD (33-4)
Assuming the system is closed, the condition for equilibrium is just:
a¹
A
+b¹
B
=c¹
C
+d¹
D
(33-5)
which becomes, if reference is made to the pure states
a(¹
A
±
(T;P)+RT loga
A
)+b(¹
B
±
(T;P)+RT loga
B
)
=c(¹
C
±
(T;P)+RT loga
C
)+d(¹
D
±
(T;P)+RT loga
D
)
(33-6)
so the condition for equilibrium becomes
a
c
C
a
d
D
a
a
A
a
b
B
=e
¡
?G
Rx±
RT
(33-7)
where the activity of a component is a
A
=°
A
X
A
which is to be determined empirically.
However,therearestandardapproximationsinwhichthesolidphasesinthereactioncanbe
considered to be pure. In this approximation, and through use the additional thermodynamic
approximations for material behavior:
MIT 3.00 Fall 2002 c ° W.C Carter 232
Standard Approximation for Solid Phase Reactions
1. Each component is in equilibrium with its vapor.
2. The vapor is treated as an ideal gas.
3. The molar free energy of the solid phase is relatively insensitive to pressure
changes.
Approximate equilibrium conditions can be obtained by practical means.
Consider the oxidation (or, the reverse, the reduction) of a pure metal:
M
(solid, pure)
+
1
2
O
2(gaseous solution)
*
)MO
(solid, pure)
(33-8)
The chemical potentials of each each component in each solid phase is in equilibrium with
the gaseous phase.
¹
M (solid)
=¹
M (gas)
and ¹
MO (solid)
=¹
MO (gas)
(33-9)
Therefore, it is appropriate to consider equilibrium in the gas phase.
Considering an ideal gas mixture
?G
Rx
=
=G
MO (gas sol)
(P =1;T)¡
1
2
G
iO
2
(gas sol)
(P =1;T)¡G
M (gas sol)
(P =1;T)
=RT log
P
M
P
1
2
O
2
P
MO
(33-10)
What is remarkable about Equation 33-10 is that it is true!
P
M
for typical metals is 10
¡8
–10
¡12
atm. P
MO
for typical oxides is 10
¡18
–10
¡24
atm. Such
tiny numbers which would be very, very di?cult to measure.
Also G
M(gas)
(P = 1;T) and G
MO(gas)
(P = 1;T) represent molar free energies that are
highly unstable with respect to forming a solid or a liquid at sub-solar temperatures.
Expressions for G
M(gas)
and G
MO(gas)
can be obtained by integrating the pressures for the
gas phase and the condensed phase:
G
M (gas sol)
(P =1;T)+RT logP
M
=G
M (solid)
(P =1;T)+
Z
P=P
M
P=1
V
M (solid)
dP
(33-11)
For almost every condensed phase the last term in Equation 33-11 is always very small com-
pared to the others, so to very good approximation:
G
M (gas sol)
(P =1;T)+RT logP
M
»
= G
M (solid)
(P =1;T) (33-12)
Putting this into Equation 33-10, the following approximation is obtained:
RT logP
1
2
O
2
=
=G
MO (solid)
(P =1;T)¡
1
2
G
O
2
(gas sol)
(P =1;T)¡G
M (solid)
(P =1;T)
(33-13)
Page 4


MIT 3.00 Fall 2002 c ° W.C Carter 230
Lecture 33
Equilibrium Conditions for Solid Solutions
Last Time
Non-Ideal Solution Models
Regular Solutions
Spinodal Decomposition
Equilibria for Reactive Solids and Vapors (Oxidation)
Consider a reaction where a solid is reacting with a gaseous component to produce a solid
phase, such as the oxidation of aluminum metal to aluminum oxide:
4
3
Al
(solid)
+O
2(gaseous solution)
*
)
2
3
Al
2
O
3(solid)
(33-1)
or, for the case of pure silicon embedded to silica dissolved in alumina:
Si
(solid, pure)
+O
2(gaseous solution)
*
)SiO
2
(in solid sol. of Al
2
O
3
-SiO
2
)
(33-2)
In the most simple cases (e.g., Eq. 33-1), it is assumed that the reactions and products are
pure (i.e., the solubility of oxygen is neglected in the solid phases.).
MIT 3.00 Fall 2002 c ° W.C Carter 231
In more realistic cases (e.g., Eq. 33-2), it should be clear that, as well as the free energies
of formation, considerations the free energy charges for forming a solution—such as those that
have been considered in previous lectures—must also be applied.
In any case, it simpli?es to divide a complex reaction into simpler steps.
For example Eq. 33-2, can be broken into two independent reactions:
reaction a: Si
(solid, pure)
+O
2(gaseous solution)
*
)SiO
2(solid, pure)
reaction b: SiO
2(solid, pure)
+Al
2
O
3(solid, pure)
*
)SiO
2(in solid solution with alumina
(33-3)
Inthisway,freeenergychangescanbecalculated(withrespecttosomestandardstate,usually
taken as a pure component at a particular temperature).
Reaction (a) in Eq. 33-3 will involve the molar free energies of formation with respect to
the pure components. These are usually tabulated in data books (e.g. the JANAF tables, the
book by O. Kubaschewski and C.B. Alcock)
Reaction (b) in Eq. 33-3 will involve the free energies of mixing were considered in the
construction of phase diagrams. Data is available for many practical systems and ThermoCalc
(software) is a method for extrapolating such data from known values and phase diagrams.
The Standard Approximation
Consider the general case:
aA+bB
*
)cC+dD (33-4)
Assuming the system is closed, the condition for equilibrium is just:
a¹
A
+b¹
B
=c¹
C
+d¹
D
(33-5)
which becomes, if reference is made to the pure states
a(¹
A
±
(T;P)+RT loga
A
)+b(¹
B
±
(T;P)+RT loga
B
)
=c(¹
C
±
(T;P)+RT loga
C
)+d(¹
D
±
(T;P)+RT loga
D
)
(33-6)
so the condition for equilibrium becomes
a
c
C
a
d
D
a
a
A
a
b
B
=e
¡
?G
Rx±
RT
(33-7)
where the activity of a component is a
A
=°
A
X
A
which is to be determined empirically.
However,therearestandardapproximationsinwhichthesolidphasesinthereactioncanbe
considered to be pure. In this approximation, and through use the additional thermodynamic
approximations for material behavior:
MIT 3.00 Fall 2002 c ° W.C Carter 232
Standard Approximation for Solid Phase Reactions
1. Each component is in equilibrium with its vapor.
2. The vapor is treated as an ideal gas.
3. The molar free energy of the solid phase is relatively insensitive to pressure
changes.
Approximate equilibrium conditions can be obtained by practical means.
Consider the oxidation (or, the reverse, the reduction) of a pure metal:
M
(solid, pure)
+
1
2
O
2(gaseous solution)
*
)MO
(solid, pure)
(33-8)
The chemical potentials of each each component in each solid phase is in equilibrium with
the gaseous phase.
¹
M (solid)
=¹
M (gas)
and ¹
MO (solid)
=¹
MO (gas)
(33-9)
Therefore, it is appropriate to consider equilibrium in the gas phase.
Considering an ideal gas mixture
?G
Rx
=
=G
MO (gas sol)
(P =1;T)¡
1
2
G
iO
2
(gas sol)
(P =1;T)¡G
M (gas sol)
(P =1;T)
=RT log
P
M
P
1
2
O
2
P
MO
(33-10)
What is remarkable about Equation 33-10 is that it is true!
P
M
for typical metals is 10
¡8
–10
¡12
atm. P
MO
for typical oxides is 10
¡18
–10
¡24
atm. Such
tiny numbers which would be very, very di?cult to measure.
Also G
M(gas)
(P = 1;T) and G
MO(gas)
(P = 1;T) represent molar free energies that are
highly unstable with respect to forming a solid or a liquid at sub-solar temperatures.
Expressions for G
M(gas)
and G
MO(gas)
can be obtained by integrating the pressures for the
gas phase and the condensed phase:
G
M (gas sol)
(P =1;T)+RT logP
M
=G
M (solid)
(P =1;T)+
Z
P=P
M
P=1
V
M (solid)
dP
(33-11)
For almost every condensed phase the last term in Equation 33-11 is always very small com-
pared to the others, so to very good approximation:
G
M (gas sol)
(P =1;T)+RT logP
M
»
= G
M (solid)
(P =1;T) (33-12)
Putting this into Equation 33-10, the following approximation is obtained:
RT logP
1
2
O
2
=
=G
MO (solid)
(P =1;T)¡
1
2
G
O
2
(gas sol)
(P =1;T)¡G
M (solid)
(P =1;T)
(33-13)
MIT 3.00 Fall 2002 c ° W.C Carter 233
In other words,
To good approximation, the activities of the pure solid reactants or prod-
ucts can be replaced with unity (a = 1) and the partial pressures of the
gaseous phases can be determined by the equilibrium expression.
An Example of a Gaseous Reaction with Pure Condensed Phase
Consider the reduction of SiO
2
to Si by heating it at P =1 atm. At what temperature will
the reduction take place and what will be the pressure of CO (gas) if the reaction takes place
at 740
±
C?
The pertinent reaction is
3SiO
2(solid, pure)
+4C
(graphite)
*
)2CO
(gas sol)
+2CO
2(gas sol)
+3Si (33-14)
The following reactions are tabulated:
Reaction ?G
rx
(T;P =1atm)(J=(mole O
2
))
Si
(solid)
+O
2(gas sol)
*
)SiO
2(solid, pure)
¡94556+174T(700¡1700
±
K)
2C
(graphite)
+O
2(gas sol)
*
)2CO
(gas sol)
¡223400¡175:3T(298¡2500
±
K)
C
(solid)
+O
2(gas sol)
*
)CO
2(gas sol)
¡394100¡0:84T(298¡2000
±
K)
Notice that:
1. Theexpressionsforthemolarfreeenergiesofreactionstaketheform: ?G=?H¡T?S
where ?H and ?S are treated as independent of T (This is approximately true if the
molar heat capacities are nearly the same.).
2. The appearance of one mole of gas is associated with an entropy production of about
175
J
±
K
.
The reaction of interest can be written as:
¡3£(First Reaction)+(Second Reaction)+2£(Third Reaction)
or
3SiO
2(solid, pure)
+4C
(graphite)
*
)2CO
(gas)
+2CO
2(gas)
+3Si
?G
Rx
(T;P =1atm)=¡728000¡700T (J=(moleO
2
))
(33-15)
so that
P
2
CO
2
P
2
CO
=e
¡?G
R
x
RT
(33-16)
Page 5


MIT 3.00 Fall 2002 c ° W.C Carter 230
Lecture 33
Equilibrium Conditions for Solid Solutions
Last Time
Non-Ideal Solution Models
Regular Solutions
Spinodal Decomposition
Equilibria for Reactive Solids and Vapors (Oxidation)
Consider a reaction where a solid is reacting with a gaseous component to produce a solid
phase, such as the oxidation of aluminum metal to aluminum oxide:
4
3
Al
(solid)
+O
2(gaseous solution)
*
)
2
3
Al
2
O
3(solid)
(33-1)
or, for the case of pure silicon embedded to silica dissolved in alumina:
Si
(solid, pure)
+O
2(gaseous solution)
*
)SiO
2
(in solid sol. of Al
2
O
3
-SiO
2
)
(33-2)
In the most simple cases (e.g., Eq. 33-1), it is assumed that the reactions and products are
pure (i.e., the solubility of oxygen is neglected in the solid phases.).
MIT 3.00 Fall 2002 c ° W.C Carter 231
In more realistic cases (e.g., Eq. 33-2), it should be clear that, as well as the free energies
of formation, considerations the free energy charges for forming a solution—such as those that
have been considered in previous lectures—must also be applied.
In any case, it simpli?es to divide a complex reaction into simpler steps.
For example Eq. 33-2, can be broken into two independent reactions:
reaction a: Si
(solid, pure)
+O
2(gaseous solution)
*
)SiO
2(solid, pure)
reaction b: SiO
2(solid, pure)
+Al
2
O
3(solid, pure)
*
)SiO
2(in solid solution with alumina
(33-3)
Inthisway,freeenergychangescanbecalculated(withrespecttosomestandardstate,usually
taken as a pure component at a particular temperature).
Reaction (a) in Eq. 33-3 will involve the molar free energies of formation with respect to
the pure components. These are usually tabulated in data books (e.g. the JANAF tables, the
book by O. Kubaschewski and C.B. Alcock)
Reaction (b) in Eq. 33-3 will involve the free energies of mixing were considered in the
construction of phase diagrams. Data is available for many practical systems and ThermoCalc
(software) is a method for extrapolating such data from known values and phase diagrams.
The Standard Approximation
Consider the general case:
aA+bB
*
)cC+dD (33-4)
Assuming the system is closed, the condition for equilibrium is just:
a¹
A
+b¹
B
=c¹
C
+d¹
D
(33-5)
which becomes, if reference is made to the pure states
a(¹
A
±
(T;P)+RT loga
A
)+b(¹
B
±
(T;P)+RT loga
B
)
=c(¹
C
±
(T;P)+RT loga
C
)+d(¹
D
±
(T;P)+RT loga
D
)
(33-6)
so the condition for equilibrium becomes
a
c
C
a
d
D
a
a
A
a
b
B
=e
¡
?G
Rx±
RT
(33-7)
where the activity of a component is a
A
=°
A
X
A
which is to be determined empirically.
However,therearestandardapproximationsinwhichthesolidphasesinthereactioncanbe
considered to be pure. In this approximation, and through use the additional thermodynamic
approximations for material behavior:
MIT 3.00 Fall 2002 c ° W.C Carter 232
Standard Approximation for Solid Phase Reactions
1. Each component is in equilibrium with its vapor.
2. The vapor is treated as an ideal gas.
3. The molar free energy of the solid phase is relatively insensitive to pressure
changes.
Approximate equilibrium conditions can be obtained by practical means.
Consider the oxidation (or, the reverse, the reduction) of a pure metal:
M
(solid, pure)
+
1
2
O
2(gaseous solution)
*
)MO
(solid, pure)
(33-8)
The chemical potentials of each each component in each solid phase is in equilibrium with
the gaseous phase.
¹
M (solid)
=¹
M (gas)
and ¹
MO (solid)
=¹
MO (gas)
(33-9)
Therefore, it is appropriate to consider equilibrium in the gas phase.
Considering an ideal gas mixture
?G
Rx
=
=G
MO (gas sol)
(P =1;T)¡
1
2
G
iO
2
(gas sol)
(P =1;T)¡G
M (gas sol)
(P =1;T)
=RT log
P
M
P
1
2
O
2
P
MO
(33-10)
What is remarkable about Equation 33-10 is that it is true!
P
M
for typical metals is 10
¡8
–10
¡12
atm. P
MO
for typical oxides is 10
¡18
–10
¡24
atm. Such
tiny numbers which would be very, very di?cult to measure.
Also G
M(gas)
(P = 1;T) and G
MO(gas)
(P = 1;T) represent molar free energies that are
highly unstable with respect to forming a solid or a liquid at sub-solar temperatures.
Expressions for G
M(gas)
and G
MO(gas)
can be obtained by integrating the pressures for the
gas phase and the condensed phase:
G
M (gas sol)
(P =1;T)+RT logP
M
=G
M (solid)
(P =1;T)+
Z
P=P
M
P=1
V
M (solid)
dP
(33-11)
For almost every condensed phase the last term in Equation 33-11 is always very small com-
pared to the others, so to very good approximation:
G
M (gas sol)
(P =1;T)+RT logP
M
»
= G
M (solid)
(P =1;T) (33-12)
Putting this into Equation 33-10, the following approximation is obtained:
RT logP
1
2
O
2
=
=G
MO (solid)
(P =1;T)¡
1
2
G
O
2
(gas sol)
(P =1;T)¡G
M (solid)
(P =1;T)
(33-13)
MIT 3.00 Fall 2002 c ° W.C Carter 233
In other words,
To good approximation, the activities of the pure solid reactants or prod-
ucts can be replaced with unity (a = 1) and the partial pressures of the
gaseous phases can be determined by the equilibrium expression.
An Example of a Gaseous Reaction with Pure Condensed Phase
Consider the reduction of SiO
2
to Si by heating it at P =1 atm. At what temperature will
the reduction take place and what will be the pressure of CO (gas) if the reaction takes place
at 740
±
C?
The pertinent reaction is
3SiO
2(solid, pure)
+4C
(graphite)
*
)2CO
(gas sol)
+2CO
2(gas sol)
+3Si (33-14)
The following reactions are tabulated:
Reaction ?G
rx
(T;P =1atm)(J=(mole O
2
))
Si
(solid)
+O
2(gas sol)
*
)SiO
2(solid, pure)
¡94556+174T(700¡1700
±
K)
2C
(graphite)
+O
2(gas sol)
*
)2CO
(gas sol)
¡223400¡175:3T(298¡2500
±
K)
C
(solid)
+O
2(gas sol)
*
)CO
2(gas sol)
¡394100¡0:84T(298¡2000
±
K)
Notice that:
1. Theexpressionsforthemolarfreeenergiesofreactionstaketheform: ?G=?H¡T?S
where ?H and ?S are treated as independent of T (This is approximately true if the
molar heat capacities are nearly the same.).
2. The appearance of one mole of gas is associated with an entropy production of about
175
J
±
K
.
The reaction of interest can be written as:
¡3£(First Reaction)+(Second Reaction)+2£(Third Reaction)
or
3SiO
2(solid, pure)
+4C
(graphite)
*
)2CO
(gas)
+2CO
2(gas)
+3Si
?G
Rx
(T;P =1atm)=¡728000¡700T (J=(moleO
2
))
(33-15)
so that
P
2
CO
2
P
2
CO
=e
¡?G
R
x
RT
(33-16)
MIT 3.00 Fall 2002 c ° W.C Carter 234
Since ?G
Rx
<0 at all T, the reaction will favor the products.
The total pressure is given by
P
CO
+P
CO
2
+P
O
2
=1 (33-17)
Question: WhyisP
O
2
included? Becauseonecan’tbalanceareactionbetweencarbonmonoxide
and carbon dioxide without it.
For
C
(graphite)
+O
2(gas)
+
*
)CO
2(gas)
?G
Rx(I)
(T =1040;P =1atm)=¡395000(J=(moleO
2
))
(33-18)
So
P
O
2(gas)
P
CO
2(gas)
=e
¡
395000
(8:3)(1040)
=small number (33-19)
for
2C
(graphite)
+O
2(gas)
+
*
)2CO
(gas)
?G
Rx(II)
(T =1040;P =1atm)=¡405700(J=(moleO
2
))
(33-20)
P
O
2(gas)
P
2
CO
(gas)
=e
¡
405700
(8:3)(1040)
=small number (33-21)
We conclude that P
O
2
is small compared to P
CO
and P
CO
2
P
CO
+P
CO
2
¼1 (33-22)
Therefore:
P
CO
2
P
2
CO
=e
¡
?G
Rx(I)
¡?G
Rx(II)
RT
=3:45 (33-23)
1¡P
CO
P
2
CO
=3:45 =) P
CO
=0:41 (33-24)
Question: Will the fraction of CO go up or down with temperature?
Ellingham Diagrams
Thermodynamic data for the oxidation of a number of common metals can be usefully and
graphically codi?ed in an Ellingham Diagram (Gaskell, page 272).
For example:
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