KJM5120 and KJM9120 Defects and Reactions
Ch 3. Thermodynamics
Truls Norby
Department of Chemistry
University of Oslo
Centre for Materials Science
and Nanotechnology (SMN)
FERMIO
Oslo Research Park
(Forskningsparken)
[email protected]
http://folk.uio.no/trulsn
Defect thermodynamics
•
The main aim of this chapter is to learn to evaluate how the defect
concentrations vary with temperature and component activities (e.g. pO2).
•
The tool for this is the mass action law, also referred to as equilibrium
coefficients and constants.
•
The chapter has an introduction to relevant thermodynamics. This may be
trivial to some, and too brief and inaccessible for others. You may choose
your level of approach based on your background and aspirations.
•
BUT: Regardless of your understanding of the thermodynamics, learning
or accepting mass action law equilibrium constant expressions and how to
write them up from a reaction equation is mandatory!
•
AND: Being able to combine such an equilibrium constant expression with
a simplified electroneutrality condition is also mandatory.
•
These are the two central elements required as a minimum to use defect
chemistry.
Equilibrium in defect chemical reactions
The compendium recalls the first and second laws of
thermodynamics and how they have lead to the understanding
that a system is at equilibrium (no net reaction in any direction)
when the Gibbs energy G = H – TS is at a minimum, i.e. when
dG = dH – TdS = 0. The compendium furthermore uses it on a
simple example of vacancies in an elemental solid.
It furthermore introduces a derivation of the mass action law and
equilibrium constants based on thermodynamic arguments.
Here, we will concentrate on an alternative derivation, namely
from kinetics. This will help understand that equilibrium does not
mean that nothing happens, just that reactions in both directions
happen at equal rates.
Equilibrium in a gas phase reaction:
•
Consider the gas phase reaction
2 H 2O( g ) = 2 H 2 ( g ) + O2 ( g )
•
The reactants have a total standard
Gibbs energy of GR0. They react via
some activated complex with energy
GA0 to arrive at the product with
energy GP0.
•
From the example, we see that GP0 >
GR0 so if you mix products and
reactants in similar amounts, it is
clear that the reaction will proceed
backwards; H2O(g) is the most stable
species. We are still interested to
calculate the ratio of products over
reactants at equilibrium coefficient
This example, where the reactant is more stable
than the products, is typical also for defect
formation reactions. It is important to realise that
even if a reaction is not favourable, it will still
happen, but only to a small degree. We shall see
that it proceeds, but stops at equilibrium already
when small amounts of the products are formed.
Equilibrium in a gas phase reaction
2 H 2O( g ) = 2 H 2 ( g ) + O2 ( g )
•
For forward reaction two H2O molecules have to react. The
chance for this to happen in the ideal case is proportional to
[H 2O][H 2 O] = [H 2 O]2 ∝ p H2 2O = a H2 2O
•
Moreover, it is proportional to the fraction of reactants that
have enough thermal energy to overcome the energy barrier.
Thus
0
0
0
−
(G
−
G
)
−
∆G
A
R
FA
rF = r0 aH2 2O exp
= r0 aH2 2O exp
RT
RT
•
Similarly, for the backwards reaction to happen, one oxygen
molecule and two hydrogen molecules have to collide, and the
backwards energy barrier to the activated complex must be
overcome:
0
− (G A0 − GP0 )
− ∆GBA
2
rB = r a aO2 exp
= r0 aH 2 aO2 exp
RT
RT
2
0 H2
Equilibrium in a gas phase reaction
2 H 2O( g ) = 2 H 2 ( g ) + O2 ( g )
•
The net reaction rate is:
rN = rF − rB
•
At equilibrium, the net reaction rate is zero:
rN = 0 ⇒ rF = rB
•
Thus, at equilibrium, the forward and backward rates are equal.
•
The equal rates are dependent on the energies of activation.
•
However, since the energies are different, the equal rates will have to be
accomplished by shifting the ratio of products and reactants so that the
difference in activation energy is counteracted by numbers of reactants. This
is the principle of the mass action law.
Equilibrium in a gas phase reaction
2 H 2O( g ) = 2 H 2 ( g ) + O2 ( g )
•
We insert to get
0
0
−
∆G
−
∆G
FA
BA
rF = rB ⇒ r0 aH2 2O exp
= r0 aH2 2 aO2 exp
RT
RT
•
Rearranging with respect to the equilibrium coefficient:
aH2 2O
2
H2
a aO2
•
= K FR
0
0
0
− ( ∆GBA
− ∆GFA
)
− ∆GFR
= exp
= exp
RT
RT
Thus, from considering collisions of the species, we
get the relation between the mass action ratio of the
activities and the Gibbs energy change of the forward
reaction.
Equilibrium in a gas phase reaction
2 H 2O( g ) = 2 H 2 ( g ) + O2 ( g )
•
Some general remarks:
•
The rate factor r0 and the energy of the activated
complex are important for rF and rB, but not for the
equilibrium coefficient K;
•
K only depends on the difference in energy between
the end products and starting reactants.
•
It is thus irrelevant to discuss what the activated
complex is…the reaction can go through a number of
steps, and yet it is only the energies of the start and
end species that matters to the equilibrium of the
reaction stated.
Equilibrium in a gas phase reaction
•
Our case
2 H 2O( g ) = 2 H 2 ( g ) + O2 ( g )
aH2 2O
2
H2
a aO2
•
= K FR
0
− ∆GFR
= exp
RT
is one example of a mass action equilibrium coefficient. It is often
exemplified for a more general reaction between a molecules of A and b
molecules of B to form c molecules of C and d molecules of D:
aA + bB = cC + dD
c d
C D
a b
A B
− ∆G 0
a a
= K = exp
a a
RT
Equilibrium based on the mass action law
What we have done till now has been an attempt to show that
chemical equilibrium is a result of forward and backward
reactions happening at the same rate, due to a shift in the ratio
of activities of the reactants and products.
This shift – or ratio – is called the equilibrium coefficient.
From the collision aspect of it it is called the law of mass action.
It applies to all chemical reactions, also defect chemical
reactions, where the derivation based on mass action may be
more or less obvious.
The compendium derives the equilibrium coefficients based
purely on thermodynamics, in which the absolute and net rates
become hidden, but where the statistical thermodynamics aspect
of defects becomes better visualised.
Just a word on equilibrium constants and
equilibrium coefficients
•
The K is often called a mass action or equilibrium constant because
it has the important mission to tell us that a particular ratio of
products and reactants is maintained constant at equilibrium.
•
However, the value changes with temperature. This is because the
Gibbs energy is a balance between the relatively constant enthalpy
change and the temperature dependent decrease in energy given
by the entropy change:
∆G 0 = ∆H 0 − T∆S 0
K = exp
•
− ∆G 0
RT
= exp
∆S 0
R
exp
− ∆H 0
RT
Therefore, K’s are not constant (except at each temperature) and
are therefore often instead referred to as coefficients.
Examples of defect equilibria in stoiciometric oxides
and how dependencies on T are derived
We will here, and in the compendium, go through
many examples.
However, be aware that the principles are taught in the
first example – on Schottky defects in MO – so take
that first example seriously. If you don’t understand
that one, you won’t understand the rest.
Schottky defects in MO
•
We start by writing the the relevant defect
formation reaction:
0 = vO•• + v M//
•
We then write its equilibrium coefficient:
K S = av•• av // = X v•• X v //
O
Activities a
M
O
M
[vO•• ] [vM// ]
=
[O] [M]
For point defects, activities
are expressed in terms of
site fractions X
The site fraction is the
concentration of defects over
the concentration of sites
Schottky defects in MO
• K’s are often simplified. There are various reasons why:
– Because you sometimes can do it properly;
– Because the simplification is a reasonable approximation;
– Because you are not interested in the difference between the full and
simplified K (this often means that you disregard the possibility to assess
the entropy change);
– Because neither the full nor simplified make much sense in terms of
entropy, so they are equally useful, and then we may well choose the
simplest.
• If we express concentrations in molar fractions (mol/mol MO),
then [M] = [O] = 1, and we may simplify to
[vO•• ] [vM// ]
KS =
= [vO•• ][vM// ]
[O] [M]
Schottky defects in MO
•
NOTE: An equilibrium coefficient expression is always valid (at equilibrium):
K S = [vO•• ][vM// ]
•
Thus the product of the concentrations of oxygen and metal vacancies is always
constant (at constant T). We may well stress this by instead writing:
[vO•• ][vM// ] = K S
•
Anyway, note again that this is an “eternal truth”…we have not made any
choices, we have simply selected an arbitrary reaction and stated that it must
follow the law of mass action, hence the K.
•
While KS represents information about the system, we have two unknowns,
namely the two defect concentrations, so this is not enough. We need one more
piece of independent input.
Schottky defects in MO
•
The second piece of input is the electroneutrality expression. If the two defects of
the Schottky pair are the dominating defects, we may write
2[vO•• ] = 2[vM// ]
which of course simplifies to
[vO•• ] = [vM// ]
•
It is now important to understand that this is NOT an “eternal truth”…the
electroneutrality statement is a choice: We choose that these are the dominating
defects.
•
The next step is to combine the two sets of information; we insert the
electroneutrality into the equilibrium coefficient:
[vO•• ][vM// ] = K S
[vO•• ] = [vM// ]
[vM// ]2 = K S
//
M
[v ] = K
1/ 2
S
[vO•• ] = [vM// ] = K S1 / 2
Voila! That’s it. We have now found the
expression for the concentration of the
defects. In this case, they are only a
function of KS.
Schottky defects in MO
•
From the general temperature dependency of K,
∆S S0
− ∆H S0
K S = exp
exp
R
RT
•
we obtain
••
O
//
M
[v ] = [v ] = K
•
1/ 2
S
∆S S0
− ∆H S0
= exp
exp
2R
2 RT
The square root and number
2 arise from the reaction
containing 2 defects.
ln or log defect concentrations vs 1/T (van ‘t Hoff plots):
∆S S0 − ∆H S0 1
ln[v ] = ln[v ] =
+
2R
2R T
••
O
//
M
log[vO•• ] = log[vM// ] =
0
S
0
S
∆S
− ∆H 1
+
2 R ln 10 2 R ln 10 T
ln10=2.303
Schottky defects in MO
•
van ‘t Hoff plot
•
Standard entropy and
enthalpy changes can
be found from intercept
with y axis and slope,
respectively, after
multiplication with 2R
and -2R.
•
log [ ] plots can be more
intelligible, but require
the additional
multiplications with ln 10
= 2.303.
∆SS0/2R
ln [ ]
-∆HS0/2R
[vO..]=[vM//]
1/T
Before we move on…
•
•
Note that:
–
The solution we found assumes that the two Schottky defects are dominating.
–
The standard entropy and enthalpy changes of the Schottky reaction refer to the
reaction when the reactants and products are in the standard state. For our defects,
that means that the site fraction is unity! We recall that this is a hypothetical state, but
nevertheless the state we have agreed on as standard. Therefore, the entropy as
derived and used here is only valid if the defect concentrations are entered (plotted) in
units of mole fraction (which in MO is the same as site fraction). We may also recall
that the entropy change represents the change in vibrational entropy.
–
The model also assumes ideality, i.e. that the activities of defects are proportional to
their concentrations. It is a dilute solution case.
Frenkel defect pair
–
Read the compendium text…..similar to the Schottky case.
Intrinsic ionisation of electronic defects
•
For localised defects (valence defects), read the compendium.
•
For conduction band electrons and valence band holes, the relevant
reaction is
0 = e/ + h•
•
The equilibrium coefficient may be written
K i = ae / ah• =
•
•
[e / ] [h • ]
N C NV
=
n p
N C NV
Here, the activities of electrons and holes are expressed in terms of
the fraction of their concentration over the density of states of the
conduction and valence bands, respectively. The reason is that
electrons behave quantum-mechanically and therefore populate
different energy states rather than different sites.
The standard state is according to this: n = NC and p = NV
Intrinsic ionisation of electronic defects
•
If we apply the concepts of standard Gibbs energy, entropy, and
enthalpy changes as before, we obtain
− ∆Gi0
∆S i0
− ∆H i0
n p
Ki =
= exp
= exp
exp
N C NV
RT
R
RT
•
•
However, this is not commonly adopted or adoptable.
In physics it is instead more common to use simply:
/
i
/
•
K = [e ][h ] = n p = N C NV exp
•
− Eg
RT
This states that the product of n and p is constant at a given
temperature, as expected for the equilibrium coefficient for the
reaction. However, the concept of activity is not applied, as standard
states for electronic defects are not commonly defined. For this
reason, we here use a prime on the Ki/ to signify the difference to a
“normal” K from which the entropy could have been derived.
Intrinsic ionisation of electronic defects
•
From
− ∆Gi0
∆S i0
− ∆H i0
n p
Ki =
= exp
= exp
exp
N C NV
RT
R
RT
and
K i/ = [e / ][h• ] = n p = N C NV exp
− Eg
RT
we see that the band gap Eg is to a first approximation the Gibbs energy
change of the intrinsic ionisation, which in turn consists mainly of the
enthalpy change.
•
We shall not enter into the finer details or of the differences here, just
stress that np = constant at a given temperature. Always!
•
Physicists mostly use Eg/kT with Eg in eV per electron, while chemists
often use Eg/RT (or ∆G0/RT) with Eg in J per mole electrons. This is a
trivial conversion (factor 1 eV = 96485 J/mol = 96.485 kJ/mol).
Intrinsic ionisation of electronic defects
•
The density of states can under certain assumptions be approximated by
 8πm kT 

N C = 
 h

*
e
2
3/ 2
 8πm kT 

NV = 
 h

*
h
2
•
•
•
3/ 2
They thus contain the effective masses me* and mh* of electrons and holes,
respectively. These are known or are often assumed close to the electron
rest mass.
NC and Nv have units of m-3.
Note the T3/2 temperature dependencies of NC and NV.
Intrinsic ionisation of electronic defects
•
If we choose that electrons and holes dominate the defect structure;
n= p
•
We insert into the equilibrium coefficient expression and get
n p = n 2 = K i/ = N C NV exp
n = p=K
•
•
/ 1/ 2
i
1/ 2
= ( N C NV )
− Eg
RT
exp
− Eg
2 RT
A logarithmic plot of n or p vs 1/T will thus have a slope that seems to
reflect Eg/2 as the apparent enthalpy.
Because of the temperature dependencies of the density of states it
should however be more appropriate to plot nT-3/2 or pT-3/2 vs 1/T to
obtain a slope that reflects Eg/2 correctly.
Examples of defect equilibria in non-stoichiometric
oxides and how dependencies on T and pO2 are derived
The next examples go by the same methodology as
before…only now oxygen becomes a reactant or
product. The defect concentrations become dependent
on pO2 in addition to temperature.
Oxygen deficient oxides
•
The compendium starts with an example comprising electrons
represented as valence defects.
•
Here we start directly with conduction band electrons.
•
Oxygen vacancies are formed according to
This big expression
may seem unnecessary,
but is meant to help you
understand…
OOx = vO•• + 2e / + 12 O2 ( g )
K vO =
av•• ae2/ aO1/22( g )
O
aO x
O
=
2
1/ 2
 pO2 


0
 pO 
 2
[OOx ]
[O]
[v ]  n 


[O]  N C 
••
O
••
O
x
O
[v
=
[O
] n 


]  NC 
2
1/ 2
 pO2 


0
 pO 
 2
Oxygen deficient oxides
• It is common for most purposes to neglect the division by NC,
to assume [OOx]=1 and to remove pO20 =1 bar, so that we get
/
K vO
= N C2 K vO = [vO•• ]n 2 pO1/22
• We are going to express many equilibria the same simplified
way, so it is important to understand how from the reaction
OOx = vO•• + 2e / + 12 O2 ( g )
to write the equilibrium coefficient expression
/
K vO
= [vO•• ]n 2 pO1/22
Oxygen deficient oxides
• We now choose to assume that the oxygen vacancies and
electrons are the two dominating defects. The electroneutrality
then reads
2[vO•• ] = n
• We now insert this into the equilibrium coefficient and get
/
K vO
= 4[vO•• ]3 pO1/22
• We finally solve with respect to the concentration of defects:
/ 1 / 3 −1 / 6
[vO•• ] = ( 14 K vO
) pO2
/ 1 / 3 −1 / 6
n = 2[vO•• ] = (2 K vO
) pO2
Oxygen deficient oxides
• We split KvO into a preexponential and the enthalpy term:
/ 1 / 3 −1 / 6
n = 2[vO•• ] = (2 K vO
) pO2
0
−
∆
H
/
1/ 3
−1 / 6
vO
= (2 K vO
)
exp
p
,0
O2
3RT
• From this, a plot of the logarithm of the defect concentrations vs
1/T will give lines with slope of –∆HvO0/3R.
• The number 3 relates to the formation of 3 defects in the defect
reaction
Oxygen deficient oxides
/ 1 / 3 −1 / 6
n = 2[vO•• ] = (2 K vO
) pO2
• By taking the logarithm:
/
log n = log 2 + log[vO•• ] = 13 log(2 K vO
) − 16 pO2
• we see that a plot of logn vs logpO2 gives
a straight line with a slope of -1/6.
• This kind of plot is called a Brouwer
diagram
• Note that log[vO..] is a parrallel line log2 =
0.30 units lower.
Oxygen deficient oxides
•
What if also neutral and singly charged oxygen vacancies are important?
OOx = vOx + 12 O2 ( g )
KvOx
vOx = vO• + e /
KvO1
vO• = vO•• + e /
KvO2
OOx = vO•• + 2e / + 12 O2 ( g )
KvO = KvOx KvO1 KvO2
n = [vO• ] + 2[vO•• ]
•
Total electroneutrality
•
May be solved analytically or as simplified cases, see compendium text.
Main lesson here for now is that a reaction may be split up into two or
more steps. Alternatively, that individual reactions may be summed and
equilibrium coefficients multiplied.
Metal excess oxide M2O3
•
Defect reaction:
3
2
•
OOx + M Mx = M i••• + 3e / + 34 O2 ( g )
Equilibrium coefficient and its simplification:
/
K Mi
= [M i••• ]n3 pO3 /2 4 [OOx ] - 3 / 2 [M Mx ] -1 = [M i••• ]n3 pO3 /2 4
•
If these defects are dominating, the electronutrality and its insertion
into equilibrium coefficient:
/
3[ M i••• ] = n = (3K Mi
)1 / 4 pO−32 / 16
•
•
Temperature dependency contains –∆HMi0/4R. (4 defects)
pO2-3/16 dependency (slope of -3/16 in Brouwer diagram).
Both oxygen deficiency and metal excess
Now we will take an example with two point defects,
namely oxygen vacancies and metal interstitials, both
compensating electrons.
We will learn a way of determining the full Brouwer
diagram for all defects. This is an important
methodology, and this is where you learn it!
Both oxygen deficiency and metal excess
•
Consider the oxide MO2. It has both oxygen deficiency and metal
excess and may as such be written M1+xO2-y.
•
We assume that the oxygen vacancies are doubly charged, We
furthermore assume that the metal interstitials are not fully charged,
only doubly charged.
•
The formation of the defects can be written:
•
OOx = vO•• + 2e / + 12 O2 ( g )
/
K vO
= [vO•• ]n 2 pO1/22
2OOx + M Mx = M i•• + 2e / + O2 ( g )
/
••
2
K Mi
2 = [ M i ]n pO2
Full electroneutrality:
n = 2[vO•• ] + 2[ M i•• ]
Instead of solving the full situation analytically or
numerically, we are now going to solve each simplified
combination of the electroneutrality condition.
Both oxygen deficiency and metal excess
•
We first choose one simplified electroneutrality condition:
•
Assume first that we are in a region where oxygen vacancies
dominate:
[vO•• ] >> [ M i•• ]
•
We know from earlier that the electroneutrality then is
n = 2[vO•• ]
and that insertion of this into the equilibrium coefficient gives
/ 1 / 3 −1 / 6
n = 2[vO•• ] = (2 K vO
) pO2
Both oxygen deficiency and metal excess
•
We next insert this behaviour of one of the dominating defects into an
expression relating it to a minority defect. The equilibrium coefficient
for metal interstitials contains concentrations of both electrons and
metal interstitials, and is suitable:
/
••
2
••
/
2 / 3 −1 / 3
••
/
2/ 3 2/ 3
K Mi
=
[M
]n
p
=
[M
](
2
K
)
p
p
=
[M
](
2
K
)
pO2
2
i
O2
i
vO
O2
O2
i
vO
•
We solve with respect to the metal interstitials:
/
/
−2 / 3 −2 / 3
2
[M i•• ] = K Mi
(
K
)
pO2
2
vO
•
We have thus found that in the region where electrons and oxygen
vacancies dominate and have pO2-1/6 dependencies, minority doubly
charged metal interstitials in MO2 have a pO2-2/3 dependence.
Both oxygen deficiency and metal excess
•
The situation we have
considered is illustrated in this
figure, right hand part.
•
Since the minority metal
interstitials have a steeper
negative pO2 dependence
than the majority defects, the
former will sooner or later
catch up and take over as
positive defects at lower pO2.
•
In Brouwer methodology, we
do not calculate the transition
region, we move directly to
the new situation where the
new defect has taken over, at
the cost of another.
Both oxygen deficiency and metal excess
•
We now move to the other simplified limiting
electroneutrality:
[ M i•• ] >> [vO•• ]
•
n = 2[ M i•• ]
The reaction is
2OOx + M Mx = M i•• + 2e / + O2 ( g )
•
/
2
••
K Mi
=
[
M
]n
pO2
2
i
Inserting the electroneutrality gives
/
1/ 3 −1 / 3
n = 2[ M i•• ] = (2 K Mi
)
pO2
2
•
Inserting this into the equilibrium coefficient of electrons
and the now minority oxygen vacancies gives
/
/
−2 / 3 1 / 6
[vO•• ] = K vO
(2 K Mi
)
pO2
2
Brouwer diagrams
•
The diagram we have developed
here is a Brouwer diagram.
•
It shows schematically the
behaviour of majority and minority
defects within regions of simplified
limiting electroneutrality.
•
The transitions between regions
are normally made sharp.
•
The changeover point between
two defects is where they both
contribute equally much in the
electroneutrality
Rules for the methodology we have used
1. Write down the full electroneutrality of all defects you intend to consider.
2. Write a number of chemical reactions and equilibrium coefficients that
relate the defects to each other. If there are n defects there must be n-1
independent reactions (meaning that no reaction must be a sum of other
reactions). The nth reaction is the electroneutrality itself.
3. Choose one simplest possible set (normally a pair) of defects and
formulate the simplified limiting electroneutrality assuming these are
dominating.
4. Insert this into an equilibrium coefficient that contains these defects, and
solve with respect to the dominating defect concentrations.
5. Insert the result from 4 into another equilibrium to determine the
behaviour of a minority defect. If there are more minority defects,
continue to insert and determine the behaviour of them all.
6. Extrapolate along e.g. pO2 to find which minority defect that takes over.
Based on this, go back to Step 3, formulate a new simplified
electroneutrality condition. Repeat this cycle in both directions of e.g.
pO2 to explore all possible combinations of defects.
Metal deficiency and oxygen excess
For these cases, see the compendium text.
Oxides with both oxygen deficiency and excess
The oxide we will use as example has oxygen
vacancies and electrons at low pO2, and oxygen
interstitials and holes at high pO2.
At intermediate pO2 a stoichiometric defect disorder
must be dominating.
We will derive Brouwer diagrams quantitatively to
understand the defect structure.
Oxides with both oxygen deficiency and
excess
•
We will only consider fully charged defects.
•
The point defects are necessarily oxygen vacancies and oxygen
interstitials.
•
Electronic defects – electrons and holes - are required to have
dominatingly oxygen deficiency or oxygen excess.
•
The full electroneutrality condition is thus:
2[vO•• ] + p = 2[Oi// ] + n
Oxides with both oxygen deficiency and
excess
2[vO•• ] + p = 2[Oi// ] + n
•
We will show how to develop
this Brouwer diagram, i.e. log
defect concentrations vs log
pO2 at constant temperature
•
We assume essentially no
knowledge of the actual
equilibrium constants, and
the diagram is thus just
schematical.
•
Try to sketch your own
vesrion of it as we proceed
Oxides with both oxygen deficiency and
excess
2[vO•• ] + p = 2[Oi// ] + n
•
We have learned to write a number of defect reactions and simplified
equilibrium coefficient expressions for the formation these defects:
OOx + vix = vO•• + Oi//
/
0=e +h
•
OOx = vO•• + 2e / + 12 O2 ( g )
1
2
O2 ( g ) + vOx = Oi// + 2h •
However, we now have 1+4=5 equations for the 4
defects, so all four equilibria are not going to be
needed, and are not independent. For instance:
K AF = [Oi// ][vO•• ]
K i/ = np
/
K vO
= [vO•• ]n 2 pO1/22
K Oi/ = [Oi// ]p 2 pO−12 / 2
K
2
AF
/
K vO
K Oi/
=
K i/
Oxides with both oxygen deficiency and
excess
•
We now choose two of the defects
as dominating; one positive and
one negative, of course.
•
We start by the well-known
situation of dominating oxygen
vacancies and electrons
2[vO•• ], n >> p, 2[Oi// ]
/ 1 / 3 −1 / 6
n = 2[vO•• ] = ( 2 K vO
) pO2
Oxides with both oxygen deficiency and
excess
•
We next insert this into other of the
equilibrium expressions in order to find
out how the minority defects vary.
/ 1 / 3 −1 / 6
n = 2[vO•• ] = ( 2 K vO
) pO2
•
We first consider the intrinsic
electronic equilibrium, and see that
holes and electrons vary inversely to
each other:
/
K
K i/ = np ⇒ p = i
n
K i/
/
p=
= K i/ (2 K vO
) −1/ 3 pO1/26
n
Note: The level where we insert p is arbitrary. Next we will
insert [Oi//]…we choose to insert that much lower than p.
Oxides with both oxygen deficiency and
excess
•
Next we will solve for [Oi//]. For this we
can use the anti-Frenkel equilibrium
K AF = [Oi// ][vO•• ]
and insert [vO..] from
/ 1 / 3 −1 / 6
n = 2[vO•• ] = ( 2 K vO
) pO2
or we use the equilibrium for oxygen
excess
K Oi/ = [Oi// ]p 2 pO−12 / 2
and insert p from
/
p = K i/ (2 K vO
) −1/ 3 pO1/26
•
In any case we obtain [Oi//]∝pO21/6
Oxides with both oxygen deficiency and
excess
•
•
The first minority defect to take over
as dominating is in our case holes
taking over for oxygen vacancies.
The new pair of dominating defects
and the new electroneutrality
condition are
p, n >> 2[vO•• ], 2[Oi// ]
n= p=K
•
/1/ 2
i
Inserting this into the reduction and
oxidation equilibria (involving
electronic defects and point defects)
yields new dependencies for the point
defects.
Oxides with both oxygen deficiency and
excess
•
Finally, oxygen interstitials take over
from electrons, and the situation to
the right in the figure can be derived.
2[Oi// ], p >> 2[vO•• ], n
Oxides with both oxygen deficiency and
excess
•
•
We can learn many general lessons from
this Brouwer diagram and its derivation.
The constancy of products
K i/ = np
•
K AF = [Oi// ][vO•• ]
enables quick establishment of some
inverse slopes.
This diagram was drawn with the choice
that
K i/ >> K AF
•
•
Any simple oxide basically has three
regions like this: Oxygen deficiency,
stoichiometry, and oxygen excess.
Although the material is essentially
stoichiometric when n=p, it is only exactly
stoichiometric when also [vO..]=[Oi//].
Oxides with both oxygen deficiency and
excess
•
Here, we show diagram obtained
if we assume
K i/ >> K AF
so that for instance oxygen
interstitials would be placed
above holes as minority defects in
the left hand part.
•
The diagram is much the same,
except that now the anti-Frenkel
pair dominates the stoichiometric
oxide.
•
Note that n∝pO2-1/4 and p∝pO21/4
when the concentrations of point
defects are independent of pO2.
Concluding remarks
•
We have worked on the relations between ∆G, ∆S, ∆H, ∆G0, ∆S0, ∆H0
and K for reactions.
–
•
You have learnt to and must be able to write equilibrium coefficient
expressions for a given reaction
–
•
•
•
•
•
You may have obtained variable understanding…
Precise or with simplifications are both OK…
You have learnt to choose a simple electroneutrality and insert it into K
to obtain expressions for the dominating defects.
You have learnt typical ways to plot T- and pO2-dependencies; van ‘t
Hoff and Brouwer diagrams.
You have learnt to insert the resulting expression for a majority defect
into an equilibrium coefficient expression for a minority defect, in order
to find how the minority defect behaves.
You have learnt to plot the majority and minority defects in Brouwer
diagrams that represent schematically the behaviour of all defects in
regions of simplified limiting electroneutralities.
By this, you have learnt the main quantitative tools in defect chemistry.