KJM and KJM Defects and Reactions

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					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)
truls.norby@kjemi.uio.no       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

     2H 2O( g )  2H 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   This example, where the reactant is more stable
     backwards; H2O(g) is the most stable   than the products, is typical also for defect
     species. We are still interested to    formation reactions. It is important to realise that
     calculate the ratio of products over   even if a reaction is not favourable, it will still
     reactants at equilibrium coefficient   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

       2H 2O( g )  2H 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 2O]  [H 2O]2  pH 2O  aH 2O
                                2       2


•   Moreover, it is proportional to the fraction of reactants that
    have enough thermal energy to overcome the energy barrier.
    Thus
                        (GA  GR )
                           0    0
                                                    ΔGFA
                                                        0
     rF  r0 aH 2O exp
              2
                                     r0 aH 2O exp
                                          2

                           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:
                      (GA  GP )
                         0    0
                                                     ΔGBA
                                                         0
    rB  r a aO2 exp
            2
          0 H2                     r0 aH 2 aO2 exp
                                        2

                         RT                           RT
Equilibrium in a gas phase reaction

    2H 2O( g )  2H 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

          2H 2O( g )  2H 2 ( g )  O2 ( g )
•    We insert to get

                            ΔGFA
                                0
                                                     ΔGBA
                                                         0
    rF  rB  r0 aH 2O exp
                  2
                                   r0 aH 2 aO2 exp
                                        2

                             RT                       RT
•    Rearranging with respect to the equilibrium coefficient:


                               ( ΔGBA  ΔGFA )        ΔGFR
       2                            0      0               0
     a H 2O
     2
               K FR     exp                    exp
    a aO2
     H2                              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

      2H 2O( g )  2H 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

      2H 2O( g )  2H 2 ( g )  O2 ( g )

                               ΔGFR
        2                          0
      a H 2O
      2
                K FR    exp
     a aO2
      H2                        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
        aC aD             ΔG      0

          a b
                K  exp
        a A aB            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

               G 0           S 0          H 0
    K  exp             exp          exp
               RT               R            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  vM
                //




•    We then write its equilibrium coefficient:

                                             
                                           [vO ] [vM ]
                                                    //
     K S  av av //  X v X v //     
               O    M          O     M
                                            [O] [M]

    Activities a        For point defects, activities     The site fraction is the
                        are expressed in terms of         concentration of defects over
                        site fractions X                  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 ]
                                                         //


     [ v M ]2  K S
         //

                                            Voila! That‟s it. We have now found the
     [v ]  K
        //
        M
                   1/ 2
                   S                        expression for the concentration of the
                                            defects. In this case, they are only a
       
     [vO ]  [vM ]  K S / 2
                //      1
                                            function of KS.
Schottky defects in MO



•   From the general temperature dependency of K,

              ΔSS0
                        ΔH S
                            0
    K S  exp      exp
               R         RT
•   we obtain
                                                             The square root and number
                                      S S
                                           0
                                                  H S0
     [v ]  [ v ]  K
       O
                 //
                 M
                           1/ 2
                           S       exp      exp             2 arise from the reaction
                                        2R        2 RT       containing 2 defects.

•   ln or log defect concentrations vs 1/T (van „t Hoff plots):

                      S S  H S 1
                           0      0
     ln[ v ]  ln[v ] 
            O
                      //
                      M      
                        2R     2R T
                                                             ln10=2.303
                                     S 0
                                               H 1 0
           
     log[ vO ]  log[vM ] 
                       //               S
                                                    S
                                   2 R ln 10 2 R ln 10 T
Schottky defects in MO



•   van „t Hoff plot

•   Standard entropy and
    enthalpy changes can         ΔSS0/2R
    be found from intercept                     -ΔHS0/2R
    with y axis and slope,
    respectively, after                ln [ ]       [vO..]=[vM//]
    multiplication with 2R
    and -2R.

•   log [ ] plots can be more
    intelligible, but require
    the additional                                  1/T
    multiplications with ln 10
    = 2.303.
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

                         [e / ] [h ]        n p
      K i  ae / ah                   
                         NC NV              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

             n p          Gi0       Si0      H i0
       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:

                  
                                         Eg
       K  [e ][h ]  n p  NC NV exp
         i
          /   /

                                        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            n p          Gi0       Si0      H i0
              Ki          exp         exp      exp
                   N C NV        RT           R         RT

                                                     Eg
    and       Ki/  [e / ][h ]  n p  NC NV exp
                                                    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 ΔG0/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

                       3/ 2
           8m kT 
                 *
     NC  
           h
                 e
                 2
                   
                   
                  

                        3/ 2
           8m kT 
                  *
     NV  
           h
                  h
                  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

                                               Eg
        n p  n 2  K i/  N C NV exp
                                              RT

                    / 1/ 2
                                                     Eg
        n  pK    i          ( NC NV )
                                       1/ 2
                                              exp
                                                    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
              
        OO  vO  2e /  1 O2 ( g )
         x                                                                            may seem unnecessary,
                          2                                                           but is meant to help you
                                                                                      understand…

                                                          1/ 2
                                                     pO2 
                                                     2
                                        [v ]  n 
                                         
                                                     0 
                                         O
                                             N 
                                             
                                         [O]  C    pO                                      1/ 2
                                                                                       pO2 
                                                                                  2
                        1
             av ae2/ aO/22( g )                   2
                                                                    
                                                                   [v    ] n 
                                                                                          
    K vO                                                               N 
                O                                                   O
                    aO x                          x
                                               [OO ]               [Ox
                                                                     O   ] C 
                                                                              
                                                                                       pO
                                                                                         0 
                                                                                       2
                        O

                                                [O]
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 C K vO  [vO ]n 2 pO/22
        /      2                   1



• We are going to express many equilibria the same simplified
  way, so it is important to understand how from the reaction
           
     OO  vO  2e /  1 O2 ( g )
      x
                       2

  to write the equilibrium coefficient expression
               
      K vO  [vO ]n 2 pO/22
        /               1
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 pO/22
      /              1


• We finally solve with respect to the concentration of defects:
                            
    [vO ]  ( 1 K vO )1/ 3 pO1/ 6
               4
                   /
                               2

                                 
     n  2[vO ]  (2 K vO )1/ 3 pO1/ 6
                        /
                                    2
Oxygen deficient oxides



• We split KvO into a preexponential and the enthalpy term:

                                                            H vO 1/ 6
                                                                0
                            
n  2[vO ]  (2 K vO )1/ 3 pO1/ 6
                   /
                                      (2 K vO,0 )1/ 3 exp
                                            /
                                                                  pO2
                               2
                                                            3RT

• From this, a plot of the logarithm of the defect concentrations vs
  1/T will give lines with slope of –ΔHvO0/3R.

• The number 3 relates to the formation of 3 defects in the defect
  reaction
Oxygen deficient oxides

                                
    n  2[vO ]  (2 K vO )1/ 3 pO1/ 6
                       /
                                   2


• By taking the logarithm:

                          
    log n  log 2  log[ vO ]  1 log( 2 K vO )  1 pO2
                                 3
                                            /
                                                   6


• 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?

    OO  vO  1 O2 ( g )
     x    x
              2
                                        KvOx

          
    vO  vO  e /
     x
                                        KvO1

         
    vO  vO  e /                      KvO2

           
     OO  vO  2e /  1 O2 ( g )
      x
                       2
                                        KvO = KvOx KvO1 KvO2

•   Total electroneutrality                          
                                       n  [vO ]  2[vO ]

•   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   OO  M M  M i  3e /  3 O2 ( g )
         x     x
                                   4


•   Equilibrium coefficient and its simplification:

    K Mi  [M i ]n3 pO/2 4 [OO ] -3 / 2 [M M ] -1  [M i ]n3 pO/2 4
      /                 3       x             x                     3



•   If these defects are dominating, the electronutrality and its insertion
    into equilibrium coefficient:
                                     
    3[ M i ]  n  (3K Mi )1 / 4 pO3 / 16
                          /
                                       2



•   Temperature dependency contains –ΔHMi0/4R. (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:
          
    OO  vO  2e /  1 O2 ( g )
     x
                      2
                                                                      
                                                             K vO  [vO ]n 2 pO/22
                                                               /               1



    2OO  M M  M i  2e /  O2 ( g )
      x     x
                                                             K Mi 2  [ M i ]n 2 pO2
                                                               /




•   Full electroneutrality:
                                          Instead of solving the full situation analytically or
           
    n  2[vO ]  2[M i ]               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

                                 
     n  2[vO ]  (2 K vO )1/ 3 pO1/ 6
                        /
                                    2
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:
                                                       
    K Mi 2  [M i ]n2 pO2  [M i ]( 2 K vO )2 / 3 pO1/ 3 pO2  [M i ]( 2 K vO )2 / 3 pO2/ 3
      /                                     /
                                                         2
                                                                                 /          2




•   We solve with respect to the metal interstitials:
                                        
     [M i ]  K Mi 2( 2 K vO )2 / 3 pO2 / 3
                  /         /
                                          2




•   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

    2OO  M M  M i  2e/  O2 ( g )
      x     x
                                                            K Mi 2  [ M i ]n 2 pO2
                                                              /




•   Inserting the electroneutrality gives

                                      
    n  2[ M i ]  (2 K Mi 2 )1/ 3 pO1/ 3
                          /
                                        2



•   Inserting this into the equilibrium coefficient of electrons
    and the now minority oxygen vacancies gives

       
     [vO ]  K vO (2 K Mi 2 ) 2 / 3 pO/26
                /       /              1
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:
                  
      OO  vix  vO  Oi//
       x                                                              
                                                      K AF  [Oi// ][vO ]

      0e h   /    
                                                      K i/  np

      OO  vO  2e /  1 O2 ( g )
       x    
                        2
                                                               
                                                      K vO  [vO ]n 2 pO/22
                                                        /               1


                                                                         
       1
       2   O2 ( g )  vO  Oi//  2h
                       x
                                                      K Oi  [Oi// ]p 2 pO1 / 2
                                                        /
                                                                           2



However, we now have 1+4=5 equations for the 4                     /     /
                                                                 K vO K Oi
defects, so all four equilibria are not going to be
needed, and are not independent. For instance:
                                                      K   2
                                                          AF   
                                                                    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// ]

                                 
    n  2[vO ]  ( 2 K vO )1/ 3 pO1/ 6
                        /
                                    2
    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.
                                   
      n  2[vO ]  ( 2 K vO )1/ 3 pO1/ 6
                          /
                                      2

•    We first consider the intrinsic
     electronic equilibrium, and see that
     holes and electrons vary inversely to
     each other:

                      K i/
       Ki/  np  p 
                      n
           K i/
        p       K i/ (2 K vO ) 1/ 3 pO/26
                            /           1

           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
                                  
     n  2[vO ]  ( 2 K vO )1/ 3 pO1/ 6
                         /
                                     2


     or we use the equilibrium for oxygen
     excess
                        
     K Oi  [Oi// ]p 2 pO1 / 2
       /
                          2


     and insert p from

      p  K i/ (2 K vO ) 1/ 3 pO/26
                    /           1




•    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// ]
                     / 1/ 2
       n pK       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

     Ki/  np
                                       
                       K AF  [Oi// ][vO ]
     enables quick establishment of some
     inverse slopes.
•    This diagram was drawn with the choice
     that
              Ki/  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
       Ki/  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 ΔG, ΔS, ΔH, ΔG0, ΔS0, ΔH0
    and K for reactions.
     –   You may have obtained variable understanding…
•   You have learnt to and must be able to write equilibrium coefficient
    expressions for a given reaction
     –   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.

				
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