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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 TS 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 8m kT * NC h e 2 3/ 2 8m 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 pK 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 ] 0e 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 pK 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 npO2-1/4 and ppO21/4 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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