4. CHEMICAL EQUILIBRIUM
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1 4. CHEMICL EQUILIBRIUM Must deal with systems containing more than one species (i.e., a mixture). How do concentrations of species adjust at equilibrium? Non reacting mixture : Ni, Fe, Cr (steal) O 2, N 2 (in air) reacting mixture : O 2 (g) + 2H 2 2H 2 O ( as phase equilibrium ) X (g) Surf. X = impurity it in solid X (dis) X(g) X(dissolved) X X 1
2 Distribution between phase Defect equilibrium Surf. null Vacancy + surface atom Phase equilibrium UO2 (s) UO2 (g) (Sublimation) Heterogeneous equilibrium Zr(s) + O2(g) ZrO2(s) 2
3 Criterion for Equilibrium du Single component : U( S, V, N ) = N u( s, v) U S V, N ds U V S, N dv TdS Multiconponent : U( S, V, N1, N2, ) PdV U U du du ds dv dni S V i dni du Or TdS Similarly V, N S, N S, V, Ni PdV i idn i df d SdT SdT PdV VdP i i idn i idn i 3
4 chemical l potential i U Ni S, V, Ni F Ni T, V, Ni Ni T, P, Ni For an isolated system(u and V constant), the condition of equilibrium ds 0 i dni 0 U, V i Valid whatever restraints [ e.g. (S,V), (T,V), (T,P) ] Using d equation d T, P 0 Equivalent statements of chemical equilibrium 4
5 Equilibrium Vacancy Concentration (Seat.6.3) Thermodynamic Process P P Extra volume = Ω =Volume per atom Surface Form a vacancy Consider defected crystal as a mixture of N atoms and Nv vacancies On Ns = N + Nv sides. 5
6 (Nv) = ibbs free energy of solid with N atoms (constant) and Nv vacant lattice sites. ( NV ) (0 ) NV gv TS mix Perfect lattice of N atoms Due to making one vacancy Ways of arranging Nv vacancies on N+Nv sites or N atoms on Ns sites gv =hv TSv Hv = εv +pω εv = energy of formation of vacancy P = external pressure(or normal stress component) on crystal Sv = excess entropy of vacancy formation (due to alternation of x vibrational modes around vacancy fromn frequency ) to 6
7 Entropy of Mixing Zmix = Number ways of arranging N objects in Ns cells start filling empty sites 1 st atom in any of Ns sites 2 nd atom in any of Ns -1 sites N th atom in any of Ns -(N-1) sites ll configurations of same energy, at Ei=0 in Z defined on no. of of ways doing this Ns( Ns1) [ Ns( N 1)] Ns! ( Ns N)! 7
8 But, this counts too many for Ns=6, N=3 C B C B B C B C C B B C The six are really only one : Divide by number of permutations among N objects = N! mix N S! N! ( N N)! Entropy of system S dln Zmix S kln Z kt kln Z dt mix mix mix V Entropy of mixing S mix 8
9 ln Z ln N! ln N! ln( N N)! mix S S N ln N N N ln N N ( N N )ln( N N ) N N S S S S S S NS N N ( NS N)ln Nln NS NS N V N V N ln Nln NS NS S mix N V N NV ln Nln k NS NS 9
10 Terms in Nvgv eq Nv 0 Nv (0) -TSmix 10
11 Criterion of Equilibrium d dn V T, P 0 d ds mix N N V gv kt gv ktln 0 dnv dnv NV N V N N V eq g v / kt e XV site fraction vacancies ( ) p sv / k v/ kt kt X e e e V From microscopic model, sv kln 0 For s P E ev X e copper at K v / 4 0, 0, 1, RT V V V 10 ( 1300 ) (~100kJ/mole) oe) For P 0, X X ( p 0) V V 11
12 Interstitial P P Volume change = - Ω Surface Form a interstitial Following Similar derivation X si i Ni N N i e e s / R / RT kln 0 i i e s RT Vibration around interstitial εi = Energy of formation of interstitial(~3ev) 12
13 For si = 0,,p = 0, εi = 3eV ( 300KJ/mole ) Xi Very much smaller that Xv For p > 0, Xi > Xi(p=0) Pressure aids formation of interstitials. Opposes formation of vacancies Pressure (stress) effect 8 N m m P => σ : as high as 700Mpa(fracture) 1.4 kt 23 N m Ω = m 3 /atom K K k = J/K-atom e T = 1500K 13
14 Interstitial Schottkyt defects : Maintain local electrical neutrality a) b) Surface c) nion vacancy Cation vacancy + - Cation vacancy 14
15 Equilibrium : Intrinsic Schottky Defects M = cation (+) X = anion (-) Nsm = number of cation lattice sites Nsx = number of anion lattice sites For MX type (NaCl), Nsm = Nsx Nvm = number of cation vacancies Nvx = number of anion vacancies Electrical Neutrality Nsm Nvm = Nsx Nvx Nvm = Nvx = Nv No. of cations of +1 charge No. of anions of -1 charge Vacancy pairs 15
16 (Nvx, Nvm) = (0,0) + Nvm єvm +Nvx єvx kt ln Z mix Z mix = Zm x Zx Have neglected p effect and set Svx=Svn=0Svn Pairs : (Nv) = (0) + Nv(εvm + εvx ) - N s! 2kT ln[ ] (N N )!N! s v v, εs energy of formation of Schottky defect 16
17 d dn v 0 x vx x vy x v e - /2RT s eneral Case dopant (impurity or host ion of different valence) forced Vacancy (EXTRINSIC)
18 q = ion charge filing of site cation : Nsm = NDm + Nm + Nvm dopant ions host inos vacant sites Nsx = Nx + Nvx anion Electrical neutrality qdndm + qn Nm = qx Nx q divide by Nsm = Nsx, use qx = qm and site filling Eqs: D ( 1) X Dm X vx X vm q m fraction impurity it on cation lattice. 18
19 Zm Zx Z ( N vm N m N D m )! (N vx N x )! N! N! N! N! N! vm m Dm vx x as before No.ofwaysofarranging3typesofobjectsinNsm site (Nvx,Nvn) = (0,0) + Nvnεvn +Nvxεvx - RT ln Z Since Nvn and N Nvx (or equivalently, Xvm and X Xvx) ) are related by electrical neutrality condition, there is really only one independent variable. d dn vm dn vm dln Z m d ln Z x 0 ( ) vm vx - kt [ + ] dn dn dn dn dn vx vx vx vm vx εs ln x vm ln x vx x vm x vx e / kt s Solve simultaneously with electrical neutrality 19
20 x vm x vx x x im ix e e pm / RT pm px / RT cation Frenkel anion Frenkel Use with electrical neutrality condition. 20
21 pplications : of chemical Potential criterion i dn i i Pure solid or liquid : 0 Molar ibbs free energy at tamp T ctivity Coef. Solid or liquid solution : RT ln( c ) Mole fraction or molarity Ideal gas, Pure or in mixture : RT ln P Pressure or partial pressure in atmosphere. 21
22 Phase Equilibrium Phase Equilibrium 0 dn dn (g) N N N BUT (s) or, dn dn Pure Solid : (g) (s) RTlnP RT / - (vap) e P (s) (g) (vap) e P (s) (g) (vap) BUT (vap) (vap) (vap) S T H RT H R / / S (vap) (vap) e e P (Clausius Clapyron) 22
23 Solid (or liquid) solution : (s) RTlnC (g) RTlnP P C e - (vap) / RT P 1, C mole fraction P C P 23
24 Homogeneous Chemical Reaction a(g) + bb(g) cc(g) dn B dn B C dn C 0 BUT, by stoichiometry of reaction : dn B b dn a and dn C c a dn a b B c C Equilibrium constant RT ln P P c C a b PB H TS (c C a b Standard state free energy of reaction B ) S H R RT K P e e 24
25 Example 1 Co ( g ) O 2 ( g 2 ) CO 2 ( g ) S, J / mole K H, RJ / mole PCO K P exp( ) exp( 3 P P R R( T /10 CO O 2 ) ) R 8.319J/mol - K gas constant S negative because fewer molecules of gas on right than on left. Hence, fewer degrees of freedom, does not favor formation of CO2 H negative favor reaction to produce CO2 T P CO K, P P CO 1, K P e e atm P O 2 atm 25
26 Heterogeneous Reactions Zr(s) + O2(g) ZrO2(s) (Metal and Oxide not mixable ) ZrO2 ( S ) O2 ( g ) Zr ( S ) ZrO 2 O 2 RTlnPO 2 Zr RTlnPO 2 ( ZrO 2 O2 Zr ) (very Negative) 26
27 27
28 ibbs phase rule : F = degree of freedon ( variable that can be independently changed) C=no.ofcomponents P=no.ofphases Example F = C P H2O(g) C=1 (H2O) P=2 (g+l) F=1, Vary T, p=p follow H2O(L) s g L C=1 P=3 F=0, ll fixed T(Triple point) 28
29 ZrO2 O2 C=2 (Zr, 0) P=2 (Two solids) Zr(s) p>>po2 p p,no gas phase p O f ( T, p) 2 Very slight effect Phase Diagrams H2O UF6 p (atm) 218 p (atm) 1 Ice(s) B.P. Water(L) 1.5 T.P. 1 Steam(g) B.P. (S) T.P. (L) (g)
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