3. Be able to derive the chemical equilibrium constants from statistical mechanics.

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1 Lecture #17 1 Lecture 17 Objectves: 1. Notaton of chemcal reactons 2. General equlbrum 3. Be able to derve the chemcal equlbrum constants from statstcal mechancs. 4. Identfy how nondeal behavor can be accounted for n chemcal reactons. 1. Notaton of chemcal reactons: Consder the chemcal reacton Cl 2 +H 2 2HCl The numbers n front of the chemcal formulae are the stochometrc coeffcents, whch are always postve. We can use the coeffcents to defne the stochometrc numbers, ν. By conventon we take the sgn to be postve for products (terms on the rght, negatve for reactants (terms on the left, and zero for nert speces. Hence, ν Cl2 = ν H2 = 1 and ν HCl = 2. Thus, we can wrte a sngle chemcal reacton as ν M = 0 where M s the chemcal formula for component. What are the stochometrc numbers for 3A+2B 4C? We can wrte a relatonshp for the number of moles of a component present at a gven tme, and the ntal number of moles of that component present before the reacton takes place. The relatonshp s cast n terms of the extent of reacton, ξ. n = n 0 +ν ξ 2. General Equlbrum Formulaton: At constant pressure and temperature the crteron for general equlbrum s a mnmum n the Gbbs free energy: π π mn G = n k µ k n k,x k = n k x k µ k k=1 k=1 where n k s the number of moles n phase k, x k s the mole fracton of component n phase k, and µ k s the chemcal potental of component n phase k. Note that the only varables we have avalable to manpulate are the total number of moles and mole fractons for each phase. There are also some constrants on the above equaton whch must be taken nto account. The constrants are:

2 Lecture #17 2 (a Materal balance n = π n k = k=1 (b Non-negatvty n k 0 π π R n k 0 + ν l ξ l k=1 k=1 l=1 (k = 1,...,π ( = 1,...,C 1 x k 0 ( = 1,...,C; k = 1,...,π If we hold the temperature and the volume constant nstead then the equlbrum condton s a mnmum n the Helmholtz free energy: mn n k,x k A = mn{ PV + µ k n k } n k,x k Consder a sngle phase, sngle chemcal reacton. The Gbbs free energy s gven by G = n µ = n 0 µ + ν ξµ Then note that dn = ν dξ We want to fnd G = 0 ( = 1,C n but, T,P dg = SdT +VdP + µ dn = SdT +VdP + µ ν dξ At constant T,P we have dg = µ ν dξ So to fnd the mnmum we need to fnd G = 0 ξ T,P Hence, the condton for chemcal equlbrum s ν µ = 0. Ths s true for any chemcal reacton at equlbrum.

3 Lecture # Reactons n deal gases The standard free energy equlbrum constant n classcal thermodynamcs s RT lnk eq = ν µ = G rxn where superscrpt refers to the chemcals n ther standard states. What s meant by standard states? t s generally taken to refer to unt pressure (.e., 1 bar or unt fugacty at the temperature of the reacton. G rxn s the change n free energy of reacton per mole of the molecules n ther standard states, G rxn = ν G = ν µ, where we have used the fact that for a pure flud G = µ. Recall that n classcal thermodynamcs we defne K eq n terms of fugactes, ( ˆf ν K eq = f where f s the standard state fugacty. For an deal gas the equlbrum constant gves a drect measure of the equlbrum converson through the mole fractons. K eq = y p ν P The goal s to compute K eq from statstcal mechancs. To do ths we need an expresson for µ from statstcal mechancs. Recall that lnq µ = kt N V,T Invokng the sem-classcal partton functon, where Q = ZqN nt Λ 3N N! q nt = q e q v q r Assume that at the standard state the molecules are deal gases. Ths s usually a good approxmaton at hgh temperatures and pressures of 1 bar. Then Z = V N and we can wrte ( V µ q nt, = kt ln, NΛ 3 where V s the volume n the standard state. Recognzng that V /N = kt/p we have G = N A µ ktqnt, = RT ln( P Λ 3,

4 Lecture #17 4 where N A s Avogadro s number and we have recognzed that the molar Gbbs free energy s just N A tmes the molecular (not molar chemcal potental of the pure flud. Thus, K eq = ν qnt, kt = P Λ 3 ( kt P ( ν qnt, Λ 3 ν where ν = ν and P = 1 bar. 4. Example: Assocaton n sodum vapor. Consder the gas phase assocaton reacton 2Na Na 2 The equlbrum constant can be wrtten as K(T = P Na 2 P P 2 Na = P (q nt,na2 /Λ 3 Na 2 kt (q nt,na /Λ 3 Na 2 The electronc ground state of Na s doubly degenerate (ω = 2. q nt,na2 2πmNa2 kt 3/2 T exp(d0 /kt Λ 3 = Na 2 h 2 2Θ r 1 exp( Θ v /T q nt,na Λ 3 Na 2πmNa kt 3/2 = q elec h 2 Pluggng n the numbers and usng T = 1000 K we get K(T = 1000 = The expermental value s Nondeal mxture behavor n flud phase reactons. (a Recall that at equlbrum ν µ = 0. Ths expresson s vald for deal and nondeal solutons. For any real (or deal flud we can wrte the chemcal potental as ˆf µ = µ +RT ln = µ +RT lna, f where µ s a functon of temperature only and a = x γ and γ s the actvty coeffcent. Let µ on a per mole bass be gven by µ = RT ln ( ktqnt, P Λ 3

5 Lecture #17 5 Then the equlbrum equaton becomes ν (µ +RT lna = ν µ + ν RT lna = 0 ktqnt, ν µ = RT ν ln P Λ 3 +RT ν lna = 0 ν ktqnt, ln P Λ 3 = ln(a ν [ C ν ] ( C ktqnt, ln P Λ 3 = ln a ν K(T = C a ν Now you can calculate K(T from statstcal mechancs as before, but use some lqud phase actvty coeffcent model for a to correct for nondeal behavor to fnd the composton. 6. Gas phase nondealtes. For reactons n nondeal gases we use the fugacty to relate the equlbrum constant to the compostons: ( ˆf ν K = f If we assume unt fugacty as the reference state ths becomes K = (ˆf ν = K f Introducng the fugacty coeffcents we have K = ν (ˆφ y P = KˆφK y P ν where Kˆφ = ˆφ ν 7. Condensed phases: Substtute a = γ x to get K eq = (γ x ν = γ ν x ν = K γ K x

6 Lecture #17 6 If you know K eq and have an expresson for γ you can use ths equaton to solve for the equlbrum composton n a nondeal lqud phase reacton. The problem s nonlnear n the mole fractons and must be solved numercally. If on the other hand you know the equlbrum constant and can measure several equlbrum compostons you can ft the parameters for an actvty coeffcent model. Walas shows how to do ths.