Solutions and Mixtures Chapter 15 # Components > 1 Lattice Model  Thermody. Properties of Mixing (S,U,F,  )

I. Entropy of Mixing Translational Entropy of Mixing Assume N lattice sites filled completely with N A A molecules and N B B molecules so that N = N A + N B. Then W = N!/( N A ! N B !) See Ch 6  S mix = k ℓn W = - k(N A ℓn x A + N B ℓn x B ) = -Nk (x A ℓn x A + x B ℓn x B ) Eqn 15.2, 3 Ex 15.1

II. Energy of Mixing (1) Assume ideal soln, then,  U mix = 0 and  F mix = -T  S mix If the soln is not ideal, then  U mix = sum of contact interactions of noncovalent bonds of nearest neighbor pairs  0. U = m AA w AA + m AB w AB + m BB w BB where m IJ = # I-J bonds and w IJ = I-J contact energies. All w terms < 0

Energy of Mixing (2) Define m AA and m AB = f(m AB, N A, N B ) Each lattice site has z sides, so z N A = total number of contacts for all A Then zN A = 2m AA + m AB. And zN B = 2m BB + m AB. Solve for m AA and m BB Use to find U = (zw AA )N A /2 + (zw BB )N B /2 + ([w AB - ½ (w AA + w BB )] m AB ) Eqn 15.8

Energy of Mixing (3) To simplify Eqn 15.8, use the Bragg- Williams or mean-field approximation to find average. Note that this is a simplification and we assume that this average is a good approx to the actual situation. (i.e. one distribution dominates vs using a distribution of m AB values.

Energy of Mixing (4) Mean-Field Approximation Assume A and B are mixed randomly. Then the probability of finding a B next to an A ≈ z x(1-x) where p A = x = x A and (1- x) = p B. Then m AB = z Nx(1-x). Plug into Eqn 15.8 for U to get final eqn for U = Eqn U = (zw AA )N A /2 + (zw BB )N B /2 + kT  AB N A N B /N

Energy of Mixing (5) Exchange Parameter U = (zw AA )N A /2 + (zw BB )N B /2 + kT  AB N A N B /N  AB = energy cost of replacing A in pure A with B; similarly for B.  AB = exchange parameter = - ℓn K exch  U mix = RT  AB x A x B

Energy of Mixing (6) Exchange Parameter  AB = exchange parameter = - ℓn K exch A + B ↔ mixing  eq. constant K exch  U mix = RT  AB x A x B  AB can be > 0 (AB interactions weaker than AA and BB); little mixing and  U mix more positive, K exch smaller  AB can be < 0 (AB stronger than AA and BB), …

III. Free Energy of Mixing (1) F = U – TS = [Eqn 15.11] – T [Eqn 15.2] = Eqn Pure A + pure B  mixed A + B has  F mix = F(N A + N B ) – F(N A, 0) – F(0,N B ) Note that F(N A, 0) = ½ x w AA N A  F mix = [x ln x + (1-x) ln(1-x) +  AB x(1-x)]NkTEqn This eqn describes a regular solution.

Free Energy of Mixing (2) If  F mix > 0, minimal mixing to form a soln. If  F mix < 0, then a soln forms If soln separates into 2 phases, Eqn does not apply. Ex 15.2

IV.Chemical Potentials and Mixing  A = (  F/  N A ) NB,T = kT ℓn x A + zw AA /2 + kT  AB (1-x A ) 2 = kT ℓn x A + corections due to AA interactions and exchange parameter. Eqn Also  =  0 + kT ℓn  x where  = activity coefficient.  x = effective mol fraction.

V. Free Energy of Creating Surface Area Consider interface or boundary between 2 condensed phases A and B.  AB = interfacial tension = free energy cost of increasing the interfacial area between A and B. Calculate  AB using the lattice model.

Surface Area (2) Assume (Fig 15.7) –A and B are the same size –N A = # A molecules and N B = B molecules –interface consists of n A and n B molecules in contact with each other –bulk molecules have z A nearest neighbors –surface A molecules have (z-1) A nearest neighbors

Surface Area (3) U = Σ n i w ij = term for A in bulk + term for A at surface + term for AB interactions + term for B in bulk + term for B at surface Then U = Eqn = F since S = 0 Let A = total area of interface = na Let a = area per molecule exposed to surface Then  AB = (  F/  A) NB,NA,T = (  F/  n) (  n/  A)   AB = [w AB – ½ (w AA + w BB )]/a

Surface Area (4) Then  AB = (  F/  A) NB,NA,T = (  F/  n) (  n/  A) = [w AB – ½ (w AA + w BB )]/a  AB = (kT/za)  AB Eqn 15.22; see Eqn If there are no B molecules Eqn reduces to Eqn  AB = - w AA /2a Ex 15.3 (mixing is not favorable, see p. 273)

Surface Area (5) Assumptions –Mean field approximation for distribution –Only translational contributions to S, U, F and μ are included. –What about rot, vib, electronic? We assume that in mixing, only translational (location) and intermolecular interactions change. –Then  F mix = F(N A + N B ) – F(N A, 0) – F(0,N B ) = NkT[x ln x + (1-x) ln (1-x) +  AB x(1-x)]

Surface Area (6) However, if chemical rxns occur, rot, vib and elec must be included.