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Solutions and Mixtures

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 .

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Solutions and Mixtures

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

  2. I. Entropy of Mixing • Translational Entropy of Mixing • Assume N lattice sites filled completely with NA A molecules and NB B molecules so that N = NA + NB. • Then W = N!/( NA! NB!) See Ch 6 • Smix = k ℓn W = - k(NAℓn xA + NBℓn xB) = -Nk (xAℓn xA + xBℓn xB) Eqn 15.2, 3 • Ex 15.1

  3. II. Energy of Mixing (1) • Assume ideal soln, then, Umix= 0 and Fmix= -TSmix • If the soln is not ideal, then Umix = sum of contact interactions of noncovalent bonds of nearest neighbor pairs  0. • U = mAA wAA + mAB wAB + mBB wBB where mIJ = # I-J bonds and wIJ = I-J contact energies. All w terms < 0

  4. Energy of Mixing (2) • Define mAA and mAB = f(mAB, NA, NB) • Each lattice site has z sides, so z NA = total number of contacts for all A • Then zNA = 2mAA + mAB. And zNB = 2mBB + mAB . Solve for mAA and mBB • Use to find U = (zwAA)NA/2 + (zwBB)NB/2 + ([wAB- ½ (wAA + wBB)] mAB) Eqn 15.8

  5. Energy of Mixing (3) • To simplify Eqn 15.8, use the Bragg-Williams or mean-field approximation to find average <mAB>. 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 mAB values.

  6. 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 pA = x = xA and (1-x) = pB. • Then mAB = z Nx(1-x). Plug into Eqn 15.8 for U to get final eqn for U = Eqn 15.10. • U = (zwAA)NA/2 + (zwBB)NB/2 + kT ABNANB/N

  7. Energy of Mixing (5)Exchange Parameter • U = (zwAA)NA/2 + (zwBB)NB/2 + kT ABNANB/N • AB = energy cost of replacing A in pure A with B; similarly for B. • AB = exchange parameter = - ℓn Kexch • Umix = RT ABxA xB

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

  9. III. Free Energy of Mixing (1) • F = U – TS = [Eqn 15.11] – T [Eqn 15.2] = Eqn 15.12 • Pure A + pure B  mixed A + B has Fmix = F(NA + NB) – F(NA, 0) – F(0,NB) • Note that F(NA, 0) = ½ x wAANA • Fmix= [x ln x + (1-x) ln(1-x) + AB x(1-x)]NkT Eqn 15.14 • This eqn describes a regular solution.

  10. Free Energy of Mixing (2) • If Fmix > 0, minimal mixing to form a soln. • If Fmix < 0, then a soln forms • If soln separates into 2 phases, Eqn 15.14 does not apply. • Ex 15.2

  11. IV. Chemical Potentials and Mixing • A = (F/NA)NB,T = kT ℓn xA + zwAA/2 + kTAB (1-xA)2 = kT ℓn xA + corections due to AA interactions and exchange parameter. Eqn 15.15 • Also  = 0 + kT ℓn x where  = activity coefficient. x = effective mol fraction.

  12. 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.

  13. Surface Area (2) • Assume (Fig 15.7) • A and B are the same size • NA = # A molecules and NB = 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

  14. Surface Area (3) • U = Σ ni wij = 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 15.19 = 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 = [wAB – ½ (wAA + wBB)]/a

  15. Surface Area (4) • Then AB = (F/A)NB,NA,T = (F/n) (n/A) = [wAB – ½ (wAA + wBB)]/a • AB = (kT/za) AB Eqn 15.22; see Eqn 15.11 • If there are no B molecules Eqn 15.22 reduces to Eqn 14.28 AB = - wAA /2a • Ex 15.3 (mixing is not favorable, see p. 273)

  16. 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 Fmix = F(NA + NB) – F(NA, 0) – F(0,NB) = NkT[x ln x + (1-x) ln (1-x) + AB x(1-x)]

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

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