Solutions and Mixtures
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Transcript of 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
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(NA ℓn xA + NB ℓn xB) = -Nk (xA ℓn xA + xB ℓn xB) Eqn 15.2, 3
• Ex 15.1
II. Energy of Mixing (1)
• Assume ideal soln, then, Umix= 0 and Fmix= -TSmix
• 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
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
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.
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 AB NANB/N
Energy of Mixing (5)Exchange Parameter
• U = (zwAA)NA/2 + (zwBB)NB/2 + kT AB NANB/N
AB = energy cost of replacing A in pure A with B; similarly for B.
AB = exchange parameter = - ℓn Kexch
Umix = RT AB xA xB
Energy of Mixing (6)Exchange Parameter
AB = exchange parameter = - ℓn Kexch
• A + B ↔ mixing eq. constant Kexch
Umix = RT AB xA 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), …
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.
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
IV. Chemical Potentials and Mixing
A = (F/NA)NB,T = kT ℓn xA + zwAA/2 + kTAB (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.
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
– 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
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
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)
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)]
Surface Area (6)
• However, if chemical rxns occur, rot, vib and elec must be included.