Ideal (solid) solutions M total sites Entopy (=entropy of mixing) Two component system ΔS = −kB (xA ln xA + xB ln xB ) M xA + xB =1 € Multi component system ΔS A € = = −kB ∑ x j ln x j M j = B constraint ∑ x j =1 € j € € € Ideal (solid) solutions M total sites Entopy (=entropy of mixing) Two component system ΔS = −kB (xA ln xA + xB ln xB ) M xA + xB =1 € Multi component system ΔS A € = = −kB ∑ x j ln x j M j = B constraint ∑ x j =1 € j € € € Free energy of mixing for ideal solutions ΔG mix = RT(xA ln xA + xB ln xB ) €Δ G mix Chemical potentials RT 0 ∂ΔG mix µB − µB = ΔG mix (xB ) + xB ∂xB = RT ln xB € 0 µA − µA = RT ln xA = RT ln(1− x ) € B Ideal solution behaves Roaultian! xB € € Regular solution theory ΔG mix = ΔHmix − TΔSmix Bragg‐Williams approximation (or mean‐field approximation) € Washed‐out mixture Enthalpy of mixing 0 0 ΔHmix ≈ ΔE mix = E mix − xA E A − xB E B = wxA xB 2 2 E mix = wAA xA + wBB xB + 2wAB xA xB € € Regular solution theory ΔG mix = wxA xB + RT(xA ln xA + xB ln xB ) 1 w = Z w − (w + w ) AB 2 AA BB € ΔHmix € ΔG mix RT € −TΔSmix € xB € € Regular solution theory ΔG mix = wxA xB + RT(xA ln xA + xB ln xB ) Chemical potentials 0 2 2 µA − µA = w(1− xA ) + RT ln xA = w(xB ) + RT ln(1− xB ) € 0 2 µB − µB = w(1− xB ) + RT ln xB € Activity € 0 w w µB − µB = RT lnaB (1−x )2 +ln x (1−x )2 RT B B RT B w aB = e = xBe lna = (1− x )2 + ln x B RT B B € Fundamental flaw in Regular Solution Theory: The approximation in R. S. T. is that the probability of observing different configurations is the same. This is not true in reality! € € Entropy induced order (not disorder) Consider a system composed of long rod‐like particles and “solvent” (small particles). What is the entropy if mixing in this case? Assume that you have NR rods and Nm solvent molecules, so that the total volume is Vtot = NR VR + Nm Vm . Solvent Rod Hints in the next page Hints: you should consider two systems. One in which your “lattice” boxes are big enough so that the rods can freely rotate, and calculate the total entropy of the system. The boxes should look like the area in the upper left corner of the sketch. Remember that you want to include the rotational part also. The second “lattice” you have to consider is a lattice where you impose the direction of the rods, such as that shown in the lower right corner, and calculate the entropy of this system. What you will find is two different forms for the Free Energy of the system, and above a certain rod length and density of rods the order phase will be preferred. Hints: you should consider two systems. One in which your “lattice” boxes are big enough so that the rods can freely rotate, and calculate the total entropy of the system. The boxes should look like the area in the upper left corner of the sketch. Remember that you want to include the rotational part also. The second “lattice” you have to consider is a lattice where you impose the direction of the rods, such as that shown in the lower right corner, and calculate the entropy of this system. What you will find is two different forms for the Free Energy of the system, and above a certain rod length and density of rods the order phase will be preferred. This is an example that simple intelligent reasoning and math is much better than complicated proofs. It almost always offers a much simpler and clearer picture of the fundamentals. .