KiaL = Cia/CiLOctanol-waterKiow = Cio/CiwSolid-waterFugacitys of liquidsPhase Transfer ProcessesExcess Free Energy, Excess Enthalpy and Excess EntropyEnvr 735, Chapter 3, - Intermolecular forces and partitioning- Free energies and equilibrium partitioning- chemical potential- fugacity- activity coef.- phase transfer- activity coef and fugacities- more on free energies and equilibrium constants1Much of this class deals with the partitioning of an organic compound i between two phasesA+B CKeq = [C]/{[A][B]}Keq = [ iphase1]/[ iphase 2]When we deal with air liquid partitioningKiaL = Cia/CiLOctanol-waterKiow = Cio/CiwSolid-waterKid = Cis/CiwWe will find that often for classes of compoundslog Kid = a log Kiow + bWhy???2For a compound to move between one phase and another, the intermolecular forces that hold a molecule in one phase need to be broken and others reformed in the other phaseSimply this can be represented as:1:i:1 + 2:2  1:1 + 2:i:2 (absorption)if the phase change is from molecule i in phase 1 to the interface or surface between 1 and 2, then1:i:1 + 1:2  1:1 + 1:i:2 (adsorption)What is the nature of the bonds that are being broken or formed??1. Nonspecific interactions (van der Walls interactions)3a. related to a compound’s polarizability () or the extent to whichan uneven electron distribution resultsin response to an imposed electronic field on timescales of10-15 sec; the intermolecular attraction energy is related to the product of the s of the interacting set of atoms…London dispersive energiesb. dipole-induced interactions (Debye energies) resulting from electron distribution differences in onemolecule (carbon and oxygen bond) inducing a charge distribution in the adjacent molecule. The strength of the interaction should be a function of the dipole moment, = qr ,in the “dipole” molecule, times the polarizability of the“charge induced molecule.c. dipole-dipole interactions: strength of attraction proportional to 1 x 242. Specific interactions: intermolecular attractions between electron rich and electron poor sites ofcorresponding molecules hydrogen bonding between the electron poor hydrogen of a carbon hydrogen bond and the unpaired oxygen electrons in an adjacent molecule…electron donor or acceptor interactionsIn the absence of electron donor or accepter interactions, London dispersive energies can be used to characterized the attractions of many molecules to their surroundings with respect to equilibrium partitioningConsider a molecule moving from the gas to a liquid phase, 1i (g) + 1:1(L)  1:i:1 (L)5when i dissolves in solvent 1, the dispersive attraction energy per interaction , disp g is given as (Israelachvile, 1992) as a function of polarizability ,, and the 1st ionization energies, I, of compounds i and solvent 1; I= Ii+ I1 /( Ii I1) disp g = -(3/2) I i 1 /(4 0)2Visible light has frequencies (and its changing electric fields) on the order of 10-15cycles /sec. A material’s ability to respond to light is related to its index of refraction, nDi, and nDi is related to that material’s polarizability via the Lorenz-Lorenz relationship6i /(4 0)= [n2Di -1]/ [n2Di +2]x(3Mi/4Na)Assuming spherical molecules, and the induced temporary dipoles distances are diameters of the molecules (see page 64 of text)d i s p p e rD i DDg Innxnni n t e r a c t i o nD i 3 2 5 61212221212/for a mole of interactions we need to consider the total surface area (TSA) of the solvated molecule and the contact area (CA) it has with solvent molecules d i s p AD i DDG N T S A C A Innxnnf o r o n e m o l eD i / /3 2 5 61212221212Since NA, CA, 3 and I are relatively constant,7If the equilibrium is dominated by dispersive forces, this free energy, disp G can be related to the equilibrium of this process by dispG = - RT ln Keq where Keq= [ iphase1]/[ iphase 2]For an organic gas in equilibrium with a pure liquid the equilibrium is:KiaL= Ciasat/CiL = Mi p*iL/[iLRT]d i s pD i DDG c o n s t x T S Annxnnf o r o n e m o l eD i .2212121212disp G = - RT ln Keq8we should be able to plotcalculated ln KiaL = M p*iL/[iLRT]vs.T S AnnxnnD i DD2212121212D ifor a pure solvent interacting with the gas phase, i=19Table 3.1 page 651011Figure 3.4page 69 new textPartitioning in air-pure solvent vs. index of refraction termPartitioning in Hexane?Partitioning in water?12Figure 3.6 page 71 air-hexane, top, air-water, bottom 13Chapter 3, then uses thermodynamics to quantify molecular energies and equilibrium partitioning Section 3.3 starts with:NinGjn,P,T)(ii1andiNi)n...n,nn,T,p(nGNi121How do we get to these equations and what do they mean??14Chapter 3 The First LawU2 - U1 = q - w workchange ininternal energy heatof an objectreservoirobjectb-U = q1-w1-U = q2-w2-a15For example one gram of H2O at 25oC is evaporated and condensed; the condensed gram of water at 25oC will have the same internal energy as it did previously.If only pV work is done and the pressure of the system is constant wrev =  pdVWhat is the work of a reversible expansion of a mole of an ideal gas at 0oC from 2.24 to 22.4 liters?pdVWVVrev21pV = nRTW nRT dV V nRTVVrevVV 1221/ ln ( )Wrev = 1mole x1.987 cal K-1 mole-1 x 273 K x 2.303 log (22.4/2.24)Wrev = 1.25 Kcal mole-116Internal energy, heat and workWhat is the energy required to vaporize water at 100oC???when one mole of water is vaporized at 100oC the work is w = p V = RT = 1.987 cal K-1 mole-1 x 373.15Kw= 741.4 cal mole-1The energy or heat required to vaporize water at 100oC requires energy to separate the liquid molecules;that is 529.7 cal g-1 q = 18.02 g mole-1 x 539 cal g-1 = 9725 cal mole-1 ;For a mole of water, the internal energy U = q - w U= 9725 cal mole-1 - 741 cal mole-1 U = 8984 cal mole-117Enthalpy-U = q - pV) at constant pressureq= (U2 + pV2) - (U1+ pV1)We define U + pV as the enthalpy, Hq = H2-H1 = Hor the heat adsorbed in a process at constantpressureThere are usually two types of calorimetric experiments usedto determine heat, one at const volume (no PV work, so U=q)and one at constant pressure.The heat of combustion of CO in a constant vol. calorimeter is –67.37 kcal mol-1. Calculate the enthalpy of combustion at const. pressure.CO(g) + ½ O2  CO2(g)Work done is (n2-n1)RT,