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Review Chapter 2 know general structures Chapter 3 Much of this class deals with the partitioning of an organic compound i between two phases A+B C Keq = [C]/{[A][B]} Keq = [ iphase1]/[ iphase 2] When we deal with air liquid partitioning KiaL = Cia/CiL Octanol-water Kiow = Cio/Ciw Solid-water Kid = Cis/Ciw We will find that often for classes of compounds log Kid = a log Kiow + b Why??? disp Gfor one mole n2Di 1 n2D1 1 N A TSA / CA3I / 256 2 x 2 nDi 2 nD1 2 The concept of free energy comes from the need to simultaneously deal with the enthalpy energy and entropy of a system at constant temp and pressure G= H -TS Equilibrium Constants From free energies of reactants and products o G RT ln (PC )c (PD )d (PA )a (PB )b Equilibrium Constants and Temperature (G/T)P= -S d(ln K eq ) dT and (G/T)P= -S Ho RT 2 log K eq Ho (1 / T) const 2.303R We than defined chemical potentials N i ( nGi )T , P , n i 1 j G i /n i = i From the Gibbs Duhem equation we ultimately showed 2i = 1i +RT ln f2i/f1i fi pure liquid = i pure liquid piL* Phase Transfer Processes Consider a compound, i ,which is dissolved in two liquids which are immiscible like water and hexane. RT ln fi hx /fiL*(pure liquid) = RT lnfi H2O /fiL*(pure liquid) fi hx = fi H2O We then went on to define an excess free energy of solution GiE1 RT ln i 1 and GEi1 = RTln i1= HEi1-TSEi1 HEi1 is the particle molar excess enthalpy of solution and SEi1 is the partial excess entropy Chapter 4 We derived ln pio Hi 1 const R T then introduced concept of sub-cooled liquid vapor pressures At the boiling point is vapSTb constant? H const slope = S T Discussed estimation techniques for S and Tb to be used in vapor pressures calculation techniques ln p *iL SvapTb R * ln piL 19(1 * ln piL ln PiS* [ 1 .8 ( 1 Tb T ) 0.8(ln b )] T T Tb ) 8.5(ln Tb )] T T ( S fus ) (Tm Tamb ) R Tamb WE then looked at LC and GC techniques to measure solid and liquid vapor pressures. Chapter 6 _ pi / Ci = i p*i V l = KiHl Wash out ratios or W Usually defined as the conc. in rain/conc. In air W = Ciw/Cia = 1/Kiaw and calculated how fast rain could remove gas phase organic contaminants from the atmosphere based on their air water partitioning or KiH we then introduced the concept of saturated Henry’s law values sat K iH p sat i sat * pl Vw sat w C iw and then asked how different iw were ?? and iwsat also KsatiH could be calculated from p*i and Csatiw values from the appendix we looked at the influence of salts, where the Henry’s law for salt water is K iH ,w , salt K i s [ salt ] sat K iH 10 ============================== Example Problem: Consider a well sealed flask with 100 ml of H2O and 900 ml air. At equilibrium estimate the amount of chlorobenzene in the air and in the water if the sum (total) in both phases is 10 g. fw = the fraction in the water phase fw = chlorobenzene mass in water/total mass fw C w Vw 1 1 C w Vw Ca Va 1 Ca Va 1 K Va H Vw C w Vw Estimation techniques for Henry’s Law values Experimental Measurement techniques Used fugacities to model an multiphase environmental system (Donald Mackay ES&T, 1979) Chapter 5 Used the Hansen and UNIFAC techniques to estimate activity coefficients Xiwsat = 1/iwsat (liquids) Xiwsat = 1/iwsat Csatiw = * pis * piL * p is Csatiw(L) * piL (solids) * p Xiwsat = 1/iwsat ig* piL (gases) Molar volumes could be estimated from Mw/density and Csatiw (L), from regressions of molar volume As an example we estimated Csatiw (L), satiwand GEiw for di-n-butyl phthalate on page 1206, -log Csatiw= 4.36 Csatiw = Xi / Vmix= 1 /i Vmix GEiw= RTlnI = 3483 J/ (molK) We then discussed solubility in terms of the cavity –iceberg theory Effects of co-solvents on dissolved organics in an organic/water solution sat fcN( h: w h:c )HSA Xmix log sat 2 . 303 RT X w HSA = Hydrophobic surface area, h:w=hydrophobic interfacial energy, h:c=hydrophobic interfacial energy where the solute contacts the organic, fc = volume fract. of organic Chapter 7 Kisw = Cis / Ciw solvent- water system Kiow = Cio / Ciw octanol-water system Water saturated octanol Vmix =(0.79)(0.16) + (0.21)(0.018) = 0.12 L mol-1 We observed that log Kihw = 1.21 log Kiow -0.43 ln Kiow= -a ln Csatiw+ b log Kiow = a log t +b Bioaccumulation and octanol water, Kiow [ trout lipid] = 1.5x10-7 moles /g fish lipid gas particle partitioning in the air-particle system Kp = Kiow/Kiaw x const Chapter 8 ' + [ H ] [ A ] H+ * log K ia log [HA] pKia = -log [Kia] At ambient pH values (4-10) low pKia compounds will be present in natural waters in their ionized (dissociated form) as their anions. log {[A-]/[HA]} = pH – pK*ia (Henderson- Hasselbach equation) at what point does pH = pKia? described the amount present in the acid (unionized or dissociated) form a 1 pH-pK ia 1+10 example, 2-nitrophenol has a pKia of 7.17. At a pH of 7, what is the fraction in the undissociated acid form? Organic Bases (BH+ is the conjugate acid of B) Kia H'+ [H+ ] B' [B] BH [BH+ ] from log {[A-]/[HA]} = pH – pKia we calculated that the addition of small amts. of a strong acid will not overly affect the pH of the natural water. The Hammett Correlation log Ka= log KaH +I for benzoic acid systems for other organic groups, log (Ka / KaH) = i looked at Nicotine partitioning