Chapter 6And this is a flux tooAn acid Rain exampleChapter 6 fugacity homework problem 3Chapter 6What is Henry’s law?? pi= p*iL Xilpi / Xil = p*iLfor non ideal solutions (low solubility)pi= i Xil p*idividing by _Vl or the molar volume of the mixture (sometimes called Vmix)X/Vl = Concentration= Cipi / _Vl = i Ci p*i pi / Ci = i p*i _Vl = const ??? = KiHl ; If air – water K iH Is the product of i, p*i Vmix a constant??Is it different for different compounds??Does it vary with temperature???Does it change with concentration?Does it change with salt or ionic content?How do we measure it?Chapter 6 Henry’s Law1iwiiHCpK traditionally11 molelitersatmlitersmolesatmKwiiHiwiaiawCCK(dimensionless Henry’s law const.)RTiHKRTV/niHKiaCipiHKiaCiHiiaiawKpCK ---2If we go to the Appendix (p.1200, new book) and look at for Henry’s law values for air-water, we see -log Kiaw ;, p*iL and Ciwsat are referenced to their states. how are these -log Kiaw values computed?Ideally, since KiH= pi*/ CiwsatIf we go to a unit-less form, Kiaw, where Kiaw = KiH/RTSo, log Kiaw= log { pi*/RT } –log CiwsatFor anthracene the Appendix has the following data:log pi*= -3.01(Pa) -log Ciwsat = 6.60 -log Kiaw =2.81atm = 101,308 pascalsso 1st we need to get - log pi*in atm10-3.01 Pa/{ 101335 Pa/atm} = 9.646 x10-9 atmto change pi* in atm. into Cair; pi*V= nRT; Cair = pi*/RTR = 0.082 L atm./mole ; T = 298 K ; this gives Cair=3.94x10-10 moles/literair ; log pi*/RT= -9.40-logKiaw= -log { pi*/RT } +log Ciwsat-logKiaw = 9.40 -6.60= 2.80 (and this is the book value)3The old book is a lot cleaner; It gives -log poL and -log pos directly in atmospheresand –log Cwsat , -log Cssat are in moles/liter and –log KH inliters atm/mol So to get log KH in the appendix (p 621)of the old book for anthracene in (old book, p. 621) logKH = log pLo – log Cwsat or logKH = log pso – log Cssat for a liquid anthracenelog KH = -6.11+ 4.48= -1.63for a sold anthracenelog KH = -8.1+6.46= -1.64KH/RT = Kiaw (in new book); and -log Kiaw= 3.034going back to Henry’s lawiwiiHCpK As Henry’s law values increase there is a tendency for higher gas phase concentrations over water i.e. partitioning is toward airfor high vapor pressure compounds the fugacity in the gas phase is highfi = i Xifi* pure liquid(fi*pure liquid = p*i pure liquid)High activity(i) coefs. favor partitioning to the gas phase i.e. Lower KiH and lower ‘s favor the liquid phase. Polar compounds?5Figure 6.2 page 111 (old book)6Wash out ratios or W and how fast does the atmosphere clean up during a rainUsually defined as the conc. in rain/conc. In airW = Ciw/Cia = 1/KiawW x Cia = conc in the rain, Ciw , with units of moles/ cc wateror Ciw in units of moles i /cc = moles i/g H20The rain can be viewed as a flux and has an intensity I, with units of grams of rain sec-1 cm-2so now ------------------------------ I x Ciw = g rain sec-1 cm-2 x molesi/g H20Since W = Ciw/Cia = 1/Kiaw and Ciw = 1/Kiaw x Cia I x 1/Kiaw x Cia = moles of i from the atmosphere hitting the surface of the earth in the rain per sec-1 cm-2And this is a flux too7We will learn in Chapter 20,Flux / (conc x depth ) = 1st order rate constant inA = Ao e-ktSo if you know the rain intensity, Kiaw and the height of the atmosphere, you can estimate how fast the atmosphere will “clean” up with a given rain intensity???_______________Flux = I x 1/Kiaw x Cia = moles of i from the atmosphere hitting the surface of the earth per sec-1 cm-2If the mixing height of the atmosphere is 300 mand we have a rain that gives an 1” of water in 2 hoursI = 2.5g cm-2 /(2x60x60 sec) = 3.47x10-4 g cm-2 sec-1Kiaw phenol = 2x10-5krate constat = I x 1/Kiaw x Cia / (Cia x30,000 cm)in units of 1/secin units of 1/sec = 0.00059 sec-18C/Co = e-kt ; t = 2 hours = 2x60x60 sec C/Co = 0.0145 or 98.5% of the phenol will be cleaned out in the air in the rainHow do different Henry’s law values impact this calculation?9Concentration effects on KiHCiw = Xi / VwVw = molar vol. H2OwliwwiwliwiwiwiiHVpVpCpK**/Under dilute conditions KiH is directly proportional to the: activity coef. saturated vapor pressure molar volume of water10What is the effect of concentration on KiH?P*iawaterorganicat saturation the vapor pressure pi = p*iw pi = i Xi p*i pure liquidsatiwisatisatiwsatiwppX11*w*lsatsatsatsatVpCpKwiwiiHThe question becomes how does KiHsat differ from KiH ?If the activity coef. changes with increases in concentration of Ciw then KHsat will change?Why?11The old book suggests from benzene partitioning data, that little difference may exist between KiHsat and KiH.For benzene K’iaw = (Cair/Ciw) a difference of <4% was observed between saturated and dilute water solutions…. This means that KiH can sometimes be approx. from KiHsat and estimated from satsatsatiwiiHCpK ExampleIf the Ciwsat for chlorobenzene = 4.3x10-3 mol/L at 25oCand p*iL = 1.6x10-2 atm what is the KiHmol/Latm.L/molx.atmx.CpKKsatsatsatiHiwiLiH-631034106132150298082063111.KxKmolLatm.molLatm.RTKKiHiaw----A simple way of changing iw into iwsat (this does not always work)loglog( )iisatisatx1212w*iLsatsatiwsatsatVpCpKwiwiHfor infinitely dilute solutionsw*iLiwwiwiLiwiwiwiiHVpV/*pCpK 13Comparison of iw and iwsatiw -logCiwsat Ciw satiwsat(Tab 5.2) (p618) mol/L 1/(CsatVmix)(old book)benzene 2400 1.64 0.0229 2425toluene 12000 2.25 0.0056 9879chlorobenz 19000 2.35 0.00447 12437hexCl-benz 9.8E+8 5.56 2.75E-6 2.0E+7octanol 37000 2.35 0.00447 18656Why are iw values sometimes greater than iwsat?14Effect of TemperatureconstTRHplnivap*Li1by analogy constTRHxlnEsatiwiw1mixsatiwsatiwVxC  so substitutingexcess heat of solution)Vmix(EsatwconstTRHClniw1satIwiLiHCPiK*HEiwiVAPsatiHconstTRHHKln +15page 115, Table 6.1 vapHi- HEiW = awHiHHenry16Figure 6.3 page 116 (old book)17What are the effects of salts?in Chapter 5 the relationship between a saturated solution in water vs. sea water is discussed(Setschenow, 1889)totssatsalt,iwsatiw]salt[KCClogilet’s say we want to calculate the equilibrium distribution of anthracene in sea water, ie KiH w,saltif we transform Setschenow’s