Chapter 6 Henry S Law

What is Henry’s law??* pi= p iL Xil* pi / Xil = p iL for non ideal solutions (low solubility)* pi= i Xil p i _ dividing by V l or the molar volume of the mixture (sometimes called Vmix)X/Vl = Concentration= Ci_ * pi / V l = i Ci p i * _ pi / Ci = i p i V l = 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? 1 Chatper 6 Henry’s Law pi KiH  Ciw traditionally atmi 1 KiH  1  atmlitersmole molesliters wCia Kiaw  (dimensionless Henry’s law const.) CiwCia Cia KiH Cia KiH KiH Kiaw     p n / V RT RT pi i KiH2As Henry’s law values increase there is a tendency for higher gas phase concentrations over water i.e. partitioning is toward air for high vapor pressure compounds the fugacity in the gas phase is high* fi = 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?3 4 Wash out ratios or W and how fast does the atmosphere clean up during a rain Usually 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 water or Ciw in units of moles i /cc = moles i/g H20The rain has an intensity, I with units of grams of rain sec-1 cm-2 so now -1 -2 I x Ciw = g rain sec cm x moles i/g H20Since W = Ciw/Cia = 1/Kiaw 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-2 And this is a flux5 We will learn in the old book Chapter 10,Flux / (conc x depth ) = 1st order rate constant inC = Co 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 m and we have a rain that gives an 1” of water in 2 hours I = 2.5g cm-2 /(2x60x60 sec) = 3.47x10-4 g cm-2 sec-1-5 Kiaw phenol = 2x10 krate constat = I x 1/Kiaw x Cia / (Cia x30,000 cm) in units of 1/sec6 in units of 1/sec = 0.00059 sec-1C/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 rain7 Concentration effects on KiHCiw = Xi / Vw Vw = molar vol. H2O* pi  iw iw pl * KiH     iw pl Vw Ciw iw /VwUnder dilute conditions KiH is directly proportional to the: activity coef.  saturated vapor pressure  molar volume of water8 What is the effect of concentration on KiH? * P iW water organic* at saturation the vapor pressure pi = p iw * pi = i Xi p i pure liquid sat sat 1 pi 1 X iw  sat *  sat  iw pi  iw psat sat i sat * K   pl Vw iH Csat w iw sat The question becomes how does KiH differ from KiH ?9 If the activity coef. changes with increases in sat concentration of Ciw then KH will change?Why?The old book suggests from benzene partitioning data, sat that little difference may exist between KiH 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 psat sat sat i KiH and estimated from K  iH Csat iwExample sat -3 o If the Ciw for chlorobenzene = 4.3x10 mol/L at 25 C* -2 and p iL = 1.6x10 atm what is the KiH psat 2 sat iL 1.6x10 atm KiH K    3.6atm L / mol iH Csat 4.3x103 mol / L iw10 K 3.6atm L mol 1 K  iH   0.15 iaw RT 0.082atm L mol 1K 1x298K sat A simple way of changing iw into iw (this does not always work) log sat log   i i sat 2 (1 xi ) psat K sat  iw sat p* V iH sat w iL w Ciw for infinitely dilute solutions pi iw iw p *iL * KiH    iw piLVw Ciw iw / Vw11  sat Comparison of iw and iw sat sat sat iw -logCiw Ciw iw(Tab 5.2) (p618) mol/L 1/(CsatVmix) (old book) benzene 2400 1.64 0.0229 2425 toluene 12000 2.25 0.0056 9879 chlorobenz 19000 2.35 0.00447 12437 hexCl-benz 9.8E+8 5.56 2.75E-6 2.0E+7 octanol 37000 2.35 0.00447 18656 sat Why are iw values sometimes greater than iw ?12 Effect of Temperature* vapHi 1 ln piL    const R T by analogy E H 1 ln x sat   iw  const iw R T sat sat xiw Ciw  so substituting Vmix excess heat of solution E H 1 lnCsat   iw  const w R T (Vmix )* PIl KiH  sat CIw H  HE ln K sat   VAP i iw  const iH RT H13 E vapHi- H iW = awHiHHenry14 15 What are the effects of salts?(Setschenow, 1889) sat  Ciw s log sat K i [ salt ]tot Ciw ,salt  let’s say we want to calculate the equilibrium distribution of anthracene in sea water, ie KiH w,salt if we transform Setschenow’s equation sat sat s logC iw ,salt logCiw Ki [ salt ]tot s sat sat Ki [ salt ]tot C iw ,salt Ciw 10 the Henry’s law for salt water is * * p s iL piL sat Ki [ salt ] KiH ,w ,salt  sat  s KiH 10 C sat Ki [ salt ] iw ,salt Ciw 10 s for anthracene Ki = 0.3, assume [salt] = 0.5 M and KiH = 0.078 atm L mol-1 (0.3)x(0.5) -1 so KIH,w,salt= 0.078x10 =0.11 atm L mol16 17 Estimating Henry’s Law values Hine and Mookerjee (1975)Log Kiaw =nj x functional groupi OH for phenol there are(older data) (new data)6 aromatic carbons at: -.33/carbon -0.264 5 aromatic C-H groups, at: 0.21/group +0.154 and one C-O group at: 0.74 -0.596 (C-OH) and one OH group at: -3.21 -3.23218 (old data) log K’H = 6x(-.33)+5(.21)+0.74+(-3.21) = -3.40(New data) log Kiaw= -4.64K’H = 0.0004 ; new book Kiaw= 0.000023 sat from p*iL / C w= 0.0004119 20 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 massCwVw 1 1 fw    CwVw  CaVa CaVa Va 1 1KH CwVw VwUsing the Hine and Mookerjee Cl K’H = Kiaw= 0.1622 fw = 1/{(1+0.1662)900/100}=0.4121 the concentration in the aqueous phase Cw isCw = fw Mtot / VwCw = 0.41x10g /0.1L = 41 g/LwaterCa= 0.59x10g /0.9 L = 6.6g/Lair22 Experimental Measurements 1. air toluene McAuliffe (1971)CiwVwv fract in H2O = CiaVg  CiwVwvVwv = vol of water for dilute systemsKiaw= Cia/Ciw = Dg,w( a gas/water part. coef.)23 Vwv fract in H2O= KiawVg Vwv each time we take a step n Cia,n= (fact in H2O) Ciw,o Kiaw taking the logs of both sides and substituting for fract in H2O and remembering that Kiaw=DgwV wv +log (C D logCia,n  n log iw o gw) Kiaw Vg Vwv24 25 26 Using fugacities to model environmental systems (Donald Mackay ES&T, 1979) Consider the phase equilibrium of five environmental compartments. Is it possible to tell where an environmental pollutant will concentrate? A B C E D where A= air, B= lake, C= Soil, D= Sediment, E= biota and suspended solidsWhen a system is at equilibrium the escaping tendencies in each phase are equal fA = fB = fC = fD = fE27 For Example: oxygen in water at 0.3 mol/m3 and in air at 8mol/m3 exert the same escaping tendency of 0.2 atm and are thus in equilibrium with the same fugacity.1. Fugacitys are linearly related to conc. oxygen in water at 0.03 mol/m3 exerts a fugacity of one tenth the fugacity of 0.3mol/m3. fA = fB = fC = fD = fEFugacities can be translated into concentrations fi Zi = C where Z is called a fugacity capacity value ------3. the mass Mtotal = Ci Vi = fi Zi Vi if the system is at equilibriumMtotal = fi  Zi ViMi = fi Zi VI4. Calculating Z values28 Zi fi = Ci; Zi= C/fIn air f is equal to the partial pressure,pi piV = nRT, pi = Cair RT, so Ziair = 1/RT at 298K , RT= 0.082 liter atm K-1 mol-1x298KRT= 0.025 m3 atm mol-1------In water pi = KiH Ciw and Ciw = Z fiw pi = KiH Ziw fiw Ziw = pi /{fw KH}= 1/KHWe will use a representative value of -4 3 -1 KiH= 1x10 m atm mol29 On soils, sediment, and suspended solidsCwi + S ----> CisCiS KiwS  ; Cis = KisxCiwxS CiW xSCis =Zi sp x fis and Ciw = pi /KiH Zi sp = KiwS x 1/KiH x pi x S/fis = Ki sp x S/ KiHFor suspended solids at 1,000 mg/m3 and -4 3 3 -1 -3 a Ki sp of 10 m /mg, Zsp= 10 mol atm mFor sediment and soils at 2x109 mg/m3 and -5 3 9 -1 -3 a Kiws of 5x10 m /mg, Zs,s= 10 mol atm mFor Aquatic BiotaZB = B  y  Kiow/KiH -6 3 3 where B is the volB/vol H20= 5x10 m /m , 5 4 y=octanol fract. of B = 0.2, Kiow=10 ; ZB=10------30 Let’s look at the Equilibrium Distribution of a toxic compound -10 3 with an atmospheric concentration of 4 x 10 mol/m .(fi x Zi = C and Mi = fi Zi Vi)Z Vol fi M % g/m3. (m3) (atm) (moles) air 40 1010 10-11 4 0.35 water 104 106 10-11 10-1 0.01 10-5 s solids 103 106 10-11 10-2 0.001 0.01 Sed 109 104 10-11 102 9.1 0.05 Soil 109 105 10-11 103 90.5 0.5 Aq biota 104 106 10-11 10-1 0.01 0.231

Chapter 6 Henry S Law

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