Chapter 6 Henry S Law s1

Chapter 6What 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?1 Does it change with salt or ionic content? How do we measure it?Chapter 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 KiH2 If we go to the Appendix (p.1200, new book) and look at for Henry’s law values for air-water, we see -log Kiaw ;, * sat p iL and Ciw are referenced to their states. how are these -log Kiaw values computed?* sat Ideally, since KiH= pi / CiwIf we go to a unit-less form, Kiaw, where Kiaw = KiH/RT* sat So, log Kiaw= log { pi /RT } –log CiwFor anthracene the Appendix has the following data:* sat log pi = -3.01(Pa) -log Ciw = 6.60 -log Kiaw =2.81atm = 101,308 pascals st * so 1 we need to get - log pi in atm10-3.01 Pa/{ 101335 Pa/atm} = 9.646 x10-9 atm* * * to change pi in atm. into Cair; pi V= nRT; Cair = pi /RTR = 0.082 L atm./mole ; T = 298 K ; this gives -10 * Cair=3.94x10 moles/literair ; log pi /RT= -9.40* sat -logKiaw= -log { pi /RT } +log Ciw-logKiaw = 9.40 -6.60= 2.80 (and this is the book value)3 o The old book is a lot cleaner; It gives -log p L and -log o p s directly in atmospheres sat sat and –log Cw , -log Cs are in moles/liter and –log KH in liters atm/mol So to get log KH in the appendix (p 621)of the old book for anthracene in (old book, p. 621) o sat logKH = log pL – log Cw or o sat logKH = log ps – log Cs for a liquid anthracene log KH = -6.11+ 4.48= -1.63 for a sold anthracene log KH = -8.1+6.46= -1.64KH/RT = Kiaw (in new book); and -log Kiaw= 3.034 going back to Henry’s law pi KiH  CiwAs 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?5 Figure 6.2 page 111 (old book)6 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 can be viewed as a flux and 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 and Ciw = 1/Kiaw x CiaI 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 flux too7 We 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 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/sec8 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 rainHow do different Henry’s law values impact this calculation?9 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 water10 What is the effect of concentration on KiH? * P ia 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 ?11 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 iw12 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 / Vw13  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 ?14 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 )* PiiL KiH  sat CIwE sat  H  H ln K   VAP i iw  const + iH RT H15 E page 115, Table 6.1 vapHi- H iW = awHiHHenry16 Figure 6.3 page 116 (old book)17 What are the effects of salts? in Chapter 5 the relationship between a saturated solution in water vs. sea water is discussed(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 18 and KiH = 0.078 atm L mol-1 (0.3)x(0.5) -1 so KIH,w,salt= 0.078x10 =0.11 atm L mol19 Table 6.2 p 117 (old book)20 Estimating Henry’s Law values Hine and Mookerjee (1975)Log Kiaw =nj x functional groupi OH for phenol there are(old book) (new book) p 206 Table 6.2 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 -3.232(old book) log K’H = 6x(-.33)+5(.21)+0.74+(-3.21) = -3.40(New book) log Kiaw= -4.64K’H = 0.0004 ; new book Kiaw= 0.000023 sat from p*iL / C w= 0.0004121 22 From Chapter 3 (Discontented Socrates) There are linear free energy techniques that permit the estimation of equilibrium constants base on molecular structure This is possible if one assumes that the overall free energy of phase transfer is related to the linear combinations of the free energies related to the individual parts of the molecule that are involved in the transfer.12Gi = 12Gparts of i + special interaction termsLogK i12 = LogKparts of i12 + special interaction termsReturning to main body of the lecture discuss a technique that is often used in the book Mass= concentration x volumeWe often become interested in the fractional mass in one phase23 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 massC V 1 1 f  iw iw   w C V V CiwViw  CiaVia ia ia ia 1 1Kiaw CiwViw ViwUsing the Hine and Mookerjee Cl K’H = Kiaw= 0.1622 fw = 1/{(1+0.1662)900/100}=0.41 the concentration in the aqueous phase Ciw isCiw = fiw Mtot / ViwCiw = 0.41x10g /0.1L = 41 g/LwaterCia= 0.59x10g /0.9 L = 6.6g/Lair24 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.)Vwv fract in H2O=  const. KiawVg VwvFor then next time step,If Vg and Vwv are equal;Ciw,1 = Ciw,o x fract in H2O;25 because Ciw,1 = Ciw,o x Ciw,1/(Ciw,1+Cai,1) n Hence 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 VwvFigure 6.4 page 119 (old book)26 2. Mackay and co-workers experimental KH technique using a stripping apparatus o C w initially bubblesIF we take this as a CSTR the conc. of Cw some time, t, after we start the bubbles is o -kt Cw = C w e if we were just flowing in clean water instead of bubbles into some volume of water Vw o -f/Vw t Cw = C w eCw time to take into account the gas that is stripping, f, the flow of water is replaced with Kiaw x flow rate bubbles * o - Kiaw V /Vw t Cw = C w e g 27 Table 6.3, p12028 An acid Rain exampleAtmospheric acidity of “pure” rainCO2 +H2O --> CO2H2O (1) for reaction 1, KiH(CO2) = pCO2 / CO2H2OCO2H2O dissociates in water + - CO2H2O  + H [HCO3 ] (2) the equilibrium const K2 for this reaction is:- + - + [HCO3 ] [H ] [HCO3 ] [H ] KHCO2 K2 = = CO2H2O pCO2 bicarbonate reacts to -  -2 + HCO3  CO3 +H (3)-2 + -2 + [CO3 ] [H ] [CO3 ] [H ] KH(CO2) K3 = = - [HCO3 ] K2 pCO229 we now have expressions for each carbon form and we -2 - could add up [CO3 ] + HCO3 ]+ [CO2H2O] and set this to [CO2]T p CO2  K1 K1K2  [CO2 ]T  1   K   2 HCO2  [H ] [H ] An additional condition of ions in solution is that there be electrical neutrality, ie.- - -2 [H+] = [OH ]+[HCO3 ]+2[CO3 ]  H2O H  OH 0)- + [HCO3 ] [H ] KH(CO2) K2 = 2) pCO2-2 + [CO3 ] [H ] KH(CO2) K3 = 3) K2 pCO2If we substitute for each of the ions in the electro-neutrality equation 30 K K p 2K K K p [K ] H(CO 1 CO2 H(CO 1 2 CO2 [H+ ] W  2)  2) [H+ ] [H+ ] [H+ ]2AT a given temperature, KH(CO2) K1,K2 Kw are known. For pCO2 = 330 ppm, it can be shown that at 283oK the pH will be ~5.6. This is often the value cited for “pure” rain water.The above equation can be numerically solved in most spread sheets!, by moving the [H+] on the left to the right side of the equation. 31 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 = fE32 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 = Ci 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 values33 Zi fi = Ci; Zi= C/fIn air f is equal to the partial pressure, pi piV = nRT, so fi = 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 and we said in air pi = fia so Ziw = pi /{fiw KiH}= 1/KiHWe will use a representative value of -4 3 -1 KiH= 1x10 m atm mol34 On soils, sediment, and suspended solidsCwi + S ----> CisCiS Kid = KiwS  ; Cis = KiwsxCiwxS CiW xSCis =Zi sp x fis and Ciw = pi /KiH Zi sp = KiwS x 1/KiH x pi x S/fis = Ki wS 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 Kid = 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------35 Let’s look at the Equilibrium Distribution of a toxic compound with an atmospheric concentration of 4 x 10-10 mol/m3. (fi x Zi = 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 0.236 Chapter 6 fugacity homework problem 3 3. A Fugacity problem (first some additional theory )-3 - In a compartment, Ci, the rate of decay in moles m year 1 for a given process, j, is rateCi = kj[Ci] where kj = bio degradation photolysis hydrolysis oxidation advection etc. rateCi= kB[Ci]+ kP[Ci]+ kH[Ci]+ kOX[Ci]+ kA[Ci] rateCi = [Ci] kj = [Ci] kT in moles per year the total rate in a compartment of volume Vi is: rateTi = [Ci] kT Vi if the system is at steady state in each compartment, the total input rate for all the compartments in moles/ year will equal the amount reacted in moles/year37 I = Ci]kT Vi) = Zi fi kT Vi) and I = fi  Zi kT Vi); why? so fi= I /Zi kT Vi)Assume three compartments: air, water and sediment (sed) in equilibrium with one another and a total input rate of 1000 moles/year goes into the entire system. From the following rate constant data (years-1), calculate for each compartment the fugacity, the resulting concentration (moles/m3), the total decay rate (moles/year) and the total #moles in each compartment. Use Z values and volumes from the class example. From the total decay rate, where does most of the degradation take place? where does most of the mass end up? Use a spread sheetRate constants (years-1) Biodeg. photolysis hydrolysis oxidation advection kTAir 0 130 0 0 50 180 water 100 100 100 0 200Sed. 0.1 0 0 0 038

Chapter 6 Henry S Law s1

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