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Class objectives:

Class objectives:. Highlight some important areas in environmental chemistry present some of the common techniques that environmental chemists use to quantify process that occur in the environment It is assumed that everyone has courses in calculus and general chemistry.

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Class objectives:

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  1. Class objectives: • Highlight some important areas in environmental chemistry • present some of the common techniques that environmental chemists use to quantify process that occur in the environment • It is assumed that everyone has courses in calculus and general chemistry.

  2. Class objectives: • We will cover general topics: Global warming, Strat. O3, aerosols, photochemical smog, acid rain, etc. • Develop relationships will be used to help quantify equilibrium and kinetic processes

  3. Important Environmental Issues • Global warming and stratospheric ozone depletion • Concentration of environmental pollutants at the poles; pesticides in foods, etc. • Buildup of environmental chemicals in the oceans; contamination of soil and ground water • Particle exposure, photochemical oxidant exposure, acid deposition • Energy shortages

  4. Energy from the earth 54.4x1020 kJoules of the sun’s energy strikes the earths surface each year Sun earth • Of this ~30% is reflected back to into space (albedo) • One Joule = 4.2 calories. It takes ~2000 K- calories to feed a human each day • What fraction of the earth’s energy striking the earth, if turned into food, could feed the planet

  5. Where are the global energy reserves oil Figure 1.5 Spiro page 10 Former USSR Middle East Asia and Australia including China

  6. Fraction of US oil reserves compared to the global total (British petroleum web site, 2007)

  7. The atmospheric compartment • How much does it weigh? • Temperature and pressure • Circulation and mixing • Where did Oxygen come from • Particle emissions • Emissions of other pollutants

  8. How thin is the air at the top of Mt. Everest? • Mt. Everest is 8882 meters high or 8.88 km high • log P = -0.06 x 8.88 • P = 10-0.06x 8.88 = 0. 293 bars • Assume there are 1.01bars/atm. • This means there is < 1/3 of the air

  9. The quantity dis called the dry adiabatic lapse rate • d = - dT/dz = 9.8 oK/kilometer • If the air is saturated with water the lapse rate is often called s • Near the surface sis ~ -4 oK/km and at 6 km and –5oK/km it is ~-6K/km at 7km high

  10. Balloon temperature 1.5 1.1 height in kilometers Dry adiabatic lines 0.4 0.3 0.2 0.1 0.0 20 25 30 35 Temp in oC Mixing height in the morning

  11. What is Global Warming and how can it Change the Climate?

  12. 1979 perennial Ice coverage Nat. Geographic, Sept 2004)

  13. 2003 perennial Ice coverage

  14. Per Capita CO2 Emissions 6 5 4 3 Metric Tonnes per year 2 1 0 US Canada Germany World-avg India Australia Russia Japan China

  15. Kinetics: 1st order reactions A ---> B -d [A] /dt = krate [A] - d [A]/[A] = kratedt [A]t= [A]0 e-kt

  16. Hr Conc [A] Ln[A] 0 2.718 1 0.3 2.117 0.75 0.6 1.649 0.50 0.9 1.284 0.25 1.2 1.000 0.00 1.5 0.779 -0.25 Some time vs conc. data

  17. A plot of the ln[conc] vs. time for a 1st order reaction gives a straight line with a slope of the 1st order rate constant.

  18. ln [A]/[A]o=-k t1/2 ; ln2 /k =t1/2 2nd order reactions A + B  products dA/dt = k2nd [A][B] If B is constant kpseudo1st = k2nd [B]

  19. kpseudo1st = k2nd [B] ln2 /k =t1/2 1. constant OH radicals in the atmosphere kpseudo1st = k2nd [OH.] 2. constant pH kpseudo1st = k2nd [OH-]

  20. The Hammett Equation and rates constants DGo= DGoH + SDGoi log Ka= log KaH +Ssi so,log (Ka / KaH )= SsI and pKa = pKaH - SsI

  21. It is also possible to show that: log(krate) = logkrateH + rsm,p orlog(krate/krateH) = rsm,p

  22. What this means is for aromatics with different substituted groups, if we know the rvalue we can calculate the rate constant from the sigma (sm,p) and the hydrogen substituted rate constant. If we know the rate constant for a number of similar aromatics with different substituted groups, we can create a y=mx+b plot and solve for the slope rvalue (see example at end of Pesticide Chapter)

  23. ln [A]/[A]o=-k t1/2 ; ln2 /k =t1/2 2nd order reactions A + B  products dA/dt = k2nd [A][B] If B is constant kpseudo1st = k2nd [B]

  24. 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? where A= air, B= lake, C= Soil, D= Sediment, E= biota and suspended solids Thermodynamics How do the pollutants in the different compartments of the environment distribute? A B C D C

  25. fA = fB = fC = fD = fE Fugacities can be translated into concentrations fi Zi = C In Air: piV = nRT, p­i = Cair RT, so Zi­air = 1/RT In water :Ziw = pi /{fw KH}= 1/KH Remember we also used Henry’s law to calculate how fast the atmosphere cleans up, and in another problem fractions of a toxic in the gas and water phase of a flask In biota: Z­B = B  y  Kiow/KiH Remember octanol/water partitioning coef. to calculated bio accumulation factors.

  26. We looked at the Equilibrium Distribution of a toxic compound with an atmospheric concentration of 4 x 10-10 mol/m3.(fi x Zi = C and Mi = fi Zi Vi) Z Volfi M % mg/m3.(m3) (atm) (moles) air 40 1010 10-11 4 0.35 water 104 106 10-11 10-1 0.01 10-5s solids 103 106 10-11 10-2 0.001 0.01Sed 109 104 10-11 102 9.1 0.05Soil 109 105 10-11 103 90.5 0.5Aq biota 104 106 10-11 10-1 0.01 0.2

  27. How are the different thermodynamic parameters related? mig = moig + RT ln pi/poi mi = moi +RT ln fi/ foi for ideal liquids p1i = x1 piL* and p2i = x2 piL* for non-ideal liquids fiL = i Xipi*L (pure liquid) fihx = fi H2O

  28. Obtained the important result:giH2O=1/Xi H2O Ci = = Xi / molar volumemix the VH2O = 0.0182 L/1 mol Vmix = S Xi Vi ; typically organics have a Vi of ~0.1 L/mol Vmix0.1 Xi + 0.0182 XH2O MW/density can be used to estimate molar volume. For most organic compounds if you do not know the density, assume 1 g/ml.

  29. From the saturated concentration of an organic in water (Ciwsat)can you calculate the mole fraction and activity coefficient? Remember the toluene homework where you were given a maximum saturation concentration in water of 515 mg/liter H2O. Convert this to moles per liter which is a Ciwsat . Ciwsat = mole fraction/molar vol.

  30. It is also possible to estimate estimated Csatiw from molar volumes ln Csatiw = -a (size) +b

  31. Henry’s law= sat. vapor pressure/ (Ciwsat) ` Sat. Vapor pressure (p* iL) can be calculated from Tb (boiling points and entropy of vaporization Tb = 198 + S funtional groups log Kiow= -a log Csatiw + b’ ab’r2 hAlkanes 0.85 0.62 0.98 hPAHs 0.75 1.17 0.99 h alkylbenzenes 0.94 0.60 0.99 chlorobenzens 0.90 0.62 0.99 hPCBs 0.85 0.78 0.92 hphthalates 1.09 -0.26 1.00 hAlcohols 0.94 0.88 0.98 h

  32. Bioaccumulation and octanol water, Kiow Moli/ml water is the concentration of a toxic in the water phase (Ciw) Henry’s law= partial pressure i/ (Ciw) `

  33. Go over problems I did at the board, problems that were covered from the notes during class, and homework problems from the short and long homework sets. Look at the natural waters homework/with answer link The exam will cover thermo, vapor pressure, henry’s law, water octanol, surface and water purification, pesticides and heavy toxic metals. It will have problems and some short questions. Good luck to all

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