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.  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 Sun earth 54.4x10 20 kJoules of the sun’s energy strikes the earths surface each year Energy from the earth

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

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

10. Mixing height in the morning Balloon temperature Temp in o C 20253035 Dry adiabatic lines height in kilometers 0.0 0.1 0.2 0.3 0.4 1.1 1.5

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

13. 1979 perennial Ice coverage Nat. Geographic, Sept 2004)

14. 2003 perennial Ice coverage

15. 0 1 2 3 4 5 6 Metric Tonnes per year US Australia Canada Russia Germany Japan World-avg China India Per Capita CO 2 Emissions

16. Kinetics: 1 st order reactions A ---> B -d [A] /dt =k rate [A] - d [A]/[A] = k rate dt [A] t = [A] 0 e -kt

17. Some time vs conc. data  Hr Conc [A]Ln[A] 0 2.7181 0.3 2.1170.75 0.6 1.6490.50 0.9 1.2840.25 1.2 1.0000.00 1.5 0.779-0.25

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

19. ln [A]/[A] o =-k t 1/2 ; ln2 /k =t 1/2 2 nd order reactions A + B  products dA/dt = k 2nd [A][B] If B is constant k pseudo 1st = k 2nd [B]

20. ln2 /k =t 1/2 1. constant OH radicals in the atmosphere k pseudo 1st = k 2nd [OH. ] 2. constant pH k pseudo 1st = k 2nd [OH - ]

21. log K a = log K aH +  i so, log (K a / K aH )=   I and pK a = pK aH -   I The Hammett Equation and rates constants  G o =  G o H +   G o i

24. It is also possible to show that: log(k rate ) = log k rateH +   m,p or log(k rate /k rateH ) =   m,p

25. What this means is for aromatics with different substituted groups, if we know the  value we can calculate the rate constant from the sigma (  m,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  value (see example at end of Pesticide Chapter)

26. ln [A]/[A] o =-k t 1/2 ; ln2 /k =t 1/2 2 nd order reactions A + B  products dA/dt = k 2nd [A][B] If B is constant k pseudo 1st = k 2nd [B]

27. Thermodynamics How do the pollutants in the different compartments of the environment distribute? 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 A B C C D

28. f A = f B = f C = f D = f E Fugacities can be translated into concentrations f i Zi = C In Air: p i V = nRT, p­ i = C air RT, so Z i ­ air = 1/RT In water : Z iw = p i /{f w K H }= 1/K H In biota: Z ­B = B  y  K iow /K iH 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 Remember octanol/water partitioning coef. to calculated bio accumulation factors.

29. We looked at the Equilibrium Distribution of a toxic compound with an atmospheric concentration of 4 x 10 -10 mol/m 3.(f i x Z i = C and M i = f i Zi Vi) ZVol f iM%  g/m 3. (m 3 )(atm)(moles) air4010 10 10 -11 40.35 water10 4 10 6 10 -11 10 -1 0.0110 -5 s solids10 3 10 6 10 -11 10 -2 0.0010.01 Sed10 9 10 4 10 -11 10 2 9.10.05 Soil10 9 10 5 10 -11 10 3 90.50.5 Aq biota 10 4 10 6 10 -11 10 -1 0.010.2

30. How are the different thermodynamic parameters related?  ig =  o ig + RT ln p i /p o i  i =  o i +RT ln f i / f o i for ideal liquids p 1i = x 1 p iL * and p 2i = x 2 p iL * f iL =  i X i p i * L (pure liquid) f i hx = f i H2O for non-ideal liquids

31. Obtained the important result:  iH2O =1/ X i H2O C i = = X i / molar volume mix the V H2O = 0.0182 L/1 mol V mix =  X i V i ; typically organics have a V i of ~0.1 L/mol V mix  0.1 X i + 0.0182 X H2O MW/density can be used to estimate molar volume. For most organic compounds if you do not know the density, assume 1 g/ml.

32. From the saturated concentration of an organic in water (C iw sat ) 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 H 2 O. Convert this to moles per liter which is a C iw sat. C iw sat = mole fraction/molar vol.

33. It is also possible to estimate estimated C sat iw from molar volumes ln C sat iw = -a (size) +b

34. Sat. Vapor pressure (p * iL ) can be calculated from T b (boiling points and entropy of vaporization Tb = 198 +  funtional groups Henry’s law = sat. vapor pressure / (C iw sat ) ` log K iow = -a log C sat iw + b’ a b’r 2  Alkanes0.850.620.98  PAHs0.751.170.99  alkylbenzenes0.940.600.99 chlorobenzens0.900.620.99  PCBs0.850.780.92  phthalates1.09 -0.26 1.00  Alcohols0.940.880.98 

35. Bioaccumulation and octanol water, K iow Mol i /ml water is the concentration of a toxic in the water phase (C iw ) Henry’s law = partial pressure i / (C iw ) `

36. 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