Embed Size (px)
DESCRIPTION
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. - PowerPoint PPT Presentation
Citation preview
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.
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
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
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
54.4x1020 kJoules of the sun’s energy strikes the earths surface each year
Energy from the earth
Where are the global energy reserves
Figure 1.5 Spiro
page 10
oil
Middle East
Asia and Australia including China
Former USSR
0200
400600800
1000
12001400
1978 1988 1998 2008
109 b
arre
ls
world
US
Fraction of US oil reserves compared to the global total (British petroleum web site, 2007)
The atmospheric compartmentHow much does it weigh?
Temperature and pressure
Circulation and mixing
Where did Oxygen come from
Particle emissions
Emissions of other pollutants
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
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
The quantity d is called the dry adiabatic lapse rate
Mixing height in the morning
Balloon temperature
Temp in oC
20 25 30 35
Dry adiabaticlines
hei
ght
in
ki l
omet
ers
0.00.10.20.30.4
1.1
1.5
What is Global Warming and how can it Change the Climate?
1979 perennial Ice coverage Nat. Geographic, Sept 2004)
2003 perennial Ice coverage
0
1
2
3
4
5
6
Met
ric
To
nn
es p
er y
ear
USAustralia
CanadaRussia
GermanyJapan
World-avgChina
India
Per Capita CO2 Emissions
Kinetics: 1st order reactions
A ---> B
-d [A] /dt = krate [A]
- d [A]/[A] = kratedt
[A]t= [A]0 e
-kt
ln[ ] ,,A k tA t
A t t 0
Some time vs conc. data
H
r 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
1st order plot
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2
time in hours
ln[A
]
A plot of the ln[conc] vs. time for a1st order reaction gives a straight line witha slope of the 1st order rate constant.
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
kpseudo 1st = k2nd [B]
kpseudo 1st = k2nd [B]
ln2 /k =t1/2
1. constant OH radicals in the atmosphere
kpseudo 1st = k2nd [OH.]
2. constant pH
kpseudo 1st = k2nd [OH-]
log Ka= log KaH +i
so, log (Ka / KaH )= I
and pKa = pKaH - I
The Hammett Equation and rates constants
COOH
R
COO-
R
+H+
Go= GoH + Go
i
It is also possible to show that:
log(krate) = log krateH + m,p
or log(krate/krateH) = m,p
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)
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
kpseudo 1st = k2nd [B]
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
AB
CCD
fA = fB = fC = fD = fE
Fugacities can be translated into concentrationsfi Zi = C
In Air: piV = nRT, p i = Cair RT, so Zi air = 1/RT
In water : Ziw = pi /{fw KH}= 1/KH
In biota: Z B = B y Kiow/KiH
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.
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 Vol fi M % g/m3.
(m3) (atm) (moles)
air 40 1010 10-11 4 0.35water 104 106 10-11 10-1 0.01 10-5
s 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 0.010.2
How are the different thermodynamic parameters related?
ig = oig + RT ln pi/p
oi
i = oi +RT ln fi/ f
oi
for ideal liquids p1i = x1 piL
* and p2i = x2 piL*
fiL = i Xipi*L (pure liquid)
fi hx = fi H2O
for non-ideal liquids
Obtained the important result: iH2O=1/ Xi H2O
Ci = = Xi / molar volumemix
the VH2O = 0.0182 L/1 mol
Vmix = 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.
From the saturated concentration of an organic in water (Ciw
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 H2O. Convert this to moles per liter which is a Ciw
sat .
Ciwsat = mole fraction/molar vol.
It is also possible to estimate estimated Csatiw from molar
volumes
ln Csatiw = -a (size) +b
Sat. Vapor pressure (p* iL) can be calculated from Tb
(boiling points and entropy of vaporization
Tb = 198 + funtional groups
Henry’s law= sat. vapor pressure/ (Ciwsat) `
log Kiow= -a log Csatiw + b’
a b’ r2
Alkanes 0.85 0.62 0.98PAHs 0.75 1.17 0.99 alkylbenzenes 0.94 0.60 0.99 chlorobenzens 0.90 0.62 0.99PCBs 0.85 0.78 0.92phthalates 1.09 -0.26 1.00Alcohols 0.94 0.88 0.98
)]()( ln.ln *
T
T
T
Tp bbiL 58119
)(.log.}/
/{log aK
mlwatermol
fishwetgmolBCF iow
i
i 700850
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) `
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
