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