materials to remedy environmental pollutions

Materials To Remedy

Environmental Pollutions

Graduate Course (445.686)

Seoul National University, 2008/1

Part 0. Environmental Pollutions

Chapter 1. Environmental Chemistry:

a global perspectives

Chapter 1. Environmental Chemistry

• Complications in the use of units for the concentration of the composition

- ppm (parts per million)(cf. ppt, ppb): g of solutes per million milliliters of solution

=

E.g. 10 ppm aq. NH4OH

=10 mgNH4+L-1 = 10 x 10-3 gNH4

+L-1

=10 x 10-3/18.0 molNH4+L-1

= 5.56 x 10-4 mol NH4+L-1

1.1 Environmental composition: Aqueous solutions

11 or mgLmLg


E.g. 10 ppm aq. NH4OH

= 5.56 x 10-4 mol NH4+L-1

= 5.56 x 10-4 mol NO3-L-1 (1:1 conversion)

= 5.56 x 10-4 x 62 x 103 mg NO3-L-1

= 34 ppm NO3-

( is due to molar mass difference between NH4+ and NO3

-)

= 34 mgNO3-L-1 =34 x 14/62 mgNL-1

= 7.7 mgNL-1 = 7.7 ppm(this is one way of avoiding complication abovementioned)



• If molar concentration units are used, there will never be any confusion.

• Another limitation of ‘parts per…’:

they give no indication of the concentration of the reactive groups

E.g. Assumption : TOC in a forest stream = 9.0 mgCL-1

COOH/g of humic material in water = 4 mmolg-1

Carbon contents in humic material = 50 %

Then,

TOC = 9 x 10-3 x 100/50 g humic material/ L water

= 9 x 10-6 x 100/50 g humic material/L

x 4 x 10-3 mol COOH/g humic material

= 8 x 10-3 mol COOH/L water



• Mass fraction of various kinds are used for solids without any ambiguity.

E.g. global avg. conc. of Fe: 4.1 wt%

V: 160 g/g = 160 ppm.

1.1 Environmental composition: Solids

1.1 Environmental composition: Gases

•Molar fraction or percentage conc. (usu. mixing ratio) are used for

gases. vol. fraction partial pressure

•For gases at low conc., ‘parts per…’ family of units is frequently used.

E.g. ppmv (parts per million by volume)

•Mass/vol, moles/vol, molecules/vol. are frequently used.

E.g. O2 in air, mixing ratio = 20.95% = 9.35 x 10-3 mol/L = 0.299 g/L


• In many cases, the species distribution of the species, particularly

existing in more than one form is more important.

E.g. carbonate species and mercury in water

1.1 Environmental composition: Species distribution

a: oceans with 0.56 mol/L Cl- content

b: well water with 2.7 x 10-4 mol/L Cl-


•Four principal compartments of the environments

-The atmosphere (the gaseous environment)

-The hydrosphere (the liquid, essentially aqueous environment)

-The terrestrial (solid) environment

-The biosphere (the living environment)

E.g. Water cycle

1.2 Chemical Processes

Since all processes are at steady state,

the residence time of water in the atmosphere

can then be determined by

days 6.90262.0

kgy 1096.4

kg 103.1

out)or (in flux

atmosphere in theamount statesteady

timeresidence

1-17

16

y


• Human activity effects on the natural processes

E.g. The Bhopal disaster in India (3 December, 1984)

- massive release of methyl isocyanate (CH3-N=C=O) gas from the storage

tank

- killed 3,000 persons, 300,000 persons affected by exposure

- MIC is a starting material for carbaryl that is carbamate pesticide

- Usu. stored under refrigeration in an underground

- Causes: technological problems (T) + human activity factors (H)

T: failure of cooling unit, leakage of water into the tank, loss of nitrogen

pressure above the MIC, failure of safety device, etc.

H: political, organizational, and human factors, etc.

1.3 Anthropogenic (human) effects

Part 1. The Earth’s Atmosphere

Chapter 2. The Earth’s Atmosphere

Chapter 2. The earth’s atmosphere

• Major components: following mixing ratios are const. up to an altitude of

ab. 80 km due to the kinetic energy of gas molecules, being sufficiently

big enough to overcome any gravitational forces

2.1 Regions of the atmosphere: General Introduction

Avg. molar mass of the

atmosphere,aM

196.28

222222

-

COCOArArOONNa

gmol

fMfMfMfMM

Total mass of the atmosphere

Earth) theof (radius m1037.6

ms81.9 Pa,101325

kg105.27

kg]in [4

6

2-0

18

20

r

gP

g

rPM atm


• Four sections of the atmosphere: depending on the T change direction,

troposphere (대류권), stratosphere (성층권), mesosphere (중간층),

thermosphere (열권)

Pressure changes with altitude

RTghM

haePP /0

h=altitude in meter

Ph=pressure at given altitude in Pa

2.1 Regions of the atmosphere: General Introduction


• In thermosphere (열권) : nearly vacuum at ab. the altitude of 100 km

• For one photon,

• For 1 mol of photons,

2.1 Regions of the atmosphere: Chemical composition

J/photon 100.2

m 10100

ms 100.3Js106.6

18

9-

-1834

hcE

kJ/mol 1200106.0100.2 2318 E

kJ/mol 945H 2Nnm) 126( N 0

2 h

kJ/mol 498H 2Onm) 240( O 0

2 h

kJ/mol 1500H e Nnm) 80( N 0-

22 h

kJ/mol 1310H eOnm) 91( O 0- h

• In mesosphere (중간층): some amount of O3 that absorbs solar radiation

OOnm) 320230( O **

23 h


2.1 Regions of the atmosphere: The troposphere

• Convection currents and winds cause constant movement of the air


2.2 Reactions and calculations in atmospheric chemistry:

Thermodynamic calculations

• For

• At 25 C, : uneasy reaction at R.T.

• At 1500 C (I.e. in a cylinder of an internal combustion engine at the time of

ignition),

• Then the equilibrium constant is

• Since , if after complete burning, in the

compressed cylinder gas at 2500 C, then after some reaction between O2 and N2

(meaning the loss of partial pressure, x, of each)

• if x650 kPa, then

• Therefore, PNO=2x=1.4 kPa I.e. NO ratio in hot cylinder =

)( NO2)(O)(N 22 ggg

kJ/mol1.173)55.86(22 00 fGG

kJ/mol 2.111)2(2 0000

)(

0

2773 22 ONNONOf SSSTHG

0080.082.42773//314.8

/111200ln

0

P

TP K

KKmolJ

molJ

RT

GK

)/)(/(

)/(00

20

22PPPP

PPK

ON

NOP kPa 0.1 kPa, 650

22 ON PP

0080.0)0.1)(650(

4 2

xx

x66.002.52.54 2 xxx

ppmv 2200ppmv10kPa650

kPa 4.1 6



Kinetic calculations

• For the reverse reaction of

• At 25 C, if atmospheric conc. of NO=100 ppbv

• Then the initial rate of decomposition of NO at R.T. is

• The half-life, : 1028 times longer than the age of the

Earth

• This means there should be other reactions must be involved in controlling

the atmospheric conc. of NO -> Think what?

/mol/sm 106.2 )( NO2)(O)(N 3/1021.36

222

4 Tekggg

mol/s/m 103.4 341

2

k

sNOk i /mol/m 102.7][ 3522

2

yNOk

ti

38

2

2/1 108.1][

1



Photochemical reactions

• For , reaction rate=

• Since the production rxn=consumption rxn,

• Therefore

ONOh 2NO

][]][[ 212 NOfNOROOk

1

22 ][

][

][

f

ROOk

NO

NO

][ 21 NOf


Reactions involving free radicals

OH2OHO

OOO

OOO

ONONO

2

3

**

23

32

2

h

MM

h

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