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Lecture No. 3

Process FundamentalsChapter 3, p.81

1. Chemistry Review

A. Organic Chemistry

1.) General

• Organic chemistry is the study of compounds containing carbon which make up muchof hazardous waste.

•The vast majority of chemical bonds for organic molecules are covalent bonds(electron are shared between atoms) and are represented by dashes connecting theatoms in a molecule. The "octet" rule says that atoms must be bonded together byTWO (one pair) of electrons; one dash is used for each pair of electrons. A secondrule is that carbon in an organic molecule can have a maximum of 4 bonds.

• Hydrocarbons are a subset of organic compounds that are composed solely of hydrogen and carbon.

2.) Hydrocarbons-

• Hydrocarbons are divided into two categories: aromatics which are a ring structureand aliphatic which contain open carbon chains or rings.

• Aliphatics include: alkanes, alkenes, alkynes and their derivatives.• Alkanes are hydrocarbons that contain only single bonds between carbon atoms and

have the general formula Cn H2n+2 where n is an integer. For example, n=1 gives

CH4 which is methane; n=4 gives C4H10 which is butane. See T3-1, p.83.

• Alkyl radicals are alkanes with a terminal hydrogen atom missing. For example, pentyl

CH3-CH2-CH2-CH2-CH2 whereas pentane would be CH3-CH2-CH2-CH2-CH3. Note that the last hydrogen atom at the end is missing. R is used to represent anunspecified alkyl radical. An alkane may be represented by R-H.

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Lecture No.3, Process Fundamentals, Page No. 2

• Example:Given: Draw Pentane and Pentyl

C

H

H

C

H

H

C

H

H

C

H

H

H HC

H

H

C

H

H

C

H

H

C

H

H

C

H

H

H C

H

H

PentylPentane

• Normal alkanes are straight chained. Branched alkanes have alkyl groups attached to anon-terminal carbon.

• Nomenclature for a branched alkane is as follows:- The longest straight carbon chain yields the base name.- Number the carbons in the base chain. The direction of naming 1, 2, 3 of the

carbons start from the end in the base chain that will give the lowest number to thecarbon that contains the alkyl group.

- Place the carbon number determined in the previous step before the alkyl group'sname. The last name is the base name.

• Example:Given: Draw 2-butyl-pentane

C

H

H

C

H

H

C

H

H

C

H

H HC

H

H

C

C

H

H H

2-butyl-pentane

H H butyl group

second carbon

basic pentane molecule

• Benzene is the simplest aromatic. PAHs (Polyaromatic hydrocarbons) are groups of aromatic rings containing only carbon and hydrogen. PAHs are two or more benzenerings fused together with at least two carbons in common. p.86.

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B. Chemical Characteristics

• Organic chemicals exist in 3 categories based on volatility:• Volatile organics (VOCs) - High vapor pressure that allows them to evaporate

quickly.• Semivolatile organics. Rely on acid/base properties to facilitate their isolation from a

water sample into a solvent.• Nonvolatile organics. No volatility even at elevated temperatures. Measured by the

same solvent extraction techniques as semivolatile organics.• Pesticides and herbicides. Chlorinated and organophosphates. T3-2, p.90.• Polychlorinated Biphenyls (PCBs). Widely used in electrical transformers and a severe

carcinogen.

2. Physical-Chemical Properties

• Physical-chemical properties determine how a specific chemical will interact and movein the environment and how they can be removed.

A. Solubility

• The degree to which one substance (solute) will dissolve into another substance(solvent); the combination is the solution. Measured in mg/l which equals ppm for dilute solutions of water i.e. when 1mg=1cc.

B. Vapor Pressure

• When molecules leave the liquid as a vapor via evaporation. Vapor pressure is the

pressure exerted by the vapor on the liquid at equilibrium. Measured in atmosphere =760mm.

• Raoult's Law is for an ideal binary liquid solution; in typical environmental solutions,Henry's Law tends to control. Henry' law: Under equilibrium conditions the partial

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pressure of a gas or volatile chemical above a liquid is proportional to theconcentration of the chemical in the liquid.

Pg = HCL eq.3-5 units p.93

Henry's Constant is the ratio of the concentration in the gas to the concentration in the

liquid.H=Cg/CL eq 3-6 units p.94

A dimensionless form of Henry's constant is:H' = eq. 3-7 units p.94

In the absence of tabular material Henry's constant may estimated by dividing thevapor pressure by the solubility:

H =• example:

Given: Benzene at 25°C.

Find: 1.) Henry's constant in water.2.) Calculate H', dimensionless constant

3.) Compare to the tabulated value in App. A

See appendix A, p.1040 for benzene1.) H

pvp = 95.2 mm Hg at 25°C. = 95.2 mm Hg / 760 mm/atm = .1253 atm

Solubility at 25°C =1.78 x 103 mg/l

Molecular weight (C6H6) = 78.12 g/mole = 78.12 x 103 mg/mole

S = 1.78 x 103 mg/l / 78.12 x 103 mg/mole

S = .02279 mol/l = 22.79 mol/m3

eq 3-9, p.94

H = = .1253 atm / 22.79 mol/m3

H = 5.50 x 10-3

atm.m3

/ mole

2.) H'eq 3-7, p.94H' = =

H' = .2236 (dimensionless)

3.) Calculate H using regression data from App. Aeq 3-8, p.94

H = e(A-B/T)

App. A. p. 1040A = 5.53

B = 3.19 x 103

T = 25+273.2T = 298.2 K

H = e (5.53 - (3.19 x 103)/298.2)

H = 5.70 x 10-3 atm.m3 / mole using regression data from App. AAssuming the regression H to be correct, the error in using solubility data calculated in part 1 is:

% error =

% error = .0351 = 3.51%

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C. Diffusion Coefficient

• Diffusion is the movement of contaminants in air or water from areas of highconcentration to areas of low concentration under the influence of a concentrationgradient as defined by Fick's Law:

J = -D eq.3-10 units p.95• Where tabulated D do not exist, they may be estimated on the basis of molecular weight:

D1 = D2 eq.3-11 units p.95

D. Partition Coefficients

• Partition Coefficients are empirical constants that describe how a chemical distributesitself between two media. Three coefficients are important in waste management.

• Octanol-water coefficient K ow:

K ow = Co/C eq. 3-15 units p.98

This coefficient, which ranges from 10-3

to 107

is an indication of how much of anorganic compound will partition between an organic phase and water e.g. how muchof the chemical will be taken up by aquatic organisms. Low K (<10) tend tohydrophilic and have low soil adsorption and low Bioconcentration Factor (BCFs).

• Soil water partition coefficient K p:

K p = X/C eq.3-16 units p. 98

Measures the tendency of a chemical to be adsorbed by soil or sediment.• Organic Carbon Partition Coefficient, Koc:

Koc = Cc /C eq.3-17 units p.99

Important because almost all of the adsorption of organic chemicals by a soil is due tothe organic carbon content of the soil.

E. Bioconcentration Factor (BCF)

• The amount of chemical that is likely to accumulate in aquatic organisms:BCF = Corg/C eq.3-19 units p. 99

• Example:Given: A shipment of 4,4'-DDT on its way to Egypt spilled in the river.Find: If the material becomes fully soluble in the river, what will the concentration of thechemical be in the fish, assuming they will not expire before the maximum is reached.From App. A. p. 1042

BCF = 54,000 L/kg

S = 5.50 x 10-3 mg/l = 5.50 x 10-6 µg/l (ppb)

eq 3-19, p99BCF = Corg/C

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54,000 l/kg = Corg/5.50 x 10-6 µg/l (ppb)

Corg = .297 µg/kg The chemical is easily taken up by the fish but luckily it is not very

soluble in the first place.

F. Sorption

• Sorption is the process by which a component (the sorbate or contaminant) movesfrom one phase to another across some boundary.

• In absorption the sorbate is taken into the sorbent (activated carbon).• In adsorption the process takes place at the surface.

3. Energy and Mass Balances

• This material is covered in CE431, Clean Water, and CE457, Solid Waste and in theThermodynamics course. It is imperative that the student be conversant with thismaterial.

4. Reactions and Reactors

• This material is covered in CE431, Clean Water, and CE457, Solid Waste and in theThermodynamics course. It is imperative that the student be conversant with thismaterial.

HOMEWORK Read Chapter 3, Process Fundamentals, pp.81-131Problems, p.126, 3-1, 3-2, 3-3, 3-4, 3-8, 3-10, 3-11, 3-12

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