elements of environmental engineering: thermodynamics and kinetics, third edition

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Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiAuthor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xv

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Energy, Population, and Pollution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Environmental Standards and Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 The Discipline of Environmental Engineering. . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Chemical Thermodynamics and Kinetics in Environmental

Engineering. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4.1 Applications of Thermodynamics and Kinetics . . . . . . . . . . . . . . . . 4

1.4.1.1 Equilibrium Partitioning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4.1.2 Fate and Transport Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4.1.3 Design of Separation Processes . . . . . . . . . . . . . . . . . . . . . . . 8

1.5 Units and Dimensions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.6 Structure of the Book . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Basic Chemical Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.1 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Fundamental Laws of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2.1 Zeroth Law of Thermodynamics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2.2 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2.3 Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2.4 Third Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.2.5 Enthalpy and Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.2.6 Thermodynamic Standard States: Enthalpies of Reaction,

Formation, and Combustion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.2.7 Combination of First and Second Laws . . . . . . . . . . . . . . . . . . . . . . . . 25

2.3 Chemical Equilibrium and Gibbs Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . 262.3.1 Free Energy Variation with Temperature and Pressure . . . . . . . . 26

2.4 Concept of MaximumWork . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.5 Gibbs Free Energy and Chemical Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.5.1 Gibbs–Duhem Relationship for a Single Phase . . . . . . . . . . . . . . . . 302.5.2 Standard States for Chemical Potential . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.6 Thermodynamics of Surfaces and Colloidal Systems . . . . . . . . . . . . . . . . . . 322.6.1 Surface Tension. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.6.2 Curved Interfaces and theYoung–Laplace Equation . . . . . . . . . . 332.6.3 Surface Thickness and Gibbs Dividing Surface . . . . . . . . . . . . . . . . 35

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2.6.4 Surface Thermodynamics and Gibbs Equation . . . . . . . . . . . . . . . . 362.6.5 Gibbs Adsorption Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3 Multicomponent Equilibrium Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 433.1 Ideal and Nonideal Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.1.1 Concentration Units in Environmental Engineering . . . . . . . . . . . 443.1.2 Dilute Solution Definition. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.2 Fugacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.2.1 Fugacity of Gases. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.2.2 Fugacity of Condensed Phases (Liquids and Solids) . . . . . . . . . . 463.2.3 Activities of Solutes and Activity Coefficients . . . . . . . . . . . . . . . . . 473.2.4 Ionic Strength and Activity Coefficients . . . . . . . . . . . . . . . . . . . . . . . 493.2.5 Fugacity and Environmental Models . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.3 Ideal Solutions and Dilute Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.3.1 Vapor–Liquid Equilibrium: Henry’s and Raoult’s Laws . . . . . . 55

3.3.1.1 Henry’s Law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.3.1.2 Raoult’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

3.3.2 Vapor Pressure of Organic Compounds, Clausius–ClapeyronEquation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

3.3.3 Vapor Pressure over Curved Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . 633.3.4 Liquid–Liquid Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

3.3.4.1 Octanol–Water Partition Constant . . . . . . . . . . . . . . . . . . . . 663.3.4.2 Linear Free Energy Relationships. . . . . . . . . . . . . . . . . . . . . 69

3.4 Nonideal Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 693.4.1 Activity Coefficient for Nonideal Systems . . . . . . . . . . . . . . . . . . . . . 69

3.4.1.1 Excess Functions and Activity Coefficients . . . . . . . . . . 703.4.2 Activity Coefficient and Solubility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 703.4.3 Correlations with Hydrophobicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

3.4.3.1 Special Structural Features of Water . . . . . . . . . . . . . . . . . . 733.4.3.2 Hydrophobic Hydration of Nonpolar Solutes . . . . . . . . 753.4.3.3 Hydrophobic Interactions between Solutes . . . . . . . . . . . 783.4.3.4 Hydrophilic Interactions for Solutes in Water . . . . . . . . 793.4.3.5 Molecular Theories of Solubility:

An Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803.4.4 Structure–Activity Relationships and Activity

Coefficients in Water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 823.4.5 Theoretical and Semi-Empirical Approaches to Aqueous

Solubility Prediction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 833.4.5.1 First-Generation Group Contribution Methods. . . . . . . 833.4.5.2 Excess Gibbs Free Energy Models . . . . . . . . . . . . . . . . . . . . 863.4.5.3 Second-Generation Group Contribution

Methods: The UNIFAC Method . . . . . . . . . . . . . . . . . . . . . . 863.4.6 Solubility of Inorganic Compounds in Water . . . . . . . . . . . . . . . . . . 88

3.5 Adsorption on Surfaces and Interfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

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3.5.1 Gibbs Equation for Nonionic and Ionic Systems . . . . . . . . . . . . . . 893.5.2 EquilibriumAdsorption Isotherms at Interfaces . . . . . . . . . . . . . . . 903.5.3 Adsorption at Charged Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101


4 Applications of Equilibrium Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1194.1 Air–Water Phase Equilibrium. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

4.1.1 Estimation of Henry’s Constant from GroupContributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

4.1.2 Experimental Determination of Henry’s Law Constants . . . . . . 1254.1.3 Effects of Environmental Variables on Kaw . . . . . . . . . . . . . . . . . . . . 126

4.2 Air–Water Equilibrium in Atmospheric Chemistry . . . . . . . . . . . . . . . . . . . . . 1364.2.1 Wet Deposition of Vapor Species. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1374.2.2 Wet Deposition of Aerosol-Bound Fraction . . . . . . . . . . . . . . . . . . . . 1384.2.3 Dry Deposition of Aerosol-Bound Pollutants . . . . . . . . . . . . . . . . . . 1434.2.4 Dry Deposition Flux of Gases from the Atmosphere . . . . . . . . . . 1444.2.5 Thermodynamics of Aqueous Droplets

in the Atmosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1474.2.6 Air/Water Equilibrium in Waste Treatment Systems . . . . . . . . . . 150

4.3 Soil–Water and Sediment–Water Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . 1524.3.1 Partitioning into Soils and Sediments fromWater . . . . . . . . . . . . . 1534.3.2 Adsorption of Metal Ions on Soils and Sediments . . . . . . . . . . . . . 1554.3.3 Adsorption of Organic Molecules on Soils

and Sediments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1594.4 Biota/Water Partition Constant (Bioconcentration Factor) . . . . . . . . . . . . 1644.5 Air-to-Aerosol Partition Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1674.6 Air-to-Vegetation Partition Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1714.7 Adsorption on Activated Carbon for Wastewater Treatment . . . . . . . . . . . 172Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

5 Concepts from Chemical Reaction Kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1895.1 Progress toward Equilibrium in a Chemical Reaction. . . . . . . . . . . . . . . . . . 1905.2 Reaction Rate, Order, and Rate Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1935.3 Kinetic Rate Laws. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

5.3.1 Isolation Method. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1955.3.2 Initial Rate Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1955.3.3 Integrated Rate Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

5.3.3.1 Reversible Reactions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1985.3.3.2 Series Reactions and Steady-State Approximation . . . 201

5.3.4 Parallel Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2075.4 Activation Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

5.4.1 Activated Complex Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2105.4.2 Effect of Solvent on Reaction Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2135.4.3 Linear Free Energy Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

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5.5 Reaction Mechanisms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2165.5.1 Chain Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216

5.6 Reactions in Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2195.6.1 Effects of Ionic Strength on Rate Constants . . . . . . . . . . . . . . . . . . . 219

5.7 Environmental Catalysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2215.7.1 Mechanisms and Rate Expressions for Catalyzed

Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2225.7.2 Homogeneous Catalysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2245.7.3 Heterogeneous Catalysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2285.7.4 General Mechanisms of Surface Catalysis . . . . . . . . . . . . . . . . . . . . . 2285.7.5 Autocatalysis in Environmental Reactions . . . . . . . . . . . . . . . . . . . . . 234

5.8 Redox Reactions in Environmental Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 2375.8.1 Rates of Redox Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240

5.9 Environmental Photochemical Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2445.10 Enzyme Catalysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247

5.10.1 Michaelis–Menten Kinetics and Monod Kinetics. . . . . . . . . . . . . . 247Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264

6 Applications of Chemical Kinetics in Environmental Systems . . . . . . . . . . 2676.1 Types of Reactors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

6.1.1 Ideal Reactors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2676.1.1.1 Batch Reactor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2696.1.1.2 Continuous-Flow Stirred Tank Reactor . . . . . . . . . . . . . . . 2696.1.1.3 Plug-Flow Reactor (PFR) or Tubular Reactor . . . . . . . . 2706.1.1.4 Design Equations for CSTR and PFR . . . . . . . . . . . . . . . . 2716.1.1.5 Relationship between Steady State

and Equilibrium for a CSTR . . . . . . . . . . . . . . . . . . . . . . . . . . 2766.1.2 Nonideal Reactors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

6.1.2.1 Dispersion Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2776.1.2.2 Tanks-in-Series Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

6.1.3 Dispersion and Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2806.1.4 Reaction in a Heterogeneous Medium. . . . . . . . . . . . . . . . . . . . . . . . . . 281

6.1.4.1 Kinetics and Transport atFluid–Fluid Interfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

6.1.5 Diffusion and Reaction in a Porous Medium. . . . . . . . . . . . . . . . . . . 2866.2 The Water Environment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290

6.2.1 Fate and Transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2906.2.1.1 Chemicals in Lakes and Oceans. . . . . . . . . . . . . . . . . . . . . . . 2906.2.1.2 Chemicals in Surface Waters . . . . . . . . . . . . . . . . . . . . . . . . . . 2936.2.1.3 Biochemical Oxygen Demand in

Natural Streams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2956.2.2 Water Pollution Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299

6.2.2.1 Air Stripping in Aeration Basins . . . . . . . . . . . . . . . . . . . . . . 2996.2.2.2 Oxidation Reactor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303

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6.2.2.3 Photochemical Reactions and WastewaterTreatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309

6.2.2.4 Photochemical Reactions in Natural Waters . . . . . . . . . . 3116.3 The Air Environment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313

6.3.1 F&T Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3136.3.1.1 Box Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3136.3.1.2 Dispersion Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319

6.3.2 Air Pollution Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3246.3.2.1 Adsorption. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3246.3.2.2 Thermal Destruction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327

6.3.3 Atmospheric Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3306.3.3.1 Reactions in Aqueous Droplets . . . . . . . . . . . . . . . . . . . . . . . 3306.3.3.2 Global Warming and Greenhouse Effect . . . . . . . . . . . . . 3376.3.3.3 Ozone in the Stratosphere and Troposphere . . . . . . . . . . 347

6.4 Soil and Sediment Environments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3576.4.1 F&T Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357

6.4.1.1 Transport in Groundwater . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3586.4.1.2 Sediment–Water Exchange of Chemicals . . . . . . . . . . . . 3646.4.1.3 Soil–Air Exchange of Chemicals . . . . . . . . . . . . . . . . . . . . . 366

6.4.2 Soil and Groundwater Treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3696.4.2.1 P&T for NAPL Removal from Groundwater . . . . . . . . . 3706.4.2.2 In Situ Soil Vapor Stripping in the Vadose Zone . . . . . 3736.4.2.3 Incineration for ex Situ Treatment of Soils

and Solid Waste . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3756.5 Applications of Chemical Kinetics in Environmental

Bioengineering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3766.5.1 Enzyme Reactors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380

6.5.1.1 Batch Reactor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3816.5.1.2 Plug-Flow Enzyme Reactor . . . . . . . . . . . . . . . . . . . . . . . . . . . 3826.5.1.3 Continuous Stirred Tank Enzyme Reactor . . . . . . . . . . . . 3836.5.1.4 Immobilized Enzyme or Cell Reactor . . . . . . . . . . . . . . . . 3876.5.1.5 In Situ Subsoil Bioremediation. . . . . . . . . . . . . . . . . . . . . . . . 391

6.5.2 Kinetics of Bioaccumulation of Chemicals in the AquaticFood Chain. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393

6.6 Applications in Green Engineering. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3986.6.1 Environmental Impact Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3996.6.2 Life Cycle Assessment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405


Appendix 1 Properties of Selected Chemicals of EnvironmentalSignificance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425

Appendix 2 Standard Free Energy, Enthalpy, and Entropy ofFormation for Compounds of EnvironmentalSignificance. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431

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x Contents

Appendix 3 Selected Fragment (bj) and Structural Factors (Bk) forOctanol–Water Partition Constant Estimation. . . . . . . . . . . . . . . . 433

Appendix 4 Concentration Units for Compartments inEnvironmental Engineering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435

Appendix 5 Dissociation Constants for Environmentally SignificantAcids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437

Appendix 6 Bond Contributions to Log Kaw for the Meylan andHoward Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439

Appendix 7 Regression Analysis (the Linear Least-SquaresMethodology). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441

Appendix 8 Error Function and Complementary Error FunctionDefinitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445

Appendix 9 Cancer Slope Factor and Inhalation Unit Risk forSelected Carcinogens . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447

Appendix 10 U.S. National Ambient Air Quality Standards . . . . . . . . . . . . . . . 449

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451

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Preface

Energy and environmental concerns are the most important issues for the currentgeneration.These problems are related to the increased demand for available resourcesbut human beings have become adept at meeting these challenges. Unfortunately, ourattempts to resolve our problems also contribute to the degradation of the environment.The consequences of environmental pollution can be disastrous, with many examplesthat demonstrate the severity of the problem. The present-day debate about globalwarming, climate change, water quality, air quality, and solid waste disposal areexamples.

Our prescription for alleviating environmental problems requires that we train stu-dents in the science and engineering of environmental issues. A well-trained cadre ofprofessionals is required to make sure that sound science is involved in the formula-tion of public policy. To this end, the present-day environmental professional needsbroad interdisciplinary training across the disciplines of physics, chemistry, biology,and engineering.

Environmental engineering is interdisciplinary. It is a challenge to develop coursesthat will provide students with a thorough, broad-based curriculum that includesevery aspect of the environmental engineering profession. Traditionally, environmen-tal engineering has been a subdiscipline of civil engineering with primary emphasison municipal wastewater treatment, sewage treatment, and landfill management prac-tices. During the second half of the twentieth century more emphasis was placed onend-of-pipe treatment in manufacturing operations to control release of water, air,and solid wastes. With the realization that environmental problems are not confinedto end-of-pipe treatment, the attention turned toward pollutant fate and transport inthe general environment, waste minimization, pollution prevention, and green engi-neering. Thus stand-alone programs in environmental engineering began to appear inmany universities.

Environmental engineers perform a variety of functions, the most critical of whichare process design for waste treatment or pollution prevention, fate and transportmodeling, green engineering, and risk assessment. Physical chemistry is an essentialelement of environmental engineering. In particular, chemical thermodynamics andchemical kinetics, the two main pillars of physical chemistry, are crucial to environ-mental engineering. Unfortunately, these topics are not covered in the environmentalengineering curricula inmost universities.Chemical engineers take separate courses inthermodynamics and kinetics. They also take several prerequisites in chemistry, suchas physical chemistry, organic chemistry, inorganic chemistry, and analytical chem-istry. Most environmental engineering programs do not require such a broad spectrumof chemistry courses, but rely on an introductory chemistry course for environmen-tal engineers. However, such a course lacks the proper depth in the two importantaspects of physical chemistry. I believe that an additional course should be taught that

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xii Preface

introduces these subjects and lays the foundation for more advanced courses in envi-ronmental process design, environmental transport modeling, green engineering, andrisk assessment. To accomplish this objective I wrote the first edition of this textbookin 1995. It is based on a course entitled “Chemical Thermodynamics and Kineticsfor Environmental Processes” that I have been offering to environmental engineeringmajors and as a single semester required course for theAccreditation Board for Engi-neering and Technology (ABET)–accredited BS degree program in environmentalengineering. It is also taught in the chemical engineering department at LouisianaState University (LSU). Students taking this course also take a fundamental ther-modynamics course in the mechanical engineering department at LSU, which dealsprimarily with the laws of thermodynamics, concepts of heat engines, and mechan-ical applications of thermodynamics. They also take an introductory environmentalengineering course preceding this.

This is an undergraduate textbook, but portions of it are also suitable for anintroductory graduate-level course. A basic understanding of physics, chemistry, andmathematics (especially differential calculus) is assumed. In writing this third edition,I have considered the comments of several instructors who have used the past editionsof the book. In response to their requests, I have eliminated redundant material andrefocused the chapters. Sections such as green engineering, biological processes, lifecycle analysis, etc., have been added. I have provided a larger number of examplesand problems in this edition. The examples are chosen to represent important appli-cations, but, since the choice is subjective, I do admit that some may be more relevantthan others. The problems are of varying levels of difficulty and they are ranked 1,2, and 3, with 1 indicating the “least difficult” and 3 indicating the “most difficult”or “advanced.” These are represented by subscripts beside the problem number. In atypical single semester course, I follow the general outline as follows:

Lecture Number Sections Covered

2 lectures 1.1–1.63 lectures 2.1–2.5, 2.6.19 lectures 3.2.1, 3.2.2, 3.2.5, 3.3.1, 3.3.2, 3.3.4.1, 3.4.1, 3.4.2, 3.5.1, 3.5.29 lectures 4.1.1–4.1.3, 4.2.1–4.2.4, 4.2.6, 4.3.1, 4.3.3, 4.4–4.66 lectures 5.2, 5.3.3, 5.4.1, 5.5.1, 5.7.1, 5.9, 5.10.114 lectures 6.1.1, 6.2.1.1–6.2.1.3, 6.2.2.2, 6.3.1.1, 6.3.3.2, 6.3.3.3, 6.4.1.2,

6.5.1.1–6.5.1.3, 6.6.2

In writing this book, I have received help and encouragement from a number ofpeople. Special thanks to Louis Thibodeaux, a friend and colleague of mine at LSU,with whom I have collaborated for over two decades in environmental engineeringresearch. I extend special thanks to David J. Wilson for being a great role model asan educator and researcher, especially for his infectious enthusiasm for teaching. Mypresent and past graduate students, in particular, Nick Ashley and R. Ravikrishna,have contributed in many ways by working out problems and pointing out errors inthe previous editions.

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Preface xiii

Thisworkwould not have been possiblewithout the unconditional love and supportfrom my family (my wife, Nisha, and my two children, Viveca and Vinay).

This book is dedicated to my beloved father, who was always there for me when Ineeded him, but did not live to see his son’s achievements, and my dear mother, whois a source of inspiration to me in my life.

Kalliat T. Valsaraj

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Author

Kalliat T. Valsaraj is the Charles and Hilda Roddeydistinguished professor and the Ike East professorin chemical engineering in the Cain Department ofChemical Engineering at Louisiana State UniversityinBatonRouge.He is currently the chair of the depart-ment of chemical engineering. He received his MScin chemistry from the Indian Institute of Technology,Madras, in 1980, and his PhD in physical chemistryfrom Vanderbilt University, Nashville, TN, in 1983.His research encompasses various areas of environ-mental, chemical, and materials engineering. He isthe author of 152 peer-reviewed journal articles, 4books, 25 book chapters, 47 reports and monographs,and 2 patents. His research has been supported by

federal agencies (the National Science Foundation, the Department of Defense, theDepartment of Energy, the Environmental ProtectionAgency, and the Department ofInterior), state agencies, and industry. He is a Fellow of the American Institute ofChemical Engineers. He is also a member of the American Chemical Society, theAmerican Association for the Advancement of Science, the American Society forEngineering Education, the Association for Environmental Engineering and ScienceProfessors, Sigma Xi, the Air and Waste Management Association, and the Soci-ety for Environmental Toxicology and Chemistry. He serves on the editorial reviewboards of several journals: Environmental Toxicology and Chemistry, Journal of theAir andWaste Management Association, Environmental Monitoring and Assessment,and Scientific Journals International (Chemical and Bioengineering Division).

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1 Introduction

Pollution is an inevitable consequence of advances in human endeavors to improvethe quality of life on this planet. As civilized societies learned to organize, humansrealized methods to purify water for drinking purposes, dispose of excrement, buildsanitary sewers, and perform municipal wastewater treatment to prevent communica-ble diseases. They also realized how air pollution can adversely affect human healthand the necessity to control the same. Thus, the history of pollution is as old as thehuman species itself.

The twentieth century has been a period of rapid technological advances, whichhas helped us to harness the available natural resources. Along with these advance-ments, we have also created myriad environmental pollution problems. Pollution isundesirable and expensive, but it is an inevitable consequence of modern life. Thereality is that we cannot eliminate pollution altogether, but we can certainly mitigateit through recycle, reuse, and reclamation.

1.1 ENERGY, POPULATION, AND POLLUTION

Energy consumption has increased dramatically as the population has exploded.Increased utilization of natural resources is necessary to sustain the various industriesthat drive the economy of industrialized nations. Unfortunately, increased industrialactivity has also produced anthropogenic pollutants. The harnessing of nuclear powerhas left us the legacy of radioactive waste. Increased agricultural activity in bothdeveloped and developing nations has been necessitated in order to sustain the bour-geoning world population. Intensive agricultural uses of pesticides and herbicideshave contributed to the pollution of our environment. Someof themajor environmentalproblems that one can cite as examples of anthropogenic origin are (a) increased car-bon dioxide and other greenhouse gases in the atmosphere, (b) depletion of the earth’sprotective ozone layer due to man-made chlorofluorocarbons, (c) acid rain due toincreased sulfur dioxide as a result of fossil fuel utilization, (d) atmospheric haze andsmog, polluted lakes, waterways, rivers, and coastal sediments, (e) contaminatedgroundwater, and (f) industrial and municipal wastes.

The environmental stress (impact) due to the needs of the population and the betterstandard of living is given by the master equation (Graedel and Allenby, 1996)

EI = P × GDP

Person× EI

GDPunit per capita, (1.1)

where EI is the environmental impact, P is the population, and GDP is the grossdomestic product.P in the above equation is unarguably increasingwith time and that too in geometric

progression. The second term on the right-hand side denotes the general aspiration

1

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2 Elements of Environmental Engineering: Thermodynamics and Kinetics

of humans for a better life, and is generally increasing as well. The third term on theright-hand side denotes the extent to which technological advances can be sustainedwithout serious environmental consequences. The third term can be made as smallas possible to limit the overall environmental impact and enable the transition to asustainable environment. Both societal and economic issues are pivotal in determiningwhether we can sustain the quality of life while at the same time mitigating theenvironmental consequences of the technologies that we adapt.

The quality of our life is inextricably linked to industrial growth and improvementsin agricultural practices. Chemicals are used in both sectors, to sustain innovations inthe industrial sector and to improve agricultural efforts at maintaining a high rate ofcrop production. Thus, the chemical manufacturing industry has been at the forefrontof both productivity and growth. During the twentieth century, the chemical manufac-turing industry has shown phenomenal growth. Overwhelming arrays of chemicalshave been produced each year. Over the whole period of human history, one estimatesuggests that approximately six million chemical compounds have been created, ofwhich only 1% are in commercial use today.Many of the compounds are important forsustaining and improving our health and well-being, since they are starting materialsfor various products that we use everyday.

1.2 ENVIRONMENTAL STANDARDS AND CRITERIA

There exist several Congressional statutes in the United States that are instrumentalin setting standards for drinking water, ambient water quality, and ambient air quality.

The Environmental Protection Agency (EPA) takes several steps before a stan-dard is set for a specific compound in a specific environment. These include factorssuch as the following: occurrence in the environment; human exposure and risks ofadverse health effects in the general population and sensitive sub-populations; ana-lytical methods of detection; technical feasibility; and impacts of regulation of watersystems, the economy, and public health.

Environmental quality standards refer to maximum contaminant concentrationsallowed for compounds in different environmental media. These concentrations areexpected to be protective of human health and useful for ecological risk management.It has long been recognized that a realistic assessment of the effects of chemicalson humans and ecosystems is mandatory for setting environmental quality standards.The implication is that our concerns for the environment should be driven by soundscience. A prudent policy with regard to environmental regulations should con-sider the weight of evidence in favor of acceptable risk against potential benefits.In fact, risk assessment should be the paradigm that is the basis for current and futureenvironmental legislation.

1.3 THE DISCIPLINE OF ENVIRONMENTAL ENGINEERING

Environmental awareness is the first step in understanding pollution problems. On theone hand, we should better understand the potential impacts, so that we can focus ourresources on themost serious problems. On the other hand, we should also understand

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Introduction 3

the technologies better, so that we can improve their effectiveness and reduce theircosts. These activities are within the discipline of environmental engineering.Environmental engineering is the study of the fate, transport, and effects of

chemicals in the natural and engineered environments and includes the formulationof options for treatment, mitigation, and prevention of pollution in both natural andengineered systems.

It has only been in the latter half of the twentieth century that environmental engi-neering grew to a mature field with depth and focus. Environmental engineering isan interdisciplinary field. It involves the applications of fundamental sciences, thatis, chemistry, physics, mathematics, and biology, to waste treatment, environmentalfate and transport of chemicals, and pollution prevention. The main pillars thatsupport an environmental engineering curriculum are physics (statics and fluids),chemistry (organic, inorganic, kinetics, thermodynamics, andmaterial and energy bal-ances), mathematics (calculus and algebra), and biology (toxicology, microbiology,and biochemistry)—see Figure 1.1. Statics, fluids, and energy and material balancesare prerequisites for several engineering disciplines (chemical, civil, petroleum, andbiological). For environmental engineering, two main pillars are from chemistry(chemical thermodynamics and kinetics). Chemical thermodynamics is that branch ofchemistry that deals with the study of the physico-chemical properties of a compoundin the three states of matter (solid, liquid, and gas). It also describes the potential fora compound to move between the various phases and the final equilibrium distribu-tion in the different phases. Chemical kinetics, on the other hand, describes the rateof movement between phases and also the rate at which a compound reacts withina phase.

Environmental engineering practice(Process design, chemical fate modeling, risk analysis, economics)

Chemistry Physics Mathematics Biology

Water treatment, air pollution, hazardous waste treatment, environmental fate and transport,

hydrology, geosciences, ecology, microbiology

Organic,kinetics,thermodynamics

Statics,fluids

Calculus

Toxicology,microbiology,biochemistry

FIGURE 1.1 Pillars of environmental engineering.

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1.4 CHEMICAL THERMODYNAMICS AND KINETICSIN ENVIRONMENTAL ENGINEERING

The environment may be conceptualized as consisting of various compartments.Table 1.1 summarizes the total mass and surface areas of various compartments inthe natural environment. The transport of materials between the compartments ona global scale depends only on forces of global nature. However, on a local scalethe partitioning and transport depend on the composition, pressure, temperature, andother variables in each compartment.

There are four distinct environmental compartments: hydrosphere, atmosphere,lithosphere, and biosphere (Figure 1.2). These may be in continuous contact with asharp boundary between them (air–water) or may be discontinuous (e.g., soil–water).In some cases, one phase will be dispersed in another (e.g., air bubbles in water,fog droplets in air, aerosols and dust particles in air, colloids suspended in water,oil droplets in water, and soap bubbles in water). Some compartments may havethe same chemical composition throughout, but differ significantly in their spatialcharacteristics (e.g., the lower troposphere versus the upper stratosphere, a stratifieddeep water body, or a highly stratified atmosphere). The biosphere, which includesall plant and animal species, is in contact with the three other compartments in theoverall scheme. Reactions and transformations occur in each phase, and the rate ofexchange of mass and energy between the compartments is a function of the extentto which the respective compartments are in nonequilibrium.

1.4.1 APPLICATIONS OF THERMODYNAMICS AND KINETICS

1.4.1.1 Equilibrium Partitioning

Chemical thermodynamics is central to the application of the equilibrium-partitioningconcept to estimate pollutant levels in environmental compartments. It assumes thatthe environmental compartments are in a state where they have reached constant

TABLE 1.1Composition of Natural Environment

Compartment Value

Air (atmosphere) (mass) 5.1 × 1018 kgWater (hydrosphere) (mass) 1.7 × 1021 kgLand (lithosphere) (mass) 1.8 × 1021 kgLand (area) 1.5 × 1014 m2

Water (area) 3.6 × 1014 m2

Source: Weast, R.C. and Astle, M.J. (Eds). 1981. CRCHandbook of Chemistry and Physics, 62nd ed.Boca Raton, FL: CRC Press, Inc.; Stumm, W.and Morgan, J.J. 1981. Aquatic Chemistry, 2nded. NewYork, NY: John Wiley & Sons, Inc.

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Introduction 5

Atmosphere

HydrosphereLithosphere

Biosphere

FIGURE 1.2 Equilibrium between various environmental compartments, reactions withincompartments, and material exchange between the compartments.

chemical composition, temperature, and pressure and have no tendency to changetheir state. Equilibrium models only give us the chemical composition within theindividual compartments and not how fast the system reached the given equilibriumstate. Although, admittedly, a time-variant (kinetic) model may have significantadvantages, there is general agreement that equilibrium partitioning is a startingpoint in this exercise. It is a fact that true equilibrium does not exist in the envi-ronment. In fact, much of what we observe in the environment occurs as a resultof the lack of equilibrium between compartments. Every system in the environmentstrives toward equilibrium as its ultimate state. Hence, the study of equilibrium is afirst approximation toward the final state of any environmental system.

1.4.1.2 Fate and Transport Modeling

The environment is a continuum in that as pollutants interact with various phases,they undergo both physical and chemical changes and are finally incorporated intothe environment. Fate models, based on the mass balance principle, are necessaryto simulate the transport between and transformations within various environmentalmedia. This is called the multimedia approach. The mass balance principle can beapplied through the use of a multimedia fate and transport model to obtain the ratesof emissions and the relative concentrations in each compartment. A number of suchmodels already exist, some of which are described in Table 1.2.

Consider a chemical that is released from a source to one of the environmentalcompartments (air,water, or soil).To assess the effect of the pollutant to the ecosystem,we need to first identify the various pathways of exposure. Figure 1.3 illustratesthe three primary pathways that are responsible for exposure from an accidentalrelease. Direct exposure routes are through inhalation from air and drinking water

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TABLE 1.2Descriptions of a Few Multimedia Fate and Transport Models that areCurrently in Use

Model Acronym Description

CalTOX A fugacity-based model to assist the California Environmental ProtectionAgency to estimate chemical fate and human exposure in the vicinity ofhazardous waste sites

ChemCAN A steady-state fugacity-based model developed for Health Canada topredict a chemical’s fate in any of the 24 regions of Canada

HAZCHEM A fugacity-based model developed as a regional scale model for theEuropean Union Member States

SimpleBOX Developed by RIVM, the Netherlands, it uses the classical concentrationconcept to compute mass balances

Source: The multi-media fate model: A vital tool for predicting the fate of chemicals, SETAC Press,1995.

from the groundwater aquifer. Indirect exposure results, for example, from ingestionof contaminated fish from a lake that has received a pollutant discharge via the soilpathway. Coupling the toxicology with a multimedia fate and transport model thusprovides a powerful tool to estimate the risk potential for both humans and the biota.Risk assessment is typically divided into the assessment of cancer and noncancerhealth effects. Animal toxicology studies or human epidemiological data are used toestablish a unit risk level for cancer.A dose–response model then relates the responseto the dosage to which the subjects are exposed. Most cancer dose–response modelsconservatively assume a no-threshold model, so that the risk is extrapolated from highto low dosages assuming that there is some response at any dosage above zero.

Air pathway

Soil pathwayReceptorsLake

Groundwater pathway

Source

FIGURE 1.3 Environmental risks and exposure pathways.

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Introduction 7

The risk alluded to is the incremental lifetime cancer risk (ILCR), and is given by

ILCR = SF × CDI, (1.2)

where SF is the slope factor for the chemical and CDI is the chronic daily intakeof the chemical by the defined exposure route. A listing of the SFs can be foundin Appendix 9. The CDI requires the use of a multimedia fate model to obtain theconcentration value that goes into its determination. Thus, knowledge of the SF andthe CDI allows the determination of the lifetime cancer risk.

If the excess cancer risk from the inhalation route is to be estimated, we can alsowrite the following equation:

Riskair = IUR × Cair, (1.3)

where IUR is the inhalation unit risk (in per μg/m3) (see Appendix 9) and Cair is theaverage exposure concentration in air. For example, for formaldehyde the unit riskis 1.3 × 10−5 μg−1/m3, and if someone is exposed to a concentration of 0.77μg/m3

over a lifetime (70 years), the lifetime risk is 1 in 105. For noncancer risk assessmentin air, we use a hazard quotient (HQ):

HQ = Cair

RfC, (1.4)

where RfC is the reference concentration (see Appendix 9 for definitions). If HQ is<1, the risk is acceptable.

Within the regulatory framework, it is nowmandatory to assess the potential harm-ful effects on humans and the environment from the use of new chemicals and fromthe continued use of existing ones. Examples are the Toxic Substances Control Act(TSCA) in the United States, the Canadian Environmental Protection Act (CEPA) inCanada, and the 7th amendment in the European Union (EU). Once risk is estab-lished, the next step will be to isolate the pollutant from the ecosystem to minimizethe risk.

In the above context, the following specific questions will need to be addressed:

(i) What is the final equilibrium state of the pollutant in the environment, that is,which of the environmental compartments is the most favorable, how muchresides in each compartment at equilibrium, and what chemical propertiesare important in determining the distribution?

(ii) How fast does the pollutant move from one compartment to another, whatis the residence time in each compartment, and how fast does it react withineach compartment or at the boundary between compartments?

The answer to the first question employs the tools of chemical thermodynamicsand the second question requires the applications of concepts from chemical kinetics.

Multimedia fate and transport (F&T) models are recognized as a necessary com-ponent for risk assessment, chemical ranking, management of hazardous waste sites,

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optimization of testing and monitoring strategies, determination of global disper-sion, and recovery times. To construct F&T models, one needs to obtain a range ofphysico-chemical parameters: thermodynamic, kinetic, and toxicological.

1.4.1.3 Design of Separation Processes

Environmental engineers design separation processes for the isolation of contaminantsfrom waste streams before they are discharged to the environment. Both physical andchemical separation processes are used in environmental engineering. For example,particulate separation from air and water involve physical separation techniques thatuse mechanisms such as aggregation, coagulation, impaction, centrifugal force, andelectromotive force, to name a few. The removal of dissolved gases and vapors fromair and water involve, on the other hand, chemical separation methods. Whereasmixing of chemicals to form a mixture is a spontaneous process, and, as we will seein Chapter 2, a thermodynamically favorable process, the reverse, namely, separationinto the component species, requires the expenditure of work. The overall objectiveof an environmental separation process is not only to isolate the pollutant, but also torecycle and reuse where possible the materials separated and separation agents thatwere used during the operation.

Invariably, environmental separation processes involve contact between two ormore phases and the exchange of material and energy between them. As an example,consider the removal of organic contaminants from water by contacting with a solidphase such as powdered activated carbon. The process has two distinct stages. In

PF PF

CFCFPR PR

(b)(a)

PF

CF

(c)

(d) (e)PF PF

CF CF

PRPR

FIGURE 1.4 General separation techniques. CF, contaminated feed, PF, pollutant-freestream, PR, pollutant-rich stream. (a) Separation by phase creation; (b) separation by phaseaddition; (c) separation by barrier; (d) separation by solid agent; and (e) separation by fieldgradient. (Modified from Seader, E.D. and Henley, E.J. 1986. Separation Process Principles.NewYork: John Wiley & Sons, Inc.)

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Introduction 9

stage I, those contaminants in water with a greater affinity for the carbon are concen-trated on the carbon by a process called adsorption. The removal depends not onlyon the rate of transport of compounds from water to carbon (the realm of kinetics),but also on the ultimate capacity of the carbon bed (the realm of thermodynamics). Instage II, we have to regenerate the medium within the reactor so that it can be reused.Moreover, in stage II, we need information on how fast the pollutant can be recov-ered from the activated carbon, and also the fate and transformation of the adsorbedpollutant (that is, the kinetic aspects).

TABLE 1.3Applications of Chemical Thermodynamics and Kinetics in EnvironmentalProcesses

Typical Equilibrium ThermodynamicProcess Representation Property Kinetic Property

Solubility in water A (pure) �A (water) Saturation solubility, C∗i Dissolution rate

Absorption inwater

A (air) �A (water) Henry’s law constant, KH Absorption rate

Precipitation fromwater

A (water) �A (crystal) Solubility product, Ksp Precipitation rate

Volatilization fromwater

A (water) �A (air) Henry’s law constant, KH Volatilization rate

Evaporation frompure liquid

A (pure) �A (vapor) Vapor pressure, P∗i Evaporation rate

Acid/basedissociation

A �A− + H+ Acidity or basicityconstant, Ka or Kb

Acidification rate

Ion exchange A+ + BX � BA + X+ Ion-exchange partitionconstant, Kexc

Ion-exchange rate

Oxidation/reduction Aox + Bred � Ared + Box Equilibrium constant, Ki Redox reaction rate

Adsorption fromwater

A (water) �A (surface) Soil/water partitionconstant, KSW

Adsorption rate

Adsorption fromair

A (air) �A (surface) Particle/air partitionconstant, KAP

Adsorption rate

Uptake by biota A (water) �A (biota) Bio-concentration factor,KBW

Rate of uptake

Uptake by plants A (air) �A (plant) Plant/air partitionconstant, KPA

Rate of uptake

Chemical reaction A+B � Products Equilibrium constant, Keq Rate of chemicalreaction

Photochemicalreaction

A+ (hν) � Products Equilibrium constant, Keq Rate of photolysis

Biodegradationreaction

A+ (enzymes) � Products Equilibrium constant, Keq Michaelis–Mentenand Monodkinetics constants

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Separation techniques can be classified broadly into five categories as shown inFigure 1.4 (Seader and Henley, 1998). In every one of these processes, the rate ofseparation is dependent on selectively increasing the rate of diffusion of the contami-nant species relative to the transfer of all other species by advection (bulk movement)within the contaminated feed. Equilibrium limits the ultimate compositions of theeffluent streams (pollutant-rich and pollutant-free). The rate of separation within thereactor is determined by the mass transfer limitations (“driving force”), whereas theextent of separation is determined by equilibrium thermodynamic factors. Both ther-modynamic and kinetic (transport) properties are thus instrumental in environmentalseparations.

The above discussions of applications in environmental engineering show thatfor both natural and engineered systems, two types of issues are paramount—anequilibrium study to describe the final distribution of pollutants within different com-partments and a kinetic study to describe the rate of transformations of chemicalswithin each compartment and the rate of movement of chemicals between compart-ments. A list of processes common to both natural and engineered systems is givenin Table 1.3. It summarizes the associated thermodynamic and kinetic properties thatare necessary for understanding each process.

In summary, we can look at chemical thermodynamics in environmental engineer-ing as the essential element by which we connect observations on simplified systemsto the segment of the environment (natural or engineered) that we are interestedin. To put it in context, Figure 1.5 appropriately places the role of thermodynam-ics in our endeavor to understand and design processes in natural and engineeredsystems.

Modelparameters

Observations(experimental

data onsimplified systems)

Thermodynamic models(chemical properties)Mathematical models

(dynamics, mass balances)

System(chemical plant,environmentalcompartment)

ReproductionEvaluation

FeedbackAdjust

DesignOptimizePredictControl

Regression

FIGURE 1.5 Role of chemical thermodynamics in understanding natural and engineeredsystems. (Modified from http://www.codata.org/codata02/04physci/rarey.pdf.)

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Introduction 11

1.5 UNITS AND DIMENSIONS

The SI (International System of Units) is the standard set of units for engineering andscience. The International Union of Pure and Applied Chemistry (IUPAC) produceda document (Cohen et al., 2007) entitled Quantities, Units and Symbols in Physi-cal Chemistry, which recommended a uniform set of units for all measurements inphysical chemistry. This document should be consulted for a more elaborate discus-sion of the various units. However, in a number of cases, environmental engineersstill prefer to use the CGS (centimeter–gram–second) system of units. Table 1.4 givesthe seven base quantities and their symbols that SI units are based on. All otherphysical quantities are called derived quantities and can be algebraically derivedfrom the seven base quantities by multiplication or division.

TABLE 1.4Base Quantities, Units, and Symbols in SI Units

Physical Quantity Name Symbol

Length Meter mMass Kilogram kgTime Second sElectric current Ampere AThermodynamic temperature Kelvin KAmount of substance Mole molLuminous intensity Candela cd

Source: Mills, I., Cvitas, T., Homann, K., Kallay, N., andKuchitsu, K. (Eds). 1988.Quantities, Units and Symbolsin Physical Chemistry, IUPACPhysical Chemistry Divi-sion. Oxford: Blackwell Scientific Publications.

TABLE 1.5Relations between SI and CGS Units for Some Derived Quantities

Derived Quantity Unit CGS Symbol Equivalent SI Unit

Force Dyne dyn 105 newtons (N)Pressure Bar bar 105 pascals (Pa)

Atmosphere atm 101,325 PaTorr torr 133.32 PaMillimeter of mercury mm Hg 133.32 PaPounds per square inch psi 6.89 × 103 Pa

Energy, work, heat Erg erg 10−7 joules (J)Calories cal 4.184 JLiter atmospheres L atm 101.325 J

Concentration Molar (mol/L) M 103 mol/m3

1mol/dm3

Viscosity Centipoise cP 10−3 kg/m s

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The physical quantity amount of substance is of paramount importance to environ-mental engineers. The SI unit for this quantity is the mole defined as the amount ofsubstance of a system that contains as many elementary entities as there are atoms in0.012 kg of carbon-12. IUPAC recommends that we should refrain from calling it the“number of moles.” Much of the published literature is based on the more familiarCGS units. The relations between the common CGS and SI units for some importantderived quantities of interest in environmental engineering are given in Table 1.5.

A description of the most common units in environmental engineering is givenin Appendix 4 and the reader should familiarize himself or herself with these beforeproceeding further.

1.6 STRUCTURE OF THE BOOK

The book is divided into six chapters. It can be broadly divided into two sections. Thefirst section (Chapters 2–4) is on chemical thermodynamics whereas the second sec-tion (Chapters 5 and 6) is on chemical reaction kinetics. Chapter 2 is an introduction tothe thermodynamics of homogeneous phases composed of single or multiple species.It also introduces the important concepts of free energy and chemical potential thatare of paramount importance in dealing with equilibrium systems in environmen-tal engineering. A concise description of surface thermodynamics is also included inChapter 2. Chapter 3 is an extension of the thermodynamics of homogeneous systemsto heterogeneous and multicomponent systems. The important concepts of activityand fugacity and nonideal solutions and gases are dealt within this chapter. Chapter 4deals with the applications of the concepts developed in Chapters 2 and 3 on air–water,soil–water, and air–soil equilibria to illustrate the concept of equilibrium partitioningbetween compartments in environmental engineering. Applications of equilibriumthermodynamics in waste treatment operations are also described. Chapter 5 gives ashort summary of the essential aspects of chemical reaction kinetics. Concepts suchas reaction rates and activation energies are introduced and discussed. The conceptsdeveloped in Chapter 5 are used to illustrate the applications of chemical kineticsin environmental waste treatment processes and biological systems in Chapter 6.Applications of reactor models and transport theory are exemplified in Chapter 6.

REFERENCES

Cohen, E.R. et al. 2007. Quantities, Units and Symbols in Physical Chemistry, 3rd ed.Cambridge, UK: The Royal Society of Chemistry (for IUPAC).

Graedel, T.E. and Allenby, B.R. 1996. Design for the Environment. Upple Saddle River, NewJersey: Prentice Hall.

Mills, I., Cvitas,T., Homann,K.,Kallay,N., andKuchitsu,K. (Eds). 1988.Quantities,Units andSymbols in Physical Chemistry, IUPAC Physical Chemistry Division. Oxford: BlackwellScientific Publications.

Seader, J.D. and Henley, E.J. 1998. Separation Process Principles. NewYork, NY: JohnWiley& Sons, Inc.

Stumm, W. and Morgan, J.J. 1981. Aquatic Chemistry, 2nd ed. NewYork, NY: John Wiley &Sons, Inc.

Weast, R.C. and Astle, M.J. (Eds). 1981. CRC Handbook of Chemistry and Physics, 62nd ed.Boca Raton, FL: CRC Press, Inc.

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2 Basic ChemicalThermodynamics

The discussion inChapter 1 indicated that knowledge of equilibriumbetween differentcompartments is important in environmental engineering. The fundamental principlesof thermodynamics are germane to the understanding of equilibrium in environmentalsystems. There are four laws of thermodynamics that epitomize the entire subject ofthermodynamics.

In this chapter the discussion of thermodynamics is limited to its basic laws, sincea number of excellent references are available elsewhere (e.g., Lewis and Randall,1961; Denbigh, 1981). Phase equilibrium, as it relates to bulk phases (single com-ponent, homogeneous), will be discussed in this chapter, followed by an expandeddiscussion of multicomponent and heterogeneous systems in Chapter 3. Since mostof the applications of thermodynamics in environmental engineering are confined toa narrow range of temperatures (∼ −40◦C to +40◦C) and atmospheric pressure, weneed not focus on extreme temperatures or pressures. A brief review of the conceptof equilibrium and the fundamental laws of thermodynamics is presented, followedby the introduction of the concepts of free energy, chemical potential, and the Gibbs–Duhem relationship. The effect of surface area on the total thermodynamic propertyof a system is negligible except when the subdivision in phases is exceedingly small,which occurs in many environmental systems. Therefore, an introductory discussionof surface thermodynamics is also included.

2.1 EQUILIBRIUM

The natural environment is a complex system. However, an environmental system canbe assessed by assuming that equilibrium exists between the different environmentalphases and that the properties are time-invariant. Equilibrium models are easy toapply since they need only a few inputs. They are sometimes grossly inappropriateand have to be replaced with kinetic models, which assume time-variant propertiesfor the phases. Kinetic models are complex and require a number of input parametersthat are poorly understood and sometimes unavailable for environmental situations.

Before beginning a discussion of the laws of thermodynamics, we need to definesome terms. To quote Lewis and Randall (1961), “Whatever part of the objectiveworld is the subject of thermodynamic discourse is customarily called a system.” Thesystemmay be separated from the surroundings by a physical boundary like the wallsof a container, or maybe concrete.A systemmay be closed if there is no mass transfereither to or from the surroundings, or open if either matter or energy can transferto and from the surroundings. For example, the Earth is an open system in contactwith the atmosphere around it with which it exchanges both matter and energy. Asystem is said to be isolated if it cannot exchange either mass or energy with its

13

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surroundings. The universe is considered an isolated system. A system is said to beadiabatic if it cannot exchange heat with its surroundings. A system is composed ofspatially uniform entities called phases. Air, water, and soil are three environmentalphases.The properties that characterize a system are defined by temperature, pressure,volume, density, composition, surface tension, viscosity, and so on. There exist twotypes of properties: intensive and extensive. Intensive properties are those that donot require any reference to the mass of the system and are nonadditive. Extensiveproperties are dependent on the mass of the system and are additive. Any change insystem properties is termed a process.

Despite its name, thermodynamics does not deal with the dynamics of a system,but its equilibrium state. The word equilibrium implies a state of balance. When asystem is in such a state that upon slight disturbances it returns to its original state,it is said to be in a state of equilibrium. This state of equilibrium is one in whichthe system is time-invariant in properties. A thermodynamic system can experiencethermal, mechanical, or chemical equilibrium.

If a system undergoes no apparent changes, it is said to be stable. However, thereare systems that are apparently stable because the rates of change within them areimperceptible. Such systems are called inert. The degree of stability of a system canvary. For example, a book placed flat on a table is at its state of rest and is stable.However, when pushed over the table to the ground it reaches another state of rest,which is also stable.

The fundamental principles of thermodynamics can be stated in terms of what arecalled system variables, modes of energy transfer, and characteristic state functions(Stumm and Morgan, 1981). The system variables are five in number, namely, tem-perature T , pressure P, volume V , moles n, and entropy S. There are only two modesof energy transfer: heat, q, transferred to a system from its surroundings; and work,w,done on the system by its surroundings. A state function is one that depends only onthe initial and final states of the system and not on the path by which the system maypass between states. Only one such characteristic state function, internal energyU, isnecessary to define the system. However, three other characteristic state functions arealso used in thermodynamics. They are Helmholtz free energy A, Gibbs free energyG, and enthalpy H .

2.2 FUNDAMENTAL LAWS OF THERMODYNAMICS

There are four fundamental laws on which the edifice of thermodynamics is built.These are called zeroth, first, second, and third laws. These laws, which lead tomodern thermodynamics, are the work of several scientists over a considerable timespan (see Table 2.1).

The unique feature of these laws is that although derived through generalizationsresulting fromnumerous experimental data, they have not yet been disproved. Einstein(1949) remarked that “. . . classical thermodynamics . . . is the only physical theory ofuniversal content concerning which I am convinced that, within the framework of theapplicability of its basic concepts, it will never be overthrown . . . .” Thermodynamicsis, as Einstein called it, “a theory of principle” based on “empirically observed general

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Basic Chemical Thermodynamics 15

TABLE 2.1Major Developments in Thermodynamics over Two Centuries

Year Person Contribution

1760–1766 Joseph Black Calorimetry1842 Julius Meyer Conversion of heat to work, and vice

versa1843–1852 James Joule Heat, work, and the first law1848–1851 William Thompson Absolute temperature scale and the

second law1854–1865 Rudolph Clausius Concept of entropy1873–1878 Josiah Willard Gibbs Chemical thermodynamics, chemical

potential, phase rule1882 H. von Helmholtz Equilibrium, free energy1884–1887 Jacobus van’t Hoff Equilibrium constant, solution theory1906 Walter Nernst Heat theorem (third law)1927 Franz Simon Third law1900–1907 Gilbert N. Lewis Fugacity, nonideality1929 W.F. Giauque Third law verification1931 Lars Onsager Nonequilibrium thermodynamics,

reciprocity relations1949 Ilya Prigogine Irreversible processes, dissipative

structures

Source: Laidler, K.J. 1993. The World of Physical Chemistry. New York, NY: Oxford UniversityPress.

properties of phenomena” that does not rely on any assumptions on “hypotheticalconstituents.” Thus, the laws of thermodynamics have stood the test of time and areprobably inviolable.

2.2.1 ZEROTH LAW OF THERMODYNAMICS

One of the fundamental system variables in thermodynamics is temperature. Thezeroth law states that if two systems are in thermal equilibrium with a third, thenthey are also in thermal equilibrium with each other. This allows us to create a “ther-mometer.” We can calibrate the change in a property, such as the length of a columnof mercury, by placing the thermometer in thermal equilibrium with a known physi-cal system at several reference points. Celsius thermometers use the reference pointsfixed as the freezing and boiling points of purewater. If we then bring the thermometerin thermal equilibrium with a human body, for example, we can determine the bodytemperature by noting the change in the thermal property. This law also establishes atemperature scale; that is, only a system at a higher temperature can lose its thermalenergy to one with a lower temperature, and not vice versa. The International Unionof Pure andApplied Chemistry (IUPAC) has adopted the kelvin as its unit of temper-ature, which is the fraction 1/273.16 of the thermodynamic temperature of the triplepoint of water (IUPAC, 1988).

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2.2.2 FIRST LAW OF THERMODYNAMICS

The universal principle of conservation of energy is the first law of thermodynamics.According to the first law, for any closed system the change in total energy (kineticenergy+potential energy+ internal energy) is the sum of the heat absorbed from itssurroundings and the work done by the system. For macroscopic changes in a closedsystem with kinetic energy EK, potential energy EP, and internal energy U, and thatwhich absorbs heat q and does work equivalent to w, the first law can be written inthe following form:

ΔEP +ΔEK +ΔU = q − w. (2.1)

For a macroscopic system at rest, we have both ΔEP and ΔEK equal to zero. If δqrepresents the infinitesimal heat absorbed from the surroundings and −δw representsthe infinitesimal work done by the system (by definition,+δw is the infinitesimalwork done on the system), then the infinitesimal change in energy dU is given by

dU = δq − δw. (2.2)

Notice that in the above equation we have used δ instead of d to remind us that q andware defined only for a given path of change; that is, the values of q andw are dependenton how we reached the particular state of the system. dU, however, is independent ofthe path taken by the system and is determined only by the initial and final states ofthe system. One measures only changes in energies as a result of a change in the stateof the system and not the absolute energies. Thus, for a system at rest and for finitemacroscopic changes in q and w, we have

ΔU = q − w. (2.3)

The work term in the above equation can involve any of the following: pressure–volume work (wPV ) and any of the other forms of shaft work, wshaft (e.g., push–pull,electrical, elastic, magnetic, surface, etc.). When the work done by a system is onlywork of expansion against an external pressure, Pext, then w is given by∗

w =∫V2V1Pext dV = PextΔV . (2.4)

For an adiabatic process, q = 0 and hence ΔU = −w, whereas for a process suchas a chemical reaction occurring in a constant-volume container, ΔV = 0 and henceΔU = q. Most environmental processes are constant-pressure processes and hencethey invariably involve PV work. An example of an adiabatic process in the environ-ment is the expansion of an air parcel and the attendant decrease in temperature asit rises through the atmosphere, leading to what are called dry and moist adiabaticlapse rates (see Example 2.1).

For open systems there is an additional contribution due to the flow of matter,dUmatter, and therefore we have

dU = δq − δw+ dUmatter. (2.5)

∗Note that we will henceforth drop the subscript on P.

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2.2.3 SECOND LAW OF THERMODYNAMICS

The second law of thermodynamics is derived from the principles of heat enginesfirst studied by Sadi Carnot. This can be best explained with reference to Figure 2.1,in which a heat engine converts heat to work. The heat engine absorbs heat Q1 fromthe hot reservoir (heat source) at temperature Thot, converts part of it to workW , anddiscards heatQ2 to the cold reservoir (sink) at temperature Tcold. The efficiency ηeff =W/Q1. By the first law, W = Q1 − Q2, further, Q ∝ T . Therefore, the efficiency ofa Carnot heat engine is given by

ηeff = Q1 − Q2

Q1= Thot − Tcold

Thot. (2.6)

From the generalizations of the concept of efficiency of heat engines for an arbitrarycycle, Carnot defined the following for a closed cycle:

∮dQ

T= 0, (2.7)

where the integral is over a path representing a reversible process from state A tostate B. Thus, Carnot realized that the function dQ/T is independent of the pathand depends only on the initial and final states. Such a function was given the nameentropy, S:

dS = dQ

T. (2.8)

The second law of thermodynamics is a general statement regarding the sponta-neous changes that are possible for a system. When a change occurs in an isolatedsystem (i.e., the universe) the total energy remains constant; however, the energy maybe distributed within the system in any possible manner. For all natural processesoccurring within the universe, any spontaneous change leads to a chaotic dispersalof the total energy. It is unlikely that a chaotically distributed energy will in timereorganize itself into the original ordered state. A large number of examples can be

Hot Cold

Thot Tcold

W

Q1 Q2

FIGURE 2.1 A Carnot heat engine.

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mentioned here and is well described in various texts (e.g.,Atkins and de Paula, 2006).To quantify this inexorable transition toward a chaotic state for the universe, the sec-ond law introduces a term called entropy, denoted by S. It is also a state function. Thechange in entropy is defined as the result of dividing the energy transferred as heat bythe absolute temperature at which the transfer took place.After a spontaneous changefor a system, the total entropy of the system plus its surroundings is greater. Clausiusenunciated the second law thus: the entropy of the universe tends to a maximum. Amore precise statement of the second law is: as a result of any spontaneous changewithin an isolated system the entropy increases.

The entropy transferred to a system by its surroundings is defined by

dSsurr = δq

T. (2.9)

The total entropy change of the isolated system, that is, the universe, is the sum of theentropy changes within the system, dSsys, and the entropy change transferred fromthe system to the surroundings, −dSsurr. Thus,

dSuniv = dSsys − dSsurr. (2.10)

Since for any permissible process within an isolated system, dSuniv ≥ 0, we havedSsys ≥ δq/T . This is called the Clausius inequality. Entropy is a measure of thequality of the stored energy; it can be appropriately identified with disorder.

2.2.4 THIRD LAW OF THERMODYNAMICS

The third law defines the value of entropy at absolute zero. It states that the entropy ofall substances is positive and becomes zero at T = 0, and does so for a perfectly crys-talline substance.Thus, S → 0 as T → 0.Although there have been several attempts,no one has ever succeeded in achieving the absolute zero of temperature. Thus, thethird law explicitly recognizes the difficulty in attaining absolute zero.

To summarize, according to the first law, energy is always conserved, and, if workhas to be done by a system, it has to absorb an equal amount of energy in the formof heat from its surroundings. Therefore, a perpetual motion machine capable ofcreating energy without expending work is a pipe dream. The permissible changesare only those for which the total energy of the isolated system, that is, the universe,is maintained constant. The first law of thermodynamics is thus a statement regardingthe permissible energy changes for a system. The second law gives the spontaneouschanges among these permissible changes. The third law recognizes the difficulty inattaining the absolute zero of temperature.

EXAMPLE 2.1 PRESSUREAND TEMPERATURE PROFILES IN THE ATMOSPHERE

Problem Statement: If atmospheric air can be considered an ideal gas, it obeys theideal gas law in the form PV = R, where V is the molar volume of air, V = M/ρ. ρ is

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the density of air (kg/m3), andM is the averagemolecular weight of air (≈29). Considera parcel of air rising from the ground (sea) level to the lower atmosphere. It experiencesa change in pressure with altitude given by the well-known relation dP(h)/dh = −ρg.Use the ideal gas law to obtain an expression for P(h), given that P0 is the atmosphericpressure at sea level.Assume that as the parcel rises it undergoes a change in volume in relation to a

decreasing pressure, but that there is no net heat exchange between it and its surround-ings, that is, it undergoes an adiabatic expansion. This is a reasonable assumption,since the size of the reservoir (ambient atmosphere) is much larger than the size of theair parcel, and any changes in the ambient temperature are imperceptible whereas thatwithin the air parcel will be substantial. Therefore, the volume expansion of the parcelleads to a decrease in temperature. This variation in temperature of dry air with heightis called the dry adiabatic lapse rate. Apply the first law relation and use the expressiondU = Cv dT (as in Section 2.2.5) to obtain an equation for the variation in temperatureof the air parcel with height.

Solution: Considering the air to be an ideal gas at any point in the atmosphere, we havethe following:

P = ρRT

M. (2.11)

Since the pressure at any point is due to the weight of the air above, we have dP(h)/dh =−ρg and hence

dP(h)

dh= −P(h)Mg

RT. (2.12)

Integrating the above equation with P(0) = P0, we obtain

P(h) = P0e−(Mhg/RT) (2.13)

The above equation gives the pressure variation with height in the atmosphere.Now we shall obtain the temperature profile in the atmosphere using the concepts

from the first law. For an adiabatic process we know that δq = 0. Hence dU = δw. Aswill be seen in Section 2.2.5, dU = CvdT . Since it is more convenient to work with Pand T as the variables rather than with P and V , we shall convert the PdV term to aform involving P and T using the ideal gas law. Thus,

d(PV) = PdV + VdP = m

MRdT . (2.14)

Hence we have for the first law expression

Cv dT =( mMRT) dP

P−( mMR)dT . (2.15)

By rearranging, we obtain

dT

dP=

m

M· RTP

Cv + m

M· R

. (2.16)

Combining the equation for dP/dh with the above, we obtain

dT

dh= − g

Cp,m, (2.17)

continued

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where Cp,m is the heat capacity per unit mass of air at constant pressure. It is definedas Cv,m + R/M. The term Γadia = g/Cp,m is called the dry adiabatic lapse rate.The student is urged to work out Problem 2.5 to get a feel for the magnitude of

temperature changes in the lower atmosphere. To quantify the atmospheric stability(the capacity of air to disperse pollutants is related to this property), one compares theprevailing (environmental) lapse rate, Γenv = −dT/dh, to the dry adiabatic lapse rate.Figure 2.2 shows the characteristic profiles for unstable, stable, and neutral atmospheres.It can be seen that for an unstable atmosphere, Γenv > Γadia , whereas for a stableatmosphere, Γenv < Γadia . For an unstable atmosphere, the less dense air parcel at thehigher altitude continues to rise, and that at the lower altitude is denser and continuesto sink. The condition is unstable since any perturbation in the vertical direction isenhanced. Pollutant dispersal is rapid in an unstable atmosphere.

NeutralGenv = Gadia

StableGenv < Gadia

UnstableGenv > Gadia

Dry adiabatic

h

T

FIGURE 2.2 Dry adiabatic and environmental lapse rates related to atmosphericstability. Neutral, stable, and unstable atmospheric conditions are shown.

EXAMPLE 2.2 HURRICANEASA HEAT ENGINE

Emanuel (2005) described the hurricane as Nature’s steam engine that obeys Carnot’senergy cycle, which is the basis for the second law of thermodynamics. The hurricanecan be considered to obey a four-step cycle, described in Figure 2.3, where a cross-section through a hurricane is shown. The axis on the left side is the central axis of thestorm. At point A near the sea surface, the air is at least hundreds of kilometers fromthe storm center. The air slowly spirals inward toward the eye wall (point B). The A to

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0 100 200

16

Distance from hurricane center/km

AB

C

D

Tcold = 200 K

Thot = 300 K

Alti

tude

/km

FIGURE 2.3 The hurricane heat engine.

Bmovement corresponds to a significant decrease in pressure. But, since the sea surfaceis an infinite heat reservoir, its temperature is constant. Unlimited heat is therefore addedto the air parcel as it moves fromA to B. This amounts to the increased humidity in theair as the seawater evaporates into the incoming air. At point B the air turns and movesupward, making the hurricane eye wall. Here the latent heat is converted into sensibleheat as water vapor condenses and pressure decreases rapidly. This is an example ofan adiabatic expansion of air. At point C (12–18 km from the surface), the adiabaticexpansion decreases the air temperature to 200K.As the air remains at this temperature,it sinks to pointD toward the tropopause,which is near isothermal conditions. Finally, theair from point D returns to pointA in a near-adiabatic compression. The thermodynamiccycle is thus complete and a perfect steam engine is created, except that the energyproduced by the hurricane shows up as the energy of the wind. Thus, we can applyCarnot’s theorem to obtain the maximum wind for an idealized hurricane.The kinetic energy dissipated by the atmosphere near the surface is given by

D ≈ CDρV2, (2.18)

where D is the rate of heat energy dissipation per unit surface area, ρ is the air density,V is the wind speed, and CD is the drag coefficient (recall fluid mechanics).The total mechanical work done by a Carnot engine is given by

W = Q

(Thot − Tcold

Thot

), (2.19)

where Q is the heat input rate. The total heat input rate per unit surface area is given by

Q = CKρVE2 + CDρV

2, (2.20)

continued

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where CK is the enthalpy exchange coefficient and E is the evaporative potential of thesea surface. E is a measure of the air–sea thermodynamic equilibrium and is related tothe greenhouse effect (see Chapter 6, Section 6.3.3.2). The larger the greenhouse effectof gases emitted to the atmosphere, the greater the E value. We now have

W = (CKρVE2 + CDρV

2)

(Thot − Tcold

Thot

). (2.21)

EquatingW to D and using algebraic manipulations, we have the final equation for thehurricane’s maximum wind velocity (Emanuel, 2003):

Vmax =√(

Thot − Tcold

Tcold

)· E. (2.22)

Note that Tcold and not Thot appears in the denominator of the above equation. This isa result of the feedback mechanisms of dissipative heating. The evaporative potentialof the sea surface is approximated by the following equation:

E ≈ CK

CD· (h∗ − h), (2.23)

where h∗ and h are, respectively, the specific enthalpy (enthalpy per unit mass) of airnear the ocean boundary surface and of the inflowing dry air in the ambient boundarylayer. The ratio CK:CD has been observed to be about 1 (Emanuel, 2005). The stu-dent is encouraged to work out Problem 2.33 to learn about the applicability of theabove equations.

2.2.5 ENTHALPY AND HEAT CAPACITY

For most chemical processes that are carried out at constant volume, the PV work iszero since dV is zero. However, there are a number of processes, especially environ-mentally relevant ones, for which the pressure is constant but the volume is not. Insuch cases, the work term is nonzero. It is appropriate for such processes to definea new term called enthalpy, denoted by the symbol H . It is also a state function thatdoes not depend on the path taken by a system to arrive at that state. If only expansionwork is considered against a constant external pressure P in Equation 2.2, then wehave the relation ΔU = q − PΔV . If the two states of the system are denoted by Aand B, then we have

(UB + PVB) − (UA + PVA) = q. (2.24)

If we define H = U + PV, then we see that ΔH = q at constant P. The importanceof H in dealing with systems involving material flow will become apparent later aswe combine several of the thermodynamic laws. We have seen thus far that the totalenergy of the system at constant volume is given byΔU = (q)V , and the total energyat constant pressure is given by ΔH = (q)P.

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Closely related to the internal energyU and enthalpyH are the terms heat capacityat constant volume Cv and heat capacity at constant pressure Cp. The concept oftemperature resulting from the third law states that the temperature differences providethe driving force for heat flow between bodies. The ratio of the amount of heattransferred to the body and the resulting temperature change is called the heat capacity.For an infinitesimal change the ratio is given by

C = dq

dT; (2.25)

Cv =(

∂U

∂T

)

V; Cp =

(∂H

∂T

)

P. (2.26)

Heat capacities are tabulated for several compounds in the literature usually as empir-ical equations. These are given as polynomial fits with temperature as the dependentvariable and are valid only within the given range of temperature.A short compilationfor some important compounds is given in Table 2.2. Note that for very small rangesof temperature, the heat capacity can be assumed to be a constant. For solids and liq-uids the heat capacities are similar in magnitude, but for gases there is a relationshipbetween the two given by

CP − CV = nR. (2.27)

2.2.6 THERMODYNAMIC STANDARD STATES AND ENTHALPIES OF REACTION,

FORMATION, AND COMBUSTION

Thermodynamic quantities are estimated at the so-called standard states. The defini-tion of the standard state is to be noted wheneverU orH or any other thermodynamicquantities are used in a calculation. For convenience, we choose some arbitrary stateof a substance at a specified temperature T , standard pressure P0, standard molalitym0, or standard concentration C0. Although the choice of P0,m0, or C0 will depend

TABLE 2.2Empirical Equations for Specific Molar Heat Capacities at ConstantPressure Cp = a + bT + cT 2 + dT 3 Applicable for 273 ≤ T ≤ 1500 K(Units: cal/K/mol)

Compound a b × 102 c × 105 d × 109

H2 6.95 −0.045 0.095 −0.208O2 6.08 0.36 −0.17 0.31H2O 7.70 0.046 0.25 −0.86CH4 4.75 1.2 0.30 −2.63CHCl3 7.61 3.46 −2.67 7.34

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on the system, the most common choices are the following (Kondepudi, 2008):

T = 298.15K, P0 = 1 atm (= 1.013 × 105 Pa), m0 = 1mol/kg,

C0 = 1mol/dm3.

For any pure substance at a given temperature, the standard state is the most stablestate (gas, liquid, or solid) at a pressure of 1 atm (= 1.013 × 105 Pa). In the gas phase,the standard state of any compound, either pure or as a component in a mixture, is thehypothetical state of ideal gas behavior at P = 1 atm. For a condensed phase (liquidor solid), the standard state, either as pure or as a component in a mixture, is thatof the pure substance in the liquid or solid form at the standard pressure of P0. Fora solute in solution, the standard state is a hypothetical state of an ideal solution atstandard concentration C0 at the standard pressure of P0.

There are three important enthalpy terms that merit discussion. These are standardenthalpy of reaction (ΔHo

r ), standard enthalpy of formation (ΔHof ), and standard

enthalpy of combustion (ΔHoc ). The standard enthalpy of reaction is the enthalpy

change for a system during a chemical reaction and is the sum of the standard enthalpyof formation of the products minus the sum of the standard enthalpy of formation ofthe reactants. The standard enthalpy of formation is the enthalpy change to produce1mol of the compound from its elements, all at the standard conditions. The standardenthalpies of reaction may also be combined in various ways. This is succinctlyexpressed in Hess’s law of heat summation: the standard enthalpy of a reaction isthe sum of the standard enthalpies of reactions into which an overall reaction maybe divided, and holds true if the referenced temperatures of each individual reactionare the same. The standard enthalpy of combustion is that required to burn or oxidize1mol of the material to a final state that contains only H2O (l) and CO2.

EXAMPLE 2.3 ENTHALPY OFAN ACID–BASE REACTION

Determine the enthalpy of reaction at 298K for the neutralization of hydrochloric acidwith sodium hydroxide:

HCl(aq) + NaOH(aq) � NaCl(aq) + H2O(aq). (2.28)

First calculate the heat of formation of each species (ΔHof ) from the heat of formation

of the constituent ions (ΔHof ). These values are obtained fromAppendix 2:

ΔHof (HCl, aq) = ΔHo

f (H+, aq) +ΔHof (Cl−, aq) = 0 − 167.2

= −167.2 kJ/mol

ΔHof (NaOH, aq) = ΔHo

f (Na+, aq) +ΔHof (OH−, aq) = −239.7 − 229.7

= −469.4 kJ/mol

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ΔHof (NaCl, aq) = ΔHo

f (Na+, aq) +ΔHof (Cl−, aq) = −239.7 − 167.2

= −406.9 kJ/mol

ΔHof (H2O, aq) = −285.8 kJ/mol.

Hence, for the overall reaction,

ΔHor = ΔHo

f (NaCl, aq) +ΔHof (H2O, aq)

−ΔHof (HCl, aq) −ΔHo

f (NaOH, aq) = −56.1 kJ/mol.

The enthalpy of reaction at temperatures other than standard temperature can beestimated from the following equation, called Kirchoff’s law, that is,

ΔHf (T) = ΔHof (To) +

∫TTo

ΔCP dT . (2.29)

2.2.7 COMBINATION OF FIRST AND SECOND LAWS

Consider a closed system, which does only PV work of expansion and for which δqis given by TdS:

dU = TdS − PdV . (2.30)

We can now define two other state functions called Gibbs free energy, G, andHelmholtz free energy, A:

G = H − TS,

A = U − TS.(2.31)

We can derive the following equations for differential changes in G and A:

dG = −SdT + VdP,

dA = −SdT + PdV .(2.32)

From the definition of enthalpy, H, we can also derive the following:

dH = TdS + VdP. (2.33)

Most environmental processes are functions of T and P, and hence Gibbs free energyis a natural state function for those cases. The above differential equations are thebasis for deriving useful physico-chemical thermodynamic relationships.

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2.3 CHEMICAL EQUILIBRIUM AND GIBBS FREE ENERGY

Gibbs free energy, G, is of significance for most environmental problems since G isa variable of temperature, T , and pressure, p. Many environmental processes occurat atmospheric pressure, which is a constant. Hence it is important to develop ourconcept of equilibrium in terms of Gibbs free energy. J.Willard Gibbs, considered bymany to be the father of modern chemical thermodynamics, invented the function. Ithas units of kilojoules (kJ). Utilizing the fundamental definition of G = H − TS =U + PV − TS, and further if P and T are constant, we have

dG = δq − δw+ PdV − TdS = δq − TdS. (2.34)

Since we know that, for spontaneous processes, δq ≤ TdS from the second law, atconstant T and P if only PV work of expansion is considered, we have the followingcriterion for any change in independent variables of the system:

dG = 0. (2.35)

For a reversible process in a closed system at constant temperature and pressure, ifonlyPV expansionwork is allowed, there is no change in theGibbs function (dG = 0)at equilibrium. Spontaneous processes will occur in such a system at constant T andP when it is not at equilibrium with a consequent decrease in free energy (dG < 0).

The above inequality for a spontaneous change represents the criteria used inequilibrium models in environmental science and engineering. For a finite change ina system at constant temperature T , we can write

ΔG = ΔH − TΔS. (2.36)

To assess whether a process in the natural environment is spontaneous or not, oneestimates the difference in Gibbs free energies between the final and initial states ofthe system, ΔG. If ΔG is negative, the process is spontaneous. In other words, thetendency of any system toward an equilibrium position at constant T and P is drivenby its desire to minimize its free energy. The free energy change for any system is,therefore, given by the inequality

ΔH − TΔS ≤ 0. (2.37)

For an exothermic process, there is a release of heat to the surroundings (enthalpychange, ΔH < 0). For an endothermic process, since ΔH > 0, TΔS has to begreater than zero and positive and larger than ΔH. Thus, an endothermic processis accompanied by a very large increase in entropy of the system.

2.3.1 FREE ENERGY VARIATION WITH TEMPERATURE AND PRESSURE

Since environmental processes occur at varying temperatures, it is useful to knowhow the criterion for spontaneity and equilibrium for chemical processes vary withtemperature. Since at constant pressure (∂G/∂T)P = −S, and S is always positive, it

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is clear that at constant pressure with increasing temperature G should decrease. If Sis large, the decrease is faster. For a finite change in the system from state a to stateb, the differential ΔG = Gb − Ga and ΔS = Sb − Sa, and hence we have

(∂(ΔG)

∂T

)

P= −ΔS. (2.38)

To arrange the above equation in a more useful manner, we need a mathematicalmanipulation, that is, first differentiating the quantity (ΔG/T) with respect to Tusing the quotient rule. This gives

∂(ΔG/T)

∂T= −ΔG

T2+ 1

T

(∂ΔG

∂T

). (2.39)

Substitution and rearrangement gives us

(∂(ΔG/T)

∂T

)

P= −ΔH

T2. (2.40)

This is the Gibbs–Helmholtz equation and is fundamental in describing the relation-ship between temperature and equilibrium in environmental processes.

Gibbs free energy is also dependent on pressure since at constant temperaturewe have (∂G/∂P)T = V . V being a positive quantity, the variation in G is alwayspositive with increasing pressure. Thus, the Gibbs free energy at any pressure P2 canbe obtained if the value at a pressure P1 is known, using the equation

G(P2) = G(P1) +P2∫

P1

V dP. (2.41)

Formost solids and liquids the volume changes are negligiblewith pressure, and henceG is only marginally dependent on pressure. However, for geochemical processes inthe core of Earth’s environment, appreciable changes in pressures as compared withsurface pressures are encountered, and hence pressure effects on Gibbs functions areimportant. In most cases, because of the huge modifications in G, most materialsundergo phase changes as they enter the Earth’s core. For gases on the other hand,significant changes in volume can occur. For example, for an ideal gas, V = nRT/P.Hence, integrating Equation 2.27 using this, we see that the change in Gibbs functionwith pressure is given by

G(P2) = G(P1) + nRT · ln(P2P1

). (2.42)

The standard Gibbs free energy of gases increases with increase in pressure.

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EXAMPLE 2.4 FREE ENERGYAND TEMPERATURE

Problem: The free energy change for a reaction is given by

ΔG0 = −RT lnKeq, (2.43)

where Keq is the equilibrium constant for a reaction (Chapter 5). If Keq for the reac-tion H2CO3(aq) → H+(aq) + HCO−

3 (aq) is 5 × 10−7 mol/L at 298K, what is the

equilibrium constant at 310K? ΔH0 is 7.6 kJ/mol.

Solution: Using the Gibbs–Helmholtz relationship, we have

∂ lnKeq

∂T= ΔH0

RT2, (2.44)

and hence

ln

(Keq(T2)

Keq(T1)

)= −ΔH

0

R·(

1

T2− 1

T1

). (2.45)

Substituting, we obtain Keq (at 310K) = 5.63 × 10−7 mol/L.Other properties such as vapor pressure, aqueous solubility, air/water partition con-

stant, and soil/water partition constant show similar changes depending on the sign andmagnitude of the corresponding standard enthalpy of the process. These are discussedin Chapter 4.

2.4 CONCEPT OF MAXIMUM WORK

We have the following equation for the total free energy change for any process:

ΔG = q − wtot + PΔV − TΔS = −wtot + PΔV (2.46)

since TΔS = q. Thus for any permissible process at constant T and P it follows that

−ΔG = wtot − PΔV = wuseful. (2.47)

Since PΔV is the expansion work done or otherwise called “wasted work,” the dif-ference between the total work and PΔV is the “useful work” that can be done bythe system. At constant temperature and pressure, the useful work that can be doneby a system on the surroundings is equivalent to the negative change in Gibbs freeenergy of a spontaneous process within the system. If the system is at equilibrium,then ΔG = 0 and hence no work is possible by the system.

EXAMPLE 2.5 CALCULATION OF MAXIMUM USEFUL WORK

Problem: Humans derive energy by processing organic polysaccharides. For example,glucose can be oxidized in the human body according to the reaction C6H12O6(s)+

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6O2(g) → 6CO2(g) + 6H2O(l). If the standard heat of combustion of glucose at roomtemperature is −2808 kJ/mol and the standard entropy of the reaction is +182 J/K/mol,calculate the maximum amount of useful non-PV work available from 1 g of glucose.Assume that the heat of combustion is independent of temperature.

Solution: Since ΔG0 = ΔH0 − TΔS0 = −2808 − (310) (0.182) = −2864 kJ/mol,T is 310K, the blood temperature of a normal human being. Hence from 1 g(≡ 0.0055mol) of glucose, the total free energy change for oxidation is−15.9 kJ. Sincethe negative of the free energy change gives the maximum non-PV work, the usefulwork available is also 15.9 kJ. Note that this is the basis for the term “number of calories”found on labels for foodstuffs.

In environmental bioengineering, microorganisms are used to process many types oforganic wastes. Similar calculations can be made of the maximum non-PV useful workperformed by these organisms, if they consume oxygen as the electron-donating species.

2.5 GIBBS FREE ENERGY AND CHEMICAL POTENTIAL

The equations described thus far are applicable to homogeneous systems composed ofa single component. In environmental engineering, one is concerned with multicom-ponents in each phase. Since this chapter is devoted to single-phase thermodynamics,the focus here is only on single and multiple components in a homogeneous singlephase. The discussion of multiphase equilibrium is reserved for Chapter 3. As wehave already seen, the fundamental equation for Gibbs free energy of a system is

dG = −SdT + VdP. (2.48)

In other words, G = G(T ,P). If the phase is composed of several components, thenthis equation must be modified to reflect the fact that a change in the number ofmoles in the system changes the Gibbs free energy of the system. This means Gis a function of the number of moles of each species present in the system. HenceG = G(T ,P, n1, . . . , nj). The total differential for G then takes the form

dG =(

∂G

∂P

)

T ,ni

dP +(

∂G

∂T

)

P,ni

dT +∑i

(∂G

∂ni

)

P,T ,nj �=idni. (2.49)

In the above equation, we recognize that

(∂G

∂P

)

T ,ni

= V ,

(∂G

∂T

)

P,ni

= −S (2.50)

while the third partial differential is a new term given the name chemical potential byGibbs in his formalism of thermodynamics. It is given the symbol μi, is applicableto each species in the system, and has units of kJ/mol:

(∂G

∂ni

)

P,T ,nj �=i= μi. (2.51)

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Chemical potential is the partial molar Gibbs free energy representing the changein the total Gibbs energy due to the addition of a differential amount of species i toa finite amount of solution at constant temperature and pressure. For a single puresubstance the chemical potential is the same as its molar Gibbs energy. Thus, thechemical potential of benzene in pure water is different from that in a mixture ofwater and alcohol. As its very name indicates, μi is an indicator of the potential for amolecule (e.g., movement from one phase to another or a chemical reaction). Thus,it is analogous to a hydrostatic potential for liquid flow, an electrostatic potential forcharge flow, and a gravitational potential formechanical work. The chemical potentialis thus a kind of “chemical pressure” and is an intensive property of the system, suchas T and P. When the chemical potential of a molecule is the same in states a and b,then equilibrium is said to exist. This satisfies the criterion for equilibrium definedearlier, that is, ΔG = 0. If the chemical potential is greater in state a than in state b,then a transfer or reaction of species i occurs spontaneously to move from a to b. Thissatisfies the criterion for a spontaneous process, that is, ΔG < 0.

The above formalism suggests that at constant T and P, the total free energy of asystem is given by

G =∑i

μini. (2.52)

EXAMPLE 2.6 SIGNIFICANCE OF CHEMICAL POTENTIAL

Problem: Calculate the change in chemical potential for the vaporization of water at1 atm and 25◦C.Solution: The reaction is H2O(l) → H2O(g). The free energy of formation per mole is−237 kJ/mol for liquid water and −229 kJ/mol for water vapor. For a pure substancethe free energy per mole is the same as the chemical potential. Hence,

Δμ = −229 − (−237) = 8 kJ/mol. (2.53)

The chemical potential of the vapor is higher, indicating that there is useful workavailable. It is clear that water vapor is more “potent” compared with liquid water.

2.5.1 GIBBS–DUHEM RELATIONSHIP FOR A SINGLE PHASE

In equilibrium calculations, an important problem is the estimation of the chemi-cal potential of different components in a single phase. Upon differentiation of theexpression for total Gibbs free energy given by Equation 2.43,

dG =∑i

μi dni +∑i

ni dμi. (2.54)

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The formal definition for dG is

dG = −S dT + V dP +∑i

μi dni. (2.55)

Therefore, we have

S dT + V dP +∑i

ni dμi = 0. (2.56)

This is called the Gibbs–Duhem relationship. At constant temperature and pressurethe above equation reduces to

∑i

ni dμi = 0. (2.57)

This equation is useful in estimating the variation in the chemical potential of onecomponent with composition if the composition and variation in chemical potentialwith that of the second component are known. By dividing the equation throughoutwith the total number of moles,

∑ni, we can rewrite the above equation in terms of

mole fractions,∑xi dμi.

2.5.2 STANDARD STATES FOR CHEMICAL POTENTIAL

Let us examine the chemical potential of an ideal gas. At a constant temperature T ,

(dμi)T =(V

ni

)dPi, (2.58)

where Pi is the pressure exerted by the component i ( partial pressure) in a constantvolume V . Note that for an ideal gas V/ni = (RT/Pi), μi at any T can now bedetermined by integrating the above equation if we choose as the lower limit a startingchemical potential (μ0

i ) at the temperature T and a reference pressure P0i , that is,

μi = μ0i + RT ln

(PiP0i

). (2.59)

Thus, the chemical potential of an ideal gas is related to a physically real measurablequantity, namely, its pressure. Whereas both μ0

i and P0i are arbitrary, they may not be

chosen independent of one another, since the choice of one fixes the other automat-ically. The standard pressure chosen in most cases is P0

i = 1 atm. Thus for an idealgas the chemical potential is expressed as

μi = μ0i (T ; P

0i = 1) + RT lnPi. (2.60)

The chemical potential for an ideal solution is analogous to the above expression forthe ideal gas. It is given by

μi = μ0i (T ; x

0i = 1) + RT · ln xi, (2.61)

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where xi is the mole fraction of component i in the solution. The reference state forthe solution is pure liquid for which the mole fraction is 1.

2.6 THERMODYNAMICS OF SURFACES ANDCOLLOIDAL SYSTEMS

The boundary between any two contiguous phases is called a surface or an interface.The term “surface” is reserved for those that involve air as one of the contiguousphases. In environmental chemistry, we encounter several interfaces: liquid/gas (e.g.,air–sea, air bubbles in water, fog droplets in air, foam), liquid/liquid (e.g., oil–water),liquid/solid (e.g., sediment–water, soil–water, colloid–water), and solid/air (e.g., soil–air, aerosols in air). Often when one phase is dispersed in another at submicrondimensions (e.g., air bubbles in sea, fog droplets in air, colloids in groundwater, acti-vated carbon in wastewater treatment), very large surface areas are involved. Surfacetension accounts for the spherical shape of liquid droplets, for the ability of waterto rise in capillaries as in porous materials (e.g., soils), and for a variety of otherreactions and processes at interfaces. It is therefore important to discuss some basicsof surface and interfacial chemistry.

2.6.1 SURFACE TENSION

Surfaces and interfaces are different from bulk phases. Consider, for example, waterin contact with air (Figure 2.4a). The number density of water molecules graduallydecreases as onemoves frombulkwater into air (Figure 2.4b). Similar gradual changesin all other physical properties will be noticed as one moves from water to air acrossthe interface. At a molecular level there is no distinct dividing surface at which theliquid phase ceases and the air phase begins. The so-called interface is therefore adiffuse region where the macroscopic properties change rather gradually across acertain thickness. The definition of this thickness is compounded when we realizethat it generally depends upon the property considered. The thickness is at least a fewmolecular diameters (a few angstroms). The ambiguity with respect to the locationof an exact dividing surface makes it difficult to assign properties to an interface.However, Gibbs showed how this can be overcome.

The reason for the diffuse nature of the surface becomes clear when we considerthe forces on a water molecule at the surface. Obviously, since there is a larger numberdensity of molecules on the water-side than on the air-side, the molecule experiencesdifferent forces on either side of the interface. The pressure (force) experienced bya water molecule in the bulk is the time-averaged force exerted on it per unit areaby the surrounding water molecules and is isotropic (i.e., the same in all directions).For a molecule on the surface, however, the pressure is anisotropic (Figure 2.4a). Ithas two components, one normal to the surface, Pn, and the other tangential to thesurface, Pt . The net pressure forces in the lateral direction are substantially reducedin the interfacial region compared with the bulk region. Because fewer liquid phasemolecules than those in the bulk phase act upon the surface molecules, the surfacemolecules possess greater energy than the bulk molecules. It requires work to bringa bulk molecule to the surface since it means increasing the surface area. The net

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Air

Air

Water

Dist

ance

Water

r/molecules.m–3

(a)

(b)

FIGURE 2.4 (a) Isotropic forces on a bulk water molecule and anisotropic forces on a surfacewater molecule. (b) Number density of water molecules as a function of distance from theinterface.

pressure forces on the surface will be negative, that is, the surface is said to experiencea tension.The surface is said to have a contractile tendency, that is, it seeks tominimizethe area, and is describable in thermodynamics in terms of a property called surfaceor interfacial tension and is designated by the symbol σ. It is the force per unit lengthon the surface (N/m or, more commonly, mN/m) or, equivalently, the free energy perunit surface area (J/m2).

The molecular picture of surface tension given above explains the variation in sur-face tension between different liquids (Table 2.3). It is clear that surface tension is aproperty that depends on the strength of the intermolecular forces betweenmolecules.Metals such as Hg, Na, andAg in their liquid form have strong intermolecular attrac-tive forces resulting from metallic bonds, ionic bonds, and hydrogen bonds. Theytherefore have very high surface tension values. Molecules such as gases and liquidhydrocarbons have relatively weak van der Waals forces between the molecules andtherefore have low surface tensions. Water, on the other hand, is conspicuous in itsstrange behavior since it has an unusually high surface tension. The entire conceptof hydrophobicity, and its attendant consequences with respect to a wide variety ofcompounds in the environment, is a topic of relevance to environmental engineeringand will be discussed in Chapter 3.

2.6.2 CURVED INTERFACES AND THE YOUNG–LAPLACE EQUATION

Curved interfaces are frequently encountered in environmental engineering.Examplesare air bubbles in water, soap bubbles in water, fog droplets in air, aerosols, colloids,and particulates in air and water environments. Consider a curved interface such asan air bubble in water as shown in Figure 2.5. Surface tension is itself independent ofsurface curvature so long as the radius of curvature of the bubble is large in comparisonto the thickness of the surface layer (a few angstroms). Young (1805) argued that in

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TABLE 2.3Surface Tensions of Some Common Substances

Compound Temperature (K) σ (mN/m)

MetalsNa 403 198Ag 1373 878.5Hg 298 485.5

Inorganic saltsNaCl 1346 115NaNO3 581 116

GasesH2 20 2.01O2 77 16.5CH4 110 13.7

LiquidsH2O 298 72.13CHCl3 298 26.67CCl4 298 26.43C6H6 293 28.88CH3OH 293 22.50C8H17OH 293 27.50C6H14 293 18.40

Source: Adamson,A. 1990.Physical Chemistry of Surfaces, 4th ed. NewYork,NY: John Wiley & Sons, Inc.

order to maintain the curved interface, a finite pressure difference ought to existbetween the inside, Pi, and outside, Po, of the bubble. The work necessary to movea volume dV of water from the bulk to the air bubble is given by (Pi − Po) dV . Thisrequires an extension of the surface area of the bubble by dAσ and the work done willbe σ dAs. For a spherical bubble, dV = 4πr2 dr and dAσ = 8πr dr. Therefore,

ΔP = 2σ

r. (2.62)

This is called theYoung–Laplace equation. If we consider any curved interface thatcan be generally described by its two main radii of curvatures, R1 and R2, then theYoung–Laplace equation can be generalized as (Adamson, 1990;Adamson and Gast,1997)

ΔP = σ(

1

R1+ 1

R2

). (2.63)

The above equation is a fundamental equation of the capillary phenomenon. For aplane surface, its two radii of curvature are infinite and henceΔP is zero. Capillarityexplains the rise of liquids in small capillaries, capillary forces in soil pore spaces,

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r

r + dr

FIGURE 2.5 Cross-section of an air bubble in water.

and vapor pressure above curved interfaces such as droplets. It also plays a role indetermining the nucleation and growth rates of aerosols in the atmosphere.

2.6.3 SURFACE THICKNESS AND GIBBS DIVIDING SURFACE

As stated earlier, a surface is not a strict boundary of zero thickness nor is it only atwo-dimensional area. It has a finite thickness (a few angstroms). This poses a problemin assigning numerical values to surface properties. Fortunately, Gibbs, the architectof surface thermodynamics, came up with a simple proposition. It is appropriatelytermed the Gibbs dividing surface.

Let us consider two phases of volumes VI and VII separated by an arbitrary planedesignated a. This is strictly a mathematical dividing plane. Gibbs suggested that oneshould handle all extensive properties (E,G, S,H , n, etc.) by ascribing to the bulkphases those values that would apply if the bulk phases continued uninterrupted up tothe dividing plane. The actual values for the system as a whole and the total values forthe two bulk phases will differ by the so-called surface excess or surface deficiencyassigned to the surface region. For example, the surface excess in concentration foreach component in the system is defined as the excess number of moles in the systemover that of the sum of the number of moles in each phase. Hence,

nσi = ni − CIi VI − CII

i VII. (2.64)

The surface excess concentration (Γi, mol/m2) is defined as the ratio of nσi (mol) andthe surface area,Aσ (m2), that is,Γi = nσi /Aσ. The same equation holds for every othercomponent in the system. If we now move the plane from a to b, then the volume ofphase b decreases by (b− a) As and phase a increases by an identical value. Sincethe total number of moles in the system is the same, the new surface excess value is

Γnewi = Γold

i + (CIIi − CI

i ) · (b− a). (2.65)

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Thus Γnewi = Γold

i only if CIIi = CI

i . Thus, the location of the Gibbs dividing surfacebecomes important in the definition of surface excess. Gibbs proposed that to obtainthe surface excess of all other components in a solution, a convenient choice ofthe dividing plane is such that the surface excess of the solvent is zero. For a purecomponent system there is no surface excess. This is called the Gibbs convention.

2.6.4 SURFACE THERMODYNAMICS AND GIBBS EQUATION

Consider a system where the total external pressure P and the system temperature Tare kept constant. Let us assume that the system undergoes an increase in interfacialarea while maintaining the total number of moles constant. The overall increase infree energy of the system is the work done in increasing the surface area. This freeenergy increase per unit area is the surface tension. Hence,

σ =(

∂G

∂Aσ

)

T ,P,ni

. (2.66)

Analogous to dG for a bulk phase, we can define a surface free energy in the followingform:

dGσ = −SσdT + σdAσ +∑i

μi dnσi , (2.67)

where we have replaced σ for P and dAσ for dV . At constant T ,

dGσ = σdAσ +∑i

μi dnσi . (2.68)

The total free energy at constant temperature for the surface is given by

Gσ = σAσ +∑i

μinσi . (2.69)

Thus the complete differential obtained for Gσ is given by

dGσ = σdAσ + Aσdσ+∑

(nσi dμi + μidnσi ). (2.70)

Comparing the two expressions for dGσ, we obtain the following equation:

Aσdσ+∑i

nσi dμi = 0. (2.71)

The above equation is fundamental to surface thermodynamics and is called theGibbsequation.

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2.6.5 GIBBS ADSORPTION EQUATION

In environmental chemistry, we encounter systems composed of more than onespecies. If, for example, we consider a two-component system (solvent ≡ 1 andsolute ≡ 2), the Gibbs equation can be rewritten using the Gibbs convention thatthe surface concentration of solvent 1 with respect to solute 2, Γ1(2) = 0, is

Γ2(1) = −dσ

dμ2. (2.72)

Note that in most cases it is simply designated Γ. This is called the Gibbs adsorp-tion equation. It is the analogue of the Gibbs–Duhem equation for bulk phases. TheGibbs adsorption equation is used to calculate the surface excess concentration of asolute by determining the change in surface tension of the solvent with the chemi-cal potential of the solute in the solvent. This equation is of particular significancein environmental engineering, since it is the basis for the calculation of the amountof material adsorbed at air–water, soil (sediment)–water, and water–organic solventinterfaces.

EXAMPLE 2.7 SURFACE CONCENTRATION FROM SURFACE TENSION DATA

Problem: The following data were obtained for the surface tension of a natural plant-based surfactant in water.

Aqueous Concentration (mol/cm3) Surface Tension (erg/cm2)

1.66 × 10−8 623.33 × 10−8 601.66 × 10−7 503.33 × 10−7 451.66 × 10−6 39

The molecular weight of the surfactant is 300. Obtain the surface concentration at anaqueous concentration of 1.66 × 10−7 mol/cm3.

Solution: Since Γ2(1) = −dσ/dμ2 and, for a dilute aqueous solution of soluteconcentration Cw, dμ2 = RT lnCw, we have surface concentration Γ = Γ2(1) =−(1/RT )(dσ/d lnCw). A plot of σ versus ln Cw gives a slope of dσ/d lnCw =−6.2 erg/cm2 at Cw = 1.66 × 10−7 mol/cm3. Hence the surface concentration isΓ = −(−6.2)/(8.31 × 107 × 298) = 2.5 × 10−10 mol/cm2.

PROBLEMS

2.11 Classify the following properties as intensive or extensive. Give appro-priate explanations: temperature, entropy, pressure, volume, number ofmoles, density, internal energy, enthalpy, molar volume, mass, chemicalpotential, and Helmholtz free energy.

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2.21 Given the ideal gas law, pV = nRT, find q for an isothermal reversibleexpansion. What is the work required to compress 1mol of an ideal gasfrom 1 to 100 atm at room temperature? Express the work in J, erg, cal,and L atm units. Note that the latter three are not SI units.

2.32 For an ideal gas show that, for an adiabatic expansion, TVr is a constantand r is given by R/Cv .

2.42 For an ideal gas show that the difference between Cp and Cv is the gasconstant, R.

2.51 Calculate the rate of change in temperature per height for the rise of a1-g dry air parcel in the lower atmosphere. The actual rate is 6.5EC/km.Suggest reasons for the difference. Cp,m = 1.005 kJ/kg/◦C.

2.62 Calculate the heat of vaporization of water at 298K given its value at 393Kis 2.26 × 106 J/kg. The specific heat of water is 4.184 × 103 J/K/kg and itsheat capacity at constant pressure is 33.47 J/K/mol. Assume that the heatcapacity is constant within the range of temperature.

2.72 Find the heat capacity at constant pressure in J/K/mol for CHCl3 at 400K.Use the data from Table 2.1. Obtain the heat capacity at constant volumeat 400K.

2.82 0.5m3 of nitrogen at 400 kPa and 300K is contained in a vessel insulatedfrom the atmosphere. A heater within the device is turned on and allowedto pass a current of 2A for 5min from a 120-V source. The electrical workis done on the system to heat the nitrogen gas. Find the final temperatureof the gas in the tank. Cp for nitrogen is 1.039 kJ/kgK.

2.92 A 5-L tank contains 2mol of methane gas. The pressure of the gas isincreased from 2 to 4 atm while maintaining a constant volume of thetank.What is the total internal energy change during the process? The heatcapacity at constant pressure for methane is 34 J/molK and is constant.

2.102 A 5-L plastic Coca Cola bottle contains air at 300K and 12.5 bar gaugepressure. How much work would the gas do if you could expand the gasso that the final pressure is 1 bar isothermally and reversibly?

2.131 Supercooled water at −3◦C is frozen at atmospheric pressure. Calculatethe maximum work for this process. The density of water is 0.999 g/cm3

and that of ice is 0.917 g/cm3 at −3◦C.2.141 An ideal gas is subjected to a change in pressure from 2 to 20 atm at a

constant temperature of 323K. Calculate the change in chemical potentialfor the gas.

2.152 A reaction that is of considerable importance in nature is the transformationof methane utilizing ozone, 3CH4(g) + 4O3(g) → 3CO2(g) + 6H2O(l).The above reaction is the main source for water in the stratosphere and isaugmented by photochemical processes.

(a) Let us carry out the reaction in a laboratory at 298K. If the standardheat of formation of ozone is 142 kJ/mol and those of carbon dioxide,water, and methane are, respectively, −393, −285, and −75 kJ/molat 298K, determine the standard heat of the overall reaction at 298K.Compare this enthalpy change with the reaction of methane withoxygen in the previous problem.

(b) If the standard Gibbs free energy of the formation of ozone, carbondioxide, water, and methane are, respectively, 163, −394, −237, and−50 kJ/mol at 298K, calculate the standard free energy change ofthe overall reaction at 298K.

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(c) What will be the standard free energy change if the temperature isdecreased to 278K? Is the reaction more or less favorable at thistemperature?

2.163 Capillary rise is an important phenomenon in enhanced oil recovery, therecovery of nonaqueous-phase liquids from contaminated aquifers, the riseof sap in trees, and also in the determination of surface tension of liquids.This is based on the balance of forces between the surface tension andhydrostatic forces due to gravity.

(a) Consider a small capillary partially immersed in water. The waterrises to an equilibrium height given by h. Equate the pressure givenby theYoung–Laplace equation for the hemispherical interface withthat of the hydrostatic pressure and derive the equation σ = rgh/2Δρ,where r is the radius of the capillary, g is a gravitational constant, andΔρ is the difference in density between water and air.

(b) Soils are considered to have capillary size pores. Calculate the capil-lary rise ofwater in soil pores of diameters 100, 1000, and 10,000μm.This gives the shape of the boundary between air and water in soilpores. Repeat the calculation for a nonaqueous-phase contaminantsuch as chloroform in contact with groundwater. σ = 20mN/m1 andΔρ = 150 kg/m3.What conclusions can you draw about the shape ofthe boundary in this case?

2.171 Calculate the minimum work necessary to increase the area of the surfaceof water in a beaker from 0.001 to 0.005m2 at 298K.

2.182 A parcel of water 1 kg in mass is located 1 km above a large body ofwater. The temperature of the air, lake, and the water parcel is uniformand remains at 22◦C. What will be the change in entropy if the waterparcel descends and mixes with the lake water and reaches equilibriumwith it?

2.192 A gas container has compressed air at 5 × 105 Pa at 300K. What is themaximum useful work available from the system per kilogram of air? Theatmospheric pressure is 1 × 105 Pa at 300K?

2.202 A hydroxyl radical (OH!) plays an important role in atmosphericchemistry.

(a) It is produced mainly via the reaction of a photoexcited oxygenatom (O∗) with moisture: O∗ + H2O → 2OH•. Using the table inAppendix 2 and the fact that the heat of formation of O∗ in thegaseous state is 440 kJ/mol, obtain the enthalpy of the above reactionat standard state.

(b) OH• is an effective scavenger of other molecules, for exam-ple, OH• + NO2 → HNO3. Obtain the standard enthalpy of thereaction.

2.211 Oxides of sulfur are an important class of inorganic pollutants in the atmo-sphere resulting mainly from coal burning. For the oxidation of SO2 in airby the reaction SO2(g) + 1

2O2(g) → SO3(g), what is the free energy at298K? Use the table in Appendix 2.

2.222 The sun’s heat on a lake surface leads to the evaporation ofwater. If the heatflux from the sun is assumed to be 5 J/min/cm2, how much water will beevaporated on a clear summer day from a square meter of the lake surfacein 5 h? What volume of the atmosphere is required to hold the evaporated

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water?Assume that both the air and water remain at a constant temperatureof 308K. The heat of vaporization of liquid water at 308K is 2425 kJ/kg.

2.233 The predominant form of CaCO3 in nature is calcite. Marine shells aremade up of this form of calcium carbonate. However, there also existsanother form ofCaCO3, called aragonite. They differ in properties at 298Kas follows:

Property Calcite Aragonite

Standard free energy offormation (kJ/mol) −1128.7 −1127.7

Density (g/cm3) 2.71 2.93

(a) Is the conversion of calcite to aragonite spontaneous at 298K?(b) At what pressure will the conversion be spontaneous at 298K?

2.243 Municipal landfills contain refuse that are continually decomposed byindigenous bacteria. This produces large quantities of methane gas. Let usconsider a large landfill that contains approximately 1010 m3 of methane.Is it possible that this landfill gas, by being allowed to expand and cool toambient conditions (without combustion), can be used to produce power?In other words, you are asked to determine the useful work that can beobtained. Assume the following: the heat capacity at constant pressure is36 J/molK, pressure in the landfill is 8 atm, and temperature in the landfillis 240◦C. The molecular weight of methane is 0.016 kg/mol.

2.252 Methyl mercury (CH3)2Hg is the most easily assimilable form of mercuryby humans. It is present in both air and water in many parts of the UnitedStates and theworld. Consider the following process bywhich it converts toelemental Hg: (CH3)2Hg → C2H6 + Hg. Is the reaction favorable underambient conditions? Look up the ΔGσf for the compounds in appropriatereferences.

2.262 For the reaction H2S(g) → H2S(aq), the enthalpy of dissolution is−19.2 kJ/mol at 298 K. If the temperature is raised to 318K, by howmuch will the equilibrium constant of the reaction increase?

2.271 State whether the following statements are true or false:

1. The surface temperature in Baton Rouge was 94◦F, and 500m abovethe temperature was 70◦F on September 5, 1997. The atmosphere isstable.

2. Surface tension can be expressed in Pa/m2.3. The entropy of an ideal gas is only a function of temperature.4. The first law requires that the total energy of a system can be

conserved within the system.5. The entropy of an isolated system must be a constant.6. The Cp of an ideal gas is independent of P.7. Heat is always given by the integral of T dS.8. Chemical potential is an extensive property.9. For an adiabatic process, dU and δw are not equal.10. The heat of reaction can be obtained from the heats of formation of

reactants and products.

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11. If ΔH >0, the reaction is always possible.12. Although Ssys may increase, decrease, or remain constant, Suniv

cannot decrease.

2.282 The molar Gibbs free energy for CO2 in the gas phase is −394.4 kJ/moland that in water is−386.2 kJ/mol. Is the dissolution of CO2 a spontaneousprocess at 298K and 1 atm total pressure?

2.291 What pressure is required to boil water at 393K? The heat of vaporizationof water is 40.6 kJ/mol.

2.302 What is the pH of pure water at 37◦C? For the reaction H2O � H+ +OH−, Keq is 1 × 10−14 M2 at 25◦C and the enthalpy of dissociation is55.8 kJ/mol.

2.313 A human being is an open system and maintains a constant body tem-perature of 310K by removing excess heat via evaporation of waterthrough the skin. Consider a person weighing 175 pounds capable ofgenerating heat by digesting 100mol of glucose through the reactionC6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l). How much water willhave to be evaporated to maintain the body temperature at 310K? TheCp for water is 33.6 J/Kmol. For water,ΔH0

v = 40.5 kJ/mol. The ambienttemperature is 298K.

2.323 A fundamental energy-giving reaction for the creation of the pri-mordial living cell is proposed to be the formation of iron pyritesby the reaction FeS(s) + H2S(aq)6FeS2(s) + 2H+(aq) + 2e−(aq). Theenergy released by this reaction is used to break up CO2 andform the carbon compounds essential to life. Estimate the max-imum work obtainable from the above reaction under standardconditions.

2.332 For a hurricane (Figure 2.2) follow Example 2.2 to derive the maximumwind velocity predicted from theory. The specific enthalpy of air undermoist conditions is given by Cp,at + (Cp,wt + hw,e)xw, where Cp,a =1.006 kJ/kg◦C and Cp,w = 1.84 kJ/kg◦C are, respectively, the specificheat capacity of air and water. The water content of moist air is xw =0.0203 kJ/kg. Temperature t is given in ◦C and enthalpy of water evapo-ration hw,e = 2520 kJ/kg. The surface temperature of air in contact withwarm, moist water can be taken to be 27◦C, while that of ambient dry airis 23◦C. Give the answer for velocity in miles per hour.

REFERENCES

Adamson, A. 1990. Physical Chemistry of Surfaces, 4th ed. New York, NY: John Wiley &Sons, Inc.

Adamson,A. and Gast,A.P. 1997. Physical Chemistry of Surfaces, 6th ed. NewYork, NY: JohnWiley & Sons, Inc.

Atkins, P.W. and de Paula, J. 2006. Physical Chemistry, 8th ed. NewYork, NY:W H Freeman.Denbigh, K. 1981. The Principles of Chemical Equilibrium, 4th ed. New York: Cambridge

University Press.Einstein,A. 1949.Autobiographical notes. In: P.A. Schlipp (Ed.),Albert Einstein: Philosopher-

Scientist. Evanston, IL: Library of Living Philosophers.Emanuel, K. 2003. Tropical cyclones. Annual Review of Earth and Planetary Sciences 31,

75–104.

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Emanuel, K.E. 2005. Divine Wind: The History and Science of Hurricanes. New York, NY:Oxford University Press.

IUPAC. 1988. Quantities, Units and Symbols in Physical Chemistry. Oxford, UK: BlackwellScientific Publications.

Kondepudi, D. 2008. Introduction to Modern Thermodynamics. NewYork, NY: JohnWiley &Sons, Inc.

Laidler, K.J. 1993. TheWorld of Physical Chemistry. NewYork, NY: Oxford University Press.Lewis, G.N. and Randall, M. 1961. Thermodynamics, 2nd ed. New York, NY: McGraw-Hill

Book Co.Stumm,W. and Morgan, J.M. 1981. Aquatic Chemistry, 2nd ed. NewYork, NY: JohnWiley &

Sons, Inc.Young, T. 1805.Miscellaneous Works. London, England: J Murray.

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3 MulticomponentEquilibriumThermodynamics

Environmental systems are inherently complex and involve several phases, eachcontaining many components. It is a characteristic of nature, and amply demonstratedthrough the science of thermodynamics, that when two or more phases are in contactthey tend to interact with each other via exchange of matter and/or energy. The phasesinteract till a state of equilibrium is reached. Two new concepts have to be introduced,namely, fugacity and activity, to describe multi-component heterogeneous systemsthat display varying degrees of nonideality.

There are three steps involved in understanding complex heterogeneous multi-component systems: (i) translation of the real problem into an abstract world ofmathematics, (ii) solution to themathematical problem, and (iii) projection of the solu-tion to the real world in terms of meaningful and measurable parameters (Prausnitz,Lichtenthaler, and de Azevedo, 1999). The chemical potential discussed in Chapter2 is the appropriate mathematical abstraction to the physical problem. The definitionof equilibrium in terms of chemical potential is the framework for the solution of thephysical problem in the abstract world of mathematics.

Knowledge of heterogeneous multi-component equilibrium is essential in design-ing processes for environmental separations. Separation efficiency is driven by thetendency of a system to move toward equilibrium from a state of disequilibrium.

As noted in Chapter 2, the property called chemical potential introduced by Gibbsis a highly useful mathematical abstraction to physical reality. The chemical potentialis only measured indirectly. Hence, a new term was defined called fugacity.We beginwith a discussion of the fugacity (Lewis and Randall, 1961).

3.1 IDEAL AND NONIDEAL FLUIDS

The distinction between ideal and real gases is straightforward. If molecules in a gasdo not interact with one another, it is considered “ideal.” Themolecules in an ideal gashave no excluded volumes. The pressure–volume–temperature (P–V–T ) relationshipfor an ideal gas is given by the well-known “ideal gas law”:

PV = nRT . (3.1)

We know that most gases are not ideal since a gas cannot be cooled to zero volumeand, even at moderate densities, the molecules interact with one another. Moleculesin real gases have definite excluded volumes. The P–V–T relationship for a real gas

43

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is the so-called virial equation of state,

PV

RT= 1 + B(T)

V+ C(T)

V2+ · · · . (3.2)

The terms B(T) and C(T) are called second and third virial coefficients.The definition of an ideal solution is different from that of an ideal gas. For solu-

tions one cannot neglect intermolecular forces. Consider a solvent (≡A) containing asolute (≡B). The possible solute–solvent interactions are: A–A, A–B, and B–B. Thesolution is considered ideal if the three forces are identical and if there is no volumechange or enthalpy change during mixing. There also exist situations in which thereare so few B molecules among a large number of A molecules that B–B interactionsare negligible. These are considered ideally dilute solutions. For these solutions, wecan apply the limiting conditions xA → 1, xB → 0.The limiting condition is a thermo-dynamic definition of an ideal dilute solution.A practical definition of dilute solutionin environmental engineering is somewhat less stringent than the thermodynamicdefinition.

3.1.1 CONCENTRATION UNITS IN ENVIRONMENTAL ENGINEERING

Environmental concentrations (air, soil, and water) are expressed in a variety of units(Appendix 4). Traditional units such as parts per million (ppm), parts per billion(ppb), and parts per trillion (ppt) are still in use, although these units are inexact andof dubious applicability. Chemical thermodynamicists prefer towork inmole fraction,molality, and molarity units.Mole fraction is defined as the ratio of the moles of a solute to the total moles of

all species including the solvent, that is,

xi = ni∑i ni

. (3.3)

Molarity is the moles of solute per liter (dm3) of the solution.Molality is the moles of solute per kilogram of solvent.

The different concentration units are shown in Table 3.1.

TABLE 3.1Concentration Units in Environmental Chemistry

Phase Conventional Units Preferred SI Units

Water or organic solvents mg/L water mol/m3 water and mol/dm3

waterAir μg/m3 air mol/m3 airSoil or sediment mg/kg solid mol/kg solid

Note: In the aqueous phase, mg/L is equivalent to ppm; in the air phase, anotherequivalent unit to μg/m3 is parts per million by volume (ppmv). See alsoAppendix 4.

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Multicomponent Equilibrium Thermodynamics 45

3.1.2 DILUTE SOLUTION DEFINITION

Let us consider a dilute solution of A in solvent B. If we consider a fixed volume V(m3) of ideal dilute solution, then addition of A and removal of B does not affect thetotal volume, and hence the above expression can be rewritten in terms of molarity as

xA = CA

CA + CB. (3.4)

For dilute solutions CA � CB, and hence xA ∼ CA/CB. For aqueous solutions CBis the molar density of water, which we designate ρw(=55.5 mol/dm3). Thus, molefraction is proportional to molarity in a dilute aqueous solution. Relationships similarto the above can be derived for solutions in gases and solids.

EXAMPLE 3.1 CALCULATION OF CONCENTRATIONS INAN ENVIRONMENTAL

MATRIX

Problem statement: Calculate the mole fraction, molarity, and molality of the followingsolutes in water: (a) ethanol, 2 g in 100mLwater; (b) chloroform, 0.7 g in 100mLwater;(c) benzene, 0.1 g in 100mL water; and (d) hexachlorobenzene, 5 × 10−7 g in 100mLwater.

Solution: First obtain the density andmolecular weight of the compounds from standardCRC tables. Then calculate the volume of each compound and therefore the total volumeof the solution. The final results are tabulated below:

MoleMolecular Molarity Fraction Molality

Compound ρ Weight (g/cm3) (mol/dm3) (mol/kg)

Ethanol 0.79 46 0.420 7.5 × 10−3 0.430Chloroform 1.48 119 0.058 1.0 × 10−3 0.058Benzene 0.87 78 0.013 2.3 × 10−4 0.013Hexachlorobenzene 1.57 285 1.7 × 10−8 3.1 × 10−10 1.7 × 10−8

Note that only for ethanol the molality is different from molarity. At very low molefractions, molarity and molality are identical. Since the molarities are all less than2.773 mol/dm3, the solutions can be considered ideal.

3.2 FUGACITY

The concept of fugacity is an invaluable tool in constructing equilibrium modelsfor the F&T of chemicals in the environment (Mackay, 1991). Fugacity is derivedfrom the Latin word fugere, which literally means to flee. Thus, fugacity measures theescaping or fleeing tendency of amolecule froma phase. If the fugacity of a compoundis the same in two phases, then the molecule is said to be in equilibrium with bothphases. Chemical potential, although a useful quantity, is somewhat awkward to usesince it can only be estimated indirectly. For example, in Chapter 2 we noted that

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μi varies nonlinearly with Pi for gases and with xi for solutions (Section 2.5.2). Forreal, nonideal gases and solutions, the expressions for chemical potentials have tobe modified and the concept of fugacity as corrected pressure was introduced. Aswill be shown later the fugacity is directly related to equilibrium and can be obtainedexperimentally.

3.2.1 FUGACITY OF GASES

Fugacity is defined as an idealized pressure, f gi , for a real gas such that the expression

for chemical potential can be written as

μgi = μg0

i + RT ln

(f gi

f g0i

). (3.5)

The standard state fugacity is f g0i = 1 atm. For an ideal gas mixture, fugacity is the

same as partial pressure, and f gi = Pi.

Fugacity coefficient, χi = f gi /Pi, indicates the degree of nonideality of the gas

mixture. Note that Pi = yiPT, where yi denotes the mole fraction of i in the gas phaseand PT is the total pressure (1 atm for most environmental calculations). Fugacity hasdimensions of pressure and is linearly related to pressure. The ratio f g

i /f g0i is termed

activity, ai.

3.2.2 FUGACITY OF CONDENSED PHASES (LIQUIDS AND SOLIDS)

The concept of fugacity can be extended to condensed phases such as liquids andsolids. Both liquids and solids exert vapor pressure, and hence their escaping tendency(fugacity) can be evaluated similar to that of a gas. If the fugacity of the saturatedvapor at temperature T and its saturated vapor pressure P∗

i is denoted by f ∗i and thefugacity of the condensed phase (liquid l or solid s) is denoted by f ci , then the followingequation can be derived (Prausnitz, Lichtenthaler, and de Azevedo, 1999):

f ci = P∗

i χi exp

(∫PP∗i

V ci

RTdP

), (3.6)

where V ci is the partial molar volume of i in the condensed phase c, and χi = f ∗i /P∗

i .The fugacity coefficient corrects for the departure from ideal gas behavior. The secondcorrection, which is the exponential factor, is called the Poynting correction andindicates that the pressure P of the condensed phase (liquid or solid) is different fromthe condensed-phase saturation pressurePs

i . The Poynting correction is nearly 1 underthe low-pressure conditions of a few atmospheres encountered in most environmentalengineering calculations. Thus we have for both pure solids and liquids,

f ci ≈ P∗

i . (3.7)

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3.2.3 ACTIVITIES OF SOLUTES AND ACTIVITY COEFFICIENTS

The discussion of nonideal solutions (liquids and solids) is identical to that of the realgases. For real gases partial pressure should be replaced with the corrected pressure,namely fugacity in the expression for chemical potential. For nonideal solutions, asimilar substitution was suggested for mole fraction xi. This nondimensional quantitywas called activity ai. Thus we have

μli = μl0

i + RT ln

(f li

f l0i

). (3.8)

The termwithin the brackets is the activity,which is defined as the ratio of the fugacityto the fugacity at some chosen standard state. For an ideal solution ai = xi. Hencef li = xiP∗

i .An activity coefficient γi (also dimensionless) can be defined to show the departure

from ideality for solutions:

γi = aixi. (3.9)

A few words are now in order regarding the appropriate standard state chemicalpotentials. There are two standard conventions in chemical thermodynamics.

In the first convention (I), the standard state of each compound i is taken to be itspure liquid at the temperature T and pressure P of the solution.

μl0i = μ∗

i . (3.10)

In this conventionlimxi→1

γi → 1. (3.11)

In the second convention (II), the standard state is that of the pure solvent (designatedA) at temperature T and pressure P of the solution.

μl0A = μ∗

A. (3.12)

We then have the following limit:

limxA→1

γA → 1, limxi→0

γi → 1. (3.13)

Table 3.2 summarizes the definitions of fugacity and activity coefficients for mixturesof real gases and solutions (liquids or solids).

EXAMPLE 3.2 CALCULATION OF CHEMICAL POTENTIALS USING DIFFERENT

CONVENTIONS

Problem: Given the activity coefficient for chloroform measured using the two con-ventions described above in a chloroform/benzene mixture at 50◦C, calculate thecorresponding chemical potentials.

continued

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Activity Coefficient (γi )

Mole Fraction of Chloroform (xi ) Convention I Convention II

0.40 1.25 0.70

Solution: The standard state chemical potential is not given. Hence we can onlydetermine the difference in chemical potentials, Δμi = RT ln(γi · xi).Convention I:

Δμi = (8.314 J/molK)(323K) ln(1.25 · 0.4) = 240 J/mol.

Convention II:

Δμi = (8.314 J/molK)(323K) ln(0.7 · 0.4) = −383 J/mol.

EXAMPLE 3.3 CALCULATION OF THE ACTIVITY OF WATER IN SEAWATER

Problem statement: If the vapor pressure of a sample of seawater is 19.02 kPa at 291K,calculate the activity of water in the solution.

Solution: The vapor pressure of pure water at 291K is 19.38 kPa (CRC Handbook ofChemistry and Physics). Hence, awater = 19.02/19.38 = 0.9814.

TABLE 3.2Standard States, Fugacity, and Activity Coefficients for Real GasMixtures and Solutions

Fugacity Coefficient orActivity Coefficient Standard States

Real gas mixture χi = fgiPi

= fgi

yiPT

fgi → Pi as PT → 0χi → 1 as PT → 0

Real solutions ai = γixi = f li

f l0iγi → 1 as xi → 0 (convention I)

γi → 1 as xi → 1 (convention II)

Solid mixture ai = γixi = f sif si 0

f l0i = P∗i

f s0i = P∗

i

Notes: (1) For gases μgi = μg0

i + RT ln(f

gi /f

g0i

), where f g0i = 1 atm, and for solu-

tions μli = μl0i + RT ln(f li /f

l0i

).

(2) P∗i denotes saturated vapor pressure and PT is the total pressure.

(3) For a gaseous mixture f gi = y i f

g,purei and is called the Lewis–Randall rule,

that is, the fugacity of i in a mixture is the product of its mole fraction in the gasmixture and the fugacity of pure gaseous component at the same temperatureand pressure.

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3.2.4 IONIC STRENGTH AND ACTIVITY COEFFICIENTS

The activities of dissociating (ionic) and nondissociating (neutral) species in aqueoussolutions are influenced by the ionic strength of the solution. Ionic strength is thecombined effect of all ionic species in water.

The ionic strength (denoted I) is given by

I = 1

2·∑i

miz2i , (3.14)

where mi is the molality of species i (mol/kg) and zi is the charge of species i in solu-tion. Ionic strength has units of mol/kg. For example, if we consider a 1:1 electrolyte(z+ = 1; z− = −1), then I = 0.5(m+ + m−) = m. Note that I is always a positivequantity and is additive for each species in solution.

The activity of dissociating (ionic) species is related to I through theDebye–Huckelequation, which is described in Section 3.4.3.5. It is based on the fact that long-rangeand Coulombic forces between ions are primarily responsible for departures fromideality in solutions. It suffices to summarize the final equation in this context. TheDebye–Huckel theory has beenmodified and extended to higher ionic strength values.These are summarized in Table 3.3 (Pankow, 1992; Stumm and Morgan, 1996).

For nondissociating (neutral organic) species, the effect of ionic strength I onactivity coefficient is given by the McDevit–Long theory, which predicts an equationwhose general form is log γi = kI, where k depends on the type of ionic species insolution and also on the temperature and pressure. k is specific to the species i ofconcern. Because of the dependence of k on both the nature of the species i and theions, a universal equation for the effect of I on the activity coefficients of neutralspecies is not likely. Positive k values indicate that salts tend to make the solvent less

TABLE 3.3Relationships between Ionic Strength (I) and Mean Ionic ActivityCoefficient (γ±)

Name Equation Range of I

Debye–Huckel log γ± = −Az2i√I I < 10−2.3

Extended Debye–Huckel log γ± = −Az2i( √

I

1 + B√I

)I < 10−1

Guntelberg log γ± = −Az2i( √

I

1 + √I

)I < 10−1

Davies log γ± = −Az2i[( √

I

1 + √I

)− 0.2I

]I < 0.5

Source: FromPankow, J.F. 1991.AquaticChemistryConcepts. Chelsea,MI:LewisPublishers.Note: Parameters A and B depend on temperature and di-electric constant of the liquid. For

water at 298K, A = 0.51 and B = 0.33. The parameter a is an ion size parameter andis listed by Pankow (1992). Note that z2i = |z+| |z−|.

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TABLE 3.4ϕ Values for Some Organic Compounds in Seawater

Molar VolumeOrganic Compound (cm3/mol) ϕ

Naphthalene 125 0.00242Biphenyl 149 0.00276Phenanthrene 182 0.00213Dodecane 228 0.000962Tetradecane 259 0.000964Hexadecane 292 0.00233Octadecane 327 0.00290Eicosane 358 0.00190Hexacane 456 0.00488

Source: From Aquan Yeun, M., Mackay, D., and Shiu, W.Y. 1979.Journal of Chemical and Engineering Data 24, 30–34.

favorable for the solute. This is called the salting-out process and is the basis of thewidely used concept of purifying organic compounds by crystallization from theirmother liquor. Since ions tend to bind water molecules in their hydration layer, theymake less water molecules available for solubilizing organics and hence the organicstend to fall out of water. The opposite effect of salting in is caused when k is negative.AquanYeun, Mackay, and Shiu (1979) suggested that the effect of ionic strength onthe activity coefficient γi of the neutral organic species i can be correlated using thefollowing form of the Setschenow equation:

log

(γi

γ0

)= ΦViCs, (3.15)

where γ0 is the activity coefficient of the neutral solute species i in pure water.Φ is a parameter that depends on the partial molar volume of the salt in solution(Vo) and the molar volume of the liquid salt (Vs). In the above equation Vi is thepartial molar volume of the organic solute species in solution (cm3/mol) and Csis the molar concentration (mol/dm3) of the salt in solution. The values of ϕ forseveral hydrocarbon compounds found in seawater are listed in Table 3.4. The val-ues are found to range from a low of 0.000962 for dodecane to as high as 0.00488for hexadecane. A mean value of 0.0025 is used in most estimation of activitycoefficients. Most environmental waters have salt concentrations varying from 0 to0.5mol/dm3, where the latter is the mean seawater salt concentration. In seawater,the increase in activity coefficient over that of pure water is seen to be from 1.2 to 2.3for compounds of molar volumes ranging from 7.5 × 10−5 to 2.0 × 10−4 m3/mol.The ionic strength effects are therefore not entirely negligible for saline waters.

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EXAMPLE 3.4 MEAN IONIC ACTIVITY COEFFICIENT CALCULATION

Problem statement: Determine γ± for a 0.002molal solution of NaCl in water at 298K.

Solution: For a 1:1 electrolyte such as NaCl, m+ = m− = m and hence I =0.002mol/kg. Since I < 0.002, we can use the Debye–Huckel limiting law, log γ± =−(0.509)(0.002)1/2 = −0.023, γ± = 0.949.

3.2.5 FUGACITY AND ENVIRONMENTAL MODELS

Fugacity is useful to estimate the tendency of molecules to partition into the variousenvironmental compartments (Mackay, 1979). Since fugacity is identical to partialpressure in ideal gases, it is also related to the vapor pressures of liquids and solids.Fugacity is directly measurable (e.g., at low pressures fugacity and partial pressure ofan ideal gas are the same) and since it is linearly related to partial pressure or concen-tration, it is a better criterion for equilibrium than the elusive chemical potential. Infact, the criterion of equal chemical potential for equilibrium can be replaced withoutloss of generality with the criterion of equal fugacity between phases. If an aqueoussolution of a compound i is brought into contact with a given volume of air, the speciesi will transfer from water into air until the following criterion is established:

μw0i + RT ln

(f wi

f w0i

)= μa0

i + RT ln

(f ai

f a0i

). (3.16)

If we choose the same standard states for both phases, we have the following criterionfor equilibrium:

f wi = f a

i . (3.17)

The above equality will hold even when the standard states are so chosen that they areat the same temperature, but at different pressures and compositions. We then havean exact relation between the standard states, that is,

μw0i − μa0

i = RT ln

(f w0i

f a0i

). (3.18)

The equality of fugacity can be generalized for many different phases in equilibriumwith one another and containing multi-components. We now have three equivalentcriteria for equilibrium between phases as described in Table 3.5.

Mackay (1991) proposed a term fugacity capacity, Z , that related fugacity (of anyphase expressed in Pa) to concentration (expressed in mol/m3). Thus

Z(mol/m3/Pa) = C(mol/m3)

f (Pa). (3.19)

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TABLE 3.5Criteria for Equilibrium between Two Phases aand b

Property Criteria

Gibbs free energy Gai = Gb

i

Chemical potential μai = μb

i

Fugacity f ai = f b

i

Note: In terms of chemical potentials, at constant temperature andeither at constant V or P, the criteria for irreversible andreversible processes are:

Δμi < 0 (irreversible, spontaneous process)Δμi = 0 (reversible, equilibrium process)Δμi > 0 (nonspontaneous process).

TABLE 3.6Definition of Fugacity Capacities for EnvironmentalCompartments

Compartment Definition of Z (mol/m3/Pa)

Air 1/RTWater 1/Kaw

Soil or sediment Kswρs/Kaw

Biota Kbwρb/Kaw

Source: FromMackay, D. 1991.Multimedia EnvironmentalModels. Chelsea,MI: Lewis Publishers.

Note: R is the gas constant (=8.314 Pam3/mol/K), Kaw is Henry’s con-stant for species (Pam3/mol), Ksw is the partition constant for speciesbetween the soil or sediment and water (dm3/kg), Kbw is the biocon-centration factor (dm3/kg), and ρs and ρb are the densities (kg/dm3) ofsoil/sediment and biota, respectively. 1 dm3 = 1 L.

The value of Z depends on a number of factors, such as identity of the solute, nature ofthe environmental compartment, temperature T , and pressure P. A fugacity capacitycan be defined for each environmental compartment (Table 3.6). In order to obtain thevalue of Z , a knowledge of other equilibrium relationships between phases (partitioncoefficients) is required. These relationships will be described in detail in Chap-ter 4. Suffice it to say at this point that the partition coefficients are to be eitherexperimentally determined or estimated from correlations.

Once the fugacity capacities are known for individual compartments, the mean Zvalue can be determined by multiplying the respective Zj value with the volume ofthe compartment (Vj) and summing over all compartments. If the total mass of thecompound or chemical input or inventory in all the compartments is known, then the

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fugacity of the compound is given by

f =∑

j Mj∑j VjZj

. (3.20)

This fugacity is common to all phases at equilibrium. Hence, the concentration ineach compartment is given by the following equation:

Cj = f · Zj. (3.21)

More sophisticated calculations suitable for realistic situations involving time-dependent inflow and outflow of chemicals into various compartments have beenproposed and discussed in detail by Mackay (1991). The student is referred to thisexcellent reference source for more details. The purpose of this section has been toimpress upon the student how the concept of fugacity can be applied to environmentalmodeling. The following example should illustrate the concept.

EXAMPLE 3.5 FUGACITYMODEL (LEVEL I) FORENVIRONMENTALPARTITIONING

Problem statement: Consider an evaluative environment consisting of air, water, soil,and sediment. The volumes of the phases are as follows: air = 6 × 109 m3, water =7 × 106 m3, soil = 4.5 × 104 m3, and sediment = 2.1 × 104 m3. Determine the equi-librium distribution of a hydrophobic pollutant such as pyrene in this four-compartmentmodel. The properties for pyrene are as follows: Henry’s constant = 0.9 Pam3/mol;Kd(soil) = 1.23 × 103 L/kg; ρs (for sediment and soil) = 1.5 × 10−2 kg/L; andKd(sed) = 2.05 × 103 L/kg. Let the temperature be 300K and the total inventory ofpyrene be 1000mol.

Solution: The first step is to calculate the fugacity capacity Z as follows:

Air: Z1 = 1

RT= 1

(8.314 × 300)= 4 × 10−4.

Water: Z2 = 1

H= 1

0.89= 1.1.

Soil: Z3 = Kdρs

H=(1.23 × 103

) (1.5 × 10−2

)

0.89= 20.7.

Sediment: Z4 =(2.05 × 103

) (1.5 × 10−2

)

0.89= 34.5.

The second step is to calculate the fugacity f :

f = 1000

1.2 × 107= 8.5 × 10−5 Pa.

continued

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The last step is to calculate the concentrations in each phase, Cj:

Cair = f Z1 = 3.4 × 10−8 mol/m3,

Cwater = f Z2 = 9.4 × 10−5 mol/m3,

Csoil = f Z3 = 1.7 × 10−3 mol/m3,

Csediment = f Z4 = 2.9 × 10−3 mol/m3.

We can now calculate the total moles of pyrene in each compartment, mj:

mair = CairVair = 205mol,

mwater = CwaterVwater = 658mol,

msoil = CsoilVsoil = 76mol,

msediment = CsedimentVsediment = 61mol.

Thus, the largest fraction (65.8%) of pyrene is in water. The next largest fraction(20.5%) resides in air. Both sediment and soil environments contain less than 10%of pyrene each.

If there is inflow and outflow from the evaluative environment, and steady stateexists, we have to use a level II fugacity calculation. If Gj (m3/h) represents both theinflow and outflow rates from compartment j, then the total influx rate I (mol/h) isrelated to the fugacity:

f = I∑j GjZj

. (3.22)

The concentrations are then given by Cj = f Zj and mass by mj = CjVj. The totalinflux rate is given by I = E + GjC0

j , where E is the total emission rate (mol/h) from

the environment and C0j is the influent concentration (mol/m3) in the stream.

EXAMPLE 3.6 LEVEL II FUGACITY CALCULATION

For the environment consisting of 104 m3 air, 1000m3 of water, and 1 m3 sediment,an air flow rate of 100m3/h, a water inflow of 1m3/h, and an overall emission rateof 2mol/h are known for pyrene. The influent concentration is 0.1mol/m3 in air and1mol/m3 in water. Given Z values of 10−4 for air, 0.1 for water, and 1 for sediment,calculate the concentration in each compartment.Total influx I = 2(mol/h) + 100(m3/h)0.1(mol/m3) + 1(m3/h)1(mol/m3) =

13 mol/h. Fugacity f = 13/[(100)(10−4) + (1)(1)] =13 Pa. Cair = (13)(0.0001) =0.0013 mol/m3, Cwater = (13)(0.1) = 1.3mol/m3, and Csed = (13)(1) = 13mol/m3.

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3.3 IDEAL SOLUTIONS AND DILUTE SOLUTIONS

3.3.1 VAPOR–LIQUID EQUILIBRIUM: HENRY’S AND RAOULT’S LAWS

There are two important relationships that pertain to the equilibrium between anideal solution and its vapor. These are called Henry’s law and Raoult’s law. We havealready alluded to these laws, although not explicitly, in the selection of standard statesfor activity coefficients.These two laws have significant applications in environmentalengineering, which are discussed in Chapter 4.

3.3.1.1 Henry’s Law

On a purely experimental basis, the English chemist William Henry noted that theamount of a gas dissolved in a liquid was proportional to the gas partial pressure abovethe liquid. This proportionality constant is called Henry’s constant and is denoted asKH. It has units of pressure (atmosphere, Torr, or Pascal).

Pi = KH · xi (3.23)

Henry’s constant is expressed in a variety of other units. If the liquid and gaseous con-centrations are expressed on a molar basis (mol/dm3), then a dimensionless Henry’sconstant (Kaw) can be obtained. If both the liquid and gas-phase concentrations areexpressed as mole fractions, another dimensionless value can be obtained for Henry’sconstant (KX). If the gas-phase concentration is denoted in pressure units (Pa) andthe liquid-phase concentration is in molarity (mol/dm3), a different unit for Henry’sconstant (K ′′

aw, Pa dm3/mol) can be obtained. One should be very careful in noting the

correct units for Henry’s constants obtained from the literature. Table 3.7 summarizesthese definitions and their interrelationships.

The constantKH not only depends on the nature of species i but also on temperatureand pressure. The criterion for equilibrium is that the gas-phase fugacity of i and theliquid-phase fugacity of i should be the same.

f ai = f w

i . (3.24)

TABLE 3.7Henry’s Constant Definitions for Vapor–Liquid Equilibriumand Their Interrelationships

Definition Units Relationship to KH

KH = Pi/xi Pa —Kaw = Cia/Ciw Dimensionless Kaw = (Vw/RT)KHK ′aw = yi/xi Dimensionless K ′

aw = (Va/RT)KHK ′′aw = Pi/Ciw Pam3/mol K ′′

aw = Vw · KHNote: Vw is the partial molar volume of water (= 0.018 L/mol at 298K) and Va the partial

molar volume of air (= 22.4 L/mol at 298K), Cia and Ciw are, respectively, themolar concentrations of solute in the air and water, and yi the mole fraction ofsolute i in the air.

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Therefore,

Pi = γixi f w0i . (3.25)

Hence we have

KH = Pixi

= γi f w0i . (3.26)

In the above equation, f w0i is the standard state liquid fugacity. Thus, if one knows

the activity coefficient of i in the liquid phase and the standard state fugacity, thenHenry’s law constant can be obtained. If the liquid phase is ideal and the more generalfugacity is used instead of partial pressure ( f ai = f w

i ), then we can write a generalexpression for KH as follows:

KH = limxi→0

f ai

xi. (3.27)

The most general definition of Henry’s law is

KH = yiχiPTγixi

. (3.28)

The above definition recognizes nonideality in both liquid and gas phases.

3.3.1.2 Raoult’s Law

Raoult’s law is an important relationship that describes the behavior of ideal solutions.Its applicability is predicated on the generally similar characters of both solute andsolvents, and hence a mixture of the two behaves ideally over the entire range of molefractions. Let us consider a solvent i in a mixture in equilibrium with its vapor,

f vapori = f liquid mixture

i , (3.29)

Pi = γixiP∗i . (3.30)

Noting that for a ideal mixture γI = 1, we have

xi = PiP∗i. (3.31)

This is called Raoult’s law.Figure 3.1 is a description of the applicability of Henry’s and Raoult’s laws for

mixtures. As shown in the figure, when the mole fraction of all solutes is very small,the solvent obeys Raoult’s law while the solute obeys Henry’s law. Solute obeyingboth of the laws is ideal in nature. Over the range when the solvent obeys Raoult’slaw, the solute will obey Henry’s law.

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Ideal solutionRaoult’s law

Nonideal solutionIdeal dilutesolutionHenry’s law

Pi = KH xi

Pi = Pi*xii

P i

Xi

FIGURE 3.1 Illustration of Henry’s law and Raoult’s law. For a component that is pure (thesolvent) and behaves ideally it follows Raoult’s law, where the partial pressure is proportionalto the mole fraction.When it is the minor component (the solute) in a dilute solution, its partialpressure is again proportional to its mole fraction, but has a different proportionality constant,which is the Henry’s law.

EXAMPLE 3.7 DIFFERENT UNITS FOR HENRY’S LAW CONSTANT

The Henry’s law constant (molar concentration ratio, Kaw) for benzene is 0.225.Calculate the value in other units (KX, KH, and K ′

aw).Use Table 3.5. Note Vw = 0.018 L/mol and Va = 22.4 L/mol. Hence, KH = (0.225)

(0.08205 ∗ 98/0.018) = 306 atm (= 3.1 × 107 Pa), KX = 22.4 ∗ 306/(0.082205 ∗298) = 280, and K ′

aw = 0.018 ∗ 306 = 5.5 L atm/mol (= 5.6 × 105 Pa dm3/mol).

EXAMPLE 3.8 ACTIVITY COEFFICIENTAND RAOULT’S LAW

Consider a mixture of chloroform and propanone for which the partial pressureof chloroform at a liquid mole fraction of 0.2 is 35Torr. If the pure compound vaporpressure of chloroform is 293Torr, calculate the activity coefficient of chloroform inthe mixture.

ai = PiP∗i

= 35

293= 0.12.

Hence

γi = aixi

= 0.12

0.20= 0.597.

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3.3.2 VAPOR PRESSURE OF ORGANIC COMPOUNDS, CLAUSIUS–CLAPEYRON

EQUATION

The definition of vapor pressure P∗i is based on the equilibrium between a pure

component and its vapor. It is the equilibrium pressure of the vapor in contactwith its condensed phase (that is, a liquid or solid). If a pure liquid is in equilib-rium with its vapor, one intuitively pictures a static system. From a macroscopicpoint of view this is indeed correct. But from a molecular point of view the sit-uation is far from serene. In fact, there is continuous interchange of molecules atthe surface, which is in a state of dynamic equilibrium. Temperature will greatlyinfluence the dynamic equilibrium and hence P∗

i is sensitive to temperature. More-over, it should be obvious that since the intermolecular forces are vastly differentfor different compounds, the range of P∗

i should be large. For typical compounds ofenvironmental significance, the range is between 10−12 and 1 bar at room temperature(see Appendix 1).

In order to formulate the thermodynamic relationships involving P∗i for solids

and liquids, we shall first study how a pure condensed phase (e.g., water) behavesas the pressure and temperature are varied. This variation is usually represented ona P–V–T plot called a phase diagram. Figure 3.2 is the phase diagram for water.Each line in the diagram is a representation of the equilibrium between the adja-cent phases. For example, line AC is the equilibrium curve between the vapor andliquid phases. Point A is called the triple point, which is the co-existence pointof all three phases (ice, liquid water, and water vapor) in equilibrium. By defini-tion the triple point of water is at 273.16K. The pressure at this point for water is

WaterIce

Vapor

Sub-cooled liquid

B

B'A

CCritical point

Tb = 372.98 KTm = 273.0004 KTt = 273.01 K

Pres

sure

/atm 1

Temperature/K

FIGURE 3.2 Schematic of the phase diagram for water.

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4.585Torr. The boiling point of the liquid Tb at a given pressure is the temperatureat which P = P∗

l while the normal boiling point is that at which P∗l = 1 atm. BA

is the solid–vapor equilibrium line. Similar to the normal boiling point for liquids,the normal melting point is the temperature at which P∗

s = 1 atm. Beyond point C,liquid and vapor phases cannot co-exist in equilibrium; this is called the criticalpoint. The critical temperature, Tc, critical pressure, Pc, and critical volume, Vc, areunique to a compound. For water Tc is 647K and Pc is 218 atm. The phase abovethe critical point of a compound is called super-critical state. Extending the lineBA beyond A to B′A, one obtains the hypothetical sub-cooled liquid state, whichis important in estimating the solubility of a solid in a liquid. The ratio of thesub-cooled liquid vapor pressure to the solid vapor pressure is the fugacity of thesolid.

The equilibrium at any point along the equilibrium P–T line in Figure 3.6 canbe described in terms of chemical potentials. If I and II represent two phases inequilibrium, then we have the following criteria:

−SIdT + V IdP = −SIIdT + V IIdP. (3.32)

Hence

dP

dT= ΔS

ΔV= ΔH

TΔV, (3.33)

where ΔS and ΔV are the entropy and volume changes for the phase transitions Iand II. We also used the definition of entropy change, ΔS = ΔH/T .

For molar changes we then have the following equation:

dP

dT= ΔHm

TΔVm. (3.34)

This is the Clausius–Clapeyron equation. For both vapor–liquid and vapor–solidequilibrium it gives the change in vapor pressure with temperature (Table 3.8). SinceΔHm andΔVm are positive for liquid → vapor and solid → vapor transitions, dP/dTis always positive, that is, both liquid and solid vapor pressure increase with temper-ature. For both liquid → vapor and solid → vapor transitions, the molar volume ofthe gas is much larger than that of the liquid or solid. Hence, ΔVm ≈ Vg

m = RT/P.Thus

dlnP

dT= ΔHm

RT2. (3.35)

The quantity ΔHm is not independent of T . However, over small ranges oftemperature, it can be assumed to be constant. Then we can integrate the aboveequation to obtain

ln

(P2P1

)= −ΔHm

R

(1

T2− 1

T1

). (3.36)

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TABLE 3.8Antoine Constants for Some EnvironmentallySignificant Compounds

Compound Range of T (◦C) A B C

Organic compoundsChloroform −35–61 6.493 929.4 196.0Benzene 8–103 6.905 1211.0 220.8Biphenyl 69–271 7.245 1998.7 202.7Naphthalene 86–250 7.010 1733.7 201.8Tetrachloroethylene 37–120 6.976 1386.9 217.5Pyrene 200–395 5.618 1122.0 15.2p-Dichlorobenzene 95–174 7.020 1590.9 210.2Pentafluorobenzene 49–94 7.036 1254.0 216.0Inorganic compoundsNitrogen 7.345 322.2 269.9Carbon dioxide 9.810 1347.7 273.0Hydrogen peroxide 7.969 1886.7 220.6Ammonia 9.963 1617.9 272.5Sulfur dioxide 7.282 999.9 237.2

IfP1 = 1 atm (as formost environmental engineering problems), T1 = Tb, the normalboiling point of the liquid, and hence

lnP∗ = −ΔHm

RT+ ΔHm

RTb. (3.37)

Within our approximation if we plot ln P versus 1/T , we should get a straight linewith a slope ΔHm/R and a constant intercept of ΔHm/RTb, that is, vapor pressuresof liquids and solids vary as ln P∗

i ≈ A− B/T . If the temperature range is significantsuch that ΔH �= constant, then the acceptable form of fitting the ln P∗

i versus T datais given by the Antoine equation.

lnP∗i = A− B

t + C, (3.38)

where t is usually expressed in Celsius. A, B, and C are constants specific for acompound. Table 3.8 lists the values for some selected compounds.

For the solid–vapor equilibrium involving the direct transition from solid to vapor(sublimation), the heat of sublimation is composed of two parts, that is, the heat ofmelting (solid to sub-cooled liquid) and the heat of vaporization of the sub-cooledliquid. If the temperature over which the vapor pressure of a solid is being monitoredincludes the melting point (Tm) of the solid, then the ln P∗

i versus 1/T curve willshow a change in slope at Tm. We consider the pure liquid as the reference phasefor chemical potential and fugacity when calculating free energy changes for phasetransfer processes. For compounds that are solids at room temperature, the appropriatechoice is the hypothetical standard state of the sub-cooled liquid.

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TABLE 3.9Thermodynamic Transfer Functions for Methane from DifferentSolvents (A) to Solvent B (Water, W) at 298 K

Solvent A ΔG0i (kJ/mol) ΔH0

i (kJ/mol) ΔS0i (kJ/mol)

Cyclohexane 7.61 −9.95 −17.561,4-Dioxane 6.05 −11.89 −17.94Methanol 6.68 −7.97 −14.65Ethanol 6.72 −8.18 −14.901-Propanol 6.69 −8.89 −15.561-Butanol 6.57 −7.14 −13.721-Pentanol 6.48 −8.35 −14.85

Source: From Franks, F. 1983.Water. London, England: The Royal Society of Chemistry;Ben Naim, A. 1980. Hydrophobic Interactions. NewYork: Plenum Press.

The entropy of vaporizationΔSv is nearly constant (≈88 J/mol/K) for a variety ofcompounds.This is calledTrouton’s rule. The constancy of the entropy of vaporizationis because the standard boiling points of liquids are roughly equal fractions of theircritical temperatures. Most liquids behave alike not only at their critical temperaturesbut also at equal fractions of their critical temperatures. Hence different liquids shouldhave about the same entropy of vaporization at their normal boiling points. Therefore,we have a convenient way of estimating the molar heat of vaporization of a liquidfrom ΔSv = ΔHv/Tb = 88 J/mol/K. Since the molar enthalpy of vaporization is anapproximate measure of the intermolecular forces in the liquid, Kistiakowsky (1923)obtained the following equation:

ΔSv = 36.6 + 8.3 ln Tb. (3.39)

Notice from Table 3.9 that Trouton’s rule fails for highly polar liquids and for liquidsthat have Tb < 150K.

For a considerable number of compounds of environmental significance, partic-ularly high molecular weight compounds, reliable vapor pressure measurements arelacking. As a consequence, we have to resort to correlations with molecular andstructural parameters.

Using the Clausius–Clapeyron equation as the starting point and the constancy ofheat capacity, Cp, and Trouton’s rule, the following equation can be derived for thevapor pressure of liquids, P∗

i (Schwarzenbach, Gschwend, and Imboden, 1993):

lnP∗i = 19

(1 − Tb

T

)+ 8.5 ln

(Tb

T

), (3.40)

where P∗i is in atmospheres.

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10010–110–210–310–410–510–610–710–810–910–1010–11

10–1

10–2

10–3

10–4

10–5

10–6

10–7

10–8

10–9

10–10

100

Benzanthracene

Anthracene

Phenanthrene

Lindane

Tetrachloroethylene

n-decaneAniline

NaphthalenePhenol

Benzene

Calculated vapor pressure / kPa

Expe

rimen

tal v

apor

pre

ssur

e/kP

a

FIGURE 3.3 A parity plot of the experimental and predicted vapor pressure of selected liquidand solid organicmolecules. (FromSchwarzenbach, R.P., Gschwend, P.M., and Imboden, D.M.1993. Environmental Organic Chemistry. NewYork: Wiley.)

For solids, the vapor pressure P∗i is related to the sub-cooled liquid vapor pressure,

P∗s(l), as follows (Prausnitz et al., 1980; Mackay, 1991):

ln

(P∗i

P∗s(l)

)= −6.8

(Tm

T− 1

). (3.41)

Figure 3.3 is a comparison of experimental vapor pressures plotted against thepredicted values using the equations above for some of the compounds.

EXAMPLE 3.9 VAPOR PRESSURE ESTIMATION

Estimate the vapor pressure of benzene (a liquid) and naphthalene (a solid) at roomtemperature (25◦C).For benzene, which is a liquid (Tb = 353K), ln P∗

i = 19(1 − 353/298) + 8.5 ln(353/298) = −3.50 + 1.44 = −2.06. Hence vapor pressure is 0.12 atm.

For naphthalene, which is a solid (Tm = 353K, Tb = 491K), ln (P∗i /P

∗s(l)) =

−6.8(353/298 − 1) =−1.255. For T<Tb, ln P∗s(l) = 19(1 − 491/298) + 8.5 ln(491/

298) = −8.06. P∗s(l) = 3.1 × 10−4 atm. Hence P∗

i = 8.8 × 10−5 atm.

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3.3.3 VAPOR PRESSURE OVER CURVED SURFACES

Consider the formation of fog droplets in the atmosphere, and condensation (nucle-ation) of small clusters leading to the formation of aerosols, cloud droplets, andraindrops. All of these involve highly curved interfaces. The development of equi-librium thermodynamic quantities (free energy and chemical potential) for thesesystems involves modifications to the vapor pressure relationships for solutes dis-tributed between a liquid and vapor phase. The vapor pressure over a curved surfaceis dependent on its radius of curvature. Let us consider the curved surface of a liquid incontact with its vapor. FromChapter 2we have at constant T , the following expressionfor the molar free energy change:

ΔG = VmΔP = Vm · σ(1

r1+ 1

r2

), (3.42)

where Vm is the molar volume of the liquid and r1 and r2 are the principal radii of cur-vature of the surface. Over a plane surface we can use theYoung–Laplace equation forΔP from Chapter 2. Further the chemical potential between the curved surface (withvapor pressureP∗c

i ) and a plane surface (with saturation vapor pressureP∗i ) is given by

ΔG = μci − μi = RT ln

(P∗ci

P∗i

). (3.43)

Equating the two free energy differences, we have the following equation:

P∗ci

P∗i

= exp

[σVm

RT

(1

r1+ 1

r2

)]. (3.44)

In environmental engineering, we are particularly interested in spherical surfaces(e.g., fog, rain, cloud, and mist) for which r1 = r2 = r. Hence we have

P∗ci

P∗i

= exp

[2

r· σVm

RT

]. (3.45)

This is the Kelvin equation, which gives the vapor pressure over a curved surface, P∗ci

relative to that over a plane surface, P∗i given the surface tension of the liquid, radius

of the drop, and temperature. For a solid crystal in equilibrium with a liquid also theKelvin equation applies if the vapor pressures are replaced with the activity of thesolute in the solvent.

Consider the case of water, the most ubiquitous of phases encountered in envi-ronmental engineering. The surface tension of water at 298K is 72mN/m and itsmolar volume is 18 × 10−6 m3/mol. Figure 3.4 shows the value of P∗c

i /P∗i for vari-

ous size water drops. When r ≥ 1000 nm the normal vapor pressure is not affected,whereas for r ≤ 100 nm there is an appreciable increase in vapor pressure. For liquidsof large molar volume and surface tension, the effect becomes even more significant.An example is mercury, which vaporizes rapidly when comminuted. The conclusion

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0.00 20 40 60 80 100 120

0.5

1.0

1.5

2.0

2.5

3.0

Droplet in airMeniscus in a capillary

Radius/μm

P c* /P

*

FIGURE 3.4 Application of the Kelvin equation for water droplets in air and water confinedin a capillary.

is that in atmospheric chemistry and water chemistry, for very small sizes the effectof radius on vapor pressure should not be neglected.

The Kelvin effect has been experimentally verified for a number of liquids downto dimensions as small as 30Å (Israelchvili, 1992). It provides the basic mechanismfor the super-saturation of vapors. The nucleation and formation of clusters from thevapor phase starts with small nuclei that grow to macroscopic size in stages. Thepresence of dust or other foreign particles augments the early stages of nucleation.In the absence of dust, the enhanced vapor pressure over curved surfaces provides anenergy barrier, and hence the early stage of nucleation will require activation energy.These and other implications of the Kelvin equation in environmental engineeringwill become clear when we discuss the theory of nucleation of atmospheric particlesin Chapter 4.

Now, consider the reverse situation of vapor pressure of liquids confined in smallcapillaries or pore spaces such as soils and sediments (Figure 3.5). The situation isopposite to that of the liquid drops mentioned above. The curvature of the surface isof opposite sign, and the vapor pressure is reduced relative to that at a flat surface.Therefore we have

P∗ci

P∗i

= exp

[−2

r

σVm

RT

]. (3.46)

Figure 3.6 also shows this relationship for different pore diameters. Liquids that wetthe solid will therefore condense into pores at pressures below the equilibrium vaporpressure corresponding to a plane surface. This is termed capillary condensation and

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r

FIGURE 3.5 The meniscus for water confined in a capillary.

is an important process in soil matrices. It is important in understanding the infiltrationof nonaqueous phase liquids in to sub-surface soil. The phenomenon is also importantin understanding the nucleation of bubbles in a liquid. To support a vapor bubble ofradius r in water, the pressure must exceed that of the hydrostatic pressure by 2σ/r.For a 100 nm radius bubble in water at room temperature, this gives a pressure of14.6 atm. To nucleate a bubble of zero radius (i.e., to start boiling) therefore, we needinfinite pressure. This is one of the reasons for the significant super-heating requiredfor boiling liquids.

0.00320 0.00325 0.00330 0.00335 0.00340 0.003450

2000

4000

6000

8000

10000

12000

CCl4Cl2CH–CH3

Cl3C–CH3

(1/T)/K–1

γ i*

FIGURE 3.6 Variation of activity coefficients at infinite dilution in water for typical environ-mentally significant compounds versus temperature. (From Tse, G., Orbey, H., and Sandler,S.I. 1992. Environmental Science and Technology 26, 2017–2022.)

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3.3.4 LIQUID–LIQUID EQUILIBRIUM

3.3.4.1 Octanol–Water Partition Constant

If two liquid phases are in contact, and a solute is present in both, then at equi-librium we have a distribution of solute between the two phases consistent withequal chemical potentials or fugacity values. In environmental engineering, liquid–liquid equilibrium is common: for example, the distribution of organic chemicals inthe water environment where a third phase (e.g., oil) is present, such as occurringduring oil spills at sea and inland waterways, floating oils in wastewater treatmentplants, and sub-surface spills in contact with groundwater. Solvent extraction is awell-known operation in environmental engineering separation processes. Since theconcept of like dissolves like is mostly true, it should be expected that most organiccompounds would have a greater affinity for organic solvents and substrates. A spe-cific liquid–liquid system (octanol–water) has special relevance to environmentalengineering.

Consider a solute i that is distributed between two solvents (octanol ≡o and water≡w). At equilibrium the solute i should have equal fugacity in both phases. Thus

f oi = f w

i , (3.47)

the total number of moles of i in the system. We have

xoi γ

oi f

loi = xw

i γwi f

loi , (3.48)

and therefore

K∗ow = xo

i

xwi

= γwi

γoi

(3.49)

is the partition constant for a solute between two phases defined as the ratio of molefractions.

A large body of literature exists on the partitioning of a variety of environmen-tally significant compounds between the organic solvent, 1-octanol, and water (Leo,Hansch, and Elkins, 1971). The octanol–water partition constant defined in terms ofthe ratio of the molar concentrations of solute i in both phases is designated Kow:

Kow = CioCiw

= xoi Vw

xwi Vo

=(γwi

γoi

)(Vw

Vo

), (3.50)

where Vw and Vo are the partial molar volumes of water and 1-octanol, respectively.The availability of such a large database on Kow is not entirely fortuitous. It haslong been a practice in pharmaceutical sciences to seek correlations of the variousproperties of a drugwith itsKow. 1-Octanol appears tomimic the lipid content of biotaverywell.A similar reasoning led to its acceptance as a descriptor of chemical behaviorin the environment. 1-Octanol has the same ratio of carbon to oxygen as the lipidsand represents satisfactorily the organic matter content in soils and sediments. It is

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TABLE 3.10Values of γw

i and γoi forTypical Organic Compounds at 298 K

Compound γoi γw

i

Benzene 2.83 2.4 × 103

Toluene 3.18 1.2 × 104

Naphthalene 4.15 1.4 × 105

Biphenyl 5.30 4.2 × 105

p-Dichlorobenzene 3.54 6.1 × 104

Pyrene 8.66 9.6 × 106

Chloroform 1.40 8.6 × 102

Carbon tetrachloride 3.83 1.0 × 104

Source: FromMackay, D. 1982.Volatilization of Organic Pollutants fromWater.EPA Report No: 600/3-82-019, NTIS No: PB 82-230939. Springfield,VA: National Technical Information Service; Chiou, C.T. 1981. Parti-tion coefficient and water solubility in environmental chemistry. In: J.Saxena and F. Fisher (eds), Hazard Assessment of Chemicals, Vol. 1.New York: Academic Press; Yalkowsky, S.H. and Banerjee, S. 1992.Aqueous Solubilities—Methods of Estimation for Organic Compounds.NewYork: Marcel Dekker.

also available in pure form readily and is only sparingly soluble in water. Appendix 1lists the log Kow values for a variety of compounds.

It is important to note that generally large Kow values are associated with com-pounds that have low affinity with the aqueous phase. This becomes clear when onenotes that most solutes behave ideally in octanol, and hence γo

i varies only slightly(from about 1 to 10) whereas γw

i varies over several orders of magnitude (0.1–107)(see Table 3.10). Since both Vo and Vw are constants, the variation in Kow is entirelydue to variations in γw

i . In other words, Kow is a measure of the relative nonidealityof the solute in water as compared to that in octanol. Hence Kow is taken to be ameasure of the hydrophobicity or the incompatibility of the solute with water.Kow values in the literature are reported at “room temperature.” This means the

temperature is 298K with occasional variability of about 5◦. It is important to notethat the temperature dependence is nearly negligible for these temperature variations.

Theremay be caseswhen reliable experimentalKow values are not available. Underthese circumstances it is possible to estimate Kow from basic structural parameters ofthe molecule (Lyman, Reehl, and Rosenblatt, 1990). Langmuir (1925) first suggestedthat the interaction of a molecule with a solvent can be obtained by summing theinteractions of each fragment of the molecule with the solvent. The same principlewas extended to octanol–water partition ratios byHansch andLeo (1979).Thismethodinvolves assigning values of interaction parameters to the various fragments that makeup a molecule. For example, an alkane molecule (ethane, CH3–CH3) is composed oftwo –CH3 groups, each contributing equally toward the Kow of the molecule. Thereare two parts to this type of calculation: a fragment constant (b) and a structural factor(B). It is presumed that from chemical to chemical these constants are the same for a

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specific sub-unit and that they are additive. Hence we can write

logKow =∑j

bj +∑k

Bk . (3.51)

The fragment constants are fundamental to the sub-unit jwhile the structural factorsrelate to specific intermolecular forces between the sub-units. The value of bj for aspecific subunit will be different based on which other atom or sub-unit it is attachedto. For example, a −Cl atom attached to an alkane C has a different bj from the oneattached to an aromatic C. Furthermore, a −Cl attached to an unsaturated −C≡ unitwill be different from the one attached to a−C= unit. In effect theπ-electron cloud ofthe−C/unit reacts differently from the one in a−C= unit. Substituents attached to anaromatic C unit typically make less of a contribution to logKow than those attached toan alkane C atom. One should also expect widely different contributions if a nonpolargroup such as−CH3 replaces the−Cl atom in themolecule.The intermolecular forcesamong fragments and subunits are characterized by the B factor. The more complexthe stereochemistry of the molecule, the less contribution it makes toward logKow.The interactions between polar moieties in themolecule give rise to electronic factors.Increasing polarity invariably decreases the contribution toward logKow.

For almost any new compound manufactured today, one can obtain logKowin this fashion. Computer software exists to compute logKow using this methodwithout even knowing the structure of the molecule. The reliability of the methodis, however, questionable for structurally complex molecules. Nevertheless, it is anacceptable method of estimation for many environmentally significant compoundsfor which the structures are well established. Appendix 4 lists the b and B factors forsome typical fragments. A detailed listing of the parameters is available in Lymanet al. (1990).

Since experimental values are always the most useful ones, another approach toobtaining logKow exists that utilizes the logKow of known compounds. This is doneby adding or subtracting appropriate b and B values from the parent compound.

logKow(new) = logKow(parent) ±∑j

bj ±∑k

Bk . (3.52)

There are a set of specific rules that must be followed to estimate the octanol–waterpartition coefficients for a given molecule based on the specific values of b and Blisted in the literature. The books by Hansch and Leo (1979) or Lyman et al. (1990)give more details on complex structures and a greater depth of discussion.

EXAMPLE 3.10 ESTIMATION OF LOG kow USING FRAGMENT CONSTANT

METHOD

Let us determine the logKow values of the following compounds: hexane, chloroben-zene, cyclopentane, and chlorobiphenyl using data from Appendix 3. The firstinformation we need in each case is the correct molecular structure.

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Hexane: H3C–(CH2)4–CH3. We have six C atoms contributing 6bc and 14 H atomscontributing 14bH. Therefore Σbj = 6(0.20) + 14(0.23) = 4.42.If there are n bonds, they contribute to (n−1)Bb factors. In this case, there are five C–Cbonds and hence ΣBk = (5 − 1)(−0.12) = −0.48. Hence logKow = 4.42 − 0.48 =3.94. The experimental value is 4.11. Hence we have a 4% error in our estimation.

Chlorobenzene: C6H5Cl. Since the logKow value of benzene (C6H6) is reported to be2.13, we only need to subtract the fragment constant for 1H bonded to an aromatic ring(bϕH) and add that for one Cl atom bonded to an aromatic ring (bϕH). Thus logKow =2.13 − 0.23 + 0.94 = 2.84. The experimental value is 2.98. The error is 5%.

Cyclopentane: C5H10. Fragment factors: 5bc + 10bH = 3.30. Structural factors:(5 − 1)Bb = −0.36. Hence logKow = 2.94. The experimental value is 3.00.

Chlorobiphenyl:H5C6–C6H4Cl. Since for biphenyl the value is 4.09, for chlorobiphenylit is 4.09 − 0.23 + 0.94 = 4.8.

3.3.4.2 Linear Free Energy Relationships

It is important to note that other solvent–water partition constants can be related toKow (Collander, 1951). The partition constant between octanol and water is relatedto other partition constants as

logKsw = a logKow + b, (3.53)

where a and b are constants. This is called a linear free energy relationship (LFER).The slope a is a measure of the relative variability of the activity coefficient of the

solute in the two solvents. The intercept b is a constant and can be regarded as the valueof logKsw for a hypothetical compound whose logKow is zero. A number of solvent–water partition constants have been related to Kow. If the solvent is similar in natureto octanol, a high degree of correlation can be expected. It should be rememberedthat using Kow as the reference emphasizes that it is the high activity coefficientcompounds in water that account for the unique partitioning behavior. The changesin logKow and logKsw track one another only for homologous series of compounds.If the compounds vary in their characteristics (polarity and stereochemistry), then thelinear relationship will be less than satisfactory. The situation then calls for separateLFERs for classes of compounds that resemble one another. In other words, a singlecorrelation for all types of compounds is untenable.

3.4 NONIDEAL SOLUTIONS

This section is devoted to nonideal behavior that is common to environmental systems.

3.4.1 ACTIVITY COEFFICIENT FOR NONIDEAL SYSTEMS

Mixtures of real fluids (gases or liquids) do not form ideal solutions, although similarfluids approach ideal behavior. All nonelectrolytes at their infinite dilution limit in

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solution follow ideal behavior.Asmentioned in the previous sections, the activity coef-ficientγi represents nonideal behavior in aqueous systems.The activity coefficient canbeobtained experimentally, andwhere it cannot be determineddirectly, chemical engi-neers have devised theoretical models to compute the activity coefficients from cor-relations with other solute parameters (e.g., surface area, volume, and octanol–waterpartition constants) or group-interaction parameters (e.g., UNIFAC and NRTL).

3.4.1.1 Excess Functions and Activity Coefficients

Apart from the activity coefficient that is a measure of nonideality, another quantitythat is also indicative of the same is the excess partial molar Gibbs energy denotedby gE and defined as Gactual − Gideal. This is an experimentally accessible quantity.Other thermodynamic functions such as excess molar enthalpy and entropy can alsobe defined in a similar fashion. By definition all excess functions are zero for xi = 0and xi = 1. Some authors use excess Gibbs function as a more appropriate measureof deviation from ideality than activity coefficient. The relationship between excessfunction and activity coefficient is easily derived. For component i in solution, theexcess molar Gibbs free energy is defined as

gEi = RT ln

(fi(actual)

fi(ideal)

)= RT ln γi, (3.54)

where fi(actual) = xiγi f l0i and fi(ideal) = xi f l0i .The excess molar Gibbs function for solution is then given by

gE =∑i

xigEi = RT

∑i

xi ln γi. (3.55)

EXAMPLE 3.11 EXCESS GIBBS ENERGY OF SOLUTION

A solution of benzene has a benzene mole fraction of 1 × 10−5 and a benzene activitycoefficient of 2400. What is the excess Gibbs free energy of benzene in solution?gEi = RT ln γi = 8.314 × 10−3 × (kJ/Kmol) × 298 (K)7.78 = 19.3 kJ/mol.

3.4.2 ACTIVITY COEFFICIENT AND SOLUBILITY

Solubility in liquids (especially water) is of special relevance in environmental engi-neering since water is themost ubiquitous of all solvents on earth. Hence considerableeffort has gone into elucidating the solubility relationships of most gases, liquids, andsolids in water. Water is called a universal solvent and it richly deserves that name.Some compounds are easily soluble in water to very high concentrations; some arecompletely miscible with water at all proportions, and some have very limited solu-bility in water. These extremes of solubility are of special interest to an environmentalchemist or engineer. As we will see in the next section, the structure of water stillevokes considerable debate.

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The solubility of gases in liquids is obtained from Henry’s law.Let us now consider a sparingly soluble liquid (say a hydrocarbon H) in contact

with water. The solubility of the liquid hydrocarbon in water may be considered tobe an equilibrium between a pure phase (H) and an aqueous phase (W). Applying thecriterion of equal fugacity at equilibrium,

f wi = f H

i ,

γwi x

wi f

l0i = γH

i xHi f

l0i ,

(3.56)

where f l0i is the pure component reference fugacity of i at system temperature. Forpure H we have xH

i = 1 and γHi = 1. Noting that xw

i = x∗i , the saturation solubility in

water, we can write

x∗i = 1

γ∗i. (3.57)

The above equation states that the activity coefficient of a saturated solution of asparingly soluble compound in water is the reciprocal of its saturation mole fractionsolubility in water.

For the compound i which is only sparingly soluble in water, we also have thefollowing Henry’s law expression:

KH = γ∗i P

∗i = 1

x∗i

· P∗i . (3.58)

For solutes that are solids, the basic condition of equal fugacity still holds atequilibrium. If solid (s) is a pure species in contact with water (w),

f s0i = γwi x

wi f

l0i (3.59)

where f s0i is the fugacity of pure solid i and f l0i is the fugacity of pure liquid i. Thisis the hypothetical sub-cooled liquid state for compounds that are solids at roomtemperature (see the discussion on phase diagrams). Thus the saturation solubility ofi in water is given by

x∗i =

(1

γ∗i

)(f s0if l0i

)=(

1

γ∗i

)exp

[−ΔHm

R

{1

T− 1

Tm

}]. (3.60)

The ratio of fugacity coefficients was obtained from Prausnitz et al. (1999). In theabove equation,ΔHm is the molar enthalpy of fusion or melting (J/mol) and Tm is themelting point of the solid (K). Since according to Trouton’s rule ΔSm = ΔHm/Tm,and for most organic compounds ΔSm/R ∼ 13.6 entropy units, we have

x∗i =

(1

γ∗i

)exp

[6.8

(1 − T

Tm

)]. (3.61)

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For sparingly soluble organic compounds in water, it is possible to equate γ∗i with

γ∞i , the so-called infinite dilution activity coefficient.The condition of infinite dilution

is the limit as xi → 0. For a large number of sparingly soluble organics in water,this condition is satisfied even at their saturation solubility. Hence the two activitycoefficients are indistinguishable. Themole fraction at saturation formost compoundsof environmental significance lies between 10−6 and 10−3. The mole fraction of 10−6

is usually taken as the practical limit of infinite dilution. For those compounds forwhich the solubility values are very large, γ∗

i is significantly different from γ∞i .

Using the Gibbs–Helmholtz relation obtained in Chapter 2 and replacing the Gibbsfree energy change in that expression by the excess molar Gibbs free energy of dis-solution of liquid i, we can derive the following fundamental relationship for activitycoefficient:

d ln γ∗i

dT= − hE

i

RT2, (3.62)

where hEi is called the excess molar enthalpy of solution for component i. If the excess

molar enthalpy of dissolution is constant over a small range of temperature, we shouldget a linear relationship

ln γ∗i = hE

i

RT+ c, (3.63)

where c is a constant. Since no phase change is involved for liquid solutes, the excessenthalpy is identical to the enthalpy change for solution. Experimental evidence for theeffect of temperature on the solubility of several organic liquids inwater shows that thedecrease in activity coefficient (increase in solubility) over a 20◦ rise in temperatureis in the range 1–1.2 (Tse, Orbey, and Sandler, 1992). This is shown in Figure 3.7.For some of the compounds the value of activity coefficient decreases slightly withincrease in temperature, particularly if the excess molar enthalpy is near zero orchanges sign within the narrow range of temperature. In conclusion, wemay state thatwithin the narrow range of temperatures encountered in the environment, the activitycoefficients of liquid solutes in water do not change appreciably with temperature.

For compounds that are solids or gases at the temperature of dissolution in water,the total enthalpy of solution will include an additional term resulting from a phasechange (sub-cooled liquids for solid solutes and super-heated fluids for gases) thatwill have to be added to the excess enthalpy of solution. Formost solids, the additionalterm for phase change (enthalpy of melting) will dominate and hence the effect oftemperature on activity coefficient will become significant. The result for naphthalene(a solid at room temperature) is shown in Figure 3.7. The mole fraction solubility ofnaphthalene at 281K was 2.52 × 10−6 whereas at 305K it increased to 5.09 × 10−6.Thus, in environmental engineering the effects of temperature on activity coefficientsfor solids in water cannot be ignored.

3.4.3 CORRELATIONS WITH HYDROPHOBICITY

Organic compounds of low aqueous solubility have generally large activity coeffi-cients. In order to explain the large activity coefficients, we need a knowledge of thestructure of water in both the liquid and solid states. The following discussion focuses

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FIGURE 3.7 Three-dimensional ice lattice structure.

on the special structural features of water that are of relevance in understanding thehydration of neutral and polar compounds.

3.4.3.1 Special Structural Features of Water

Water is an inorganic compound. It is the only inorganic compound that exists onearth in all three physical forms, namely, gas, liquid, and solid. It is a remarkablesolvent and there is practically no compound on earth that is insoluble in water. Largemolecules such as proteins are hydrated with water molecules. Small molecules suchas helium and argon (rare gases) have low solubilities in water. Ionic compounds (e.g.,NaCl) are highly soluble in water. Particulate and suspended materials in water existin the atmosphere as aerosols, fog, and mist. In spite of its ubiquitous nature in ourenvironment, the structure of water is far from being completely resolved.

Water displays anomalous properties. It expands in volume on freezing unlike theother compounds. It exists over a wide range of temperature as a liquid that indicatesthat long-range intermolecular forces are dominant. That the molar enthalpy of fusion(5.98 kJ/mol) is only 15% of the molar enthalpy of vaporization (40.5 kJ/mol) isindicative of the fact that water retains much of its ordered ice-like structure even asa liquid, which disappears only when it is boiled. The melting point, boiling point,and the enthalpy of vaporization are unexpectedly high for a compound of such alow molecular weight. Not that these high values are unusual, but they are mostlyexhibited by metallic and ionic crystals. Although ice is less dense than liquid water,the isothermal compressibility of liquid water is remarkably low, indicating that the

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repulsive forces in liquid water are of low magnitudes. The surface tension of water(72mN/m), an indication of the forces at the surface of the liquid, is also unusuallyhigh. The most anomalous property is its heat capacity. For a compound of such lowmolecular weight it is abnormally high. It is also interesting that the heat capacityreduces to about half when water is frozen or when it is boiled. It is this particularproperty of liquid water that helps maintain the ocean as a vast storehouse of energy.Although the P–V–T relationship for water is abnormal giving rise to the above-mentioned anomalies, its transport properties such as diffusion constant and viscosityare similar to those of most molecular liquids.

We, therefore, conclude that peculiar intermolecular forces are at play to givewaterits anomalous properties. To gain an understanding, we need to focus on the structureof water and the ice lattice. Figure 3.8 shows the structure of an ice lattice whereinthe central water molecule is tetrahedrally linked to four other water molecules. TheO–H intramolecular bond distance is 0.10 nm whereas the intermolecular O–H bonddistance is only 0.176 nm. At first glance, it may seem that the O–H bond is alsocovalent in nature. It is indeed larger than the true covalent O–H bond distance of0.1 nm, but it is smaller than the combined van der Waals radii of the two molecules(0.26 nm). Bonds that have these intermediate characters are called hydrogen bonds.For many years it was thought that the H-bond had a predominant covalent nature.However, more recently it has been convincingly shown that it is an electrostaticinteraction. The H atom in water is covalently linked to the parent O atom, but entersinto an electrostatic linkagewith the neighboringO atom. Thus H acts as amediator inbonding between two electronegativeOatoms and is represented asO–H–O.Althoughnot covalent, the H-bonds show some characteristics of weak covalent bonds; forexample, they have bond energies of 10–40 kJ/mol and are directional in nature.Actualcovalent bonds have energies of the order of 500 kJ/mol andweak van derWaals bonds

FIGURE 3.8 A solute molecule within the tetrahedral cage of water molecules.

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are only of the order of 1 kJ/mol. The directional character of the H-bond in waterallows it to form a three-dimensional structure wherein the tetrahedral structure ofthe ice lattice is propagated in all three dimensions.

Although the actual structure of liquid water is still under debate, it is generallyaccepted that the tetrahedral geometry of the ice lattice is maintained in liquid waterto a large extent shown by infrared (IR) and Raman spectroscopic investigations. Anaccepted model for water is called the ST2 model (Stillinger and Rahman, 1974).In this model each H atom in the tetrahedron covalently bonded to an O carry a netcharge of +0.24e (e is the electron unit) each, while the two H atoms on the oppositeside of the O atom that participate in H-bonds carry a net compensating charge of−0.24e each. The H–O–H bond angle is 109◦. Molecular computer simulations ofthe ST2 model confirm that the tetrahedral coordination of the O atom with other Hatoms is the cause for the many unusual properties of water. In liquid water althoughthe ice-like structure is retained to a large extent, it is disordered and somewhat openand labile. The interesting fact is that the number of nearest neighbors in the lat-tice in liquid water increases to five on the average, whereas the average number ofH-bonds per molecule decreases to 3.5. The mean lifetime of a H-bond in liquid wateris estimated to be 10−11 s. Generally, it can be said that only a tetrahedral structurein liquid water can give rise to this open three-dimensional structure, and it is thisproperty more than even the H-bonds themselves that imparts strange properties to alow molecular weight compound such as water.

3.4.3.2 Hydrophobic Hydration of Nonpolar Solutes

The introduction of a solute changes the intermolecular forces in water. The solute–water interactions are called hydration forces. Most of our current understanding ofhydration phenomena is based on indirect experimental evidence. An understandingof hydration (or more generally solvation) is by starting with a thermodynamic cycle,which incorporates all of the so-called standard thermodynamic transfer functions.It is useful and instructive to compare the thermodynamic functions for transfer ofa solute molecule (say i) at a specified standard state from solvent A to solvent B.Solvent B is water while solventA can be any other liquid or even air. Both solventsAand B can be either pure or mixtures. Let us consider a solute (methane) that is trans-ferred from different organic solvents (A) to solvent water (B/W). The values of molarfree energy, enthalpy, and entropy have been obtained and tabulated (Franks, 1983).These values are shown in Table 3.9. The free energy change is positive in all thecases. It is less positive for a more nonpolar solvent (e.g., methanol versus cyclohex-ane). However, the enthalpy change in all the cases is negative, indicating exothermictransfer of the molecule from solvent to water, that is, the enthalpic contribution to thetransfer of methane to water is highly favorable. The entropic contribution is, how-ever, positive in all these cases. Thus, we arrive at the remarkable conclusion that thedissolution of a nonpolar solute such as methane in water is entropically unfavorable.Methane is not the only solute for which this behavior has been observed. Most non-polar compounds or slightly polar compounds of environmental significance showthis feature. A majority of polar compounds that have only one polar group such asalcohols, amines, ketones, and ethers also show this behavior.

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0.00310 0.00315 0.00320 0.00325 0.00330 0.00335 0.00340 0.00345 0.003506

8

10

12

14

16

18

20

Heptachlor

Benzene

(1/T)/K–1

–ln

γ i*

FIGURE 3.9 A plot of ln(1/x∗i ) versus 1/T for heptachlor and benzene.

How can we explain the large entropic contribution to solution of nonpolar com-pounds in water? The first convincing explanation was given by Frank and Evans(1945), and further expanded by Franks (1983) and Israelchvili (1992). The essentialtenets of the theory are summarized here. Most nonpolar compounds are incapable offorming H-bonds with water. This means that in the presence of such a solute, watermolecules lose some of their H-bonds among themselves. The charges on the tetra-hedral water structure will then have to be pointed away from the foreign moleculeso that at least some of the H-bonds can be re-established. If the solute molecule isof very small size, this may be possible without loss of any H-bonds, since water hasan open flexible structure. If the molecule is large, thanks to the ability of the tetrahe-drally coordinated water to rearrange themselves there is, in fact, more local orderingamong the water molecules in the hydration layer (Figure 3.9). Thus, the introductionof a nonpolar or a polar molecule with a nonpolar residue will reduce the degrees offreedom for the water molecules surrounding it. Spatial, orientational, and dynamicdegrees of freedom of water molecules are all reduced. As stated earlier, the averagenumber of H-bonds per molecule in water is about 3.5. In the presence of a nonpolarsolute this increases to four. If there is local ordering around the solute, we shouldexpect from thermodynamics that the local entropy of the solvent is reduced; this isclearly unfavorable. Such an interaction is called hydrophobic hydration. It shouldbe remembered that in transferring to water, although the solute entropy increases,the solvent entropy decrease is so much larger that it more than offsets the former;hence the overall entropy of the process is negative. None of the present theories ofsolute–solvent interactions can handle this phenomenon satisfactorily. In Table 3.11the entropy contribution is about 60–70% of the overall free energy of transfer.

Table 3.11 shows the thermodynamic functions of solution of several gases fromthe vapor phase into water. Once again, we see that the entropic contribution towardthe overall free energy of dissolution is large. The incompatibility of nonpolar and

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TABLE 3.11Thermodynamic Functions for Transfer of NonpolarMolecules from theVapor Phase (A) toWater (W) at 298 K

Solute ΔG0i (kJ/mol) ΔH0

i (kJ/mol) ΔS0i (kJ/mol)

Methane 26.15 −12.76 −38.91Ethane 25.22 −16.65 −41.89Propane 26.02 −23.85 −49.87n-Butane 26.52 −25.10 −51.62

Source: From Nemethy, G. and Scheraga, H.A. 1962. Journal of ChemicalPhysics 36, 3401–3417.

slightly polar compounds with water is called the hydrophobic effect; this propertyis characterized by the activity coefficient of the solute in water or the octanol–waterpartition constant of the solute.

The entropic contribution toward the free energy of solution increases as the solutesize increases. It has become evident from the data that although for smallermoleculesthe entropic contribution predominates, for largermolecules the enthalpic contributionis equally important in making the excess free energy of solution positive.

EXAMPLE 3.12 EXCESS THERMODYNAMIC FUNCTIONS OF SOLUTION OF LARGE

HYDROPHOBIC MOLECULES IN WATER

(a) The following data were obtained by Biggar and Riggs (1974) for the aqueoussolubility of a chlorinated insecticide, namely, heptachlor. It has a molecular weight of373 and a melting point of 368K.

Temperature t (◦C) Solubility (μg/L)

15 10025 18035 31545 490

The enthalpy of melting of heptachlor is 16.1 kJ/mol. Calculate the excess functions forsolution of heptachlor in water at 298K.

Since heptachlor is a solid at 298K, we need to properly account for the enthalpy ofmelting of the solid. The equation for solid solubility given above should be used. Thedata required are x∗i and 1/T .

1/T/(1/K) x∗i

0.00347 4.82 × 10−9

0.00335 8.68 × 10−9

0.00324 1.52 × 10−8

0.00314 2.36 × 10−8

continued

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A plot of ln(1/x∗i ) versus (1/T) is then obtained (Figure 3.18). The slope of the plot is4845K, the intercept is 2.328, and the correlation coefficient is 0.999. From the slopewe obtain the total enthalpy change, ΔHs

i , as 4845 × 8.314 × 10−3 = 40.3 kJ/mol.

The excess enthalpy of solution is then given by hEi = 40.3 − 16.1 = 24.2 kJ/mol. The

excess free energy at 298K is given by gEi = RT ln γ∗i = 42 kJ/mol. The contribution

from excess entropy is TSEi = −17.8 kJ/mol. The entropy of solution is, therefore,

SEi = −59.8 J/Kmol.(b) The solubility data for benzene are given as follows (May et al., 1983).

T (K) x∗i

290.05 4.062 × 10−4

291.75 4.073 × 10−4

298.15 4.129 × 10−4

298.95 4.193 × 10−4

Benzene has a melting point of 5.5◦C and a boiling point of 80.1◦C. Its molecularweight is 78.1. Determine the excess functions of solution of benzene in water.

Since benzene is a liquid at the temperatures given, we can ignore theenthalpy contribution from phase changes. Hence the equation for solubility ofliquids given earlier can be used. A plot of ln(1/x∗

i ) versus (1/T) can be made(Figure 3.20). The slope is 261.8K, the intercept is 6.907 and the correlationcoefficient obtained is 0.861. The excess enthalpy of solution is then given byhEi = (261.8)(8.314 × 10−3) = 2.2 kJ/mol. The free energy of solution at 298.15Kis gE

i = 18.9 kJ/mol. Hence the excess entropy of solution of benzene is SEi =

−56 J/Kmol.It should be noted that in addition to the unfavorable entropy contribution, the

enthalpy contribution also contributes toward the unfavorable excess Gibbs freeenergy of solution for large molecules in water.

3.4.3.3 Hydrophobic Interactions between Solutes

Having discussed the hydration phenomena of a nonpolar solute in water, it is natu-ral to ask what are the consequences of bringing two such solutes near one anotherin water. This process is called hydrophobic interaction. Since the introduction ofany nonpolar residue in water is an unfavorable process, it is likely that a hydropho-bic compound in water will seek out another of its own kind to interact with. Inother words, the solute–solute interaction for hydrophobic compounds in water is anattempt to partially offset the entropically unfavorable hydration process. Accord-ingly, the free energy, enthalpy, and entropy of hydrophobic interactions should allbe of opposite sign to that of hydrophobic hydration. It is also true that hydrophobicinteractions for two solutes in water are even larger than in free space (vacuum).For example, Israelchvili (1992) calculated that for two methane molecules in free

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space, the interaction energy is −2.5 × 10−21 J whereas in water it increases to−14 × 10−21 J. It should be emphasized that what we classify as hydrophobic inter-action is not a true bond between solute species in water, although earlier thoughtson this led to the formulation of terms like hydrophobic bonds that are mislead-ing. The type of interactions between two solutes in water occurs via overlappinghydration layers and is of much longer range than any other types of bonds. Ther-modynamic parameters for hydrophobic interactions are not extensive because of theinherent low solubilities of hydrophobic compounds. Tucker and Christian (1979)determined that the interactions between benzene molecules in water to form a dimeris about −8.4 kJ/mol, whereas Ben Naim (1980) calculated a value of −8.5 kJ/molbetween two methane molecules in water.A satisfactory theory of hydrophobic inter-action for sparingly soluble organics in water is still lacking and progress is beingmade in this field. The same hydrophobic interactions play a major role in theformation of associated structures from surfactants in water, biological membranestructures, and conformations of proteins in biological fluids. For these cases ther-modynamic parameters are well established and hydrophobic interactions are prettywell understood.

3.4.3.4 Hydrophilic Interactions for Solutes in Water

Noninteracting solutes in water experience repulsive forces among one another. Polarcompounds, in particular, exhibit this behavior. They are mostly hygroscopic andtend to incorporate water when left exposed in humid environments. Many of thestrongly hydrated ions, and some zwitter ions (those that have both anionic andcationic characteristics), are strongly hydrophilic. Some nonpolar compounds alsoshow this behavior if they contain electronegative atoms capable of interacting withthe H-bonds in water. In contrast to the structuring of water imposed by a hydropho-bic solute, the hydrophilic compounds tend to disorder the water molecules aroundthem. As an example, urea dissolved in water tends to make the water environmentso different that it can unfold a hydrophobic protein molecule in water. In summary,for polar molecules in water, the contributions to free energies are determined pri-marily by solvation effects with water giving a much more favorable conformationthan most other organic solvents.

Electrolytes in water provide multiple species (ions) that interact differently withwater. Aqueous electrolytes behave nonideally even at very low concentrations.Fortunately, the limiting behavior (infinitely dilute solution) for electrolytes is wellunderstood and is called the Debye–Huckel theory. The final result from this theorywas presented in Section 3.2.4, where the effects of ionic strength on the mean activ-ity coefficient of ions and on the activity coefficient of neutral solutes in water werediscussed.

Using the Debye–Huckel theory, the definition of mean activity coefficient is(Bockris and Reddy, 1970)

lnγ± = −NA(z+z−)e2

2εRT· κ, (3.64)

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where 1/κ is the Debye length given by

κ2 = 4p

εkT

∑i

n0i zie2. (3.65)

Rewriting n0i in terms of molality we have

∑i

n0i zie2 = NAe2

1000

∑i

miz2i . (3.66)

Since ionic strength I = (1/2)Σmi · z2i , we have κ = BI1/2 where B = (8πNAe2/1000 εkT)1/2. Therefore we have the familiar equation given in Section 3.2.4 asfollows:

log γ± = −A(z+z−) · I1/2, (3.67)

where A = [(NA.e2)/(2.303 2εRT)]B. Note that A and B are functions of temperatureand most importantly the dielectric constant of the solvent. For water at 298K, A =0.511 (L/mol)1/2 and B = 0.3291Å−1(L/mol)1/2. The above equation is called theDebye–Huckel limiting law and has been confirmed for a number of dilute solutions ofelectrolytes. Many applications in environmental engineering involve dilute solutionsand this is an adequate equation for those situations. The theory is untenable at highNaCl concentrations (i.e., high I values or small values of 1/κ). Attempts have beenmade to correct theDebye–Huckel law to higher ionic strengths.Without enumeratingthe detailed discussion of these attempts, it is sufficient to summarize the final results(see Section 3.2.4 and Table 3.3).

3.4.3.5 Molecular Theories of Solubility: An Overview

The molecular model for the dissolution of nonpolar solutes in water starts by con-sidering the energy associated with the various stages of bringing a molecule fromanother phase (gas or liquid) into water. The Gibbs free energy for the process hastwo components (Figure 3.10). The first stage involves the formation of a cavity inwater capable of accommodating the solute. The solute is then placed in the cavity,whereupon it establishes the requisite solute–solvent interaction. The water then rear-ranges around the solute so as tomaximize its favorable disposition with respect to thesolute. This concept of solubility was suggested byUhlig (1937) and Eley (1939). Thefree energy for solution ΔG is the sum of the cavity term Gc and the solute–solventinteraction term Gt. The work required to stretch a surface is the work done againstthe surface tension of the solvent. Therefore, the work required to form a cavity inwater should be the product of the cavity area and surface tension of water. Thus Gcis directly proportional to surface tension. This was verified by Saylor and Battino(1958) for a nonpolar solute, argon. Choi, Jhon, and Eyring (1970) showed that infact it is not the bulk surface tension of the solvent that should be considered butthe microscopic surface tension of the highly curved cavity. The surface tension of a

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Solute transfer

Cavity formation Solute accommodation andsolute–solvent interactions

ΔG = Gc+ Gi

FIGURE 3.10 Molecular description of the solubility of a nonpolar solute in water. The firststage is the formation of a cavity to accommodate the solute and the second stage involvesestablishing molecular interactions between the solute and water molecules.

microscopic cavity was found to be approximately one third of the value for a planesurface.

Hermann (1972) obtained the free energy associated with the formation of acavity in water and the introduction of various hydrocarbons into water. Table 3.12summarizes some of his calculations. It gives the values of Gc and Gt for a vari-ety of hydrocarbons in water. The agreement between the experimental values andpredicted values is satisfactory considering the fact that an approximate structurefor water was assumed. The most important aspect is that the values of cavitationfree energies are all positive whereas those for the interactions are all negative. Themajor contribution toward the unfavorable free energy of dissolution of hydrophobicmolecules results from the work that has to be done against perturbing the structureof water. Practically all molecules have interaction energies that are favorable towarddissolution. Thus the term hydrophobic molecule is a misnomer. It is not that themolecules have any phobia toward water but, in fact, it is the water that rejects thesolute molecule.

EXAMPLE 3.13 HENRY’S CONSTANT FROM FREE ENERGY OF SOLUTION

Kaw = exp

(ΔG

RT

). (3.68)

Knowledge of the free energy of transfer of a mole of solute i from the gas phaseto water from theory allows a direct estimation of Henry’s constant, Kaw, from theequation. For methane the free energy change at 298K is 6.7 kJ/mol. Hence Henry’sconstant is estimated to be 14.9. The experimental value is 28.6.

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TABLE 3.12Free Energy Values for the Dissolution of Hydrocarbons in Water

Cavity Surface Gc Gt ΔGcalc ΔGexptCompound Area, Å2 (kJ/mol) (kJ/mol) (kJ/mol) (kJ/mol)

Methane 122.7 22.8 −16.1 6.7 8.1Ethane 153.1 31.7 −24.3 7.4 7.5Propane 180.0 39.8 −31.7 8.2 8.3n-Butane 207.0 48.4 −39.0 9.4 9.0n-Pentane 234.0 57.0 −46.2 10.8 9.82,2,3-Trimethylpentane 288.5 75.1 −62.7 12.4 11.9Cyclopentane 207.1 48.4 −41.3 7.0 5.0Cyclohexane 224.9 54.1 −47.2 6.9 5.2

Source: From Hermann, R.B. 1972. Journal of Physical Chemistry 76, 2754–2759.

3.4.4 STRUCTURE-ACTIVITY RELATIONSHIPS AND ACTIVITY

COEFFICIENTS IN WATER

This section deals with different methods by which activity coefficients can beestimated from solute structural parameters. In cases where reliable experimentalvalues are not available, there will arise the need to obtain estimates of activitycoefficients from correlations. Since activity coefficients are inversely related tomole fraction solubility (or molar solubility), it is enough to know one param-eter to obtain the other. From time to time in this section, we will use theseinterchangeably.

The activity coefficients of solutes in water can be related to several solute parame-ters such as molecular surface area, molar volume, octanol–water partition constants,and normal boiling points. This technique of estimation is called the structure-activityrelationship (SAR). These correlations are given in Table 3.13.

EXAMPLE 3.14 ESTIMATING ACTIVITY COEFFICIENT FROM MOLECULAR AREA

Assess the applicability of the equation γ∗i = exp(ΔGm/RT) for the solubility ofbenzene in water. Explain any discrepancy.

Let us determine the molecular surface area of benzene by assuming a sphericalshape for the molecule. This is clearly an approximation. The molecular radius can bedetermined from the molar volume as r = (3M/4πρN)1/3 = 3.9Å. Hence, the surfacearea of amolecule is given by 4πr2 = 149Å2. Since σi/w for the benzene–water system

is 35mN/m, we obtain ΔGm = 23 kJ/mol. Hence γ∗i = 3.35 × 105.Now let us use the actual surface area of benzene obtained via a computer calculation.

The value is 109.5Å2. We then obtain ΔGm = 23 kJ/mol. Hence γ∗i = 1.11 × 104.The actual experimental value at 298K is 2.43 × 103.

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Note that if the free energy were 19.3 kJ/mol instead of the calculated value of23 kJ/mol, we would have predicted the actual experimental value of the activity coef-ficient. The above calculation shows how sensitive the estimates are to the free energyof solution due to its exponential dependence. The free energy is a function of bothsurface energy and molecular area, and hence both of these contribute to uncertaintyin the free energy calculation. The dilemma arises as to what is the exact value of sur-face energy that should be used, since at small radius of curvature the surface energyis dependent on curvature. The concept of surface tension does not hold for a singlemolecule; however, the concept of interfacial energy remains valid even for an isolatedmolecule (Israelchvili, 1992). Sinanoglu (1981) showed that clusters of gaseous car-bon tetrachloride composed of even as small as seven molecules show surface energiessimilar to those of a planar macroscopic surface.

If we use the correlation given in Table 3.13, the solubility in water for benzene isC∗i = 0.0467mol/dm3, which gives a γ∗i = 1.2 × 103.

EXAMPLE 3.15 AQUEOUS SOLUBILITY FROM MOLECULAR PARAMETERS

Calculate the aqueous solubility of naphthalene from molecular area.For naphthalene Am is 156.76Å2. Hence logC∗

i = −0.0282(156.76) + 1.42 −0.0095(80.3) = −3.76.C∗

i = 0.00017mol/L.

EXAMPLE 3.16 CALCULATIONS OF AQUEOUS SOLUBILITY FROM Kow

(a) Naphthalene: Melting point = 80.3◦C (a solid at room temperature), logKow =3.25. Hence from Table 3.13 we have − logC∗

i = (0.880)(3.25) + (0.01)(80.3) +0.012 = 3.675. Hence C∗

i = 2.11 × 10−4 mol/dm3 = 27mg/dm3. The experimental

value is 30mg/dm3. Hence the error in the estimation is −10%.(b) 1,2,4-Trichlorobenzene: Melting point = 17◦C (a liquid at room temperature),logKow = 4.1. Hence from Table 3.13 we have − logC∗

i = (0.987)(4.1) + 0.718 =4.76. Hence C∗t

i = 1.72 × 10−5 mol/dm3 = 3.1mg/dm3. The experimental value is31mg/dm3. The error in the estimation is −90%. Such large errors are not unusual andhence care should be taken in accepting the estimated solubilities at their face values.

3.4.5 THEORETICAL AND SEMI-EMPIRICAL APPROACHES TO AQUEOUS

SOLUBILITY PREDICTION

3.4.5.1 First-Generation Group Contribution Methods

Langmuir (1925) suggested that each group in a molecule interacts separately withthe solvent molecules around it, and hence the overall interaction can be determined

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TABLE 3.13Correlations for Aqueous Solubility Using SARs

Compound Class Type a1 a2 a3 r2

Solubility and molecular surface area, Am(Å2) : log C∗i = a1 + a2Am + a3tm

Aliphatic alcohols (l) 3.80 −0.0317 — 0.986Aliphatic hydrocarbons (l) 0.73 −0.0323 — 0.980Alkylbenzenes (l) 2.77 −0.0184 — 0.988Polycyclic aromatics (s) 1.42 −0.0282 −0.0095 0.988Halogenated benzenes (s) 3.29 −0.0422 −0.0103 0.997Polychlorinated biphenyls (s) 1.21 −0.0354 −0.0099 0.996

Compound Class Type a b c r2

Solubility and molecular volumelog C∗

i = a + bVm, where volume is in 0.01 Å3

Variety of compounds Liquids 1.22 −2.91 — 0.506log C∗

i = a + bVm + cVH, where VH accounts for the polarityVariety of compounds Liquids 0.72 −3.73 4.10 0.960log C∗

i = a + bVm + ctm, where tm is the melting point in ◦CPAHs Solids and liquids 3.00 −0.024 −0.010 0.975

Solubility and octanol–water partition constants: log C∗i = a log Kow + btm + c

Alcohols 1.113 — −0.926 0.935Ketones 1.229 — −0.720 0.960Esters 1.013 — −0.520 0.980Ethers 1.182 — −0.935 0.880Alkyl halides 1.221 — −0.832 0.861Alkynes 1.294 — −1.043 0.908Alkenes 1.294 — −0.248 0.970Benzene and derivatives 0.996 — −0.339 0.951Alkanes 1.237 — +0.248 0.908Halobenzenes 0.987 0.0095 0.718 0.990Polynuclear aromatics 0.880 0.01 0.012 0.979PCBs 1.000 0.01 0.020 —Alkyl benzoates 1.14 0.005 −0.51 0.991Mixed drugs 1.13 0.012 −1.02 0.955Priority pollutants 1.12 0.017 −0.455 0.960Acids, bases, and neutrals 0.99 0.01 −0.47 0.912Steroids 0.88 0.01 0.17 0.850

by summing the group contributions. The basic idea behind the group contributionapproach is that whereas there are thousands of compounds, there are only a smallnumber of functional groups that make up these compounds through a variety of per-mutations and combinations. An excellent example of such a maxim can be seen ifwe consider the variation in solubility among a homologous series of aliphatic hydro-carbons. McAuliffe (1966) reported careful studies of the solubility of homologous

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series of n-alkanes, n-alkenes, and n-dienes. For n-alkanes, each CH2 group addedto the molecule increases the logarithm of the aqueous activity coefficient of thesolute by a constant factor of 0.44 units. The increase in activity coefficients for aro-matic hydrocarbons, aliphatic acids, alcohols, and substituted phenols also showeda similar trend. Thus came the recognition that the solubility (activity coefficient)of different substituent groups on a parent compound can be separately ascertained.This would then enable one to predict the solubility of other members of the homol-ogous series. This rationalization led Prausnitz and coworkers (1999) to develop aset of group contribution parameters for obtaining the infinite dilution activity coef-ficients of aromatic and aliphatic substituent groups on benzene. They noted thatthe log γ∞

i of a series of alkylbenzenes can be expressed in the form of a linearequation, that is,

log γ∞i = a+ b · (n− 6), (3.69)

where n is the total number of carbon atoms in the molecule, (n− 6) is therefore thenumber of carbon atoms in addition to the benzene ring. The constant (designateda) represents the contribution from the parent group (in this case the aromatic ring).Additional atoms or groups attached to the aromatic ring or the aliphatic side chainsthen make specific contributions to the activity coefficient. The same general form ofthe equation (with different values of a and b) also represents the activity coefficientsof polyaromatic hydrocarbons in water.

EXAMPLE 3.17 SOLUBILITY FROM GROUP CONTRIBUTIONS

The following group contribution values were reported by Tsonopoulos and Prausnitz(1971) for aromatic and aliphatic substitutions on the benzene ring.

Substituent Position AliphaticGroup or Atom on Ring Side Chain

−Cl 0.70−Br 0.92−OH −1.70 −1.90−C=C− −0.30

The constants a and b for benzene derivatives and polyaromatic hydrocarbons are givenbelow:

Compound Class a b

Alkylbenzenes 3.39 0.58Polyaromatics 3.39 0.36

continued

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Let us estimate the infinite dilution activity coefficients of some compounds using theabove information. In each case we start with the parent group, in this case the aromaticring for which the contribution is the value of a from the linear correlation. To this wethen add the contributions of the functional groups.

(a) Ethylbenzene: For this case we use the linear correlation directly with n = 8 givinglog γ∞i = 3.39 + 2 × 0.58 = 4.55.

(b) C6H5–CH2–CH2–OH: Since this is a derivative of ethylbenzene, we add to it thecontribution from an OH group giving log γ∞i =4.55–1.90 = 2.65.

(c) Hexachlorobenzene: In this case we start with the basic group, that is, the aromaticring for which the contribution is 3.39, and then add to it the contributions from thesix Cl atoms. Hence, log γ∞i = 3.39 + 6 × 0.70 = 7.59.

(d) Naphthalene: The basic building block in this case is again the benzene ring, butwe use the correlation for polyaromatics with n = 4. Hence, log γ∞i = 3.39 + 4 ×0.36 = 4.83.

The above calculation is an illustration of what is available in the existing chemicalengineering literature for estimating activity coefficients through group contributions.Themethod ofTsonopoulos and Prausnitz (1971)was chosen for illustrative purposes.It should be borne in mind that the same general method of obtaining group contri-bution methods has also been reported by other investigators. Pierotti, Deal, and Derr(1959) correlated the activity coefficient group contributions based on existing liquid–liquid and vapor–liquid equilibrium data. Wakita et al. (1986) reported a set of groupcontribution parameters for a variety of aliphatic and aromatic fragments.

The group contribution approach has inherent drawbacks that are worth noting. Itis limited by the lack of availability of the requisite fragment values. In practice, thereare several groups for which group contributions are as yet unavailable. The methodalso has limitations when it comes to distinguishing between geometric isomers ofa particular compound. Any group contribution approach is essentially approximatesince the contribution of any group in a given molecule is not always exactly the samein another molecule. Moreover, the contribution made by one group in a molecule isconstant and nonvarying only if the rest of the groups in the molecule do not exert anyinfluence on it. In fact, this is a major drawback of any group contribution scheme.

3.4.5.2 Excess Gibbs Free Energy Models

There are several equations to estimate activity coefficients of liquid mixtures basedon excess Gibbs free energy of solution. These are summarized in Table 3.14. Fordetails of their derivation, see Sandler (1999).

3.4.5.3 Second Generation Group ContributionMethods: The UNIFAC Method

The most useful and reliable method of activity coefficient estimation resulted fromthe need for chemical engineers to obtain activity coefficients of liquid mixtures. Inthe chemical process industry (CPI), the separation of components from complexmixtures is a major undertaking. It was recognized early on that a large database for

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TABLE 3.14Correlation Equations for Activity Coefficients in a Binary Liquid MixtureBased on Excess Gibbs Free Energy

Description Equation

One constant Margulesa γ1 = Ax22RT

, γ2 = Ax21RT

Two constant Margulesb RT ln γ1 = α1x22 + β1x32, RT ln γ2 = α2x21 + β2x31

van Laarc ln γ1 = α[1 + (α/β) (x1/x2)

]2 , ln γ2 = β[1 + (β/α) (x2/x1)

]2

Flory–Hugginsd ln γ1= ln

(φ1

x1

)+(1 − 1

m

)φ2 + χφ22, ln γ2= ln

(φ2

x2

)+(m − 1)φ1+χφ21

Wilsone ln γ1 = − ln (x1 +Λ12x2) + x2

(Λ12

x1 +Λ12x2− Λ21

Λ21x1 + x2

)

NRTLf ln γ1 = x22

[τ21

(G21

x1 + x2G21

)2+ τ12G12

(x2 + x1G12)2

]

aOnly applicable for liquid mixtures when the constituents (1 and 2) are both of equal size, shape, andchemical properties.

bαi = A+ (3 − 1)i+1B and βi = 4(−1)iB; i = 1 or 2.cα = 2q1a12, β = 2q2a12. a12 is the van Laar interaction parameter, q1 and q2 are liquid molar volumes,and are listed in Perry’s Chemical Engineer’s Handbook. Note also that as x1 → 0, ln γ1 = ln γ∞1 ,and as x2 → 0, ln γ2 = ln γ∞2 .

dϕ1 = x1/(x1 + mx2) and ϕ2 = mx2/(x1 + mx2) are volume fractions, m = v2/v1. χ is the Floryparameter.

eΛ12 and Λ21 are given in (1964). Journal of the American Chemical Society 86, 127.fG12,G21, τ12, and τ21 are given in (1968). AIChE Journal 14, 135.

activity coefficients of a vast number of binary and ternary liquid–liquid and vapor–liquid systems already exists. At the same time there also exists a sound theory ofliquid mixtures based on the Guggenheim quasi-chemical approximation originatingin statistical thermodynamics.This is calledUNIversalQUAsiChemical (UNIQUAC)equation. This theory incorporates the solvent cavity formation and solute–solventinteractions and is useful for establishing group contribution correlations where theindependent variables are not the concentrations of the molecules themselves butthose of the functional groups. The basic tenet of this theory was combined with thelarge database on activity coefficients to obtain group contribution parameters for avariety of molecular groups. This approach was called UNIversal Functional groupActivity Coefficient (UNIFAC) method. Although developed by chemical engineers,it has of late found extensive applications in environmental engineering.

The UNIFAC method is especially suitable for activity coefficients of complexmixtures. It was developed for nonelectrolyte systems and should be used only forsuch systems. Most of the data for group contributions were obtained at high molefractions, and therefore extrapolation to very small mole fractions (infinitely dilute)for environmental engineering calculations should be made with caution. Arbuckle

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(1983) noted that when the mole fractions were very small, UNIFAC under-predictedthe solubilities. Banerjee andHoward (1988) also noted this and appropriate empiricalcorrections have been suggested. In spite of the above observations, it is generally rec-ognized that for most complex mixtures, UNIFAC can provide conservative estimatesof activity coefficients. In fact, its remarkable versatility is evident when mixturesare considered. As more and more data on functional group parameters in UNIFACbecome available, this may eventually replace most other methods of estimation ofactivity coefficients.

Since UNIFAC activity coefficients of compounds in octanol can also be similarlydetermined, it is possible to obtain directly UNIFAC predictions of octanol–waterpartition constants. It has been shown that adequate corrections should be made forthemutual solubilities of octanol andwater if UNIFACpredictions are to be attemptedin this manner.

It is unnecessary to perform detailed calculations nowadays, since sophisticatedcomputer programs are available to calculate UNIFAC activity coefficients forcomplex mixtures.

3.4.6 SOLUBILITY OF INORGANIC COMPOUNDS IN WATER

One area that we have not considered is the solubility of metal oxides, hydroxides,and other inorganic salts. As discussed in Chapter 2, because of the preponderanceof organic over inorganic compounds, there is a general bias toward the study ofenvironmentally significant organic compounds. As a result, less of an emphasis hasbeen placed on the study of inorganic reactions in environmental chemistry. Perhapsthe most significant aspect of environmental inorganic chemistry is the study of theaqueous solubility of inorganic compounds. As observed earlier, the water moleculepossesses partial charges at its O and H atoms, which give rise to a permanent dipolemoment of 1.84 × 10−8 esu. The partial polarity of water molecules allows it toexert attractive or repulsive forces toward other charged particles and ions in thevicinity. The dipoles between water molecules are attracted to one another by theH-bonds. In the presence of an ion, the water dipoles near the ion orient themselvessuch that a slight displacement of the oppositely charged parts of each moleculeoccurs, thereby lowering the potential energy of water. In other words, the H-bondsbetween water molecules are slightly weakened, facilitating the entry of the ion intothe water structure. This discussion points to the importance of understanding thecharge distribution around the central ion in water (see Section 3.4.3.5).

There are important items to be understood if the chemistry ofmetals,metal oxides,and hydroxides in the water environment has to be studied. The development of theseideas requires some knowledge of the kinetics of reactions and reaction equilibria.Elaborate discussion of ionization at mineral–water interfaces, complexation, effectsof pH and other ions, and other reactions are relegated to Chapter 5.

3.5 ADSORPTION ON SURFACES AND INTERFACES

In Chapter 2, the thermodynamics of surfaces and interfaces were explored usingthe concepts of surface excess properties defined by Gibbs. The Gibbs equation is

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the fundamental equation of surface thermodynamics that forms the basis for allof the relationships between surfaces or interfaces and the bulk phases in equilib-rium. Examples of phase boundaries of relevance in environmental chemistry includethe following: air/water, soil (sediment)/water, soil(sediment)/air, colloids/water, andatmospheric particulate (aerosols, fog)/air. Typically the relationship between thebulk-phase and surface-phase concentrations is given by an adsorption isotherm.

3.5.1 GIBBS EQUATION FOR NONIONIC AND IONIC SYSTEMS

Let us apply the Gibbs equation to two different systems, that is, compounds that areneutral (nonionic) and those that are ionic (dissociating).

Let us consider a binary system (solute/i, solvent (water)/w) for which the Gibbsadsorption equation was derived in Chapter 2.

Γi = −dσwa

dμwi, (3.70)

where the surface excess is defined relative to a zero surface excess of solvent (water).σwa represents the surface tensionofwater. For a solid–water interfacewhere the solutedissolved in water, the surface tension is replaced by the interfacial tension of thesolid–water boundary. For a solid–air interface, the analogous term is the interfacialtension (energy) of the solid–air boundary.

If the solute in water is nondissociating (neutral and nonionic), then the equationfor chemical potential is

μwi = μw0

i + RT ln ai, (3.71)

where ai is the activity of solute i in water. Using the above expression, we obtain thefollowing

Γi = − 1

RT· dσwa

d ln ai. (3.72)

If the solute dissociates in solution to give two or more species in the aqueous phase,then the above equation has to be modified

Γi = − 1

nRT

dσwa

d ln a±, (3.73)

where n is the number of dissociated ionic species in water resulting from the solute.Note the difference between the expressions for ionic and nonionic systems. In bothcases, the surface excess can be determined by obtaining the solution surface tension(interfacial energy in the case of solids) as a function of the activity of solute i insolution. If (dσ/d ln a) > 0, then Γi is negative and we have net depletion at thesurface. If, on the other hand, (dσ/d ln a) < 0, then Γi is positive and we have a netsurface excess (positive adsorption) on the surface.

Although the Gibbs equation applies theoretically to solid–water interfaces aswell, the direct determination of the interfacial tension at the solid–water boundary isimpractical. However, adsorption of both molecules and ions at this boundary leads

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to a decrease in surface energy of the solid. In principle, however, there is a relatedterm called adhesion tension, Asw = σsa − σsw (s stands for solid), which can be usedto replace the interfacial tension term. The solid–air interfacial tension is assumed tobe a constant, and hence dAsw = −dσsw appears in the Gibbs adsorption equation.

3.5.2 EQUILIBRIUM ADSORPTION ISOTHERMS AT INTERFACES

The equilibrium between the surface or interface and the bulk phase is determinedby the same general principle of equality of fugacity (or chemical potential) as wediscussed earlier for bulk phases. Some common types of adsorption equilibria inenvironmental engineering and their examples are shown in Figure 3.11.

At equilibrium, we can write f Γi = f li for interface/liquid equilibrium and f Γi = f gifor interface–gas equilibrium. The superscript Γ represents the interface. In eithercase this gives a relation between the surface concentration Γi and the bulk-phaseconcentration Ci (or activity ai). This relationship is called an adsorption isotherm,and is a convenientway to display experimental adsorption data. To utilize the equalityof chemical potentials, we need to obtain an expression for f Γi in terms of Γi. Let usconsider the surface as a two-dimensional liquid phase with both solvent moleculesand solute molecules in it. The solute i which has an excess concentration on thesurface is assumed to be dilute. The appropriate expression for fugacity is then

f Γi = θiγΓi f Γ0i , (3.74)

where θi is the fractional surface coverage of solute i, that is, θi = Γi/Γmaxi , with

Γmaxi is the maximum surface excess for solute i. Note that the standard fugacity is

defined as the chemical potential of the adsorbate when θi = 1. It has to be so definedsince the liquid near the surface is different from that of the bulk liquid. Note thatif the surface phase was considered nonideal, then we need to replace the Γs in θ

Liquid

Gas

Solid

Gas

Solid

Liquid

fig = fi

Γ fil = fiΓfig = fi

Γ= fil

Air-seaAir bubble in waterFog in airRain drops in air

Soil-waterSediment-waterActivated carbon-waterMetal oxid in waterColloids in water

Aerosol in airSoil-airActivated Carbon

FIGURE 3.11 Examples of adsorption equilibria encountered in natural and engineeredenvironmental systems.

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by the respective surface activities to get aΓ/aoΓ. Thus, the analogy of the definition

of fugacity of surface states of molecules with those in the bulk is evident. For theadsorbate in equilibrium with the liquid phase, we obtain

θiγΓi fΓ0i = xiγ

li fl0i . (3.75)

For ideal dilute solutions, the activity coefficients are unity. Hence we have

θi

xi= f l0if Γ0i

= K ′Γl. (3.76)

Further for dilute solutions, xi = CiVl, we have upon rearranging

Γi = KΓlCi. (3.77)

This is called the linear (Henry’s) adsorption equation with the linear adsorptionconstant (units of length) given by KΓl.

In a number of environmental systems, where dilute solutions are considered, theabove equation is used to represent adsorption from the liquid phase on both liquidand solid surfaces. If a solid–gas interface is consideredwhere adsorption occurs fromthe gas phase, the concentration (Γmax

i ) in the above equation may be replaced bythe corresponding partial pressure. If we define a surface layer thickness, δ (unit oflength), and express both Γi and Γmax

i as concentration units (CΓ = Γi/δ), we canexpress the linear adsorption constant as a dimensionless value, Kads. However, sincethe definition of an interface thickness is difficult, the above approach is less useful.

It is conventional, both in soil chemistry and in environmental engineering, toexpress adsorbed-phase concentrations on the solid surface in terms of amountadsorbed per mass of the solid. This means that the surface concentrations Γi andΓmi , which are in moles per unit area of the solid, are converted to moles per gram of

solid, which is the product Wads = Γiam, where am is the surface area per unit massof the solid. This definition then changes the units of the linear adsorption constant,designated Kads, which is expressed in volume per unit mass of solid.

Wads = KadsCi. (3.78)

The linear adsorption constant is not applicable for many situations for a varietyof reasons. At high concentrations of molecules on the surface, the assumptionof no adsorbate–adsorbate interactions fails. Lateral interactions between adsorbedmolecules necessitate that we assume a limited space-filling model for the adsorbedphase. This can be easily achieved by assuming that the expression for surface cov-erage be written as (θi/1 − θi) to account for the removal of an equivalent amount ofsolvent from the interface to accommodate the adsorbed solute i. In other words, theadsorption of a solute i on the surface can be considered as an exchange process h thesolute and solvent molecules. Equating fugacity as above, for the liquid–gas interfacewith the liquid chemical potential, we obtain

θi

1 − θi = xi ·(f l0if Γ0i

)= xi · K ′′

Γw. (3.79)

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Noting further the dilute solution approximation for xi, we have

θi

1 − θi = KLang · Ci. (3.80)

The adsorption isotherm given above is called the Langmuir adsorption isotherm. Itis particularly useful in a number of situations to represent the adsorption data. Inthe case of the solid–gas interface, the concentration term is replaced by the partialpressure of solute i in the gas phase. In fact, Langmuir first suggested this equationto represent gas-phase adsorption data in catalysis.

It is important that one should know how to use experimental data to obtain appro-priate adsorption isotherm parameters. In the case of a linear adsorption isotherm, it isobvious that a plot ofΓi versusCi should be linear with a slope ofKΓl. However, whena plot of the same is made for a Langmuir isotherm, a linear behavior is displayedat low Ci and approaches an asymptotic value at high Ci values. The linear region ischaracterized by ΓI = KLangCi, and .we have Γi = Γm

i (Figure 3.12). In practice, forthe Langmuir isotherm, one plots 1/Γi versus 1/Ci, the slope of which gives an inter-cept of (KLangΓ

mi )−1 and an intercept of (Γm

i )−1. If one takes the intercept over theslope value, one obtains directly KLang. Just the mere fact that a given set of data fitsthe Langmuir plot does not necessarily mean that the adsorption mechanism followsthat of the Langmuir isotherm. On the contrary, other mechanisms such as surfacecomplex formation or precipitation may also lead to similar plots.

An empirical relationship that represents any set of data on adsorption at low con-centrations is called the Freundlich adsorption isotherm. Apart from its universalityin data representation, it was thought to have little theoretical value for a longtime. Recently, it has been shown that it can be derived theoretically by consideringthe heterogeneous nature of adsorption sites (Adamson, 1990; Sposito, 1984). TheFreundlich isotherm is expressed as follows:

Γi = KFreun (Ci)1/n, (3.81)

Linear isotherm Langmuir isotherm

Slope KH

Slope KLang

Ciw/mol ∙ cm–3

Γ i /

mol

∙ cm

–2

Γ i /

mol

∙ cm

–2

Ciw/mol ∙ cm–3

Γim

FIGURE 3.12 Schematic of linear and Langmuir adsorption isotherms.

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where KFreun and 1/n are empirically adjusted parameters. KFreun is an indicator ofthe adsorption capacity and 1/n is an indicator of the adsorption intensity. If n is 1,there is no distinction between the Freundlich and linear adsorption isotherms.

EXAMPLE 3.18 USE OF ADSORPTION ISOTHERMS TO ANALYZE EXPERIMENTAL

DATA

Three important examples are chosen to illustrate the use of adsorption equations toanalyze experimental data in environmental engineering.The first example is a solid–liquid system and is an important component in the

design of an activated carbon reactor for wastewater treatment. The requisite first stepis to obtain the isotherm data for a compound from aqueous solution onto granularactivated carbon (GAC) in batch shaker flasks. In these experiments, a known amountof the pollutant is left in contact with a known weight of GAC under stirred conditionsfor an extended period of time and the amount of pollutant left in the aqueous phase atequilibrium is determined using chromatography or other methods. Dobbs and Cohen(1980) produced extensive isotherm data at 298 K on a number of priority pollutants.Consider the case of an insecticide (chlordane) on activated carbon.

Amount Adsorbed Equilibrium Aqueous-Phase(mg/g Carbon) Concentration (mg/L)

87 0.13279 0.06164 0.02653 0.007143 0.003231 0.002922 0.002118 0.001612 0.000611 0.0005

The molecular weight of chlordane is 409. Hence the above data can be divided by themolecular weight to obtain Γi in mol/g and Ci in mol/L. Since it is not clear as to whichisotherm will best represent these data, we shall try both Langmuir and Freundlichisotherms.

Figure 3.13a is a Langmuir plot of 1/ΓI versus 1/Ci. The fit to the data at least in themid-region of the isotherm appears good, although at low 1/ΓI the percent deviation isconsiderable. The correlation coefficient (r2) is 0.944. The slope is 4.03 × 10−5 with astandard error of 3.46 × 10−6, and the intercept is 6655 with a standard error of 2992.Hence, KLang = 1.65 × 108 L/mol and Γmi = 1.5 × 10−4 mol/g.

Figure 3.13b is a Freundlich plot of log Γi versus logCi. The correlation coefficientis (r2) 0.897. The intercept is −1.0821 with a standard error of 0.112, and the slope is0.381 with a standard error of 0.045. Hence KFreun = 0.0827 and 1/n = 0.381.Although r2 appears to be better for the Langmuir plot, the errors involved in the esti-

mated slopes and intercepts are considerably larger. The adsorption isotherms obtainedfrom these parameters are shown in Figure 3.14. Both the isotherms fit the data rather

continued

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satisfactorily at low Ci values, whereas significant deviations are observed at high Civalues. The Langmuir isotherm severely under-predicts adsorption at high Ci values,whereas the Freundlich isotherm severely over-predicts the adsorption in the sameregion. Thus neither of the isotherms can be used over the entire region of concentrations

4.00E + 4(a) (b) –8–8.1–8.2–8.3–8.4–8.5–8.6–8.7–8.8–8.9

–9–9

3.50E + 4

3.00E + 4

2.50E + 4

2.00E + 4

1.50E + 4

1.00E + 4

5.00E + 3

0.00E + 0

1/Ci (L/mol)

log Ci

0.00

E +

01.

00E

+ 8

2.00

E +

83.

00E

+ 8

4.00

E +

85.

00E

+ 8

6.00

E +

87.

00E

+ 8

8.00

E +

89.

00E

+ 8

1/Γ

(mol

/g)

log

Γ

–8.5 –8 –7.5 –7 –6.5 –6

FIGURE 3.13 Langmuir (a) and Freundlich (b) isotherm plots for the adsorption ofan insecticide (chlordane) on GAC.

Ci/mol ∙ L–1

0.0 5.0e-8 1.0e-7 1.5e-7 2.0e-7 2.5e-7 3.0e-7 3.5e-7

Γ/m

ol ∙ g

–1

0.00000

0.00005

0.00010

0.00015

0.00020

0.00025

0.00030

LangmuirFreundlichExperimental

FIGURE 3.14 Adsorption isotherms for chlordane on GAC.

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in this particular case. It therefore does not sufficiently justify choosing the Langmuirmodel.Very small errors in the data can throw the prediction of the adsorptionmaximumoff by as much as 50%, and hence the isotherm fit may be weakened considerably. Ifseveralmeasurements are available at each concentrationvalue, then anF-statistic canbedone on each point and the lack-of-fit sum-of-squares determined to obtain quantitativereinforcement of the observations. The necessity for multiple determinations of multi-point adsorption data needs no emphasis.The second example is that of a liquid–gas system. In partitioning experiments at

the air–water interface, it is important to obtain the ability of surface-active moleculesto enrich at the interface. I will choose, as an example, the adsorption of n-butanolat the air–water interface. The measurement is carried out by determining the surfacepressure Π = σ0aw − σaw and using the Gibbs equation to get Γi. The following datawere obtained at 298K (Kipling, 1965).

Equilibrium Aqueous Adsorbed ConcentrationConcentration Ci (mol/L) Γi (mol/cm2)

0.0132 1.26 × 10−10

0.0264 2.190.0536 3.570.1050 4.730.2110 5.330.4330 5.850.8540 6.15

A plot of 1/Γi versus 1/Ci was made. The slope of the plot was 8.5 × 107 and theinterceptwas 1.41 × 109 with an r2 of 0.9972.Hence,Γm

i = 7.1 × 10−10 mol/cm2, andKLang = 16.6 L/mol. Figure 3.15 represents the experimental data and the Langmuirisotherm fit to the data. There is good agreement, although at large Ci values theadsorption is somewhat over-predicted. The maximum adsorption capacity, Γm

i canbe used to obtain the area occupied by a molecule on the surface, am = 1/(NΓm

i ) =23 × 10−16 cm2. The total molecular surface area Am is 114 × 10−16 cm2. The lat-ter area is closer to the cross-sectional area occupied by the −(CH2)n− group on thesurface when oriented perpendicular to the surface. Therefore, n-butanol is likely ori-ented normal to the surface with its OH group in the water and the long chain alkylgroup away from the water surface. Most other surfactants also occupy such an orien-tation at the air–water interface at low concentrations. One can explain this by invokingthe concepts of hydrophobicity and hydrophobic interactions described in the earliersection.The third example is the adsorption of two neutral organic compounds, namely, 1,2-

dichloroethane (DCE) and 1,1,1-trichloroethane (TCE) on soils. These compounds aresignificant soil and groundwater pollutants resulting from leaking storage tanks andimproperly buried hazardous waste. The data are from Chiou, Peters, and Freed (1979)and are for a typical silty loam soil.

continued

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7.0E–10

6.0E–10

5.0E–10

4.0E–10

3.0E–10

2.0E–10

1.0E–10

0.0E+100 0.1 0.2 0.3 0.4 0.5

Ci/mol/L0.6

Calculated

Experimental

0.7 0.8 0.9 1

Γ/m

ol/c

m2

FIGURE 3.15 Adsorption of n-butanol at the air–water interface.

Equilibrium Aqueous AmountConcentration (μg/L) Adsorbed (μg/g)

TCE DCE TCE DCE1.0E05 1001.5E05 2803.0E05 4503.7E05 6004.0E05 7405.2E05 880

3.7E05 1009.8E05 3001.4E06 3502.3E06 700

It is generally true in environmental engineering literature that most data for sparinglysoluble neutral organics is plotted as linear isotherms. The linear isotherm concept issaid to hold for these compounds even up to their saturation solubilities in water. Forthe above compounds, the saturation solubilities are 8.4E06 for DCE and 1.3E06 forTCE. A plot of the amount adsorbed versus the equilibrium solution concentration canbe made for both solutes (Figure 3.16). The linearity is maintained in both cases withcorrelation coefficients of 0.9758 and 0.9682 for DCE andTCE, respectively. The linearadsorption constants are 0.291 and 1.685 L/kg for DCE and TCE, respectively. SinceTCE is less soluble and more hydrophobic (cf. the discussion of hydrophobicity in the

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700

600

500

400

300

200

100

0

0.0E

+ 0

5.0E

+ 5

1.0E

+ 6

1.5E

+ 6

Ci/μg/L

Wi/μ

g /g

2.0E

+ 6

2.5E

+ 6

900

800

700

600

500

400

300

200

100

0

0.0E

+ 0

1.0E

+ 5

2.0E

+ 5

3.0E

+ 5

4.0E

+ 5

5.0E

+ 5

6.0E

+ 5

Ci/μg/ L

Wi/μ

g/g

1, 1, 1-Trichloroethane1, 2-Dichloroethane

FIGURE 3.16 Linear adsorption isotherms for two organic compounds on a typicalsoil. (Data from Chiou, C.T., Peters, L.J., and Fread,V.H. 1979. Science 206, 831–832.)

earlier section), it is no surprise that it has a large adsorption constant. For a majorityof hydrophobic organic compounds, the linear adsorption isotherms on both soils andsediments afford a good representation of adsorption data.

It is well documented in both the surface chemistry and environmental chemistryliterature that various forms of adsorption isotherms are possible.Most of the isothermshapes can be explained starting from what is known as the Brauner–Emmett–Teller(BET) isotherm. The Langmuir isotherm limits the adsorption capacity to a singlemolecular layer on the surface. The BET approach relaxes the assumption in theLangmuir model and suggests that adsorption need not be restricted to a singlemonolayer and that any given layer need not be complete before the subsequentlayers are formed. The first-layer adsorption on the surface occurs with an energyof adsorption equivalent to the heat of adsorption of a monolayer just as in the caseof the Langmuir isotherm. The adsorption of subsequent layers on the monolayeroccurs via vapor condensation. Figure 3.17 is a schematic of multi-layer formation.A multi-layer can be formed in the case of adsorption of solutes from the gas phaseand of solutes from the liquid phase onto both solid and liquid surfaces. The the-ory underlying the BET equation is well founded and can be derived either froma purely kinetic point of view or from a statistical thermodynamic point of view(Adamson, 1990). Without elaborating on the derivation, we shall accept the finalform of the BET equation. If it is assumed that (i) the number of layers on thesurface is n = 64, and (ii) the energy of adsorption for any molecule in all layers

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Solute in the bulk phase

Solid surface

MonolayerMultilayers

n = 1n = 2

n = ∞

FIGURE 3.17 A schematic of the adsorption of a compound from either the gas phase orliquid phase on to a solid. Simultaneous formation of mono- and multilayers of solute on thesurface is shown. Note that the same phenomenon can also occur on a liquid surface instead ofthe solid surface. The BET isotherm applies in either case.

n = 2, 3, 4, . . . , is the same, but is different for a molecule in the layer n = 1, oneobtains

θi = Γi

Γmi= KBETψi

(1 −ψi) {1 + (KBET − 1) ·ψi} , (3.82)

where Γmi is the monolayer capacity and ψi = Pi/P∗

i for adsorption from the gasphase and ψi = Ci/C∗

i for adsorption from the liquid phase. P∗i is the saturation

vapor pressure of solute i in the gas phase, whereas C∗i is the saturated concen-

tration of solute i in the liquid phase. Note that θi can now be greater than 1,indicating multi-layer adsorption. KBET is the BET adsorption constant. Figure 3.18illustrates the isotherm shapes for different values of KBET. For values of 10 and100, a clear transition from a region of monolayer saturation coverage to multi-layeris evident. These isotherms are characterized as Type II. The Langmuir isothermis characterized as a Type I isotherm. When the value of KBET is very small (0.1and 1 in Figure 3.18), there is no such clear transition from monolayer to multi-layer coverage. These isotherms are called Type III isotherms. There are numerousexamples in environmental engineering where Type II isotherms have been observed;examples include the adsorption of volatile organic compounds (VOCs) on soils(Chiou and Shoup, 1985; Valsaraj and Thibodeaux, 1988) and aerosol particulates(Thibodeaux et al., 1991). Type III isotherms are also common in environmentalengineering, for example, the adsorption of hydrocarbon vapors on water surfaces(Hartkopf and Karger, 1973; Valsaraj, 1988) and in soil–water systems (Pennel et al.,1992).

The BET isotherm is often used to obtain the surface areas of soils, sediments, andother solid surfaces by monitoring the adsorption of nitrogen and other inert gases.The method has also enjoyed use in the environmental engineering area. The BET

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

10

9

8

7

6

5

Thet

a

yi

4

3

2

1

0

0.01

1

100

Type III

Type II

FIGURE 3.18 Different shapes of the BET isotherm. Note that the shapes vary with theisotherm constant, KB.

equation can be recast into the following form:

1

Γi· ψi

1 −ψi = 1

KBETΓmi

+ (KBET − 1)

KBET·ψi, (3.83)

so that both KBET and Γmi can be obtained from the slope and intercept of a linear

regression of the LHS versus yi. The linear region of the plot typically lies betweenψi values of 0.05 and 0.3 and extrapolation below or above this limit should beapproached with caution. If the specific area of the adsorbate, σm, is known, then thetotal surface area of the adsorbent can be obtained from the equation

σmNΓmi = Sa. (3.84)

Pure nitrogen, argon, or butane with specific surface areas of 16.2, 13.8, and 18.1 ×10−16 cm2/molecule are used for this purpose.

EXAMPLE 3.19 USE OF THE BET EQUATION

Poe et al. (1988) reported that the vapor adsorption of an organic compound (ethyl ether)on a typical dry soil (Weller soil fromArkansas) showed distinct multi-layer formation.The following data were obtained:

continued

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Relative Partial Pressure ψi Ether Adsorbed (mg/g soil)

0.10 6.740.28 13.270.44 15.320.56 18.040.63 23.490.80 32.280.90 46.09

Use the above data to obtain (i) the BET isotherm constants and (ii) the surface area ofthe soil (in m2/g).

Since the data are given directly in terms ofψi and Γi, a plot of (1/Γi)(ψi/1 −ψi)versus ψi can be made as shown in Figure 3.19a. Only the data between ψi valuesof 0.01 and 0.63 were considered for this plot. A good linear fit with an r2 of 0.9776was obtained. The slope S was 0.1142 and the intercept I was 0.00211. Hence Γm

i =1/(S + I) = 8.59mg/g and KBET = (S/I) + 1 = 55.1. Using these parameters, theBET isotherm was constructed and compared with the experimental plot of Γi versusψi. An excellent fit to the data was observed as in Figure 3.19b. The transition froma monolayer to a multi-layer adsorption is clearly evident in this example.

From the value of Γmi , one can calculate the total surface area of the soil if the

molecular area occupied by ether on the surface is known. In order to do this, we shallfirst express Γm

i as 8.59 × 10−3/74 = 1.16 × 10−4 mol/g. The surface area occupiedby a molecule of ether in the monolayer is given by Am = ϕ(M/ρN)2/3, where ϕ isthe packing factor (1.091 if there are 12 nearest neighbors in the monolayer), M isthe molecular weight of ether (= 74), ρ is the liquid density of ether (= 0.71 g/cm3),and N is the Avogadro number (= 6.023 × 1023). In this example, substituting these

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7yi

0.08(a) (b)

0.07

0.06

0.05

0.04

0.03

0..02

0.01

0

(1/Γ

)*( y i

/1–y

i )

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9yi

90

80

70

60

50

40

30

20

10

0

Γ i /m

g / g

FIGURE 3.19 (a) Linear BET plot to obtain isotherm constants and (b) BET isotherm andthe experimental points.

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values, we obtain Am = 28 × 10−16 cm2. Therefore, the total surface area of the soilSa = Γm

i NAm = 20m2/g. For large polyatomic molecules, ϕ will be considerablydifferent from 1.091. Moreover, for ionic compounds the lateral interactions betweenthe adsorbateswill affect the arrangements in themonolayer. It is also known that ioniccompounds will affect the interlayer spacing in the soil, and hence will over-estimatethe surface areas. For these reasons, ideally, the BET surface areas are measured viathe adsorption of neutral molecules (e.g., nitrogen and argon).

3.5.3 ADSORPTION AT CHARGED SURFACES

There are numerous examples in environmental engineering where charged (elec-trified) interfaces are important. Most reactions in the soil–water environment aremediated by the surface charges on oxides and hydroxides in soil. Colloids used forsettling particulates in wastewater treatment work on the principle that modifyingthe charge distribution on particulates facilitates flocculation. Surface charges on airbubbles and colloids help in removing metal ions by foam flotation. The aggregationand settling of atmospheric particulates and the design of air pollution control devicesalso involve charged interfaces. Most geochemical processes involve the adsorptionand/or complexation of ions and organic compounds at the solid/water interface. Itis therefore obvious that a quantitative understanding of adsorption at charged sur-faces is imperative in the context of this chapter. Two notable books deal extensivelywith these aspects (Morel and Herring, 1993; Stumm, 1993). In this section, we shallexplore how the distribution of charges in the vicinity of an electrified interfacewill becomputed. The applications of this to the problems cited are delegated to Chapter 4. Inthe aqueous environment, the problem of applying these simple principles of adsorp-tion leads to a more complex theory that incorporates chemical binding to adsorptionsites besides the electrical interactions. This is called the surface complexation model.We shall discuss some of this in Chapter 4 and other reaction kinetics applicationsin Chapter 6.

For the most part the approach is similar to the one used to obtain the chargedistribution around an ion. The Poisson equation from electrostatics is combinedwith the Boltzmann equation from thermodynamics to obtain the charge distributionfunction. This is known as the Guoy–Chapman theory. The essential difference isthat here we are concerned with the distribution of charged particles near a surfaceinstead of near an ion. For simplicity, we choose an infinite planar surface as shownin Figure 3.20. The electrical potential at the surface, Ψ0, is known. The planarapproximation for the surface holds in most cases when the size of the particlesnear the charged surface is much larger than the distance over which the particle–surface interactions occur in the solution. As shown in Figure 3.20, we have a doublelayer of charges: a localized charge density near the surface compensated for by thecharge density in the solution. This is called the diffuse-double-layer model. The finalequation for charge distribution is (Bockris and Reddy, 1970)

eu/2 = eu0/2 + 1 + (eu0/2 − 1)

e−κx

eu0/2 + 1 − (eu02 − 1)

e−κx , (3.85)

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+

+

++

+

+

+

+

+

++ +

+

+

+

Surface potential, Ψ0

Stern layer

Diffuse layer, Ψ(x)

FIGURE 3.20 Distribution of charges in a solution near a negatively charged surface.

where u = zeΨ/kT and κ is the inverse of the Debye length defined earlier. For verysmall values of Ψ0 (−25mV or less for a 1:1 electrolyte), that is, u0 � 1, we canderive the simple relationship,

Ψ = Ψ0e−κx . (3.86)

The significance of κ is easily seen in this equation. It is the value of κ at whichΨ = Ψ0/e, and hence 1/κ is taken as the effective double-layer thickness.

If u0 � 1, that is, Ψ0 � 25mV (for a 1:1 electrolyte), the result is

Ψ = 4kT

ze· e−κx , (3.87)

which tends to a value of Ψ0 = 4kT/ze as x → 1/κ.The expression for Ψ obtained above can be related to the surface charge density

σ through the equation

σ = ε

4π

∫∞

0

d2Ψ

dx2dx =

√2n0εkT

πsinh

(u02

). (3.88)

For small values of u0 we get σ = (εκ/4π) Ψ0, which is similar to that for a chargedparallel plate condenser (remember undergraduate electrostatics!).

If the definition of Debye length is cast in terms of the following equation,κ2 = (4πe2/εkT)Σniz2i , then the above equation becomes generally applicable toany type of electrolyte. Note that the ionic strength I is related to Σniz2i , sinceni = (NACi/1000) with Ci in mol/L. It is useful to understand how the value of

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Ψ changes with κ, which in turn is affected by the ionic strength I . Since I isproportional to κ, this suggests that at high ionic strength the compressed doublelayer near the surface reduces the distance over which long-range interactions dueto the surface potential is felt in solution. This is the basis for flocculation as awastewater treatment process to remove particles and colloids. It is also the basisfor designing some air sampling devices that work on the principle of electro-static precipitation. A number of other applications exist, which we shall explorein Chapter 4.

The Guoy–Chapman theory is elegant and easy to understand, but it has a majordrawback. Since it considers particles as point charges, it fails when x → r, the radiusof a charged particle. Stern modified the inherent assumption of zero volume ofparticles assuming that near the surface there is a region excluded for other particles.In other words, there are a number of ions “stuck” on the surface that have to bebrought into solution before other ones from solution replace them. Thus, the drop-off in potential near the surface is very gradual and almost flat till x reaches r, beyondwhich theGuoy–Chapman theory applies. The region is called the Stern layer. Furthermodifications to this approach have been made, but are beyond the scope of this book.Interested students are encouraged to consult Bockris and Reddy (1970) for furtherdetails.

EXAMPLE 3.20 CALCULATION OF DOUBLE-LAYER THICKNESS

Calculate the double-layer thickness around a colloidal silica particle in a 0.001MNaCl aqueous solution at 298K.κ2 = (4πe2/εkT)Σniz

2i . We know that Σniz2i = 2n, e = 4.802 × 10−10 C, ε =

78.5, k = 1.38 × 10−16 erg/molecule K, and T = 298K. Noting the relationshipbetween Ci and ni given earlier, we can write κ2 = 1.08 × 1015Ci, hence κ = 3.28 ×107,C1/2 = 1.038 × 106 cm−1. Hence double-layer thickness 1/κ = 9.63 × 10−7 cm(= 96.3Å).

PROBLEMS

3.11 Calculate the mole fraction, molarity, and molality of each compound inan aqueous mixture containing 4 g of ethanol, 0.6 g of chloroform, 0.1 gof benzene, and 5 × 10−7 g of hexachlorobenzene in 200mL of water.

3.21 A solution of benzene in water contains 0.002mole fraction of benzene.The total volume of the solution is 100mL and has a density of 0.95 g/cm3.What is the molality of benzene in solution?

3.32 Given the partial pressure of a solution of KCl (4.8molal) is 20.22Torr at298K and that of pure water at 298K is 23.76Torr, calculate the activityand activity coefficient of water in solution.

3.42 Give that μ0 = −386 kJ/mol for carbon dioxide on the molality scale,calculate the standard chemical potential for carbon dioxide on the molefraction and the molarity scale.

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3.51 Estimate the mean ionic activity coefficients of the following aqueoussolutions:

(a) 0.001molal KCl; (b) 0.01molal CaCl2; and (c) 0.1molal Na3PO4.

3.62 Consider a total evaluative environment of volume 10 × 1010 m3 withvolume fractions distributed as follows: air 99.88%; water 0.11%; soil0.00075%; sediment 0.00035%; and biota 0.00005%. Consider twocompounds benzene and DDT (a pesticide) with properties as follows:

Property Benzene DDT

K ′aw (Pam3/mol) 557 2.28

Ksw (L/kg) 1.1 12,700Kbw (L/kg) 6.7 77,400

The temperature is 310K and the densities of soil and sediment are1500 kg/m3 while that of the biota is 1000 kg/m3. Determine the concen-trations and mass of each compound in the various compartments usingthe fugacity level I model. Comment on your results.

3.72 Confined disposal facilities (CDFs) are storage facilities for contaminatedsediments. Being exposed to air, the contaminated sediment is a source ofVOCs to the atmosphere. Consider a two-compartment evaluative environ-ment of volume 5 × 105 m3, with the following volume fractions: air 60%and sediment 40%. Consider benzene for which K ′

aw = 557 Pam3/moland Ksw = 5 L/kg. If the temperature is 298K, and the density of sedi-ment is 1200 kg/m3, determine the mass fraction of benzene in air usingthe fugacity level I model.

3.73 Mercury exists in the environment in a variety of forms. Organic mer-cury exists predominantly as dimethyl mercury, (CH3)2Hg which is easilyassimilated by the biota. Hence fish in the lakes of several states in theUnited States are known to contain high concentrations of Hg. Dimethylmercury has the following properties: Kaw = 0.31 (molar concentrationratio), Ksw = 7.5 L/kg, Kbw = 10 L/kg. Consider an evaluative environ-ment of 109 m3 of air in contact with 105 m3 of water in a lake containing103 m3 of sediment and10m3 of biota.The sediment density is 1000 kg/m3

and that of biota is 1500 kg/m3.

(a) What is the fraction accumulated by the biota in the lake? Use thefugacity level I model.

(b) If an influent feed rate of 100m3/h of air with a concentration of0.1mol/m3 is assumed, calculate the fraction assimilated by the biotausing the fugacity level II model.

3.83 Dimethylsiloxane (DMS) is a prevalent compound in various soaps, sham-poos, and shaving creams. It is expected to pose problems when it getsinto the natural environment through sewage and eventually distributesto all compartments. Consider a 100m2 area of sewage treatment unitthat contains 10m height of atmosphere above 1m of a water columnat 300K. The water contains 5% by volume of solids and 10% by vol-ume of bioorganisms. Consider the following properties: Henry’s constant

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125 Pam3/mol; logKow = 5.1; logKsw = 2.9; logKbw = 3.1. The solidsdensity is 0.015 kg/L and the bioorganism density is 1 kg/L. Determinewhich compartment will have the highest amount of DMS?

3.92 The problem of oxygen solubility in water is important in water qualityengineering. Reaeration of streams and lakes is important for maintainingan acceptable oxygen level for aquatic organisms to thrive.

(a) The solubility of oxygen in water at 1 bar pressure and 298K is givenas 1.13 × 10−3 molal. Calculate the free energy of solution of oxygenin water.

(b) It is known that atmospheric air has about 78% by volume of nitrogenand 21% by volume of oxygen. If the Henry’s constants (KH ) foroxygen and nitrogen are 2.9 × 107 and 5.7 × 107 Torr, respectively,at 293K, calculate the mass of oxygen and nitrogen in 1 kg of pristinewater at atmospheric pressure.

(c) If Raoult’s law is obeyed by the solvent (water), what will be the par-tial pressure of water above the solution at 98K. The vapor pressureof pure water is 23.756Torr at 298K.

3.102 The Henry’s constants for oxygen at three different temperatures are givenbelow:

T (K) KH (Torr)

293 2.9 × 107

298 3.2 × 107

303 3.5 × 107

(a) Calculate the heat of solution of oxygen in water assuming nodependence on temperature between 293 and 303K.

(b) Estimate the standard molar entropy of solution at 298K.

3.113 The Henry’s constants for ethylbenzene at different temperatures are givenbelow:

T (K) K ′aw (atm m3/mol)

283 0.00326288 0.00451293 0.00601298 0.00788303 0.01050

(a) Express Henry’s constant in various forms (KH, Kaw,K ′′aw, and Kx)

at 298K.(b) Calculate the enthalpy of solution for ethylbenzene.

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3.123 For naphthalene the following vapor pressure data were obtained from theliterature (Sato et al., 1986; Schwarzenbach et al., 1993).

T (K) Vapor Pressure (Pa)

298.5 10.7301.5 14.9304.5 19.2307.3 25.9314.8 52.8318.5 71.1322.1 97.5322.6 101325.0 124330.9 200358.0 1275393.0 5078423.0 12756

(a) Plot the log of vapor pressure as a function of 1/T and explain thebehavior.

(b) Estimate from the data the heat of sublimation of naphthalene and theheat of melting (fusion) of solid naphthalene to sub-cooled liquid.

(c) Estimate the vapor pressure of naphthalene over the temperaturerange given using the equations from the text and compare with theactual values.

3.132 The enthalpy of vaporization of chloroform at its boiling point 313.5K is28.2 kJ/mol at 25.3 kPa (atmospheric pressure).

(a) Estimate the rate of change of vapor pressure with temperature at theboiling point.

(b) Determine the boiling point of chloroform at a pressure of 10 kPa.(c) What will be its vapor pressure at 316K?

3.143 (a) The normal boiling point of benzene is 353.1K. Estimate the boilingpoint of benzene in a vacuum still at 2.6 kPa pressure. Use Trouton’srule.

(b) The vapor pressure of solid benzene is 0.3 kPa at 243K and 3.2 kPaat 273K. The vapor pressure of liquid benzene is 6.2 kPa at 283Kand 15.8 kPa at 303K. Estimate the heat of fusion of benzene.

3.153 0.5m3 of nitrogen at 400 kPa and 27◦C is contained in a vessel kept insu-lated from the atmosphere. A heater within the device is turned on andallowed to pass a current of 2A for 5min from a 120V source. Thus, elec-trical work is done to heat the nitrogen gas. Find the final temperature fornitrogen in the tank. CP for nitrogen gas is 1.039 kJ kg/K.

3.162 What pressure is required to boil water at 393K? The enthalpy ofvaporization of water is 40.6 kJ/mol.

3.173 During the flooding of a residential area of NewOrleans byHurricane Kat-rina on August 29, 2005, well-mixed floodwaters containing suspendedLake Pontchartrain sediments entered the homes and businesses of NewOrleans.After the floodwaters stagnated, sediment settled out of the watercolumn and was deposited on floors and other interiors of the homes,

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where it remained till the floodwaters were pumped out of the city twoweeks later. Work by Ashley et al. (2007) provided evidence that sedi-ment deposited inside the homes had contaminant concentrations muchlarger than that deposited outside. One such example was chlordane, acommon organochlorine pesticide, detected in the sediment inside thehomes. A concern arises about the potential for this species to volatilizefrom the sediment and be found inside the vapor phase of the homes.Additionally high concentrations of mold spores represents another sinkfor these particles from the gas phase and returning residents and firstresponders can inhale both the air inside the homes and the aerosolizedmold spores. Using a level I fugacity model determine the concentra-tions of chlordane present in both the gas phase (μg/m3) and the aerosolphase (mg/kg of mold) inside the home. For the aerosol phase the fugac-

ity capacity can be obtained from ZQ = 10(logKoa

+ log fom

−11.91)ZAρQ ×109. Note that logKoa = 8.872, fom = 0.60, ρQ = 5.31 × 10−18 kg/m3.Chlordane properties are as follows: molecular weight = 409.8, K ′

aw =4.91 Pam3/mol, and Ksw = 4265 L/kg. Sediment density is 2.5 g/cm3.Volumes of air, aerosol, and sediment are 401. 6m3, 6.7 × 10−8 m3, and0.329m3, respectively. (Note: This problem was provided by NicholasAshley.)

3.182 The equilibrium constant for the dissociation reaction A2 ↔ 2A is 6 ×10−12 atm at 600K and 1 × 10−7 atm at 800K. Calculate the standardenthalpy change for this reaction assuming that the enthalpy is constant.What is the entropy change at 600K?

3.192 In a chemodynamics laboratory an experiment involving the activity ofsediment dwelling worms in a small aquarium was found to be seriouslyaffected for a few days. The problem was traced to bacterial growth aris-ing from dirty glassware. Bacteria can survive even in boiling water atatmospheric pressure. However, literature data suggested that they can beeffectively destroyed at 393K. What pressure is required to boil waterat 393K to clean the glassware? The heat of vaporization of water is40.6 kJ/mol.

3.202 Obtain the enthalpy of vaporization of methanol. The vapor pressures atdifferent temperatures are given below.

Temperature (◦C) Vapor Pressure (mm Hg)

15 74.120 97.625 127.230 164.2

3.211 Can an assemblage of liquid droplets be in true equilibrium with its bulkliquid? Explain your answer.

3.222 The formation of fog is accompanied by a simultaneous decrease in tem-perature and an increase in relative humidity. If water in the atmosphereis cooled rapidly to 298K, a degree of super-saturation is reached whenwater drops begin to nucleate. It has been observed that for this to hap-pen the vapor pressure of water has to be at least 12.7 kPa. What will bethe radius of a water droplet formed under super-saturation? How many

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water molecules will be in the droplet? Conventional wisdom says that thenumber of water molecules on a drop should be less than the total number.Calculate the ratio of water molecules on the surface to the number in thedroplet. Comment on your result. Molecular area of water = 33.5Å2.

3.231 The value of Kow as a ratio of molar concentrations is always smaller thanthat expressed as ratio of mole fractions. Explain why?

3.243 Using the values from Appendix 3 and the rules given in Lyman et al.(1990), obtain the logKow for the following compounds. Obtain the correctstructures for the compounds from standard organic chemistry texts:

(i) Pentachlorophenol,(ii) Hexachlorobenzene,(iii) p,p′-DDT,(iv) Lindane,(v) Phenanthrene,(vi) Isopropyl alcohol

3.252 The melting point of naphthalene is 353.6K. Its solubility in water at roomtemperature has been measured to be 32mg/L.

(a) Calculate the activity coefficient of naphthalene in water.(b) What is its ideal solubility in water?

3.262 The solubility of anthracene in water at different temperatures has beenreported by May et al. (1977) as follows:

t (◦C) Solubility (μg/kg)

5.2 12.710.0 17.514.1 22.218.3 29.122.4 37.224.6 43.4

Determine the excess thermodynamic functions for the solution ofanthracene in water at 25◦C. ΔHm is 28.8 kJ/mol.

3.271 The cavity surface area for accommodating benzene in water is 240.7Å2,whereas the molecular area of benzene is only 110.5Å2. Can you explainthe difference?

3.282 Calculate the aqueous solubility of hexachlorobenzene in water usingappropriate correlations and calculations involving (a) molecular area, (b)molecular volume, and (c) molar volume.

3.292 The compound 2,2′,3,3′-tetrachlorobiphenyl is a component of Aroclor,a major contaminant found in sedimentary porewater of New Bedfordharbor, MA. In order to perform F&T estimates on this compound, weneed, first and foremost, its aqueous solubility in seawater.

(a) Use the fragment constant approach to obtain Kow.(b) Obtain the aqueous solubility in ordinary water from the appropriate

Kow correlation.(c) Check the uncertainty in the prediction using alternate methods such

as molecular area (225.6Å2) and molar volume (268 cm3/mol).

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3.302 The mean activity coefficients of KCl in water at 25◦C are given below:

Molality of KCl Γ∀0.1 0.770.2 0.720.3 0.690.5 0.650.6 0.640.8 0.621 0.60

Test whether the above data are in compliance with the Debye–Huckellimiting law.

3.312 State whether the following statements are true or false:

(a) Octanol–water partition constant is a measure of the aqueous activityof a species.

(b) Excess free energy of dissolution for a large hydrophobic moleculeis positive since the excess enthalpy is always positive.

(c) Methanol does not increase the solubility of dichlorobenzene inwater.(d) At equilibrium the standard state chemical potentials for a species i

in air and water are the same.(e) Fugacity is a measure of the degree to which equilibrium is estab-

lished for a species in a given environmental compartment.(f) If two aqueous solutions containing different nonvolatile solutes

exhibit the same vapor pressure at the same temperature, the activitiesof water are identical in both solutions.

(g) If two liquids (toluene and water) are not completely miscible withone another, then a mixture of the two can never be at equilibrium.

3.322 Chlorpyrifos is an insecticide. It is a chlorinated heterocyclic compoundand is only sparingly soluble in water. Its properties are tm = 42◦C,M =350.6, logKow = 5.11, and, Kaw = 4.16 × 10−3 atmP/mol. Estimate itsvapor pressure using only the given data.

3.332 Calculate the infinite dilution activity coefficient of pentachlorobenzene(C6HCl5) in water. It is a solid at room temperature and has a meltingpoint of 86◦C and a molecular weight of 250.3. It has a logKow of 4.65.

3.341 Match the equation on the right with the appropriate statement on the left.Only one equation applies to each statement.

Raoult’s Law (A) Gc + GtHenry’s constant (B) RT Pn γiOctanol–water partition constant (C) (dPn P/dT) = ΔH/RT2

Ionic strength (D) KC1/m

Gibbs excess free energy (E) fwi = f a

iFreundlich isotherm (F) 0.5 3miz2iClausius–Clapeyron equation (G) P∗

i /C∗i

Free energy of solution (H) Pi/P∗i

(I) KC(J) H + TS

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3.352 A mixture of polystyrene (2) in toluene (1) is said to follow theFlory–Huggins model for activity coefficients. Given that ν2 =1000ν1 and χ = 0.6, obtain γ1 and γ2 as a function of x2 at 298K.

3.362 A mixture of methyl acetate (1) and methanol (2) has a Margules constantA = 1.06RT . Obtain a plot of the activity coefficient of methanol as afunction of its mole fraction at 298K. Use the one constant Margulesequation.

3.372 If seawater at 293K has the following composition, what is the total ionicstrength of seawater?

Electrolyte Molality

NaCl 0.46MgSO4 0.019MgCl2 0.034CaSO4 0.009

3.383 One gram of a soil (total surface area 10m2/g) from a hazardous wastesite in North Baton Rouge, Louisiana, was determined to contain 1.5mgof chloroform (molecular surface area 24Δ2) at 298K. The soil samplewas placed in a closed vessel and the partial pressure of chloroform wasmeasured to be 10 kPa at 298K. The heat of adsorption of chloroformon the soil from the vapor phase was estimated to be −5 kJ/mol. If theLangmuir adsorption isotherm is assumed, what is the adsorption constantat 298K?

3.393 The following data were reported by Roy et al. (1991) for the adsorptionof herbicide (2,4-dichlorophenoxy acetic acid designated 2,4D) on a typ-ical Louisiana soil. Five grams of soil were equilibrated with 200mL ofaqueous solution containing known concentrations of 2,4D and kept vig-orously shaken for 48 h. The final equilibrium concentrations of 2,4D inwater were measured using a liquid chromatograph.

Final Equilibrium Aqueous Initial Spike AqueousConcentration (mg/L) Concentration (mg/L)

1.0 1.12.9 3.16.1 6.214.8 15.045.3 48.9101.9 105.3148.0 152.8239.5 255.0

Based on appropriate statistics, suggest which of the three isotherms(linear, Langmuir, or Freundlich) best represents the data.

3.402 The following adsorption data were reported by Thoma (1994) for theadsorption of a polyaromatic hydrocarbon (pyrene) on a local sediment(University lake, Baton Rouge, Louisiana).

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Equilibrium Aqueous AdsorbedConcentration (μg/L) Concentration (ng/g)

5.1 35.67.0 51.89.7 51.913.4 67.213.0 70.9

The molecular weight of pyrene is 202 and its aqueous solubility is130μg/L. Obtain the linear partition constant (in L/kg) for pyrene betweenthe sediment and water. What is the adsorption free energy for pyrene onthis sediment?

3.412 The following data were obtained by Poe et al. (1988) for the adsorp-tion of an aromatic pollutant (benzene) on an Arkansas soil (Weller soil,Fayetteville, AR). Plot the data as a BET isotherm and obtain the BETconstants. From the BET constants obtain the surface area of the soil andthe adsorption energy for benzene on this soil.

Relative Vapor Pressure Amount Adsorbed (mg/g)

0.107 7.070.139 10.680.253 14.180.392 18.320.466 20.890.499 22.15

3.422 The following data were reported by Karger (1971) for the adsorption ofhydrocarbon (n-nonane) vapor on a water surface at 12.5◦C.

Relative Partial Amount AdsorbedPressure (mol/cm2)

0.07 0.20 × 10−11

0.12 0.360.17 0.480.22 0.600.25 0.780.30 0.940.34 1.120.38 1.260.43 1.480.47 1.680.52 1.880.56 2.10

What type of an isotherm do these data represent? Obtain the isothermconstants and the free energy of adsorption. The reported free energy ofadsorption of 12.5◦C is −4.7 kcal/mol.

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3.431 The following two observations have the same explanation. River water isoften muddy, whereas seawater is not. At the confluence of a river withthe sea, a large silt deposit is always observed. Based on the principles ofcharged interfaces, provide a qualitative explanation for these observations.

3.442 Determine the ionic strength and corresponding Debye lengths for thefollowing solutions: 0.005M sodium sulfate, 0.001M calcium sulfate, and0.001M ferrous ammonium sulfate.

3.452 Typical concentrations of ions in natural waters are given below. Calculatethe ionic strength and Debye lengths in these solutions:

Concentration (mmol/kg)

Medium Na+ K+ Ca2+ Mg2+ Cl− SO2−4 NO−

3 F−

Seawater 468 10 10.3 53 546 28 0.05 0.07Rainwater 0.27 0.06 0.37 0.17 0.22 0.12 0.02 0.005Fogwater 0.08 – 0.2 0.08 0.2 0.3 – 1

3.462 Trichloroethylene is used as a dry cleaning fluid and is an environmentallysignificant air and water pollutant. Obtain the enthalpy of vaporizationfrom the following vapor pressure data:

Temperature (◦C) Vapor Pressure (mm Hg)

20 56.825 72.630 91.5

3.472 Estimate the vapor pressure of (a) chlorobenzene (a liquid) and (b)1,3,5-trichlorobenzene (a solid) at a temperature of 25◦C. The meltingpoint of trichlorobenzene is 63◦C. Compare with the reported values inAppendix 1.

3.482 Alkylethoxylates (AE) are nonionic surfactants that find widespread usein cleaning, industrial, and personal care products. As such their envi-ronmental mobility is predicated upon a knowledge of their adsorptionto sediments. Consider n-alkylethers of poly(ethylene glycol) of generalformula CH3–(CH2)n−1–(OCH2CH2)xOH, that is, AnEx . If n = 13 andx = 3, it has a molecular weight of 332. The following data were obtainedfor adsorption of A13E3 to a sediment (EPA 12):

Mass in Aqueous Solution (nM) Mass Adsorbed (μmol/kg)

175 25270 45350 80450 100

Determinewhether theLangmuir or Freundlich isothermbest fits the abovedata.

3.493 DDT is a pollutant that canbe transported through the atmosphere to remotelocations. The sub-cooled liquid vapor pressure (below 110◦C) and liquid

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vapor pressure of p,p′-DDT at different temperatures are given below:

t (◦C) P∗i (Torr)

25 2.58 × 10−6

90 2.25 × 10−3

100 5.18 × 10−3

110 0.0114120 0.0243130 0.0497140 0.0984150 0.189

The melting point of DDT is 109◦C. (A) Obtain from the above data theenthalpy of vaporization of DDT. (B) Estimate the vapor pressure of solidDDT at 25◦C.

3.503 An aqueous solution containing 2.9 × 10−4 mole fraction of CO2 isin equilibrium with pure CO2 vapor at 101.325 kPa total pressure and333K. Calculate the mole fraction of CO2 in water if the total pressure is10132.5 kPa at 333K. Neglect the effect of pressure on Henry’s constant.Also assume that the vapor phase behaves ideally, whereas the aqueousphase does not.

3.512 Given the following data on the aqueous solubility of SO2 at 298K,calculate Henry’s constant in various forms:

xwi Pi (atm)

3.3 × 10−4 0.00665.7 × 10−4 0.0132.3 × 10−3 0.0654.2 × 10−3 0.1317.4 × 10−3 0.2631.0 × 10−2 0.395

3.523 Oil field waste, considered nonhazardous, is transported in a tanker truckof volume 16m3 and disposed in landfarms. The total waste is composedof 8m3 of water over 1m3 of waste oil. The waste oil is sour and thereis the possibility of hydrogen sulfide generation. Use the fugacity level Imodel to determine the percent hydrogen sulfide in the vapor phase insidethe truck. Hydrogen sulfide properties: KH = 560 atm, waste oil–waterpartition constant = 10, waste density = 1.1 g/cm3.

3.533 Freshwater lakes receive water from the oceans through evaporation bysunlight. Consider the ocean to be made entirely of NaCl of 0.4M in ionicstrength and the lake to be 0.001M in MgCl2. Compute the free energyrequired to transfer 1mol of pure water from ocean to lake. Assume aconstant temperature of 298K.

3.542 The Henry’s constant for phenanthrene at 20◦C was measured as2.9 Pam3/mol and its enthalpy of volatilization was determined to be41 kJ/mol. What is the Henry’s constant at 25◦C? The melting point ofphenanthrene is 100.5◦C.

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3.552 A solution of benzene and toluene behaves ideally at 25◦C. The pure com-ponent vapor pressure of benzene and toluene are respectively 12 kPa and3.8 kPa at 298K. Consider a spill of the mixture in a poorly ventilatedlaboratory. What is the gas-phase mole fraction of benzene in the room?Use Raoult’s law. The mixture is 60% by weight of benzene.

3.562 Five grams of soil contaminated with 1,4-dichlorobenzene from a localSuperfund site were placed in contact with 40mL of clean distilled waterin a closed vessel for 24 h at 298K. The solution was stirred and after72 h the soil was separated by centrifugation at 12,000 rpm. The aqueousconcentration was determined to be 35μg/L. If the partition constant wasestimated to be 13P/kg, what is the original contaminant loading on thesoil?

3.572 Estimate the activity coefficient of oxygen in the following solutions at298K: (a) 0.01M NaCl; (b) seawater (see problem 3.69 for composition);and (c) wastewater containing 0.001MNa2SO4 and 0.001M CaSO4. Usethe McDevit–Long equation from the text. ϕ = 0.0025. Oxygen solubilityin pure water is 8.4mg/P at 298K and 1 atm total pressure.

3.582 The Henry’s constant for methyl mercury at 0◦C is 0.15 and at 25◦C is0.31. Estimate the enthalpy of dissolution of gaseous methyl mercury inwater.

3.593 In pure water the solubility of benzo-[a]-pyrene varies as follows:

t (◦C) C∗i (nmol/L)

8.0 2.6412.4 3.0016.7 3.7420.9 4.6125.0 6.09

The melting point of the compound is 179◦C. Calculate the excess entropyof solution at 298K.

3.602 Siloxanes are used in the manufacture of a number of household andindustrial products for everyday use. It is therefore of importance to knowits distribution in the environment. Consider octamethyl cyclotetrasilox-ane (D4) whose properties are given below: Kaw = 0.50, logKow =5.09, logKsw = 2.87. For an evaluative environment of total volume109 m3 which is 50% air, 40% water, and 10% sediment, which com-partment will have the highest fraction of D4? Use the fugacity level Imodel.

3.612 Atmospheric reactions in hydrometeors (rain, fog, mist, and dew) producealdehydes andketones.Consider the conversionof pyruvic acid to acetalde-hyde in fog droplets. The reaction is CH3–C(O)–COOH (g) → CH3CHO(g) + CO2 (g). (a) Obtain the standard free energy change for the reactionat 298K. (b) What is the equilibrium constant for the reaction at 298K?

3.622 It is known that when deep sea divers come up rapidly from their dives,they get what is called “bends.” This is due to the formation of bubbleson nitrogen in their blood serum (approximated as water). If the Henry’sconstant for nitrogen is 8.6 × 103 atm,what is the concentration of nitrogenin the blood serum of a diver who came up from 200 ft below sea level?

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3.632 Acid rain occurs by dissolution of SO3 in aqueous droplets. SO2 from coalburning power plants is the culprit. The reaction is SO2(g) + 0.5O2(g) →SO3 (g).What is the equilibrium constant for this reaction at 298K? If thepartial pressure of O2 is 0.21 atm, what is the equilibrium ratio of partialpressures of SO3 to SO2 at 298K?

3.643 Consider a 0.1mole fraction of ammonia solution in water at 298K. Usethe ideal solution approximation to obtain the partial pressure of water inthis system. If the actual measured vapor pressure is 3.0 kPa, obtain theactivity and activity coefficient of water in the solution.

3.652 Find the fugacity of a liquid A in water. The concentration of A in wateris 1mg/L. The molecular weight of A is 78. The vapor pressure of A is0.02 atm. Assume that A behaves ideally in water.

3.662 For the reaction H2S(g) → H2S (aq), the enthalpy of dissolution is−19.2 kJ/mol at 298K. If the temperature is raised to 318K, by howmuchwill the equilibrium constant of the reaction change?

3.673 Consider the flooding of New Orleans by Hurricane Katrina. The flood-water from Lake Pontchartrain extended over a square area of 10 miles× 10miles to a depth of 3m. Assume that the sediment from the lakeformed a thin layer under the floodwater extending over the entirecity to a depth of 1 ft. The atmosphere above the city extends to acompletely mixed depth of 1 km. The temperature is 300K. Gasolinecomponents (benzene, toluene, ethylbenzene, and xylene) were observedin the floodwaters and originated from submerged vehicles and gaso-line pumping stations. Consider the component ethylbenzene. Apply alevel I fugacity model to obtain the fraction of ethylbenzene presentin the floodwater. Henry’s constant for ethylbenzene is 851 Pam3/mol,sediment–water partition constant is 1.4 × 104 L/kg, and sediment densityis 0.015 kg/L.

3.683 Wetlands can be useful in phytoremediation of contaminants throughuptake of pollutants by plants.A wetland consists of soil, plants, and waterin equilibriumwith the air above. Consider an evaluative environment withthe following volumes: air = 107 m3, water = 105 m3, soil = 103 m3,and plants = 103 m3. Consider a chlorinated compound, TCE with thefollowing properties: Henry’s constant 3090 kPL/mol, soil–water parti-tion constant = 15 L/kg, plant–water partition constant = 300 L/kg, soildensity = 0.01 kg/L, and plant density = 0.1 kg/L. Evaluate the fractionof pollutants in the plants.

3.692 The equilibrium constant, K ′eq for the dissociation of bromine into atoms:

Br2 → Br• + Br• is 6 × 10−12 at 600Kand 1 × 10−7 at 800K.Calculatethe standard enthalpy change for this reaction assuming that it is constantin the temperature range. What is the entropy change at 600K?

3.703 During the flooding of a residential area of New Orleans by HurricaneKatrina, well-mixed floodwaters containing suspended Lake Pontchar-train sediments entered the homes and businesses of New Orleans. Afterthe floodwaters stagnated, sediment settled out of the water column andwas deposited on the floors and other interior surfaces of homes, whereit remained after floodwaters were finally pumped out of the city twoweeks later.A recent paper byAshley et al. [Chemosphere 70: 833 (2008)]provides evidence that contamination concentration in sediment deposited

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inside homes can bemuch higher than from samples collected fromoutsidehomes within the same neighborhoods. One such example is chlordane, acommon organochlorine pesticide, detected in the sediment in homes athigh concentrations. A concern arises about the potential for this speciesto volatilize from the sediment and be found in the vapor phase insidehomes. Additionally, the high concentration of mold spores representsanother sink for these particles from the gas phase, and returning residentsand first responders can inhale both air inside the home and aerosolizedmold spores. Using a level I fugacity model, determine the concentrationof chlordane present in both the gas phase (μg/m3) and the aerosol phase(mg/kgmold spores) inside homes.Assume 100 g of chlordane as the basis.For the aerosolized phase, the fugacity capacity is given by the followingequation:

ZQ = 10(logKoa+log fom−11.91)ZairρQ × 109.

Data required are fom = 0.60 (fraction of organic matter in the aerosol),aerosol (mold spore) density, ρQ = 5.31 × 10−18 kg/m3, T = 291.3K,

sediment density = 2.5 g/cm3.Volumes of phases:Vair = 401.6m3,VQ =6.7 × 10−8 m3, Vsediment = 0.329m3. For chlordane: molecular weightis 409.8, logKow = 5.08, octanol–air partition constant, logKoa = 8.872,density = 1.6 g/cm3, K ′

aw = 4.91 Pam3/mol, sediment–water partitionconstant, KSW = 4265 L/kg.

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4 Applicationsof EquilibriumThermodynamics

Chemical thermodynamics has numerous applications in environmental engineering.Wewill discuss specific examples that are relevant in understanding (a) the fate, trans-port, and transformations of chemicals in air, water, sediment, and soil environments,and (b) the design of waste treatment and control operations. The discussion willprimarily involve the equilibrium partitioning of chemicals between different phases.In general, the distribution of species i between any two phases A and B at a giventemperature is given by

KAB(T) = CiACiB

, (4.1)

whereCiB andCiA are concentrations of component i in phases B andA, respectively,at equilibrium.

4.1 AIR–WATER PHASE EQUILIBRIUM

In a number of situations in the environment, the two phases air and water co-exist.The largest area of contact is between these two phases in the natural environment.Exchange of mass and heat across this interface is extensive. In some cases onephase will be dispersed in the other. A large number of waste treatment operationsinvolve comminuting either of the phases to maximize the surface area and masstransfer. Firstly, we shall discuss the area of fate and transport modeling (chemo-dynamics). Secondly, we will discuss an area within the realm of separation processesfor waste treatment.

Consider the exchange of a compound at the air–water interface (Figure 4.1).Compounds transfer between air and water through volatilization and absorption.Gas bubbles transport materials from the ocean floor to the atmosphere. Ejectionsof particulates attached to the bubbles occur upon bubble bursting at the interface.Dissolved compounds and particulates in air can be deposited on sea or land viaattachment or dissolution in fog, mist, and rain. As discussed in Chapter 2, Henry’slaw describes the equilibrium at this interface. The air–water equilibrium constantKaw at any temperature is given by

Kaw(T) = CiaCiw

. (4.2)

119

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Atmosphere CloudDry deposition(particles)Volatilization/absorption

Water

Sediment

Bedrock

Wet deposition(particles + gas)

Bubble ebullition

FIGURE 4.1 Exchange of chemicals between the water and the atmosphere. Equilibrium isimportant in many cases such as volatilization–absorption, wet deposition, dry deposition, andgas bubble transport.

EXAMPLE 4.1 USE OF AIR–WATER PARTITION CONSTANT

Water from an oil-field waste pond contains hydrogen sulfide at a concentration of500mg/L at an ambient temperature of 298K. What will be the maximum equilibriumconcentration of H2S above the solution?At 298K, Kaw for H2S is 0.41 (Appendix 1). Ciw = 500mg/L = 0.5/34 =

0.0147mol/L. HenceCia = (0.41)(0.0147) = 0.006mol/L. Partial pressure in air,Pi =CiaRT = 0.147 atm.

Many waste treatment separation processes also involve air–water exchange ofchemicals (Figure 4.2). Examples include wastewater aeration ponds and surfaceimpoundments in which surface and submerged aerators are used to transfer oxygento the water, and to remove volatile organic and inorganic compounds from waterusing flotation. Volatile organics are removed using a tower where air and waterare passed countercurrent to one another through a packed bed. The operations arenonequilibrium processes, but where, the tendency of the system to attain equilibriumprovides the driving force for separation.

The transfer of material (say i) from one phase to the other (e.g., water to air) isdue to a gradient in concentration of i between the phases (Figure 4.3). This flow ofmaterial i from water to air is termed flux and is defined as the moles of i passing aunit area of interface per unit time. The equation for flux is

Ni = Kw(Ciw − C∗

iw

), (4.3)

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Applications of Equilibrium Thermodynamics 121

Wastewater treatment pond(submerged aerator)

Water

Water

Water

Air

Water

Air

Air

AirAirWater

Bubble air stripper towerPacked tower air stripper

FIGURE 4.2 Examples of air–water contact and equilibrium in separation processes. In eachof the three cases, contact between air and water is important.

where C∗iw is the concentration of i in water that would be in equilibrium with air.

This is given by the air–water distribution constant

C∗iw = Cia

Kaw. (4.4)

The maximum overall flux from water to air will be attained when Cia = 0. Thus,Nmaxi = KwCiw. If Ciw = 0, then the flux will be from air to water. Thus, if the sign

is positive the flux is from water to air, and if negative it is from air to water. Ifthe bulk air and water phases are in equilibrium, then Ciw = Cia/Kaw and the netoverall flux Ni = 0. Thus, the flow of material between the two phases will continueuntil the concentrations of the two bulk phases reach their equilibrium values asdetermined by the air–water equilibrium constant.At equilibrium, the net overall fluxis zero, and the water-to-air and air-to-water diffusion rates are equal but oppositein direction.

The flux expression given in Equation 4.3 above is predicated upon the assumptionthat compound i diffuses through a stagnant water film to a stagnant air film across aninterface of zero volume (Figure 4.3). Equilibrium is assumed to exist at the interface,but the twobulk phases are not in equilibriumand have uniformconcentrations beyondthe film thickness, δ. The term (Ciw − Cia/Kaw) is called the concentration drivingforce for mass transfer. The overall mass transfer coefficient Kw includes the valueof the equilibrium constant in the following form (see Section 6.1.4.1 in Chapter 6):

1

Kw= 1

kw+ 1

kaKaw, (4.5)

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fi w

fi aFugacity profile

fi

Ciw

Concentration profile Ciweq

Ciaeq

Cia

Airfilm

Waterfilm

Bulk water Bulk airInterface

FIGURE 4.3 Film theory of mass transfer of solutes between air and water. The above dia-gram is for the volatilization of a compound from water. The concentration profile shows adiscontinuity at the interface, whereas the fugacity profile does not. The overall resistance tomass transfer is composed of four individual resistances to diffusion of chemical from waterto air. For more details, see also Section 6.1.4.1.

where kw and ka are the individual film mass transfer coefficients on the water andair side, respectively. Each term in the above equation corresponds to a resistance tomass transfer. 1/Kw is the total resistance, whereas 1/kw is the resistance to diffusionthrough the water film (δw). 1/kaKaw is the resistance provided by the air film (δa).

EXAMPLE 4.2 OVERALLAND INDIVIDUAL MASS TRANSFER COEFFICIENTS

For evaporation of benzene from water, kw = 7.5 × 10−5 m/s and ka = 6 × 10−3 m/s.Estimate the overall mass transfer coefficient Kw. What is the percent resistance in theair side of the interface?At 298K, Kaw = 0.23 (from Appendix 1). Hence 1/Kw = 1/kw + 1/(kaKaw) =

1.4 × 104 and Kw = 7.1 × 10−5 m/s. Percent resistance in the air film =(Kw/kaKaw) × 100 = 5.1%.Benzene volatilization is thereforewater phase controlled.Note that if Kaw is small, percent air-side resistance will increase.

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Quantitative determination of the mass transfer rates thus requires estimationof the equilibrium partition constant between air and water phases. Note thatit appears both in the driving force and in the overall mass transfer coefficientexpressions.

Note from Figure 4.3 that there exists a discontinuity in concentrations at the inter-face. However, fugacity changes gradually without discontinuity. Since the fugacityin air is different from that in water, a concentration difference between the two phasesexists and the system moves toward equilibrium, where f w

i = f Γi = f ai . From the dis-

cussion in Chapter 3 on fugacity capacity, it can be shown (Mackay, 1991) that themass transfer flux can also be written in terms of fugacity as

Ni = Kw,f(f wi − f a

i

),

where Ni is in mol/m2/s as defined earlier, and Kw,f is given in terms of fugacitycapacities (Zw = 1/K ′

aw and Za = 1/RT) as

1

Kw,f= 1

kwZw+ 1

kaZa. (4.6)

Note that Kw,f = Kw/K ′aw with Kw as in Equation 4.5. The definition of K ′

aw is givenin Chapter 3. The net flux is identified as the algebraic sum of the net volatilizationfromwater,Kw,f f w

i , and the net absorption rate from air,Kw,f f ai .When these two rates

are equal the system is at equilibrium. The term f wi − f a

i is appropriately called thedeparture from equilibrium.

A variety of methods can be employed to obtain Henry’s constants for both organicand inorganic compounds in air–water systems. The following are the descriptions ofsome of the more common methods:

4.1.1 ESTIMATION OF HENRY’S CONSTANT FROM GROUP CONTRIBUTIONS

Hine and Mookerjee (1975) suggested one method of estimating the air–water par-tition constant by summing bond or group contributions in a molecule. Meylan andHoward (1992) recently refined the method. The method is analogous to the onediscussed in Chapter 3 for the determination of log Kow. It relies on the assumptionthat the free energy of transfer of a molecule between air and water is an additivefunction of the various groups or bonds present in the molecule. The correlation wasdeveloped to obtain the reciprocal of the air–water partition constant defined in thisbook. Thus the method gives 1/Kaw. An extensive listing of the bond and group con-tributions is given in Hine and Mookerjee (1975). It is best to illustrate this methodthrough examples.

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EXAMPLE 4.3 Kaw FROMA BOND CONTRIBUTION SCHEME

Given the following bond contributions to log K ′aw, calculate Kaw for (a) benzene, (b)

hexachlorobenzene, and (c) chloroform.

Bond log K ′aw

H–Car −0.154Car–Car 0.264H–Cal −0.119Cl–Car −0.024Cl–Cal 0.333

The symbols above are self-explanatory. The subscript ar denotes aromatic while alrefers to aliphatic.

(a) In benzene there are six Car–Car bonds and six H–Car bonds. Thus −logKaw= 6(−0.154) + 6(0.264) = 0.66. Hence Kaw = 0.219. The experimental value is0.231.

(b) In hexachlorobenzene there are six Car–Car bonds and six Cl–Car bonds.Hence−logKaw = 6(0.264) + 6(−0.024) = 1.46.Kaw = 0.038.The experimen-tal value is 0.002.

(c) In chloroform we have one H–Cal and three Cl–Cal bonds. Hence −logKaw =−0.119 + 3(0.333) = 0.87. Kaw = 0.134. The experimental value is 0.175.

The bond contribution scheme suffers from a major disadvantage in that interactionsbetween bonds are usually not considered. This will be a serious drawback when polarbonds are involved.

EXAMPLE 4.4 Kaw FROMA GROUP CONTRIBUTION SCHEME

Estimate the Henry’s constant for benzene and hexachlorobenzene using the followinggroup contributions:

Group −log Kaw

Car–H(Car)2 0.11Car–(C)–(Car)2 0.70Car–Cl(Car)2 0.18

For benzene we have six Car–H(Car)2 groups. Hence −logKaw = 6(0.11) = 0.66.Kaw = 0.218. The experimental value is 0.231. The prediction is only marginally betterthan the bond contribution scheme.

For hexachlorobenzene we have six Car–Cl(Car)2. Hence −logKaw = 6(0.18) =1.08. Kaw = 0.083. The experimental value is 0.002. The agreement is poor and is nobetter than that for the bond contribution scheme.

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Since the group contribution scheme of Hine and Mookerjee is of limitedapplicability for compounds with multiple polar groups, Meyland and Howard (1992)proposed an alternative, but more reliable group contribution scheme whereby

logKaw =∑i

aiqi +∑j

bjQj. (4.7)

Appendix 6 lists qi and Qj values for several bonds.

4.1.2 EXPERIMENTAL DETERMINATION OF HENRY’S LAW CONSTANTS

In Chapter 3, we noted that

Kaw =(Vw

RT

)·(P∗i

x∗i

), (4.8)

which means that Kaw can be obtained from a ratio of the saturated vapor pressureand aqueous solubility of a sparingly soluble compound.

Based on the above premise, there are several methods for Kaw measurementsdescribed in the literature. The different methods fall generally into three categories:(1) from a ratio of vapor pressure and aqueous solubility that is independently mea-sured; (2) static methods; and (3) mechanical recirculation methods. In the firstcategory of methods, the errors in independent measurements of vapor pressure andaqueous solubility lead to additive errors in Kaw. In the second category of meth-ods, because of difficulties associated with the simultaneous precise measurementsof concentrations in both air and water, they are restricted to compounds with largevapor pressures and aqueous solubilities. Such methods are ideally suited for solu-ble solutes, such as CO2, SO2, and several volatile organics. The EPICS method ofGossett (1987) and the direct measurement technique of Leighton and Calo (1981)fall under this category. In reality, both categories 1 and 2 methods should be superiorto the third since they require the measurement in both phases which would allowmass balance closure if the initial mass of organic solute introduced into the systemis accurately known. The errors in these methods arise directly from the concentra-tion measurements. The third class of methods (e.g., batch air stripping, wetted wallcolumn, and fog chamber) suffers from a serious deficiency, and that is the necessityto ascertain the approach to equilibrium between phases.

Several investigators have compiled Henry’s constants for a variety of compoundsof environmental interest (e.g., Mackay and Shiu, 1981; Ashworth et al., 1988). Thevalues reported show considerable scatter for a single compound. Table 4.1 is anexample of the degree of agreement reported by different workers. It is reasonableto expect a great deal of scatter, especially for those compounds that have low vaporpressures and aqueous solubilities. The standard deviation in Hc is only 2% of themean for benzene, whereas it is 12% of the mean for chloroform and 17% of themean for carbon tetrachloride. The standard deviation is 67% of the mean for lindane,which has a low Hc value. For most environmental purposes, it has been suggestedthat a reasonable standard error in Hc is about 5–10% (Mackay and Shiu, 1981).

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TABLE 4.1Comparison of Kaw Values from the Literature

Technique Reference∗ Chloroform Carbon Tetrachloride Benzene Lindane

Equilibrium 1 0.175 1.229 0.22 –Batch stripping 2 0.141 1.258 0.231 –Multiple equilibration 3 0.125 0.975 – –Vapor pressure 4 0.153 0.807 0.222 1.33 × 10−2

and solubility 8 – – – 8.41 × 10−4 to2.18 × 10−3

Direct measurement 5 0.153 1.14 0.227 –Wetted wall 6 – – – 3.43 × 10−3

Fog chamber 7 – – – 3.54 × 10−3

Kaw (mean) 0.149 1.082 0.225 7.17 × 10−3

Standard deviation 0.018 0.189 0.005 4.82 × 10−3

∗(1) Ashworth et al. (1988); (2) Warner et al. (1987); (3) Munz and Roberts (1986); (4) Mackay andShiu (1981) and Suntio et al. (1987); (5) Leighton and Calo (1978); (6) Fendinger and Glotfelty (1988);(7) Fendinger et al. (1989).

4.1.3 EFFECTS OF ENVIRONMENTAL VARIABLES ON Kaw

The air–water partition constants estimated at room temperature using binary aqueoussystems (solute +water) are not particularly applicable to natural systems (wastewater,atmospheric moisture, seawater, etc.). Several environmental variables affect Kaw indifferent ways. In this section, we briefly review four important variables and theireffects on air–water partition constants. These are (i) temperature, (ii) presence ofother miscible solvents, (iii) presence of colloids, and (iv) variations in pH.

As noted above, since Kaw is a ratio of P∗i to x∗

i , the effects of temperature onKaw are directly related to the variations in vapor pressure and aqueous solubilitywith temperature. As per the Clausius–Clapeyron equation, both P∗

i and x∗i show

logarithmic dependence on temperature. Hence, the variation inKaw with temperatureis given by the equation

lnKaw(T) = A− B

T, (4.9)

where A and B are constants for a given compound. The values for A and B for avariety of compounds have been tabulated (Table 4.2). In reality, the above equationis a method whereby single values of air–water partition constants at a standardtemperature T0 are extrapolated over a modest range using the solute enthalpy ofsolution. A thermodynamics expression associated with van’t Hoff relates the air–water partition constants at two temperatures (Smith and Harvey, 2007). The implicitassumption is that the enthalpy of solution is constant over the temperature range andthis is valid only over a narrow range of temperature (≈20K).

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TABLE 4.2Temperature Dependence of Kaw

Kaw = 1RT exp

(A − B

T

)

Compound Temperature Range (◦C) A B

Benzene 10–30 5.534 3194Toluene 10–30 5.133 3024Ethylbenzene 10–30 11.92 4994o-Xylene 10–30 5.541 3220Chloroform 10–30 11.41 50301,1,1-Trichloroethane 10–30 7.351 3399Trichloroethylene 10–30 7.845 3702

Kaw = a + bt + ct2

Compound Temperature Range (◦C) a b c

Benzene 0–50 0.0763 0.00211 0.000162Toluene 0–50 0.115 −0.00474 0.000466Ethylbenzene 5–45 0.05 0.00487 0.000250o-Xylene 15–45 0.0353 0.00444 0.000131Chloroform 0–60 0.0394 0.00486 –1,1,1-Trichloroethane 0–35 0.204 0.0182 0.000173Trichloroethylene 5–45 0.151 −0.00597 0.000680

Source: FromAshworth, R.A., Howe, G.B., Mullins, M.E., and Rogers, T.N. 1988. Journal of HazardousMaterials 18, 25–36.

Note: R = 8.205 × 10−5 atmm3/Kmol, T is in K, and t is in ◦C.

Over a wide range of temperature, Kaw is correlated to T using the followingpolynomial, Kaw(T) = a+ bT + cT2, where a, b, and c for several compounds arelisted in Table 4.2.

EXAMPLE 4.5 Kaw AND TEMPERATURE

Determine Kaw for benzene at 30◦C using Table 4.2.Using Kaw = (1/(8.205 × 10−5 × 303) ∗ exp(5.534 − (3194/303)) = 0.267Using Kaw = 0.0763 + 0.00211 ∗ 30 + 0.000162 ∗ 900 = 0.285.

Wastewaters and contaminated waters have varied composition and often containother organic solvents that are miscible with water. These solvent+water mixturesaffect the overall solubility of organic solutes thus varying x∗

i in the equation for Kaw.There is evidence in the literature that the effects do not become predominant unlesssignificant concentrations (>10wt%) of a very soluble co-solvent are present in natu-ral systems.An important paper in the literature in this regard is the one by Munz andRoberts (1987), where they evaluated the Henry’s constants of three chloromethanes(chloroform, carbon tetrachloride, and hexachloroethane) often encountered inwastewaters and atmosphericwater.Munz andRoberts (1987) observed no significant

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TABLE 4.3Measured Air–Water Partition Constants for Selected Compoundsin Natural Waters

Compound Matrix Composition Kaw Reference

Toluene Distilled water 0.244 Yurteri et al. (1987)Tap water 0.231“Creek” water 0.251

Chloroform Distilled water 0.125 Nicholson et al. (1984)Natural lake water 0.121

CH3CCl3 Distilled water 0.53 Hunter-Smith et al. (1983)Seawater 0.94

CHClBr2 Distilled water 0.036 Nicholson et al. (1984)Natural lake water 0.034

CHBr3 Distilled water 0.018 Nicholson et al. (1984)Natural lake water 0.017

Hexachlorobenzene Distilled water 0.054 Brownawell (1986)Seawater 0.070

2,4,4′-Trichlorobiphenyl Distilled water 0.00595 Brownawell (1986)Seawater 0.00885

effects on Henry’s constants in the presence of co-solvents such as methanol and iso-propanol up to concentrations as large as 10 g/L (i.e., co-solvent mole fraction of0.5 × 10−3). It was observed that the effects were pronounced for very hydrophobicsolutes.A compound that interacts better with the co-solvent exhibits the effect at lowco-solvent concentrations than does a compound that is more hydrophilic. In gen-eral, for any hydrophobic solute, the presence of a co-solvent increases the aqueoussolubility and lowers Kaw. These conclusions have been substantiated by others whodetermined the Henry’s constants of volatile solutes such as toluene and chloroformin distilled, tap, and natural lake waters. Table 4.3 is a summary of these observations.

The effects of inorganic salts on air–water partition constants are manifest throughtheir effects on the infinite dilution activity coefficients that are described in Chap-ter 3. Most hydrophobic organic compounds show the salting-out behavior, wherebythe mole fraction solubility in water is decreased (or activity coefficient is increased).Consequently, the air–water partition constant decreases in the presence of salts. Anumber of investigators have found that, for example, the partition constant for chlori-natedVOCs such as chlorofluorocarbons (CFCs) and chlorobiphenyls in seawater are20–80% greater than the values determined in distilled or deionized water (Table 4.3).

EXAMPLE 4.6 Kaw AND IONIC STRENGTH

Determine Kaw for phenanthrene in an industrial wastewater with 1M total ionconcentration at 292K.

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In water, without added salts, Kaw = Cia/Ciw. Cia is not affected by ionic strength inwater, but Cw

i is modified using the McDevit–Long theory (Section 3.2.4). Let Cwwi

represent the new wastewater concentration. Then newKaw = Cia/Cwwi . ByMcDevit–

Long theory, γwwi = γw

i exp(φVHCs). Since γwi 4 1/xw

i 4 1/Ciw, we can write Ciw =Cwwi exp(φVHCs). Therefore Kaw = K0

aw exp(φVHCs). From Table 3.3, for phenan-threne, φVH = (0.00213)(182) = 0.38. From Appendix 1, K0

aw = 3.16 kPa dm3/mol.Hence Kaw = 3.16 exp(0.38 × 1) = 4.62.

Colloids and particulates are ubiquitous in both natural waters and wastewaters.They are characterized by their size, which can range from a few nanometers tothousands of nanometers. Examples of colloidal material encountered in natural andwastewaters are given in Table 4.4. Organic colloids are characterized by sizes 103 nmor less. They are dispersed phases composed of high molecular weight (>1000)macromolecules of plant origin. They are composed of C, H, and O with traces ofN and S. They possess ionizable groups (OH or COOH) and are macro-ions. Theseare classified as dissolved organic compounds (DOCs) and range in concentrationsfrom a few mg/L in oceans to as large as 200mg/L in peaty catchments or swamps,and are known to give a distinct brownish tinge to water. DOCs are also observedat concentrations ranging from 10 to 200mg/L in atmospheric water (fogwater andrainwater). DOCs are known to complex with inorganic metal ions and bind organicpollutants. The concentration of metals and organic compounds bound to a singlemacromolecule can be large (Wijayaratne and Means, 1984). The effective solubil-ities of both inorganic and organic species can, therefore, be several times larger inthe presence of DOCs.

Hydrophobic pollutants will preferentially associate with DOCs since they pro-vide an organic medium shielding the pollutants from interactions with water. Ithas been observed that the association of organic compounds is correlated to theirhydrophobicity (Kow). The more hydrophobic the compound, the greater its sorp-tion to DOCs. Chiou et al. (1986, 1987) determined the effect of natural humic

TABLE 4.4Sizes of Organic and Inorganic Particulatesin Natural and Wastewaters

Type of Particle Size (nm)

Organic macromolecules (humics) 1–10Virus 10–100Oxides (iron and aluminum) 10–1000Clays 10–1000Bacteria 103–104

Soil particles 103–106

Calcium carbonate, silica 104–106

Note: 1 nm = 10−3 μm.

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acids (extracted from soil) on the aqueous solubility of several chlorinated organics(1,2,4,-trichlorobenzene, lindane, and several polychlorinated biphenyls). Haas andKaplan (1985) reported a similar study on the solubility of toluene in water containinghumic acids. Carter and Suffett (1982) determined the effect of humic acids on theaqueous solubility of several polyaromatic hydrocarbons. Boehm and Quinn (1973)reported the enhanced solubility of several hydrocarbons in seawater brought about bythe presence of DOCs. The effects were observed to be specific to the type of DOCs.Commercial humic acids behaved differently from marine or soil humic acids.

In general, a larger solubility was always noted in the presence of DOCs in water.Chiou et al. (1986) showed that ifC∗

i represents the solubility of the organic compoundin pure water and Capp

i is the apparent solubility in DOC-rich water, then

Cappi

C∗i

= 1 + CCKcw, (4.10)

where CC is the DOC concentration (g/mL) in water and Kcw is the equilibriumpartition constant for the organic solute between DOC and water (expressed in mL/g).The values of Kcw for several compounds are given in Table 4.5.

SinceKaw is inversely proportional to the aqueous solubility, the air–water partitionconstant in the presence of DOCs will be given by

K∗aw = Kaw

1 + CCKcw. (4.11)

It can therefore be seen that the effects on air–water partition constants for a compoundof relatively high aqueous solubility will be less than that for one with a very lowaqueous solubility.

Mackay et al. (1982) noted that when commercially available humic acid or fulvicacid was used in the aqueous solutions, up to concentrations of 54mg/L, the air–water partition constant for naphthalene showed a reduction of about 7 × 10−5 forevery 1mg/L of the colloid added. It was concluded that although natural organic

TABLE 4.5Partition Constants for Pollutants between DOC and Water

Compound Type of DOC Log KC Reference

p,p′-DDT Soil humic 5.06 Chiou et al. (1986)p,p′-DDT Aldrich humic acid 5.56 Chiou et al. (1987)1,2,3-Trichlorobenzene Soil humic 3.00 Chiou et al. (1987)Toluene Aldrich humic acid 3.3 Haas and Kaplan (1985)Lindane Soil humic 2.7 Chiou et al. (1986)Pyrene Soil humic 4.9–5.5 Gauthier et al. (1986)2,4,4′-PCB Soil humic 4.24 Chiou et al. (1987)Anthracene Soil humic 4.92 Carter and Suffett (1982)Fluoranthene Soil humic 5.32 Acha and Rehbun (1992)

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colloids reduce the value of air–water partition constant, it is probably negligiblefor most of the soluble organics at typical environmental concentrations of DOCs.Callaway et al. (1984) reported a reduction in Kaw for two low-molecular-weightorganic compounds, namely, chloroform and trichloroethene (TCE) in the presenceof humic acid. The experiments involvedmeasuring the concentrations of theVOCs inclosed vessels after equilibration of a known volume of the aqueous phase containinghumic acid (10wt%). It was then compared with a system without humic acid. Theratio K∗

aw : Kaw was 0.39 and 0.79 for chloroform, the different values correspondingto the pretreatment procedure used for preparing the humic acid solution. The ratiofor TCE was 0.28. Yurteri et al. (1987) observed that the ratio of air–water partitionconstants decreased to 0.94 for toluene in the presence of 10mg/L humic acid inwater, whereas it was 0.99 for 5mg/L humic acid in water. A significant observationwas that the combined effects of inorganic salts, surfactants, and humic acids on thepartition constant were far more significant than the effect of any single componentin wastewater samples.

Municipal wastewaters in many regions of the world contain high concentrationsof detergents (e.g., see Takada and Ishiwatari, 1987). Linear alkyl benzene sulfonates(LABs) and linear alkyl sulfates (LAS) are common constituents of synthetic deter-gents. These and other surfactants behave in a manner similar to the DOCs describedabove. They tend to increase the aqueous solubility of many organic and inorganicspecies.

Valsaraj and Thibodeaux (1988) studied the effects of surfactants (both anionicand cationic) on the air–water partition constants of low-molecular-weight hydro-carbons of environmental interest. The effects of two anionic surfactants (sodiumdodecylbenzene sulfate, DDS, and sodium dodecylbenzenesulfonate, DDBS) andone cationic surfactant (hexadecyltrimethyl ammonium bromide, HTAB) were stud-ied. The reduction in air–water partition constant from that in pure water wasdeduced. It was observed that the partition constant for chloroform decreased to0.77 times its value in pure water in the presence of 42mM of DDBS, and it fur-ther decreased to 0.66 times its pure water value in the presence of 82mM ofDDBS. There was no decrease till the critical micellar concentration (CMC) of DDBSwas exceeded. The results with DDS and HTAB micelles were similar to that withDDBS.

EXAMPLE 4.7 Kaw AND DOC

For anthracene, what will be the concentration in water in the presence of 1000mg/Lof a DOC?

From Table 4.5, Kcw for anthracene is 104.92 L/kg. Given CC is 1000mg/L =10−3 kg/L, the increase in aqueous-phase concentration is Capp

i /C∗i = 1 + CCKcw =

16.8. Since C∗i = 3.3 × 10−5 mol/L, Capp

i = 5.6 × 10−4 mol/L. Moreover, note thatsince Capp

i /C∗i = x

appi /x∗i = γ∗i /γapp

i , the activity coefficient is 16.8 times lower in thepresence of DOC.

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Many inorganic gases (e.g., carbon dioxide, sulfur dioxide, and ammonia) thatdissolve in water undergo chemical reactions depending on the pH of the solution.For example, SO2 dissolving in rainwater gives rise to sulfuric acid, which is the causeof acid rain. Many organic species also undergo reactions with water depending onthe pH. For example, phenol is converted into phenolate ion under alkaline conditions.

The ionization of water gives rise to hydrogen and hydroxide ions:

H2O � H+ + OH−. (4.12)

At equilibrium, in pure water, the concentrations of both ions are equal and have avalue 1 × 10−7 M. The equilibrium constant for the reaction is Kw = [H+][OH−] =1 × 10−14 M2. The pH of pure water is given by −log[H+] = 7.0. Pure rainwateris slightly acidic with a pH of 5.6. The dissolution of gases and organic compoundsthat ionize in solution can change the pH of water. This process can also change theequilibrium constant for dissolution of gases in water given by the Henry’s law.

Let us consider SO2 dissolution in water. The transfer of SO2 from gas to water isgiven by the following equilibrium reaction:

SO2(g) + H2O(l) � SO2 · H2O(l). (4.13)

The reaction equilibrium is discussed in Chapter 5 (see Section 5.6.2). For the abovereaction, the equilibrium constant is given by

Keq = [SO2 · H2O]l

PSO2 [H2O]l. (4.14)

Since the concentration of water is a constant, it can be incorporated into the equilib-rium constant. If we express the [SO2 · H2O]l in molar concentration, we have a newequilibrium constant,

K∗eq = [SO2 × H2O]l

PSO2

(4.15)

expressed in units of mol/L/atm. Note that this is the inverse of the conventionaldefinition of the air–water partition constant.

K ′aw = PSO2

[SO2 × H2O]l= 1

K∗eq

(4.16)

The complex species SO2 · H2O(l) can undergo subsequent ionization to producebisulfite, HSO−

3 and sulfite, SO2−3 ions according to the following reactions:

SO2 · H2O(l) � HSO−3 (l) + H+(l) (4.17)

and

HSO−3 (l) � SO2−

3 (l) + H+(l). (4.18)

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The respective equilibrium constants for the reactions can be written in terms ofHenry’s constants for SO2 as follows:

Ks1 =[HSO−

3

]l

[H+]

l

[SO2 · H2O]l= [HSO−

3

]l

[H+]

l · K′aw

PSO2

(4.19)

Ks2 =[SO2−

3

]l

[H+]2

l · K ′aw

Ks1PSO2

. (4.20)

The total concentration of SO2 that exists in the various forms [SO2.H2O]l, [HSO−3 ]l,

and [SO2−3 ]l is given by

[SO2]T = PSO2

K ′aw

·[1 + Ks1[

H+]l

+ Ks1Ks2[H+]2

l

]. (4.21)

The apparent air–water partition constant, K∗aw, defined by PSO2

/[SO2]T is given by

K∗aw = K ′

aw[1 + (Ks1

/ [H+]

l

)+(Ks1Ks2

/ [H+]2

l

)] . (4.22)

Thus as [H+]l increases (or pH decreases), more and more of SO2 appears in solutionin the form of bisulfite and sulfite ions. Therefore the apparent partition constantdecreases. For the SO2 dissolution, Ks1 = 0.0129M and Ks2 = 6.014 × 10−8 M(Seinfeld, 1986).

The dissolution of CO2 in water is similar to that of SO2 in that three differentreaction equilibria are possible. These and the respective equilibrium constants aregiven below:

CO2(g) + H2O(l) � CO2 · H2O(l); K ′aw = PCO2

[CO2 · H2O]l, (4.23)

CO2 · H2O(l) � H+(l) + HCO−3 (l); Kc1 =

[HCO−

3

]l

[H+]

l

[CO2 · H2O]l, (4.24)

HCO−3 (l) � H+(l) + CO2−

3 (l); Kc2 =[CO2−

3

]l

[H+]

l[HCO−

3

]l

. (4.25)

The apparent air–water partition constant is given by

K∗aw = K ′

aw[1 + (Kc1

/ [H+]

l

)+(Kc1Kc2

/ [H+]2

l

)] . (4.26)

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Just as for the SO2 example, increasing pH decreases the air–water partition con-stant for the CO2 case also. The values of the dissociation constants are Kc1 =4.28 × 10−7 M and Kc2 = 4.687 × 10−11 (Seinfeld and Pandis, 1998).

The effect of pH on the dissolution of ammonia in water is opposite to the ones forCO2 and SO2. NH3 dissolves in water to form ammonium hydroxide, which furtherdissociates to give NH+

4 and OH− ions. Thus, the solution is made more alkaline bythe presence of ammonia. As pH increases and the solution becomes more alkaline,more of ammonia remains as the gaseous species and hence its solubility in waterdecreases. Therefore the effective Henry’s constant will increase with increasing pH.The reactions of relevance here are the following:

NH3(g) + H2O(l) � NH3 · H2O(l); K ′aw = PNH3

[NH3 · H2O]l, (4.27)

NH3 · H2O(l) � NH+4 (l) + OH−(l); Ka1 =

[NH+

4

]l

[OH−]

l

[NH3 × H2O]l. (4.28)

Substituting for [OH−] = Kw/[H+] in the second of the above equations, one obtainsthe following expression for total ammonia,

[NH3]T = PNH3

K ′aw

·[1 + Ka1

Kw[H+]l

], (4.29)

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00 1 2 3 4 5

pH

K * aw

/K ′ aw

6 7 8 9 10

SO2

CO2

NH3

FIGURE 4.4 Effect of pH on the air–water partition constant of CO2, SO2, and NH3.

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and hence the apparent air–water partition constant is

K∗aw = K ′

aw[1 + (Ka1/Kw) [H+]l

] . (4.30)

From the above equation, it is clear that as [H+]l decreases (or the solution becomesmore alkaline), the apparent air–water partition constant increases. The value of Ka1is 1.709 × 10−5 M (Seinfeld and Pandis, 1998).

The effect of pH on the ratio of air–water partition constants for the three speciesCO2, SO2, and NH3 is shown in Figure 4.4.

EXAMPLE 4.8 CO2 EQUILIBRIUM BETWEEN AIR AND WATER IN A CLOSED

SYSTEM

Consider a 1mM solution of sodium bicarbonate in an aqueous phase in a closed vesselof total volume 1L at 298K. Let the ratio of gas to aqueous volume be 1:100. If thepH of the solution is 8.5, what is the equilibrium concentration of CO2 in the gasphase? To solve this problem, the primary relationship we need is the equilibriumbetween air and water for CO2 given by Henry’s law. Kaw = 31.6 atm/mol/L at 298K.Hence,

[CO2]g[CO2 · H2O]w = K ′

awRT

= 1.29. (4.31)

CO2 in the gas phase is generated from the sodium bicarbonate in solution and followsthe reaction equilibria mentioned earlier in this chapter. The total CO2 in the aqueousphase is given by

[CO2

]T,w = [CO2 · H2O

]w +

[HCO−

3

]w

+[CO2−

3

]w. (4.32)

Using the expressions for Kc1 and Kc2 given in the text, we can show that [CO2 ·H2O]w = α0 [CO2]T,w, where

α0 = 1

1 + (Kc1/ [

H+]l

)+(Kc1Kc2

/ [H+]2

l

) (4.33)

is the aqueous-phase mole fraction of [CO2 · H2O]w species. Similar expressions canbe derived for the other species [HCO−

3 ]w and [CO2−3 ]w also (see Stumm and Morgan,

1996). Since pH = 8.5, [H+] = 3.2 × 10−9 M. UsingKc1 = 4.3 × 10−7 M andKc2 =4.7 × 10−11 M, we then have αo = 7.3 × 10−3. Hence we have [CO2.H2O]w = 7.3 ×10−3 [CO2]T,w. A total mass balance requires that in the given closed system, the total

continued

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CO2 mass arising from the initial concentration of NaHCO3 should equal [CO2]T,w inthe aqueous phase at equilibrium plus the [CO2]g in the gas phase. Thus,

Vw(1 × 10−3) = Vw[CO2]T,w + VG[CO2]g, (4.34)

1 × 10−3 =[CO2 · H2O

]w

7.3 × 10−3+ 1

100· (1.29 × [CO2 · H2

]w

). (4.35)

Therefore [CO2 · H2O]w = 7.3 × 10−6 M and [CO2]g = 9.4 × 10−6 M.

EXAMPLE 4.9 BENZENE CONCENTRATION NEARA WASTEWATER LAGOON

The atmosphere above a wastewater lagoon contained 0.01μg/P of benzene. Theaverage surface water concentration in the lagoon was measured as 10μg/P. If thetemperature of air and surface water was 20◦C, in what direction is the flux of benzene?Cai = 0.01μg/L, Cw

i = 10μg/L. Kaw at 298K is 0.23 (fromAppendix 1). Using thetemperature variation for benzene (Table 4.2), we obtain at 293K, Kaw = 0.23. Thedriving force for flux is therefore (0.01 − 0.01 × 10−9/0.23) = 0.01 g/cm3. Since theflux is positive, we conclude that the impoundment acts as a source of benzene to the air.

4.2 AIR–WATER EQUILIBRIUM IN ATMOSPHERIC CHEMISTRY

Atmospheric moisture is present in the form of hydrometeors ( fog, rain, clouddroplets, aerosols, and hydrosols).An aerosol is a stable suspension of solid or liquidparticles or both in air, whereas a hydrosol is a stable suspension of particles in water.Aerosols provide the medium for a variety of atmospheric reactions. In the atmo-sphere, solar radiation is absorbed and scattered by both particles and gas moleculesthat comprise aerosols and hydrosols. Atmospheric aerosol mass is rarely lower than1μg/m3 close to the earth’s surface and occasionally increases to 100μg/m3 in urbanareas. The concentration of most trace gases in the atmosphere is in the vicinity of103–104 μg/m3. Thus, except for some rare cases (such as rural areas), the trace gasesin the atmosphere far exceed the aerosol concentrations. Aerosols, however, play alarger role in the hydrologic cycle by providing condensation nuclei for both fog andcloud formation. Sea spray from the ocean surface provides a rich source of liquiddroplets that upon evaporation of water form sea-salt crystals. Some of the liquiddroplets arise from condensation of organic vapors when the vapor pressure exceedsthe saturation point. An example of this type of atmospheric droplets is the aerosolgenerated from incomplete combustion of wood or agricultural residues.

The primary focus of interest for us in these cases is the equilibrium betweenthe dispersed phase and the continuous phase that contain trace gases (organic andinorganic).We wish to analyze the extent of removal of molecules from the gas phaseby various processes.

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TABLE 4.6Properties of Atmospheric Particles and Droplets

Liquid Water TypicalNature of Surface Area, Content, AtmosphericDroplet Size (μm) ST (m2/m3) L (m2/m3 ofair) Lifetime, τ

Aerosols 10−2–10 −1 × 10−3 10−11–10−10 4–7 daysFog droplets 1–10 −8 × 10−4 5 × 10−8–5 × 10−7 3 hCloud drops 10–102 −2 × 10−1 10−7–10−6 7 hRaindrops 102–103 −5 × 10−4 10−7–10−6 3–15minSnowflakes 103–105 −0.3 15–50min

Source: From Seinfeld, J.H. 1986. Atmospheric Chemistry and Physics of Air Pollution. NewYork:Wiley;Gill P.S., Graedel, T.E., and Wechsler, C.J. 1983. Reviews of Geophysics and Space Physics21, 903–920; and Graedel, T.E. and Crutzen, P.C. 1993. Atmospheric Change: An Earth SystemPerspective. NewYork: W.H. Freeman Co.

Table 4.6 lists the typical properties of atmospheric droplets and particles. Typicalaerosol and fog droplets are smaller than 10μm in diameter. Cloud drops, raindrops,and snowflakes have larger diameters. The surface areas of most particles are inthe range of 10−4–10−1 m2/m3, except for snowflakes that have large surface areas.Small particles (e.g., aerosols and fog) have relatively large atmospheric lifetimescompared with large particles. Large particles settle out faster through sedimentationand gravity settling. Small particles (colloids) are kept in suspension by frequentcollisions resulting from their Brownian motion. Large particles (raindrops, clouddrops, and snowflakes) have high liquid water content than aerosols and fog. Thewater content of aerosols depends largely on the relative humidity of air.

Many organic compounds (volatile or semi-volatile) as well as inorganic com-pounds are transported over long distances and appear even in the remote arctic regionsvia dispersion through the troposphere. Compounds volatilize from their sources inthe temperate and tropical regions and are transported through the atmosphere to theoceans and polar caps. Pollutants exist both in the gaseous form (≡G) and bound toparticulates (≡P). A large fraction of the material is transported in the gaseous form,which exchange directly with the earth and oceans via dry deposition; the flux beingJDry(G). Gaseous materials are also solubilized and absorbed by hydrometeors andscavenged to the earth. This is called wet deposition. Materials bound to aerosols inthe atmosphere are also scavenged by hydrometeors, which is also called wet depo-sition. The total flux due to wet deposition is JWet(G+P). The aerosols grow in sizeand settle to the earth by dry deposition and the flux is JDry(P). A schematic of theseexchange processes is given in Figure 4.5. The total flux is

JTotal = JWet + JDry = JWet(G+P) + JDry(P) + JDry(G). (4.36)

4.2.1 WET DEPOSITION OF VAPOR SPECIES

The concentration of a gaseous species i, scavenged by a hydrometeor at equilibrium,is given by the ratio of concentration per m3 of water to the concentration per m3 of

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Atmospheric pollutants

Gaseous Particle-bound

Direct depositionto surfaces

Washout with hydrometeors

Direct particledeposition to surfacesJdry(P)Jdry(G) Jwet(G+P)

FIGURE 4.5 Mechanisms of deposition of pollutants from the atmosphere.

air. This is called the washout ratio, Wg, from the gas phase. It is easy to recognizethat for equilibrium scavenging Wg = 1/Kaw.

For closed systems the total concentration of a species is fixed. If CiT representsthe total concentration of species i per unit volume of air, it is distributed between theair and water content θL (g/m3), and hence

CiT = Cia + θL · Ciw. (4.37)

Applying Henry’s law to the aqueous concentration, Ciw = Cia/Kaw, we obtain thefollowing equation for the fraction of species i in the aqueous phase:

φiw = θL

Kaw + θL. (4.38)

For compounds with lowHenry’s constant,Kaw, a significant fraction of species iwillexist in the aqueous phase. With increasing θL, a greater amount of water becomesavailable to solubilize species i and hence the value ofφw

i increases. The concentrationof species i in the aqueous phase is then given by

Ciw = CiTθL + Kaw

. (4.39)

4.2.2 WET DEPOSITION OF AEROSOL-BOUND FRACTION

Compounds that exist in the gaseous form in air are in equilibrium with those that areloosely bound to atmospheric aerosols; this is called the exchangeable fraction. Theremaining fraction is incorporated within the particle matrix and is tightly bound ornonexchangeable and does not participate in the equilibrium with the gas phase. Theextent to which the exchangeable molecules are associated with the aerosol is directlyrelated to the partial pressure of the compound. Junge (1977) provided an equationrelating the adsorbed fraction (φiP) to the total surface area of the aerosol (ST) andthe vapor pressure of the compound (P∗

i ).

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Assume that adsorption on aerosols is governed by a BET isotherm

Γi

Γmi= KBET

(1 − yi)[1 + (KBET − 1) yi

] . (4.40)

Since yi � 1,

Γi = Γmi KBETPiP∗i. (4.41)

Using the ideal gas law andMi the molar mass of compound i, we obtain

Γi = Γmi KBETCiaRT

MiP∗i. (4.42)

Consider a total suspended particle concentration in air, Cs; then the fraction of i inthe solids is

φiP = ΓiCs

ΓiCs + Cia. (4.43)

Now, Cs = Av/Am, where Av is the total surface area per unit volume of air and Amis the surface area of one particle. Hence,

φiP = ρiAv

ρiAv + P∗i, (4.44)

where ρi is given by AιKBETRT/MιAμ. It is characteristic of the compound anddepends on the compound molar mass and the surface concentration for monolayercoverage of the compound (Pankow, 1987). P∗

i is the subcooled liquid vapor pressurefor compounds that are solids at room temperature. Compounds that have large vaporpressures tend to have small values ofφp

i ,whereas compounds of small vapor pressuresare predominantly associated with the aerosols. φp

i is small for P∗i > 10−6 mmHg.

In a clean atmosphere, therefore, many of the hydrophobic compounds such as PCBcongeners, DDT, and Hg should exist in the vapor phase. A range of vapor pressuresfrom 10−4 to 10−8 mmHg are observed for environmentally significant compounds,and those with P∗

i > 10−4 mmHgmust be predominantly in the vapor phase whereasthose with P∗

i < 10−8 mmHg must exist almost entirely in the particulate phase. Thefraction occurring in the aerosol phase generally increases with increasing molecularweight and decreasing P∗

i for a homologous series of compounds (e.g., n-alkanes).For a compound that is removed by both gas and aerosol scavenging by rain, cloud,

or fog droplets, the following is the overall expression for washout ratio:

WT = Wg (1 − φiP) +WPφiP. (4.45)

WP is the aerosol washout ratio, which is a function of meteorological factors andparticle size.A raindrop reaches equilibrium with the surrounding atmosphere withina 10-m fall, and hence the washout of vapors may be viewed as an equilibrium

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partitioning process. Because of their small sizes and long atmospheric residencetimes, fog droplets can also be considered to attain equilibrium with the atmosphererather quickly.

Using the above equation for washout ratio, the overall flux of pollutants to thesurface (in mol/m2/s) by wet deposition can be obtained as

Jwet(G+P) = RI ·WT · Cia, (4.46)

where RI is the rainfall intensity (cm/h) and Cia is the measured concentration of iin air.

If Kaw is large, then the vapor uptake into droplets is negligible and only theaerosol particulate fraction is removed by wet deposition. A number of environ-mentally significant hydrophobic organic compounds fall in this category. Examplesare n-alkanes, polychlorinated biphenyls, chlorinated pesticides, and polyaromatichydrocarbons. Compounds with small Kaw values are removed mostly via gas scav-enging by droplets. Examples of this category include phenols, low-molecular-weightchlorobenzenes, and phthalate esters. A number of investigators have reported exper-imentally determined values of bothWg andWp for a variety of metals and organics.The range of values observed for typical pollutants in some regions of the world issummarized in Table 4.7. Aerosols have overall washout ratios approximately 106.Aerosols with washout ratios less than 105 are generally considered to be insolubleand relatively young in that their dimensions are of the order of 0.1μm. It has been

TABLE 4.7Range of Washout Ratios for Typical Pollutants

Compound Wg Wp

n-Alkanes 1.3–2.2 × 104 Portland, OR3.5–5.8 × 105 College Station, TX4.5–1.6 × 106 Norfolk, VA

PAHs 1.9 × 102–1.8 × 104 2 × 103–1.1 × 104 Portland, OR1.4–2.5 × 105 Isle Royale

Phenanthrene 64 ± 46 1.1 × 106 Chespeake Bay, MDPyrene 390 ± 280 8.5 × 105 AMetals 2.0 × 105–1 × 106 Ontario, Canada

1 × 105–5 × 105 Northeast U.S.Mercury, Hg0 10–100Radionuclides 2 × 105–2 × 106

PCDDs/PSDFs 1 × 104–1 × 105

PCBs 1 × 102–2 × 106 (rain)5 × 104–4 × 106 (snow)

Sulfate, SO2−4 6 × 104–5 × 106

Source: From Bidleman, T.F. 1998. Environmental Science and Technology 22, 361–367; Ligocki,M.P., Leuenberger, C., and Pankow, J.F. 1985. Atmospheric Environment 19, 1609–1617;Baker, J.T. 1997. Atmospheric Deposition of Contaminate to the Great Lakes and CoastedWaters. Pensacola, FL: SETAC Press.

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shown that for most nonreactive organic compounds such as PAHs in Table 4.7, thevapor scavenging washout ratios indicated that equilibrium was attained between thefalling raindrops and the gas phase within about a 10-m fall through the atmosphere(Ligocki, Leuenberger, and Pankow, 1985). Hence, these values can be satisfactorilypredicted from the temperature-corrected Henry’s constants for the compounds. Theaerosol scavenging washout ratios of organic compounds were observed to be similarto the washout ratios of fine-particle elements such as Pb, Zn, As, and V.

EXAMPLE 4.10 A TYPICAL CALCULATION OF WASHOUT RATIOS

Given typical values of particle washout ratios, Wp = 2 × 105, aerosol surface area,Av = 3.5 × 10−6 cm−1 of air and the constantρi = 1.7 × 10−4 atm cm.Let us estimatethe total washout ratios and the relative contributions of gas and aerosol scavenging forbenzene, benzo-[a]-pyrene and γ-hexachlorocyclohexane (HCH). Let us also assumethat ρi is approximately a constant between these compounds.The necessary properties are given below.

Compound P∗i (atm) Kaw (−)

Benzene 0.125 0.228γ-HCH 3.1 × 10−7 2.0 × 10−5

Benzo-[a]-pyrene 1.2 × 10−10 2.3 × 10−6

The calculated values of φPi and washout ratios are as follows.

Compound φpi Wg(1 − φ

pi ) Wpφ

pi WT

Benzene 4.7 × 10−9 4 9 × 10−4 4γ-HCH 1.6 × 10−3 5 × 104 3 × 103 5.3 × 104

Benzo-[a]-pyrene 0.83 7 × 104 2 × 105 2.7 × 105

For benzene the deposition is wholly via gas scavenging. For γ-HCH, although thevapor scavenging dominates, there is a small contribution from the aerosol fraction.However, for benzo-[a]-pyrene, the contribution is almost completely from the particle(aerosol)-scavenging process.

Fog forms close to the ground, where most of the gases and aerosols are con-centrated. It is formed as a result of a decrease in air temperature and an increasein relative humidity, whereby water vapor from the atmosphere condenses on tinyaerosol particles. Fog droplets have diameters generally between 1 and 10μm. Sincea fog droplet is approximately 1/100th of a raindrop, it is more concentrated than rain.In addition to its importance as a site for chemical reactions, fog exerts a significantinfluence as a scavenger of both organic and inorganicmaterials from the near-surfaceatmosphere. Hence, it can have a significant effect on human health and vegetation.

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Since fog forms close to the earth’s surface, its dissipation leaves particles in the near-surface atmosphere that are highly concentrated with pollutant. This observation hassignificant implications for the long-range transport and deposition of environmentalpollutants (Munger et al., 1983). Knowledge of fogwater chemistry also has relevanceto the ambient acid formation and acidic precipitation problems. SO2 oxidation reac-tions increase the acidity of fog. Fog droplets have greater deposition rates than drygases and aerosols. Most of the liquid water content in fog is lost to the surface. Thus,in regions of heavy fog events, calculation of dry deposition and rain fluxes may notaccount for the actual deposition of pollutants. Historically, fog has been a cause forsevere human health effects such as those of the infamous London fog of 1952, whichcaused close to 10,000 deaths in 4 months.

EXAMPLE 4.11 TOTAL DEPOSITION OFA PESTICIDE BY FOG

The concentration of a pesticide (chlorpyrifos) in air from a foggy atmosphere in Parlier,CA, was 17.2 ng/m3 at 25◦C. The aerosol particle concentration in air was negligible.If the moisture deposition rate from fog is 0.2mm/h and it lasts for 4 h, how muchchlorpyrifos will be deposited over 1 acre of land surface covered by fog?The Henry’s constant for chlorpyrifos is Kaw = 1.7 × 10−4 at 25◦C. Cg

i =17.2 ng/m3. Particle deposition is zero; hence only gaseous species need be consid-ered. Wi = 1/Kaw. Flux to surface is Ji = RICia/Kaw = 2 × 10−12 g/cm2 h. Amountdeposited = (2 × 10−12) (4) (4.04 × 107) = 3.27 × 10−4 g.

Droplets larger than about 10μm contribute the bulk of liquid water in fogs.However, droplets smaller than this will be more concentrated in pollutants. Largeconcentrations of colloidal organic matter (between 10 and 200 mg/L) are sometimesobserved in fogwater. These colloids arise from dissolved humics and fulvics andare surface active (Capel, Leuenberger, and Giger, 1991). The fogwater collected inCalifornia was observed to contain a variety of inorganic ions (NO−

3 , SO2−4 , H+,

and NH+4 ), which were also the major components of aerosols collected from the

same region (Munger et al., 1983). This showed that highly concentrated fogwaterappeared to result from highly concentrated aerosols. Likewise, dissipation of highlyconcentrated fog resulted in highly reactive and concentrated aerosols. This points tothe fact that fog is formed via condensation on aerosol nuclei. The chemical compo-sition of fog is not only determined by the composition of their condensation nuclei(e.g., ammonium sulfate or trace metal-rich dust), but also by the gases absorbed intofog droplets. High concentrations of formaldehyde (−0.5mM) were also observedin California fog. Many other inorganic species such as Na+, K+, Ca2+, and Mg2+have also been identified in the fog collected from different regions of the world. Fogis an efficient scavenger of both gaseous and particulate forms of S(IV) and N(III).The high millimolar concentrations of metal ions in fogwater and the resultant metalion catalysis for the transformation of S(IV) to S(VI) in a foggy atmosphere are inter-esting research areas. A number of investigators have shown that fogwater contains

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high concentrations of polyaromatic hydrocarbons and chlorinated pesticides (Capelet al., 1991; Glotfelty, Seiber, and Liljedahl, 1987).

4.2.3 DRY DEPOSITION OF AEROSOL-BOUND POLLUTANTS

The transfer of aerosol-bound pollutants directly from air to water is called drydeposition. The deposition flux is parameterized as

JDry(P) = Vd · Cia, (4.47)

where Vd is the deposition velocity (cm/s). Typical deposition velocities are given inTable 4.8. Vd is composed of three resistance terms:

Vd = 1

ral+ 1

rbl+ 1

rs,

where ral is the aerodynamic layer resistance, rbl is the boundary layer resistance,and rs is the surface resistance. Vd is affected by environmental factors, such as typeof aerosol, particle size, wind velocity, surface roughness, atmospheric stability, andtemperature. An excellent summary is provided by Davidson and Wu (1992).

TABLE 4.8Typical Dry Deposition Velocities for Gasesand Aerosols

Compound Range of Vd (cm/s)

SO2(g) 0.3–1.6Nox(g) 0.01–0.5HCl(g) 0.6–0.8O3(g) 0.01–1.5

SO2−4 0.01–1.2

NH+4 0.05–2.0

Cl− 1–5Pb 0.1–1.0Hg0 0.02–0.11Ca, Mg, Fe, Mn 0.3–3.0Al, Mn, V 0.03–3.0PCB, DDT 0.19–1.0Fine particles (1μm) 0.1–1.2

Source: From Davidson, C.I. and Wu, Y.L. 1992. AcidPrecipitation—Sources, Deposition and Canopy Inter-action, Vol. 3, pp. 103–216. New York: Springer;Eisenreich, S.J., Looney, B.B., and Thornton, J.D.Environmental Science and Technology 5, 30–38, 1981.

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4.2.4 DRY DEPOSITION FLUX OF GASES FROM THE ATMOSPHERE

Gaseous molecules can exchange with the aqueous phase via dissolution and/orvolatilization. This dissolution is driven by the concentration gradient between thegas and water phases. At the beginning of this chapter (Section 4.1), the concept of atwo-film theory of mass transfer across the air–water interface was introduced. Theflux of solute i (mass transferred per unit area per unit time) across the interface isgiven by

JDry(G) = Kw ·(CiaKaw

− Ciw

), (4.48)

where the overall mass transfer coefficient Kw is given by

1

Kw= 1

kw+ 1

kaKaw. (4.49)

The equilibrium air–water partition constant appears in both the concentration drivingforce and mass transfer coefficient terms. The overall mass transfer coefficient iscomposed of two terms: an aqueous-phase coefficient kw and an air-phase coefficientka. The reciprocal of the transfer coefficient, kw, is the resistance to mass transfer inthe aqueous film, whereas the reciprocal of kaKaw is the mass transfer resistance inthe gaseous film. If 1/kw � 1/kaKaw, the solute transfer is said to be aqueous phasecontrolled and Kw ≈ kw. On the other hand, if 1/kw � 1/kaKaw, then the solutetransfer is said to be gas phase controlled and Kw ≈ kaKaw. A number of gaseouscompounds of interest (e.g., oxygen, CFCs, methane, and carbon monoxide) are saidto be liquid phase controlled. So also are many of theVOCs (e.g., chloroform, carbontetrachloride, perchloroethylene, and 1,2-dichloroethane). Some highly chlorinatedcompounds (e.g., diledrin, pentachlorophenol, and lindane) are predominantly gasphase controlled. A general rule of thumb is that compounds with Kaw ≥ 0.2 areliquid (aqueous) phase controlled, whereas those with Kaw ≤ 2 × 10−4 are gas phasecontrolled. Table 4.9 lists the values of liquid- and gas-phasemass transfer coefficientsfor some compounds of environmental significance. For water crossing the interfacefrom the bulk phase into air, there exists no resistance in the liquid phase and hence it iscompletely gas phase controlled.Amean value of ka for water was estimated as 8.3 ×10−3 m/s (Liss and Slater, 1974). For all other gas-phase-controlled chemicals, onecan obtain ka values bymultiplying the above value for water with the ratio of the gas-phase diffusivity of water and the specific gas. Generally, the mass transfer coefficientis a function of the wind velocity over the water body (atmospheric turbulence). Inthe case of those species that dissociate in aqueous solution, associate with or reactwith other species (such as colloids, ions, and othermacromolecules), amore complexdependence of the transfer rate on the aqueous chemistry has been observed. For thesecompounds only the truly dissolved, unassociated fraction in water will equilibratewith the vapor in air.Therefore, the effects of the various parameters on the equilibriumpartitioning (Henry’s law) for compounds that we studied in Section 4.1.1 are of directrelevance in this context.

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TABLE 4.9Air–Water Partition Constants and Air–Water Mass Transfer Coefficientsfor Selected Pollutants in the Natural Environment

Compound Kaw (−) kw (m/s) ka (m/s)

Inorganic

Sulfur dioxide 0.02 9.6 × 10−2 4.4 × 10−3

Carbon monoxide 50 5.5 × 10−5 8.8 × 10−3

Nitrous oxide 1.6 5.5 × 10−5 5.3 × 10−3

Ozone 3 4.9 × 10−4 5.1 × 10−3

Ammonia 7.3 × 10−4 6.7 × 10−5 8.5 × 10−3

Oxygen 25.6 5.5 × 10−5 –

Water – – 8.4 × 10−3

Organic

Methane 42 5.5 × 10−5 8.8 × 10−3

Trichlorofluoromethane 5 3.1 × 10−5 3.0 × 10−3

Chloroform 0.183 4.2 × 10−5 3.2 × 10−3

Carbon tetrachloride 0.96 5.2 × 10−5 4.1 × 10−3

Benzene 0.23 7.5 × 10−5 6.0 × 10−3

Dichlorobenzene 0.069 3.5 × 10−5 4.5 × 10−3

Lindane 2 × 10−5 3.8 × 10−5 3.1 × 10−3

DDT 1.6 × 10−3 3.6 × 10−5 2.8 × 10−3

Naphthalene 0.021 3.6 × 10−5 4.6 × 10−3

Phenanthrene 1.6 × 10−3 5.0 × 10−5 4.0 × 10−3

Pentachlorophenol 1.1 × 10−5 3.3 × 10−5 3.2 × 10−3

PCB (Aroclor 1254) 4.4 × 10−4 1.7 × 10−2 1.9

Source: From Warneck, P. 1986. Chemistry of the Natural Atmosphere. New York: Academic Press;Eisenreich, S.J., Looney, B.B., and Thornton, J.D. 1981. Environmental Science and Tech-nology 5, 30–38; Mackay, D. and Leinonen, P.J. 1975. Environmental Science Technology 9,1178–1180; Liss, P.S., and Slater, P.G. 1974. Nature, 247, 181–184.

Note: All values are at 298K.

EXAMPLE 4.12 FLUX OF SO2 IN THE ENVIRONMENT

The mean concentration of SO2 in the atmosphere has been determined to be approx-imately 3μg/m−3. SO2 is fairly rapidly oxidized in the slightly alkaline environmentof seawater, and hence its concentration in surface ocean waters is approximately zero.Using the data from Table 4.10 for kw, ka and the Henry’s law constant Kaw, we haveKw = {(1/kw) + (1/kaKaw)}−1 = 9.0 × 10−5 m/s. Since Ciw = 0, we have for theflux JDry(G) = 9.0 × 105−{3/0.02} = 0.013μg/m2/s. The flux is directed from the

atmosphere to the ocean. If the total surface area of all oceans is 3.6 × 1014 m2, thetotal annual flux would be 1.5 × 1014 g, which is the same order of magnitude as theestimated emissions from fossil fuel burning (Liss and Slater, 1974).

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TABLE 4.10A Typical Louisiana Sediment/Soil Composition

Component/Property Sediment Soil

Bulk density (g/cm3) 1.54 1.5

Particle density (g/cm3) 2.63

Bed porosity (cm3/cm3) 0.59 0.4

Organic carbon, φoc (humics, fulvics, etc.) (%) 2.1 2.3

Sand (%) 27 57.8

Silt (%) 37 25.5

Clay (%) 34 16.7

Iron (g/kg) 13 23

Aluminum (g/kg) 12 25

Calcium (g/kg) 3 1

Magnesium (g/kg) 2 0.4

Manganese (mg/kg) 206 100

Phosphorus (mg/kg) 527 166

Zinc (mg/kg) 86 54

Sodium (mg/kg) 77 29

Lead (mg/kg) 56 35

Chromium (mg/kg) 20 23

Copper (mg/kg) 17 20

Nickel (mg/kg) 12 15

Note: Sediment sample obtained from BayouManchac, LA. Soil sample obtained fromNorth Baton Rouge, Louisiana.

EXAMPLE 4.13 FLUX OF Hg0 IN THE ENVIRONMENT

In the equatorial Pacific Ocean, Kim and Fitzgerald (1986) reported that the principalspecies of mercury in the atmosphere is elemental Hg0. It was shown that a significantfraction of the dissolved mercury in the surface water was elemental Hg0. The con-centration in the surface water, Ciw, was 30 fM (= 30 × 10−15 mol/L = 6 ng/m3). Themean atmospheric concentration was 1 ng/m3. The Henry’s constant at 25◦C for Hg0

is 0.37. The overall mass transfer coefficient Kw = 5.1m/day. The flux of Hg0 is there-fore given by JDry(G) = 5.1[1/0.37 − 6] = −16.8 ng/m2/day.The annual flux ofHg0 is

given by −16.8 × 365 = −6.1 × 103 ng/m2 = −6.1μg/m2. The flux is negative, andhence is directed from the ocean to the atmosphere. The total flux from 1.1 × 1013 m2

surface area of the equatorial Pacific Ocean is 0.67 × 108 g/year. The total flux of Hg0

to air from all regions of the world combined is 8 × 109 g/year. Hence, the flux fromthe equatorial Pacific Ocean is only 0.08% of the total.

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EXAMPLE 4.14 FLUX OF CO2 BETWEEN THE ATMOSPHEREAND SEAWATER

Broeker and Peng (1974) estimated that the mean Kw for CO2 is 11 cm/h in seawater at20◦C. This is representative of the world’s oceans. The maximum rate of CO2 transfercan be obtained by assuming thatCiw ≈ 0, that is, CO2 is rapidly assimilated in the sur-facewaters of the oceans. SincePi for CO2 is 0.003 atm, the value ofCia = 0.003/RT =1.25 × 10−7 mol/cm3. Kaw for CO2 is 1.29. Hence Cia/Kaw = 9.7 × 10−8 mol/cm3.The flux is therefore JDry(G) = 11(9.7 × 10−8) = −1 × 10−6 mol/cm2 h. The flux isfrom the atmosphere to seawater.

4.2.5 THERMODYNAMICS OF AQUEOUS DROPLETS IN THE ATMOSPHERE

In Section 3.3.4, it was noted that for pure liquids, the vapor pressure above a curvedair–water interface is larger than that over a flat interface. This was termed the Kelvineffect. The Kelvin equation representing this effect was derived as

ln

(P

P∗

)= 2σaw

r· Vw

RT, (4.50)

where r is the radius of the droplet, P and P∗ are the vapor pressures over the curvedinterface and the flat surface, respectively, and Vw is the molar volume of water.

Let us now consider the aqueous droplet to contain a nonvolatile species i withmolar volume Vi. If the number of moles of i is ni and that of water is nw, thetotal volume of the drop (4/3)πr3 = niVi + nwVw. The mole fraction of water in thedroplet is given by xw = nw/(ni + nw). Using these relations, one can write

xw = 1[1 + (niVw/(4/3)πr3 − niVi

)] . (4.51)

If the Raoult’s law convention is applied to the case of the flat air–water interface(refer to Section 3.2.3), the vapor pressure of water above the solution will be given by

Pw = γwxwP∗w, (4.52)

where P∗w is the vapor pressure of water over the flat interface. The Kelvin equation

now takes the form (Seinfeld and Pandis, 1998)

ln

(Pw

γwxwP∗w

)= 2σaw

r

Vw

RT. (4.53)

Substituting for xw, we can rearrange the above equation, and use the dilute solutiondefinition, that is, (4/3)πr3 � niVi. Furthermore, for a dilute solution accordingto the Raoult’s law convention, the activity coefficient γw = 1. Therefore, the finalequation is

ln

(Pw

P∗w

)= B1

r− B2r3

,

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where B1 = 2σawVw/RT and B2 = 3niVw/4π. Note that B1 (inμm)≈ 0.66/T and B2(in μm3) ≈ 3.44 × 1013νms/Ms, where ν is the number of ions resulting from solutedissociation, ms is the mass (g) per particle of solute, andMs is the molecular weight(g/mol) (Seinfeld and Pandis, 1998).

Note the difference between the equation for a pure water droplet and that for anaqueous solution droplet.Whereas for a pure water droplet there is a gradual approachof Pw → P∗

w as r is increased, in the case of an aqueous solution droplet Pw can eitherincrease or decrease with r depending on the magnitude of the second term, B2/r3,which results solely from the solute effects and is a function of its mole number,ms/Ms. When the two terms on the right-hand side become equal, Pw = P∗

w; theradius at which this is achieved is called the potential radius and is given by

r∗ =(3

8

niπ

RT

σaw

)1/2

. (4.54)

The maximum value of ln(Pw/P∗w) will be reached when the derivative with respect

to the radius goes to zero, which gives a critical radius

rc =√3B1B2

= √3r∗. (4.55)

A typical plot of Pw/P∗w for both pure water droplets and solution droplets (solute

being nonvolatile, e.g., an inorganic salt or an organic compound) is shown inFigure 4.6.We shall assume that the concentration of i is low enough that σaw remainsunaffected. If the compound is surface active, σaw will be lower and hence the Kelvinterm will be even smaller. This can be the case especially for many of the hydropho-bic compounds of environmental interest (Perona, 1992). Although electrolytes alsoaffect the surface tension ofwater, their influence ismostly<20%within the observedranges of concentrations.

For pure water droplets, the curve shows a steep decrease with increasing r. For thesolution droplet, there is an initial steeply rising portion till rc is reached. This resultsfrom the solute effects. Beyond this point, the Kelvin term dominates and the slowdecrease in Pw/P∗

w is apparent. Curves such as these are called Kohler curves. Theyare useful in estimating the size of an aqueous droplet given the relative humidity(Pw/P∗

w ) and the concentration of the solute.Consider a fixed ratio of Pw/P∗

w. For this fixed value there are two possible radii,one less than rc and the other greater than rc. For r < rc, the drop is at its equilibriumstate. If it adds anywater, its equilibriumvapor pressurewill be larger than the ambientvalue and it therefore quickly loses that water through evaporation and reverts to itsoriginal equilibrium. If r > rc, the drop will have a lower than ambient vapor pressureupon gaining more water molecules. Therefore, it will continue to accumulate waterand grow in size. If the value of Pw/P∗

w is greater than the maximum value, whateverbe the size of the drop, it always has lower than ambient vapor pressure and henceit continues to grow in size. A drop that has crossed this threshold is said to beactivated. With increasing radius of the drop, the height of the maximum decreases

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rrc

r*

(B1/r) – (B2/r 3)

For aqueous solution droplet

(2σawVw/r RT )Kelvin equation for pure water droplet

ln(P

w/P

w*)

FIGURE 4.6 Kohler curves for pure water droplet and aqueous solution droplet in air. Notethe differences in the shape of the curve when water contains a solute.

and hence the larger particles are activated preferentially. In fact, this is how fogand cloud droplets that are −10μm in diameter are formed from aerosols that areonly ≈0.1μm in diameter. In continental aerosols, the larger particles are generallyefficient condensation points for cloud droplets. The small (Aitken) particles havenegligible contribution toward cloud condensation.

For compounds that do not dissociate (e.g., neutral organics), the value of ni thatappears in the equation for B2 is straightforward to estimate. For dissociating species(e.g., inorganic salts), J2 should include both the dissociated and undissociated speciesin solution. If the degree of dissolution is vi (e.g., for HCl themaximumvalue is 2) andthe initial mass ismi, then ni = νimi/Mi, whereMi is the molecular weight. Generallydissociation leads to small values ofPw/P∗

w. If only a portion of the species i is solublein the aqueous phase (nsol

i ) and a portion is insoluble (ninsoli ), then particle growth will

be retarded somewhat sinceB2 is now defined as (3/4π)(nsoli + ninsol

i )Vw. This makesthe solute effect more pronounced than the previous case when all of i was soluble inthe aqueous phase (i.e., ni in the previous equation is equal to nsol

i ). The experimentalverification of these different cases is reported in the literature (Warneck, 1986).

EXAMPLE 4.15 VAPOR PRESSUREABOVE AQUEOUS DROPLETS

Determine the vapor pressure over (i) a 1μm aqueous droplet containing 5 × 10−13 gof NaCl per particle at 298K, and (ii) a 1μm of pure water droplet.

continued

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(i) For NaCl, v = 2, ms = 0.1, Ms = 58. B1 = 0.66/298 = 2.2 × 10−3 μm, B2 =3.44 × 1013 (2) (5 × 10−13)/58 = 0.55(μm)3. ln(Pw/P∗

w) = (2.2 × 10−3) −(0.55/13) = −0.55.HencePw/P∗

w = 0.57. SinceP∗w = 223.75,Pw = 13.7mmHg.

(ii) σa/w = 72 erg/cm2, Vw = 18 cm3/mol, r = 1 × 10−4 cm, R = 8.314 × 107 erg/

K mol, T = 298K. Hence ln(Pw/P∗w) = (2)(72)(18)/(1 × 10−4 × 8.314 ×

107 × 298) = 1.04 × 10−3. Hence Pw/P∗w = 1.001. Therefore Pw = P∗

w =23.75mmHg.

4.2.6 AIR/WATER EQUILIBRIUM IN WASTE TREATMENT SYSTEMS

The equations and methods of calculation of flux in the atmosphere, as described inSection 4.2.4, also apply to the estimation of air emissions from wastewater storageimpoundments, settling ponds, and treatment basins (Springer et al., 1986). Waste-waters are placed temporarily in open confined aboveground facilitiesmade of earthenmaterial. In some cases, the storage of wastewater in such impoundments is for thepurpose of settling of solids by gravity. In other cases, wastewater treatment usingmechanical agitation or submerged aerators is conducted in these impoundmentsfor the purposes of neutralization, chemical reaction, biochemical oxidation, andevaporation. All these activities cause a significant fraction of the species from waterto volatilize to the atmosphere. Natural forces in the environment can dominate thetransfer process. Examples of these are wind, temperature, humidity, solar radiation,ice cover, and so on at the particular site. All of these effects have to be properlyaccounted for if accurate air emissions of VOCs are to be made.

Consider a surface impoundment as shown in Figure 4.7. The natural variable thathas the most influence on air emissions is wind over the water body. Both kw and kadepend on wind velocity.An average overall mass transfer coefficient Kw over a 24-hperiod can be estimated. The flux of material from water will be given by

Ni = Kw ·(Ciw − Cia

Kaw

). (4.56)

If the two phases are at equilibrium, Ciw = Cia/Kaw, and the net overall flux iszero. In other words, at equilibrium the rates of both volatilization and absorptionare the same, but opposite in direction. The air movement over the water body due towind-induced friction on the surface is transferred to deeper water layers, thereby set-ting up local circulations in bulkwater and enhancing exchange between air andwater.Thus the wind-induced turbulence tends to delay approach to equilibrium betweenthe phases. The effects of wind are far greater on ka than on kw, and hence depend-ing on which one of these two controls mass transfer the overall effect on Kw willbe different.

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Wind

Turbulent mass transfer

Oxygenabsorption

VOCvolatilization

Surface aerator

Water

Air

Surface impoundment

FIGURE 4.7 Transport of a VOC from a surface impoundment. Note the two distinct areasof mass transfer, the turbulent zone where surface aerators enhance the volatilization, andthe quiescent area where natural processes drive the evaporation of the chemical from water.Surface winds carry the VOC away from the source. Simultaneous to volatilization of VOCsfrom water, absorption of oxygen from the atmosphere also occurs.

EXAMPLE 4.16 FLUX OF BENZENE FROMA TYPICAL SURFACE IMPOUNDMENT

A surface impoundment of total area 2 × 104 m2 is 5m deep and is well mixed by asubmerged aerator. The average concentration of benzene in water was 10μg/L, and thebackground air concentration was 0.1μg/L. If the induced mass transfer is describedby kw = 50m/h and ka = 2500m/h, estimate the mass of benzene emitted during an8-h period.

1/Kw = 1/kw + 1/kaKaw = 1/50 + 1/(0.23)(2500) = 1/46 h/m. Hence Kw =46m/h. Ni = 4600(10 × 10−9 − 0.1 × 10−9/0.23) = 4.4 × 10−5 g/cm2 h. Amountemitted in an 8-h period = 4.4 × 10−5 (104) (2 × 104)(8) = 70,400 g = 70.4 kg.The above calculation is for a given volume of water where there is no additional

input of benzene with time. This is called a batch process. In practice, most surfaceimpoundments receive a continuous recharge of material via a given volumetric inflowand outflow of wastewater. This is called a continuous process. We shall revisit theseaspects later in Chapter 6, where applications of kinetics in environmental reactors willbe discussed.

Another wastewater treatment process that involves air–water contact is flotation.Flotation is the process where tiny air bubbles are forced into a tall water column(Figure 4.2). Bubbles can be formed either by pressurizing or depressurizing thesolution (dissolved air flotation) or by introducing air under pressure through a smallorifice in the form of tiny bubbles (induced air flotation). The purpose of bubble

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aeration is to increase the dissolved oxygen level in water. Oxygen enhancement isessential for biological treatment of wastewaters and activated sludge and also for theprecipitation of iron and manganese from wastewaters. An illustrative example forthe case of absorption of oxygen into water is given below.

EXAMPLE 4.17 OXYGEN ABSORPTION INA WASTEWATER TANK

The oxygen levels in wastewater will be lower than its equilibrium value because ofhigh oxygen demands. Improved oxygen levels in water can be achieved by bubbling air(oxygen) through fine bubble porous diffusers (see Figure 4.2). This is called submergedaeration. Large aeration units can be assumed to be homogeneous in oxygen concen-tration due to the turbulent liquid mixing within the water. The oxygen absorption rate(air to water) is given by

Ni = (Kwa) ·(C

eqiw − Ciw

), (4.57)

where Ceqiw is the equilibrium saturation value of oxygen in water (mol/m3). This is

given by the air–water partition constant, Ceqiw = Pi/K ′

aw · Ciw is the concentration ofoxygen in bulk water (mol/m3). To account for other wastewater constituents (e.g.,DOCs, surfactants, and co-solvents), Ciw in wastewater can be related to the Ciw in tapwater by a factor β for which a value of 0.95 is recommended. Kw is the overall liquid-phase mass transfer coefficient for oxygen (m/s) and a is called the specific air–waterinterfacial area provided by the air bubbles. It has units of m2/m3 and is given by (6/d)εg, where d is the average bubble diameter at the sparger and εg is the void fraction.The void fraction is defined as the ratio of air volume to the total volume (air + water)at any instant in the aeration vessel. It is a difficult quantity to measure. Hence Kwais lumped together and values reported. KO2a is temperature dependent. Eckenfelder(1989) gives (Kwa) = (Kwa)20◦Cθ(t−20), where θ is a parameter between 1.015 and1.04 for most wastewater systems. Kwa is characteristic of both the specific physicaland chemical characteristics of thewastewater. It is generally estimated in the laboratoryby performing experiments on the specific wastewater sample (see Eckenfelder, 1989).

In a specific bubble aeration device, Kwa was determined to be 0.8 h−1 for awastewater sample at 298K. At 298K Pi in air is 0.2 atm. The air–water parti-tion constant for oxygen at 298K is 0.774 atm/m3/mol. The latter two parametersgive us equilibrium concentration of oxygen in water, Ceq

iw = 0.26mol/m−3(= 2.6 ×10−4 mol/L). If the wastewater was completely mixed (homogeneous) and if it hadzero DOC, then the maximum rate of oxygen absorption is Nmax

i = 0.208molm3/h.If the total volume of the wastewater is 28,000 L (= 1000 ft3), the oxygen absorbedwill be 5.8mol/h (= 93 g/h).

4.3 SOIL–WATER AND SEDIMENT–WATER EQUILIBRIUM

Contaminated sediments in the marine or freshwater environment exist in the UnitedStates and other parts of the world. A few examples in the United States are the

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Great Lakes and New Bedford Harbor contaminated with PCBs, Indiana Harborcontaminated with PAHs, and Hudson River sediment contaminated with metals andorganic compounds. Developing countries also suffer extensive contamination ofmajor rivers and tributaries. There are many areas in the United States and elsewhere,where extensive sediment contamination has led to an outright ban on fishing andrecreation. Some well-known examples of sediment contamination are the oil spill inPrince William Sound near Alaska from the Exxon-Valdez oil tanker disaster and theKuwait oil fires after the First Persian Gulf War.

Sediments generally accumulate contaminants over a period of time. Over timeas inputs cease, the previously contaminated sediments become sources of pollu-tants to the water column. Contaminants enter the marine food chain and furtherbioaccumulate in birds and mammals. The soil–water environment is another areaof significance. In the United States and other parts of the world, groundwater isa major source of clean water. Improperly buried waste (e.g., old landfills, leakingunderground petroleum storage tanks) releases contaminants that leak into the ground-water. Inadvertent solvent spills and intentional dumping also create major threats tothe limited clean water supplies in the world. In the groundwater environment, theequilibrium between the soil and porewater is the issue of concern.

4.3.1 PARTITIONING INTO SOILS AND SEDIMENTS FROM WATER

The soil and sediment environments can be characterized by a solid phase composedmainly of mineral matter (kaolinite, illite, and montmorillonite), organic macro-molecules (humic and fulvic acids of both plant and animal origin), and porewater.The mineral matter is composed of oxides of Al, Si, and Fe. Typical Louisiana soiland sediment have compositions given in Table 4.10. The sand, silt, and clay frac-tions predominate whereas the total organic carbon fraction is only a small fractionof the total.

A soil or sediment has micro-, meso-, and macro-pores that contain porewater.Hydrogen and hydroxyl ions in porewater control the pH. Different metal ions con-trol the ionic strength of the porewater. Both pH and ionic strength are important incontrolling the dissolution and precipitation reactions at the mineral–water inter-faces. The most important process as far as transport of organic and metal ionsin the sedimentary and subsurface soil environment is concerned is sorption. Bysorption we mean both adsorption on the surface and absorption (physical encapsu-lation) within the solids. Metal ions and organic molecules in porewater will establishsorption equilibria with the solid phase. The mineral matter in the soil developscoatings of organic matter, thereby providing both an inorganic mineral surface andan organic coated surface for sorption. Figure 4.8 is a schematic of the differentsorption sites.

The mineral surfaces (oxides of Al, Si, and Fe) are electrically charged andhave electrical double layers arising from the ions present in porewater. Waterstrongly interacts with the charged surfaces and adopts a three-dimensional hydrogen-bonded network near the surface in which some water molecules are in directcontact with the metal oxide and others attached via H-bonds to other molecules

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Soil mineral

Soil porewaterAdsorbedpollutant

Soil mineral

Soil mineral

Soil organic matter

FIGURE 4.8 Partitioning of a solute between soil and porewater. Various niches are shownwhere pollutants reside in the soil pore.

(Thiel, 1991). The strongly oriented water molecules at the surface have propertiesdifferent from that of bulk water. The near-surface water molecules are sometimesreferred to as vicinal water. Drost-Hansen (1965) has summarized the properties ofvicinal water.

Typically, a metal oxide such as SiO2 is strongly hydrated and possesses Si–OH(silanol) groups. The surface charge density in this case is dependent on the pH ofthe adjacent porewater. Metal ions can enter into exchange reactions with the silanolgroups forming surface complexes (Stumm, 1993).

The adsorption of compounds on soil or sediment surfaces occurs through a varietyof mechanisms, each of which can be assumed to act independently. The overalladsorption energy is given by

ΔGads =∑n

ΔGn, (4.58)

where eachΔGn represents the contribution from the specific nth mechanism. Thesemechanisms include (i) interactions with the electrical double layer, (ii) ion exchangeincluding protonation followed by ion exchange, (iii) coordination with surface metalcations, (iv) ion–dipole interactions, (v) hydrogen bonding, and (vi) hydrophobicinteractions. The first two mechanisms are of relevance only for the adsorption ofionizable species. Coordination with surface cations is important only for compounds(e.g., amines) that are excellent electron donors compared towater. Ion–dipole interac-tions between an uncharged molecule and a charged surface are negligible in aqueoussolution. Only compounds with greater hydrogen bonding potential than water canhave significant hydrogen bondingmechanisms.Most neutral nonpolar compounds ofhydrophobic character will interact through the last mechanism, namely, hydrophobicinteractions.

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4.3.2 ADSORPTION OF METAL IONS ON SOILS AND SEDIMENTS

Figure 4.9a represents a mineral surface (silica) where an electron-deficient siliconatom and groups of electron-rich oxygen atoms are presented to the aqueous phase.The surface hydroxyl groups are similar to water molecules in that both can formhydrogen bonds. The adsorption energy for water on a silica surface is −500mJ/m2.The energy decreases as the surfacewatermolecules are progressively replaced.Wateris therefore more favored by most mineral surfaces. There are two possible orienta-tions ofwatermolecules on amineral surface as depicted in Figure 4.9b.An interestingconsequence of these orientations is that water molecules in direct contact with themetal oxide can adopt specific, favorable conformations and can simultaneously par-ticipate in the tetrahedral three-dimensional network that gives it the special featuresnoted earlier.

The strongly polar mineral surface possesses a surface charge density and a doublelayer near it. The surface charge distribution can be obtained from the application ofthe Poisson–Boltzmann equation. It is given by

σe =(2εRTI

πF2

)1/2

sinh

(zFψ

2RT

), (4.59)

where z is the valence of ions in the electrolyte (e.g., for KCl, z = 1), ε is the dielectricconstant of water (= 7.2 × 10−10 C/V/m at 298K), I is the ionic strength (mol/m3),R is the gas constant (8.3 J/K/mol), F is the Faraday constant (96,485C/mol), and ψis the surface potential (J/C/V). The surface charge density is determined by the pH ofthe adjacent solution. Therefore, both H+ and OH− are called potential determiningions. Based on the reactivities of these ions with the surface hydroxyl groups, the

Si

Si

O

OH

HO

H

HSiO2

Water

Oxygen in the 2nd layer(Water)

(b)

(a)

Oxygen in 1st layer

(Metal oxide)

FIGURE 4.9 A typical mineral surface (silica) (a) where an electron-deficient silicon atomand groups of electron-rich oxygen atoms are presented to the aqueous phase (b).

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following two acid–base surface exchange reactions can be written as

≡M−OH+2 ≡M−OH + H+

≡M−OH ≡M−O_ + H+ (4.60)

with equilibrium constants given by

K1 = [≡M−OH][H+][≡M−OH+

2 ] , K2 = [≡M−O−][H+][≡M−OH] , (4.61)

The OH bonds in bulk water have an intrinsic activity derived solely from the disso-ciation of water molecules, whereas the activity of the surface OH groups involvedin the above surface acid equilibrium constants will introduce the additional work(energy) required to bring an H+ ion from the bulk to the surface, which is given bythe electrostatic potential of the surface. Thus,

K1 = K int r1 exp

(Fψ

kT

),

K2 = K int r2 exp

(Fψ

kT

).

(4.62)

The intrinsic terms quantify the extent of H+ exchangewhen the surface is uncharged.With increasing pH, the surface dissociation increases since more of the surface H+ions are consumed, and hence increasingly it becomes difficult to remove hydrogenions from the surface. The above equations represent this effect mathematically.

EXAMPLE 4.18 INTRINSICAND CONDITIONAL EQUILIBRIUM CONSTANTS

Derive the equations for the conditional equilibrium constants in terms of the intrinsicconstants given above.For the two ionization reactions of concern, the conditional equilibrium constants areK1and K2 as given by the equations earlier. The activity coefficients of surface-adsorbedspecies are assumed to be equal. The equilibrium constants are conditional since theydepend on the surface ionization, which further depends on the pH. The total free energyof sorption is composed of two terms: ΔG0

tot = ΔG0intr +ΔzFψ0, where the second

term is the “intrinsic” free energy term. The third term represents the “electrostatic”term that gives the electrical work required to move ions through an interfacial potentialgradient.Δz is the change in charge of the surface species upon sorption. SinceΔG0 =−RT ln K , we can obtain

Kintr = Kcond exp

(ΔFzψ0

RT

). (4.63)

Hence, we derive the equation for K1 and K2.

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It is clear from the above discussion that the surface charge density is the differ-ence in surface concentration (mol/m2) of ≡M–OH+

2 and ≡M–O− moieties on thesurface. Thus,

σe = [≡M−OH+2 ] − [≡M−O−]. (4.64)

The above equation gives the overall alkalinity or acidity of the surface. If thespecies concentrations are expressed in mol/kg, then the overall charge balance canbe written as

CA − CB + [OH−] − [H+] = {≡M−OH−2 } − {≡M−O−}, (4.65)

where CA and CB are the added aqueous concentrations of a strong acid and a strongbase, respectively. [OH−] and [H+] are the aqueous concentrations of base and acidadded. Note that { } denotes mol/kg, whereas [ ] represents mol/m2 for M–OH+

2 andM–O− species.

When σe is zero, there are equal surface concentrations of both species. The pH atwhich this happens is called pHpzc, the point of zero charge. At the pHpzc, the surfacepotential, Ψ → 0. Now, one can relate the pHpzc to the intrinsic surface activities byequating the concentrations of the surface species, [≡M–OH+

2 ] and [≡M–O−].

pHpzc = 1

2

(pK intr

1 + pK intr2

). (4.66)

If pH < pHpzc, the surface has a net positive charge, whereas for pH > pHpzc, thesurface is negatively charged.

Getting back to the equation for σe and utilizing the equations for intrinsicequilibrium constants, one obtains

σe = [≡M−OH]

[[H+]

K intr1

exp

(−Fψ

RT

)− K intr

2[H+] exp

(Fψ

RT

)]. (4.67)

We can now substitute in the above expression the equation forΨ in terms of σe andget an explicit equation relating the surface hydroxyl concentration to σe through ionicstrength I . A listing of the characteristic values of K intr

1 , K intr2 , and pHpzc for several

common oxideminerals is given inTable 4.11. In the presence of other ions in aqueoussolution (in addition to H+ and OH−), the influence of the specifically sorbed ionson the surface introduces some complexity into this analysis. These concepts havebeen discussed by Stumm (1993) and should be consulted for an in-depth discussion.The mathematical treatment of surface charges and surface ionization come under theso-called surface complexation models (Stumm andMorgan, 1996). The metal cationadsorption data over a wide range of pH values can be represented by an equation ofthe form

lnD = a+ b · pH, (4.68)

where D = KswCs = φsorb/φsoln. Ksw is the distribution (partition) constant betweenthe solid and water, Cs is the mass of adsorbent per unit volume of aqueous solution,

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TABLE 4.11Surface Properties of Important Oxide Minerals Present in theEnvironment

Surface Composition Surface Area pK intrs1 pK intr

s2 pHpzc

Quartz SiO2 0.14 −3 7 2Geothite FeO(OH) 46 6 9 7.5Alumina Al2O3 15 7 10 8.5Iron oxide Fe(OH)3 600 7 9 8Gibbsite Al(OH)3 120 5 8 6.5Na-montmorillonite Na3Al7Si11O30(OH)6 600–800 — — 2.5Kaolinite Al2Si2O5(OH)4 7–30 — — 4.6Illite KAl3Si3O10(OH)2 65–100 — — —Vermiculite 600–800 — — —Muscovite 60–100 — — —

Source: From Schwarzenbach, R.P. et al. 1993. Environmental Organic Chemistry. New York:Wiley; Sparks, D. 1999. Environmental Soil Chemistry. NewYork: Academic Press.

φsorb is the ratio of moles of metal adsorbed to the fixed initial moles added, andφsoln = 1 − φsorb is the fraction of metal unadsorbed. A plot of ln D versus pH iscalled theKurbatov plot. The values of a and b have physical significance. If half of theadded metal is adsorbed, thenD = 1 and pH at this point is designated pH50 = −a/b.Then we have (

dφsorb

d pH

)

pH50

= b

4(4.69)

or

lnD = b · (pH − pH50). (4.70)

Hence,

φsorb = 1

1 + exp[−b · (pH − pH50

)] . (4.71)

Thus, the evaluation of a and b from the Kurbatov plot gives us the fraction ofmetal adsorbed at any given pH.

There are other aspects that should also be considered in understanding theexchange of metal ions with clay minerals, since with surface exchange reactions, theforces of bonding within the clay structure will also physically distort their size andshape. In other words, extensive changes in the activity coefficients of the adsorbedions are possible. Expansion or contraction of clayswill complicate the simple activitycoefficient relationships. Such general relationships are, heretofore, lacking. Generaltrends have been noted and verified, and hence qualitative relationships are possible.The simple models described above should serve to exemplify the general approachto the study of exchangeable metal ions on soils and sediments.

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4.3.3 ADSORPTION OF ORGANIC MOLECULES ON SOILS AND SEDIMENTS

As is evident from Table 4.10, a typical sediment or soil will have an overwhelm-ing fraction of sand, silt, or clay and only a small fraction of organic material. Thetotal organic fraction is characterized mainly by humic and fulvic acids. They existbound to the mineral matter through Coulombic, ligand exchange, and hydrophobicinteractions. Some of the species are weakly bound and some are strongly bound.The weakly bound fraction enters into dynamic exchange equilibrium between thesolid (mineral) surfaces and the adjacent solution phase. In some cases, this relation-ship can be characterized by a Langmuir-type isotherm (Thoma, 1992; Murphy etal., 1990). The adsorbed concentration rapidly reaches a maximum. The differencesin the maxima for different surfaces are striking. The sorption of humic substances,for example, will depend not only on the site concentrations but also on its relativeaffinity with iron and aluminol hydroxyl groups. It is known that adsorption of DOCson hydrous alumina is a pH-dependent process with maximum adsorption betweenpH 5 and 6, and decreasing adsorption at higher pH (Davis and Gloor, 1981). As aconsequence of this change, alumina coagulates more rapidly at pH values close tothe adsorption maximum. It is also known that increasing molecular weight of a DOCleads to increased adsorption on alumina. In other words, adsorption of DOCs onalumina is a fractionation process.An adsorbed organic film such as DOCs (humic orfulvic compounds) can alter the properties of the underlying solid and can present asurface conducive for chemical adsorption of other organic compounds (Figure 4.10).

As discussed in the previous section,mineralmatter has a large propensity forwatermolecules because of their polar character. Nonpolar organicmolecules (e.g., alkanes,polychlorinated biphenyls, pesticides, and aromatic hydrocarbons) have to displacethe existing water molecules before adsorption can occur on mineral surfaces. Forexample, on a Ru(001) metal surface, Thiel (1991) found that the adsorption of waterfrom the gas phase involved a bond strength of 46 kJ/mol for water and 37 kJ/molfor cyclohexane. The adsorption bond strengths are comparable. However, the areaoccupied by one cyclohexane molecule is equivalent to about seven water molecules.

1009080706050

HA

adso

rbed

(um

ol/m

2 )

40302010

00 1

HA in solution (umol/mL)20.5 1.5 2.5

HA onhaematite

HA onKaolinite

16

15.8

15.6

15.4

Sedi

men

t org

anic

carb

on (m

g/g)

15.2

150 10 20 30 40

Dissolved organic carbon (mg/L)50 60 70 80

OC on EPA 5Sediment

(a) (b)

FIGURE 4.10 (a) Sorption of humic acid on hematite and kaolinite; (b) sorption of organiccarbon on EPA5 sediment.

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Hence, a meaningful comparison should be based on the energy change per unit area.On this basis, the replacement of cyclohexane on the surface by water involves anenthalpy decrease of about 21–23 kJ/mol. The entropy change was relatively small forboth molecules. Thus the enthalpy change per unit area is the major driving force forthe displacement of cyclohexane from the metal surface. Such large enthalpy changesare possible for water molecules because of their special tetrahedral coordination andH-bonding tendency, which are lacking in cyclohexane molecules.

The adsorption of several organic molecules on solids from their aqueous solutionsis reported in the literature. The data are invariably reported asmass of solute (sorbate)adsorbed per unit mass of solid sorbent,

Ksw (L/kg) = Wi (mg sorbate/kg sorbent)

Ciw (mg sorbate/L solution)=(WOCi +Wmin

i

)

Ciw. (4.72)

The overall partition constant, Ksw, has two contributions – one from mineral matter(min) and the other from organic matter (oc). Thus,

Ksw = Kocsw + Kmin

sw . (4.73)

If foc is the fraction of organic carbon on the solid (g/g) and the mineral surface area isgiven by Sa (m2/kg), Koc (L/kg), and Kmin (mg/m2) represent the partition constantsfor a compound normalized to oc and min, respectively.

Koc = Kocsw

foc, Kmin = Kmin

sw

Sa(4.74)

Schwarzenbach Gschwend, and Imboden (1993) suggest that for hydrophobic com-pounds, which are driven to themineral surface via exclusion from thewater structure,an LFER should exist between Kmin and γ∗

i . The form of the LFER is

logKmin = a log γ∗i

+ b. (4.75)

Based on a limited set of data, this relationship appears to hold for a particular sorbentadsorbing different nonpolar compounds under identical conditions of pH and ionicstrength (Mader, Goss, and Eisenreich, 1997).

The amount adsorbed on mineral surfaces is indeed very small. However, if themineral surface is coated with even a small percent of organic macromolecules, theadsorption capacity is enhanced by several orders ofmagnitude.Mineralmatter devel-ops these coatings rapidly and hence natural soil and sediment can be considered toprovide dual sorbent sites. The bare mineral surfaces are characterized by low Kmin.The organic matter coating the mineral provides a highly compatible medium withwhich a neutral hydrophobic compound can associate and thereby limit its interactionwith water.

The organic matter in sediments and soils is composed of approximately one halfof the total carbon that is directly measured as organic carbon. Hence, sorption isoften keyed to organic carbon rather than organic matter through the approximate

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relationship φom . 2φoc. Karickhoff, Brown, and Scott (1979) demonstrated that thelinear isotherm described above was valid for several organic compounds over a widerange of aqueous concentrations. The linear isotherm was found to be valid up toapproximately 50% of the aqueous solubility of the compound. The isotherms werereversible and showed only a 15% decrease in Ksw at an ion (NaCl) concentrationof 20mg/mL. They also demonstrated the linear relationship between Ksw and focfor a polyaromatic hydrocarbon (pyrene). Further, Means et al. (1980) extended thisrelationship to several PAHs on sediments and soils. The slope of the plot of Ksw(kg sorbate/kg sorbent) versus foc is a constant for a given compound on various soilsand sediments. Thus, we have

Ksw = Koc foc. (4.76)

The slope Koc is therefore a convenient way of characterizing the sorption of aparticular hydrophobic compound.

EXAMPLE 4.19 OBTAINING Koc FROM EXPERIMENTAL DATA

One gram of soil in 500mL of aqueous solution was spiked with 10mg/L of an organiccompound. After equilibration for 48 h, the aqueous concentration of the compoundwas 1mg/L. If the soil organic carbon content was 0.02, obtain Ksw and Koc for thecompound.

ws = 1 g, Vw = 500mL,Cw,0i = 10mg/mL Ciw = 1mg/L, foc = 0.02.

Wi/ws = (0.01 − 0.001)0.5/1 = 0.0045 g/g. Ksw = (Wi/ws)/Ciw = 4.5 L/kg.

Koc = Ksw/foc = 225 L/g = 2.25 × 105 L/kg.

The partitioning of a solute from water to soil organic carbon is similar to thepartitioning to octanol. This leads to an LFER between Koc and Kow. As discussedearlier, the octanol–water partition constant is given by

Kow =(γiw

γio

)(V∗

w

V∗o

). (4.77)

Since the partitioning into the organic fraction of the mineral matter can be describedsimilarly, we can write

Koc =(γiwc

γicw

)(V∗

w

V∗c

)1

ρc, (4.78)

whereγiwc andγicw are the respective activity coefficients of solute i inwater saturatedwith organic matter (humus) and humic saturated with water, V∗

w and V∗c are the

respective molar volumes, and ρc is the density of the organic matter so that Koc isexpressed in L/kg. Using the equation for Kow to substitute for V∗

w, we obtain

Koc = C1Kow

(γio

γicw

)(γiwc

γiw

). (4.79)

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Further, if γiwc = γiw, we have

Koc = C1Kowγio

γicw. (4.80)

It has been observed that most solutes behave nonideally in the octanol phase tothe extent that γo

i is given by (Curtis et al., 1994)

γio = 1.2K0.16ow . (4.81)

Therefore,

logKoc = 1.16 logKow + logC2 − log γicw, (4.82)

where C2 = 1.2C1. Hence linear correlations between log Koc and log Kow are pre-dicted. For example, Curtis et al. (1994) obtained the following linear relationship foradsorption on the natural organic matter of soils

logKoc = 0.92 logKow − 0.23. (4.83)

Table 4.12 displays the relationship between log Koc and log Kow. Table 4.13 liststhe available correlations between logKmin and log γ∗

i . Note that γ∗i is related directly

to log Kow (Section 3.4.4.2).

EXAMPLE 4.20 DETERMINING Ksw FROM Kow

A soil from a Superfund site in Baton Rouge, Louisiana, was found to have the followingproperties: clay 30%, sand 22%, silt 47%, and organic carbon content 1.13%. Estimatethe soil–water partition constant for 1,2-dichlorobenzene on this soil.

For 1,2-dichlorobenzene, log Kow is 3.39. Hence, log Koc = 0.92(3.39) − 0.23 =2.89. Koc = 774. Ksw = Koc φoc = (774)(0.0113) = 8.7P/kg.

EXAMPLE 4.21 TIME OF TRAVEL OFA POLLUTANT IN GROUNDWATER

The same approach as described above is also used to describe the movement of pollu-tants in the subsurface groundwater (Weber,McGinley, andKatz, 1991). Sorption retardsthe velocity of pollutant movement in groundwater (up) in relation to the velocity ofthe groundwater itself (u0). This can be expressed as

u0up

= 1 + ρb

εKsw = RF, (4.84)

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where RF is the retardation factor. For compounds that are strongly sorbing, up � u0and the pollutant concentration front is slowed down considerably. For compounds thatare nonsorbing (such as chloride ions), u0 = up and no retardation is seen. McCartyet al. (1992) report the results of an experiment in which the retardation of various halo-genated compounds present in a reclaimed municipal wastewater was injected into anaquifer in PaloAlto, CA. The fractional breakthrough in an observation well downfieldwas obtained for three adsorbing pollutants (chloroform, bromoform, and chloroben-zene) and a nonsorbing tracer (chloride ion). The results are shown in Figure 4.11. Thefield-measured retardation factors were 6 for chloroform and bromoform, and 33 forchlorobenzene. Clearly the greater the hydrophobicity of the pollutant, the slower itsmovement in the aquifer. Retardation is an important process in groundwater for twomain reasons. Firstly, if an aquifer were to become polluted with compounds of dif-fering hydrophobicity, they would tend to appear in a down-gradient well at differenttimes in accordance with their retardation factors. This would make the concentrationsand nature of water at the observation well quite distinct from the original contami-nation, and hence identification of the pollution source will be difficult. Secondly, theretardation factor will give us an idea of how much material is on the solid phase andhow much is in free water, and therefore develops appropriate remediation alternativesfor the restoration of both the groundwater and the aerial extent of the contaminatedaquifer.

TABLE 4.12Correlations between log Koc and log Kow for Various Compoundsof Environmental Significance

log Koc = a + b log KowCompound Class B a r2 References

Pesticides 0.544 1.377 0.74 Kenaga and Goring (1980)Aromatics (PAHs) 0.937 −0.006 0.95 Lyman et al. (1982)Aromatics (PAHs ) 1.00 −0.21 1.00 Karickhoff et al. (1979)Herbicides 0.94 0.02 — Lyman et al. (1982)Insecticides, fungicides 1.029 −0.18 0.91 Rao and Davidson (1980)Phenyl ureas and carbamates 0.524 0.855 0.84 Briggs (1973)Chlorinated phenols 0.82 0.02 0.98 Schellenberg et al. (1984)Chlorobenzenes, PCBs 0.904 −0.779 0.989 Chiou et al. (1983)PAHs 1.00 −0.317 0.98 Means et al. (1980)Polychlorinated biphenyls 0.72 0.49 0.96 Schwarzenbach and Westall

(1981)

Note: Koc is in L/kg or cm3/g.

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TABLE 4.13Correlations between log Kmin and log γ∗

i

log Kmin = a + b log γ∗iCompound Class Solid a b r2 References

Chlorobenzenes, PAHs, biphenyls α-Al2O3 −10.68 0.70 0.94 [1]Chlorobenzenes, PAHs biphenyls α-Fe2O3 −11.39 0.98 0.92 [1]PAHs Kaolin, glass, alumina −14.8 1.74 0.97 [2]Various HOCs Kaolinite −12.0 1.37 [3]Various HOCs Silica −12.5 1.37 [3]

Source: From [1] Mader et al. (1997); [2] Backhus (1990); [3] Schwarzenbach et al. (1993).Note: Kmin is in mL/m2.

10

05

00 10 20

ChlorideFrac

tiona

l bre

akth

roug

h

ChloroformBromoformChlorobenzene

Volume (103 m3)30

FIGURE 4.11 Sequential breakthrough of solutes at an observation well during the PaloAlto groundwater recharge study. (Reprinted from Roberts et al. 1982. Water Research16, 1025–1035; with kind permission from Elsevier Science Limited, The Boulevard,Kidlington, UK.)

4.4 BIOTA/WATER PARTITION CONSTANT(BIOCONCENTRATION FACTOR)

There are a number of receptors for a pollutant released into the environment. Riskassessment is the process of understanding and minimizing the effects of pollu-tants on receptors. To illustrate this, let us consider the sediment–water environment.Transport from contaminated sediments to the overlying water column exposes themarine species to pollutants. The marine animals accumulate these chemicals and the

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contaminants then make their way into the food chain of higher animals, includinghumans. This is the process of bioaccumulation.

The uptake of organic chemicals by aquatic species can be modeled at differentlevels of complexity. The uptake depends on factors such as exposure route (dermal,ingestion by mouth, and inhalation), and loss through digestion and defecation. Ananimal can also imbibe chemicals through its prey that has been exposed to the chem-ical. Although rate-based models may be better suited to describe these phenomena,thermodynamic models have been traditionally used to obtain first-order estimatesof the extent of bioaccumulation (Mackay, 1982). It has been suggested that abioticspecies are in near-equilibrium conditions in most circumstances. Hence, it is appro-priate to discuss briefly the thermodynamic basis for modeling the bioaccumulationphenomena.

In its simplest form, a partition coefficient (also called a bioconcentration factor,Kbw) is used to define the concentration level of a pollutant in an aquatic speciesrelative to that in water. Since the major accumulation of a pollutant in an animaloccurs in its lipid fraction, it is customary to express the concentration on a lipidweight basis.

The general equation for partitioning between the organism and water is given by

Kbw = CiBCiw

, (4.85)

where CiB is the animal concentration (mg/kg) and Ciw is the aqueous concentration(mg/L) at equilibrium. Ifwe assume that the organism is comprised of j compartments,each of concentration I given by Cij, and with a volume fraction ηj, then we can writethe total moles of solute i in the organism as

∑j

mij =∑j

CijηjV , (4.86)

where V is the total organism volume. Thus, we have

CiB =∑

j mij

V=∑j

Cijηj. (4.87)

For the aqueous phase we have

Ciw = xiwVw

. (4.88)

At equilibrium, the fugacity in all j compartmentswould be equal to that in the aqueousphase, f wi = f ji . For any compartment j,

f ji = xijγijf0i = CijVjγijf

0i , (4.89)

where f 0i is the reference fugacity on the Raoult’s law basis, Vj = ηjV . Thus,

Cij =(f jif 0i

)(1

Vjγij

)(4.90)

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or

CiB =∑j

(f jif 0i

)(1

Vjγij

)ηj. (4.91)

For the aqueous phase we have

f wi = xiwγiw f 0i = CiwVwγiw f 0i (4.92)

or

Ciw =(f wi

f 0i

)(1

Vwγiw

). (4.93)

Now since f wi = f ji , we obtain

KBw = CiBCiw

= Vwγiw∑j

ηj

Vjγij. (4.94)

Since the dominant accumulation of hydrophobic solutes in an organism occurs in itslipid content, we can write

KBW =(Vw

VL

)·(γiw

γiL

)· ηL, (4.95)

where L refers to the lipid phase. Organisms with high lipid content (ηL) should havehigh KBW values.

Since γiw is directly proportional to the octanol–water partition coefficient, Kow,we can expect an LFER relationship between KBW and Kow. Using the definition ofKow given previously along with Equation 4.95, we obtain

KBW

Kow= ηL

(γio

γiL

)(Vo

VL

). (4.96)

For compounds that have similar volume fraction of lipids (ηL) and similar ratiosof activity coefficients (γio/γiw), the ratio KBW/Kow should be fairly constant. Thissuggests that a linear one-constant correlation should suffice:

logKBW = a logKow + b. (4.97)

Such a correlation was tested and confirmed by Mackay (1982). The correlationdeveloped was for a restricted set of compounds, namely, those with log Kow < 6,nonionizable, and those with small KB values. The overall fit to the experimentaldata was

logKBW = logKow − 1.32, r2 = 0.95. (4.98)

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TABLE 4.14log KBW−log Kow Correlations

log KBW = a log Kow + bChemical Class a b r2 Species

Various 0.76 −0.23 0.823 Fathead minnow, bluegill,trout

Ether, chlorinated compounds 0.542 +0.124 0.899 TroutPesticides, PAHs, PCBs 0.85 −0.70 0.897 Bluegill, minnow, troutHalogenated hydrocarbons,halobenzenes, PCBs,diphenyl oxides, P-pesticides,acids, ethers, anilines

0.935 −1.495 0.757 Various

Acridines 0.819 −1.146 0.995 Daphnia pulex

Source: From Veith, G.D. et al. 1980. An evaluation of using partition coefficients and water solubilityto estimate the bioconcentration factors for organic chemicals in fish. ASTM SPE 707, ASTM,Philadelphia, PA, pp. 116–129; Neely,W.B. et al. 1974. Environmental Science and Technology8, 1113–1116; Kenaga, E.E. and Goring, C.A.I. 1980. Relationship between water solubility,soil sorption, octanol-water partitioning, and concentration of chemicals in biota, ASTM SPE707, pp. 78–115; Southworth, G.R. et al. 1978.Water Research 12, 973–977.

Thus KBW = 0.048 Kow. The implication is that fish is about 5% lipid or it behavesas if it is about 5% octanol by volume. The above correlation can give estimatesof the partitioning of hydrophobic organic compounds into biota provided (i) theequilibrium assumption is valid, and (ii) the nonlipid contributions are negligible. Itshould be borne in mind, however, that the correlation is of dubious applicability fortiny organic species such as plankton, which have very large area-to-volume ratiosand hence surface adsorption may be a dominant mechanism of partitioning. Otheravailable correlations are given in Table 4.14.

EXAMPLE 4.22 BIOCONCENTRATION FACTOR FORA POLLUTANT

A fish that weighs 3 lb resides in water contaminated with biphenyl at a concentrationof 5mg/L. What is the equilibrium concentration in the fish?

For biphenyl, logKow = 4.09. Hence, logKBW = 4.09 − 1.32 = 2.77. KBW =589 L/kg. Hence, CiB = 589 (5) = 2944mg/kg and mass in the fish = (2944) (3)(0.45) = 3974mg = 3.97 g.

4.5 AIR-TO-AEROSOL PARTITION CONSTANT

Particulates (aerosols) in air adsorb volatile and semi-volatile compounds from theatmosphere in accordance with the expression derived by Junge (1977). IfWi (μg/m3

air) is the amount of a solute associated with total suspended particulate concen-tration Csp (μg/m3 air), and Cia (ng/m3) is the concentration in the adjoining air in

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equilibrium with it, then the partition constant between air and particulates is

KAP = Cia(Wi/Csp)

. (4.99)

Note that KAP has units of ng/m3.Traditionally, aerosols are collected using a high-volume air sampler into which a

large volume of air is pulled through a glass fiber filter that retains particulates andsubsequently through a tenax bed that retains the vapors. The filter-retained materialis taken to be Wi and the adsorbent-retained solute is taken to be equivalent to Cia.In our earlier discussion on partitioning into aerosols, we had established that thefraction adsorbed to particulates in air (φp

i ) is determined by the subcooled liquidvapor pressure of the compound (P∗

s(l)). Compounds with small P∗s(l) showed large

values of φpi . This means that the value of KAP will be correspondingly large.

KAP =(1 − φp

i

φpi

)Csp. (4.100)

The value of KAP has been found to be a sensitive function of temperature. Manyinvestigators have collected field data and developed correlations of the form

logKAP = m

T+ b, (4.101)

where m and b are constants for a particular compound. Yamasaki, Kuwata, andMiyamoto (1982), Pankow (1987), Bidleman (1988), and Subramanyam et al. (1994)have reported correlations (Table 4.15). Pankow (1987) showed theoretically that theconstants in the above equation are given by

m = − ΔHdes

2.303R+ Ta

4.606, (4.102)

b = log

(2.75 × 105(Ma/Ta)

1/2

Apt0

)− 1

4.606, (4.103)

where ΔHdes is the enthalpy of desorption from the surface (kcal/mol), Ta is themidpoint of the ambient temperature range considered (K), Ap is the specific surfacearea of the aerosol (cm2/μg), and t0 is the characteristic molecular vibration time(10−13–10−12 s). Since the value of b is only weakly dependent on molecular weight,M, the compound specificity on KAP appears through the slope m where ΔHdes ischaracteristic of the compound. At a given T , KAP decreases as ΔHdes increases. Ingeneral, one can write

KAP = 109P∗s(l)

NsApRT exp((ΔHdes −ΔHv)/RT), (4.104)

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TABLE 4.15Relationships between KAP and T for Some Organics in Ambient Air

log KAP = m/T + bCompound m b r2 Reference

α-HCH −2755 14.286 0.574 Bidleman and Foreman (1987)Hexachlorobenzene −3328 16.117 0.687Aroclor 1254 −4686 19.428 0.885Chlordane −4995 21.010 0.901p,p′-DDE −5114 21.048 0.881p,p′-DDT −5870 22.824 0.885Fluoranthene −5180 20.80 0.682 Keller and Bidleman (1984)Fluoranthene −4420 18.52 0.805 Yamasaki et al. (1982)Pyrene −4510 18.48 0.695 Keller and Bidleman (1984)Pyrene −4180 17.55 0.796 Yamasaki et al. (1982)Fluoranthene −4393 21.41 Subramanyam et al. (1994)Fluoranthenea −6040 25.68 Subramanyam et al. (1994)Phenanthrene −3423 19.02 Subramanyam et al. (1994)

aAnannular denudermethodwasused to reduce sampling artifacts resulting fromahigh-volume samplingprocedure. All other reported data in the table use the Hi-Vol sampling procedure. Bidleman and co-workers obtained data from Columbia, SC,Yamasaki et al. from Tokyo, Japan, and Subramanyam et al.from Baton Rouge, Louisiana.

where P∗s(l) is the subcooled liquid vapor pressure (atm), Ns is the number of moles of

adsorption sites per cm2 of aerosol (mol/cm2), andΔHv is the enthalpy of vaporizationof the liquid (kcal/mol). R is the gas constant (= 83 cm3 atm/molK). In most casesΔHdes −ΔHv, and hence the above equation simplifies to

KAP = 1.6 × 104P∗s(l)

NsAp, (4.105)

whereP∗s(l) is expressed inmmHg. IfKAP is expressed inμg/m3 (amore conventional

unit) denoted by the symbol K∗AP, we need to multiply by 0.001.

EXAMPLE 4.23 AIR-TO-PARTICULATE PARTITIONING OF A POLYAROMATIC

HYDROCARBON

Calculate the air–particulate partition constant for a pollutant inBatonRouge,Louisiana.A typical value of specific area, av, for aerosols collected from Baton Rouge air wasestimated at 5 × 10−4 m−1. The average particulate concentration, Csp, in summertime in Baton Rouge is 40μg/m3. Therefore Ap = av/Csp = 1.25 × 10−5 m2/μg. Forphysical adsorption of neutral compounds, Pankow (1987) observed thatNs is generally

continued

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compound independent and has an average value of 4 × 10−10 mol/cm2. For phenan-threne at an ambient temperature of 298K, the sub-cooled liquid vapor pressure is5 × 10−7 atm. HenceKAP at 298K for phenanthrene in aerosols is 1.6 × 108 ng/m3, orK∗

AP is 1.6 × 105 μg/m3. log K∗AP = 5.2. The experimental value is 5.7 (Subramanyam

et al., 1994).

Earlier in this chapter we considered Junge’s equation

φPi = ρiav

P∗s(l) + ρiav

. (4.106)

Pankow (1987) showed that within a given class of compounds KAP can be related tothe volatility of the species or the subcooled vapor pressure of the compound P∗

s(l)

logKAP = a1 logP∗s(l) + b1, (4.107)

where b1 is temperature independent. Many investigators have shown that the abovecorrelation is useful in estimating KAP (Table 4.16). The slope a1 ≈1 in most cases.

In the atmosphere, φpi is a function of the relative humidity. Since water competes

effectively with organicmolecules for sorption sites on the aerosol, it reduces the frac-tion of adsorbed organic compound. Experimental data support the effect (Pankow,Storey, and Yamasaki, 1993). The modified equation for air-to-aerosol partitioningwill be

logKAP = m

T+ b− log f (xw), (4.108)

TABLE 4.16Relationship between KAP and P∗

s(l)

log KAP = a′ log P∗s(l) + b′

Compound Class a′ b′ Location

PAHs 0.8821 5368 Portland, OR0.760 5100 Denver, CO0.694 4610 Chicago, IL0.631 4610 London, UK1.04 5950 Osaka, Japan

PCBs 0.610 4740 Bayreuth, Germany0.726 5180 Chicago, IL0.946 5860 Denver, CO

Chlorinated pesticides 0.740 5760 Brazzaville, Congo0.610 4740 Bayreuth, Germany

Source: FromFalconer, R.L. andBidleman, T.F. 1997. In: J.L. Baker (Ed.),Atmospheric Depositionof Contaminants to the Great Lakes and Coastal Waters. Pensacola, FL: SETAC Press.

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where m and b are the same constants described earlier.

f (xw) = 1

1 + KLangxw, (4.109)

whereKLang is the Langmuir adsorption constant for water on aerosols. Notice that asxw → 0, f (xw) → 1. As xw increases, f (xw) decreases and therefore with increasingrelative humidity log KAP increases.

4.6 AIR-TO-VEGETATION PARTITION CONSTANT

Amajor portion of the land area (80%) on earth is covered by vegetation. As such thesurface area covered byvegetation cannot be overlooked as an environmental compart-ment. Plants also take up nutrients and organic compounds from the soil/sedimentenvironments. Thus plants participate in the cycling of both inorganic and organiccompounds in the environment. A large number of processes, both intra- and extra-cellular, are identified near the root zones and leaves of plants. In the water–soilenvironment a number of organic compounds are imbibed through the roots andenzymatically degraded within the plants. In the air environment, a number of studieshave revealed that plant–air exchange of organic chemicals plays a major role in thelong-range transport and deposition of air pollutants.

Wax or lipid layers exist to prevent excessive evapo-transpiration from mostplant surfaces exposed to air. The combination of high surface area and thepresence of wax/lipid suggests the high partitioning of hydrophobic organic com-pounds into vegetation. As in the case of the bioconcentration factor, we define avegetation–atmosphere partition coefficient, KVA:

KVA = WV

fLCia, (4.110)

whereWv is the concentration in vegetation (ng/g dry weight), fL is the lipid contentof vegetation (mg/g dry weight), and Cia is the atmospheric concentration (ng/m3).KVA has units of m3 air/mg lipid.A dimensionless partition coefficient, K∗

VA, can alsobe obtained if we use an air density of 1.19 × 106 mg/m3 at 298K.

Just asKAP for aerosolswas related to temperature,KVA has the functional relation-ship: ln KVA = A/T + B. The intercept B is common for a given class of compounds.For PAHs, B was −35.95 (Simonich and Hites, 1994). Values of A for several PAHsare listed in Table 4.17.

Several investigators have shown that just as KAP is correlated to P∗s(P), so can we

relate KVA to P∗s(P). A better correlating parameter is Koa, the octanol–air partition

constant. A few of these correlations are shown in Table 4.17 for PCBs.Whereas Simonich and Hites (1994) attributed all of the seasonal variations inKVA

to the temperature variations, Komp and McLachlan (1997) have argued that this isdue more to a combination of effects such as growth dilution, decrease in dry mattercontent of leaves in late autumn, and erosion of HOCs from the surface of vegetation.These preliminary studies have shown the difficulties in understanding the complexinteractions of HOCs with the plant–air system.

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TABLE 4.17Correlations of KVA versus 1/T for Various Compounds

PAHs: ln KVA = (A/T ) + BCompound A B

Pyrene 10,227 −35.95Phenanthrene 9840 −35.95Anthracene 9773 −35.95Fluoranthene 10,209 −35.95Benz[a]anthracene 10,822 −35.95Benzo[a]pyrene 10,988 −35.95

PCBs: log K ∗VA = (A/T ) + B

Compound A B r2

2,2′,5-PCB 3688 −7.119 0.9932,2′,5,5′-PCB 4524 −9.307 0.9922,2′,3,5′,6-PCB 4795 −9.843 0.9852,2′,4,4′,5,5′-PCB 6095 −13.135 0.9742,2′,3,3′,5,=5′,6,6′-PCB 5685 −11.557 0.960

Source: From Simonich, S.L. and Hites, R.A. 1994. Environmental Science and Technology 28,939–943; Komp, P. and McLachlan, M.S. 1997. Environmental Science and Technology31, 886–890.

Note: Note that KVA has units of m3/mg of lipid and K∗VA is dimensionless.

EXAMPLE 4.24 ESTIMATION OF UPTAKE OF ATMOSPHERIC POLLUTANTS

BY VEGETATION

Estimate the equilibrium concentration of pyrene on a sugar maple leaf with a lipidcontent of 0.016 g/g. The air concentration is 10 ng/m3.

From Table 4.17, at 298K, ln KVA = (10, 227/298) − 35.95 = −1.63. HenceKVA = 0.196m3/mg lipid.Wv = KVAφLC

ai = (0.196) (16) (10) = 31 ng/g.

4.7 ADSORPTION ON ACTIVATED CARBON FORWASTEWATER TREATMENT

Activated carbon has a high capacity to adsorb organic compounds from both gasand liquid streams. It is perhaps the earliest known sorbent used primarily to removecolor and odor from wastewater. A large number of organic compounds have beeninvestigated in relation to their affinity toward activated carbon (Dobbs and Cohen,1980). The data were correlated to a Freundlich isotherm and the constant KF and ndetermined. Table 4.18 is for a selected number of organic compounds.

Activated carbon is available in both powdered and granular forms. The powderedform can be sieved through a 100 mesh sieve. The granular form is designated 12/20,20/40, or 8/30. 12/20 means it will pass through a standard mesh size 12 screen, but

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TABLE 4.18KF and n for Selected Organic Compounds on GAC

Compound KF 1/n

Benzene 1.0 1.6Toluene 26.1 0.44Ethylbenzene 53 0.79Chlorobenzene 91 0.99Chloroform 2.6 0.73Carbon tetrachloride 11.1 0.831,2-Dichloroethane 3.5 0.83Trichloroethylene 28.0 0.62Tetrachloroethylene 50.8 0.56Aldrin 651 0.92Hexachlorobenzene 450 0.60

Source: From Dobbs, R.A. and Cohen, J.M. 1980. Carbon adsorption isothermsfor toxic organics. Report No: EPA-600/8-80-023, U.E. EnvironmentalProtection Agency, Cincinnati, OH.

Note: The carbon used was Filtrasorb 300 (Calgon Corporation).

will not pass through a 20-size screen. The granular form is less expensive and ismore easily regenerated, and hence is the choice in most wastewater treatment plants.

Activated carbon treatment is accomplished either in a continuousmode by flowingwater over a packed bed of carbon (see Figure 4.12a) or in batch (fill-and-draw) modewhere a given amount of activated carbon is kept in contact with a given volume ofwater for a specified period of time (see Figure 4.12b). In the continuous process theexit concentration slowly reaches the inlet concentrationwhen the adsorption capacityof carbon is exceeded. The bed is subsequently regenerated or replaced.

Activated carbon(a)(b)

Feed EffluentContinuous mode

operationBatch modeoperation

FIGURE 4.12 (a) Continuous mode fixed-bed operation of activated carbon adsorption forwastewater treatment. Water is passed over a bed of carbon. (b) A batch mode of operationwherein a fixed weight of carbon is kept in contact with a given volume of aqueous phase.After treatment the water is replaced. This is also called the fill-and-draw or cyclic, fixed-bedbatch operation. Once the carbon is exhausted it is also replaced with a fresh batch.

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EXAMPLE 4.25 ACTIVATED CARBON FOR TREATINGA WASTEWATER STREAM

At an industrial site in North Baton Rouge, Louisiana, it is proposed that the con-taminated groundwater be pumped up to the surface and taken through an activatedcarbon unit before being discharged into a nearby lagoon. This is generally referredto as pump-and-treat (P&T) technology. The primary compound in the groundwateris hexachlorobutadiene (HxBD) at a concentration of 1000μg/L. The production rateof groundwater is 200 gal/min for 10 h of operation. It is desired to achieve an efflu-ent concentration of HxBD below the wastewater discharge limit of 27μg/L. A batchadsorption system is contemplated to ease the load on the downstream incinerator, whichis planned for further destruction of the organics in the effluent air before discharge.Estimate (i) the carbon dosage required to achieve the desired level of effluent quality,(ii) the amount of HxBD removed per day, and (iii) the mass and volume of carbonrequired per day. The carbon bulk density is 20 lb/ft3.A standard batch shaker flask experiment conducted gave the following adsorption

data for HxBD on a GAC (Dobbs and Cohen, 1980).

Ciw (mg/L) Wi (mg/g carbon)

0.098 93.80.027 50.60.013 34.20.007 25.80.002 17.3

2.5

2

1.5

1

0.5

log

Wi

0–3 –2 –1–2.5 –1.5

log Ciw

0–0.5

Isotherm Experimental

FIGURE 4.13 The Freundlich isotherm fit for the adsorption of hexachlorobutadienefrom water onto GAC.

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Plot the isotherm data as log Wi versus Ciw to fit a Freundlich isotherm (Figure 4.13).The values of KF and n are 245 and 0.44, respectively (correlation coefficientis 0.989).

(i) Using the Freundlich isotherm determine the adsorption capacity of the carbonfor the required effluent concentration, that is, 27μg/L. This gives wi = 245(0.027)0.44 = 50mg HxBD/g carbon. Notice that fortuitously this is one of theexperimental points in the batch shaker flask experiments. Thus for every 0.05 g ofHxBD removed, we will need 1 g carbon.

(ii) Amount of HxBD to be removed in one day of operation is (1 − 0.027)mg/L (200) gal/min (3.785) L/gal (1.44 × 103)min/day (2.2.5 × 10−3) lb/g(0.001) g/mg = 2.34 lb/day. Hence mass of carbon required per day is 2.34/0.05 =47 lb/day.

(iii) The total volume of carbon required is 47/20 = 2.3 ft3/day.

PROBLEMS

4.12 Estimate the Henry’s constants for the following compounds using (i)the bond contribution method of Hine and Mookerjee and (ii) thegroup contribution scheme of Meylan and Howard:ammonia, toluene,1,2,4-trichlorobenzene, benzyl alcohol, pentachlorophenol, and 1,1,1-trichloroethane.

Bond log K ′aw (Hine and Mookerjee)

C−H −0.11C−Cl 0.30Car−H −0.21Car−Cl −0.11Ct−Cl 0.64Ct−H 0.00O−H 3.21N−H 1.34Car−Car 0.33C−O 1.00C−C 0.04

Note that Car denotes an aromatic carbon and Ct denotes a tertiaryC atom.

4.23 The gas-phase concentration of hydrogen peroxide and itsHenry’s constantare of critical significance in obtaining its concentration in cloud droplets.H2O2 is an abundant oxidant in the atmosphere capable of oxidizingS(IV) in atmospheric water and is thought to be the dominant mecha-nism of converting SO2 from the atmosphere into a stable condensedphase. H2O2 has been observed in rainwater even in remote maritimeregions.

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The following data were reported by Hwang and Dasgupta (1985) forthe equilibrium between air and water for hydrogen peroxide at varioustemperatures:

At 10◦C At 20◦C At 30◦C

P∗i (atm) Cw

i (mg/L) P∗i (atm) Cw

i (mg/L) P∗i (atm) C∗

i (mg/L)

2.8 × 10−7 0.08 1.1 × 10−6 0.05 8 × 10−7 0.1043.5 × 10−7 0.10 2.2 × 10−6 0.10 1 × 10−8 1.3 × 10−3

4.4 × 10−7 0.12 2.8 × 10−6 0.12 5 × 10−9 6.5 × 10−4

1 × 10−9 1.6 × 10−4

(a) Obtain the Kaw for hydrogen peroxide as a function of temperature.(b) Estimate the heat of volatilization of H2O2 from water. Clearly state

any assumptions made.

4.32 (a) The heat evolved upon dissolution of atmospheric ozone into wateris determined to be +5 kcal/mol at 298K. Given that ozone has aHenry’s constant of 9.4 × 10−3 mol/L/atm at 298K, determine itsHenry’s constant at 273K.Note that the definition ofHenry’s constantis for the process air to water.

(b) If the atmospheric ozone concentration in a polluted air is 100 ppbv,what is the concentration in equilibrium with a body of water in thatregion at 298K? Note that you need to assume no losses of ozoneother than absorption into water.

4.42 The following solubility data were obtained for an organic compound(toluene) in the presence of humic acid extracted fromawastewater sample.

Humic Acid Concentration Solubility of(mg of organic C/L) Toluene (mg/L)

0 4768 48814 50225 50439 517

Estimate the KC for toluene on the humic acid.It was determined that the concentration of humic acid in the waste-

water pond was 20 g/m3. If the pure water Henry’s constant for toluene is0.0065 atm/m3/mol, estimate the partial pressure of toluene in the air inequilibrium with the wastewater.

4.52 Consider 2L of a 2mM solution of NH4OH in a closed vessel broughtto equilibrium with 100mL of nitrogen. If the final pH of the aque-ous solution is 8, what is the mass of ammonia in the gas phase atequilibrium? The temperature is 298K. Kaw = 0.016 atm/mol/L. Ka1 =1.7 × 10−5 M. Note that you have to derive the equation for the aqueousmole fraction αNH3 · H2O.

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4.63 Calculate the pH of an aqueous solution that is in equilibrium with100 ppmv of CO2 in the gas phase. The liquid-to-gas volume is 100:1.The temperature is 298K. The necessary parameters are given in the text.

4.71 100mL of an aqueous solution containing 0.0004mM n-nonane is placedin a closed cylindrical vessel of inner diameter 5 cm and total height10 cm. What fraction of n-nonane is present at the air–water interfacein this system? If the total height of the vessel is reduced to 5.5 cm, whatis the adsorbed fraction? Assume that n-nonane does not adsorb to thevessel wall.

4.82 CFCs form an important class of atmospheric pollutants implicated in theformation of ozone hole above the polar regions. These are no longer usedas a result of the worldwide agreement to ban their production and utiliza-tion. One most common CFC was CCl3F (Freon-11) used as refrigerants.If the dry air concentration of Freon-11 is 10 × 10−6 ppmv, determineits concentration in atmospheric moisture as a function of water contentsranging from 10−9 to 10−4 in a closed system. Kaw for Freon-11 is 5.

4.92 The surface area of an average aerosol in Baton Rouge, Louisiana, is5 × 10−4 m−1 Determine the fraction of a polyaromatic hydrocarbon(phenanthrene) adsorbed on the aerosol.Assume that the value ofρi = 0.10in the Junge’s equation. The average temperature during the collectionof the aerosols was 73◦F on April 16, 1993. The average precipita-tion was 0.46′′ of water. The average particulate phase concentration ofphenanthrene measured on that day was 0.21 ng/m3. The air–particulatepartition constant for phenanthrene was related to temperature in Kelvinsthrough the equation log KAP = −3423/T + 19. Determine the flux ofphenanthrene in μg/m2/day.

4.102 The concentration of a pesticide (chlorpyrifos) measured in air from afoggy atmosphere in Parlier, CA, was 17.2 ng/m3 at about 25◦C. TheHenry’s constant for chlorpyrifos at 25◦C is 170 × 10−6 and the air–waterpartition constant is 6.4 × 10−4 at 12.5◦C. The average fog diameter is1μm. If the fog deposition rate is 0.2mm/h, and the fog lasts for 4 h, whatis the surface flux of chlorpyrifos?

4.112 Liss and Slater (1974) determined that over the Atlantic Ocean the meanatmospheric concentration of a CFC (CCl3F, Freon-11, molecular weight137) is 50 × 10−6 ppmv, whereas the mean aqueous concentration is7.76 × 10−12 cm3/mL. The overall aqueous-phase mass transfer coeffi-cient was estimated as 11 cm/h. Determine the magnitude and directionof freon flux? If the ocean surface area is 3.6 × 1014 m2, and the totalannual flux of freon is 3 × 1011 g, what fraction is contributed by theAtlantic Ocean?

4.123 (a) The aqueous solubility of hexachlorobenzene is 5μg/L. Estimate itsoctanol–water partition constant. Use this information alone to cal-culate the aqueous concentration that will be present in a sedimentsuspension containing 100mg of a sediment with an organic carbonfraction of 0.02.

(b) If the equilibrium concentration is equal to the saturation aqueoussolubility, what is the sediment concentration?

(c) If the aqueous solution in (a) also contains 0.05mol fractionof ethanol, what is the value of the sediment–water partitionconstant?

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4.132 The solubility of pyrene in an aqueous solution of polystyrene latex colloidsis given below:

[Latex], mg/L [Pyrene], mg/L

0 905 34010 44015 65020 880

Obtain the value of the partition constant for pyrene on latex col-loids. Assume that latex is 100% organic carbon and estimate the partitionconstant from all reasonable correlations. Compare your results andcomment.

4.142 In a field study, a nonadsorbing (conservative) tracer (chloride) wasinjected into a groundwater aquifer (unconfined and relatively homoge-neous) along with two adsorbing organic compounds (carbon tetrachlorideand tetrachloroethylene). A test well 5m downstream from the injectionwell was monitored for all three compounds. The average time of arrival(breakthrough time) at the monitoring well was 66 days for chloride ion,120 days for carbontetrachloride, and 217 days for tetrachloroethylene.

(a) Obtain from the above data the retardation factors for the sorbingcompounds.

(b) Obtain the field-determined partition constants. The soil organiccarbon is 0.02.

(c) ComparewithKd obtained using the correlation ofCurtis et al. (1994)given in text. Explain any difference.

4.153 The bioconcentration factor can be predicted solely from the aqueouscompatibility factor (measured as solubility). Derive an appropriate rela-tionship between the two factors. Howdoes it comparewith the predictionsusing Kow? Use pyrene and benzene as examples. For a fish that weighs1.5 lb, how much pyrene can be accumulated within its tissue?

4.162 It is desired to remove benzene from an aqueous waste stream at an indus-trial waste site that produces approximately 20,000 gallons of water perday with a pH of 5.3 and a benzene concentration of 2000μg/L. A batchadsorption system is to be used. A preliminary bench scale analysis wasconducted to get the adsorption data of benzene on two types of carbons(I and II) at pH 5.3. The data are given below:

Carbon Initial Aqueous Final EquilibriumDose (mg/L) Concentration (mg/L) Concentration in Water (mg/L)

I II96 19.8 14.2 12.5191 19.8 10.3 12.0573 19.8 6.2 11.0951 19.8 5.8 9.01910 19.8 4.0 5.0

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(i) Which isotherm (Freundlich or Langmuir) would better represent thedata?

(ii) What is the maximum adsorption capacity of carbon I and II?(iii) If the effluent water quality for benzene is 20μg/L, how much of

carbon would be required per day? Which carbon type would yourecommend?

4.172 The sediment in University Lake near Baton Rouge is known to be con-taminated with PCBs (predominantly trichlorobiphenyl) to an averageconcentration of 10μg/kg. The sediment has an foc of 0.04. Calculatethe PCB loading in a 1 lb catfish dwelling in the lake.Assume equilibriumbetween the various phases in the system.

4.182 Inorganic Hg (HgCl2, Hg(OH)2) can be deposited to a water body via wetand dry deposition. Consider the Henderson Lake, a small lake east ofLafayette, Louisiana, where an Hg advisory was issued by the LouisianaDepartment of Health and Hospitals in 1996. The lake is approximately7.2m in depth. Consider a global average atmospheric concentration ofHgCl2 = 0.1 ng/m3. Estimate the deposition rate of HgCl2 to the lake.An average annual rainfall of 0.1 cm/h can be assumed. Kaw for HgCl2 is2.9 × 10−8.

4.192 Estimate the fractional mass of a pesticide (aldrin) that is washed out byrainfall at an intensity of 1mm/h over an area 104 m2. Appendix 1 givesthe physicochemical data for aldrin.

4.203 A closed storage vessel is filled with water and ethanol (mole fraction0.002). The vessel is only 60% full. If equilibrium is established abovethe solution, what is the partial pressure of ethanol in the vapor? Doesthe vessel stand to explode? The OSHA explosion limit for ethanol is1000 ppmv.

4.213 Vapor, aerosol, and rainwater concentrations of a PCB (2,2′,4-isomer) inan urban area are given below:

Vapor: 100 pg/m3; aerosol: 0.3 pg/m3.

Estimate the total washout ratio for this compound at an ambient tem-perature of 25◦C. log Kow = 5.7, log P∗

i (atm) = −6.5. Compare thetheoretical and experimental washout ratios. The average surface area ofan aerosol is 1 × 10−6 cm2/cm3 air and ρi = 1.7 × 10−4 atm cm.

4.222 The concentration of sulfate in the atmosphere over oceans is−0.06μg/m3. If the water content of cloud is 1 g/m3, estimate the pHof cloud water. Assume that all the sulfate exists as sulfuric acid.

4.232 Based on the data in Table 4.12, estimate the percent gas-phase resis-tance for evaporation of the following chemicals from water: naphthalene,chloroform, PCB, and ammonia.

4.243 Below the water table, the groundwater aquifer is a closed system withrespect to exchange of CO2 with a separate vapor phase.Without an exter-nal source or sink for dissolved CO2 in groundwater (i.e., no carbonate)the dissolved carbon concentration should be constant. Assuming that theground water at the water table is in equilibrium with the atmosphere with

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CO2 at 10−3.5 atm partial pressure, construct a plot showing the changein CO2 with pH in the closed system.

4.252 What is the vapor pressure of aqueous solution droplets of the follow-ing diameters: 0.01, 0.05, 0.1, 0.5, and 1μm? The solution contains1 × 10−13 g of NaCl per particle at 298K. Plot the value of Pw/P∗

w asa function of droplet size.

4.263 PSBs are ubiquitous in the environment. The average air concen-tration in 1986 over Lake Superior for a PCB congener (2,3′,4,4′-tetrachlorobiphenyl) was measured as 38.7 pg/m3 and that in surfacewater was 32.1 pg/L. Obtain the magnitude and direction of air–waterexchange flux. Estimate the transfer coefficient using data in Table 4.12and Appendix 1.

4.272 Phthalate esters are used as plasticizers. Di-n-butyl phthalate (DBP) isfound both in marine waters and the atmosphere. The mean concentra-tions in the Gulf of Mexico are 94 ng/L in surface water and 0.3 ng/m3

in air. Estimate the direction and magnitude of the annual flux of DBP.The reported vapor pressure and aqueous solubility of DBP at 25◦C are1.4 × 10−5 mmHg and 3.25mg/L, respectively (Giam et al., 1978).

4.282 A facility operates a surface impoundment for oil field produced waterthat contains benzene. The permitted annual emission is 6000 kg. Thefacility operator makes sure that the impoundment is operated only duringthose atmospheric conditions of instability when there is maximum mix-ing and dispersal of benzene. Estimate the mean aqueous concentrationof benzene allowed in the impoundment. The area of the impoundmentis 104 m2.

4.292 Estimate the vapor pressure of the following compounds at 25◦C:

Compound Melting Point (◦C)

Chlordane 103CCl4 −23p,p′-DDT 108Diethyl phthalate −40.52-Methyl phenol 30.9

4.302 A sample of silica is coated with 0.1% of organic carbon. The silica hasa surface area of 20m2/g. Assume that only 10% of the area is occu-pied by organic carbon. Estimate the adsorption constant for pyrene onmodified silica. Use data from Table 4.17 for the mineral contribution toadsorption.

4.312 (i) Estimate the sorption constant of the following compoundson a soil that has an organic carbon content of 1%: pen-tachlorobenzene, phenanthrene, endrin, 2,2′-dichlorobiphenyl, andparathion.

(ii) Estimate the effect of (a) 10mg/L of DOC in water and (b) 0.1%methanol in water on the partition constant of phenanthrene on theabove soil.

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4.322 The following data are for the adsorption of a cation dodecylpyridiniumbromide on Borden sand:

log Wi (mol/kg) log Cwi (mol/L)

−4.8 −7.0−4.6 −6.5−4.3 −6.0−4.1 −5.5−3.8 −5.0−3.4 −4.5−3.1 −4.0

Determine which isotherm (linear or Freundlich) best fits the data.4.332 Contaminated sediment from a harbor is proposed to be dredged and

stored in an open confined disposal facility (CDF). The sediment concen-trations of Aroclor 1242 and 1254 are 687 and 446mg/kg, respectively.The sediment suspension in water is expected to be highest immediatelyafter loading the CDF. The highest suspended sediment concentrationexpected using the selected dredged head is 490mg/L. The area of theCDF is 1.3 × 105 m2. What is the expected air emission of Aroclor1242. Properties for Aroclor 1242 and 1254 should be obtained from theliterature.

4.342 On April 17, 1992, a high-volume sampler was used to obtain the follow-ing atmospheric data in Baton Rouge, Louisiana: Average air temperature71◦F, concentration of fluoranthene on particulates in air = 0.51 ng/m3,particulate concentration = 100μg/m3. Estimate the dry (vapor) depo-sition rate of fluoranthene. A deposition velocity of 0.1m/s can beassumed.

4.352 The mean concentration of formaldehyde in air is 0.4 ppb, whereas that inseawater is 40 nM.What is the direction and magnitude of flux? The parti-tion constant for formaldehyde between seawater and air can be estimatedfrom –log K ′

aw = −6.7 + 3069/T , where K ′aw is in mol/L atm.

4.362 What is the fraction of each of the following compounds partitioned fromthe atmosphere to a sugar maple leaf? Assume an ambient temperatureof 25◦C and a typical lipid content of 0.01 g/g. The air concentration ofeach compound can be assumed to be 10 ng/m3: naphthalene, phenan-threne, anthracene, and pyrene. Plot your results versus log Koa andcomment.

4.373 Sodium azide (NaN3) is used as a component of air bags in automobiles(=106 kg in 1995).When dissolved in water it gives rise to volatile hydra-zoic acid, HN3. It is a weak acid and the reaction HN3 � H+ + N−

3 hasa pKa value of 4.65 at 25◦C. The intrinsic Henry’s constant Kaw for theneutral species is 0.0034 at 25◦C (Betterton and Robinson, 1997).

(a) What is the apparent air–water partition constant at pH 7?(b) The enthalpy of solution is −31 kJ/mol. Calculate the air–water

partition constant at 4◦C.(c) In a wastewater at 25◦C, if the azide concentration is 0.1mM at pH

6.5, will the gas phase HN3 exceed the threshold limit value (TLV)of 0.1 ppmv?

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4.382 Aspill of 1,2-dichloroethane (DCE)occurred in a chemical plant producingpesticides. Because of the delay in cleanup some of the material seepedinto the groundwater flowing at a velocity of 1.5m/d. The soil at the sitehas an organic carbon content of 0.001, a porosity of 0.3, and a bulk densityof 1.4 g/cm3. The nearest community that receives drinking water from theaquifer is located 500m away in the direction of groundwater flow. Howlong will it take for the chemical to show up in the community drinkingwater? Assume no biodegradation of DCA.

4.393 Kaw values for p,p′-DDT (a pesticide) at various temperatures wereobtained for deionized water and saline water (Cetin et al., 2006). Fromthe data calculate the following:

(i) ΔH0a→w, (ii) ΔS

0a→w, and (iii) Setschenow constant for salting out.

Kaw/[−]

T (◦C) Deionized Water Saline Water (0.5 M NaCl)

5 0.19 ± 0.04 1.1715 0.52 1.9 ± 0.6320 0.69 ± 0.15 —25 0.97 ± 0.27 3.4 ± 0.6435 2.8 ± 0.39 5.5

4.402 Hexabromobenzene (HBB) is found along with flame-retardant materialssuch as in carpets, bedding, insulation, and furniture accessories. It is foundas a contaminant in soils and sediments in ourwaterways. It has a structuralformula of C6Br6, a molecular weight of 557.52, and a melting point of327◦C. (a) Determine the log Kow of HBB using a group contributionmethod and a derivative method. (b) From the data in (a) determine theaqueous solubility of BHH expressed in mol/L. (c) From the above datadetermine the sediment–water partition constant Ksw using an appropriatecorrelation. The organic fraction of the soil is 0.05; neglect the mineralcontribution to adsorption. (d) Ten grams of soil with 10μg/kg of HBBare mixed with 100mL of water and agitated for 24 h. The solution wasallowed to reach equilibrium and the solid fraction separated. It was foundto contain HBB at a concentration of 5μg/kg. What is the concentrationof HBB in the water that was separated?

4.413 2,3-Dichlorophenol is a contaminant resulting from wood and pulp indus-try operations. It has a molecular weight of 163, a melting point of 45◦C,and a boiling point of 210◦C. (a) Obtain the octanol–water partition con-stant of the compound; (b) from (a) obtain the aqueous solubility of thecompound; (c) obtain the vapor pressure of the compound from an appro-priate correlation; (d) using data from (b) and (c) obtain the Henry’sconstant, expressed in dimensionless molar ratio; and (e) a wastewaterstream of 100m2 and a depth of 1m is contaminated with the compoundat a concentration of 1μg/L. A measurement of flux to air showed 10μgin a chemical trap obtained in 1 h of sampling. What is the concentrationof the chemical in ambient air above the stream? Note that the individualwater- and air-side mass transfer coefficients for the chemical are 1 ×10−5 m/s and 0.005m/s, respectively.

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4.422 Consider 100m3 of lake water in equilibrium with 100m3 of air. A 1 lbcatfish resides in the lake water. The air above the lake is found to havetraces of a pesticide (lindane) at a concentration of 1μg/m3. Determinethe concentration of the pesticide in the catfish.

4.433 Ethylbenzene is an aromatic hydrocarbon that is commonly found in con-taminated sediments and waters. Its molecular weight is 106, log Kow is3.15, vapor pressure is 1.3 kPa at 25◦C, aqueous solubility is 0.0016mol/Lat 25◦C, enthalpy of vaporization is 2.1 kJ/mol, and enthalpy of solutionis 46.2 kJ/mol. (a) Find its vapor pressure at 30◦C. (b) Find its aqueoussolubility at 30◦C. (c) From the above data find its Henry’s constant at30◦C and express in kPa/L/mol. (d) If the air above a lagoon has ethylben-zene at a partial pressure of 0.1 kPa at 30◦C in equilibrium with the water,what is the aqueous concentration at that temperature? (e) If the water inthe lagoon is overlying the sediment in equilibrium with it at 30◦C, and ifthe sediment organic carbon fraction is 0.02, what is the concentration ofethylbenzene in the sediment (answer in mg/kg)?

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5 Concepts from ChemicalReaction Kinetics

Environmental systems are dynamic in nature. Changes with time are important inunderstanding the F&T of chemicals and also in process design for waste treatment.The time scales of change in natural systems (weathering of rocks, atmosphericreactions, biological, or thermal-induced reactions) range from a few femtoseconds(10−15 s) to as large as billions of years. Thermodynamics does not deal withthe questions regarding time-varying properties in environmental systems. Chem-ical kinetics plays the key role in determining the time-dependent behavior ofenvironmental systems.Chemical kinetics is the study of changes in chemical properties with time due to

reaction in a system. For example, principles from chemical kinetics can be used toanswer the following questions:

1. At what rate does a pollutant disappear from or transform in an environmentalcompartment or a waste treatment system?

2. What is the concentration of the pollutant in a given compartment at a giventime?

3. At what rate will a chemical move between different environmental compart-ments, and how fast will it be transferred between phases in a waste treatmentsystem?

To answer questions (1) and (2), rates and mechanisms of reactions are needed. How-ever, question (3) requires knowledge of kinetic data as well as momentum and masstransfer information between different phases.

In this chapter, we will discuss concepts from chemical kinetics that form the basisfor the discussion of rates and mechanisms in environmental engineering. Chemicalkinetics covers a broad range of topics.At its simplest level, it involves empirical stud-ies of the effects of variables (temperature, concentration, and pressure) on variousreactions in the environment. At a slightly advanced level, it involves elucidation ofreaction mechanisms.At its most advanced level, it involves the use of powerful toolsfrom statistical and quantum mechanics to understand the molecular rearrangementsaccompanying a chemical reaction. We shall not deal with the last issue since it fallsbeyond the scope of this book. In general, the applications of chemical kinetics in envi-ronmental engineering are limited to the following: (i) experimentally establishing therelationship between concentration, temperature, and pressure in chemical reactions;(ii) using the empirical laws to arrive at the reaction mechanism; and (iii) using therate data in models for predicting the F&T of pollutants in the natural environment,and in process models for designing waste treatment systems.

189

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As mentioned in Chapter 1, processes in natural systems are generally driven bynonequilibrium conditions. True chemical equilibrium in natural environmental sys-tems is rare considering the complex and transient nature of energy andmass transportin natural systems. Although only local phenomena can be affected in some cases,they are coupled with global phenomena; hence any minor disturbance is easily prop-agated and alters the rate of approach to equilibrium (Pankow and Morgan, 1981). Ifthe rate of input of a compound equals its rate of dissipation in a system, it is saidto be at steady state. For most natural systems this occurs for long periods of timeinterrupted by periodic offsets in system inflows and outflows. Such a behavior ischaracterized as quasi-steady-state. If the concentration of a compound changes con-tinuously (either decreases or increases) with time due to reactions and/or continuouschanges in inflows and outflows, the system is said to show unsteady-state behavior.The time rate of change of concentrations of metals and organic compounds in naturalsystems can be ascertained by applying the appropriate equations for one or the otherof the above-mentioned states in environmental models.

5.1 PROGRESS TOWARD EQUILIBRIUM IN A CHEMICALREACTION

A chemical reaction is said to reach equilibrium if there is no perceptible change withtime for reactant and product concentrations. The process can then be characterizedby a unique parameter called the equilibrium constant (Keq) for the reaction. For ageneral reaction represented by the following stoichiometric equation,

aA + bB � xX + yY, (5.1)

the equilibrium constant is defined by

Keq = axXayY

aaAabB

, (5.2)

where a denotes activity. In general, a double arrow indicates a reversible reaction atequilibrium, whereas a single arrow indicates an irreversible reaction proceeding inthe indicated direction. The general stoichiometric relation that describes a chemicalreaction such as given in Equation 5.1 above is

∑i

νi Mi = 0, (5.3)

where νi is the stoichiometric coefficient of the ith species and Mi is the molecularweight of i. Note the convention that νi is positive for products and negative forreactants. At constant T and P, the free energy change due to a reaction involvingchanges dni in each species is

dG =∑i

μi dni. (5.4)

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Concepts from Chemical Reaction Kinetics 191

We can define a term called the extent of the reaction, ξ, as defined by

ni = n0i + νiξ, (5.5)

where n0i is the number of moles of i when ξ = 0 (i.e., the initial condition). Note thatwhen ξ = 1, all reactants have converted to products. From Equation 5.5 we have

dni = νi dξ. (5.6)

Therefore, the change in free energy for the reaction is given by

dG =∑i

μiνi dξ. (5.7)

As noted in Chapter 2, the quantity Σiνiμi is called the free energy change of thereaction, ΔG:

ΔG =∑i

νiμi. (5.8)

Thus, we have the relation

ΔG = dG

dξ. (5.9)

The free energy change of a reaction is the rate of change of Gibbs free energy withthe extent of the reaction. Figure 5.1 shows a typical change in free energy with theextent of reaction. Note that at equilibrium, dG/dξ = 0, that is, ΔG = 0, as requiredby the laws of thermodynamics. The entropy production is

dS = −ΔGT

dξ. (5.10)

Since μi = μ0i + RT ln ai,

ΔG =∑i

νiμ0i + RT

∑i

νi ln ai. (5.11)

This can also be written as

ΔG = ΔG0 + RT ln∏i

(ai)νi . (5.12)

For the general reaction (Equation 5.1) we have

∏i

(ai)νi = axXa

yY

aaAabB

= K . (5.13)

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Equilibrium position

= 0

0 1

G

ξ

dGdξ

FIGURE 5.1 Gibbs function variation with the extent of reaction.

Hence,

ΔG = ΔG0 + RT lnK . (5.14)

Note that only at equilibrium (i.e., ΔG = 0) does K equal Keq, at which point wehave

ΔG0 = −RT lnKeq. (5.15)

K is called the reaction quotient and Keq the equilibrium constant. Therefore, we canwrite

ΔG = dG

dξ= RT ln

(K

Keq

). (5.16)

The approach to equilibrium for a chemical reaction is measured by the above equa-tion. The activity of a compound is given by a = γ[i], where [i] represents theconcentration and γ is the activity coefficient. Note that in this chapter [i] representsthe molar concentration Ci (mol/L) for reactions in solution or the partial pressuresPi (atm or kPa) for reactions in the gas phase. Hence,

Keq = [X]x[Y]y[A]a[B]b

γ xXγyY

γaAγbB

. (5.17)

From the relation for ΔG or dG/dξ, one can conclude that if ΔG = dG/dξ < 0,K < Keq, the reaction will be spontaneous and will proceed from left to right in Equa-tion 5.1, whereas ifΔG = dG/dξ > 0, K > Keq, the reaction will be spontaneous in

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the reverse direction. The driving force for the chemical change is thus ΔG. We alsoknow that ΔG = ΔH − TΔS and for standard conditions ΔG0 = ΔH0 − TΔS0. Ifa reaction has attained equilibrium,ΔG = 0, that is,ΔH = TΔS. If |ΔH| > |TΔS|,ΔG > 0, whereas if |ΔH| < |TΔS|,ΔG < 0. However, both the sign and magnitudeof ΔH and ΔS terms together determine the value of ΔG and hence the spontaneityof a reaction.

EXAMPLE 5.1 DEFINITIONS OF EXTENT OF REACTIONAND FRACTIONAL

CONVERSION

For a reaction,A −→ B + C, (5.18)

we define the extent of reaction ξ such that

dξ = dnA

−1= dnB

1= dnC

1. (5.19)

Generalizing, we have

ξ = ni − n0iνi

, (5.20)

where n0i represents the initial conditions in the closed system.We can also define a fractional conversion χ such that it is 0 (no reaction) and 1

(reaction completed) through the equation

nA = n0A(1 − χ). (5.21)

Note that 0 < χ < 1 only if A is chosen as the limiting reactant, which disappearscompletely when χ = 1. For any species in the closed system we can then write thefollowing equation:

n0Aχ = ni − n0iνi

. (5.22)

5.2 REACTION RATE, ORDER, AND RATE CONSTANT

Since ni = n0i + νiξ, we havedξ = 1

νi· dni. (5.23)

As long as the above equation represents a single reaction, it is immaterial as to thereferenced species i. However, if the reaction occurs in a series of steps, then the rateat which one species is consumed will be different from the rate of production ofanother, and hence the rate has to be specified in concert with the species it refersto. For example, consider the reaction H2 + (1/2)O2 →H2O, for which the extent ofreaction is

dξ = dnH2O

(1)= dnO2

−(1/2)= dnH2

(−1). (5.24)

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Note that the appropriate use of sign for νi assures that ξ is always positive. Thus therate can be expressed either as the decrease in moles of the reactant with time or asthe increase in moles of product with time. The rate of the reaction is related to theextent of the reaction by

r = 1

V

dξ

dt= 1

V

1

νi

dnidt

= 1

νi

d[i]dt

, (5.25)

where [i] = ni/V is the concentration. It is important to bear in mind that, since theabove definitions encompass all macroscopic changes in concentration of a givenspecies with time, the rate of a reaction at equilibrium should be zero. Thus theentire realm of chemical kinetics is geared toward understanding how fast a systemapproaches equilibrium.

In the study of chemical reaction kinetics, the first step is obtaining the functionalrelationship between the rate of change in concentration of one of the species andthe concentration of all other species involved. Such a relationship is called a rateequation. It is obtained through a series of experiments designed to study the effects ofeach species concentration on the reaction rate. For a general stoichiometric equationof the type aA + bB → cC + dD, the empirically derived rate expression is written as

r = 1

(−a)d[A]

dt= k[A]α[B]β[C]γ[D]δ, (5.26)

where the rate r is expressed as the disappearance of A. The rate is always a positivequantity and has units of concentration per unit time (e.g., mol/dm3/s or mol/L/s). Forgas-phase reactions, partial pressure replaces concentration and the rate is in pressureper unit time (e.g., kPa/s or atm/s).

The order of the above reaction is n = α+ β+ γ+ δ with the reaction beingtermedαth order inA, βth order in B, γth order in C, and so on. Note that stoichiometryand order are not the same. The order of a reaction is more complicated to ascertainif the rate equation involves concentrations of A, B, and so on in the denominator aswell. Such situations are encountered if a reaction proceeds in several steps whereonly some of the species take part in each step. The order should be distinguished fromthe total number of molecules involved in the reaction; this is called the molecularityof the reaction. The constant k in the above rate expression is called the specific rateor rate constant. It is numerically the same as the rate if all reactants are present atunit concentrations. The unit of k will depend on the concentration units. Generally,it has dimensions of (concentration)(1−α−β−γ−δ)(time)−1.

5.3 KINETIC RATE LAWS

Any experiment designed to obtain a rate expression will require that one follows thechange in concentration of a species with time. Starting from an initial (t = 0) valueof [i]0 the concentration will decay with time to its equilibrium value [i]∞. This iswhat is represented by the empirical rate expression given earlier. The experimentsare designed to obtain the rate constant and the order of the reaction. This is achievedvia the following methods.

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5.3.1 ISOLATION METHOD

Consider a reaction involving two reactants A and B. Let the rate be r = k[A][B],where the overall order of the reaction is two. If, however, [B] � [A], throughoutthe reaction, [B] remains constant in relation to [A]. Hence k[B] ∼ constant(k′) andthe rate is r = k′[A]. This is called the pseudo-first-order rate. If the rate were morecomplicated such as, for example,

r = k1[A]2k2[B] + k3

, (5.27)

we have r = k′[A]2, where k′ is called the pseudo-second-order rate. The dependenceof r on each reactant can be isolated in turn to obtain the overall rate law.

5.3.2 INITIAL RATE METHOD

This is used in conjunction with the isolationmethod described above. The velocity orrate of an nth-order reaction withA isolatedmay be generally expressed as r = k[A]n.Hence we have log r = log k + n log [A]. The slope of the plot of log r versus log [A]will give n. This is most conveniently accomplished by measuring initial rate atdifferent initial concentrations. In Figure 5.2a, the slope of [A] versus time as t → 0gives the initial rate, r0. The log of initial rate is then plotted versus the log of initialconcentration to obtain the slope, n (Figure 5.2b).

5.3.3 INTEGRATED RATE LAWS

The most common method of obtaining the order and rate of a reaction is the methodof integrated rate laws.The initial rates do not often portray the full rate law, especially

t

Initial rate

[A]

(a) (b)

log k

nlog

r 0

log [A]0

FIGURE 5.2 (a) Concentration versus time for various [A]0 values. (b) Logarithm of initialrate versus logarithm of initial concentrations. The slope is the order of the reaction and theintercept gives the logarithm of the rate constant for the reaction.

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if the reactions are complex andoccur in several steps. Figure 5.3 represents the changein concentrations of reactantA and product B for a first-order reaction such asA → B.The derivative −d[A]/dt is the rate of disappearance ofA with time (this is also equalto d[B]/dt at any time t). Since the reaction is first order in A, we have

rA = 1

νA

d[A]dt

= −d[A]dt

= k[A], (5.28)

since νA = −1. The rate is directly proportional to [A]. Since at t = 0, [A] = [A]0, aconstant, one can integrate the above expression to get

ln

( [A][A]0

)= −kt. (5.29)

If the reaction is nth order in A, that is, nA → B, the integrated rate law is

(1

n− 1

)·(

1

[A]n−1− 1

[A]n−10

)= kt. (5.30)

Thus for a first-order reaction a plot of ln[A]/[A]0 versus twill give k, the rate constant.Similar plots can be made for other values of n. By finding the most appropriateintegrated rate expression to fit a given data, both n and k can be obtained.

If the reaction involves two or more components, the expression will be different.For example, if a reaction is second order (first order in A and first order in B),A + B → products, then −d[A]/dt = k[A][B]. If x is the concentration of A that hasreacted, then [A] = [A]0 − x and [B] = [B]0 − x. Hence

−d[A]dt

= dx

dt= k([A]0 − x)([B]0 − x). (5.31)

[B]

[A] = [A0] – [B]

[A0]

[I]

t / min

FIGURE 5.3 Change in concentrations of A and B for the first-order reaction A → B.

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The integrated rate law is

(1

[A]0 − [B]0)ln

( [B]0([A]0 − x)

[A]0([B]0 − x)

)= kt. (5.32)

A plot of the term on the left-hand side versus t would give k as the slope.Physical chemists have investigated a large variety of possible kinetic rate expres-

sions over the years and the integrated rate laws have been tabulated in the literature(Laidler, 1965; Moore and Pearson, 1981). Table 5.1 lists some of the rate laws mostfrequently encountered in environmental engineering.

An important parameter that is useful in analyzing rate data is the half-life of areactant, t1/2. This is defined as the time required for the conversion of one half ofthe reactant to products. For a first-order reaction this is (ln 2)/k and is independentof [A]0. For a second-order reaction the half-life is 1/(k[A]0) and is inversely propor-tional to [A]0. Similarly, for all higher-order reactions appropriate half-lives can bedetermined.

TABLE 5.1Integrated Rate Laws Encountered in Environmental Systems

Reaction Type Order Rate Law t1/2

A → B + · · · 0kt = x [A]0/2k1kt = ln([A]0/[A]0 − x) (ln 2)/k

≥2kt = {1/n− 1}{1/([A]0 − x)n−1 − 1/([A]n−10 } (2n−1 − 1)/(n− 1)k[A]n−1

0

A + B → C + D + · · · kt = 1

[B]0 − [A]0 · ln[ [A]0([B]0 − x)

[B]0([A]0 − x)

]1/k[B]0

A � B kbt = [A]eq

[A]0 · ln( [B]eq

[A] − [A]eq

); kf = kbKeq

A + B � C + D kf t = [B]eq

2[A]0([A]0 − [B]eq)· ln[ [B]([A]0 − 2[B]eq) + [A]0[B]eq

[A]0([B]eq − [B])]

Ak1−−→ B

k2−→ X [A] = [A]0e−k1t

[B] = [A]0k1k2 − k1

(e−k1t − e−k2t

)

[C] = [A]0k2 − k1

[k2(1 − e−k1t) − k1(1 − e−k2t)

]

Ak1−−→ C [A] = [A]0e−(k1+k2)t

Ak2−−→ D [C] = k1[A]0

k1 + k2

[1 − e−(k1+k2)t

]

[D] = k2[A]0k1 + k2

[1 − e−(k1+k2)t

]

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EXAMPLE 5.2 REACTION RATE CONSTANT

The rate of loss of a VOC (chloroform) from water in an open beaker is said to be afirst-order process. The concentration in water was measured at various times as givenbelow:

t (min) [A]/[A]0

0 120 0.540 0.2360 0.1180 0.05100 0.03

Find the rate constant for the loss of chloroform from water.First obtain a plot of ln [A]/[A]0 versus t as in Figure 5.4. The slope of the plot

is 0.035min−1 with a correlation coefficient of 0.997. Hence the rate constant is0.035min−1.

–4

–3

–2

–1

0

1

0

y = –0.036153x

ln([A

]/[A

] 0)

t/min20 40 60 80 100 120

FIGURE 5.4 Experimental points and data fit to a first-order reaction rate law.

5.3.3.1 Reversible Reactions

The reaction rate laws described above discount the possibility of reverse reactions andwill fail to give the overall rate of a process near equilibrium conditions. In the naturalenvironment such reactions are common. When the product concentration becomessignificant, the reverse reaction will also become significant near equilibrium. This

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is in keeping with the principle of microscopic reversibility enunciated by Tolman(1927), which states that at equilibrium the rate of the forward reaction is the sameas that of the backward reaction. As an example, let us choose the reaction

Akf�kb

B, (5.33)

where the forward and backward reactions are both first order. The rate of the forwardreaction is rf = kf [A] and that of the backward reaction is rb = kb[B]. The net rate ofchange in [A] is that due to the decrease inA by the forward reaction and the increasein the same by the reverse reaction. Thus

rA = −d[A]dt

= kf [A] − kb[B]. (5.34)

If [A]0 is the initial concentration of A, then by the mass conservation principle[A]0 = [A] + [B] at all times (t > 0). Therefore, we have

d[A]dt

= −(kf + kb)[A] + kb[A]0. (5.35)

This is a first-order ordinary differential equation, which can be easily solved to obtain

[A] = [A]0 ·{kb + kf e−(kf+kb)t

kb + kf

}. (5.36)

As t → ∞, [A] → [A]eq and [B]→ [B]′eq.

[A]eq = [A]0 kb

kf + kb,

[B]eq = [A] − [A]eq = [A]0 kf

kf + kb.

(5.37)

The ratio [B]eq/[A]eq is the equilibrium constant of the reaction, Keq. It is importantto note that

Keq = [B]eq

[A]eq= kf

kb. (5.38)

The connection between thermodynamics and kinetics becomes apparent. In practice,for most environmental processes, if one of the rate constants is known, then the othercan be inferred from the equilibrium constant. It should be noted that whereas theratio kf/kb describes the final equilibrium position, the sum (kf + kb) determines howfast equilibrium is established.

An example of a reversible reaction is the exchange of compounds between soiland water. Previously, we showed that this equilibrium is characterized by a partitioncoefficient Ksw. Consider the transfer as a reversible reaction

Awaterkf�kb

Asoil. (5.39)

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The equilibrium constant is

Keq = kf

kb= [A]s

[A]w= WA

CAw= Ksw. (5.40)

Similar analogies also apply to air–soil, aerosol–air, and biota–water partition con-stants. Thus, an equilibrium partition constant is a ratio of the forward and backwardrate constants for the processes. A large value of Ksw implies either a large value ofkf or a small value of kb.

EXAMPLE 5.3 REVERSIBLE REACTION

A reaction A � B is said to occur with a forward rate constant, kf = 0.1 h−1. Theconcentration of A monitored with time is given:

t (h) [A] (mM)

0 11 0.95 0.6510 0.4815 0.40100 0.33500 0.33

Find Keq and kb.As tΠ4, [A]eq = [A]0 (kb/kf + kb). Since [A]eq = 0.33, kb/kf + kb = 0.33/1 =

0.33. Hence kb = 0.05 h−1 and Keq = 0.1/0.05 = 2. [B]eq = [A]0 (kf/kf + kb) =0.67. Figure 5.5 plots the change in [A] and [B] with t.

–0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 100 200 300 400 500 600 700

[A]/mM[B]/mM

[A]/m

M

t /hr

FIGURE 5.5 Change in concentrations of A and B for a reversible reaction A � B.

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EXAMPLE 5.4 EQUILIBRIUM CONSTANT FROM STANDARD THERMODYNAMIC

DATA

Find the equilibrium constant at 298K for the reaction: H2S(g)⇔H2S(aq) fromthe standard free energy of formation for the compounds

G0f (kJ/mol)

H2S(g) −33.6H2S(aq) −27.9

The overall free energy change for the reaction is the difference between the G0f

of the product and reactant. ΔG0f = −27.9 + 33.6 = +5.7 kJ/mol. Hence, Keq = exp

[−5.7/(8.314 × 10−3)(298)] = 0.1.Note thatKeq = kf/kb = [H2S]w/PH2S = 1/K ′aw

as defined previously in Chapter 3. Hence the Henry’s constant K ′aw = 9.98 L atm/mol.

FromAppendix 1, the value is 8.3 L atm/mol.

5.3.3.2 Series Reactions and Steady-State Approximation

A large number of reactions in the environment occur either in series or in paral-lel. Reaction in series is of particular relevance to us since it introduces both theconcepts of steady state and rate-determining steps that are of importance in envi-ronmental chemical kinetics. Let us consider the conversion of A to C through anintermediate B:

Ak1−→ B

k2−→ C. (5.41)

The rates of production of A, B, and C are given by

d[A]dt

= −k1[A],d[B]

dt= k1[A] − k2[B], (5.42)

d[C]dt

= k2[B].

Throughout the reaction, we have mass conservation such that [A] + [B] + [C] =[A]0. The initial conditions are at t = 0, [A] = [A]0, and [B] = [C] = 0. Solvingthe equations in succession and making use of the above mass balance and initialconditions one obtains, after some manipulation, the values of [A], [B], and [C] asgiven in Table 5.1. Figure 5.6 is a plot of [A], [B], and [C] with time for repre-sentative values of k1 = 1min−1 and k2 = 0.2min−1. We note that the value of [A]decreases to zero in an exponential manner. The value of [B] goes through a max-imum and then falls off to zero. The value of [C], however, shows a slow initial

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0.2

0

0.4

0.6

0.8

1

1.2[A]/[A0][B]/[A0][C]/[A0]

[i]/[A

0]

t/min0 2 4 6 8 10

FIGURE 5.6 Change in concentrations of A, B, and C with time for a series reaction A →B → C.

growth (which is termed the induction period) followed by an exponential increaseto [A]0.

In those cases where there are several intermediates involved in a reaction, suchas occurs in most chemical reactions in air and water environments, the derivation ofthe overall rate expression will not be quite as straightforward as described above.The intermediate (e.g., B above) is necessarily of low concentration and is assumedto be constant during the reaction. This is called the pseudo-steady-state approxi-mation (PSSA). It allows us to set d[B]/dt to zero. Thus, at pseudo-steady-state wecan obtain

−k2[B∗] + k1[A] = 0,

[B∗] = k1k2

[A] = k1k2

[A]0e−k1t .(5.43)

Hence

[C] = [A]0(1 − e−k1t). (5.44)

Comparing with the exact equations for [B] and [C] given in Table 5.1, we observethat the exact solutions approximate the steady-state solution only if k2 � k:1 in otherwords, when the reactivity of B is so large that it has little time to accumulate. Thedifference between the exact and approximate solutions can be used to estimate thedeparture from steady state.

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EXAMPLE 5.5 SERIES REACTIONS

The consumption of oxygen (oxygen deficit) in natural streams occurs due to biologicaloxidation of organic matter. This is called the biochemical oxygen demand (BOD). Theoxygen deficit in water is alleviated by dissolution of oxygen from the atmosphere.These two processes can be characterized by a series reaction of the form

organic matterk1−−−−−→

oxidationoxygen deficit

k2−−−−−−−→oxygenation

oxygen restoration,

Ak1−→ B

k2−→ C.

(5.45)

In this case, the initial conditions are somewhat different from what was discussedearlier.

Here at t = 0, [A] = [A]0, [B] = [B]0, and [C] = [C]0. The solution to the problemis given by

[A] = [A]0e−k1t ,

[B] = k1k2 − k1

[A]0(e−k1t − e−k2t) + [B]0e−k2t , (5.46)

[C] = [C]0{1 − (k2e−k1t + k1e−k2t)

(k2 − k1)

}+ [B]0(1 − e−k2t) + [C]0.

Given a value of k1 = 0.1 d−1, k2 = 0.5 d−1, initial organic matter concentration[A]0 = 10mg/L, and initial oxygen deficit [B]0 = 3mg/L, after 24 h (1 day), the oxygendeficit will be [B] = (0.1)(10/0.4)(e−0.1 − e−0.5) + (3)e−0.5 = 2.56mg/L.

A frequently encountered reaction type in environmental engineering is one inwhich a pre-equilibrium step precedes the product formation.The steady-state conceptis particularly useful in analyzing such a reaction.

Akf1�kb1

Bk2−→ C. (5.47)

Most enzyme reactions follow this scheme. It is also of interest in many homogeneousand heterogeneous reactions (in both the soil and sediment environments), that is,those that occur at interfaces. For the above reaction

d[B]dt

= kf1[A] − kb1[B] − k2[B]. (5.48)

Using the pseudo-steady-state approximation we can set d[B]/dt = 0. Hence

[B∗] =(

kf1

kb1 + k2

)[A]. (5.49)

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For species A we obtain

d[A]dt

= −kf1[A] + kb1[B∗] = −(

k2kf1

kb2 + k2

)[A] (5.50)

and for species C we have

d[C]dt

= k2[B∗] =(

k2kf1

kb1 + k2

)[A]. (5.51)

These expressions can be readily integrated with the appropriate boundary conditionsto obtain the concentrations of species A and C. If kb1 � k2,

d[C]dt

= k2Keq[A], (5.52)

where Keq = kf1/kb1 is the equilibrium constant for the first step in the reaction. Thisoccurs if the intermediate B formed is converted to A much more rapidly than to C.The rate of the reaction is then controlled by the value of k2. Thus B → C is said to bethe rate-determining step.Another situation is encountered if k2 � kb1, in which cased[C]/dt = kf1[A]. This happens if the intermediate is rapidly converted to C. Thenkf1 determines the rate. The rate-determining step is then said to be the equilibriumreaction A � B.

The following are illustrations of how the concepts of integrated rate laws can beused to analyze particular environmental reaction schemes. An example from waterchemistry and another one from air chemistry are chosen.

EXAMPLE 5.6 SOLUTION OF INORGANIC GASES IN WATER

A reaction of environmental relevance is the dissolution of an inorganic gas (e.g., CO2)in water. The reaction proceeds in steps. The important step is the hydration of CO2followed by dissolution into HCO−

3 species in water. Stumm and Morgan (1996) andButler (1982) have analyzed this reaction, and we shall adopt their approach here. Inanalyzing these reactions, we shall consider water to be in excess such that its concen-tration does not make any contribution toward the overall rate. The overall hydrationreaction can be written as

CO2(aq) + H2Okf�kb

HCO−3 (aq) + H+(aq), (5.53)

with kf = 0.03 s−1 and kb = 7 × 104 mol/L/s. Since H2O concentration is constant, we

shall consider the functional reaction to be of the formAkf�kb

B + C, whereA represents

CO2, B represents HCO−3 , and C represents H+. Laidler (1965) provides the integrated

rate law for the above reaction:( [B]eq

2[A]0 − [B]eq

)ln

{ [A]0[B]eq + [B]([A]0 − [B]eq)

[A]0([B]eq − [B])}

= kf t, (5.54)

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where [B]eq is the concentration of B at equilibrium. If the pH of a sample ofwater is 5.7, then [H+] = [C]eq = 2 × 10−6 mol/L. By stoichiometry [HCO−

3 ]eq =[B]eq = [C]eq = 2 × 10−6 mol/L. If the closed system considered has an initial CO2 of5 × 10−5 mol/L, then [A]0 = [CO2(aq)] = 5 × 10−5 mol/L. Then we have [B] = 2 ×10−6[(e1.5t − 1)/(e1.5t − 0.96)] and [A] = [A]0 − [B] = 5 × 10−5 − [B]. If the equi-librium pH is 5, then [B]eq = 1 × 10−5 mol/L, and hence [B] = 1 × 10−5[(e0.27t −1)/(e0.27t − 0.8)] and [A] = 5 × 10−5 − [B]. Figure 5.7 gives the concentration ofaqueous CO2 with time as it is being converted to HCO−

3 in the system. The func-tional dependence on pH is shown. Equilibrium values of CO2(aq) is pH dependentand reaches 48μM in ≈ 0.7 s at a pH of 5.7 and 40μM in ≈14 s at a pH of 5. If thefinal equilibrium pH is to increase, more CO2(aq) has to be consumed and hence theconcentration falls to a lower equilibrium value.

In the case of SO2 solution in water, we have

SO2(aq) + H2Okf�kb

HSO−3 (aq) + H+(aq), (5.55)

where kf = 3.4 × 10−6 s−1 and kb = 2 × 108 mol/L/s. Notice first of all that kf in thiscase is much larger than for CO2, and hence a virtually instantaneous reaction can beexpected. Let the initial SO2 concentration be 5 × 10−5 mol/L. A similar analysis asabove for CO2 can be carried out to determine the approach to equilibrium for SO2.Figure 5.8 displays the result at two pH values of 5.7 and 5.0. The striking differencesin time to equilibrium from those for CO2 dissolution reactions are evident. At a pH of5.7 the equilibrium value is reached in ≈6 ns, whereas at a pH of 5.0 the characteristictime is ≈0.1 μ s.

4 10–5

4.2 10–5

4.4 10–5

4.6 10–5

4.8 10–5

5 10–5

5.2 10–5

0 5 10 15 20 25

pH = 5.7pH = 5.0[C

O2]

/mol

/L

t/s

FIGURE 5.7 Kinetics of solution of CO2 in water at different pH values.

continued

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4 10–52.5 10–82 10–81.5 10–81 10–85 10–9

4.2 10–5

4.4 10–5

4.6 10–5

4.8 10–5

5 10–5

5.2 10–5

0

pH = 5.7pH = 5.0[S

O2]

/mol

/L

t/s

FIGURE 5.8 Kinetics of solution of SO2 in water at different pH values.

EXAMPLE 5.7 ATMOSPHERIC CHEMICAL REACTIONS

A large variety of reactions between organic molecules in the atmosphere occur throughmediation by N2 or O2−, which are the dominant species in ambient air. If A and Brepresent two reactants and Z represents either N2 or O2, then the general reactionscheme consists of the following steps:

A + Bk1�k−1

A − −B,

A − −B + Zk2−→ AB + Z.

(5.56)

The above is an example of a series reaction with a pre-equilibrium step discussedearlier. The method of solution is similar. A − −B represents an excited state of theABspecies, which is the final product. Typically these excited intermediates are producedby photo or thermal excitation. This short-lived intermediate transfers its energy to Z(N2 or O2) to form the stable AB complex. Thus the overall reaction scheme is A +B + Z

k′−→AB + Z. Each step in the reaction above is called an elementary reaction.Complex reactions are composed of many such elementary reactions.

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The rate of disappearance of each species is given below:

− d[A]dt

= k1[A][B] − k−1[A − −B],d[A − −B]

dt= k1[A][B] − k−1[A − −B] − k2[A − −B][Z], (5.57)

d[AB]dt

= k2[A − −B][Z].

To simplify the analysis we make use of the concept of steady state for [A − −B].Thus, d[A − −B]/dt = 0 and hence [A − −B]∗ = k1[A][B]/(k−1 + k2[Z]). Thus wehave the following differential equation for [A − −B]:

d[A − −B]dt

= k1k2[A][B]k−1 + k2[Z] [Z] = k′[A][B], (5.58)

with k′ = k1k2[Z]/(k−1 + k2[Z]) is a constant since [Z] is in excess and varies little.As [Z]→0, we have the low-pressure limit, k′′0 = (k1k2/k−1)[Z] = k′0[Z]. In the high-pressure limit, [Z] is very large, and k′4 = k1 and is independent of [Z]. From k′0 and k′4one can obtain k′ = k′′0 [1 + (k′′0/k′∞)]−1. Table 5.2 lists the rate constants for a typicalatmospheric chemical reaction. Note that the values are a factor of two lower understratospheric conditions. The general solution to the ordinary differential equation aboveis the same as that for reaction A + B → products as in Table 5.1. From the integratedrate law one can obtain the concentration–time profile for species A in the atmosphere.

TABLE 5.2Low- and High-Temperature Limiting k Values for the Atmospheric

Reaction OH + SO2Z−→ HOSO2

[Z](molecule/ k ′

0/(cm6/ k ′∞/(cm3/ k ′/(cm3/T (K) P (Torr) s/cm3) molecule2/s) molecule/s) molecule/s)

300(troposphere)

760 2.4 × 1019 (3.0 ± 1.5) × 10−31 (2.0 ± 1.5) × 10−12 1.1 × 10−12

−219(stratosphere)

−39 1.7 × 1018 8.7 × 10−31 2.0 × 10−12 5.2 × 10−13

Source: Finlayson-Pitts, B.J. and Pitts, J.N. 1986. Atmospheric Chemistry. NewYork, NY:John Wiley & Sons, Inc.

5.3.4 PARALLEL REACTIONS

In environmental reactions there also exist caseswhere amolecule can simultaneouslyparticipate in several reactions. For example,

Ak1−→ C,

Ak2−→ D.

(5.59)

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In this case, the rate of disappearance of A is given by

r = −d[A]dt

= k1[A] + k2[A],d[C]dt

= k1[A],d[D]dt

= k2[A].

(5.60)

Solutions for this case are also given in Table 5.1. First, both C and D increaseexponentially with a rate constant (k1 + k2), and A decreases exponentially with arate constant (k1 + k2). Secondly, the ratio of products, [C]/[D] = k1/k2 at all timesis called the branching ratio.

5.4 ACTIVATION ENERGY

The rates of reactions encountered in nature are very sensitive to temperature. Ingeneral, an increase in temperature of 10◦ causes an approximate doubling of the rateconstant. In the nineteenth century, the Swedish chemist Svänte Arrhenius proposedan empirical equation based on a large number of experimental observations. This iscalled the Arrhenius equation

k = Ae−(Ea/RT), (5.61)

whereA is called the pre-exponential factor andEa the activation energy. The equationis also written in an alternative form by combining the two terms

k∞ e−(ΔG†/RT), (5.62)

where ΔG† is called the Gibbs activation energy. In this form k bears a strongresemblance to the equilibrium constant Keq, which is a function of the Gibbs freeenergy. The integrated form of the Arrhenius equation is

ln

(k2k1

)= −Ea

R

(1

T2− 1

T1

). (5.63)

In general, a plot of ln k versus 1/T will give −Ea/R as the slope and lnA as theintercept.The activation energy is interpreted as theminimumenergy that the reactantsmust possess in order to convert to products. In the gas phase, according to the kinetictheory of collisions, reactions are said to occur if two molecules have enough energywhen they collide. Although a large number of collisions do occur per second, onlya fraction leads to a chemical reaction. The excess energy during those collisionsthat lead to a reaction is equivalent to the activation energy, and hence the termexp(−Ea/RT) is interpreted as the fraction of collisions with large enough energyto lead to reactions. The pre-exponential factor A is interpreted as the number of

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collisions that occur irrespective of the energy. The product A exp(−Ea/RT) is ameasure of the productive collisions.

EXAMPLE 5.8 ACTIVATION ENERGY FOR THE DECOMPOSITION OFAN ORGANIC

MOLECULE IN WATER

Consider the organic molecule (dibromosuccinic acid) in water. Its rate of decomposi-tion is a first-order process, for which the following rate constants were determined atvarying temperatures:

T (K ) k hr−1

323 1.08 × 10−4

343 7.34 × 10−4

362 45.4 × 10−4

374 138 × 10−4

A plot of ln k versus 1/T can be made as shown in Figure 5.9. The straight line hasa slope −Ea/R of −11,480 and an intercept of 26.34. The correlation coefficient is0.9985. Therefore, Ea = 95 kJ/mol and A = 2.76 × 1011 hr−1 for the given reaction.

–10

–9

–8

–7

–6

–5

–4

0.0026 0.0027 0.0028 0.0029 0.003 0.0031

y = 26.349 – 11480x R = 0.99934

ln (k

/h–1

)

(1/T)/(1/K)

FIGURE 5.9 Temperature dependence of the rate constant for the decompositionof dibromosuccinic acid in aqueous solution (data from Laidler, 1965). The plot ofln k versus 1/T is the linearized form of the Arrhenius equation. The slope gives thelogarithm of the pre-exponential factor (ln A) and the slope is −Ea/R, where Ea is theactivation energy.

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5.4.1 ACTIVATED COMPLEX THEORY

The Arrhenius equation can be interpreted on a molecular basis. The theoreticalfoundation is based on the so-called activated complex theory (ACT ).

We know that for the reactants to be converted to products, there should be a generaldecrease in the total energy of the system. For example, let us consider a bimoleculargas-phase reaction where an H atom approaches an I2 molecule to form HI. When Hand I2 are far apart, the total potential energy is that of the two species H and I2.WhenH nears I2, the I–I bond is stretched and an H–I bond begins to form. A stage willbe reached when the H–I–I complex will be at its maximum potential energy and istermed the activated complex. This is termed the transition state. A slight stretchingof the I–I bond at this stage will simultaneously lead to an infinitesimal compressionof the H–I bond and the formation of the H–I molecule with the release of the I atom.The total potential energy of the HI and I species together will be less than that of theH and I2 species that we started with. The progression of the reaction is representedby a particular position along the reaction path (i.e., intermolecular distance) and istermed the reaction coordinate. A plot of potential energy versus reaction coordinateis called a potential energy surface and is shown in Figure 5.10. The transition stateis characterized by such a state of closeness and distortion of reactant configurationsthat even a small perturbation will send them downhill toward the products. There isa distinct possibility that some of the molecules in the activated complex may revertto the reactants, but those that follow the path to the products will inevitably be ina different configuration from where they started. In actuality, the potential energysurface is multidimensional, depending on the number of intermolecular distances

Reactant

Activated complex

Product

Reaction coordinate

Pote

ntia

l ene

rgy

FIGURE 5.10 Formation of the activated complex in a reaction.

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involved. Generally, simplifications are made to visualize the potential energy surfaceas a three-dimensional plot.

The ACT presumes that an equilibrium between the reactants (A and B) and anactivated complex (AB)† is established. This complex further undergoes unimoleculardecay into product P:

A + Bk†�(AB)†

K†−→ P. (5.64)

The ACT tacitly assumes that even when the reactants and products are not atequilibrium, the activated complex is always in equilibrium with the reactants.

The rate of the reaction is the rate of decomposition of the activated complex.Hence

r = k†PAB† . (5.65)

However, since equilibrium exists between A, B, and (AB)†

K† = PAB†

PAPB(5.66)

with units of pressure−1 since the reaction occurs in the gas phase. Thus

r = k†K†PAPB = k∗PAPB, (5.67)

where k∗ denotes the second-order reaction rate constant and is equal to k†K†. Therate constant for the decay of the activated complex is proportional to the frequencyof vibration of the activated complex along the reaction coordinate. The rate constantk† is therefore given by

k† = κ · ν, (5.68)

where κ is called the transmission coefficient, which in most cases is ≈1. ν is thefrequency of vibration and is equal to kBT/h, where kB is the Boltzmann constantand h is the Planck’s constant. ν has a value of 6 × 1012 s−1 at 300K.

Concepts from statistical thermodynamics (which is beyond the scope of thistextbook) can be used to obtain K†. A general expression is (Laidler, 1965)

K† = QAB†

QAQBe−(E0/RT), (5.69)

whereQ is the partition function, which is obtained directly frommolecular properties(vibration, rotation, and translation energies). E0 is the difference between the zero-point energy of the activated complex and the reactants. It is the energy to be attainedby the reactants at 0K to react, and hence is the activation energy at 0K. Thus

k∗ = κkBT

h

QAB†

QAQBe−(E0/RT). (5.70)

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Note that K† is the pressure units-based equilibrium constant. If molar concen-trations are used instead of partial pressures, the appropriate conversion has to beapplied. It should also be noted that the partition functions are proportional to Tn, andhence

k∗ ≈ aTne−(E0/RT). (5.71)

Therefore, we have the following equation:

d ln k∗

dT= E0 + nRT

RT2. (5.72)

The experimental activation energy, Ea was defined earlier

d ln k∗

dT= Ea

RT2. (5.73)

Hence, Ea = Eo + nRT . This gives the relationship between the zero-point activationenergy and the experimental activation energy.

The statistical mechanical expressions of the ACT lead to significant difficultiessince the structure of the activated complex is frequently unknown. This has led to amore general approach in which the activation process is considered on the basis ofthermodynamic functions. Since K† is the equilibrium constant, we can write

ΔG† = −RT lnK† (5.74)

as the Gibbs free energy of activation. Hence, we have

k∗ = κkBT

he−(ΔG†/RT). (5.75)

We can further obtain the components of ΔG†, namely, the enthalpy of activationΔH† and the entropy of activation ΔS†. Hence

k∗ = κkBT

heΔS

†/Re−(ΔH†/RT). (5.76)

If k∗ is expressed in L/mol/s (or dm3/mol/s), then the standard state for bothΔH† andΔS† is 1mol/L (or 1mol/dm3). The experimental activation energy is related toΔH†

as per the equation Ea = ΔH† − PΔV† + RT . For unimolecular gas-phase reactionsΔV† is zero, and for reactions in solutions ΔV† is negligible. Hence, we have

k∗ = eκkBT

heΔS

†/Re−(Ea/RT). (5.77)

In terms of the Arrhenius equation, A = 2 × 1013eΔS†/R

L/mol/s at 298K. For abimolecular reaction in the gas phase, PΔV† = Δn†RT = −RT and Ea = ΔH† +2RT , and

k∗ = e2κkBT

heΔS

†/Re−(Ea/RT). (5.78)

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EXAMPLE 5.9 ACTIVATION PARAMETERS FORA GAS PHASE REACTION

The bimolecular decomposition of NO2 by the following gas-phase reaction is of sig-nificance in smog formation and in combustion chemistry of pollutants: 2NO2 →2NO + O2. The reaction rate is r = k∗P2NO2

. An experiment was designed to measurethe rate constant as a function of T :

T (K) k∗ (L/mol s)

600 0.46700 9.7800 1301000 3130

A plot of ln k∗ versus 1/T gives a slope of −13,310 and an intercept of 21.38with a correlation coefficient of 0.9994. Therefore, Ea = 110 kJ/mol and A = 1.9 ×109 L/mol s. For a bimolecular gas-phase reactionΔH† = Ea − 2RT.At 600K,ΔH† =110 − 2(8.314 × 600)/1000 = 100 kJ/mol. Assume that the transmission coefficientκ = 1. Then, we have from the equation for rate constant, k∗ = 0.46 = e2(1.2 ×1013)eΔS

†/R(2.65 × 10−10), from which ΔS† = −90 J/molK. Note that in general,ifΔS† is negative the formation of the activated complex is less probable and the reac-tion is slow. IfΔS† is positive the activated complex is more probable and the reaction isfaster. This is so since exp(ΔS†/R) is a factor that determines whether a reaction occursfaster or slower than normal.At 600K, ΔG† = ΔH† − TΔS† = 164 kJ/mol. This is the positive free energy

barrier that NO2 must climb to react. The value of A calculated from ΔS† is A =e(kT/h) exp(−ΔS†/R) = 6.6 × 108 L/mol s. The value of K† = exp(−ΔG†/RT) =5.3 × 10−15 L/mol. Note that the overall equilibrium standard free energy of thereaction will be ΔG0 = 2(86.5) − 2(51.3) = 71 kJ/mol. Hence, ΔG† ≈ 2.3ΔG0.

5.4.2 EFFECT OF SOLVENT ON REACTION RATES

The ACT can also be applied to reactions in solutions, but requires consideration ofadditional factors. Consider, for example, the reaction A + B � C + D that occursboth in the gas phase and in solution. The ratio of equilibrium constants will be

Ksoleq

Kgaseq

= KHAKHB

KHCKHD

(RT

V0

)ΔnγAγB

γCγD, (5.79)

where KHi is Henry’s constant (=Pi/xi),V0 is the volume per mole of solution, andΔn is the change in moles of substance during the reaction. γ denotes the activitycoefficient in solution. It can be argued that as per the above equation, reactions insolution will be favored if the reactants are more volatile than products and vice versaif products are more volatile. The primary interaction in the solution phase is thatwith the solvent molecules for both reactants and products on account of the close-packed liquid structure; whereas in the gas phase no such solvent-mediated effectsare possible.

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Consider the following activated complex formation in solution: A + B �(AB)† → P. The equilibrium constant for activated complex formation in the solutionphase must account for nonidealities due to the solvent phase. Hence

K†sol = [AB†]

[A][B]γAB†

γAγB(5.80)

and the rate in solution is

rsol = kBT

h· [AB†] = kBT

hK†

sol[A][B]γAγB

γAB†. (5.81)

The rate constant for the reaction in solution is therefore given by

ksol = kBT

hK†

solγAγB

γAB†. (5.82)

The activity coefficients are referred to the standard state of infinite dilution forsolutes. The ratio of rate constants for solution and gas-phase reactions is

ksol

kgas= KHAKHB

KH,AB

(V0RT

)γAγB

γAB†. (5.83)

For a unimolecular reaction both γA and γ†AB are similar if the reactant and activatedcomplex are similar in nature, and hence ksol ≈ kgas. Examples of these cases aboundin the environmental engineering literature.

The above discussion presupposes that the solventmerelymodifies the interactionsbetween the species. In these cases, since the solvent concentration is in excess ofthe reactants, it provides a medium for reaction. Hence it will not appear in the rateexpression. If the solvent molecule participates directly in the reaction, its concen-tration will appear in the rate equation. It can also play a role in catalyzing reactions.In solution, unlike the gas phase, the reaction must proceed in steps: (i) diffusion ofreactants toward each other, (ii) actual chemical reaction, and (iii) diffusion of prod-ucts away from one another. In most cases, steps (i) and (iii) have activation energiesof the order of 20 kJ, which is much smaller than the activation energy for step (ii).Hence, diffusion is rarely the rate-limiting step in solution reactions. If the rate isdependent on either step (i) or (iii), then reaction will show an effect on the solventviscosity.

We can rewrite the equation for ksol as

ksol = k0γAγB

γAB†, (5.84)

where k0 is the rate constant when γ→ 1 (ideal solution). The dependence of γ onsolvent type is best represented by the Scatchard–Hildebrand equation

RT ln γi = Vi(δi − δs)2, (5.85)

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where Vi is the molar volume of solute (A, B, or AB†), δ2i = ΔEi/Vi is the internalpressure or the cohesive energy density, and δs is the solvent internal pressure. If δis the same for both solvent and solute, solubility maximum is observed. Values forδ are listed elsewhere (e.g., Moore and Pearson, 1981). Utilizing the above we canwrite

ln ksol = ln k0 + VA

RT(δs − δA)2 + VB

RT(δs − δB)2 + VAB†

RT(δs − δAB†)

2. (5.86)

Since VA,VB, and V†AB are similar, it is the δ term that determines the value of ln ksol.

If the internal pressures of A and B are similar to that of the solvent s, but differentfrom that of AB†, then the last term dominates and ln ksol will be lower than the idealvalue. The value of ln ksol will be high if the internal pressure for eitherA or B differsconsiderably from s, but is similar to that for AB†.

5.4.3 LINEAR FREE ENERGY RELATIONSHIPS

There exists a relationship between equilibrium constant Keq(=exp−ΔG0/RT) andrate constant k(=exp−ΔG†/RT) for several reactions. The correlation is linear andis evident for reactions in solution of the type R− X + A → products, where R is thereactive site andX a substituent that does not directly participate in the reaction.Whenln k is plotted against lnKeq for different reactions that involve only a change in X, alinear correlation is evident. The linearity signifies that as the reaction becomes ther-modynamically more favorable, the rate constant also increases. These correlationsare called LFERs and are similar to the LFERs between equilibrium constants thatwe encountered in Chapters 3 and 4 (e.g., Koc versus Kow,C∗

i versus Kow). LFERsare particularly useful in estimating k values when no such data are available for aparticular reaction. The correlations are of the type

(k′

k

)=(K ′

eq

Keq

)α, (5.87)

where k′ and K ′ are for the reaction with substituent X′. The relationship betweenfree energies is therefore

ΔG† = αΔG0 + β. (5.88)

In environmental chemistry, several LFERs have been obtained for the reactions oforganicmolecules in the water environment. These relationships are particularly valu-able in environmental engineering since they afford a predictive tool for rate constantsof several compounds for which experimental data are unavailable. Brezonik (1990),Wolfe et al. (1980a,b), Betterton, Erel, and Hoffmann (1988), and Schwarzenbach etal. (1988) have reported LFERs for the alkaline hydrolysis of several aromatic andaliphatic compounds. The hydrolysis of triaryl phosphate esters follows the HammettLFER between log k and log k0 most readily. The organophosphates and organophos-phorothionates appear to obey the LFER between log k and logKa. The primary

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amides and diphthalate esters obey the Taft LFER. The rates of photooxidation ofsubstituted phenols by single oxygen (1O2) have also been shown to obey an LFER.Although several hundred LFERs exist in the organic chemistry literature, few havedirect applicability in environmental engineering since (a) the compounds are notof relevance to environmental engineers, (b) the solvents used to develop these arenot representative of environmental matrices, or (c) considerable uncertainty existsin these predictions because they are based on limited data. More work is certainlywarranted in this area. Nonetheless, at least order of magnitude estimates of reactionrates are possible using these relationships as starting points.

5.5 REACTION MECHANISMS

Reactions in the environment are complex in nature, consisting of several steps.Empirical rate laws determined through experiments can give us ideas about theunderlying mechanisms of these complex reactions (Moore and Pearson, 1981). Itis best illustrated using a reaction which admittedly has only limited significance inenvironmental engineering. The reaction is the formation of hydrogen halides fromits elements. The hydrogen halides do play a central role in the destruction of ozonein the stratosphere through their involvement in the reaction of ozone with CFCs.Several introductory physical chemistry textbooks use this reaction as an illustrationof complex reactions. Hence, only a cursory look at the essential concept is intended,namely that of the deduction of reaction mechanism from a knowledge of the rate lawfor the formation of a hydrogen halide.

5.5.1 CHAIN REACTIONS

In the early part of the twentieth century Bodenstein and Lind (1907) studied in greatdetail the reaction

H2 + Br2 � 2HBr. (5.89)

It was observed that the rate expression at the initial stages of the reaction where[HBr] � [H2] and [Br2] was

r = d[HBr]dt

= k′[H2][Br2]1/2. (5.90)

At later times the rate was

r = d[HBr]dt

= k′′[H2][Br2]1/21 + (k′′[HBr][Br2]) . (5.91)

Thus as time progressed the rate was inhibited byHBr formed during the reaction. Thefractional orders in the rate expression invariably indicate a chain reaction involvingfree radicals. A chain reaction is one in which an intermediate compound is gener-ated and consumed, which initiates a series of several other reactions leading to thefinal product. A chain reaction involves an initiation step, a propagation step, and atermination step.

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The initiation step involves decomposition of one of the reactants. Usually thiswill occur for the reactant with the smallest bond energy. For example, between H2and Br2, with dissociation energies of 430 and 190 kJ/mol, respectively, Br2 willdissociate easily as

Br2k1−→ 2Br. (5.92)

The propagation step consists of the following reactions:

H2 + Brk2−→ HBr + H,

H + Br2k3−→ HBr + Br,

(5.93)

where it should be noted that the Br radical is consumed and regenerated. In complexreactions of this type HBr can also react with H to give H2 and Br:

H + HBrk4−→ H2 + Br. (5.94)

This is called an inhibition step since the H radical is consumed by a reaction otherthan by chain termination. The reason why HBr reacts with H and not with Br is thatthe former is an exothermic reaction (67 kJ/mol) whereas the latter is an endothermicreaction (170 kJ/mol). The termination step occurs by the reaction

2Br + Mk5−→ Br2 + M, (5.95)

where M is a third body that absorbs the energy of recombination and thereby helpsto terminate the chain.

To derive the overall rate expression we must first note that the concentrations ofthe intermediates H and Br radicals are at steady state, and hence

d[H]dt

= 0 = k2[Br][H2] − k3[H][Br2] − k4[H][HBr] (5.96)

and

d[Br]dt

= 0 = 2k1[Br2] − k2[Br][H2] − k3[H][Br2] − k4[H][HBr] − 2k5[Br]2.(5.97)

Upon solving these equations simultaneously, we get

[Br] =(k1[Br2]k5

)1/2

(5.98)

and

[H] = k2(k1/k5)1/2[H2][Br2]1/2k3[Br2] − k4[HBr] . (5.99)

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The overall rate of production of HBr is

d[HBr]dt

= 2k2(k1/k5)1/2[H2][Br2]1/21 + [(k4/k3) · ([HBr]/[Br2])] . (5.100)

Note that this expression gives the correct dependencies at both initial times and latertimes. The constants k′′ and k′′′ in the earlier equation can be identified as related tothe individual rate constants.

Chain reactions such as the above are of significance in environmental engineeringand will be frequently encountered in atmospheric (gas-phase) reaction chem-istry (Seinfeld, 1986), in catalytic reactions in wastewater treatment, heterogeneouscatalysis of atmospheric solution chemistry (Hoffmann, 1990), and in combustionengineering. An interesting example of a free radical chain reaction discussed byHoffmann (1990) is the oxidation of S(IV) by Fe(III), which is a prevalent metal inatmospheric particles.

Fe3+ + SO2−3 −→ Fe2+ + SO•−

3 . (5.101)

EXAMPLE 5.10 CHAIN REACTION FOR THE OXIDATION OF ORGANIC

COMPOUNDS IN NATURAL WATERS

An example of a chain reaction is the oxidation of organic compounds by peroxidesin water, sediment, and atmospheric environments (Ernestova et al., 1992). Let RHrepresent an organic compound and AB an initiator (e.g., hydrogen peroxide, metalsalts, or organic azo compounds). H2O2 is an excellent oxidant in natural water. Thereare at least three possible initiation steps in this case:

ABhν,T−−−→ A• + B•,

RH +A• fast−−→ R• +AH,

RH + O2 −→ R• + HOO•.

(5.102)

The initiation can be induced by sunlight, ionizing radiation (cosmic rays for example),acoustic waves, and temperature fluctuations. Dissolved gases such as ozone can alsoinitiate free radicals. In shallow water bodies, the principal factor is sunlight. H2O2 hasbeen recognized as an important component in the self-purification of contaminatednatural waters. It also provides OH radical in the atmosphere that reacts with mostother organics, which leads to increased oxidation, aqueous solubility, and scavengingof organics. The hydroxyl radical is appropriately termed the atmosphere’s detergent.The last reaction given above is an initiation of oxygen in the absence of any AB.

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The propagation steps are

R• + O2k1−−→ ROO•,

ROO• + RHk2−−→ ROOH + R•.

(5.103)

Thus the radical R• is reformed in the last propagation step. These steps can be repeatedseveral times, depending on the light or thermal energies available for initiation. Theonly termination steps are radical recombinations such as

ROO• + ROO• k1−−→ Pr,

ROO• + R• −→ Pr,

R• + R• −→ Pr.

(5.104)

If the pollutant concentration in the water column is low, the most likely terminationstep is the first one listed above. The rate of oxidation of the organic compound is givenby Equation 5.102 above:

r = d[RH]dt

= −k2[ROO•][RH]. (5.105)

In natural waters ROO• concentration may reach steady-state concentrations of∼10−9 mol/L.Applying the pseudo-steady-state approximation for the peroxide radicalwe obtain.

If we denote the rate of peroxide formation as rPER = k1[R•][O2], then we can write

r = −k2(rPER

kt

)1/2[RH] = k[RH]. (5.106)

In most natural waters the rate of peroxide formation rPER is constant. Hence k is apseudo-first-order rate constant. There exist a few classes of compounds (e.g., polyaro-matic hydrocarbons and nitrosoamines) that are subject to free radical oxidation in bothnatural waters and atmospheric moisture (Schnoor, 1992).

5.6 REACTIONS IN SOLUTIONS

5.6.1 EFFECTS OF IONIC STRENGTH ON RATE CONSTANTS

Reactions in the hydrosphere occur in the presence of many different ions. Hence,the ionic strength influences the rate of a reaction.

TheACT states that the rate of the reaction is r = k†[AB†]with the thermodynamicequilibrium constant,K† = a(AB†)/a(A)a(B), where a is the activity of a species. Thus

K† = [AB†][A][B]

γAB†

γAγB= [AB†]

[A][B]K†γ , (5.107)

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where K†γ is the ratio of activity coefficients. The overall rate expression in terms

of K†γ is

r = k†K†

K†γ

[A][B] = k[A][B]. (5.108)

If we denote the rate constant at zero ionic strength (I = 0) as k0, then

k = K0

K†γ

. (5.109)

In Section 3.4, we noted that the activity coefficient of a solute i in a dilute elec-trolyte solution is given by the Debye–Huckel limiting law, ln γ = −AzizjI1/2, whereI is the ionic strength. For the present case, we have two components A and B andhence

lnK†γ = A[z2A + z2B − (zA + zB)]1/2 = −2zAzBI

1/2 (5.110)

where (zA + zB) is the charge of the activated complex, AB†. Hence

ln

(k

k0

)= 2zAzBI

1/2. (5.111)

If zA and zB are of the same charge, then increasing I increases the rate constant,whereas for ions of opposite charge, the rate constant decreases with increasing I . Ifeither of the species is uncharged, I will have no effect on the rate constant. Theseeffects are called kinetic salt effects. Using the value of A = 0.51 derived in Section3.4., the approximate dependence of I on the rate constant can be readily estimatedusing the following equation:

ln

(k

k0

)= 1.02zAzBI

1/2. (5.112)

Hence a slope of the plot of ln k versus I1/2 should give a slope of 1.02 zAzB.Note that one can also substitute other relationships for γ as given in Table 3.2 andobtain the appropriate relationship between rate constant and ionic strength that areapplicable at higher values of I .

The ionic strength effect on rate constants will become significant only for I >

0.001M. For rainwater I is small, whereas for lakes and rivers it is close to the abovevalue. For atmospheric moisture (fog and cloud) and for seawater I exceed 0.001M.Wastewater also has values of I > 0.001M. Only for these latter systems does thedependence of I on k become significant.

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EXAMPLE 5.11 EFFECT OF IONIC STRENGTH ON RATE CONSTANT

The ionic strength affects a bimolecular reaction rate constant as follows:

I (mol/kg) k (mol/L/s)

2.5 × 10−3 1.053.7 × 10−3 1.124.5 × 10−3 1.166.5 × 10−3 1.188.5 × 10−3 1.26

A plot of the rate constant versus I gives as intercept k0 = 0.992 L/mol/s withr2 = 0.936 and slope = 31.5. A plot of ln(k/k0) versus I1/2 gives a straight linewith slope 2.2 and a correlation coefficient r2 = 0.756. The slope 2.2 = 1.02 zAzB.If one of the ions A has zA = −1 (e.g., OH−), the other ion B must have acharge zB = −2. Thus, we can infer the charge of the ions involved in the activatedcomplex.

5.7 ENVIRONMENTAL CATALYSIS

The rates of environmental reactions are influenced by organic or inorganic species,solid particles, and liquid surfaces. The resulting change in reaction rates is calledcatalysis and the entities responsible for the charge are called catalysts. Catalysis isprevalent in the natural environment (water, air, and soil) and also in waste treatmentand pollution prevention processes. If the process occurs such that the catalysts are inthe same phase as the reactants, it is termed homogeneous catalysis. Some reactionsare, however, affected by the presence of a separate phase (e.g., solid particles inwater); these are called heterogeneous catalysis. Most surface reactions, in one wayor another, belong to the latter category. A list of typical examples in environmentalengineering is given in Table 5.3.

Catalysts participate in a reaction, but are regenerated in the system such thatthere is no net concentration change. The equilibrium constant Keq = kf/kb remainsunchanged. Therefore, it must be noted that both kf and kb are influenced to thesame extent. For most environmental catalysis reactions, the general rate expressionwill be

r = [ f ([A], [B], . . .)] · [X] + f ′([A], [B], . . .), (5.113)

where [X] represents the catalyst concentration f ([A], [B], . . .) and f ′([A], [B], . . .)denote the dependence on the substrate concentration. As [X] → 0, r →f ′([A], [B], . . .) and in some cases since f ′([A], [B], . . .) = 0, r → 0.

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TABLE 5.3Examples of Homogeneous and Heterogeneous Catalysisin Environmental Engineering

Catalysis Type Reaction Applications

Homogeneous Oxidation of S(IV) by H2O2 Atmospheric chemistry,aquatic chemistry, andwaste treatment

Homogeneous andheterogeneous

Acid and base hydrolysis ofpesticides and esters

Aquatic, soil, and sedimentchemistry

Homogeneous Enzyme-catalyzedbiodegradation

Aquatic, soil chemistry, andwaste treatment

Homogeneous andheterogeneous

NOx formation in combustionreactors

Atmospheric chemistry andhazardous wasteincinerators

Homogeneous Ozone destruction ingas phase

Stratospheric ozonechemistry

Heterogeneous Production ofhydrochlorofluoro-carbons

Manufacture of CFCreplacement chemicals

Heterogeneous Hydroxylation of N2O overzeolite catalysts

Oxidative removal of N2Owhich contributes togreenhouse effect

Heterogeneous Oxidation of organics inwater on TiO2

Removing pollutants from oilslicks

Heterogeneous Dehalogenation of pesticidesusing membrane catalystsand bacteria

Hazardous waste treatment ofsoils

5.7.1 MECHANISMS AND RATE EXPRESSIONS FOR CATALYZED REACTIONS

For environmental catalysis, the primary step is the formation of a complex Z betweencatalyst X and substrate A.

X +Akf�kb

Z +Y. (5.114)

The complex Z further reacts with another reactant W (e.g., the solvent) to give thedesired products:

Z +Wk′−→ P + X. (5.115)

The products consist of the regenerated catalyst and other reaction products. Notethat the second step is a nonequilibrium reaction. As we will see in the next section,Z is a surface-adsorbed complex for heterogeneous catalysis.Y andW do not exist inthat case. Acids or bases catalyze many reactions in the environment. In these cases,X can be H+ or OH−. If X≡H+, then Z is the conjugate base of A and the reaction isacid catalyzed, whereas if X≡OH−, then Z is the conjugate acid ofA and the reactionis said to be base catalyzed.

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The overall equilibrium constant for the formation of complex Z is

Keq = kf

kb= [Z][Y]

[X][A] . (5.116)

If we start with initial concentrations [X]0 and [A]0, and [Z] is the concentration ofthe intermediate complex, then [X] = [X]0 − [Z] and [A] = [A]0 − [Z]. We can nowobtain an expression that is only dependent on [Z] and [Y]:

Keq = [Z][Y]([X]0 − [Z])([A]0 − [Z]) . (5.117)

If, for example, [A]0 � [X]0, then

[Z] = Keq[X]0[A]0Keq[A]0 + [Y] (5.118)

and the rate of product formation is

r = k′[Z][W] = k′ Keq[X]0[A]0Keq[A]0 + [Y] [W]. (5.119)

Atmospheric chemical reactions follow the above rate expressionwhenY is absent andW is either O2 or N2 (see Example 5.7). In those cases where Keq[A]0 � [Y], r = k′[X]0[W] and the rate is linear in [X]0 and independent of [A]0. Inmany environmentalsystems where acid or base catalysis prevails, the condition of interest is Keq[A]0 �[Y]. The rate is then proportional to the first power in [A]0.

A different reaction rate will ensue if we start with a high catalyst concentration,[X]0 � [A]0. We then have [X] = [X]0, [A] = [A]0 − [Z], and hence the followingrate:

r = k′ Keq[X]0[A]0Keq[X]0 + [Y] . (5.120)

The catalyst concentration enters the rate expression in a distinctly nonlinear fashion.If the second reaction (dissipation of the complex Z) is extremely fast, then the

rate of dissipation can be handled using a pseudo-steady-state approximation. Thus,

d[Z]dt

= 0 = kf [X][A] − kb[Z][Y] − k′[Z][W]. (5.121)

Since [X] = [X]0 − [Z], [A] = [A]0 − [Z], [Z] is small, and [Z]2 is even smaller,

[Z] = kf [X]0[A]0kf([X]0 + [A]0) + kb[Y] + k′[W] (5.122)

and the rate is

r = kf [X]0[A]0kf([X]0 + [A]0) + kb[Y] + k′[W]k

′[W]. (5.123)

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When either [X]0 or [A]0 is low, the rate is linear in both. At high concentrations, therate is independent of both [X] and [A].

Now that we have seen how a catalyst affects the reaction rate, it is easy tounderstand how it affects the energy of the reaction. As an example, consider thedecomposition of hydrogen peroxide in the aqueous phase. Under normal con-ditions the activation energy is ∼76 kJ/mol. This reduces to ∼57 kJ/mol in thepresence of a little bromide in the aqueous phase. The enhancement in rate isexp[(−57 + 76)/RT ] = 2140. In some cases k′ � kb, and hence the first energybarrier E1 is rate controlling; this is termed an Arrhenius complex mechanism. In thesecond case kb � k′, and a second barrier controls the rate, which is termed the van’tHoff complex mechanism. In either case the catalyst provides an alternate route oflow energy for the reaction to occur.

5.7.2 HOMOGENEOUS CATALYSIS

Homogeneously catalyzed gas phase reactions that do not involve chain reactions haveno generalmechanisms.An example of such a reaction in atmospheric chemistry is thecombination of NO and Cl2 catalyzed by Br2 molecules. The reaction mechanism is

2NO + Br2kf�kb

2NOBr

2NOBr + Cl2k′−→ 2NOCl + Br2.

(5.124)

Bromine is regenerated in the process, and is therefore the catalyst in this reaction.If the second reaction is rate-controlling the rate is given by

r = k′ kf

kb[NO]2[Cl2][Br2]. (5.125)

The experimentally determined rate expression is in agreement with the above.Another gas phase homogeneous catalysis of significance in atmospheric chemistry

is the decomposition of ozone in the upper atmosphere catalyzed by oxides of nitrogenand other chlorine-containing compounds such as Freon:

2O3catalyst−−−−→ 3O2. (5.126)

The above reaction has serious consequences since ozone plays a significant rolein moderating the amount of UV light that reaches the earth. We shall discuss thisreaction in detail in Chapter 6.

In the aqueous environment, and particularly in surface waters and soil–sedimentporewaters, an important reaction is the hydrolysis of organic pollutants such as alkylhalide, ester, aromatic acid ester, amide, carbamate, and so on. Many pesticides andherbicides are also hydrolyzable. The extent of hydrolysis plays an important role indeciding how nature tends to cleanse itself.

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Mabey and Mill (1978) reviewed the environmental hydrolysis of several organiccompounds. pH was found to have the most profound effect on the hydrolysis rate.Both acids and bases were found to catalyze the reaction rates. The following discus-sion is on the specific acid–base catalysis, where the acid is H+ and the base is OH−.If a catalyzed reaction is carried out at a high [H+] such that [OH−] is negligible, therate of the reaction will be directly proportional to [H+] and [A]:

r = k′H[H+][A]. (5.127)

For a constant pH, [H+] is constant, and therefore we have a pseudo-first-orderreaction in [A], the rate of which is given by

r = kH[A]. (5.128)

Similarly, the rate of a base-catalyzed reaction is given by

r = kOH[A]. (5.129)

For an uncatalyzed reaction the rate is

r = k0[A]. (5.130)

The overall rate of an acid–base catalyzed reaction is therefore

r = (k0 + kH + kOH)[A] = k[A]. (5.131)

The overall rate constant can also be written in terms of [H+] and [OH−] as follows:

k = k0 + k′H[H+] + k′

OHKw

[H+] . (5.132)

For general acid–base catalysis, we have

k = k0 + k′H[H+] + k′

OHKw

[H+] + k′acid[acid] + k′

base[base].

The shape of the log k versus [H+] curve is shown in Figure 5.11. If k′H[H

+] is large,then k = k′

H[H+] and the slope of the curve is −1. This is the left limb of the curve

in Figure 5.11. The right limb has a slope of +1. In some cases, the transition regionwhere k is independent of pH is not clear. This happens if k0 � (Kwk′

Hk′OH). The

term IAB in the figure denotes the pH at which acid and base catalysis rates areequal. Hence IAB = (1/2) log(k′

Hk′OH/Kw). The rates of both acid and base-catalyzed

reactions are dependent on the nature of the substituent and active moieties on thereactant (Schwarzenbach, Gschwend, and Imboden, 1993).

The mechanisms of acid or base catalysis will depend on the nature of the moietyW as defined in the general reaction scheme earlier. W can be a solvent molecule,another base, or an acid in the aqueous phase. The catalyst X can be either an acid or

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log k0 IAN INB

IAB

Slope = –1 Slope = +1

Actual curve

log

k

pH

FIGURE 5.11 Variation in acid–base hydrolysis rate with pH for organic compounds in theenvironment.

a base. Hence, there exist several permutations that should be considered in derivingrate equations for acid–base catalysis. Moore and Pearson (1981) enumerate ninesuch mechanisms. Table 5.4 lists these reaction mechanisms, the rate expressions,and some examples from environmental engineering.

As noticed in Section 5.4.2, the specific hydrolysis rate constants can be estimatedusing an LFER relationship. For example,Wolfe, Zepp, and Paris (1978) showed thatlog kB for base-catalyzed hydrolysis of N-phenyl carbamates is related to the pKa ofthe alcohol group.

An aspect of homogeneous catalysis that we have not considered thus far is theaction of enzymes (X = enzyme). This is an important aspect of environmentalbioengineering and is relegated to Chapter 6 where rates of enzyme reactions areconsidered.

EXAMPLE 5.12 OBTAINING ko, kA, AND kB FROM RATE DATA

The reaction α glucose 6β glucose is called mutarotation. Acids and bases catalyze it.At 291K, the following first-order rate constants were obtained for the process usingacetic acid in an aqueous solution containing 0.02M sodium acetate.

Acetic acid (mol/L) 0.02 0.105 0.199k (min−1) 1.36 × 10−4 1.40 × 10−4 1.46 × 10−4

In the general expression for k, both kH and kOH are negligible under these conditions.kB is also negligible under these conditions. Hence, k = k0 + kacid[acid]. A plot of k

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TABLE 5.4Different Types of Acid–Base Hydrolysis Mechanisms in EnvironmentalChemistry

Type Reaction Rate Expression Examples

I S + H+ � SH+ kKeq[S][H+][R] Ester, amide, and etherhydrolysisSH+ +W

slow−−−→ PII SH + H+ � HSH+ kKeqKa[BH+][HA] Hydrolysis of

alkyl-benzoimides,keto–enol changes

HSH+ + Bslow−−−→ BH+ + SH

III HS + HA � HS · HA kKeq[HS][HA][B] Mutarotation of glucose

HS · HA + Bslow−−−→ P

IV S + HA � S · HA kKeq[S][HA][R] General acid catalysis,hydration of aldehydesS · HA + R → P

V S− + HAslow−−−→ SH +A− k[S−][HA] Decomposition of

diazoacetateSHfast−−→ P

VI HS + B � S− + BH+ kKeq[SH][R][OH−]/Kb Caisen condensation

S− + Rslow−−−→ P

VII HS + Bslow−−−→ S− + BH+ k[HS][B] General base catalysis

S− fast−−→ products

VIII R + S � T [T][∑

i kι[Bι] +∑j kj[HAj]]

Aromatic substitutions

IX HS + B � B · HS kKeq[B][HS][R] General base catalysis,ester hydrolysisB · HS slow−−−→ P

Source: From Moore, J.W. and Pearson, R.G. 1981. Kinetics and Mechanism, 3rd ed. New York:Wiley-Interscience.

Note: S represents reactant, B is the general base, HA is the general acid, and R is a reactant whetheracid or base.

versus [acid] gives as intercept, k0 = 1.35 × 10−4 min−1, and slope, kacid = 5.6 ×10−5 L/molmin. The correlation coefficient is 0.992. This is a general method ofobtaining k. For catalysis by different species, each k value can be isolated as givenabove.

EXAMPLE 5.13 EFFECTOFSUSPENDEDSEDIMENT (SOIL) ONTHEHOMOGENEOUSLY

CATALYZED HYDROLYSIS OF ORGANIC COMPOUNDS

Since atmospheric moisture (e.g., fogwater and rain) and lake and river water containsuspended solids, the rate of hydrolysis of organics is likely to be influenced by them.This example will illustrate the effect for some organic compounds.

Many organic compounds that are common pollutants (e.g., pesticides—malathionand dichloro-diphenyl-trichloroethane (DDT)) have low reactivity and are knownto associate with colloidal matter that has high organic carbon content (see Chapter 4).

continued

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These organic compounds generally do not enter into chemical reactionswith the colloidthat acts as an inert sink for organic pollutants. The concentration of a compound trulydissolved in water will be considerably small if organic compounds adsorb to colloids(particulates). If the total amount of an organic in water is Wi and Wci is the massadsorbed from a solution containing ρs mass of a sorbent per unit volume of solution,then from the equations developed in Section 4.3.2, we have

Wci

Wi= 1

1 + ρsKsw.

There are two competing rates to consider—the organic adsorption rate to the sorbentand its hydrolysis rate in water. In most cases, the rate at which the organic is adsorbedto the sorbent is relatively fast compared with the rate of hydrolysis. Hence hydrolysis isthe rate-limiting step.The rate constant for hydrolysiswill be reduced by (1 + ρsKsw)−1

since the concentration of a truly dissolved organic that can hydrolyze will be reducedto the same extent. Hence the modified rate constant is k∗ = k(1 + ρsKsw)−1, where kis the hydrolysis rate constant in the absence of solids.

Consider the hydrolysis of DDT.At 298K, k = 2.95 × 10−8 s−1 at a pH of 8 (Wolfeet al., 1977). If Ksw for DDT on a sediment is 104 L/kg, then for a ρs of 10−5 g/cm3,k∗ = 2.3 × 10−8 s−1, whereas for a ρs of 10−3 g/cm3, k∗ = 2.3 × 10−9 s−1. Thehydrolysis of DDT in slightly alkaline solutions is dramatically reduced in the presenceof solid sorbents. These calculations are valid only if the sorbents do not participate asheterogeneous catalysts as discussed in the next section.

5.7.3 HETEROGENEOUS CATALYSIS

In the atmosphere, hydrosphere, and lithosphere there are numerous reactions thatoccur at the surfaces of solids or liquids. These gas–solid, gas–liquid, and solid–liquid interface reactions are influenced by the nature and property of the surface.For example, the hydrolysis rates of organic esters and ethers are accelerated in thepresence of sediment particles. Particle-mediated catalysis plays a large role in manyatmospheric photochemical reactions. Many waste treatment processes also rely onreactions at surfaces. Someexamples are catalysts for air pollution control.Removal ofvolatile organics from automobile exhaust involves the use of sophisticated catalysts.Reactions such as the scrubbing of stack gases using solvents can be accelerated ifthe gaseous species reacts at the gas–liquid interface. Thus, the removal of ammoniaby gas scrubbing is enhanced if the solution is slightly acidic.

5.7.4 GENERAL MECHANISMS OF SURFACE CATALYSIS

Reactions on surfaces differ considerably from those in bulk phases.A surface reactioninvolves a series of successive steps. These are shown schematically in Figure 5.12.The first step, namely, bulk phase diffusion, is generally fast. Hence it is unlikely to bea rate-determining step, except for diffusion in solutions. It is difficult to differentiatebetween steps 2, 3, and 4. For example, a molecule can simultaneously react with thesurface as it adsorbs. Hence the entire process of adsorption, reaction, and desorptionis usually considered a single rate-limiting step.

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1 5

A

Z P

P

1-Diffusion of reactant2-Surface adsorption3-Surface reaction4-Product desorption5-Diffusion of product

2 43

Solid surface

FIGURE 5.12 Schematic of the steps in a heterogeneous reaction on the surface of a solid.

A heterogeneous surface reaction mechanism involves postulating that a molecule(A) becomes adsorbed on a surface (X), which further becomes an activated complex(Z) that then breaks down to give the product (P). This is the basis of the Langmuir–Hinshelwood mechanism for heterogeneous surface reactions.

A + X � Z −→ X + P. (5.133)

If two species A and B are involved we have the following scheme:

A + X � Z1,

B + X � Z2,

Z1 + Z2 −→ X + P.

(5.134)

The above scheme requires that two species be adsorbed on adjacent surface sites. Insome cases, only one (say B) gets adsorbed which then reacts with a gaseous species(say A) to give the products. This is the Langmuir–Rideal mechanism.

For the Langmuir–Hinshelwood mechanism, the rate of the reaction depends onthe surface concentration of A. The Langmuir isotherm for adsorption from the gasphase gives the surface coverage of A,

θA = KLang,APA

1 + KLang,APA, (5.135)

The rate of conversion of the adsorbed complex to products is

r = kθA = kKLang,APA

1 + KLang,APA. (5.136)

At high pressures, KLang,APA � 1, r → k and is independent of the concentrationof A. At low pressures, KLang,APA � 1 and r → k KLang,APA and the rate is firstorder in A.

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The above formalism also allows the elucidation of reaction rates when more thanone species is involved in the reaction. The Langmuir isotherm for two competingspecies A and B gives

θA = KLang,APA

1 + KLang,APA + KLang,BPB,

θB = KLang,BPB

1 + KLang,APA + KLang,BPB.

(5.137)

If the heterogeneous catalysis is bimolecular involving both adsorbed species A andB (Langmuir–Hinshelwood mechanism), we have the following overall rate:

r = kθAθB = kKLang,AKLang,BPAPB

(1 + KLang,APA + KLang,BPB)2. (5.138)

Note that the above equation indicates the competition for surface sites between Aand B. As a consequence, if PA is held constant, the rate will go through a maximumas PB is varied.

If the heterogeneous catalysis is bimolecular involving adsorbed B reacting withthe gas-phase species A (Langmuir–Rideal mechanism), we have the following rate:

r = kθBPA = kKLang,BPAPB

(1 + KLang,APA + KLang,BPB). (5.139)

By replacing pressure PA with the aqueous concentration [A], we obtain the rate ofheterogeneous surface catalysis in solutions.

Let us now consider the energetics of heterogeneous catalysis. Consider the rateof the unimolecular reaction at low pressures:

r = kKLang,APA = k′PA, (5.140)

where k′ is the overall first-order rate constant. Note that the first-order rate constantk varies with T according to the Arrhenius expression

d ln k

dT= Ea

RT2. (5.141)

Similarly, as discussed in Section 3.5, the Langmuir adsorption constant KLang,A alsohas a relationship with T ,

d lnKLang,A

dT= − qads

RT2, (5.142)

where qads is the heat evolved during adsorption. Hence

d ln k′

dT= (Ea − qads)

RT2= E′

a

RT2. (5.143)

The true activation energy E′a is smaller than Ea by the quantity qads as shown

schematically in Figure 5.13. For the case of high pressures, Ea is the same as E′a.

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A

Adsorbed A

P

Activated complex

Ea-qads

qads

Ea

Reaction co-ordinate

Pote

ntia

l ene

rgy

FIGURE 5.13 Potential energy surface for a heterogeneous reaction where the pressure ofthe reactant in the gas phase is low.

At low PA, only a few A are adsorbed and these need an energy Ea − qads to crossthe barrier. At high PA, most of the A is on the surface and the system needs to over-come a larger barrier to form products. Most environmental reactions in the naturalenvironment fall in the category of low pressure or concentration reactions and havepotential energy diagrams such as those depicted in Figure 5.13.

EXAMPLE 5.14 HETEROGENEOUS CATALYSIS OF ESTER HYDROLYSIS IN WATER

An example of heterogeneous catalysis in the environment is the metal oxide (min-eral surface)-catalyzed hydrolysis of esters. Stone (1989) discussed an environmentallyimportant reaction, namely, the influence of alumina on the base hydrolysis ofmonophenyl terephthalate (MPT) in aqueous solution. Phthalate esters are ever-presentpollutants in wastewater and atmospheric moisture. The hydrolysis in a homogeneoussystem without alumina follows the reaction MPT− + OH− → PhT + PhOH, wherePhT is phenyl phthalate and PhOH is phenol. The rate is given by

r = d[MPT−]dt

= kbase[OH−][MPT−],

with the base hydrolysis rate constant kbase = 0.241 L/mol s at 298K in bufferedsolutions (pH 7.6–9.4). At pH > pKa = 3.4, MPT exists mostly as MPT−, which isthe species of interest throughout the pH range of hydrolysis. The addition of a smallconcentration of a heterogeneous surface (alumina) increases the rate of hydrolysis sub-stantially (Figure 5.14). The rate increased with increasing pH and [Al2O3], whereasthe ionic strength adversely affected the rate. It was shown that the adsorption of MPT−

continued

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0

0.2

0.4

0.6

0.8

1

0

0.2

0.4

0.6

0.8

1

0 50 100 150 200 250 300

0 50 100 150 200 250 300

log (

[MPT

]/[M

PT] 0

)

time/h

pH = 5.69, Blank

pH = 5.71, 10.3 g/L alumina

FIGURE 5.14 Heterogeneous catalysis of MPT by alumina in the aqueous phase.The square symbol represents particle-free solution and the triangular symbol representsparticle-laden solution. Reaction was conducted in 0.003M acetate buffer. (From Stone,A.T. 1989. Journal of Colloid and Interface Science 127, 429–441.)

on alumina surface enhanced the hydrolysis rate thus providing an additional pathwayfor the reaction. The overall reaction rate can be written as

r = −d[MPT−]dt

= kbase[OH−][MPT−] + kΓKadsΓOH−ΓMPT− , (5.144)

where Kads is the adsorbed complex formation constant for MPT− on alumina, Γ−OH

is the OH− concentration in the diffuse layer, and ΓMPT− is the adsorbed MPT−concentration (obtained from the Poisson–Boltzmann equation, Section 3.5).

ΓOH−ΓMPT− = [OH−][MPT−]e(F(ψs+ψd)/RT). (5.145)

Utilizing the above equation, we have the overall rate constant for hydrolysis, k∗H:

k∗H =[kΓKadse

(F(ψs+ψd)/RT) + kbase

][OH−]. (5.146)

Thus k∗H is substantially larger than kbase. At a constant pH the exponential termdecreases with ionic strength. Similarly, due to competition from other ions in solution,Kads for MPT− also decreases. Thus, k∗H will decrease with increasing I . The aboveequation also predicts a maximum in the rate with pH. With increasing pH, ΓOH−increases and ΓMPT− decreases. These opposing effects should cancel each other atsome pH where the maximum in rate is observed.

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EXAMPLE 5.15 SILICADISSOLUTIONINWATER—AHETEROGENEOUSREACTION

Mineral weathering is the process of congruent dissolution (without the formationof new phases) and incongruent dissolution (where new solid phases are formed).An example of the latter is an aluminosilicate mineral and an example of the formeris a carbonate mineral. The origin of major ions in seawater over geologic times canbe explained as a result of these two processes. The discussion of heterogeneous solidreactions in this section has a direct bearing on this topic. The five steps involved in theheterogeneous reaction that we alluded to earlier occur over long periods of time andcover large surface areas (of sediments) and volumes (of water) and proceed inexorablytoward equilibrium. To illustrate this process, let us consider the dissolution of silicathat forms the most prevalent mineral on earth. From a geochemistry point of view, itoccurs near shores (surface sediments) at ambient temperatures and in the interior ofthe earth at exorbitant temperatures. Stumm and Morgan (1996) identified four majorspecies of silica—quartz, α- and β-cristobolite, and amorphous silica in the order ofstability. The relevant hydration reaction of interest is

SiO2(s) + 2H2Okdiss�kprec

Si(OH)4(aq) (5.147)

with an equilibrium constant

Keq = [Si(OH)4]γSi(OH)4

aH2O · aSiO2(s). (5.148)

Solid SiO2 has unit activity (aSiO2 = 1), and since the solution is dilute (aH2O = 1).We also note that the activity coefficient of Si(OH)4 is one. Thus

Keq � [Si(OH)4]. (5.149)

Hence, the equilibrium constant is the molar solubility of silica. At the ambienttemperature (298K) and a natural pH of 9.5, the Si(OH)4 remains undissociated. Sincethe above reaction is reversible, the net rate of change of Si(OH)4 concentration is thebalance between dissolution (rate constant, kdiss) and precipitation (rate constant, kprec).Therefore, the following equation was proposed (Rimstidt and Barnes, 1980; Brezonik,1994):

−d[Si(OH)4]dt

= As

Vw(kdiss[SiO2][H2O] − kprec[Si(OH)4]). (5.150)

In dilute solutions we have [SiO2][H2O] = 1, using the definition Keq = kdiss/kprec,and noting that β = [Si(OH)4]/Keq is the saturation ratio in the aqueous phase

−dβ

dt= As

Vwkprec(1 − β), (5.151)

where As/Vw is the surface area of silica per unit volume of water.The above differential equation can be integrated using the initial condition that at

t = 0, β = 0 to obtainβ = 1 − e−((As/Vw) · kprec)t . (5.152)

continued

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TABLE 5.5Equilibrium and Rate Constants for SiO2 Dissolution Reactions at 298 K

Type ln Keq ln kd (s−1) ln kp (s−1)

Quartz −9.11 −30.81 −21.70α-Cristobolite −7.71 −29.41 −21.70β-Cristobolite −6.72 −28.44 −21.70Amorphous silica −6.25 −27.61 −21.40

Source: From Stumm, W. and Morgan, J.M. 1981. Aquatic Chemistry, 3rd ed. NewYork, NY:John Wiley & Sons, Inc.; Brezonik, P.L. 1994. Chemical Kinetics and Process Dynamicsin Aquatic Systems. Boca Raton, FL: Lewis Publishers/CRC Press, Inc.

It is obvious from the above equation that a plot of β versus t gives Askprec/Vw from thetime constant for the exponential fit. Thus kprec can be obtained, from which kdiss =kprec/Keq can be determined. In experimental systems, the initial stages of dissolutionare dominated by the faster solubility of the exposed surface silica layer. The equilibriumand rate constants for silica dissolution obtained from different sources for the varioussilica forms is tabulated in Table 5.5.

5.7.5 AUTOCATALYSIS IN ENVIRONMENTAL REACTIONS

In some environmental chemical reactions some of the products of the reaction act ascatalysts. A bimolecular autocatalysis reaction is represented as A + B → 2B. Therate expression is r = −d[A]/dt = k[A][B]. Let us define the progress of the reactionby ξ. If at any time [A]0 − [A] = [B] − [B]0 such that [B] = [A]0 + [B]0 − [A] =[A]0 + [B]0 − ξ,

−dξ

dt= kξ(([A]0 + [B]0 − ξ) . (5.153)

The above equation can be integrated to obtain

[B] = [A]0 + [B]01 + ([A]0/[B]0) · e−k([A]0+[B]0)t . (5.154)

Note from the above that at t = 0, [B] = [B]0. For [B]0, the only condition is that[B] = 0 for all t. However, if [B]0 �= 0 at t = 0, [B] will slowly increase; this istermed the induction period. The value of [B] increases continuously and reaches itsmaximumvalue of [A]0 + [B]0 as t → ∞.The characteristic S-shaped curve shown inFigure 5.14 is characteristic of autocatalytic reactions in the environment. Oscillatoryreactions such as the oxidation of malonic acid by bromate and catalyzed by ceriumions (otherwise called the Belousov–Zhabotinsky reaction) are classic examples ofautocatalysis.

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[B]max= [A]0 + [B]0

t* where (d[B]/dt) is maximum

t

[B]0

[B]

FIGURE 5.14 General shape of a concentration profile for an autocatalytic reaction.

EXAMPLE 5.16 AUTOCATALYSIS IN NATURAL WATERS—Mn(II) OXIDATION

The oxidation of Mn(II) in aqueous solutions is known to be base catalyzed,

Mn(II)OH−−−−→O2

MnOx(s) (5.155)

with the rate law

r1 = −d[Mn(II)]dt

= k′[OH−]2PO2 [Mn(II)]. (5.156)

At constant pH and PO2 , the reaction follows pseudo-first-order kinetics with a rateconstant, k = k′[OH−]2PO2 . Thus

r1 = k[Mn(II)]. (5.157)

The oxidation of Mn(II) gives MnOx in the solution, which acts as an auto-catalyzingagent.

MnOx(S) + [Mn(II)] O2−−→ 2MnOx(S). (5.158)

The rate of the above heterogeneous base catalysis is

r2 = −d[Mn(II)]dt

= k′′PO2

Kads

[H]+ [Mn(II)][MnOx], (5.159)

where [MnOx] = [Mn(II)]0 − [Mn(II)], and Kads is the adsorption constant for Mn(II)on the MnOx catalyst. At constant pH and [O2] the rate is

continued

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r2 = k∗[Mn(II)][MnOx]. (5.160)

The overall rate of the auto-catalyzed reaction is

r = −d[Mn(II)]dt

= (k + k∗[MnOx])[Mn(II)]. (5.161)

Upon integration using the boundary condition [MnOx]0 = 0 we obtain

[MnOx]([Mn(II)]0 − [MnOx]) = k

(k∗[Mn(II)]0 + k)

[e((k+k∗[Mn(II)]0)t) − 1

]. (5.162)

If k∗[Mn(II)]0 > k, we can approximate the above equation and rearrange to obtain

ln

( [MnOx][Mn(II)]0 − [MnOx]

)= ln

(k

k∗[Mn(II)]0

)+ [k + k∗[Mn(II)]0]t. (5.163)

Thus, a plot of ln[[MnOx]/([Mn(II)]0 − [MnOx])

]versus t will give at suffi-

ciently large t values, a straight line slope of k + k∗[Mn(II)]0, and an intercept ofln(k/k∗[Mn(II)]0) from which the values of k and k∗ can be ascertained. The data forMn(II) oxidation at a constant partial pressure of oxygen and different pH values werereported by Morgan and Stumm (1964), and are plotted in Figure 5.16. The linear fit tothe data shows the appropriateness of the rate mechanism given above.

y = 0.0212x – 2.0956R2 = 0.9912

–2.5

–2

–1.5

–1

–0.5

0

0 20 40 60 80 100t / h

Log

{[P]/(

[A] 0

–[P]

)}

FIGURE 5.16 Kinetics of autooxidation of Mn(II) in alkaline solutions at pH = 9.8at 298K. [A]0 = 8 × 10−5 M; PO2 = 1 atm. (FromMorgan, J.J. and Stumm,W. 1964.Proceedings of the Second International Water Pollution Research Conference, Tokyo,Elmsford, NY: Pergamon Press.)

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5.8 REDOX REACTIONS IN ENVIRONMENTAL SYSTEMS

The transfer of electrons (e−) between compounds is an important aspect of severalreactions in the environment. If a compound accepts an e− it is said to undergo reduc-tion, and the process of donation of an e− is called oxidation. These come under theumbrella term redox reactions. The study of redox systems is the central aim of thediscipline called electrochemistry, which is a specialized branch of physical chem-istry. Most photo-assisted and biochemical degradation of organic compounds in thenatural environment and hazardous waste treatment processes include e− mediation.Redox reactions are also prevalent in the dark and hence can occur in the subsurfaceenvironment as well. As a result of redox processes remarkable differences in theproperties of surficial and deep sediment layers are observed. Since the surface sed-iment is in the aerobic (oxygen-rich) zone it is easily oxidized. This layer containsseveral important species such as O2, CO2, SO

2−4 , NO−

3 , and oxides of iron. In theaerobic zone most substances are rapidly oxidized. The redox potential (as shall bediscussed in the subsequent paragraphs) is always positive in the aerobic zone andranges from +0.4 to +0.1V. The deeper sediments are anaerobic and have redoxpotentials between−1 and−2.5V. It contains predominantly reduced species such asH2S, NH3, CH4, and several organic compounds. The redox potential discontinuity(RPD) zone separates the aerobic and anaerobic zones. The RPD zone, where thereis a decrease in redox potential from +1 to −1V has a very short depth. Processesoccur at different rates in these zones and are mediated by the biota that inhabits thearea. The microbes that are present in both aerobic and anaerobic zones can act asintermediaries in e− exchange between compounds.

Redox processes are generally slower that most other chemical reactions and sys-tems involving them are in disequilibrium. Microbial mediated e− exchange plays alarge role in the bio-remediation of contaminated groundwater aquifers.

The transfer of e− can be understood via an inventory of the so-called oxidationstates of reactants and products. The oxidation state of an atom in a molecule is thecharge associated with that atom if the ion or molecule were to be dissociated. Theoxidation state of a monoatomic species is its electron charge. The sum of oxidationstates is zero for a molecule, but for an ion it is equal to its charge. In the naturalenvironment, compounds that undergo redox reactions are comprised of those thathave C, N, or S atoms. For redox reactions to occur in the environment, there has tobe a source and a sink for e− in the system. It has been shown through both laboratoryand field observations that even the most recalcitrant (refractory) organic compoundscan undergo redox reactions. Let us take a simple example, namely, chloroform thatundergoes a transformation to methylene chloride as follows:

CHCl3 + H+ + 2e− −→ CH2Cl2 + Cl−. (5.164)

In the reactant CHCl3, the oxidation state of C is +2. The addition of H+ and theremoval of Cl− reduce the oxidation state of C to 0 in the product (CH2Cl2). Sincea reduction in oxidation state and the release of a Cl species has been brought aboutsimultaneously it is called a reductive dechlorination process. In complex environ-mental matrices, it is often difficult to clearly delineate the exact source or sinkfor e−.

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Formally, there is an analogy between the transfer of e− in a redox process and thetransfer of H+ in an acid–base reaction. It is useful to first understand how the electronactivity in solutions is represented. A redox reaction can be generally represented asthe sum of two half-cell reactions, one where the e− is accepted (reduction) and theother where it is donated (oxidation). For example, a redox reaction involving theoxidation of Zn by Cu2+ can be represented as

Cu2+(aq) + 2e− −→ Cu(s) (Reduction),

Zn(s) −→ Zn2+(aq) + 2e− (Oxidation),

Cu2+(aq) + Zn(s) −→ Cu(s) + Zn2+(aq) (Redox).

(5.165)

It is useful to represent each as a reduction reaction and then the overall process isthe difference between the two:

Cu2+(aq) + 2e− → Cu(s) (Reduction),

Zn2+(aq) + 2e− → Zn(s) (Reduction),

Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) (Redox).

(5.166)

Each half-reaction is denoted as Ox/Red and is represented by Ox + ne− → Red.The equilibrium quotient for the reaction is K = [Red]/[Ox][e−]n.

Analogous to the definition of H+ activity, pH = − log[H+], we can define theelectron activity [e−] using pe = − log[e−]. Thus for the above redox reaction,

pe = pe0 − 1

nlog

( [Red][Ox]

)= pe0 + 1

nlog

( [Ox][Red]

), (5.167)

where pe0 = (1/n) logK is the electron activity at unit activities of [Red] and [Ox]species. The above definition is generally applicable to any reaction involving e−such as

∑i νiAi + ne− = 0, where vi is the stoichiometric coefficient (positive for

reactants, negative for products) and for which we have

pe = pe0 + 1

nlog

(∏i

(Ai)vi

). (5.168)

EXAMPLE 5.17 pe OF RAINWATER

Neglecting all other ions, we have the following equation driving the pe of rain-water: (1/2)O2(g) + 2H+ + 2e− → H2O(l). Since the activity of pure water is 1,we can write pe = pe0 + (1/2) log[P0.5O2

[H+]2], where pe0 = (1/2) logK with K =1/P0.5O2

[H+]2[e−]2 = 1041. If atmospheric partial pressure of O2(=0.21 atm) is usedand a pH of 5.6 in rainwater is considered, pe = 14.7.

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The pe value can be obtained experimentally from the electrode potential of a redoxreaction, EH (in volts). The subscript H denotes that it is measured on a hydrogenscale. Consider the redox reaction, Ox + ne− → Red. The Gibbs function for thereaction is given by ΔG = −nFEH, which is the work required to move n electronsfrom the anode to the cathode of an electrochemical cell. EH is the electrode potential(Volts) on a hydrogen scale, that is, assuming the reduction of H+ to H2 has a zeroreduction potential. F is the Faraday constant (96,485Coulombs/mole). The standardGibbs energy, ΔG0 is defined as equal to −nFE0H. Since we know that pe = pe0 +(1/n) log([Ox]/[Red]), and pe0 = (1/n) logK = −ΔG0/2.303RT, we have

−ΔG = −ΔG0 + 2.303RT

nlog

[Ox][Red] (5.169)

or

EH = E0H + 2.303RT

nlog

[Ox][Red] , (5.170)

which is called the Nernst equation for electrochemical cells. It is also apparent fromthe above that

pe = FEH

2.303RT. (5.171)

Thus, pe is obtained from EH, which is easily measured using electrochemicalcells. Some examples of EH values for redox reactions of environmental interest aregiven in Table 5.6.

For redox equilibria in natural waters, it is convenient to assign activities of H+and OH− values that are applicable to neutral water. The pe0 values relative to E0

H arenow designated pe0(w) relative to E0

H(w). The two are related through the ion productof water pe0(w) = pe0 + (nH/2)logKw, where nH denotes the number of protonsexchanged per mole of electrons. This type of characterization of pe0 allows one tograde the oxidizing capacity of ions at a specified pH (namely, that of neutral water,7). Thus a compound of higher pe0(w) will oxidize one with a lower pe0(w).

EXAMPLE 5.18 CALCULATION OF E0H AND pe0(w) FORA HALF-CELL REACTION

Consider the half-cell reaction, SO2−4 + 9H+ + 8e− � HS− + 4H2O. The standard

free energy of the reaction isΔG0 = ΔG0H = G0

f (HS−, aq) + 4G0f (H2O) − G0

f (SO2−4 ,

aq), since both H+ and e− have zero G0f by convention. Therefore ΔG0

H = 12 +4(−237) − (−744) = −192 kJ/mol. Hence E0H = −ΔG0

H/nF = (192 × 1000)/(8 ×96485) = +0.24V.To obtainEH at any other pH (say 7), we can use theNernst equationto obtain EH at a pH of 7, EH = E0H + (0.059/8) log([SO2−

4 ][H+]9/[HS−][H2O]6).If all species are at their standard states of unit activities, except [H+] = 10−7 M,we have, E0H(w) = +0.24 + (0.0074) log[10−7]9 = −0.22V. Hence pe0(w) =E0H(w)/0.059 = −3.73.

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Since the transfer of protons (acid–base reactions) and electrons (redox reactions)in environmental processes are closely linked, it is convenient to relate the pe andpH of the reactions. It can provide information as to when either of these reactionspredominate. The following discussion is a brief annotation of the salient aspects ofpe–pH diagrams. The reader is referred to Stumm and Morgan (1996) for furtherdetails.

5.8.1 RATES OF REDOX REACTIONS

The rate law for a general redox reaction of the form Aox + Bred → Ared + Boxis given by r = kAB[Aox][Bred]. As an example, let us consider the homogeneousoxidation of Fe(II) in water at a given partial pressure of O2 above the solution. Therate law is

r = −d[Fe(II)]dt

= k[Fe(II)]PO2 . (5.172)

For fixed PO2 , the rate is r = kobs[Fe(II)], where kobs is a pseudo–first-order rateconstant. It has been shown that the rate derives contributions from three species ofFe(II), namely, Fe2+, FeOH+, and Fe(OH)2 such that

kobs[Fe(II)] = k0[Fe2+] + k1[FeOH+] + k2[Fe(OH)2]. (5.173)

For a general metal-ligand addition reaction proceeding as follows:

M + L � ML + L � ML2 · · · � MLn, (5.174)

TABLE 5.6E 0

H and pe0 Values for Selected Reactions of EnvironmentalSignificance

Reaction E0H (V) pe0

O2(g) + 4H+ + 4e− � 2H2O +1.22 +20.62Fe3+ + e− � Fe2+ +0.77 +13.012NO−

3 + 12H+ + 10e− � N2(g) + 6H2O +1.24 +20.96CHCl3 + H+ + 2e− � CH2Cl2 + Cl− +0.97 +16.44CH3OH + 2H+ + 2e− � CH4(g) + H2O +0.58 +9.88S(s) + 2H+ + 2e− � H2S(aq) +0.17 +2.89α− FeOOH(s) + HCO−

3 + 2H+ + e− � FeCO3 (s) + 2H2O −0.04 −0.80

SO2−4 + 10H+ + 4e− � H2S(g) + H2O +0.31 +5.25

SO2−4 + 9H+ + 8e− � HS− + 4H2O +0.25 +4.25

N2(g) + 8H+ + 6e− � 2NH+4 +0.27 +4.68

2H+ + 2e− � H2(g) 0.00 0.006CO2(g) + 24H+ + 24e− � C6H12O6 + 6H2O −0.01 −0.17CO2(g) + 4H+ + 4e− � CH2O + H2O −0.07 −1.20CO2(g) + H+ + 2e− � HCOO− −0.28 −4.83

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we can define an equilibrium constant for each step

Keq,i = [MLi][MLi−1][L] (5.175)

and a stability constant for each complex

βi = [MLi][M][L]i . (5.176)

Using these definitions one can write for the Fe(II) system discussed here (seeMorel and Herring [1993] for details)

kobs = k0 + [k1(K1Kw/[H+])] + [k2(β2K2w/[H+]2)]

[1 + (K1Kw/[H+]) + (β2K2w/[H+]2)] . (5.177)

Typical values of k0, k1, and k2 at 298K are 1 × 10−8 s−1, 3.2 × 10−2 s−1, and1 × 104 s−1, respectively. Since ko � k1 � k2, we can see why predominantlyhydroxo species of Fe(II) are formed. Competition for ligands (e.g., Cl− and SO2−

4 )

will significantly lower the rate of oxidation. Therefore, iron oxidation is far morefavorable in freshwaters than in openmarine systems.Reported rate constant inmarinesystems are a 100 times lower than in freshwater systems.

The redox reaction as discussed above proceeds with transfer of electrons betweenmolecules. There are two knownmechanisms of electron transfer between the oxidantand reductant. Both involve the formation of an activated complex. The distinguishingfeature is the type of activated complex. In the first variety, the hydration shells ofthe two ions interpenetrate each other sharing a common solvent molecule. This iscalled the inner sphere (IS) complex. In the second type, the two hydration shells areseparated by one or more solvent molecules, this being termed the outer sphere (OS)complex. In the OS complex, the solvent will mediate the electron transfer. The twomechanisms can be distinguished from one another since, in the IS case, the rateswill depend on the type of ligand forming the bridged complex. Electron transferreactions are generally slower than other reactions because of the rearrangements andorientations of the solvent molecule required for the reaction to proceed.

A formal theory of e− transfer in reactions is provided by the well-knownMarcustheory. The basic tenet of the theory is that the overall free energy of activation forelectron transfer is made up of three components (Marcus, 1963). The first involvesthe electrostatic potential (ΔG†

elec) required to bring two ions together. The second

is the free energy for restructuring the solvent around each ion (ΔG†solv). The last

and final term arises from the distortions in bonds between ligands in products andreactants (ΔG†

lig).

ΔG†AB = ΔG†

elec +ΔG†solv +ΔG†

lig. (5.178)

Marcus derived expressions for the free energy terms and combined them to give theoverall free energy. This resulted in the following equation for the rate of electrontransfer

kAB = κkTh

· e−(ΔG†AB/RT). (5.179)

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Marcus also derived the following relationship between individual ionic interactionsrelating kAB to kAA and kBB

kAB = (kABkBBkAB f )1/2 (5.180)

where ln f = (1/4)(lnKAB)2/ ln(kAAkBB/Z2). Z is the collision frequency betweenuncharged A and B (≈1012 L/mol s). From the expression for kAB, we can obtain

ΔG†AB = 1

2

(ΔG†

AA +ΔG†BB +ΔG0

AB

)− 1

2RT ln f . (5.181)

Formost reactionsΔG0AB ∼ 0.Note that kAA and kBB are sodefined that they represent

the following redox reactions

Aox +A∗red � A∗

red +Aox,

Box + B∗red � B∗

red + Box.(5.182)

The above equations represent self-exchange reactions. The overall redox process hasKAB as the equilibrium constant

KAB = [Ared] + [Box][Aox][Bred] . (5.183)

Since for one-electron exchange reactions we have already seen that

lnKAB = (E0HA − E0

HB)

0.059= pe0A − pe0B, (5.184)

we can compare the equilibrium constant calculated usingMarcus theory to that deter-mined experimentally. Good agreement is generally observed between the observedand predicted values, lending validity to the Marcus relationship.

The limiting case of the Marcus relationship leads to a simple LFER betweenkAB and KAB for OS electron transfer reactions of the type AoxL + Bred → AredL +Box. If a plot of ln kAB versus lnKAB is made, a linear relationship with slope of0.5 is observed, provided f ≈ 1 and ΔG0

AB is small representing near equilibriumconditions. For very endergonic reactions, the slope is ≈1. Several examples of suchrelationships in environmental reactions have been established (Wehrli, 1990). Severalauto-oxidation reactions of interest in environmental science were considered byWehrli (1990) and data tabulated for both kAB and KAB for reactions of the typeAred + O2 → Aox + O−

2 . Figure 5.17 is a plot showing the unit slope for the LFERinvolving different redox couples.

It is appropriate at this stage to summarize the various LFERs that we have dis-cussed so far for the prediction of rate constants and equilibrium constants in a varietyof contexts. These are compiled in Table 5.7. Armed with a knowledge of thesecategories of LFERs, one should be able to predict the rates ofmany common environ-mental reactions and/or the equilibrium constants for various partitioning processes.This can be especially useful if only few values are available within a group.

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–6

–4

–2

0

2

4

6

8

–20 –15 –10 –5 0log KAB

Fe3+/Fe2+

VO2+/VO(OH)+

Fe(OH)2+/Fe(OH)+

Fe(OH)2+/Fe(OH)2

CuCl+/CuCl

Cu2+/Cu+

FIGURE 5.17 Marcus free energy relationships for the oxidation of several ions of envi-ronmental interest. (From Wehrli, B. 1990. In: W. Stumm (Ed.), Aquatic Chemical Kinetics,pp. 311–336. NewYork, NY: John Wiley & Sons, Inc.)

TABLE 5.7A Summary of LFERs in Environmental Engineering

Type Application Special Feature

Kinetic Rate ConstantsBrönsted Acid and base catalysis, hydrolysis,

association, and dissociationreactions

log ka or kb related to logKa or Kb

Hammet σ p- and m-substituted aromatichydrolysis, enzyme catalysis

Substituent effects on organic reactions

Taft σ∗ Hydrolysis and other reactions foraliphatic compounds

Steric effects on substituents

Marcus Electron transfer (OS), metal ionauto-oxidations

log kAB related to logKAB

Equilibrium Partition ConstantsC∗i − Kow Aqueous solubility logC∗

i related to logKow

Koc − Kow Soil–water partition constant logKoc related to logKow

KBW − Kow Bioconcentration factor logKBW related to logKow

Kmic − Kow Solute solubility in surfactantmicelles

logKmic related to logKow

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5.9 ENVIRONMENTAL PHOTOCHEMICAL REACTIONS

Solar radiation is the most abundant form of energy on earth. All regions of the sun’sspectrum (UV, visible, and infrared) reach the earth’s atmosphere. However, only asmall fraction of it is absorbed by water or land, while the rest is reflected, absorbed,or dissipated as heat. Several reactions are initiated in the natural environment as aresult of the absorbed radiation. The term photochemistry refers to these transforma-tions. Photochemistry is basic to the world we live in. Plants depend on solar energyfor photosynthesis. Photolytic bacteria draw solar energy for conversion of organicmolecules to other products. Absorption of photoenergy by organic carbon in naturalwaters leads to the development of color in lakes and rivers. Redox reactions in theaquatic environment are initiated by absorption of light energy. Reactive free radi-cals (OH•, NO•

3, and HO•2) are formed by photochemical reactions in the atmosphere

where they react with other species.There are two fundamental laws in photochemistry:

1. Grotthus–Draper law: Only light absorbed by a system can cause chemicaltransformations.

2. Stark–Einstein law: Only one quantum of light is absorbed per molecule ofabsorbing and reacting species that disappear.

There are two important laws that are derived from the Grotthus–Draper law. Thefirst is called the Lambert’s law, which states that the fraction of incident radiationabsorbed by a transparent medium is independent of the intensity of the incident light,and that successive layers in the medium absorb the same fraction of incident light.The second is called the Beer’s law, which states that absorption of incident light isdirectly proportional to the concentration of the absorbing species in the medium. Bycombining the two laws we can express the ratio of change in absorption to the totalincident radiation as follows:

dI

I= ανCi dz. (5.185)

The above equation is called the Beer–Lambert law. αv is proportionality constant(m2/mole). dz is the increment in thickness of themediumperpendicular to the incidentradiation. ν is the frequency of the incident light. If I = I0 at z = 0, we can obtainupon integration

logI0I

= ενCiZ , (5.186)

where εv = αv/2.303 is called themolar extinction coefficient (m2/mole). It is specificto a specific frequency ν, or wavelength λ = 1/ν. Ci is expressed in mole/m3, andZ is expressed in m. If Ci is in mole/dm3, then Z is expressed in mm. log(I0/I) iscalled the absorbance Aλ. I is expressed in J/m2 s. The amount of light absorbed bythe medium is

Iabs = I0 − I = I(1 − 10−εvCiZ). (5.187)

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If several components are absorbing simultaneously in the sample, then

Iabs = I0 − I = I(1 − 10−∑i εvCiZ

). (5.188)

The Stark–Einstein law explicitly identifies that only molecules that are electron-ically excited take part in photochemical reactions, that is, those that are chemically“active” must be distinguished from those that are “excited” by photons. Excitedspecies can lose energy by non-chemical pathways and via thermal reactions. The effi-ciency of a photochemical reaction was defined by Einstein as quantum efficiency, φ,

φ = number of molecules formed

number of quanta absorbed. (5.189)

This concept can be extended to all physical and chemical processes following theabsorption of light. It can therefore be identified as a means of keeping tabs on thepartitioning of absorbed quanta into the various modes.

Consider a molecule B that received a quantum of light energy to form the excitedspecies B∗

B + hv −→ B∗. (5.190)

The absorption spectrum for the molecule is the plot of absorbance (A) versus wave-length (λ) or frequency (ν). Since the absorbance depends on the nature of thefunctional groups in the molecule that are photoexcited, the absorption spectrumof each compound can be considered to be its fingerprint. These functional groupsare called chromophores. The wavelength at which maximum absorption is possi-ble (designated λmax) and the corresponding ε are listed in standard handbooks (CRCHandbook of Chemistry and Physics, 1994). Organic compounds with fused aromaticrings or unsaturated heteroatom functionalities have generally high absorbances.

Amolecule in its excited state can undergo four main types of primary photochem-ical processes.

Fluorescence: B∗ −→ B + hν′,

Dissociation: B∗ −→ P1 + P2,

Quenching: B∗ + M −→ B + M,

Energy transfer: B∗ + C −→ B + C∗,

Reaction: B∗ + C −→ S1 + S2.

Each of the processes given above has aφi associatedwith it. The value of the quantumefficiency is a function of the wavelength and denoted by φλ. In the atmosphere(troposphere) the range of wavelengths of interest is between 280 and 730 nm, sincemost of the UV wavelengths <280 nm are blocked by the stratospheric ozone layer.Beyond 730 nm no reactions of interest take place. In the aquatic environment (lakes,rivers, and oceans) the upper range is rarely above 400 nm. The general form of the

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photolysis reaction is represented as

B + hv −→ P1 + S1. (5.191)

The rate of the photochemical reaction is

−d[B]dt

= J[B], (5.192)

where J is the first-order photochemical rate constant. Since the reaction occurs overthe entire wavelength range λ1 to λ2, we have to obtain an average rate constant.

The photolysis rate constant J can be expressed in terms of the absorption cross-section σλ(T), which is a function of both λ and T . The rate constant is

J =∫λ2λ1

σ(λ)φ(λ)Iλdλ (5.193)

or

J =λ2∑λ1

σλiϕλi IλΔλ, (5.194)

where the overbar denotes the values at mid range centered at λi in the intervalΔλ. Inusing the above expression one should separately evaluate σλi and Iλ. Normally theshortest wavelength for photochemistry in the atmosphere is 290 nm. Finlayson-Pittsand Pitts (2000) list the values of the absorption cross-section σλi and the actinideflux Iλ for many atmospheric reactions. The values of J for a number of reactions arelisted in Table 5.8.

The following two examples will illustrate the use of the above equations fordetermining the photolytic rate constants in water and air environments.

EXAMPLE 5.19 RATE CONSTANT FOR PHOTOLYSIS IN THE ATMOSPHERE

For the case of nitrogen dioxide photolysis in air, we have the following data:

λ (nm) σλi (cm2) ϕλi (Photons/cm2 s) IλΔλ σλiϕλi IλΔλ

300 2E−19 0.98 1.6E+13 3.14E−06310 3E−19 0.972 2.81E+14 8.19E−05320 3.5E−19 0.964 7.19E+14 0.000243330 4E−19 0.956 1.29E+15 0.000494340 4E−19 0.948 1.42E+15 0.000537350 5E−19 0.94 1.59E+15 0.000748360 5.5E−19 0.932 1.66E+15 0.000852370 6E−19 0.924 2.07E+15 0.001146380 6E−19 0.916 2.02E+15 0.001110390 6E−19 0.908 2.06E+15 0.001122400 6E−19 0.8 2.81E+15 0.001349410 6E−19 0.17 3.56E+15 0.000363420 6E−19 0.03 3.7E+15 6.65E−05

Sum: 0.008114

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Hence, J = 0.0081 s−1 and t1/2 = 85 s. In this case since IλiΔλ changeswith the zenithangle, the rate constant varies diurnally and hence NO2 decomposition by photolysisalso varies diurnally.

5.10 ENZYME CATALYSIS

5.10.1 MICHAELIS–MENTEN KINETICS AND MONOD KINETICS

A knowledge of the biochemical reaction kinetics will allow us to predict not only thereaction rates but also the present and future concentrations of a pollutant involved

TABLE 5.8Photolytic Rate Constants for Compounds in Airand Water

Air EnvironmentReaction J (s−1)

O3 → O2 + O(1D) 10−5 at 10 km, 10−3 at 40 km

NO2hν−→ NO + O 0.008 (surface), 0.01 (30 km)

NO3hν−→ NO + O2 0.016

NO3hν−→ NO2 + O 0.19

CH3COCH3hν−→ CH3 + CH3CO 12.4 × 10−6

HCHOhν−→ HCO + O 10.1 × 10−6

CO2hν−→ CO + O

(3P)

2.2 × 10−8

Water EnvironmentCompound J (s−1)

(PAHs)Naphthalene 2.7 × 10−6

Pyrene 2.8 × 10−4

Anthracene 2.6 × 10−4

Chrysene 4.4 × 10−5

(Pesticides)Malathion 1.3 × 10−5

Sevin 7.3 × 10−7

Trifluralin 2.0 × 10−4

Mirex 2.2 × 10−8

Parathion 8.0 × 10−7

Source: Seinfeld, J.H. and Pandis, S.N. 1998. Atmospheric Chemistry and Physics of Air Pollution.NewYork,NY: JohnWiley&Sons, Inc.; Finlayson-Pitts,B.J. andPitts, J.N. 1986.AtmosphericChemistry. NewYork, NY: John Wiley & Sons, Inc.; Lyman, W.J. et al. 1982. Handbook ofChemical Property Estimation Methods. New York: McGraw Hill Book Co.; Warneck, P.1988. Chemistry of the Natural Atmosphere. NewYork: Academic Press.

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in a natural biochemical process. It will also give us information as to how onecan influence a reaction to obtain a desired product in exclusion of other undesiredproducts. Biochemical reactions can be thermodynamically feasible if the overall freeenergy change is negative. However, even if ΔG is negative, not all of the reactionswill occur at appreciable rates to be of any use in a living cell unless other criteriaare satisfied. The answer to this is catalysis caused by the enzymes present in allliving organisms. Enzymes are a class of proteins that are polymeric chains of aminoacids held together by peptide bonds. Chemical forces arrange these amino acids ina special three-dimensional pattern. The enzymes are designed to bind a substrate toa particular site where the conversion to the product species occurs. This active siteis designed to accommodate specific amino acids in a given arrangement for optimalcatalysis. Enzyme catalysis is, therefore very specific, but quite fast. In general, thereexists a complementary structure relationship between the active site of the enzymeand the molecule upon which the enzyme acts. Two models represent this. The firstone is the lock-and-key model, where the substrate and enzyme fit together like ajigsaw puzzle (Figure 5.18a). In the second model called the induced fit model, theactive site wraps around the substrate to adopt the required conformation to bringabout catalysis (Figure 5.18b).

The rate of an enzyme-catalyzed reaction can be obtained by following theconversion of the substrate into products. The basic mechanism is as follows:

E + Sk1�k−1

E − Sk2−→ E + P, (5.195)

where E is the enzyme, S the substrate, and P the product. Note that the above mech-anism is a special case of the general catalysis mechanism described in Section 5.7.

Lock-and-key model

S

S

E E

S

S

EE

Induced fit model

FIGURE 5.18 Different models of enzyme–substrate interactions.

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The enzyme–substrate complex E–S is assumed to be at pseudo-steady-state. Conser-vation of enzyme concentration is evident from the above mechanism, and we have[E]0 = [E] + [E − S]. For the substrate, we have [S]0 = [S] + [P], with the caveatthat [E] ≈ [E − S] � [S]. The rate of formation of the product is given by

r = d[P]dt

= −d[S]dt

= k2[E − S]. (5.196)

The rate of formation of [E-S] is given by

d[E − S]dt

= k1([E]0 − [E − S])[S] − (k−1 + k2)[E − S]. (5.197)

Applying the pseudo steady-state approximation,

[E − S]ss = k1[E]0[S]k1[S] + k−1 + K2

. (5.198)

The expression for the rate can now be written as

r = Vmax[S]Km + [S] (5.199)

with Vmax = k2 [E]0 andKm = (k−1 + k2)/k1. The equation given above is called theMichaelis–Menten equation. Note the similarity of the equation to that derived for

Vmax

Vmax/2

[S]1/2 = Km

Slope = Vmax[S]/Km

[S]/mol·m–3

r/m

ol·m

–3 ·s–1

FIGURE 5.19 Variation in an enzyme-catalyzed reaction rate with substrate concentration asper the Michaelis–Menten equation.

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heterogeneous catalysis obeying the Langmuir isotherm. Figure 5.19 represents theequation. The value of r used is generally the initial reaction rate. The limiting condi-tions of the equation are of special interest.When [S] � Km, that is, at high substrateconcentration, r → Vmax, themaximum rate achieved.At low substrate concentration,[S] � Km, r → (Vmax/Km)[S], that is, the rate is directly proportional to the substrateconcentration. It is seen from the rate equation that when r → Vmax/2,Km → [S]1/2,which is the substrate concentration at which the enzyme reaction rate is one halfof the maximum rate. The parameters Vmax and Km therefore characterize the kinet-ics of a reaction catalyzed by enzymes. The following example is an illustration ofhow these constants can be obtained from experimental data on enzyme-catalyzedsubstrate conversions.

EXAMPLE 5.20 ESTIMATIONOFTHEMICHAELIS–MENTENKINETICPARAMETERS

FROM EXPERIMENTAL DATA

The enzymatic degradation of a pollutant to CO2 and H2O was measured over a rangeof pollutant concentration [S] (mol/L). The initial rates are given below:

[S] (mol/L) r (mM/min)

0.002 3.30.005 6.60.010 10.10.017 12.40.050 16.6

Obtain the Michaelis–Menten constants.

TheMichaelis–Menten equation can be recast in three different forms to obtain linearfits of the experimental data whereby the values of Km and Vmax can be determinedfrom the slopes and intercepts of the plots. These are:

a. Lineweaver–Burke plot: A plot of 1/r versus 1/[S]:1

r= Km

Vmax

1

[S] + 1

Vmax. (5.200)

The slope is Km/Vmax and the intercept is 1/Vmax.b. Langmuir plot: A plot of [S]/r versus [S]:

[S]r

= 1

Vmax[S] + Km

Vmax. (5.201)

The slope is 1/Vmax and the intercept is Km/Vmax.c. Eadie–Hofstee plot: A plot of r versus r/[S]:

r = −Kmr

[S] + Vmax. (5.202)

The slope is −Km and the intercept is Vmax.

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The three plots for the above data are shown in Figure 5.20. The values of Kmand Vmax obtained are given below:

Method r2 Km/mM Vmax/mM/min

Lineweaver–Burke 0.9999 10.1 20.0Langmuir 0.9998 10.0 19.9Eadie–Hofstee 0.9986 10.0 19.9

Often the Lineweaver–Burke plot is preferred since it gives a direct relationshipbetween the independent variable [S] and the dependent variable r. There is one short-coming, that is, as [S] → 0, 1/r → ∞, and hence is inappropriate at low [S]. The valueof the Eadie–Hofstee plot lies in the fact that it gives equal weight to all points unlikethe Lineweaver–Burke plot. In the present case, the Lineweaver–Burke plot appears tobe the best.

0.05

0.1

0.15

0.2

0.25

0.3

0.35(a) (b)

(c)

0 100 200 300 400 500 600

Lineweaver-Burke Plot

y = 0.049933 + 0.00050611x R2

= 0.99991

1/r/

(mM

/min

–1)

1/[S]/M–1

0.0005

0.001

0.0015

0.002

0.0025

0.003

0.0035

0 0.01 0.02 0.03 0.04 0.05 0.06

Langmir Plot

y = 0.00050409 + 0.050194x R2

= 0.99988

[S]/r

/(M/m

M.m

in–1

)

[S]/M

2

4

6

8

10

12

14

16

18

200 400 600 800 1000 1200 1400 1600 1800

Eadie-Hofstee plot

y = 19.932 - 0.010049x R2

= 0.99859

r/m

M.m

in–1

r/[S]/mM/M.min–1

FIGURE 5.20 Estimation of Michaelis–Menten parameters using different methods.(a) Lineweaver–Burke plot, (b) Langmuir plot, and (c) Eadie–Hofstee plot.

Michaelis–Menten kinetics considers the case where the number of living cellsproducing the enzymes is so large that little or no increase in cell number occurs. Inother words, the Michaelis–Menten law is applicable for a no-growth situation with

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a given fixed enzyme concentration. We know that food metabolism in biologicalspecies leads to growth and reproduction. As cells grow rapidly and multiply innumber, the concentration of the enzyme and the substrate degradation rate will vary.These aspects come under growth kinetics or cell kinetics.

Generally the growth of microorganisms occurs in phases. Figure 5.21 is a typicalgrowth curve. A measure of the microbial growth is the cell number density (numberof cells per unit volume) denoted [X]. During the initial stage the change in celldensity is zero for some time. This is the lag phase. Subsequent to the lag phase isthe phase of growth and cell division. This is called the accelerated growth phase.This is followed by the exponential growth phase in which the growth rate increasesexponentially. Further the growth rate starts to plateau. This is called the deceleratedgrowth phase. This leads to a constant maximum cell density in what is called thestationary phase. As time progresses, the cells deplete the nutrient source and beginto die. The cell density rapidly decreases during the final stage called the deathphase.

The rate of cell growth per unit time is represented by the following equation:

rg = d[X]dt

= ζ[X], (5.203)

where ζ is the specific growth rate (h−1). The effect of substrate concentration [S] onζ is described by the Monod equation:

ζ = ζmax[S]Ks + [S] . (5.204)

Note that when [S] � Ks, ζ ∝ [S] and when [S] � Ks, ζ = ζmax. The above equa-tion is an empirical expression and states that if the nutrient concentration [S] reaches

Lag phase

Log

cell

num

ber (

dens

ity)

Constantgrowthphase

Acceleratedgrowthphase

Deathphase

Exponentialgrowth phase

t/h

FIGURE 5.21 A typical growth curve for microorganisms.

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a large value, the specific growth rate plateaus out. However, this is not always thecase. In spite of this, theMonod kinetics is a simplified expression for growth kineticsat low substrate concentrations.

Introducing the Monod equation in the overall expression we get

rg = d[X]dt

= ζmax[S]Ks + [S] [X]. (5.205)

When [S] � Ks, we have

rg = ζmax

Ks[S][X] = k′′[S][X], (5.206)

where k′′ is a pseudo second-order rate constant. This expression is an adequaterepresentation for the growth rate of microorganisms in the soil and groundwaterenvironments (McCarty et al., 1992). k′′ has units of L/mol h (or dm3/mol h). Notethat although Monod kinetics for cell growth and the Michaelis–Menten kinetics forenzyme catalysis give rise to similar expressions, they do differ in the added term [X]in the expression for Monod kinetics.

The Michaelis–Menten expression gives the rate at which a chemical transfor-mation occurs if the enzyme was separated from the organisms and the reactionconducted in solution. If we represent the rate r as the change in moles of substrateper unit organism in a give volume per time, then R = r[X]0 is the moles of chemicaltransformed per unit volume per unit time. [X]0 is the initial number of organismsper volume of solution. If r is given by the Michaelis–Menten expression with Vmaxin moles per organism per time, we have

R = −d[S]dt

= Vmax[S]Ks + [S] [X]0. (5.207)

The cell yield yc is defined as the ratio of mass of new organisms produced to themass of chemical degraded:

yc = (d[X]/dt)−(d[S]/dt) . (5.208)

Note that the maximum specific growth rate ζmax is the product of the maximum rateof degradation (Vmax) and yc:

ζmax = Vmaxyc. (5.209)

Hence we can write the Monod equation as

−d[S]dt

= Vmax[S]Ks + [S] [X]. (5.210)

If we identifyKm withKs and [X] with [X]0, one can infer theMonod kinetics expres-sion directly from the Michaelis–Menten kinetics expression. This corroborates the

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earlier statement that the Monod no-growth kinetics and Michaelis–Menten kineticsare identical.

To consider the fact that as organisms grow they also decay at a rate kdec(h−1),one rewrites the Monod equation as

d[X]dt

=(ζmax[S]Ks + [S]

)[X] − kdec[X]. (5.211)

At low values of [S], there is no net increase in organism density since its growth isbalanced by its decay. Thus d[X]/dt = 0, and

[S]min = Kskdec

ζmax − kdec(5.212)

represents the minimum concentration of substrate that can maintain a given micro-bial population. However, the organism may find another compound with a different[S]min that can sustain another level of population. This compound is called aco-metabolite. In the natural environment this type of biodegradation kinetics is oftenobserved.

PROBLEMS

5.12 If the standard enthalpy and entropy of dissolution of naphthalene in waterare 9.9 kJ/mol and 59 J/mol/K, estimate the equilibrium partial pressureof naphthalene over an aqueous solution containing 1 × 10−4 mol/L ofnaphthalene.

5.22 The rate of formation of a pollutant NO(g) by the reaction 2NO2 →2NO(g) + O2(g) is 1 × 10−5 mol/L/s. Obtain the rate of the reaction andthe rate of dissipation of NO2. IfΔG

0 for the given reaction is 35 kJ/mol,what is the concentration of NO2 in ppmv in air that has 21% oxygen at1 atm. The total concentration of NO2 is 10 ppmv.

5.31 For the reaction A + 2B → 3C, assume that it follows the stoichiomet-ric relation given. What is the order of the reaction in B? Write downthe expression for the rate of disappearance of B and the rate of appearanceof A.

5.41 For the following hypothetical reaction, A2 + B2 → 2AB, the reciprocalof the plot of [A] (in moles per 22.4 L) versus time has a slope of−5 whenbothA2 and B2 are at unit concentrations. Determine the rate constant forthe disappearance of B2 in units of m3/mol/s and m3/s.

5.52 Derive the equation for the half-life of an nth-order reaction, nA → C.Would you be able to determine the order of a reaction if the only dataavailable are different initial concentrations of A and the correspondingt1/2 values? If so, how?

5.61 The reaction 2A → C proceeds with a rate constant of 5 × 10−4 mol/L/s:(i) Calculate the half-life of the reaction when C0

A is 2mol/L. (ii) Howlong will it take for CA to change from 5 to 1mol/L?

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5.72 The decomposition of a species A in air occurs via the following reaction:A +A → C.The pressure ofC as a function of twasmonitored in a reactor.

t (min) pC (atm)

0 020 0.0140 0.0560 0.1090 0.12120 0.144 0.16

Obtain (i) the order of the reaction, (ii) the rate constant, and (iii) thehalf-life of A.

5.82 For the following autocatalytic reaction scheme for conversion of Ato C determine the rate of formation of C using the pseudo-steady-stateapproximation:

Ak1−→ B,

B + Ck2−→ C +A.

5.93 Sulfur in the +4 oxidation state participates in a number of atmosphericreactions. S(IV) represents the total concentration in the+4oxidation state:[S(IV)] = [SO2] + [SO2H2O] + [HSO−

3 ]. The second-order reaction ofS(IV) with a component A is given by A + S(IV) → products. (a) Whatis the rate expression for the process? (b) Rewrite the above expression interms of the partial pressures of the species.

5.101 A first-order reaction is 30% complete in 45min.What is the rate constantin s−1? How long will it take for the reaction to be 90% complete?

5.113 An important reaction in both atmospheric chemistry and wastewa-ter treatment is the conversion of ozone to oxygen. In the gas phasethe reaction is 2O3(g) � 3O2(g). This proceeds in steps as follows:

Step 1: O3 + Mka�k′

a

O2 + O + Mand Step 2: O + O3kb−→ 2O2.M is a third

species.

(i) Using the steady-state approximation, obtain an expression forthe decomposition rate of ozone,

(ii) The following data were obtained for the thermal decompositionof ozone in aqueous acidic solutions (Sehestad et al., 1991) at304K in the presence of [HClO4] = 0.01M:

[O2] (μM) [O3] (μM) Initial Rate Constant (s−1)

10 200 1.3 × 10−4

400 200 7.0 × 10−5

1000 200 4.7 × 10−5

1000 95 2.4 × 10−5

1000 380 8.0 × 10−5

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What can you infer regarding themechanismof decomposition of ozonein solution?

5.122 The oxidation of polyaromatic hydrocarbons is an important reac-tion in atmospheric chemistry. The reaction between benzene andoxygen was studied by Atkinson and Pitts (1975) at varioustemperatures.

T (K) k (L/mol/s)

300 1.44 × 10−7

341 3.03 × 10−7

392 6.9 × 10−7

Obtain, Ea, for the above reaction.

5.132 The atmospheric reaction between methane (CH4) and hydroxyl radical(OH!) is of importance in understanding the fate of several hydrocarbonpollutants in the air environment. An activation energy of 15 kJ/mol and arate constant of 3 × 10−12 cm3/molecule/s were determined in laboratoryexperiments for the reaction. Using the temperature profiles in the atmo-sphere (Chapter 2), determine the rate constants for this reaction at variousaltitudes. The approximate distances of the various layers from groundlevel are (i) atmospheric boundary layer (= 0.1 km), (ii) lower troposphere(= 1 km), (iii) middle troposphere (= 6 km), and (iv) lower stratosphere(= 20 km).

5.143 The thermal decomposition of alkane (e.g., ethane) is of importance inenvironmental engineering (e.g., combustion kinetics of petroleum hydro-carbons). A chain reaction mechanism is proposed. Identify the initiation,propagation, and termination steps in the reaction. Using the steady-stateapproximation obtain the overall rate of decomposition of ethane.

C2H6k1−−→ 2CH•

3,

CH•3 + C2H6

k2−−→ C2H•5 + CH4,

C2H•5

k3−−→ C2H4 + H•,

H• + C2H6k4−−→ C2H

•5 + H2,

2C2H•5

k5−−→ n− C4H10,

2C2H•5

k6−−→ C2H4 + C2H6.

5.153 The concentrations of some major ions in natural waters are given below.

Concentration (mmol/kg)

Medium Na+ K+ Ca2+ Mg2+ Cl− SO2−4 NO−

3 F−

Seawater 468 10 10.3 53 546 28 0.05 0.07Riverwater 0.27 0.06 0.37 0.17 0.22 0.12 0.02 0.005Fogwater 0.08 — 0.2 0.08 0.2 0.3 — 1

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The following two reactions are of importance in both atmosphericmoisture and seawater:

(a) oxidation of H2S by peroxymonosulfate (an important oxi-dant in clouds), HSO−

5 + SO2−5 → 2SO2−

4 + O2 + H+, with k =0.1 L/mol/s at 298K in distilled water and

(b) oxidation of nitrite by ozone to form nitrate, NO−2 + O3 → NO−

3 +O2, with k = 1.5 × 10−5 L/mol/s at 298K in distilled water.

Estimate the rate constants of the above reactions in the natural watersgiven above.

5.162 The following rate data were reported by Sung and Morgan (1980) for thehomogeneous oxidation of Fe(II) to Fe(III) in aqueous solution:

I (M) k (L2/mol2/atm/min)

0.009 4.0 × 1013

0.012 3.1 × 1013

0.020 2.9 × 1013

0.040 2.2 × 1013

0.060 1.8 × 1013

0.110 1.2 × 1013

Obtain the product of the charges of the species involved in the reaction.Tamura et al. (1976) proposed the following rate-limiting step for the reac-tion: FeOH+ + O2OH

− → Fe(OH)+2 + O−2 . Does your result support

this contention?5.173 Corrosion inwastewater collection systems by sulfuric acid generated from

H2S is a nuisance. Oxidation of H2S by O2 is therefore an important issue.The rate of this reaction is accelerated by catalysts. Kotranarou and Hoff-mann (1991) discussed the use of CoIITSP (a tetrasulfophthalocyanine)as a possible catalyst for auto-oxidation of H2S in wastewater. The rateconstant for the reaction was of the form k = k′[O2]/(K + [O2]). Thefollowing reaction scheme was suggested:

CoII(TSP)4−2 � 2CoII(TSP)2−,

CoII(TSP)2− + HS− � HSCoII(TSP)3−,

HSCoII(TSP)3− + O2 � HSCoII(TSP)(O−•2 )3−,

HSCoII(TSP)(O−•2 )3− + HS− � HSCoII(TSP)(O−•

2 )(HS)4−,

HSCoII(TSP)(O−•2 )(HS)4− slow−−−→ HSCoII(TSP)(O2−

2 )5−S0 + H,

HSCoII(TSP)(O2−2 )5− + 2H+ fast−−→ HSCoII(TSP)3− + H2O2.

Obtain the rate expression for−d[S(−II)]/dt, where S(−II) is the reducedform of sulfur in the above reaction scheme. Simplify the expression forthe pseudo-first-order kinetics to show that the rate constant has the generalform given above.

5.183 Different isomers of HCH are abundant pesticides in the world’s oceansand lakes. They are also known to be present in atmospheric particles and

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in the sedimentary environment. Ngabe, Bidleman, and Falconer (1993)studied the hydrolysis rate of the α-isomer using buffered distilled waterwhere all other processes including biotransformations and photo-assistedprocesses were suppressed. The data obtained were as follows:

T (K) pH k (min−1)

278 9.04 5.01 × 10−7

288 9.01 4.19 × 10−6

303 9.01 9.74 × 10−5

318 9.01 6.45 × 10−4

318 7.87 4.86 × 10−5

318 7.01 7.52 × 10−6

Given that α-HCH is resistant to acid attack, use the above data to obtain(i) the specific neutral and base-catalyzed hydrolysis rate constants, (ii)the relative contribution of each mechanism, and (iii) the activation energyfor the base-catalyzed reaction.

If a local lake in Baton Rouge, Louisiana, has an average pH of7.2 and a temperature of 14EC, what is the time required to reduce theα-HCH concentration in the lake to 75% of its original value? Neglect anysedimentation or biotransformation in the lake.

5.193 Consider a body of water containing 10−4 M of a pollutant A that is incontactwith air that contains hydrogen sulfide.The dissolvedH2S convertscompoundA to its reduced form according to the equationAox + H2S →Ared with an EσH of 0.4V. Given that the total gas+ aqueous concentration

of hydrogen sulfide is [H2S]T10−2 M and that the pKa of H2S is 7 at

298K, obtain the following: (i) the redox potential at a pH of 8.5 (note:EσH H2S(aq)) = +0.14V at 298K and (ii) the fraction of Aox convertedat pH 8.5.

5.203 Chlorine dioxide (ClO2) is used as an oxidant in wastewater treatmentand drinking water disinfection processes. It is preferred over other oxi-dants since it produces no chloroform and very few other chloroorganics.It also has its limitations since one of its reduced forms (chlorite) is ablood poison at high concentrations. The reaction of ClO2 with a pol-lutant A follows the stoichiometric equation: ClO2 + αA → products,with α = (ΔA)/(ΔClO2), and follows the rate law −d[ClO2]/dt = ktot[ClO2]

n[A0]n. In a series of experiments Hoigne and Bader (1994) used

[A0] � 5[ClO2] to determine the rate constants for the above reactionwithseveral pollutants. The reaction with H2O2 is of particular importance.Since H2O2 dissociates with pH changes, ktot = kHA + β(kA− − kHA),with HA denoting the undissociated H2O2 and A− its dissociated formHO−

2 . β is the fraction in the dissociated form. The pKa of H2O2 is 11.7.

kHA = 0.1 L/mol s, kA = 1.3 × 10−5 L/mol s at 298Kwith [H2O2] in therange 0.3–20mM and pH between 2 and 11. Determine the half-life of a10μM H2O2 solution at a pH of 8 if the water is disinfected with 1μMof ClO2.

5.213 Barry et al. (1994) studied the oxidation kinetics of Fe(II) (a prevalentmetalin most natural waters) in an acidic alpine lake. The total rate of oxidationwas found to be a combination of the following: (i) homogeneous oxidation

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governed by

r = −d[Fe(II)]dt

=(k0 + (k1K1/[H+])+

(k2β2/[H+]2

))[Fe2+]PO2 ,

where K1 and β2 are respectively the equilibrium and stability constantsof the two major hydroxo species, Fe(OH)+ and Fe(OH)2 (refer to Sec-tion 5.8.1); (ii) an abiotic oxidation given by r = k′3A[Fe(II)][OH−]2PO2 ,where A is the area of the surface to which iron is attached; and (iii) abiotic oxidation process given by r = k4[Bac][Fe(II)][OH

−]2PO2 , with[Bac] measured as grams volatile solids per liter. The rate constants at298K were determined to be k0 = 1 × 10−8 s−1, k1 = 3.2 × 10−2 s−1,k2 = 1 × 104 s−1, k′3 = 4.9 × 1014 (L/mol)2/atm/m2/s, k4 = 8.1 × 1016

(L/mol)2/atm/s/(g volatile solids/L),K1 = 10−9.5, β2 = 10−20.6, andA =4.2m2/g. Calculate the half-life for Fe(II) in the acidic lake if the pH is5.2. The sediment showed a 7% loss of weight upon combustion. Thesuspended sediment concentration in the lake water was 1 g/L.

5.222 Determine the equilibrium constant for the following reactions at 298K.Assume unit activities of all species except [H+] = 10−6 M.

(i) PbO(s) + 2H+(aq) � Pb2+(aq) + H2O(l).(ii) 1

4CH2O + Fe(OH)3(s) + 2H+ � Fe2+ + 14CO2 + 11

4 H2O.

The first reaction is of importance in corrosion problems, whereas thesecond reaction is an important oxidation reaction occurring in soils andsediments.

5.232 Nitrogen fixation in the environment is mediated by a bacteria callednitrobacter. The conversion of N2 to NH

+4 in nature is the primary reduc-

tion reaction at pH 7. This is accompanied by the oxidation of naturalorganic matter (designated CH2O) to CO2. What are the free energychanges for the two half reactions? What is the total free energy of theredox process? How is the reaction affected in a slightly acidic lake water(pH = 6)?

5.242 Assume a linear adsorption constant Kd for a nonionic organic pol-lutant with suspended particles of concentration ρp. If the pollutanthas an average rate constant of kads in the sorbed state and kaq inthe aqueous phase for transformation to a product B, obtain the ratioof the overall reaction rate kobs to kaq. For the following condi-tions, how is the rate affected: (a) lake waters of ρp = 10mg/L; (b)bottom sediments and soils of ρp = 106 mg/L. Discuss your resultswith respect to the two compounds 1,2-DCE and p, p′-DDT.

5.252 The oxygen transfer coefficient (kla) into a biological fermentor can beobtained using the sodium sulfite method, where the following reaction is

supposed to occur: Na2SO3 + (1/2)O2Co2+−−−→ Na2SO4. Co

2+ catalyzesthe reaction. In a batch fermentor filled with 0.01m3 of 0.5M sodiumsulfite containing 0.002M Co2+ ion, fresh air is bubbled in at a constantrate. After 10min of bubbling, an aqueous sample (10 × 10−6 m3) waswithdrawn and analyzed for sodium sulfite. The final concentration ofsodium sulfite was 0.2M. Calculate kla for oxygen from the above data.Given that the temperature of the experiment was 298K and the pressurewas 1 atm, obtain the Henry’s constant for oxygen from the text.

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5.263 Sedlak and Andren (1994) showed that the aqueous-phase oxidation ofpolychlorinated biphenyls (PCBs) by OHX is influenced by the presenceof

A − Pk1−→k−1

A + P,

A + OH• k2−−→ A − OH

particles such as diatomaceous earth (represented as P). The mecha-nism representing this is given below, where A–OH is the hydroxylatedPCB. Using appropriate assumptions derive the following equation for theconcentration of A in the aqueous phase:

[A][A]0 = k1

(β− α) (e−αt − e−βt),

where αβ = k1k2[OH•] and α+ β = k1 + k−1[P] + k2[OH•]. For2,2′,4,5,5′-pentachlorobiphenyl, k1 = 4.53 × 10−4 s−1, k2 = 1.41 L/g s,and k2 = 4.6 × 10−9 cm3/mol s.At an average concentration of [OHX ] =1 × 10−14 mol/L, and [P]/[Fe] = 10μM, sketch the progress of hydroxy-lation of PCB.Repeat your calculation for [P] = 50μM.What conclusionscan your draw from your plots?

5.272 The photochemical reaction in the upper atmosphere that leads to theformation and dissipation of O2 follows the following pathway:

O2hν−−→ O(1D) + O,

O(1D) + Mk2−−→ O + M.

O(1D) is a transient state (singlet) for O atom and M is a third bodysuch as N2. The rate of the first reaction is represented as R1. If [M] =1014 molecules/cm3 and k2 = 1 × 10−25 cm6/(molecule)2 s, determinethe steady-state concentration of singlet oxygen as a function of R1.

5.282 Atmospheric photo-oxidation of propane to acetone is said to occur by thefollowing pathway:

CH3CH2CH3 + OH• −→ CH3CH•CH3 + H2O,

CH3CH•CH3 + O2

fast−−→ CH3CH2(O•2)CH3,

CH3CH2(O•2)CH3 + NO −→ NO2 + CH3CH(O•)CH3,

CH3CH(O•)CH3 + O2fast−−→ CH3COCH3 + HO•

2.

Obtain an expression for the rate of production of NO2 due to thedissipation of propane. Make appropriate assumptions.

5.293 Atmospheric chemists consider the oxidation of CH4 by OH• to be aconvenient way of estimating the concentration of hydroxyl ion in the

atmosphere: CH4 + OH• k−→ CH•3 + H2O. In so doing, it is assumed that

the hydroxyl radical concentration is constant, since it is much larger thanthe methane concentration. Given that k = 107m3/mol s at 300K and that

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methane has an atmospheric residence time of 10 years, obtain the meanOH• concentration in the atmosphere.

5.302 The acid-catalyzed decomposition of urea has a rate constant of 4.3 ×10−4 h−1 at 334K and 1.6 × 10−3 h−1 at 344K. Estimate the activationenergy and pre-exponential factor for the reaction. What is the rate at aroom temperature of 300K?

5.312 A first-order process has the following rate constants at different ionicstrengths of a 1:1 electrolyte:

I (mol/L) 0.007 0.01 0.015 0.020K (h−1) 2.06 3 5.1 7.02

If one of the ions involved in the reaction has a positive charge, what is thecharge on the other ion?

5.321 The rate constant for the decay of natural 14C radio isotope is 1.2 ×10−4 y−1. What is the half-life of 14C?

5.333 The hydrolysis of ArOCOCH3 to ArOH is base catalyzed. At values ofpH > 2 acid catalysis is unimportant. The uncatalyzed reaction rate k0 is4 × 10−5 s−1 at 20◦C.At pH = 8,what is the rate constant of the reaction?How much time will it take to reduce the concentration of the compoundto 95% of the original value in a lake of pH 8 at 20◦C?

5.343 Pentachloroethane (PCA) is spilled into the Mississippi river at a meanpH of 7.2 and a temperature of 20◦C. Haag and Mill (1988) report thefollowing parameters for the uncatalyzed and base-catalyzed reactions ofPCA:

k0 (h−1) = 4.9 × 10−8,Ea = 95 kJ/mol,

kb (h−1) = 1320,Ea = 81 kJ/mol.

How long will it take to reduce the concentration of PCA in the river to95% of its original concentration?

5.353 Loftabad et al. (1996) proposed that polyaromatic hydrocarbons (≡A) incontact with dry clay surfaces underwent oligomerization. This was inhib-ited by moisture present in the soil. The reaction scheme proposed was asfollows:

S +A � S −A,

S +W � S −W,

S −A −→ S + products.

W represents water and S represents a surface site for compoundA. Basedon the above, derive the following rate expression for the reaction of A:−rA = k0CA/(1 + Kθ), where K is the equilibrium constant for wateradsorption and θ is the water content (g/kg of soil). k0 is the apparent rateconstant in dry soil (θ = 0). State the assumptions made in deriving therate expression.

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5.362 The decomposition of N2O5 to NO2 in the gas phase occurs via thefollowing reaction scheme:

N2O5kf1�kb1

NO2 + NO3,

NO2 + NO3k2−→ NO + O2 + NO2,

NO + NO3k3−→ 2NO2.

Derive the following equation for the overall decomposition rate of N2O5by using the PSSA to show that −d[N2O5]/dt = ktot[N2O5]. Identify theexpression for ktot .

5.373 A complete and general mechanism for enzyme catalysis should include areversible decomposition of the E–S complex:

E + Sk1�k−1

E − Sk2�k−2

E + P

Derive an equation for the rate of product formation. Contrast it withthat derived in the text where reversible decomposition of E–S was notconsidered.

5.382 Some substrates can bind irreversibly to either an enzyme or the enzyme–substrate complex and interfere with enzymatic reactions. These are calledinhibitors and are of two types:

(a) Derive the rate expressions for product formation in each case andsketch the behavior as Michaelis–Menten plots.

(b) The hydrolysis of sucrose is said to be inhibited by urea. The ratedata for the same are given below:

[S] (mol/L) r (mol/L s) r (mol/L s) with [I] = 2 M

0.03 0.18 0.080.06 0.27 0.110.10 0.33 0.180.18 0.37 0.190.24 0.37 0.18

What type of an inhibitor is urea?5.391 The uptake of CO2 in water and subsequent dissociation into bicarbon-

ate ion is shown to be accelerated by an enzyme called bovine carbonicanhydrase. The following data are provided on the initial rate of thereaction:

CO2(aq) + H2O(aq)E−→ H+(aq) + HCO−

3 (aq).

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at pH 7 for different CO2(aq) concentration:

[CO2]0 (mol/dm3) r0 (mol/dm3 s)

1.25 × 10−3 3 × 10−5

2.5 × 10−3 5 × 10−5

5 × 10−3 8 × 10−5

2 × 10−2 1.5 × 10−4

Obtain the Michaelis–Menten parameters for the above reaction.5.403 The atmospheric reaction of chlorine monoxide (ClO) with NO2 proceeds

as follows:

ClO + NO2 + Mkf�kb

ClONO2 + M

with kb = 10−6.16 exp(−90.7 kJ/mol/RT) in units of cm3/molecule/s. Thehigh and low pressure limiting rate constant at 298K were obtained fromexperiments as follows:

k0 = 1.8 × 10−31 cm6/molecule2/s

k∞ = 1.5 × 10−11 cm3/molecule/s.

Obtain the effective bimolecular rate constant kf at 298K and 1 atm.5.412 A recent article (Li et al., 2008) showed that apart from the well-

known ozone dissociation that produces OH in the atmosphere, there alsoexists another pathway for OH production. This involves the reaction ofelectronically excited NO2 with H2O as follows:

NO2J1−−→hν

NO∗2,

NO∗2 + M

k2−→ NO2 + M,

NO∗2 + H2O

k3−→ OH + HONO,

HONOJ4−−→hν

OH + NO.

Apply a steady-state approximation to NO∗2 to obtain the following

equation:

rOH = 2J1k3[NO2][H2O]k3[H2O] + k2[M] .

5.422 An autocatalysis reaction is one in which the reactant is re-formed as thereaction proceeds:

Ak1−−→ B,

B +Ak2−−→ C +A.

(a) Using the pseudo-steady-state approximation, derive the rate equa-tion for the formation of the product C.

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(b) If the overall rate constant is 2 h−1, and the initial concentration ofA is 10mol/m3, what is the initial rate of formation of C?

5.432 For the reaction 2Ak−→ B,

(a) Write the differential rate equation for the dissipation of A.(b) Integrate the above expression with the initial condition that at t = 0,

[A] = [A]0, and obtain the equation for A as a function of time.(c) If the rate constant is 2 h−1, what is the half-life of the reaction?

5.443 The decomposition of di-2-methylpropan-2-yl into propanone and ethanegoes through free radical mechanisms given as follows:

(CH3)COOC(CH3)3k1−−→ 2(CH3)3CO

•,

(CH3)3CO• k2−−→ CH3COCH3 + CH•

3,

CH•3 + CH•

3k3−−→ C2H6.

Formulate the expression for the rate of production of ethane and showthat the reaction is first order in the reactant di-2-methylpropan-2-yl. Usethe pseudo-steady-state approximation and perform algebra. Rememberthat two (CH3)3CO

• radicals are formed for every one molecule of thereactant.

5.452 For the reaction A + Bk�k−1

Ck2−−→ D, obtain the expression for the rate of

production of D in terms of the reactant concentrations. What is the orderof the overall reaction?

5.463 The reaction of gas-phase ozone with dimethyl sulfide (DMS) in a glassreactor occurs via the following two-step mechanism:

O3k1−−→ wall decay,

O3 + DMSk2−−→ products.

Experiments were conducted under a large excess of DMS concentration,[DMS]0, to obtain the value of −d ln[O3]/dt (Du et al., 2007):

[DMS]0/1014 mol/cm3 −d ln[O3]/dt/105 s−1

0 0.6093.63 6.004.84 7.296.05 7.83

Derive the rate equation for the conditions stated, and from an appropriateplot of the data obtain k1 and k2.

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Barry, R.C., Schnoor, J.L., Sulzberger, B., Sigg, L., andStumm,W. 1994. Iron oxidation kineticsin an acidic alpine lake.Water Research 28, 323–333.

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Butler, J.N. 1982. Carbondioxide Equilibria and its Applications, 2nd ed. NewYork:Addison-Wesley Publishing Co.

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Hoffmann, M.R. 1990. Catalysis in aquatic environments. In: W. Stumm (Ed.), AquaticChemical Kinetics, pp. 71–112. NewYork, NY: John Wiley & Sons, Inc.

Hoigne, J. and Bader, H. 1994. Kinetics of reactions of OClO in water I. Rate constants forinorganic and organic compounds.Water Research 28, 45–55.

Kotranarou, A. and Hoffmann, M.R. 1991. Catalytic autooxidation of hydrogen sulfide inwastewater. Environmental Science and Technology 25, 1153–1160.

Laidler, K.J. 1965. Chemical Kinetics. NewYork: McGraw-Hill.Li, S., Matthews, J., and Sinha, A. 2008. Atmospheric hydroxyl radical production from

electronically excited NO2 and H2O. Science 319, 1657–1659.Loftabad, S.K., Pickard, M.A., and Gray, M.R. 1996. Reactions of polynuclear aromatic

hydrocarbons on soil. Environmental Science and Technology 30, 1145–1151.Lyman, W.J., Reehl, W.F., and Rosenblatt, D.H. 1982. Handbook of Chemical Property

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6 Applications ofChemical Kinetics inEnvironmental Systems

In Chapter 2, we discussed the differences between open and closed systems. Naturalenvironmental systems are open systems. Both material and energy are exchangedacross the system boundaries. However, it is convenient and useful to model them asclosed systems as well if we consider a fixed composition and ignore any changesin composition. Many waste treatment operations are conducted as closed systems,where exchange of energy and/or mass occurs among compartments within thesystem, but not necessarily with the surroundings.

The rate laws described in Chapter 5 are for closed systems, where the concen-tration of a pollutant changes with time, but there is no mass exchange with thesurroundings. Rarely this represents chemical kinetics in the natural environment,where we encounter time-varying inputs and outputs. The discussion that followsis derived from the chemical engineering literature (Hill, 1977; Levenspiel, 1999;Fogler, 2006).

Applications of kinetics in environmental engineering F&Tmodeling and in wastetreatment operations are numerous (Thibodeaux, 1996). The discussion will start withthe conventional reactor theory. Various types of reactors (ideal and nonideal) willbe discussed. This will be followed by several applications in the hydrosphere, atmo-sphere, and lithosphere. Finally, the chemical kinetics for the biosphere with specialrelevance to biodegradation, enzyme kinetics, and bioaccumulation will be discussed.Several problems are provided, which serve to illustrate the concepts presented withineach section.

6.1 TYPES OF REACTORS

6.1.1 IDEAL REACTORS

Chemical reactors are classified into batch and flow systems (Figure 6.1). In a batchreactor, a fixed volume (mass) of material is treated for a specified length of time.If material flows into and out of the reactor in a continuous manner, it is called acontinuous flow reactor.

To analyze a reactor, we use the fundamental concept of mass balance or conser-vation of mass. This principle dates back to 1789 when the renowned French scientist,Antöine Lavoisier, considered to be the father of modern chemistry, first enunciated it

267

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Feed Effluent

CSTR (completely mixed)

Feed

(a) (b)

(c)

Batch reactor(completely mixed)

Effluent

PFR (no mixing)

FIGURE 6.1 Different types of ideal reactors used in environmental engineering processes.(a)A batch reactor where there is uniform concentration in the reactor. (b)A CSTRwhere thereis uniform mixing and composition everywhere within the reactor and at the exit. (c) A PFRwhere there is no mixing between the earlier and later entering fluid packets.

in hisTraité élémentaire de chimie. His observations translated intoEnglish (Chemicaland Engineering News, September 12, 1994) reads as follows:

For nothing is created, either in the operations of art, or in those of nature, andone can state as a principle that in every operation there is an equal quantity ofmaterial before and after operation; that the quality and quantity of the [simple]principles are the same, and that there are nothing but changes, modifications.

We have used the conservation of mass in several examples in earlier chapters.The examples involved the equilibrium partitioning between different environmentalcompartments; in this chapterwe generalize it to chemical systems involving transportand transformations. We establish a control volume as shown in Figure 6.2, wherethe general case of a continuous system is represented. The feed rate of solute j is Fj0(moles/time). The effluent rate from the system is Fj (moles/time). If r represents therate of the reaction, then Vvjr represents the rate of generation of j (moles/time). Notethat vj is the stoichiometric coefficient of j for the reaction as shown in Chapter 5.A total mass (mole) balance within the reactor is then written as (Schmidt, 2005)

[accumulation] = [input] − [output] + [generation by reaction]

dNjdt

= Fj0 − Fj + Vvjr. (6.1)

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Applications of Chemical Kinetics in Environmental Systems 269

Feed

Reactor volumeV

Concentration[A]

Effluent

Fj 0 Fj

Accumulation Reaction

FIGURE 6.2 Mass balance for a reactor.

Note that Nj is the total moles of j in the reactor. The above equation is the basisfor reactor design. Note that if there are additional source terms within the system,they are included in the input term. Specifically, we can determine either the time orthe reactor volume required for a specified conversion of the reactants to products.

6.1.1.1 Batch Reactor

Since both Fj0 and Fj are zero in this case, we have

dNjdt

= Vvjr. (6.2)

Consider a first-order reaction A → B for which r = kCA and vA = −1. Hence,

dNA

dt= −kCAV , (6.3)

where V is the total volume of the reactor. Further, since NA = CAV in a constant-volume batch reactor, we can write

dCA

dt= −kCA. (6.4)

Thus, for this case the rate of change in concentration ofA in the reactor, CA(t), isgiven by the rate of the reaction. We can substitute the appropriate expression for therate (rA) as given in Chapter 5 and solve the resulting differential equation for CA(t).A batch reactor is an unsteady-state operation, where the concentration changes withtime, but is the same throughout the reactor.

6.1.1.2 Continuous-Flow Stirred Tank Reactor

A common type of reactor encountered in environmental engineering is thecontinuous-flow stirred tank reactor (CSTR) in which the effluent concentration is

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the same as the concentration everywhere within the reactor at all times (Figure 6.1).In other words, the feed is instantaneously mixed in the reactor. The mole balance fora pollutant j in the CSTR is

dNjdt

= Fj0 − Fj + Vvjr. (6.5)

In many cases, the above equation can be further simplified by assuming a steadystate such that dNj/dt = 0, that is, the rate of accumulation is zero. If we also stipulatethat the rate of the reaction is uniform throughout the reactor, we obtain the followingequation:

Fj = Fj0 + Vvjr. (6.6)

Thus, the volume of a CSTR is

V = (Fj − Fj0)

vjr. (6.7)

Let us consider the special case where we have a constant volumetric feed flowrate Q0 and an effluent volumetric flow rate of Qe. If CA0 and CA represent theconcentration ofA in the feed and the effluent, respectively, we canwriteFj0 = Q0Cj0and Fj = QCj. Then, we have the following equation for the reactor volume:

V = (QCj − Q0Cj0)

vjr. (6.8)

6.1.1.3 Plug-Flow Reactor (PFR) or Tubular Reactor

Another type of a continuous-flow reactor is the tubular reactor, where the reactionoccurs such that radial concentration gradient is nonexistent, but a gradient exists inthe axial direction. This is called plug flow. The flow of the fluid is one in whichno fluid element is mixed with any other either behind or ahead of it. The reactantsare continually consumed as they flow along the length of the reactor and an axialconcentration gradient develops. Therefore, the reaction rate is a function of the axialdirection. A polluted river that flows as a narrow channel can be considered to be inplug flow.

To analyze the tubular reactor, we consider a section of the cylindrical pipe oflength Δz as shown in Figure 6.3. If the rate is constant within the small-volumeelement ΔV = AcΔz and we assume steady state, the following mass balance oncompound j for the volume element results:

Fj(z) − Fj(z +Δz) + Acdzvjr = 0. (6.9)

Since ΔV = AcΔz, where Ac is the cross-sectional area, we can write the aboveequation as

Fj(z +Δz) − Fj(z)

Acdz= vjr. (6.10)

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Fj0 Fj

Fj(z) Fj (z + Δz)

Δ z

FIGURE 6.3 Schematic of a tubular or plug-flow reactor.

Taking the limit as Δz →0 and rewriting, we obtain

1

Ac

dFjdz

= vjr (6.11)

ordFjdV

= vjr. (6.12)

It can be shown that the above equation also applies to the case where the cross-sectional area varies along the length of the reactor. This is the case of a river goingthrough several narrow channels along its path.

If the reaction is first order, namely, A → B, the rate of disappearance of A isr = kCA. The stoichiometry of the reaction is such that νA = −1. If we consider aconstant-density reactor, and note thatFA = Q0CA withQ0 being the volumetric flowrate, we have

−dCA

dV= k

Q0CA. (6.13)

Integrating with the initial condition, CA = CA0 at 0, we obtain

V = Q0

kln

(CA0

CA

)(6.14)

If 50% of CA0 has to be converted into products, then V = 0.693Q0/k is the reactorvolume required.

6.1.1.4 Design Equations for CSTR and PFR

For any general stoichiometric equation of the form aA + bB → cC + dD, we canwrite the following equation on a per mole of A basis:

A + b

aB −→ c

aC + d

aD. (6.15)

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For continuous systems (CSTR and PFR), it is convenient to express the change interms of conversion of A, χ, which is defined as the fractional conversion of A atsteady state. Then the mass balance can be written as follows: molar rate ofA leavingthe reactor,FA = molar rate ofA fed into the reactor,FA0−molar rate of consumptionof A in the reactor, FA0χ.

Therefore,

FA = FA0(1 − χ). (6.16)

For a CSTR, the reactor volume is given by

V = FA0χ

r. (6.17)

For a PFR, we haved(FA0χ)

dV= r. (6.18)

Since FA0 is a constant, we can integrate the above to obtain V .

EXAMPLE 6.1 COMPARISON OF REACTOR VOLUMES FORA CSTR ANDA PFR

In both a CSTR and a PFR, rA can be expressed as a function of the conversion xA. Amost frequently observed relationship between rA and χ (for a first-order reaction) inenvironmental engineering is rA = kCA0(1 − χ). Using this relationship for a CSTR,we have

VCSTR

FA0= χ

kCA0(1 − χ).

For a PFR we obtain the following:

VPFR

FA0= 1

kCA0

∫χ0

χ

(1 − χ).

If the molar feed rate FA0 and initial feed concentrationCA0 are the same in both cases,we have

VCSTR

VPFR= χ

(1 − χ) ln(1/(1 − χ)).

If the desired conversion is χ = 0.6, then the ratio of volumes is 1.63. Thus, a 63%larger volume of the CSTR than that of a PFR is required to achieve a 60% conversion.

EXAMPLE 6.2 A CSTR MODEL FORA SURFACE IMPOUNDMENT

Surface impoundments are used at many industrial sites for the treatment of wastewater.It is generally a holding tank where water is continuously fed for treatment.

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Enhanced aeration (using surface agitation) can volatilize organic compounds fromsurface impoundments. This process also facilitates the growth of bioorganisms thatare capable of destroying the waste. Surface impoundments also perform the functionof a settling pond in which particulates and other sludge solids are separated from thewaste stream. This is called an activated sludge process. Figure 6.4 depicts the threeprimary pathways by which a pollutant is removed from a waste stream entering animpoundment.A surface impoundment can be considered to be a CSTR with complete mixing

within the aqueous phase, and the performance characteristics analyzed. If the reactoris unmixed (i.e., no backmixing) such that a distinct solute concentration gradient existsacross the reactor, the impoundmentwill have to bemodeled as a PFR.The three primaryloss mechanisms for a chemical entering the impoundment are: (i) volatilization acrossthe air–water interface, (ii) adsorption to settable solids (biomass), and (iii) chemical orbiochemical reaction. Each of these loss processes within the reactor can be consideredto be a first-order process. Chemical reactions degrade the pollutant. Exchange betweenphases (e.g., volatilization, settling) within the reactor tends to deplete the chemicalconcentration; it does not, however, change the nature of the chemical. The overallmass balance for a chemical A in the impoundment is

dNA

dt= FA0 − FA + VνArtot ,

where rtot is the total rate of loss of A by all processes within the reactor.The rate of volatilization of A is

rvoln = kvCA.

Effluent Q0CA

Feed Q0CA0

CSTR

Solidssettling

VolatilizationChemical/

biodegradation

FIGURE 6.4 Primary loss mechanisms in a surface impoundment modeled as aCSTR.

continued

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The rate of settling of A on solids in the water column is given by

rset = ksCA.

The chemical reaction rate is given by

rreact = krCA.

The total rate of loss of A from the surface impoundment is given by

rtot = rreact + rvoln + rset = (kr + kv + ks)CA = k∗CA.

The overall first-order rate constant is k∗. For a first-order loss process, νA = −1. Fora constant-density reactor, NA = VCA. If the influent and effluent feed rates are same(=Q0), we have the following mass balance equation:

VdCA

dt= Q0CA0 − Q0CA − k∗CAV .

Assuming steady state,

CA

CA0= 1

1 + (k∗(V/Q0)),

where CA0 is the concentration of A in the influent. The quantity V/Q0 has units oftime and is called themean detention time, residence time, or contact time in the reactorfor the aqueous phase and is designated τd.

0

0.2

0.4

0.6

0.8

1

0 20 40 60 80 100

C A/C

A0

td /h

CSTR

PFR

FIGURE 6.5 Comparison of exit to inlet concentration ratio in a surface impound-ment modeled as a CSTR and a PFR.

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Blackburn (1987) reported the determination of the three transport rates for 2,4-dichlorophenol in an activated sludge process in an impoundment. The impoundmenthad a total volume of 106 L with a solids (biomass) concentration of 2000mg/L. Themean detention time was 1 day. The depth of the impoundment was 1.5m. The averagebiodegradation rate kr was 0.05 h−1. The rate constant for settling of the biomass wasestimated to be 2.2 × 10−6 h−1. The rate constant for volatilization was 2 × 10−5 h−1.The overall rate constant k∗ = 0.05 h−1. Therefore, CA/CA0 = (1 + 0.05τd)−1. Thisis shown in Figure 6.5. The contribution of each mechanism toward the loss of DCPcan also be ascertained. For example, the fractional loss from biodegradation is fbio =krτd/(1 + krτd).

If the surface impoundment behaves as a PFR, then the steady-state mass balancewill be given by −(dCA/dV) = (k∗/Q0)CA, which gives CA/CA0 = exp(−k∗τd) =exp(−0.05τd). This is also shown in Figure 6.5. Note that the decay of DCP is fasterin the PFR than in the CSTR. It is clear that an ideal PFR reactor can achieve the sameremoval efficiency as that of a CSTR, but utilizing amuch smaller volume of the reactor.

EXAMPLE 6.3 COMBUSTION INCINERATORASA PFR

Combustion incinerators are used to incinerate municipal and industrial waste. Gener-ally they require high temperatures and an oxygen-rich environmentwithin the chamber.Consider a waste containing benzene being incinerated using an air stream at a velocityof 5m/s.A typical inlet temperature is 900◦C (1173K).Due to heat transfer the tempera-ture changes along the chamber length from inlet to outlet.Assume a gradient of 10◦C/m.As a consequence, the rate constant will also vary linearly with length. For benzene theArrhenius parameters are activation energy Ea = 225 kJ/mol and pre-exponential factorA = 9 × 1010 s−1.A typical length of the incinerator chamber is−10m. Hence the exittemperature will be 900 − 10 × 10 = 800◦C (1073K). The first-order rate constant forbenzene will be k(@1173K) = 8 s−1 and k(@1073K) = 1 s−1. As an approximation,let us consider an average first-order rate constant k = 4 s−1. The concentration in theexit stream can be obtained using the equation for a plug-flow reactor:CA/CA0 = e−kτ.Since τ = 10m/5 (m/s) = 2 s, CA/CA0 = 3 × 10−4. Hence the destruction efficiencyis 99.96%.

In many instances in environmental reactors, the reactor (e.g., a CSTR) may notbe at steady state at all times. Examples of these include a waste impoundment andchemical spills into waterways. In these cases, we need to obtain the solution to themass balance equation under non-steady-state conditions. For illustrative purposeswe will use a first-order reaction.

VdCA

dt= Q0 (CA0 − CA) − kCAV . (6.19)

By rearranging, we obtain

dCA

dt+(k + Q0

V

)CA = Q0

VCA0. (6.20)

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Integrating the above equation with the initial condition (at t = 0) CA = CA0, weobtain

CA(t) = 1

(1 + kτd)CA0

[1 − e−(k+(1/τd))t

]+ CA0e

−(k+(1/τd))t , (6.21)

where τd is as defined earlier. Note that as t → ∞, the above equation reduces tothe steady-state solution obtained earlier. Depending on the functionality of CA0, theresponse of the CSTR will vary. Levenspiel (1999) considered several such inputfunctions and derived the corresponding response functions.

6.1.1.5 Relationship between Steady State and Equilibrium for a CSTR

As discussed in Chapter 2, equilibrium for closed systems is the state of minimumfree energy (or zero entropy production), which is a time-invariant state. For opensystems, the equivalent time-invariant state is called the steady state. Howclosely doesa steady state resemble the equilibrium state? This question is important in assessingthe applicability of steady-state models in environmental engineering to depict thetime-invariant behavior of systems.

As an example, consider the reversible reaction

Akf�kb

B. (6.22)

Consider the reaction in an open system where both species A and B are enteringand leaving the system at a feed rate Q0 with feed concentrations CA0 and CB0,respectively. The mass balance relations are

VdCA

dt= Q0 (CA0 − CA) − kfCA + kbCB,

VdCB

dt= Q0 (CB0 − CB) − kfCA + kbCB.

(6.23)

Applying the steady-state approximation to both species, and solving the resultingequations, we obtain (

CA

CB

)

ss= 1

Keq+ 1

kfτd, (6.24)

where Keq = kf/kb and τd = V/Q0. Recalling that kf = 0.693/t1/2 for a first-orderreaction, we have (

CA

CB

)

ss= 1

Keq+ t1/2

0.693 · τd. (6.25)

Note that if τd is very large and t1/2 is very small, then (CA/CB)ss → 1/Keq. Thus,if the mean residence time is very large and the reaction is fast, the steady-state ratioapproaches that at equilibrium. For many natural environmental conditions, eventhough τd may be quite large, t1/2 is also relatively large (since kf is small) and hencesteady state and equilibrium are not identical.

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6.1.2 NONIDEAL REACTORS

The reactor designs we have discussed thus far assumed ideal flow patterns—plugflow (no mixing) as one extreme and CSTR (complete mixing) as the other.Althougha large number of reactors show ideal behavior, natural environmental reactors fallsomewhere in between the two ideal reactors. Fluid channeling, recycling, or stagna-tion points in the reactor cause the deviations. There are two models in the chemicalengineering literature that are used to explain nonideal flows in reactors. These arecalled dispersion and tanks-in-series models.

6.1.2.1 Dispersion Model

Ideal plug flow involves no intermixing between fluid packets. However, in mostcases molecular and turbulent diffusion can skew the profile as shown in Figure 6.6.As the disturbances increase, the reactor will approach the characteristic of completemixing as in a CSTR. The change in concentration is represented by Fick’s equationfor molecular diffusion written as follows:

∂C

∂t= Dax

∂2C

∂x2, (6.26)

where Dax (m2/s) is called the axial dispersion coefficient. If Dax = 0 we have plugflow (PFR), and asDax → ∞ the flow is completelymixed (CSTR).Tonondimension-alize the above equation, we express ζ = (ut + x)/L and θ = ut/L, where u representsthe fluid velocity and L is the length of the reactor. We have then (Levenspiel, 1999)

∂C

∂θ= Dax

uL

∂2C

∂ζ2− ∂C

∂ζ. (6.27)

Ideal plug flow

Nonideal plug flow

FIGURE 6.6 Schematic of the plug flow and dispersed plug-flow models.

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0

0.5

1

1.5

2

0 0.5 1 1.5 2 2.5

F(t/t

)

t/t

0.02

0.2

0.5

Dax/uL

Ideal plug flow

FIGURE 6.7 Relationship between F(t/τ) and dimensionless time for different values ofDax/uL. True plug-flow model is represented by the vertical line at dimensionless time of 1.0.

The term Dax/uL is the dispersion number.If a perfect pulse is introduced in a flowing fluid, the solution to the above equation

gives the exit concentration (Hill, 1977)

C(L, t)∫∞0 C(L, t)d(t/τ)

= 1

2VR√

(πDax/uL)(t/τ)e−((1−(t/τ))2/(4Dax/uL)·(t/τ)). (6.28)

A plot of the right-hand side of the above equation versus t/τ is given in Figure 6.7.For small values of Dax/uL, the curve approaches that of a normal Gaussian errorcurve. However, with increasing Dax/uL (>0.01), the shape changes significantlyover time. For small values of dispersion (Dax/uL < 0.01), it is possible to calculateDax by plotting log C(L, t)/C(0) versus t and obtaining the standard deviation σ fromthe data. C(0) is the influent pulse input concentration. The equation is

Dax = 1

2σ2u3

L. (6.29)

6.1.2.2 Tanks-in-Series Model

The second model for dispersion is a series of CSTRs in series. The actual reactor isthen composed of n CSTRs; the total volume is VR = nVCSTR. The average residencetime in the actual reactor is τ = VR/Q0 = nVCSTR/Q0, where Q0 is the volumet-ric flow rate into the reactor. For any reactor n in the series, the following materialbalance holds:

Cn−1Q0 = CnQ0 + VCSTRdCndt

. (6.30)

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The concentration leaving the jth reactor is given by

CjC(0)

= 1 − e−n(t/τ){1 + n

t

τ+ 1

2!(nt

τ

)2

+ · · · + 1

( j − 1)!(nt

τ

)j−1}. (6.31)

EXAMPLE 6.4 DISPERSION INA SOIL COLUMN

Chloride is a tracer used to determine the dispersivity in a soil column. It is conservativeand does not react with soil particles. A pulse input is introduced at the bottom of a soilcolumn of height 6.35 cm at an interstitial porewater velocity of 5.37 × 10−4 cm/s. Thesoil has a porosity of 0.574. The chloride concentration at the exit (top) as a function oftime is given below:

t (min) C (L, t)/C (0)

125 0.05150 0.15180 0.30200 0.40250 0.90

Determine the dispersivity in the column.

120

140

160

180

200

220

240

260

.01 .1 1 5 10 20 30 50 70 80 90 95 99 99.9 99.99

t/min

Percent

FIGURE 6.8 Probability plot of dimensionless concentration versus time to obtainthe dispersion coefficient from experimental data.

continued

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First plot log C(L, t)/C(0) as a function of t on a probability plot as shown inFigure 6.8. The standard deviation in the data is σt = 46min, which is the dif-ference between the 84th and 50th percentile points, t84% − t50%. Hence Dax =(1/2)σ2t u

3/L = 9.3 × 10−5 cm2/s.

EXAMPLE 6.5 PULSE INPUT INA RIVER

A conservative (nonreactive, nonvolatile, water-soluble) pollutant is suddenly dis-charged into a river flowing at an average speed of 0.5m/s. A monitoring station islocated 100miles downstream from the spill. The dispersion number is 0.1. Determinethe time when the concentration at the station is 50% of the input.Dax/uL = 0.1 (intermediate dispersion). τ = L/u = (100miles) (1600m/mile)/0.5

(m/s) = 3.2 × 105 s = 3.7 days. For C(L, t) = 0.5C(0),

0.5 = 1√3.14 × 0.1(t/τ)

exp

(− (1 − (t/τ))2

0.4(t/τ)

).

By trial and error t0.5/τ = 1.9. Hence t1/2 = 7 days.

6.1.3 DISPERSION AND REACTION

Let us now consider the case where the flow is nonideal, and the compound enteringthe reactor is undergoing reaction as it flows through the reactor. Let the reaction ratebe r with the stoichiometric coefficient for A being νA.

Figure 6.9 shows the reactor configuration in which we perform a material balanceover the volume between x and x +Δx. There are two separate inputs and outputs,one due to bulk flow (uCAAc) and the other due to axial dispersion (DaxAc∂CA/∂x).

Bulkflow

Dispersion

Reactionx + Δx

Bulk flow

Dispersion

x = 0 x = LxFeed, CA0

Effluent, CAe

FIGURE 6.9 Material balance in a reactor with dispersion and reaction.

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Within the reactor the compound reacts at a rate r. The overall material balance is

[accumulation] = input(bulk flow + dispersion)] − [output(bulk flow

+ dispersion)] + [reaction].At steady state, accumulation is zero, Hence,

0 =(uAcCA|x − DaxAc

dCA

dx

∣∣∣∣x

)−(uAcCA|x+Δx − DaxAc

dCA

dx

∣∣∣∣x+Δx

)

+ vArAcΔx. (6.32)

Rearranging the terms and dividing by the volume AcΔx gives

u

(CA|x − CA|x+Δx

Δx

)− Dax

((dCA/dx)|x − (dCA/dx)|x+Δx

Δx

)+ νAr = 0.

(6.33)Taking limit as Δx →0, we obtain

−u∂CA

∂x+ Dax

∂2CA

∂x2+ vAr = 0. (6.34)

The above equation and its variations appear in many cases in environmental engi-neering. The first term on the left-hand side is called the advection or convection term,the second is the dispersion term, and the last is the reaction term. For a first-orderreaction A → B, the equation can be written in the following form:

Daxd2CA

dx2− u

dCA

dx− kCA = 0. (6.35)

To make the above equation dimensionless, we use the following transformations:z = x/L and dz = dx/L with L being the length of the tube. Hence, we have

d2CA

dz2− PeL

dCA

dz− DaLCA = 0, (6.36)

where PeL = uL/Dax is called the Peclet number and DaL = kL2/Dax is called theDamköhler number. If mixing is rapid, that is, Dax/uL is very small or uL/Dax islarge, the axial dispersion term will be negligible and the system will approach plug-flow behavior. This is a good approximation for analyzing contaminant dynamics insurface waters. In the atmosphere, since the dispersion is large, we have to considerthe entire advective–dispersion equation.

6.1.4 REACTION IN A HETEROGENEOUS MEDIUM

In many cases, the reactant is transported from onemedium to another where it reacts.A good example is gaseous NH3 that dissolves and reacts with an aqueous acidic

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solution to give NH4OH. Another example is the reaction and transport of gases(CO2, O2) and volatile species in soil. As a general example, consider a compoundA in air that diffuses in air and reacts transforming to a compound B. There are threesteps involved:

Aggas diffusion−−−−−−−→ Aσ,

AσkA−→ Bσ,

Bσliquid diffusion−−−−−−−−→ Baq.

(6.37)

Depending on which of the above reactions is slow, the overall rate will vary.For the heterogeneous reaction scheme considered above, the bulk gas phasewill be

well mixed with a concentration CgA, whereas the solute concentration at the interface

is CσA. The change occurs across a very thin film of thickness δ near the interface. Thediffusive flux of A to the interface is given by

nA,diff = DAL

δ

(Cg

A − CσA), (6.38)

where DAL is the diffusion constant of A in solution. DAL/δ is the film transfercoefficient forA.More appropriately we can define a general mass transfer coefficientkc to represent this term.

The first-order surface reaction rate is given by

rσ = kσCσA. (6.39)

At steady state, the flux to the interface must be the same as the reactive loss at thesurface. This gives

CσAkc = kcCgA

kc + kσ. (6.40)

The rate of the reaction at steady state is

rσ = kσkcCgA

kσ + kc. (6.41)

There are two limiting conditions for the above rate: (i) if diffusion is controlling,kσ � kc, r → kcC

gA; and (ii) if reaction is controlling, kσ � kc, rσ → kσC

gA.

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EXAMPLE 6.6 REACTION- AND DIFFUSION-LIMITED REGIONS

The transport-limitedmass transfer coefficient can be related to other system parameterssuch as flow velocity, particle size, and fluid properties using correlations, for example,the Fröessling correlation:

Sh = kcDp

DA= 2 + 0.6 Re1/2Sc1/3, (6.42)

where Sh is the Sherwood number, Re is the Reynolds number (=uDp/ν), and Sc isthe Schmidt number (=ν/DA). For Re > 25, the above equation can be simplified toobtain

Sh = 0.6Re1/2Sc1/3. (6.43)

Thus we have the following equation for kc:

kc = 0.6

(D2/3

A

ν1/6

)(u

Dp

)1/2. (6.44)

For aqueous solutions, DA ≈ 10−10 m2/s, ν = 1 × 10−6 m2/s, and kc ≈ 1.2 ×10−6(u/Dp)1/2. Using a typical environmental reaction condition,Cg

A ≈ 10−5 mol/m3

and kσA ≈ 0.1 s−1, we obtain

r = (1 × 10−6)

1 + (8.3 × 104/(u/Dp)1/2). (6.45)

10–12

10–11

10–10

10–9

10–8

10–7

10–6

0 2 × 104 4 × 104 1.2 × 1051 × 1058 × 1046 × 104

r (m

ol · c

m–3

· s–1

)

(U/DP)(1/2)

Diffusion limited

Reaction limited

FIGURE 6.10 Diffusion and reaction-limited regions for a heterogeneous reactioninvolving transport of reactants to the surface.

continued

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The above rate is plotted in Figure 6.10. There are two distinct regions. In the first, therate is proportional to (u/Dp)1/2 and is called the diffusion-limited region. In this region,r increases with increasing Dp and decreasing u. In the second region, r is independentof (u/Dp)1/2 and is called the reaction-limited region. Diffusion-limited reactions areimportant in several important environmental processes.

6.1.4.1 Kinetics and Transport at Fluid–Fluid Interfaces

Consider a gas and a liquid in contact (Figure 6.11). Let us consider a component ifrom the gas phase that exchanges with the liquid phase. This component can eitherbe reactive in the liquid phase or not. Let us first consider the base case where thereis no reaction in either phase. The two bulk fluids are completely mixed so thatthe concentration of species is C∞

Gi and C∞Li . Note that we can also represent the

gas-phase composition by a partial pressure Pi. Mixing and turbulence in the bulkphase quickly disperse the species in the solution. However, near the interface onboth sides there is insignificant mixing (low turbulence). Hence, diffusion throughthe interfacial films limits mass transfer. The interface, however, is at equilibrium andair–water partitioning equilibrium applies so that

CintGi = KawC

intLi . (6.46)

Interface(at equilibrium)

Pi

Bulk liquid(mixed)

Bulk gas(mixed)

Gas film Liquid film

CGi∞

CLi∞

CLiint

CGiint

FIGURE 6.11 Schematic of the two-film theory of mass transfer for transfer of a solute fromthe gas to the liquid, namely absorption. Note that equilibrium exists only at the arbitrarydividing plane called the interface. Mixing is complete in both bulk phases at distances awayfrom the interface.

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The rate of mass transfer from gas to liquid through the gas film is given by

r1 = kG

(C∞

Gi − CintGi

)(6.47)

and that through the liquid film by

r2 = kL

(Cint

Li − C∞Li

). (6.48)

At steady state, the rate at which the mass reaching the interface from the gas sideshould equal that leaving through the liquid film, that is, r1 = r2,

kG

(C∞

Gi − CintGi

)= kL

(Cint

Li − C∞Li

). (6.49)

Since CintGi and Cint

Li are not known or obtainable from experiments, we need toeliminate these using the air–water partition equilibrium to obtain

CintGi =

(kGC∞

Gi + kLC∞Li

)

(kG + (kL/Kaw)). (6.50)

Hence,

r1 = 1

((1/kL) + (1/kGKaw))

(C∞

Gi

Kaw− C∞

Li

)(6.51)

is the rate of transfer from gas to liquid (absorption). If the transfer is from liquid togas (stripping), the rate is

r1 = 1

((1/kL) + (1/kGKaw))

(C∞

Li − C∞Gi

Kaw

). (6.52)

If the rate of absorption is expressed in the form

r1 = KL

(C∞

Gi

Kaw− C∞

Li

), (6.53)

where KL is the overall mass transfer coefficient based on the bulk-phase concentra-tions, we recognize that

1

KL= 1

kL+ 1

kGKaw, (6.54)

where each term represents a resistance to mass transfer

RT = RL + RG. (6.55)

The rate can also be expressed as follows:

r1 = KG(C∞

Gi − KawC∞Li

), (6.56)

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where1

KG= 1

kG+ Kaw

kL. (6.57)

The general equation for the rate of absorption in the case of an enhancement in theliquid phase due to reaction is given by

r1 = 1

((1/kL) + (Kaw/kGE))C∞

Gi , (6.58)

where E is the enhancement in mass transfer due to reaction

E ≡ rate of uptake with reaction

rate of uptake without reaction. (6.59)

For example, consider an instantaneous reaction given byA(g) + B(aq) → products;the enhancement factor E is 1 + (DB,aq/DA,aq)(CB,aq/Cint

A ).The derivation of the two-film mass transfer rate at fluid–fluid interfaces can

be generalized to a number of other cases in environmental engineering such assoil–water and sediment–water interfaces.

6.1.5 DIFFUSION AND REACTION IN A POROUS MEDIUM

The lithosphere (e.g., soils, sediments, aerosols, activated carbon) is characterized byan important property, namely, its porous structure. Therefore, all of the accessiblearea around a particle is not exposed to the pollutants in the bulk fluid (air or water).The diffusion of pollutants within the pores will lead to a concentration gradient fromthe particle surface to the pore. The overall resistance to mass transfer from the bulkfluid to the pore will be composed of the following: (a) diffusion resistance within thethin-film boundary layer surrounding the particle; (b) diffusion resistance within thepore fluid; and (c) the final resistance from that due to the reaction at the solid–liquidboundary within the pores. This is shown schematically in Figure 6.12a and b.

If the external film diffusion controls mass transfer, the concentration gradientis C∞

A − CσA, and the diffusivity of A is the molecular diffusivity in the bulk liquidphase, DA. If internal resistance controls the mass transfer, the gradient is C∞

A CA(r),but the diffusivity is different from DA. The different diffusivity results from the factthat the solute has to diffuse through the tortuous porous space within the particle. Atortuosity factor, τ, is defined as the ratio of the actual path length between two pointsto the shortest distance between the same two points. Since only a portion of the solidparticle is available for diffusion, we have to consider the porosity ε of the medium.ε is defined as the ratio of the void volume to the total volume. Thus the bulk-phasediffusivity DA is corrected for internal diffusion by incorporating ε and τ.

DeffA = DA

ε

τ. (6.60)

For most environmental applications, ε/τ is represented by the Millington–Quirkapproximation which gives θ10/3/ε2, where θ is the volumetric content of the fluid(air or water).

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Internal diffusion,surface reaction

External filmdiffusion

Rtot = Rfilm + Rint + Rr ¥ n

R

r dr

NA (r) NA (r + dr)

(a) (c)

Surfacediffusion Solid

(b) (d)

AA A

A Pore

Porediffusion

FIGURE 6.12 Schematic of diffusion and reaction/sorption in a porous medium. (a) Internaland external resistance tomass transfer and reaction/sorptionwithin a spherical porous particle.(b)Various resistance tomass transfer and reaction/sorption. (c)Material balance on a sphericalshell. (d) Simultaneous bulk diffusion and surface diffusion within a pore.

The process of reaction occurring within the pore space can be either a surfacetransformation of A → B, or simply a change from the porewater to the adsorbedstate. We shall represent this by a general first-order surface reaction such that

rσ = kσCAσ, (6.61)

where kσ has units of length per time so that rσ can be expressed in moles/area/timeand CAσ is expressed in moles/volume.

Consider Figure 6.12c. A mass balance on the spherical shell of thicknessΔr canbe made. The diffusion of solute into the center of the sphere dictates that the fluxexpression should have a negative sign so thatNA(r) is in the direction of increasing r.The overall balance is: flux ofA in at r-flux ofA out at (r +Δr) + rate of generationofA by reaction = rate of accumulation ofA in the solid. The rate of accumulation ofA in the solid is given by ε∂CA/∂t with ε being the porosity and CA the concentrationof A per unit volume of the void space in the solid.

NA(r)4πr2|r − NA(r)4πr2|r+Δr + rσAs4πr2Δr = ε∂CA

∂t, (6.62)

where As is the internal surface area per unit volume and 4πr2Δr is the volume ofthe shell. At steady state ∂CA/∂t will be zero. Then by dividing the expression into

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4πΔr, and letting Δr → 0, we can obtain

d

dr(r2NA(r)) − r2rσAs = 0. (6.63)

The expression for NA(r) is given by Fick’s law of diffusion, NA(r) =−Deff

A (dCA/dr). The surface reaction rate rσ = kσCA(r).Hence,

1

r2d

dr

(r2dCA

dr

)− kσAs

DeffA

CA = 0. (6.64)

The above differential equation has to be solvedwith appropriate boundary conditionsto obtain CA(r).

Let the outer surface of the sphere be at a constant concentrationCAσ. This impliesthat external film diffusion is not important. Hence the first boundary condition isCA = CAσ at r = R. A second boundary condition is that the concentration at thecenter of the spherical particle is finite, that is, dCA/dr|r =0= 0. At this point, wecan introduce a set of dimensionless variables that will simplify the form of thedifferential equation. These are Ψ = CA(r)/CAσ and Λ = r/R. This gives

1

Λ2

d

dΛ

(Λ2 dΨ

dΛ

)−Φ2Ψ = 0, (6.65)

whereΦ = (kσAs/DeffA

)1/2is called the Thiele modulus. Upon inspection of the equa-

tion forΦ, we note that it is the ratio of the surface reaction rate to the rate of diffusion.Thus the Thiele modulus gives the relative importance of reaction and diffusion rates.For large Φ, the surface reaction rate is large and diffusion is rate limiting, whereasfor small Φ, surface reaction is rate limiting. The transformed boundary conditionsare Φ = 1 at Λ = 1 and dΨ/dΛ|Λ =0= 0. The solution to the above differentialequation is (Fogler, 2006; Smith, 1970)

CA

CAσ= 1

Λ

sinh(ΦΛ)

sinh(Φ). (6.66)

Figure 6.13a is a plot of CA/CAs versus Λ (i.e., r) for three values of the Thielemodulus, Φ.

In the field of catalysis it is conventional to define a related term called the overalleffectiveness factor, ξ, which is an indication of how far a molecule can diffuse withina solid before it disappears via reaction. It is defined as the ratio of the actual overallreaction rate to the rate that would result if the entire interior surface were exposedto the surface concentration CAs. It is given by (Fogler, 2006)

ξ = 3

Φ2(Φ cothΦ− 1). (6.67)

The above function is plotted in Figure 6.13b, which shows that with increasing Φ,the effectiveness factor decreases. With increasing Thiele modulus, the accessibility

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0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

C A/C

As

r/R

(a) (b)

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

zeta

phi0 2 4 6 8 10 12

FIGURE 6.13 (a) Variation in pollutant concentration with radial position for various Thielemodulus. (b) Variation of overall effectiveness factor with Thiele modules.

of the reactant to the interior surface sites is reduced and consequently the processis diffusion limited within the particle. The overall reaction rate is therefore rAσ =ξk1CAσ if the surface reaction is first order.

Let us now include the external film resistance to mass transfer. This can be intro-duced to make the problem general and also to evaluate the relative importance ofprocesses external and internal to the particle. At steady state, the net rate of transferof mass to the surface of the particle should equal the net reaction (on exterior surfaceand interior surface). Thus,

kmt(C∞

A − CAσ)Aext = rAσAint, (6.68)

where Aext is the external surface area of the particle and Aint is the internal surfacearea. Utilizing the expression for rAσ, we have upon rearranging

CAσ = kmtAextC∞A

kmtAext + ξkσAint. (6.69)

The steady-state rate of mass transport to the surface is then given by

rssmt = rAσAint = ξkσAintCAσ = ωkσAintC

∞A , (6.70)

where ω = ξ/1 + (ξ(k1Aint/kσAext)) denotes the change in the effectiveness factoras the external mass transfer becomes significant. The transfer rate is dependent onlyon the bulk phase (air or water) concentration. As kσ becomes smaller, ω decreasesand the external mass transfer resistance becomes important.

Most reactions in environmental systems occur in assemblages of porous particles.This gives rise to both macropores and micropores within the medium. The diffusion

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and kinetics in such a porous bed medium can be modeled as a catalytic reactor,with the aim of obtaining the degree of conversion at any defined position within thereactor. The analysis can be used to model a waste treatment unit operation such asactivated carbon, ion-exchange, or other physicochemical treatment processes. It canalso be applied to the modeling of F&T of pollutants in a porous medium such as asoil or sediment.

6.2 THE WATER ENVIRONMENT

In this section, wewill discuss applications of chemical kinetics and reactor models inthe water environment. The first part of the discussion will be illustrations of F&T inthe natural environment. The second part will be examples in water pollution controland treatment.

6.2.1 FATE AND TRANSPORT

6.2.1.1 Chemicals in Lakes and Oceans

A number of anthropogenic chemicals have been introduced into our lakes, rivers,and oceans. To elucidate the impact of these chemicals on marine species, birds,mammals, and humans, we need a clear understanding of their F&T in the waterenvironment.A chemical introduced into a lake, for example, is subjected to a varietyof transport, transformation, and mixing processes. There are basically two types ofprocesses: (i) transport processes such as advection, dispersion, and diffusion withinand between compartments (water, sediment, and air) and (ii) reaction processesthat chemically transform compounds via photolysis, microbiological reactions, andchemical reactions. These processes do not occur independent of one another, andin many cases are influenced by one another. Figure 6.14 is an illustration of theseprocesses (Schwarzenbach et al., 1998).

At its simplest, amodelwill consist of a one-dimensional vertical transport betweenthe sediment, water, and air. The horizontal variations in concentrations are neglected.For a deep lake this is a good approximation, whereas for shallow water bodies theapproximation fails. One can increase the spatial resolution of the model to obtainmore sophisticated models.

To obtain the time constants for the various processes involved in the model, let usfirst construct a zeroth-order chemodynamic model for the lake. In a lake we have anepilimnion and a hypolimnion. To a good first approximation, we shall consider theepilimnion as a well-mixed CSTR (Figure 6.15). The various processes that occur caneach be assumed to be a first-order loss process.A material balance for the compoundA can be written as

input = output + reaction + accumulation.

Q0CA0 = Q0CA + rtotV + VdCA

dt, (6.71)

where CA is the total concentration of compound in the epilimnion and V is thevolume of the epilimnion. Note that Q0/V = kF, the rate constant for flushing, and

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Sediment

Water

Air

Wet and drydeposition Outflow

Input fromsurface

Air-waterexchange

Vertical &Horizontal

mixing

Sediment-waterexchange

Resuspension

SorptionParticletransport

Chemical/bio/photo reactions

Sedimentation

FIGURE 6.14 Schematic of various fate and transport processes for a pollutant entering alake.

Mixing

ReactionPhotolysis

Bioreactions

Epilimnion

HypolimnionSedimentation

(Sediment)

(Atmosphere)

Q0 CA0

Q0 CA

FIGURE 6.15 A CSTR one-box model for the epilimnion of a lake.

that the overall rates will be composed of the several first-order processes listed inFigure 6.15.

rtot = (kvoln + ksed + krxn + kphoto + kbio)CA = ktotCA. (6.72)

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Therefore,

dCA

dt= kF(CA0 − CA) − ktotCA. (6.73)

Expressions for individual rate constants are given in Table 6.1. Note that a similarequation was used in Example 6.2 to illustrate the loss mechanisms in a wastewatersurface impoundment.

EXAMPLE 6.7 FATE OF TETRACHLOROETHENE IN LAKE GRIEFENSEE,SWITZERLAND

Schwarzenbach et al. (1998) have estimated the following parameters for LakeGriefensee: Average area = 2 × 106 m2, average epilimnion depth = 5m, particulatematter concentration = 2 × 10−6 kg/L, average particulate settling velocity = 2.5m/d,average Kw for tetrachloroethene (TetCE) = 0.15m/d, average load of TetCE perday to the lake = 0.1 kg/d, average flow rate of water = 2.5 × 105 m3/d. Estimate thesteady-state removal of TetCE in the lake.

Firstly, note thatTetCE is a refractory compound.Hence for the zeroth approximation,let us assume that chemical, photochemical, and biological reactions are negligible.Kaw for TetCE is 0.6 at 298K. Log Kow = 2.1. Hence log Kow = (0.92)

(2.10) − 0.23 = 1.70. Hence Koc = Kc = 50 L/kg. φpA = (50)(2 × 10−6)/[1 + (50)

(2 × 10−6)] = 1 × 10−4. kvoln = (0.15)(1)/5 = 0.03 d−1, ksed = (2.5)(1 × 10−4)/5= 5 × 10−5 d−1, kF = 2.5 × 105/(5)(2 × 106) = 0.025 d−1. At steady state, (CA0 −CA/CA) = (kvoln + ksed/kF) = 1.2. Hence 1 − CA/CA0 = 0.55.

TABLE 6.1Expressions for First-Order Rate Constants in a CSTR Modelfor the Epilimnion of a Lake

Process Rate Constant Expression

Flushing kF QF/Ve

Volatilization kvoln

(KW

he

)(1 − φp

A

)

Sedimentation ksed

(Uset

he

)φ

pA

Reaction loss krxn + kphoto + kbio Specific to chemicals

Notes: QF = feed rate, m3/d; Ve = epilimnion volume, m3; Kw = overall liquid-phase mass transfer coefficient of A, m/d; he = height of the epilimnion, m;ϕPA = fraction of A bound to particulates, KocCC/ (1 + KocCC), where Koc isthe organic carbon-based partition constant of A (L /kg), CC is the particulateorganic carbon concentration, L /kg, and Uset = average settling velocity ofparticulates, m /d.

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EXAMPLE 6.8 EVAPORATION FROMA WELL-STIRRED SURFACE

Many compounds of low solubility (such as pesticides, PCBs, and hydrocarbons) evap-orate from openwaters in lakes and oceans. For a 1m2 of area with depth Z cm, estimatethe half-life for evaporation of the following compounds: benzene, biphenyls, aldrin,and mercury.The evaporation is assumed to occur fromavolumeZ cm × 1 cm2 = Z cm3 of surface

water that is well mixed as a result of surface turbulence. A material balance over thevolume Z cm3 gives


0 = Kw(Cw − CaKaw) + 0 + ZdCw

dt, (6.74)

where the rate of loss by reaction is zero on account of the refractory nature of thechemical. Rearranging and integrating with Cw = C0 at t = 0, we obtain

Cw = Ca

Kaw+(C0 − Ca

Kaw

)exp

(−Kw

t

Z

). (6.75)

If background air concentration is negligible, Ca = 0.

Cw = C0 exp

(−Kw

t

Z

)= C0 exp(−kvolnt). (6.76)

Hence, the half-life is t1/2 = 0.693/kvoln = 0.693Z/Kw.Mackay and Leinonen (1975)give values of Kw for several compounds at 298K:

Compound P∗i (mm Hg) Kw (m/h) t1/2 (h) for Z = 1 m

Benzene 95.2 0.144 4.8Biphenyl 0.057 0.092 7.5Aldrin 6 × 10−6 3.72 × 10−3 186.3Mercury 1.3 × 10−3 0.092 7.5

Note that the half-lives of both biphenyls and mercury are the same, although theirvapor pressure varies by a factor of 2. Note also that Kw is obtained from the individualtransfer coefficients kw and ka and requires a knowledge of Kaw as well. These can beobtained by applying the diffusivity correction for kw of oxygen (20 cm/h) and ka forwater (3000 cm/h), if experimental values of individual mass transfer coefficients arenot available.

6.2.1.2 Chemicals in Surface Waters

The surfacewater in a fast-flowing stream is generally unmixed in the directionof flow,but is laterally well mixed. This suggests that a plug-flow model will be applicable inthese cases. The appropriate equation is the advection–dispersion equation with theaxial dispersion term neglected. Figure 6.16 depicts a river stretch where we apply a

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Loss to air

uCw (x + Δx)

x x + Δx

uCw (x)

Air

Water

Input

Sediment

Loss to sediment

FIGURE 6.16 Material balance on a section of a stream assuming complete mixing acrossthe width and depth of the stream.

material balance across the volumeWHΔx:


uCw(x)WH = uCw(x +Δx)WH + rWHΔx + loss to air

+ loss to sediment +WHΔxdCw

dt.

At steady state, we have dCw/dt = 0. The rate of loss to air is Kw(Cw −(Ca/Kaw))(WΔx). The rate of loss to sediment is Ks(Cw − (wi/Kaw))(WΔx). If afirst-order rate of reaction is considered, r = krCw. The overall material balance is

u ·(Cw (x +Δx) − Cw(x)

Δx

)= −Kw

H

(Cw − Ca

Kaw

)− Ks

H

(Cw − W

Ksw

)− krCw.

(6.77)Since time, t = x/u, Δt = Δx/u, and taking lim

Δx→0, we obtain the following

differential equation:

dCw

dt= −Kw

H

(Cw − Ca

Kaw

)− Ks

H

(Cw − W

Ksw

)− krCw. (6.78)

If both sediment and air concentration remain constant, we can solve the aboveequation using the initial condition, Cw = C0 at t = 0, to obtain (Reible, 1998)

Cw = C0e−αt + β

α(1 − e−αt), (6.79)

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where β = (KwCa/HKaw) + (KsW/HKsw) and α = (Kw/H) + (Ks/H) + kr. Notethat α is a composite rate constant that characterizes the loss mechanism within thewater.

EXAMPLE 6.9 LOSS OF CHLOROFORM FROMA SHALLOW STREAM

For the Amite River near Baton Rouge, Louisiana, the depth is 2m, foc is 0.02 for thesediment and has an average flow velocity of 1m/s. For an episodic spill of chloroformin the river, determine its concentration 20miles downstream of Baton Rouge.Assume that the background concentration of chloroform in air is negligible. Since

chloroform is highly water soluble, and its hydrophobicity is small, we can assumethat its concentration in the sediment will be low. Thus β is negligible. Because ofthe high stream velocity and low sorption to sediment, the water-to-air mass trans-fer is likely to dominate over water-to-sediment mass transfer. Hence α = (Kw/h) +kr . The reaction in water is hydrolysis and its rate is given by 4.2 × 10−8 h−1.Clearly, this is low and chloroform is a refractory compound. From Table 4.12,1/Kw = (1/4.2 × 10−5) + (1/(3.2 × 10−3)(0.183)). Hence Kw is 3.9 × 10−5 m/s =0.14m/h. Since t = x/u = (20miles)(1609m/mile)/(1m/s)(3600 s/h)= 8.9 h. Henceα = 0.14/2 = 0.07 h−1. Cw/C0 = exp[−(0.07)(8.9)] = 0.53. Hence, the concentra-tion at the monitoring station will be 53% of its concentration at the spill point.

6.2.1.3 Biochemical Oxygen Demand in Natural Streams

Bacteria and microorganisms decompose organic matter in wastewater via enzymecatalysis. The amount of oxygen that the bacteria need to decompose a given organiccompound to products (CO2 and H2O), and in the process produce new cells, is calledthe biochemical oxygen demand (BOD). The reaction is represented as

organic matter + O2 → CO2 + H2O + new cells.

BOD is used to characterize the extent of pollution in rivers, streams, lakes, and inwastewater under aerobic conditions. BOD is measured using a five-day test, that is,the oxygen demand in five days of a batch laboratory experiment. It is representedas BOD5. The BOD5 test is a wet oxidation experiment whereby the organisms areallowed to break down organic matter under aerobic conditions. It is a convenientmethod, but does not represent the total BOD. A 20-day test (BOD20) is consid-ered the ultimate BOD of a water body. BOD measurements are used to define thestrength of municipal wastewater, determine the treatment efficiency by observing theoxygen demand remaining in an effluentwaste stream, determine the organic pollutionstrength in surface waters, and determine the reactor size and oxygen concentrationrequired for aerobic treatment processes.

The oxygen demand, L, in water is given by a first-order reaction rate

−dL

dt= k1L. (6.80)

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Upon integration, we obtain

L = Loe−k1t , (6.81)

where Lo is called the ultimate oxygen demand for organic matter decomposition.(Note that there are additional oxygen demands for those compounds that containnitrogen, which we will ignore.) The above equation shows that the oxygen demanddecreases exponentially with time. The total oxygen demand of a sample is the sumof waste consumed in t days (BODt) and the oxygen remaining to be used after tdays. Hence,

Lo = BODt + L (6.82)

or

BODt = Lo(1 − e−k1t). (6.83)

Figure 6.17 represents a typical BOD curve. The value of k1 depends on the type ofsystem under study. Typical values range from 0.05 to 0.3 d−1. Since BOD tests arecarried out at 20◦C, k1 has to be corrected for other temperatures using the followingequation:

(k1 at θ◦C) = (k1 at 20

◦C)(1.047)(θ−20). (6.84)

Apart from the BOD, another important and related term used in wastewaterengineering is the oxygen deficit, Δ (expressed in mg/L or kg/m3), that is defined as

Δ = C∗O2

− CO2 , (6.85)

Time/days

Oxy

gen

dem

and,

BO

Dt

BODt = L0 (1 – e–k1t)

L = L0 e–k1t)

L0

FIGURE 6.17 A typical BOD curve.

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where C∗O2

is the saturation concentration of oxygen in water (mg/L or kg/m3) and

CO2 is the existing oxygen concentration in water (mg/L or kg/m3). The followingexample will clarify the relationship between Lo and Δ for a flowing stream such asa river.

EXAMPLE 6.10 OXYGEN DEFICIT INA POLLUTED NATURAL STREAM

Bioorganisms decompose organic molecules by consuming oxygen from water in theprocess. At the same time, dissolution of oxygen from air into water tends to restorethe oxygen level in a polluted natural stream. The latter process is called reaeration,and the former process is termed deoxygenation. The rate constant for deoxygenationis kd and that for reaeration is kr , both being first-order rate constants. Assume that thestream is in plug flow. The rate of increase in oxygen deficit Δ is given by

dΔ

dt= kdLoe−kdt − krΔ. (6.86)

Integrating the above equation with the initial condition, Δ = Δo at t = 0, we obtain

Δ = Δoe−kr t +(

kdLo

kr − kd

)(e−kdt − e−kr t

). (6.87)

This is the well-known Streeter–Phelps oxygen-sag equation, which describes theoxygen deficit in a polluted stream. t is the time of travel for a pollutant from itsdischarge point to the point downstream. Thus it is related to the velocity of the streamas t = y/U, where y is the downstream distance from the outfall and U is the streamvelocity.

Subtracting the value of Δ as given above from the saturated value C∗O2

gives theoxygen concentration CO2 at any location below the discharge point.The above equation can be used to obtain the critical oxygen deficit (Δc) at which

point the rate of deoxygenation exactly balances the rate of reaeration, that is, dΔ/dt =0. At this point, we have the following equation, which gives Δc.

Δc =(kd

kr

)Loe−kdtc . (6.88)

This gives theminimumdissolved oxygen concentration in the polluted stream.To obtainthe value of tc at which the value of dΔ/dt = 0, we can differentiate the expressionobtained earlier forΔ with respect to t and set it equal to zero. This gives a relationshipfor tc solely in terms of the initial oxygen deficit (Δo) as follows:

tc =(

1

kr − kd

)· ln[kr

kd

(1 − (kr − kd)

kd

Δo

Lo

)]. (6.89)

Figure 6.18 is a typical profile for CO2 as a function of time t (or distance y from thedischarge point) in a polluted stream. In any given stream, Lo is given by the BOD ofthe stream water plus that of the wastewater at the discharge point.

continued

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Lo = QwLw + QsLs

(Qw + Qs), (6.90)

whereQw andQs are volumetric flow rates ofwastewater and streamwater, respectively.The initial oxygen deficit Δo is given by

Δo = C∗O2

−(QwCw

O2+ QsCs

O2

Qw + Qs

), (6.91)

where CwO2

and CsO2

are respectively the oxygen concentration (mg/L or kg/m3) in thewastewater and the stream water just upstream of the discharge location. The threeparameters of significance obtained from the above analysis are Δ, Δc, and tc (or yc).This information gives the maximum possible pollutant concentration at the dischargepoint. To utilize the Streeter–Phelps equation, one has to estimate values for kd and krwith precision. kd is determined by obtaining the BOD at two known locations (a and b)in a stream and using the equation

kd = y

Ulog

(La

Lb

). (6.92)

The value of kr is obtained from an equation such as that of O’Connor and Dobbins:

(kr

d−1

)= 3.9

(U

H3

)1/2, (6.93)

where U is the stream velocity (m/s) and H is the mean depth (m). The constant in theequation for kr corrects for the bed roughness in a stream that affects the stream velocity.kr is affected by the presence of algae (that affect CO2 through diurnal variations from

Zone of deaeration

Co2min

Co2*

C o2 /

mg

∙ L–1

Zone of initial deficit

Zone of recovery

Clean water

Δc

Δ0

tc t or y

FIGURE 6.18 A typical Streeter–Phelps oxygen-sag curve.

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photosynthesis), surface-active substances (oil/grease at the air–water interface), andother pollutants that affect BOD. Typical values of kr vary from 0.1–0.23 d−1 in smallponds to 0.7–1.1 d−1 for swift streams. Corrections to kr and kd for the temperature ofthe stream are obtained using the following equation:

(k at θ◦C) = (k at 20◦C)(ω)(θ−20), (6.94)

where ω is a temperature coefficient that is 1.056 for kd and 1.024 for kr .Temperature influences the oxygen sag in a manner such that with increasing tem-

perature, the Δc value is reached faster. Whereas the rate of aeration decreases withtemperature, the rate of deoxygenation increases with temperature. Therefore, we cansubstantiate the faster response of the stream at higher temperature. Diurnal variationsin dissolved oxygen result from increased CO2 due to algal respiration in the daytimeand decreased dissolved oxygen levels at nighttime. Thus dissolved oxygen levels arelargest during the afternoon and least at night.

6.2.2 WATER POLLUTION CONTROL

This section describes the applications of chemical kinetics and reactor models forselected wastewater pollution control operations.

6.2.2.1 Air Stripping in Aeration Basins

Wastewater is treated in lagoons into which air is introduced in either of two ways: (a)air bubbles at the bottom of the lagoon or (b) mechanical surface aerators placedat the water surface to induce turbulence. In either case, transfer of VOCs fromwater to air with simultaneous transfer of oxygen from air to water is the objec-tive. The analysis of oxygen uptake (absorption) and VOC desorption (stripping)are complementary to one another (Matter-Muller, Gujer, and Giger, 1981; Munz,1985; Valsaraj and Thibodeaux, 1987). In this section, we will illustrate the kinet-ics of desorption of VOCs from wastewater lagoons by diffused aeration using airbubbles.

A typicalVOC stripping operation using air bubbles is depicted in Figure 6.19. Letus consider first a batch operation (Q0 = 0). If the reactor is completely mixed, theconcentration in the aqueous column Cw is the same everywhere. Consider a singleair bubble as it rises through the aqueous column. It continuously picks up solute as itmoves up and exits the column carrying the solute to the atmosphere. Obviously, thereis a trade-off between a more contaminated medium (water) and a less contaminatedmedium (air). The rate of transfer is controlled by diffusion across the thin boundarylayer surrounding the bubble. The overall water-phase mass transfer coefficient isKw.The rate expression for the mass transfer to a single gas bubble is given by

dCw

dt= Ab

VbKw(Cw − C∗

w

), (6.95)

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QLCw0

QgQg

QLCw

Hw

FIGURE 6.19 Schematic of submerged bubble aeration in a wastewater lagoon.

where C∗w is the concentration in the aqueous phase that would be in equilibrium

with the concentration associated with the bubble, Cg. Note that the effect of externalpressure on the bubble size has been neglected in deriving the above equation. Theequilibrium concentration in the vapor phase of the bubble is given by the air–waterpartitioning constant, Cg = K,awC∗

w.In an aeration apparatus, it is more convenient to obtain the specific air–water

interfacial area, av (m2 per m3 of total liquid volume). Hence Ab/Vb = 6/Db =av(Vw/Va), where Db is the average bubble diameter, Vw is the total liquid volume,and Va is the total air volume in the reactor at any time. The ratio Va/Vw is called thegas hold-up, εg. Therefore, we can write av = (6/Db)εg. Using the above definitions,we can rewrite the equation for the rate of change of concentration associated withthe bubble as

dCg

dt+ (Kwa)

(Vw

Va

)1

KawCg = (Kwa)

(Vw

Va

)Cw. (6.96)

Since the rise time of a single bubble is really small (τ ≈ a few seconds), a rea-sonable assumption is that during this time, Cw is a constant. This means we canintegrate the above equation to obtain the concentration of the pollutant associatedwith a single bubble.

Using the initial condition, CgA(τ = 0) = 0, we have

Cg(τ) = KawCw

[1 − e−((Kwa)(Vw/Va)(1/Kaw)τ)

]. (6.97)

The rate of change of pollutant concentration within the aqueous phase is given by

dCw

dt= −

(Qg

Vw

)Cg(τ), (6.98)

whereQg is the volumetric flow rate of air. The initial condition for the aqueous phaseisCw(0) = C0

w for a batch reactor.We also note that the residence time of a gas bubble

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is τ = Va/Qg. Hence upon integration, we obtain the following:

ln

(Cw

C0w

)= −

(Qg

Vw

)Kaw

[1 − e−((Kwa)(Vw/Qg)(1/Kaw))

]t = −kremt, (6.99)

where krem is the first-order removal rate constant from the aqueous phase. One candefine partial gas-phase saturation as φ′ = 1 − exp(−φHs), where

φ = (Kwa)

(Vw

Qg

)(1

KawHs

).

In this definition, Hs is the depth at which the air bubble is released in the lagoon.Thus we can rewrite the equation for krem as follows:

krem =(Qg

Vw

)Kaw(1 − e−φHs). (6.100)

Two special limiting cases are to be noted:

(a) If the exit gas is saturated and is in equilibrium with the aqueous phase,1 − exp(−φHs) → 1 and krem = (Qg/Vw)Kaw. This condition can be satis-fied if (Kwa) is large.

(b) The second limiting case is (Kwa)(Vw/Qg)(1/Kaw) � 1, for which krem =(Kwa). This represents the case when the exit gas is far from saturation. Forlarge Kaw andQg values, this limiting case will apply. This is the case in mostsurface aeration systems, where a large volumetric flow of air is in contactwith an aqueous body.

Let us now consider a continuous flow system. If the continuous flow system isin plug flow, then at steady state the ideal residence time for the aqueous phase ist = Vw/QL, and substitutingC0

w = C0 in the above equation for a batch reactor shouldgive us the appropriate equation for a PFR bubble column:

ln

(Cw

C0

)= −

(Qg

QL

)Kaw(1 − e−φHs). (6.101)

The term (Qg/QL)Kaw = S is called the separation factor in chemical engineering,and gives the maximum separation achievable if the exit air is in equilibrium with theaqueous phase.

If the mixing in the aqueous phase is large, a CSTR approach can be used to modelthe process. The overall mass balance is then given by

dCw

dt= QL

Vw(C0 − Cw) + Qg

Vw

(C0

g − Cg(τ)). (6.102)

Since the entering gas is always clean, C0g = 0. Assuming steady state, dCw/dt = 0.

Therefore, we obtainCw

C0= 1

1 + S(1 − e−φHs). (6.103)

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The net of removal in a CSTR is given by

RCSTR = 1 − Cw

C0= S(1 − e−φHs)

1 + S(1 − e−φHs), (6.104)

and for a PFR

RPFR = 1 − Cw

C0= 1 − exp(−S(1 − e−φHs)). (6.105)

A general nomograph can be drawn to obtain the removal efficiency as a functionof both S and φHs. Figure 6.20 represents the nomograph. With increasing S, Rincreases. For a given S, increased φHs increases the value of R. φHs can be large ifKwa is large. Large values of S can be realized if either Qg/QL or Kaw is large. Sincea (Vw/Va) = 6/Db (where Db is the average bubble diameter or more accurately theSauter mean bubble diameter), decreasing the bubble size could markedly improvethe aeration efficiency. The nomograph is useful in designing the size (volume) of areactor required to obtain desired removal efficiency if the values of liquid and gasflow rates are fixed for a given compound.

In some wastewater aeration ponds one encounters a thin layer of oil or a sur-factant foam above the water layer as a natural consequence of the composition ofwastewaters. The oil layer can act as an additional compartment for the accumulationof a pollutant from the aqueous phase. The solute activity gradient between the aque-ous and oil layers helps establish transport from solvent to the aqueous phase whichcompetes with the unidirectional solute transport by air bubbles from the aqueous

00 2 4 6 8 10

0.2

0.4

0.6

0.8

1

0.1110

R CST

R

S

ΦHs

FIGURE 6.20 Nomograph for steady-state removal in an aeration lagoon operated as aCSTR.Air bubbles are introduced at a depth hs below the surface of water.

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phase. The additional removal mechanism by solvent extraction has been known toeffectively increase the rate of removal from the aqueous phase. This process of flota-tion and extraction using air bubbles into an organic solvent layer is termed solventsublation. The process of solvent sublation has been investigated in bubble columnreactors (see Valsaraj, 1994; Valsaraj and Thibodeaux, 1987 for details). It has beenshown to be effective in removing several organic compounds and metal ions.

EXAMPLE 6.11 REACTOR SIZING FORA DESIRED REMOVAL IN AIR STRIPPING

1,2-Dichloroethane (DCA) from a contaminated groundwater is to be removed usingair stripping. Obtain the size of a bubble column reactor required for 80% removal ofDCA from 10,000 gallons per day of groundwater using a 1-ft-diameter column.The air–water partition constant Kaw for DCA is 0.056. QL = 10,000 gpd =

0.027m3/min. Consider a ratio of Qg/QL = 100. Hence Qg = 2.7m3/min. Since thecolumn radius rc = 0.152m, Ac = πr2c = 0.0729m2. Hence the superficial gas velo-city ug = Qg/Ac = 0.617m/s. This parameter appears in the correlation for εg and kla.Although several correlations are available for bubble column reactors, we choose theone recommended by Shah et al. (1982):

εg

(1 − εg)4= 0.2α1/8β1/12

ug

(gDc)1/2and a = 1

3Dcα1/2β1/10ε1.13g ,

where α = gD2cρL/σ and β = gD3

c/ν2L. The different parameters are g = 9.8m2/s,

ρL = 1000 kg/m3, νL = 1 × 10−6 m2/s, and σ = 0.072N/m. Using these parameters,εg/(1 − εg)4 = 2.087.A trial and error solution gives εg = 0.35. Hence, we obtain a = 525m−1. This

gives a/εg = 1500 = 6/Db, and the average bubble diameter in the column is Db =0.004m. Since the aqueous-phase diffusivity of DCA is 9.9 × 10−10 m2/s (see theWilke–Chang correlation—Reid et al., 1987), we obtain kwa = 0.076 s−1. As before1/Kwa = (1/kwa) + (1/kgaKaw), where kw and kg are individual phase mass trans-fer coefficients. For predominantly liquid-phase-controlled chemicals (such as DCA),Kwa = kwa. Thus for the given conditions S = 5.6. If the desired removal in theCSTR is RCSTR = 0.80, we have φ′ = 1 − expφHs) = 0.71, and φHs = 1.24. SinceφHs = KwaVw/KawQg, we obtain Vw = 0.033m3. Therefore, the height of the reactoris Hs = Vw/Ac = 0.45m.

If back mixing is insignificant, the bubble column will behave as a plug-flow reactor.In such a case, for RPFR = 0.80, φHs = 0.34, which gives Vw = 0.011m3, and henceHs = 0.15m. Thus if axial back mixing is avoided in the bubble column, a high degreeof removal can be obtained in small reactors. In reality, this is rarely achieved becauseat large air rates the bubbles are larger (low a values) and the axial dispersion increases.

6.2.2.2 Oxidation Reactor

In this section, we will discuss the application of a redox process for wastewatertreatment. Consider the oxidation of organic compounds by ozone. Ozone (O3) is an

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oxidant used for disinfection and organic compound removal from water. Ozone hasa large aqueous solubility (8.9mg/L) and a high reactivity (ΔG◦

f = +163 kJ/mol). Itsredox potential is high, making it a powerful oxidant.

O3 + 2H+ + 2e−E0

H−2.1V� O2 + H2O. (6.106)

Wastewater treatment using ozone requires an ozone generator, a contactor, andoff-gas treatment devices. Ozone is generated using an electric arc in an atmosphere ofoxygen. The ozone contactor system is a submerged diffuser where ozone is bubbledinto the water column immediately upon generation. The typical depth of the watercolumn is 20 ft (Ferguson, Gramith, andMcGuire, 1991).With contact times as smallas 10min in most water treatment plants, ozone can perform microbial destruction,but can only partially oxidize organic compounds. Hence, it is used in combinationwith H2O2 and UV oxidation processes (Masten and Davies, 1994). Series (cascade)reactors are required to complete the oxidation process.

The reaction of ozone in “pure” water is a radical chain reaction comprising theformation and dissipation of the powerful hydroxyl radical (OH•). The sequencebegins with the base-catalyzed dissociation of ozone

O3 + OH− → HO•2 + O•−

2 ,

HO•2 → H+ + O•−

2 .(6.107)

The above acid–base equilibrium has a pKa of 4.8. Since, per reaction two radicalsare produced, the rate constant k1 = 2 × 70 L/mol/s. The subsequent reactions leadto the regeneration of the catalyst OH− and the formation of O2 and HO•. Staehelinand Hoigne (1985) published a concise summary of the various reactions involved inthe overall scheme (Figure 6.21). The reaction involving the consumption of ozone bythe superoxide anion (O•

2) has a rate constant k2 = 1.5 × 109 L/mol/s. In the presenceof an acid, the last reaction proceeds in two steps:

O•−3 + H+ k3=5×1010L/mol/s−−−−−−−−−−−→ HO•

3,

HO•3

k4=1.4×105 L/mol/s−−−−−−−−−−−−→ HO• + O2.

(6.108)

The hydroxyl radical forms an adduct with ozone, which gives rise to HO•2

and O2 via decomposition, thus propagating the chain. The HO•2 also reacts with

the adduct forming O3, O2, and water with a rate constant k6 of approximately1010 L/mol/s.

Radical scavengers that reactwithOH• can efficiently terminate this chain reaction.In freshwater that is slightly alkaline (due to carbonate and bicarbonate species), thehydroxyl radical is efficiently scavenged by inorganic carbonate species. Hence thedecay rate of ozone is reduced. The presence of organic solutes (such as alcohols,DOC) and other species (such as phosphates) in water impacts the radical chainreaction in different ways. The following discussion is based on the scheme suggested

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O3

k2 = 1.6 × 109

O3 + O2 + H2OO3OH

k7 = 4 × 108

k6 ≈ 1010

k5 = 3 × 109

k1 = 140

k8 = 1.5 × 107

k4 = 1.4 × 105k3 = 5 × 1010

O3–•

O2–• ⇔ HO2

•

O2

O2 OH–

OH–

OH–

H+

(All rate constants arein L ∙ mol–1 ∙ s–1.)

Inhibitionreactions

HO3•

HCO3–

O2

O3CO3

2–

OH•

FIGURE 6.21 Radical chain reaction mechanism for ozone decomposition in pure water.Inhibition by carbonate alkalinity is also shown. (Reprinted from Staehelin, J. and Hoigne, J.1985. Environmental Science and Technology 19, 1206–1213. American Chemical Society.With permission.)

byStaehelin andHoigne (1985) (Figure 6.22).A soluteMcan interferewith the radicalchain reaction involving the ozonide and hydroxyl radical in four different ways: (i)Solute M can react directly with ozone to produce the ozonide radical and an M+ion with a rate constant k13. (ii) Solute M can be oxidized to Mox by direct reactionwith O3 with a rate constant k14. (iii) As discussed in Figure 6.21, the solute M canreact with the hydroxyl radical in two ways. The reaction can lead to the scavengingof OH• as in the case of CO2−

3 or HCO−3 . This step is generally represented by the

formation of the productΦ in Figure 6.22 with a rate constant k12. (iv) In some cases,the solute M reacts with OH• to form the peroxo radical ROO•, which further addsO2 to give Mox and re-forming the HO•

2. This is characterized by the rate constantsk10 and k11 in Figure 6.22.

Let us now ascribe to the above reaction scheme (Figure 6.22) a reaction rate lawby using steady-state approximations for O−•

2 and OH• species, and consider the lossof ozone by all possible reaction pathways. The rate of ozone reaction withM directlyis k14[M][O3], and the total rate of initiation of chain reactions (i.e., the formation ofO−•2 and O−•

3 ) is k13[M][O3]+2k1[OH−][O3]. The rate at which OH• is converted toHO−•

2 is k9[M][OH•], and the rate of scavenging of OH• by M is k12[M][OH•]. Therate of ozone consumption via chain reactions is given by

−d[O3]dt

= k2[O•−2

] [O3]. (6.109)

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ROO

M

H+

M MOx

M

M+

R•

MOx

M

MΦ

k1 = 140

O2–• ⇔ HO2

•

k4 = 1.4 × 105

k12

k14

k13

k9

k10

k11

k3 = 5 × 1010

k2 = 1.6 × 109

O3

O2

HO3•

O3

O3–•

OH–

O2

OH•

FIGURE 6.22 Radical chain reaction mechanism for ozone decomposition in impure water.(Reprinted from Staehelin, J. and Hoigne, J. 1985. Environmental Science and Technology19, 1206–1213. American Chemical Society. With permission.)

Applying the PSSA for O−•2 , we have

[O•−2

]ss = 2k1[OH−][O3] + k9[M][OH•]ss

k2[O3] . (6.110)

Assuming PSSA for OH•, we obtain

[OH•]ss = k13[M][O3] + k2[O•−2 ]ss[O3]

(k9 + k12)[M] . (6.111)

Solving the above equations, the steady-state concentration of the hydroxyl radical is

[OH•]ss =(2k1[OH−] + k13[M]

k12[M])

[O3]. (6.112)

Hence we have the following equation:

− 1

[O3]d[O3]dt

= 2k1[OH−](1 + k9

k12

)+ k13[M]

(k9k12

). (6.113)

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Considering all of the chain reactions for the loss of ozone, we have

− 1

[O3]gd[O3]dt

= k1[OH−] + 2k1[OH−](1 + k9

k12

)+ k13[M]

(k9k12

). (6.114)

Including the total ozone consumption via nonchain reactions as well, we can writethe above equation as

− 1

[O3]gd[O3]dt

= k1[OH−] + 2k1[OH−](1 + k9

k12

)+ k13[M]

(k9k12

)+ k14[M].

(6.115)

At constant [OH−] (or pH) and constant [M], we can write

d[O3]dt

= −ktot[O3]tot. (6.116)

Thus, the rate is pseudo-first-order with a rate constant ktot. A plot of ln[O3] versust should yield a straight line with slope ktot for a specified [OH−] and [M]. For abatch reactor, the above equation will give the rate of disappearance of ozone in thepresence of different substrates. The identity of solute Mmay be different at differentpoints along the chain. Thus, for example, the species M undergoing oxidation andforming radicals that start the chain reaction with the rate constant k13 is called aninitiator (represented by I). Those that terminate the chain by reacting with OH•are called terminators or suppressors (represented by S). Those species that reactwith OH• to reform O−•

2 with rate constant k9 are called propagators (representedby P). The direct reaction of M with ozone characterized by the rate constant k14is designated kd. Staehelin and Hoigne (1985) have identified different species inwater that perform the above functions. These are listed in Table 6.2 along with therespective rate constants. To differentiate these species correctly in the rate equation,they also generalized the above rate constant to give

ktot = kI[OH−] + (2kI[OH−] + kI[M])(1 + kp[M]

ks[S])

+ kd[M], (6.117)

where kI = k13, kd = k14, kP = k9, and kS = k12. For most hard wastewaters wherebicarbonate is the predominant scavenger of OH•, we have kI[M] � 2k1[OH•], andhence

ktot = kI[M](1 + kp[M]

ks[S])

+ kd[M]. (6.118)

Now that we have gained an understanding of the kinetics of ozone oxidation pro-cesses, we shall next see how this information can be utilized in analyzing ozonationreactors.

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TABLE 6.2Common Types of Initiators, Propagators, and Scavengers Foundin Wastewater, and Values of ktot for Ozone Depletion in WaterContaining Different Species, M

Initiator Promoter Scavenger

Hydroxyl ion Aromatic Bicarbonate ionFerrous ion Alcohols Carbonate ionDOCs (humic and fulvic acids) DOCs (humic acid)

[M] pH ktot (s−1)

0 (none) 4.0 0.15±0.02

50mM PO2−4 4.0 0.072±0.006

7μM t-BuOH 4.0 0.055±0.00450μM t-BuOH 4.0 0.02±0.002

Source: From Staehelin, J. and Hoigne, J. 1985. Environmental Science and Technology19, 1206–1213.

EXAMPLE 6.12 WASTEWATER OXIDATION USING OZONE IN A CONTINUOUS

REACTOR

Consider a diffuse aerator reactor operated in the continuous mode (Figure 6.23). Agiven influent feed rate of water (QL, m3/s) is contacted with an incoming gas streamof ozone at a concentration [O3]ing (mol/m3) at a volumetric flow rate ofQG (m3/s).Weare interested in obtaining the extent of ozone consumption within the reactor if “pure”water is used and also in the presence of other solutes. Consider a CSTR, for which amass balance for ozone in the aqueous phase within the reactor gives

VLd[O3]dt

= QL([O3]0 − [O3]) + KwaVL([O3]∗ − [O3]

)− ktotVL[O3], (6.119)

where [O3]0 = 0,Kwa is themass transfer coefficient for ozone fromgas towater, [O3]∗is the saturation concentration of ozone inwater, and ktot is the first-order decompositionof ozone as described earlier. At steady state since d[O3]/dt = 0, we have

[O3][O3]∗ = 1

1 + (QL/VL + ktot) · (1/Kwa). (6.120)

With increasing ktot , the ratio decreases indicating that the exit ozone concentrationis lower and more of ozone is consumed within the CSTR. Consider a typical ozonewastewater oxidation reactor withQL = 2500m3/h and VL = 1500m3.Kwa for ozoneis typically 0.03min−1 (Roustan et al., 1993). FromStaehelin andHoigne (1985), ktot =3min−1 at pH 4 in the presence of 50mMphosphate. Hence the ratio is [O3]ss/[O3]∗ =0.0098. If the partial pressure of ozone PO3 in the incoming gas phase is known, then

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[O3]∗ = PO3/K′′aw, where K ′′

aw = 0.082 is the Henry’s constant expressed inatmm3/mol. For a typical PO3 = 0.0075 atm, [O3]∗ = 0.092mol/m3. Hence [O3]ss =9 × 10−4 mol/m3. A mass balance for ozone over the entire reactor yields

QG

([O3]ing − [O3]out

g

)= (QL + ktotVL)[O3]ss = 4mol/min.

If QG = 1000m3/h (= 16m3/min), [O3]ing − [O3]outg = 0.24mol/m3. Since [O3]ing =

PO3/RT = 0.31mol/m3, we obtain [O3]outg = 0.07mol/m3. Hence the ozone transfer

efficiency in the reactor is 77%.

QG [O3]gout

QG [O3]gin

[O3]in = 0

[M]out

[O3]

M → Mox

water

Ozone

QL QL

[O3]

[M]in

FIGURE 6.23 Schematic of an ozone reactor for wastewater oxidation.

6.2.2.3 Photochemical Reactions and Wastewater Treatment

Photochemical reactions are useful in treating wastewater streams. An application inthis area is the use of semiconductors (e.g., TiO2) in enhancing the UV-promoted oxi-dation of organic compounds. The reaction pathway provided by TiO2 is complicated(Legrini, Oliveros, and Braun, 1993). TiO2 is a semiconductor. It has a structure thatis composed of a valence band (filled electronic level) and a conduction band (vacantelectronic level) that are separated by a band gap (see Figure 6.24). As an electronjumps from the valence band to the conduction band, a hole (positive charge) is leftbehind in the valence band. This e− jump can be brought about through excitation bylight (UV or visible). Organic molecules that are thereby oxidized can scavenge thehole left behind in the valence band. The photoexcited TiO2 with the electron–hole

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Conduction band

Band gap

h+

e–

EcEgEv

Valence band

FIGURE 6.24 Valence and conduction bands in a metal oxide semiconductor.

pair (e− + h+) is depicted as

TiO2hn−→TiO2(e

− + h+).

Two types of oxidation reactions have been noted in aqueous suspensions of TiO2.One is the electron transfer from the organicmolecule adsorbed on the surface (RXads)

TiO2(h+) + RXads −→ TiO2 + RX+•

ads,

and the second is the e− transfer to the adsorbed solvent molecule (H2Oads)

TiO2(h+) + H2Oads −→ TiO2 + HO•

ads + H+.

The above reaction appears to be of greater importance in the oxidation of organiccompounds. The abundance of water makes this a feasible reaction. Similarly,adsorbed hydroxide ion (OH•

ads) also appears to participate in the reaction

TiO2(h+) + OH−

ads → TiO2 + HO•ads.

The ever-present O2 in water provides another avenue of reaction by acting as anelectron acceptor:

TiO2(e−) + O2 → TiO2 + O−•

2 .

The addition of hydrogen peroxide has been shown to catalyze this process, presum-ably via a surface dissociation of H2O2 to OH• that subsequently oxidizes organiccompounds.

TiO2(e−) + H2O2 → TiO2 + OH− + OH•.

The above reaction is sensitive to pH changes and the electrical double-layerproperties of TiO2 in aqueous solutions. Therefore, the UV-promoted dissociation

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TABLE 6.3Photodegradation of Organics by TiO2/UV Process

Phenol, TOC reduced by 35% in 90minAcetic, benzoic, formic acids, ethanol, methanol, TOC reduced by >96% in 10minDegradation of aniline, salicylic acid, and ethanol1,2-Dimethyl-3-nitrobenzene and nitro-o-xylene from industrial wastewater,>95%TOC removed in 30min

Degradation of pentachlorophenolDegradation of other organic compounds (trichloroethylene, chlorobenzene,nitrobenzene, chlorophenols, phenols, benzene, chloroform)

Degradation of a pesticide, atrazine

Source: From Legrini, O., Oliveros, E., and Braun,A.M. 1993. Chemical Reviews,93, 671–698.

of organic compounds by TiO2 in the presence of H2O2 is significantly affected bypH and solution ionic strength.

The large-scale development of semiconductor-promoted reactions awaits opti-mal design of the photoreactor, photocatalyst, and radiation wavelengths. The criticalissue in this regard is a suitable design for the treatment of a sample that can simul-taneously absorb and scatter radiation.Work in this area appears to be promising andis summarized in Table 6.3.

6.2.2.4 Photochemical Reactions in Natural Waters

Natural waters are a mixture of several compounds, and in most cases transient radi-cals or species are produced by photochemical reactions. Zepp (1992) has summarizeda variety of transient species. Their effects on photochemical reactions are such thatthe rates of reactions can vary by orders of magnitude depending on the type and con-centration of the transient. DOCs(humic and fulvic compounds), inorganic chemicals(nitrate, nitrite, peroxides, iron, and manganese), and particulates (sediments, biota,andminerals) can produce transients in natural waters upon irradiation. The transientsparticipate in and facilitate the redox reactions of other compounds in natural waters.

There are a variety of transients that have been identified in natural waters. Someof the important ones are (i) solvated electron, (ii) triplet and singlet oxygen, (iii)superoxide ions and hydrogen peroxide, (iv) hydroxyl radicals, and (v) triplet excitedstate of dissolved organic matter. The level of steady-state concentrations of someof these species and their typical half-lives in natural waters are given in Table 6.4.Solvated electron (e−aq) is a powerful oxidant, observed in natural waters during irradi-ation. It reacts rapidly with electronegative compounds (both organic and inorganic).Its reaction with O2 is the primary pathway for the production of superoxide anionsin natural waters. The major source of e−aq is aquatic humic compounds. The quantum

yield for their production is approximately 10−5. Dissolved organic matter in naturalwaters is also known to absorb photons to generate singlet and triplet excited states.These then decay by transferring energy to dissolved oxygen to produce singlet oxy-gen. Singlet oxygen is an effective oxidant and the quantum yield for its formation is

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TABLE 6.4Photochemical Transient Species in Natural Waters

Typical Steady-stateTransient Species Source of Transient Concentration (mol/L) Typical Half-life (h)

1O2 DOM 10−14–10−12 ≈3.5e−aq DOM ≈10−11 ≈ 4.3OH• NO−

3 , NO−2 , H2O2 10−17–10−15 ≈170

O−•2 DOM 10−8–10−7 ≈2

Source: From Zepp, R.G. 1992. In: J.L. Schnoor (ed.), Fate of Pesticides and Chemicals in theEnvironment, pp. 127–140. NewYork: Wiley.

0.01–0.03 in UV and blue spectra. Superoxide ions and hydrogen peroxide are foundin both lakes and atmospheric moisture (fog and rain). They are longer lived thanother transients and are powerful oxidants for most organics. Algae and other biotaare known to quench the action of superoxide ions.

In general, the kinetics and concentrations of photochemical transients in naturalwaters can be understood through the application of the steady-state theory. Theformation of transient (Tr) is via the reaction

SO2−→hv

Tr (6.121)

with the rate rf ,Tr = φI . The transient species can either decay by itself or by reactionwith a solute molecule A:

TrkTr−→P,

Tr +Akr−→P.

(6.122)

The total rate of disappearance of the transient is

rd,Tr = kr[A][Tr] + kTr[Tr]. (6.123)

At steady state, rd,Tr = rf ,Tr, and hence

[Tr]ss = φI

kr[A] + kTr. (6.124)

The overall rate of the reaction of A is

rA = −d[A]dt

= kr

(φI

kr[A] + kTr

)[A]. (6.125)

If the concentration of A is sufficiently low, as is the case in most natural systems,then kr[A] � kTr, and hence we have

rA = kr

kTrφI[A] = k′

r[A], (6.126)

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where k′r is a pseudo-first-order rate constant. Zepp (1992) has summarized how each

of the terms comprising k′r can be obtained experimentally.

6.3 THE AIR ENVIRONMENT

In this section, the use of concepts from chemical kinetics and mass transfer theoryfor modeling the F&T of chemicals in the air environment and in the design of airpollution control devices is addressed.

6.3.1 F&T MODELS

6.3.1.1 Box Models

It is possible to explore environmental systems through the use of box models. These“models” simulate the complex behavior of a natural system by applying useful sim-plifications. For example, box models are derived from the concept of CSTRs. Eachphase is considered to be awell-mixed compartment and pollutants exchange betweenthe different compartments.

CSTRboxmodels are used inunderstanding the transport of chemicals frommobileand stationary sources (automobiles, chemical, and coal-powered power plants) tothe atmosphere, transport and fate of pollutants between air and water, sediment andwater, and between soil and air. The sediment–water system is pictorially representedin Figure 6.25a. Chemicals enter the water stream either with the inflow, or by wetand dry deposition from the atmosphere.A third pathway is the resuspension of solids

Soil/sediment

(a) (b)

Water

Air

Influent Effluent

SedimentationDissolution

Current

Drydeposition

Wetdeposition

Evaporationvolatilization

Waterlevelchange

Reactions

Soil

Loweratmosphere

Solar fluxExchange with

upper atmosphere

Effluent

Surfacesources

Deposition

Reactions(chemical/

photo)Influent

Windcurrent

FIGURE 6.25 Schematic of typical “box” models for (a) a natural aquatic stream, and (b)atmosphere. The various transport and fate processes in each case are shown.

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from the sediment beneath the stream. The chemicals dissipate from the stream byeither chemical reaction or biodegradation; or are removed with the outflow and byevaporation from the surface. They may also subside into the underlying sediment bysedimentation with solids.

A similar CSTR box model can also be developed for atmospheric transport ofchemicals (Figure 6.25b). Stationary sources such as industrial plants, waste pits,and treatment plants provide pollutant inputs to the atmosphere. Mobile sourcessuch as automobiles also contribute to pollutant influx to the atmosphere. Depositionof aerosol-associated pollutants removes pollutants from the atmosphere. Advectiveflow due to winds constitutes the inflow and outflow from the CSTR. Chemical andphotochemical reactions change concentrations within the box.

Consider awell-mixed box of downwind lengthL, crosswindwidthW , and verticalheight H. Figure 6.26 depicts the atmosphere of volume V = LWH that is homoge-neously mixed (i.e., no concentration gradients).A large urban atmosphere with goodmixing (turbulence) and a wide distribution of emission sources in the lower atmo-sphere or a small indoor volume that experiences excellent mixing and ventilationcan be represented by the above scheme. Global atmospheric models rely on theassumption that the troposphere is a well-mixed CSTR. In some cases, the northernand southern hemispheres are considered to be two separate reactors with differ-ent degrees of mixing, but linked via chemical exchange. These models are usefulin obtaining the residence times of important species in the atmosphere (Seinfeld,1986). For the case in Figure 6.26, a mass balance on species A within the controlvolume will give the following:

rate of input = rate of output + rate of reaction + rate of accumulation.

G[A]0 + wvAs[A]∞ + SAAs = G[A] + wpAs[A] + wvAs[A] + rAV + Vd[A]dt

.

(6.127)

(Surface soil)

(Air)

SArA

wpG G

W

H

[A]∞wv

[A]0 [A]

L

FIGURE 6.26 Material balance for a well-mixed box model for an urban atmosphere.

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Dividing throughout by V and noting that Q/V = U and As/V = 1/H, we obtain

d[A]dt

= U

L([A]0 − [A]) + wv

H([A]∞ − [A]) − wp

H[A] + SA

H− rA, (6.128)

where the overall loss by reaction is given by rA = k[A],U is the wind velocity (m/s),wv is the ventilation velocity away from the box (m/s), wp is the surface depositionvelocity from the atmosphere (m/s), SA is the surface source strength of A (μg/m2 s),[A]∞ is the concentration of A in the upper atmosphere (μg/m3), and [A]0 is theconcentration of A in the incoming air (μg/m3). As is the surface area (m2). Thefollowing examples illustrate the application of the above model.

EXAMPLE 6.13 A CSTR BOX MODEL FORAN URBAN AREA

Assume that the pollutant is conservative (nonreactive) and that velocities wp and wvare negligible. Then the overall mass balance reduces to the following equation:

d[A]dt

= U

L([A]0 − [A]) + SA

H. (6.129)

At steady state, d[A]/dt = 0 and the steady-state concentration in the box is given by

[A]ss = [A]0 + SA

U

L

H. (6.130)

Thus if [A]0 = 0, [A]ss is proportional to SA/U, that is, the larger the source strength,the larger the steady-state concentration in the box.

If the process is at unsteady state, then we can solve for [A] as a function of t withthe initial condition that at t = 0, [A] = [A]initial and obtain the following equation:

[A] =(SA

U

L

H+ [A]0

)(1 − e−(U/L)·t)+ [A]initiale

−(U/L)·t (6.131)

and for [A]initial = 0 and [A]0 = 0,

[A] =(SA

U

L

H

)(1 − e−(U/L)·t) , (6.132)

which states that the concentration increases toward the steady-state value as t → ∞.The rate constant is U/L. The inverse L/U is called the residence time in the box.

Urban areas receive air pollution from stationary and mobile sources. In the UnitedStates, for example, LosAngeles, CA, experiences air pollution resulting from automo-bile exhaust, whereas in Baton Rouge, Louisiana, the problem results from industrialsources. In the case of LosAngeles, a city nestled in a valley betweenmountains, we canconsider the valley to be a CSTRwhere periodic strong SantaAna winds replace the air.The prevailing winds bring with it a background level of pollutant (e.g., carbon monox-ide, CO) from surrounding areas that adds to the existing background CO in the valley.The automobiles within the city emit CO as a component of its exhaust gas. Figure 6.27schematically represents the various processes. The total source strength in moles perhour is given by STOT (mol/h) = SA(mol/m2 h) LW = rate of CO input by prevailingwinds+ rate of CO input by automobiles. The total volumetric flow rate of air through

continued

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Wind currents

CSTR

Atmospheric inversion layer

Valley

Stationarysource

Mobile source

FIGURE 6.27 A simplified well-mixed “box” model for an urban area nestledbetween mountains (e.g., Los Angeles, CA).

the valley is G(m3/h) = U(m/h)WH(m2) and U(m/h) = GL/V . Hence,

[A] = STOT

G

(1 − e−(G/V)·t)+ [A]0e−(G/V)·t . (6.133)

Let us take a specific example where V = 1 × 1012 m3, G = 5 × 1011 m3/h. CO inputby wind = 2 × 106 mol/h, CO input by automobiles = 1 × 108 mol/h, and [A]initial =

00 2 4 6 8 10

5 10–5

0.0001

0.00015

0.0002

0.00025

[A]/m

ol ∙ m

–3

t/h

NCO,auto = 0 mol ∙ h–1

NCO,auto = 1E8 mol ∙ h–1

FIGURE 6.28 Change in concentration of carbon monoxide with time in an urbanarea as a result of automobile emissions. A comparison is made with one that is devoidof automobile contributions.

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1.2 × 10−5 mol/m3 (background CO). The equation is [A] = 2.04 × 10−4(1 −e−t/2) + 1.2 × 10−5 e−t/2.The steady-state concentration [A]ss is 2.04 × 10−4 mol/m3.Figure 6.28 depicts the change in CO with time. If the contribution from automobilesis negligible, the [A] in the CSTR is continually diluted.

EXAMPLE 6.14 AN INDOOR AIR POLLUTION MODEL

In this case we consider a pollutant that may not only react, but also has a source withinthe indoor atmosphere. We disregard both wv and wp. The overall mass balance in thiscase is

d[A]dt

= U

L· ([A]0 − [A]) + SA

H− k[A]. (6.134)

The termU/L = (UWH)/(LWH) = G/V is called the air exchange rate, Ia and SA/H =(SALW)/(LWH) = STOT/V , where STOT is the surface source strength expressed inμg/s. Using the initial condition, [A] = [A]initial at t = 0, we obtain

[A] =( [(STOT/V) + [A]0Ia]

Ia + k

)(1 − e−(Ia+k)t

)+ [A]initiale

−(Ia+k)t . (6.135)

The steady-state concentration (t → ∞) is

[A]ss = [(STOT/V) + [A]0Ia]Ia + k

. (6.136)

For the special case where [A]0 = 0 and k = 0 (conservative pollutant) and whereunsteady state is applicable,

[A] =(STOT

IaV

)(1 − e−Ia t

). (6.137)

If a room of total volume V = 1000m3 where a gas range emits 50mg/h of NOxis considered for an air exchange rate Ia = 0.2 h−1, and a first-order rate of lossat 0.1 h−1, we have at steady state, [A]ss = (50/1000)/(0.2 + 0.1) = 0.16mg/m3 =160μg/m3.

EXAMPLE 6.15 A GLOBAL MIXING MODEL

Globally the stratosphere and the atmosphere are considered to be separate compart-ments (boxes) with internal mixing, but with slow exchange between the two boxes(Warneck, 1999; Reible, 1998). This is one of the reasons why nonreactive chemicals(e.g., CFCs) released to the troposphere slowly make their way into the stratospherewhere they participate in other reactions (e.g., with ozone). If the slow exchange rateis represented G (see Figure 6.29), the rate of change in concentration in each CSTR is

continued

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Stratosphere

Troposphere

Earth

G [A]st

[A]tr

FIGURE 6.29 Schematic of the model assuming well-mixed troposphere and strato-sphere with a slow exchange between the two compartments.

given as follows:

rate of accumulation = rate of input − rate of output.

Vtrd[A]trdt

= G([A]st − [A]tr),

Vstd[A]st

dt= G ([A]tr − [A]st).

(6.138)

Consider radioactive materials released through atomic weapons testing in the early1960s. The release occurred in the stratosphere where they reacted only slowly. Uponreaching the troposphere, however, they were quickly reacted away. In such a case,[A]st � [A]tr and hence

d[A]st

dt= −G

V[A]st = − 1

τst[A]st , (6.139)

where τst is the exchange (mixing) time between the stratosphere and troposphere.Upon integration with [A]st = [A]0st at t = 0, that is, when weapons testing first began,we obtain

[A]st = [A]0ste−(τ/τst). (6.140)

The concentration in the stratosphere should decline exponentially with time. Inother words, a plot of ln[A]st versus t will give 1/τst as the slope from which τst can be

TABLE 6.5Exchange (Mixing) Times between andwithin Compartments Using Data on 85Kr

Exchange Compartments Mixing Time (Years)

Stratosphere–troposphere 1.4Troposphere–troposphere 1.0Stratosphere–stratosphere 4.0

Source: FromWarneck, P. 1988. Chemistry of the NaturalAtmosphere, Academic Press, NewYork.

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ascertained. Warneck (1988) has listed values of mixing time for several radioactivetracers. A similar approach can also be used for appropriate tracers to calculate mixingtimes within the troposphere and stratosphere, if the two boxes are further subdividedinto a north and south hemisphere. The average values obtained are summarized inTable 6.5.

EXAMPLE 6.16 ATMOSPHERIC RESIDENCE TIME

Atmospheric residence time is important in atmospheric models. This tells us the dura-tion a molecule spends in the atmosphere before it is removed by either wet or drydeposition to the surface of the earth. This is based on the perfectly mixed box approach.Starting with 1 cubic volume of the lower atmosphere, we have the following massbalance (Seinfeld and Pandis, 2006; Warneck, 1999):

input rate by flow − output rate by flow + rate of production

− rate of removal = accumulation.

FinA − Fout

A + SA − RA = dmA

dt, (6.141)

wheremA is the total mass of speciesA in the unit volume of air considered. The averageresidence time is defined as τA = mA/(SA + Fout

A ) = mA/(SA + FinA ), since at steady

stateRA + FoutA = SA + Fin

A . For the entire atmosphere at steady state,FinA = Fout

A = 0and SA = RA.Atmospheric concentrations are expressed inmass per unit volume of air,CA, (μg/m3), or as a mixing ratio, ξi = CA/Ctot , where CA is the molar concentrationof A in air and Ctot is the total molar concentration of air. Note that since the ideal gaslaw applies ξA = PA/P.CA(μg/m3) and ξA (ppmv) are related as follows: ξA = 8.314(T/PMA) CA (see also Appendix 4). For sulfur, which has a total rate of production of2 × 1014 g/y and an average mixing ratio of 1 ppb, we find the total mass at steady stateto be (1 × 10−9 g/g) (5 × 1021 g), where 5 × 1021 g is the total mass of the atmosphere.Hencem = 5 × 1012 g. Thus the average residence time is τA = 5 × 1012/2 × 1014 =0.025 y. Note that if τA = G/RA and RA = kAG, we have τA = 1/kA.

6.3.1.2 Dispersion Models

When pollutant sources are not widely distributed across a large area, and mixingis insufficient, the CSTR box model described above becomes inapplicable. Exam-ples are point source emissions from smokestacks, episodic spills on the ground,explosions, or accidental releases of air pollutants. We shall use the general case of acontinuous smokestack emission to illustrate the principles of air pollution modelingfrom point sources.

We saw in Chapter 2 that temperature gradient in the atmosphere determineswhether the atmosphere is stable or unstable. The variation in temperature with heightdetermines the extent of buoyancy-driven mixing within the atmosphere. To do so, we

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consider the ideal casewhere a parcel of dry air is allowed to rise adiabatically throughthe atmosphere. The air parcel experiences a decrease in temperature as it expands inresponse to decreasing temperature with increasing elevation. This is called the adia-batic lapse rate,Γadia(=dT/dz), and is a standard value of 1◦C/100m for dry air, aswesaw in Chapter 2. For saturated air this value is slightly smaller (0.6◦C/100m). If theprevailing atmosphere has an environmental lapse rate, Γenv > Γadia, the atmosphereis considered unstable and rapid mixing occurs, thereby diluting the air pollutant. IfΓenv < Γadia, the atmosphere is stable and little or mixing occurs.

The second important parameter is the wind velocity. In general, the velocity ofwind above the surface follows a profile given by Deacon’s power law,

UzUz1

=(z

z1

)p,

where Uz is the wind velocity at height z, Uz1 is the wind velocity at height z1, andp is a positive exponent (varies between 0 and 1 in accordance with the atmosphericstability).

The above effects of temperature and wind velocity are introduced in the disper-sion models through the Pasquill stability criteria. However, we are still left with oneadditional issue, that of the terrain itself. Obviously, the wind velocity near a sur-face is strongly influenced by the type of terrain. Rough surfaces or irregular terrain(buildings or other structures) will give rise to varying wind directions and velocitiesnear the surface. Open wide terrains do not provide that much wind resistance.

The dispersion of gases and vapors in the atmosphere is influenced by the degreeof atmospheric stability, which in turn in determined by the temperature profile in theatmosphere, the wind direction, wind velocity, and surface roughness. These factorsare considered in arriving at equations to determine the dispersion of pollutants in aprevailing atmosphere. Let us consider the emission of a plume of air pollutant froma tall chimney as shown in Figure 6.30. The emission occurs from a point source andis continuous in time. The process is at steady state. The air leaving the chimney isoften at a higher temperature than the ambient air. The buoyancy of the plume causesa rise in the air parcel before it takes a more or less horizontal travel path. We applythe convective–dispersion equation that was derived in Section 6.1.3, extended to allthree Cartesian co-ordinates.We consider here a nonreactive (conservative) pollutant.Hence, we have the following equation:

U∂[A]∂x

= Dx∂2[A]∂x2

+ Dy∂2[A]∂y2

+ Dz∂2[A]∂z2

. (6.142)

For stack diffusion problems, from experience the following observations can bemade:

(i) In the x-direction, the mass transfer due to bulk motion, U∂[A]/∂x, farexceeds the dispersion, Dx∂2[A]/∂x2.

(ii) The wind speed on the average remains constant, that is, U is constant.

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z

y x

zs z1

FIGURE 6.30 Emission of a plume of air pollutant from a tall chimney.

The simplified expression is then

U∂[A]∂x

= Dy∂2[A]∂y2

+ Dz∂2[A]∂z2

. (6.143)

Upon integration and rearranging, the following solution results:

[A](x, y, z) = Qs

πUσyσzexp

[−1

2

(y2

σ2y+ (z − zs)2

σ2z

)], (6.144)

where σ2y = 2Dyx/U and σ2z = 2Dzx/U are the respective squares of the mean stan-dard deviations in y- and z-directions for the concentrations. The above equation isonly applicable for the concentration in the downwind direction up to the point inthe x-direction where the ground-level concentration (z = 0) is significant. After this“reflection” of pollutants will occur, because soil is not a pollutant sink and materialwill diffuse back to the atmosphere. This gives rise to a mathematical equivalenceof having another image source at a distance zs (Figure 6.31). A superposition of thetwo solutions gives the final general equation for pollutant concentration anywheredownwind of the stack. The concentration of vapor in the air at any point from aplume of effective height zs is given by

[A](x, y, z) = Qs

πUσyσz

[e−((zs−z)2/2σ2z ) + e−((zs+z)2/2σ2z )

]e−( y2/2σ2y ). (6.145)

The concentration is given in g/m3 if Qs is expressed in g/s. It can be converted to themore common units of μg/m3 or ppmv using appropriate conversions.

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ReflectedplumeSurface

z

Point of interestz – zs

Realsource

Virtualsource

z1

z = 0x

zs

–z1–zs

FIGURE 6.31 Reflection of a gaseous plume of a pollutant mathematically analyzed usingan imaginary source.

The standard deviations σy and σz have been correlated to both mechanical turbu-lence (wind speed at 10m height) and buoyant turbulence (time of day). These giverise to what are called Pasquill stability categories given in Table 6.6. Based on thestability class, appropriate standard deviations for any distance x downwind of thesource are obtained (Figure 6.32).

TABLE 6.6Pasquill Stability Categories

Day Night

Incoming Solar Radiation Cloud Cover

Wind Speed at Mostly Mostly10 m Height (m/s) Strong Moderate Slight Overcast Clear

<2 A A–B B E F2–3 A–B B C E F3–5 B B–C C D E5–6 C C–D D D D>6 C D D D D

Note: Class A is the most unstable and Class F is the most stable. Class B is moderately unstableand Class E is slightly stable. Class C is slightly unstable. Class D is neutral stability andshould be used for overcast conditions during day or night.

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AB

CD E

F

A ExtremelyB Moderately

F Moderately

C Slightly

E Slightly Stable

UnstableD Neutral

102

103

102

10

1103 104

Distance from source, m

Vert

ical

disp

ersio

n co

effici

ent,

σ x m

105

A ExtremelyB Moderately

F Moderately

C Slightly

E Slightly Stable

UnstableD Neutral

102

103

104

102

10

1103 104

Distance from source, mH

oriz

onta

l disp

ersio

n co

effici

ent,

σ y m

105

F

E

D

C

B

A

FIGURE 6.32 Vertical and horizontal dispersion coefficients for use in the Gaussian airdispersion model. (From Turner, D.B. 1969.Workbook of Atmospheric Dispersion Estimates.Washington, DC: HEW.)

The effective height of the stack zs(m) is the sum of the actual stack height z1(m)and the plume riseΔz(m). The plume rise can be estimated using a number of differentexpressions (Wark, Warner, and Davis, 1998).

One parameter that is often sought in air pollutionmodeling is the so-called ground-level center-line concentration

[A](x, 0, 0) = Qs

πUσyσze−(z2s /2σ2z ). (6.146)

The maximum concentration at a receptor along the ground-level center-line will begiven by substituting z2s = 2σ2z :

[A]max = 0.1171Qs

Uσyσz. (6.147)

EXAMPLE 6.17 ACCIDENTAL SPILL SCENARIO

Early morning, on a cloudy day, a tanker truck carrying benzene developed a leakspilling all of its 5000 gallons on the highway, forming a pool 50m × 50m. What isthe ground-level concentration 1 km downwind of the spill? The wind speed averaged2m/s and the ambient temperature was 25◦C.The spill was 1500m2 in area of pure benzene. Evaporation of benzene from a

pure phase is gas-phase mass transfer controlled with kg = 0.37m/s. P∗i = 0.125 atm.

Hence Qs = (0.37 cm/s) (3600 s/h)(0.125 atm)(1500 × 104 cm2)/[(8.205 × 10−5 cm3

atm/K/mol)(298K)] = 1.7 × 105 mol/h = 3.7 × 103 g/s.AtU = 2m/s, for an overcastcontinued

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morning the stability class is D. At 1000m, then σy = 70m and σz = 30m.Hence [A] (1000m, 0, 0) = (3700 g/s)/(3.14)(2m/s)(70m)(30m) = 0.280 g/m3 =280mg/m3. This is higher than the TLV of benzene (32mg/m3).

6.3.2 AIR POLLUTION CONTROL

There are two kinds of industrial air pollutants: particulates and vapors (gases). Thecontrol of these two classes of air pollutants is dependent on their mechanism of cap-ture in specific reactors. Processes that removeparticulates rely onphysical forces, thatis, gravity, impaction, electrical forces, or diffusion to bring about separation. Vaporsand gases are separated using chemical processes such as adsorption, absorption, andthermal processes. In the selection and design of any method, concepts from reactionkinetics and mass transfer play important roles. We will describe selected methodswith a view to illustrating the applications of kinetics and mass transfer theories.

Gases and vapors in polluted air streams are treated by chemical methods such asabsorption into a liquid stream (water), adsorption onto a solid adsorbent (activatedcarbon, silica, alumina), or by thermal means (incineration, combustion). We shalldiscuss each of these with the aim of illustrating the chemical kinetics and masstransfer aspects.

6.3.2.1 Adsorption

Adsorption is a surface phenomenon as we discussed in Chapter 3. Adsorption onsolids is an effective technology for pollutant removal from water or air. The designof both water treatment and air treatment is similar. Hence, our discussion in thisinstance for air pollution is the same as for water pollution. Adsorption from bothair and water is conducted in large columns packed with powdered sorbent such asactivated carbon. Feed stream containing the pollutant is brought into contact witha fresh bed of carbon. As the pollutant moves through the bed, it gets adsorbed andthe effluent air stream is free of the pollutant. Once the bed is fully saturated, thepollutant breaks through and the effluent stream has the same concentration as theinfluent feed. The progression of the bed saturation is shown in Figure 6.33. At anyinstant in the bed, there are three distinctly discernible regions. At the point wherethe pollutant stream enters, the bed is quickly saturated. This is called the saturatedzone. This zone is no longer effective in adsorption.Ahead is a zone where adsorptionis most active. This small band is called the active (adsorption) zone. Farther fromthis zone is the unused bed where there is no pollutant. Provided the feed rate isconstant, the adsorption zone slowly moves through the bed until it reaches the exit,at which stage breakthrough of the pollutant is attained. The rate of movement of theadsorption zone gives us information on the breakthrough time.

If thewave front is ideal, it will be sharp as shown in Figure 6.34. However, becauseof axial dispersion (nonideal flow), the wave front will broaden (Figure 6.34). Thevelocity of wave front advance can be obtained by determining the capacity of the bedfor a given feed rate GF at a concentration [A]F in the feed stream. A mass balance

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[A]=0 [A]=0 [A]=[A]F

Activezone

Unusedzone

Usedzone

GF; [A]F

t2t1

tB

Break point

Time or volume of gas treated

[A]F

[A]

FIGURE 6.33 Movement of the active (adsorption) zone through a fixed bed of activatedcarbon. Breakthrough is said to occur when the adsorption zone has reached the exit of thecarbon bed.

over the entire column gives the following:

mass flow rate of pollutant into the adsorption zone = rate of adsorption.

GF[A]F = MF

ρg[A]F = ρadsACVadsW0, (6.148)

where MF is the mass flow rate, ρg is the density of air, ρads is the density of theadsorbent, Vads is the adsorption zone velocity, Ac is the cross-sectional area, andW0

δ

GF[A] = 0

GF[A]F

VadsVg Vg

Zs

[A]

[A] + d[A]

dxX2X1

FIGURE 6.34 Analysis of an adsorption wave through a bed of activated carbon.

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is the mass of pollutant per unit bed weight. The general Freundlich isotherm (seeChapter 3) in the following form is found to best fit the data over a wide range ofconcentrations for activated carbon:

W0 =( [A]F

KFreun

)1/n

. (6.149)

Hence

Vads = Vg

ρadsK1/n

Freun[A](1−(1/n))F , (6.150)

where Vg = MF/ρgAc is the superficial gas velocity.Consider a differential volume of the bed Acdx.Amaterial balance on the pollutant

gives the following:

input by flow = output by flow

+ mass gained on the bed by transfer from gas to solid.

GF[A] = GF([A] + d[A]) + KmtAC([A] − [A]eq) dx, (6.151)

where [A]eq is the equilibrium value of air concentration that would correspond to theactual adsorbed concentration W in the differential volume. Kmt is the overall masstransfer coefficient (time−1) from gas to solid. Rearranging, we obtain

−Vgd[A] = Kmt([A] − KFreunWn) dx. (6.152)

Rearranging and integrating using the boundary conditions that [A] = [A]F at x = 0and [A] = 0 at x = δ,

δ∫

0

dx = −(Vg

Kmt

) 0∫

[A]F

d[A]([A] − KFreunWn)

. (6.153)

An overall mass balance over the differential volume gives [A] GF = AcWρadsVads.Utilizing the expression for vad and rearranging, we obtain KFreunWn = [A]n[A]1−nF .Hence,

δ = −(Vg

Kmt

) 0∫

[A]F

d[A]([A] − [A]n[A]1/nF

) . (6.154)

Note that as Kmt → ∞, δ→ 0 and the process becomes equilibrium controlled.Utilizing the dimensionless variable η = [A]/[A]F and Kmtδ/Vg = St, the Stantonnumber, we can write

St =1∫

0

dη

η− ηn . (6.155)

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Since the integral is undefined at η = 0 and 1, we take the limits as 0.01 (leadingedge) and 0.99 (trailing edge) so that [A] approaches 1% of the limiting value at bothlimits. We then have the following definite integral:

St = 4.505 +(

1

1 − n

)ln

[1 − (0.01)n−1

1 − (0.99)n−1

]. (6.156)

If we assume that the mass in the adsorption zone is much smaller than the mass inthe saturated zone, the breakthrough time can be obtained:

tB = HB − δVads

,

where HB is the total height of the adsorbent bed.

EXAMPLE 6.18 BREAKTHROUGH TIME FORAN ACTIVATED CARBON ADSORBER

Obtain the breakthrough time for a pollutant in a carbon bed at an air mass velocity of2.0 kg/s at 1.1 bar and 30◦C with an influent pollutant concentration of 0.008 kg/m3.The Freundlich isotherm parameters for the pollutant are KF = 500 kg/m3 and n = 2.The mass transfer coefficient is 50 s−1. The bed length is 2m with a cross-sectionalarea of 6m2 and a bulk density of 500 kg/m3.

Gasdensity,ρg = P/RT = (1.1)(29)/(303)(0.083) = 1.3 kg/m3.Vads = 2.0(500)0.5

(0.008)0.5/(1.3)(500)(6) = 1.0 × 10−3 m/s. Hence

δ = 2.0

(50)(6)(1.3)

(4.595 + 1

1.0ln

1 − (0.01)1.0

1 − (0.99)1.0

)= 0.047m.

Breakthrough time, tB = (2.0 − 0.047)/1.0 × 10−3 = 1953 s = 0.54 h.

6.3.2.2 Thermal Destruction

Organic compounds that form a large fraction of air pollutants in industrial effluentscan be oxidized in incinerators at high temperatures. These devices are also calledthermal oxidizers or afterburners. Both direct thermal oxidation and catalytic oxida-tion are practiced in the industry.A schematic of a direct thermal oxidizer for gases isshown in Figure 6.35. Thermal incineration is also used in destroying organic com-pounds in a liquid stream or sludge.A rotary kiln incinerator is used for the latter anda schematic is shown in Figure 6.36.

Cooper and Alley (1994) present the theory and design of a typical incinerator.Combustion is the chemical process of rapid reaction of oxygen with chemical com-pounds resulting in heat (Chapter 2). Most fuels are made of C and H, but may alsoinclude other elements (S, P, N, andCl).Although the exactmechanism of combustion

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Polluted air

ExhaustFuel

Combustion air

FIGURE 6.35 Schematic of a thermal incinerator (afterburner).

can be very complex, the basic overall process can be represented by the followingtwo reactions:

CnHm +(n2

+ m

4

)O2

k1−→ nCO + m

2H2O,

nCO + n

2O2

k2−→ nCO2,(6.157)

Rotary kiln

Wastedrums

Solid sludge

Ash to landfill

Air to pollutioncontrol device

Secondarycombustion

chamber

Waste(pumpable)

Waste(pumpable)

FIGURE 6.36 Schematic of a rotary kiln incinerator. (Reproduced from Buonicore, A.T.and Davis, W.T. (eds). 1992. Air Pollution Engineering Manual, p. 278. New York, NY: VanNostrand Reinhold.)

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where n and m are the number of C and H atoms in the parent hydrocarbon. Therates of reaction are rHC = k1[HC][O2], rCO = k2[CO][O2] − nk1[HC][O2], andrCO2 = k2[CO][O2]. Combustion is frequently carried out in the presence of excessair. The oxygen mole fraction is usually in the range 0.1–0.15 for a HC mole fractionof 0.001. Hence, rHC = k′

1[HC], rCO = k′2[CO] − nk′

1[HC], and rCO2 = k′2[CO]. In

Chapter 5, we saw that the above reaction rates reflect the following series reaction:

HCk′1−→ CO

k′2−→ CO2.

It is to be expected that since reaction rates are sensitive to temperature, so is theefficiency of an incinerator. Since reactions in thermal oxidizers occur at varying rates,sufficient time shouldbegiven for the reaction to go to completion.Mixing in the incin-erator should be sufficient to bring the reactants together. Thus, three important timefactors are to be considered in the designof an incinerator.These are the residence time,τres = VR/G = Zs/Vg, chemical reaction time, τrxn = 1/k, and mixing time, τmix =Z2

s /De. VR is the total reactor volume (m3), G is the volumetric waste gas flow rateat afterburner temperature (m3/s), Vg is the gas velocity (m/s), Zs is the reaction zonelength (m), andDe is the turbulent diffusivity (m2/s). The relative magnitudes of thesetimes are represented in terms of two dimensionless numbers: Peclet number, Pe =ZsVg/De, and Damköhler number, Da = Zsk/Vg. If Pe is large and Da is small, thenmixing is rate controlling in the thermal oxidizer, whereas for small Pe and large Da,chemical kinetics controls the rate of oxidation. At most temperatures used in after-burners, a reasonably moderate Vg is maintained so that mixing is not rate limiting.

The rate constant k for a reaction is related to its activation energy (Chapter 5),k = A exp(−Ea/RT). For a variety of hydrocarbons, values of A and Ea are available(Table 6.7), and hence k at any given temperature can be obtained.

TABLE 6.7Thermal Oxidation Reaction Rate Parameters

Compound A(s−1) Ea(kJ/mol)

Acrolein 3.3 × 1010 150Benzene 7.4 × 1021 4011-Butene 3.7 × 1014 243Chlorobenzene 1.3 × 1017 321Ethane 5.6 × 1014 266Ethanol 5.4 × 1011 201Hexane 6.0 × 108 143Methane 1.7 × 1011 218Methyl chloride 7.3 × 108 171Natural gas 1.6 × 1012 206Propane 5.2 × 1019 356Toluene 2.3 × 1013 236Vinyl chloride 3.6 × 1014 265

Source: From Buonicore, A.T. and Davis, W.T. (eds). 1992.Air Pollution Engineering Manual. New York, NY:Van Nostrand Reinhold.

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The reaction zone in the incinerator can be modeled as a plug-flow reactor. Theoverall destruction efficiency is given by

η = 1 − [A]exit

[A]feed= 1 − e−kτres . (6.158)

EXAMPLE 6.19 INCINERATOR FOR AIR POLLUTION CONTROL

At what air velocity should a toluene/air mixture be introduced in a 20-ft incinerator toobtain 99.99% efficiency at a temperature of 1300◦F?T = 1300◦F = (5/9)(1300) + 255.4 = 977K.FromTable 6.11,A = 2.3 × 1013 s−1

and Ea = 236 kJ/mol. Hence k = 2.3 × 1013 exp[−236/(8.31 × 10−3)(977)] =5.3 s−1. Hence E = 1 − exp(−kτres) = 0.9999, kτres = 9.21, τres = 1.74 s, and sinceτres = Zs/Vg, we have Vg = 11.5 ft/s.

6.3.3 ATMOSPHERIC PROCESSES

In this section, we shall analyze gas- and liquid-phase chemical reactions in relationto various atmospheric processes. These applications involve the use of the principlesof chemical kinetics and transport that we discussed in Chapter 5 and also in previoussections of this chapter.

6.3.3.1 Reactions in Aqueous Droplets

Acid rain is a problem in some industrialized nations (Hutterman, 1994). It has beenparticularly severe in parts of Western Europe and the northeastern United States.Similar issues have become prominent in several developing nations (e.g., India,China). In India, acid rain has defaced one of the cherished monuments of the world,the Taj Mahal. Its exquisite marble sculpture is eroding due to uncontrolled SO2emissions from local industries. In Germany, the beautiful Black Forest is severelyaffected by acid rain. In the United States, the forests of the northeast are known tobe affected by acid precipitation involving pollutants from the region’s industrial belt(Bricker andRice, 1993).Table 6.8 lists specific pollutants and average concentrationsin the northeastern United States. pH values as low as 4 with high concentrations ofSO2−

4 and NO−3 have been observed in parts of the northeastern United States.

Oxides of sulfur and nitrogen released to the atmosphere from fossil fuel com-bustion form strong acids by combining with the atmospheric moisture. These acidsinclude H2SO4 and HNO3. They react with strong bases (mainly NH3 and CaCO3)

and also associate with atmospheric dust (aerosols). Below cloud scavenging by rainand fog wash these particles to the surface with a low pH. This is called acid rain.

In the absence of strong acids in the atmosphere, the pH of rainwater is set by thedissolution of CO2 to form carbonic acid (H2CO3). The equilibrium is

CO2 + H2O � H2CO3 −→ H+ + HCO−3 . (6.159)

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TABLE 6.8Typical Constituents of Rainwater in the Northeastern United States

Rainwater Concentration Collected Average Concentration in theIon in Ithaca, NY, on July 11, 1975 (μM) Northeast United States (μM)

Sulfate 57 28 ± 4Nitrate 44 26 ± 5Ammonium 29 16 ± 5pH 3.84 4.14 ± 0.07

Source: From Park, D.H. 1980. Science 208, 1143–1145; Galloway, J.N., Likens, G.E., andEdgerton, E.S. 1976. Science, 194, 722–724.

At a constant partial pressure of carbon dioxide, PCO2 = 350 ppm = 10−3.5 atm,we have [H2CO3] = (1/K ′′

aw) PCO2 = (10−1.5)(10−3.5) = 10−5 M. Hence [H+] =Kc1 [H2CO3] = (10−6.4)(10−5) = 10−11.4 M. Thus, the pH of water is 5.7, which is1.3 units less than neutral. Natural rainwater is therefore slightly acidic.When strongacids are present that lower the pH to below 4.3, the above reaction is driven to theleft and carbonic acid makes no contribution to acidity. The components of strongacids are derived from SO2−

4 and NO−3 .

Gaseous SO2 in the atmosphere dissolves in atmospheric moisture to form sulfu-ric acid and contributes to sulfate in atmospheric particles. SO2 arises mainly fromanthropogenic emissions (burning of fossil fuels). The wet deposition of sulfate byrainfall occurs near industrial locations that emit SO2 to the air. The emission oftotal S to the atmosphere from natural sources is estimated to be ∼8.4 × 1013 g/y, ofwhich∼1.5 × 1013 g/y is attributed to gaseous SO2. Episodic events such as volcaniceruptions also contribute to the global emission of S compounds to the atmosphere;they disrupt the quasi-steady-state conditions that exist in the atmosphere.

Let us estimate what sources contribute most to H+ in the atmosphere. If only CO2is the contributing factor toward acidity, the pH of rainfall on earth will be 5.7. If theaverage annual rainfall is 70 cm/y and the surface area of the earth is 7 × 1018 cm2,this gives a total deposition of [H+] = (10−5.7)(10−3)(70)(7 × 1018) = 1.2 ×1012 mol/y. The total acidity from all natural sources is ≈8 × 1012 mol/y. In com-parison, the anthropogenic sources due to fossil fuel burning, automobile, and otherindustrial sources contribute NOx and SO2, which lead to ∼7.4 × 1012 mol of [H+]per year. It is, therefore, clear that human activities are contributing to the acidity ofour atmosphere. Over geologic time our atmosphere has been acidic, but has increasedonly slightly in acidity. Natural alkalinity resulting from NH3 tends to neutralize theacidity to some extent.The totalNH3 emission is≈3 × 1012 mol/y, and hence a reduc-tion of ≈3 × 1012 mol of [H+] per year can be attributed to reaction with NH3. Thisstill leaves ≈12 × 1012 mol [H+] that are being annually added to the atmosphere.

Gaseous SO2 is photochemically activated in the atmosphere and subsequentlyoxidized to SO3, which reacts further with water to form H2SO4:

SO2hv−−−−→ SO∗

2,

SO∗2 + O2 −→ SO3 + O,

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SO2 + O −→ SO3, (6.160)

SO3 + H2O −→ H2SO4.

There are two absorption bands for SO2: one is a weak band at 384 nm, whichgives rise to the excited triplet state of SO2, and the other is a strong absorp-tion at 294 nm, which gives rise to a higher energy excited singlet state. Thesereactions are slow and do not account for the observed rates of SO2 oxidationin the atmosphere (≈0.01–0.05 h−1). The oxidative process in the gas phase isdriven by the abundantly available highly reactive free radical species OH• in theatmosphere.

SO2 + OH• −→ . . . . . . . . . . . . . . . . . . −→ H2SO4 (6.161)

Oxidation of SO2 to sulfate in atmospheric moisture is catalyzed by species suchas ozone, hydrogen peroxide, metal ions (Fe(III), Mn(III)), and nitrogen (N(III) andNO2). The rate constants and rate expressions for several of these reactions have beenstudied by Hoffmann and co-workers and are given in Table 6.9. For pH values <4,the predominant oxidizer is H2O2, whereas for pH = 5, O3 is an order of magnitudemore powerful as an oxidizer. Only at pH ≥ 5 is the catalyzed oxidation by Fe andMn of any significance.

The following example is to illustrate how the change in pH of an open systemcan be computed as the sulfate content in the atmosphere increases. The problemcan be extended to realistic atmospheric conditions, and described by Seinfeld andPandis (2006).

TABLE 6.9Oxidation of SO2 to Sulfate by Different Speciesin the Atmosphere

Oxidizer r = d [S (IV)]

dtOzone

(k0[SO2 · H2O

]+ k1[HSO−

3

]+ k2

[SO2−

3

]) [O3]w

Hydrogen peroxide

⎛⎝ k

[H+] [HSO−

3

]

1 + K[H+]

⎞⎠ [H2O2

]w

Mn(II) k′2 [Mn(II)] [HSO−3 ]

Fe(II) k = [Fe(III)][SO2−3 ]

Note: k0 = 2.4 × 104 L/mol s, k1 = 3.7 × 105 L/mol s, k2 = 1.5 × 109 L/mol s,k = 7.5 × 107 L/mol s, K = 13 L/mol, k′2 = 3.4 × 103 L/mol s, k = 1.2 ×106 L/mol s.

Source: From Hoffmann, M.R. and Calvert, J.G. 1985. Chemical Transforma-tion Modules for Eulerian Acid Deposition Models. Volume II. TheAqueous-phase Chemistry. Boulder, CO: National Center for Atmo-spheric Research.

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EXAMPLE 6.20 EFFECT OF SO2 OXIDATION ON THE pH OF AN OPEN SYSTEM(AQUEOUS DROPLET) THAT CONTAINS HNO3, NH3, H2O2,AND O3

Consider an aqueous droplet with HNO3, NH3, H2O2, O3, and SO2 as the mainconstituents. The following initial conditions (t = 0) were chosen by Seinfeld andPandis (2005): [S(IV)]total = 5 ppb, [HNO3]total = 1 ppb, [NH3]total = 5 ppb, [O3]total= 5 ppb, [H2O2]total = 1 ppb, [S(VI)]t=0 = 0, and water content = 10−6.For an open system, we can assume that the partial pressures of the different species

remain constant. As S(IV) gets oxidized to S(VI), the new species of interest will com-prise SO2−

4 and HSO−4 .We shall indicate [S(VI)] = [SO2−

4 ] + [HSO−4 ] + [H2SO4]aq.

The electroneutrality equation is

[H+]+

[NH+

4

]=[OH−]+

[HSO−

3

]+2[SO2−

3

]+ 2

[SO2−

4

]+[HSO−

4

]+[NO−

3

].

6.2

6.0

5.8

5.6

5.4

5.2

5.0

4.8

4.6

0 10 20 30

Time, min

Closed system

Open system

pH

40 50 60

FIGURE 6.37 pH as a function of time for both open and closed systems. The condi-tions for the simulation are [S(IV)]total = 5 ppb, [NH3]total = 5 ppb, [HNO3]total =1 ppb, [O3]total = 5 ppb, [H2O2]total = 1 ppb, θ� = 10−6, pHo = 6.17. (Reprintedfrom Seinfeld, J.H. and Pandis, S.N. 1998. Atmospheric Chemistry and Physics,p. 390. NewYork: Wiley. With permission.)

continued

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From the equation for [S(VI)], we can obtain (see Seinfeld, 1986)

[SO2−

4

]≈ [S (VI)]

1 + ([H+] /Ks4) and

[HSO−

4

]≈ [S (VI)]

1 + (Ks4/[H+])

with Ks4 = 0.012mol/L.The rates of conversion to S(VI) by H2O2 and O3 are given in Table 6.9. Hence

[S (VI)]t=1min = [S (VI)]t=0 +(d [S (VI)]

dt

)

t−ΔtΔt.

We then obtain a new value for [HSO−4 ] and [SO2−

4 ]. Substituting this in the elec-troneutrality equation, we get the new [H+], and hence the new pH. Continuing in thismanner to obtain [S(VI)] at different times, we can follow the changes in pH as more ofS(IV) is converted to S(VI) by H2O2 and O3. This is provided by Seinfeld and Pandis(2006) (Figure 6.37). In 60min, the pH decreased to 5.3 from its initial value of 6.1.The assumption of an open system is questionable. In a given cloud volume, it is

unlikely that the partial pressures of the different species will remain constant overthe duration of the reaction. In such a case, one should consider the changes in par-tial pressures of NH3 and HNO3. A detailed account of this aspect of atmosphericreaction modeling is given by Seinfeld and Pandis (1998) and is beyond the scope ofthis book.

Although SO2 oxidation in the gas phase to form H2SO4 (aerosol) is a linearfunction of OH• concentration, it indirectly depends on the concentration of NOx

in the atmosphere as well. As seen above, the oxidation in the aqueous dropletalso depends on the concentration of other oxidants such as H2O2 and O3. It canbe shown that if the calculations in the above example are repeated with a differ-ent concentration of HNO3, the level of NO−

3 in the aqueous phase will influence[S(VI)] in the droplet. The conversion of NOx to HNO3 within the droplet is primar-ily a function of the photochemical reactions in the gas phase, and hence respondsdirectly to changes in NOx levels in the atmosphere. The atmospheric moisture con-tent is obviously an important factor in deciding the fraction of S(IV) converted toS(VI). With increasing moisture content, the fraction oxidized also increases. Thusa cloud with a large moisture content will have high acidity due to a large concen-tration of S(VI). In conclusion, acid rain, which is mostly a regional problem, isa complex process that depends on the prevailing local conditions, particularly thelevels of other species present in the atmosphere. Models exist to predict the acid-ity to be expected in precipitation if sources, and their strengths are identified withconfidence.

In our analysis thus far, we have only considered reactions within aqueous dropletsin the atmosphere. However, this may not always be the controlling resistance forconversion of S(IV) to S(VI). The evolution of acidity in atmospheric precipitation(rain or fog) depends on factors such as diffusion of species toward the droplet fromthe air and reaction within the droplet.

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Consider an aqueous droplet falling through the atmosphere. There are fiveprocesses that must be considered. These are (see Seinfeld, 1986; Schwartz andFreiberg, 1981):

(i) Diffusion of solute in the gas phase, characterized by the diffusionconstant, Dg.

(ii) Mass transfer across the air–water interface of the droplet characterizedby the mass transfer coefficient, kmt, and the progress toward equilibriumat the interface. The latter is characterized by the air–water equilibriumconstant, Kaw.

(iii) For a species such as SO2, its dissolution in the aqueous phase is imme-diately followed by a dissociation reaction. The dissociation is as follows:

[SO2 · H2O]aqk1�K2

H+ + HSO−3 [ ], for which the equilibrium concentration

of [SO2 · H2O]aq can be obtained (see Section 5) from

[SO2 · H2O] − [SO2 · H2O]eq

[SO2 · H2O] − [SO2 · H2O]eq= e−αt ,

where α = k1 + k−1([H+]eq + [HSO−3 ]eq).

(iv) If the droplet is not well mixed, then diffusion within the aqueous phasewill play a role. This is characterized by the diffusion constant, Dw.

(v) The final item to be considered is chemical reaction within the droplet.This is what we discussed earlier and is characterized by the reaction rateconstant,

k = − 1

[S(VI)]

d [S(VI)]

dt.

Seinfeld (1986) has analyzed each of these steps in detail. He derived an equationfor the characteristic time (τ) in each case. Table 6.10 summarizes the expressions forτ. The terms, their definitions, and typical values are also given. If the characteristictime for any one step is larger than that for the chemical reaction within the droplet,then equilibrium will not be achieved in that step and the observed rate at whichthe products are formed will be smaller than the reaction rate. This has interestingconsequences as far as acid rain is concerned.

From Table 6.21, one observes that for S(IV) oxidation in a 10-μm droplet, τr islarger than all other τ values and hence the conversion is reaction limited. In general,for all practical purposes τg and τd are small. τi, τa, or τr is then rate limiting. Unlessthe droplets are much smaller, τa is large compared with τi and τr.τr is both pH andspecies dependent.

EXAMPLE 6.21 SCAVENGING OF CONSERVATIVE POLLUTANTS BY RAIN OR FOG

Consider rain or fog drops falling through the atmosphere. Each drop reaches terminalvelocity determined by its size and scavenges pollutants as they drop to the surface.

continued

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This is called “below-cloud scavenging.” The rate of increase in concentration in thedroplet is given by the mass transfer to the drop (Seinfeld and Pandis, 2006):

πD3

P6

d [A]aq

dt= πD2

PKmt [A]g ,

d [A]aq

dt= 6

DPKmt [A]g ,

where we assume that the pollutant is irreversibly absorbed into the drop and thepollutant is nonreactive. For a falling drop at its terminal velocity, UT, we can write

d [A]aq

dt= UT

d [A]aq

dz

since UT = dz/dt. Thus,d [A]aq

dz= 6

DP

Kmt

UT[A]g.

If we assume that [A]g is constant (homogeneous atmosphere), we can obtain thefollowing equation upon integration:

[A]aq = [A]0aq + 6

DP

Kmt

UT[A]g z.

The mass scavenged (W) by a droplet of volumeπD3P/6 will be given by ([A]aq–[A]0aq)

times the droplet volume

W = πD2PKmt

UT[A]g z.

If the rainfall intensity is RI (mm/h), the number of drops falling per unit area per hourwill be

Nd

(m2/h

)= 0.006RI

Kmt [A]gUTDP

z.

Hence, the rate of removal of the pollutant by rain (fog) drops from below-cloud volumeof V is given by

−V d [A]gdt

= NdWAc.

Noting that V/Ac = z, we have

−d [A]gdt

= 0.006RIKmt

UTDP[A]g,

which upon integration gives

[A]g = [A]0g e−λt ,

whereλ = 0.006(RIKmt/UTDP) is called the scavenging ratio. The amount of pollutantscavenged from the atmosphere by rain or fog depends on the drop diameter DP. Notethat λ is inversely proportional to DP; the smaller the drop diameter, the better thescavenging efficiency. Small drops also have low UT and hence have a large residencetime in the atmosphere, which increases the mass transfer to the drop.

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TABLE 6.10Characteristic Times for S(IV) Oxidation in an Aqueous Droplet

Process Expression for τ Typical Value

Diffusion in the gas phase τg = R2/4Dg 2.5 × 10−6 sEquilibrium at the air–water interface τi = (2πMRT/α2)DwK2

aw 0.15 sDissociation in the aqueous droplet τd = [k1 + k−1([H+] + [HSO−

3 ])]−1 2 × 107 sDiffusion in the aqueous phase τa = R2/4Dw 0.025 sReaction within the aqueous droplet τr = −[S(IV)]/{d[S(IV)]/dt} ≈1.5 s

Definitions of characteristic times:τg is the time to achieve steady-state concentration in the gas phase around the droplet;τi the time to achieve local equilibrium at the interface;τd the time to achieve equilibrium for the dissociation reaction;τa the time to achieve uniform steady-state concentration in the droplet; andτr the time to convet 1/e of the reactants to products.

Note: R = 10μm, Dg = 0.1 cm2/s, k1 = 3.4 × 106 s−1, k−1 = 2 × 108 mol/L s, Dw = 10−5 cm2/s,MK2

aw/α2 = 98, pH = 4, PSO2 = 1 ppbv, PH2O2 = 1 ppbv.Source: From Seinfeld, J.H. 1986. Atmospheric Chemistry and Physics, p. 390, NewYork: Wiley.

6.3.3.2 Global Warming and Greenhouse Effect

Our atmosphere was, at one time, oxygen rich. The highly oxidative atmospherecould not sustain early life forms. Gradually, the atmosphere became oxygen depleted(nitrogen rich) and evolved into the one that we have today. The present atmosphere,conducive to life, is believed to be sustained by a symbiosis between the biota andthe various atmospheric processes. This is the central theme of the so-called Gaiahypothesis (Lovelock, 1979). In short, our planet is a gigantic experiment, whetherby design or by chance.

The composition of air is 78% (v/v) nitrogen and 20% (v/v) oxygen. All othergases together constitute the remaining 2% (v/v) of our atmosphere; these are calledtrace gases. Of the trace gases, a few are of special relevance. Rare gases such asargon do not vary in concentration to any measurable extent. Other gases such asCO2, CO, NOx , CH4, and CFCs are variable. Even though these species are at traceconcentrations, they exert profound effects on the environment.

The atmosphere does not absorb incoming energy from the sun in the visible regionof the spectrum. The stratosphere absorbs only a part of the UV radiation. The cloudand the earth’s surface reflect a large fraction of the radiation. A portion of the sun’senergy is used to heat the surface of the earth, which reradiates heat in the infraredregion of the spectrum (4–100μm).A portion of the reradiated energy (13–100μm) isabsorbed in the atmosphere, mostly by water and CO2.Water drops in the atmospherealso absorb energy in the region between 4 and 7μm. It is in the window of 7–13μmthat heat escapes freely into space (Figure 6.38).

The atmosphere acts as a greenhouse trapping moderate amounts of heat energythat is reradiated from the earth’s surface, and thus maintaining a comfortable aver-age temperature that sustains life on earth. The atmospheric CO2 and H2O are mainly

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Earth

Heat reflectedor radiatedfrom surface

Re-radiatedTo space

Solarenergy

Attenuated radiation

Radiated tosurface

Infra red radiationemitted from earth

Atmosphere

CO2, CO4, H2O, N2O

Stratosphere

Reflected radiation

Re-radiated to space

FIGURE 6.38 Solar energy absorbed and reflected in the earth’s atmosphere.

responsible for this. During the pre-industrial era, the CO2 concentration in the atmo-sphere remained fairly constant. In 1850 the average concentration was ≈270 ppmv.Since then the concentration has steadily increased. In 1957 it was 315 ppmv andin 1992 it was 356 ppmv. Figure 6.39 shows the concentration of CO2 observed atthe Mauna Lao observatory in Hawaii. The oscillations in CO2 concentration reflectthe seasonal cycles of photosynthesis and respiration by the biota in the northernhemisphere. Note the steady increase in CO2. This increase is attributed to the burn-ing of fossil fuels (coal and oil), and is therefore strictly anthropogenic in origin(Figure 6.40). Human intervention through deforestation of tropical forests also playsa role by reducing photosynthesis that fixes CO2 from the atmosphere. The increasein CO2 emissions worldwide increases the capacity of the earth’s atmosphere to trapthe outgoing radiation. The other trace gases such as CH4, CO, NOx , and CFCs alsoperform a similar function, since their concentrations have also increased steadilywith time (see Table 6.11). Presently, CO2 contributes the largest toward the green-house effect. Figure 6.41 indicates the radiative forcing effects of various greenhousegases (GHG) as predicted by the most recent evaluation by the International Panel onClimate Change (IPCC, 2007).

The ability of CO2 to initiate the greenhouse effect has been contemplated as farback as 1896 by the Swedish chemist SvänteArrhenius (of theArrhenius equation forrate constant in Chapter 5). But only in the past decade has the prospect of a globalgreenhouse effect gained attention and been recognized to have serious repercussionswithin the lifetime of the present generation. Recently, however, some good newson this front has emerged. Globally the rate of input of CFCs into the atmospherehas slowed down. This can be traced to the worldwide ban on its production and theimplementation of the Montreal Protocol.

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8

380

(a)

(b)

360

CO2 m

ixin

g ra

tio [p

pm]

Glo

bal e

miss

ion

Gtc

yr–1

O2 (

per m

eg)

δ13 C

(CO

2 ) %

–100

–200

–300

–400

–8.1

–8.0

–7.9

–7.8

–7.7

–7.6

340

320

7

6

5

41970 1975 1980 1985 1990

Year1995 2000 2005

FIGURE 6.39 Monthly average CO2 concentration observed continuously at Mauna Lao,Hawaii. (Reprinted from Solomon, S., Qin, D., Chen, Z., Marquis, M., Averyt, K.B., Tignor,M., and.Miller,H.L. (Eds) IPCC2007:ClimateChange 2007:ThePhysical Basis. Contributionof working group 1 to the fourth assessment report of the Intergovernmental Panel on ClimateChange, Cambridge, United Kingdom, and NewYork: Cambridge University Press, 996 pp.)

Let us turn our attention toward some of the natural processes that removeGHG from the atmosphere. We will focus on CO2 in the following discussion. Themost significant pathway is the dissolution of CO2 in the ocean water and finalconversion to HCO−

3 and CO2−3 . This fixing of carbon is rapid in the ocean surface

(∼500m of the upper mixed layer), and hence can be considered to be an equilib-rium process. Weathering of rocks and minerals, dissolution of CaCO3 in oceans,and increased bacterial activity in some regions also contribute to the transfer of CO2from the atmosphere to the oceans.

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–4

–2

0

2

4

6

8

Def

ores

tatio

n

Foss

il fu

el b

urni

ng

Land

-bas

ed si

nk

Oce

anic

sink

Glo

bal C

O2

/ GtC

.yr –1

FIGURE 6.40 Percent contributions to CO2 emissions.

TABLE 6.11Greenhouse Gases in the Atmosphere

Parameter CO2 (ppmv) CH4 (ppbv) CFC-12 (pptv) N2O (ppbv)

Preindustrialrevolution(1750–1800)

280 715 0 270

2005 concentration 379 ± 0.65 1774 ± 1.8 538 ± 0.18 319 ± 0.12Rate of annualaccumulation

1.8 0.015 17 0.8

Atmospheric lifetime(years)

50–200 12±3 102 120

Source Fossil fuel burning,deforestation

Rice fields, cattle,landfills, fossil fuelproduction

Aerosolpropellants,refrigerants,foams

Fertilizers,biomass burning

Total emissions,million tons/year

5500 550 ≈1 25

Source: Adapted fromHoughton et al. 1990.Climate Change: The IPCC Scientific Assessment, CambridgeUniversity Press, New York; IPCC 2007. Climate Change: The Physical Chemical Basis. NewYork: Cambridge University Press, International Panel on Climate Change.

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CO2

CH4

HalocarbonsTropospheric

Black carbonon snow

(0.01)

Land use

Stratospheric(–0.05)

Long-livedgreenhouse gases

Ozone

N2O

Stratosphericwater vapor

Surface albedo

Direct effect

Cloud albedoeffect

TotalAerosol

Linear contrails

Solar irradiance

Nat

ural

proc

esse

sH

uman

activ

ities

Total nethuman activities

Radiative forcing (Watts per square metre)–2 –1 0 1 2

Radiative forcing of climate between 1750 and 2005Radiative forcing terms

FIGURE 6.41 Radiative forcing effects of various GHG gases. (Reprinted from Solomon,S., Qin, D., Chen, Z., Marquis, M., Averyt, K.B., Tignor, M., and. Miller, H.L. (Eds) IPCC2007:Climate Change 2007: The Physical Basis. Contribution of working group 1 to the fourthassessment report of the Intergovernmental Panel on Climate Change, Cambridge, UnitedKingdom, and NewYork: Cambridge University Press, 996 pp.)

To start this discussion, we shall first consider the total carbon in solution,[CO2]tot = [CO2]aq + [HCO−

3 ] + [CO2−3 ], and the Revelle buffer factor, RB, as

([CO2]tot/PCO2)(∂PCO2/∂[CO2]tot)[Alk], where [Alk] is the alkalinity of water.This is defined (see Chapter 4) as [Alk] = [OH−] − [H+] + [HCO−

3 ] + 2[CO2−3 ].

Since we have seen in Chapter 4 that [CO2]tot = (PCO2/K′′aw)(1 + (Ka1/[H+]) +

(Ka1Ka2/[H+]2)), we have

[Alk] = Kw[H+] − [H+]+ PCO2

K ′′aw

(Ka1[H+] + 2Ka1Ka2[

H+]2). (6.162)

Rewriting the equation for the Revelle buffer factor as

RB = [CO2]tot

PCO2

·(∂PCO2/∂

[H+])

[Alk](∂ [CO2]tot /∂

[H+])

[Alk](6.163)

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and performing some algebraic manipulations, one obtains (see Butler, 1982 fordetails) (

∂PCO2

∂[H+]

)

[Alk]≈ PCO2

[H+] (6.164)

and (∂[CO2]tot

∂[H+]

)

[Alk]≈ [CO2]aq + [CO2−

3 ][H+] . (6.165)

Hence,

RB � [CO2]tot

[CO2]aq + [CO2−3 ] . (6.166)

For typical seawater samples that have an average pH of 8, we can approxi-mate [Alk] ≈ [HCO−

3 ] = 10−2.7 and [CO2]tot ≈ [HCO−3 ] = 10−2.7.Hence, [CO2] =(

10−pH/Ka1) [

HCO2−3

]= 10−4.7,

[CO2−

3

]= [HCO−

3

] (Ka2/10−pH

) = 10−3.8, and

RB ⊕ 11, compared with an experimental value of ≈ 9.5 at 298K (Butler, 1982).The equation forRB gives the change in partial pressure of CO2 required to produce

a specified change in [CO2] in seawater if [Alk] is constant. Thus if RB ⊕ 10, Butler(1982) estimated that about a 10% change in atmospheric concentration is requiredto bring about a total change of 1% in seawater CO2 concentration. He also estimatedthat the pre-industrial era (1750–1800) must have had∼145mol CO2 per m2 of oceansurface area and the mixed layer of ocean surface water must have had∼900mol/m2.A 10% increase in atmospheric CO2 ∼15mol/m2) should correspond to only a 1%(∼9mol/m2) increase in CO2 in surface water. Extrapolating, we can conclude thatin about two and one-half years, the oceans can absorb about 50% of the increasedCO2 in the atmosphere. Thus, there is a significant lag in the CO2 absorption by theworld’s oceans. This feed-forward mechanism tends to increase the atmospheric CO2concentration.

Sophisticated and complex climate models have been developed to forecast theCO2 increase, and how it could affect the climate and crops in different regions ofthe world. We shall not discuss these topics here; the student is referred to the recentdocument by IPCC (2007) for further details.

As already stated, the largest exchange of carbon (as CO2 or carbonates and organicmolecules) occurs between the ocean and the atmosphere. The exchange between thebiota and the atmosphere is equally important (see Figure 6.42). The average time thata CO2 molecule remains free in the air is∼4 years before it is taken up by the biota orocean. However, the adjustment time, that is, the time taken by atmospheric CO2 levelto reach a new equilibrium, if either the source or sink is disturbed, is∼50–200 years.The net flux into or out of oceans depends both on the partial pressure of CO2 in theatmosphere and on the concentration of total carbon in surface waters.

One can construct a simple model to relate the changes in atmospheric CO2 toincreased global production of CO2. The following example is an illustration of sucha model.

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Atmosphere597 + 165

VegetarianSoil & Debritus

2300 + 101 –140

Intermediate& Deep Ocean37,100 + 100

Reservoir sizes in GtCFluxes and Rates in GtC yr1 Surface sediment

150

LandUseChange

Rivers

Landsink

1.6

0.4 0.8

70.6 70 22.2 20

8.4

Fozzil Fuels3700-244

1.6

5039

10190.2

2.60.2

0.2

17

0.2

119.6

RespirationGPP

Weathering

Weathering

120

Surface Ocean Marime Biota

3900 + 18

FIGURE 6.42 Global carbon reservoirs and fluxes. Numbers underlined indicate accumula-tion of CO2 due to human action. Units are gigatons of carbon for reservoir sizes and gigatonsof carbon per year for fluxes. (Reprinted from Solomon, S., Qin, D., Chen, Z., Marquis, M.,Averyt, K.B., Tignor, M., and. Miller, H.L. (Eds) IPCC 2007: Climate Change 2007: ThePhysical Basis. Contribution of working group 1 to the fourth assessment report of the Intergov-ernmental Panel on Climate Change,Cambridge, United Kingdom, and NewYork: CambridgeUniversity Press, 996 pp.)

EXAMPLE 6.22 A TWO-BOX MODEL FOR THE EFFECT OF INCREASEDATMOSPHERIC CO2 ON THE WORLD’S OCEANS

McIntyre (1978) proposed a model to relate the effects of increased CO2 emissions onthe world’s oceans. Since the ocean is in theoretical equilibrium with the atmosphere,dissolved CO2 is quickly converted to HCO−

3 . Organic materials that sediment in the

ocean carry HCO−3 downward. Fresh CO2 from the atmosphere replaces the lost CO2

in the surface ocean. Plants photosynthesize and remove CO2 from the air, and areprimarily responsible for keeping the oceans undersaturated with CO2. As discussedearlier, any change in CO2 concentration is quickly buffered in the ocean. However, ifthe rate of change in CO2 exceeds the rate of establishment of equilibrium, there willexist a disequilibrium between the ocean and the atmosphere. Both natural processes(photosynthesis) and anthropogenic (fossil fuel burning) will upset the equilibrium.How does the ocean then react to this change?

McIntyre (1978) suggested that the terrestrial biosphere is in equilibrium with theatmosphere, and the marine biosphere is in equilibrium with the surface ocean (up to adepth of∼500m the ocean is completely mixed in a short period of time). He also madethe assumption that alkalinity in the ocean is only due to carbonates, and that the input ofcarbonates from rivers is balanced exactly by its precipitation as CaCO3 in sediments.The sediment is, however, not in equilibrium with the atmosphere or the surface ocean.A two-box model such as represented in Figure 6.43 can then be envisaged.

continued

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Atmosphere

Ocean surface

Δ[CO2]tot

AddedCO2

ΔPCO2

FIGURE 6.43 A two-box model for the distribution of CO2 between the atmosphereand the surface ocean.

Utilizing the expression already derived, we obtain

(∂PCO2

∂[CO2

]tot

)

[Alk]= RB

(P0CO2

[CO2]0tot

),

where P0CO2and [CO2]

0tot are pre-industrial values. In other words,

ΔPCO2

P0CO2

= RB

(D[CO2]tot[CO2

]0tot

).

Thus, if atmospheric PCO2 increases by y%, [CO2]tot increases by y/RB%. For agiven alkalinity, we can obtain the following equation: [Alk] = [CO2

]tot (α1 + 2α2) +[

OH−]− [H+], and hence for a given [H+] we can obtain [CO2]tot . Since alkalinityin oceans is also caused by borate species, we can write the following general equation:[Alk] = [CO2

]tot (α1 + 2α2) + [OH−]− [H+]+ BTα

−B , with α1 and α2 as defined

earlier. Since [CO2]tot is known for any given [Alk], we can now obtain the changesin P with changes in [CO2]tot . This is shown in Figure 6.44. For a buffer factor RB of9.7 (at 15◦C), a 10% increase in PCO2 causes a 1% change in [CO2]tot . RB increaseswith increasing PCO2 . It can be observed that if PCO2 increased from the present valueof 330–600 ppmv, RB changes to 17.4. Thus a doubling of PCO2 from its present levelleads to a 5–6% change in [CO2]tot (Stumm and Morgan, 1996).

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77Pco2

10–3

8 × 10–4

6 × 10–4

4 × 10–4

2 × 10–4

79 81 83 pH

pH

CT

present

22 × 10–3M 23 × 10–3M

Ct = [H2CO+3] = [HCO–

3] = [CO23]

24 × 10–3M

FIGURE 6.44 Effect of oceanic PCO2 upon CT and pH of the ocean water.The calculations have been made for the following conditions: seawater at 15◦C,Ptotal = 1 atm, [Alk] = constant = 2.47 × 10−3 eq L−1, [B(OH)−4 ] + [H3BO3] =4.1 × 10−4 M, 1/Ha = 4.8 × 10−2 mol/L atm., Kc1 = 8.8 × 10−7, Kc2 = 5.6 ×10−10 and KH3BO3 = 1.6 × 10−9. (Reproduced from Stumm, W. and Morgan, J.J.1996. Aquatic Chemistry, 4th ed., p. 922. NewYork: Wiley. With permission.)

In drawing further conclusions from the above analysis, one should rememberthat although the surface ocean is likely in equilibrium with the atmosphere, thedeeper water will reach equilibrium only slowly. This is brought about by mixingdue to upwelling of cold water from deeper layers and subsiding warm water towardthe bottom. This thermal inertia (lag) will likely delay the overall readjustment toequilibrium for the earth.

Now that we have learned about the increased CO2 content in the atmosphere, thenext question is how this impacts the temperature of the atmosphere.Radiative forcing (ΔF) is a term that represents the amount of heating per m2

of the surface contributed by a GHG. This is related to the temperature changecontributed by the GHG in the form ΔT = λΔF, where λ is a climate sensitivityparameter (K/W/m2). The NRC (1983) report stated that theΔT associated with CO2fluctuations is given by

ΔT = η ln(PCO2

P0CO2

), (6.167)

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Multi-model averages and assessed ranges for surface warming

6.0 A2A1BB1Year 2000 constantConcentrations20th century

©IPCC 2007: WG1-AR4

5.0

4.0

3.0

Glo

bal s

urfa

ce w

arni

ng (°

C)

2.0

1.0

0.0

–1.0

1900 2000Year

2100

B1 A1T

A1B

A2

A1F

l

B2

FIGURE 6.45 Projected ΔT for our planet as per the IPCC 2007 Assessment. (Reprintedfrom Solomon, S., Qin, D., Chen, Z., Marquis, M., Averyt, K.B., Tignor, M., and. Miller, H.L.(Eds) IPCC 2007:Climate Change 2007: The Physical Basis. Contribution of working group 1to the fourth assessment report of the Intergovernmental Panel on Climate Change,Cambridge,United Kingdom, and NewYork: Cambridge University Press, 996 pp.)

where η is a constant, PCO2 is the current partial pressure of CO2, and P0CO2

is thereference partial pressure of CO2. Thus, the increase in surface temperature is a non-linear function of PCO2 . If we define a scaling factor, ΔT

∗, which is the temperatureincrease due to a doubling of P0

CO2, we obtain ΔT∗ = η ln 2. Therefore,

ΔT =(ΔT∗

ln 2

)ln

(PCO2

P0CO2

). (6.168)

Each of the GHG will contribute a certain ΔT and hence the cumulative changeis ΔToverall =∑i(ΔT)i. The increase in temperature of the surface can have catas-trophic consequences. The most recent IPCC assessment provides several scenariosfor the projected ΔT for our planet (Figure 6.45). A rise in surface temperature canlead to changes in ocean levels. Thus areas could be flooded and islands and coastalplains can disappear. Regions that are presently the food baskets of the world can behit with severe drought, and agricultural production will be curtailed in those regions.At the same time semi-arid regions of the world may become more conducive to agri-culture. Attendant possibilities of international conflicts exist. If we are to mitigatethese likely effects, international cooperation and strong leadership are necessary tocurtail the emissions of CO2 and other GHG.

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EXAMPLE 6.23 GLOBAL TEMPERATURE CHANGE WITH CO2 INCREASE IN THE

ATMOSPHERE

The pre-industrial era partial pressure of CO2 was 280 ppmv. If a doubling of CO2 inthe atmosphere changes the atmospheric temperature by 3◦C, what is the partial pres-sure of CO2 for an observed temperature change of 1◦C?ΔT = 1◦C,ΔT∗ = 3◦C, P = 280 ppmv. Hence

1

3= 1

0.693ln

(PCO2

280

),

which gives PCO2 = 352 ppmv.

6.3.3.3 Ozone in the Stratosphere and Troposphere

In this section, we will explore the chemical kinetics of atmospheric systems thatinvolve interactions between several species. An interesting example is the chemistryof ozone, in both the upper stratosphere and the lower atmosphere (troposphere).The troposphere extends up to 16 km and the stratosphere extends up to 50 km abovethe surface of the earth. The pressure in the stratosphere decreases exponentiallywith increasing altitude. This has interesting consequences for chemical reactionsin the stratosphere. Reduced pressure leads to reduced reaction rates. The reducedtemperature in the upper troposphere impacts reactions with high activation energies;the rates of these reactions are lowered. Thus the lower troposphere is a far morereactive region than the upper troposphere.

Themajor constituents of the atmosphere are nitrogen and oxygen. Generally theseare not the dominant reactive species. Trace species such as ozone, hydroxyl radical,CO2, SO2, CH4, NOx , and CFCs are the ones that significantly impact atmosphericchemistry. We have already seen that OH• is the most reactive species and appropri-ately termed the “atmospheric detergent.” It is typically present at mixing ratios of1 to 4 × 10−14, despite the fact that the atmosphere contains 21% molecular oxygenby volume. The GHG (CO2, CH4, and CFCs) were discussed in the previous section.SO2 and NOx participate in the acidity of the atmosphere and were also discussed inan earlier section.

From the viewpoint of exploring different reaction kinetics in the stratosphereand troposphere, ozone is a good choice. The total mass of ozone in dry air is ∼3.3 ×1012 kg. It has a maximum concentration of ∼500μg/m3 in the upper stratosphere atabout 30 km height. In the troposphere it varies between 60 and 100μg/m3 in cleanair. The significance of ozone in the stratosphere lies in its ability to shield the earthfrom the harmful effects of the sun’s UV radiation. However, in the lower troposphereit is an undesirable species since it leads to the formation of smog. The U.S. federalregulations on air quality stipulate that ozone concentration greater than 235μg/m3

is harmful and can cause breathing problems and eye irritations. Several cities in theUnited States have low air quality due to nonattainment of ozone levels.

Let us consider the formation and dissipation of ozone in the upper stratosphere.At ∼30 km height, ozone forms via dissociation of molecular oxygen into O atoms

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by solar radiation (λ < 240 nm). In the presence of a third body, Z, the oxygen atomsreact with O2 to produce ozone.

O2hν−→J1

2O,

2(O + O2 + Zk2−→ O3 + Z∗). (6.169)

The overall reaction is 3O2 → 2O3. The species Z∗ in the above case representsvibrationally excitedO2 or N2 molecule. UV radiation also splits upO3 intoO2 andO.

O3hν−→J3

O2 + O. (6.170)

The above reaction is the predominant mechanism by which ozone performs thefunction of shielding the earth from harmful UV light.

Ozone also reacts with O atom to give rise to O2 as follows:

O3 + Ok4−→ 2O2. (6.171)

The above set of four reactions constitutes what is called the Chapman mechanismfor ozone formation and dissociation in the upper stratosphere.

Let us now consider the rates of formation and destruction of ozone in thestratosphere. The monoatomic O species obeys the following rate expression:

d[O]dt

= 2J1[O2] + J3[O3] − [O](k2[O2][Z] + k4[O3]), (6.172)

where Ji is the photolysis rate constant given by∑

i φi(λ)Ii(λ)σi(λ)Δλ as describedin Chapter 5.We prefer to distinguish this from the chemical rate constants expressedas k2 and k4.

For ozone the equation is

d[O3]dt

= k2[O][O2][Z] − J3[O3] − k4[O3][O]. (6.173)

At steady state we can write d[O]/dt = d[O3]/dt = 0. Hence,

[O] = 2J1[O2] + J3[O3]k2[O2][Z] + k4[O3] ,

[O3] = k2[O][O2][Z]J3 + k4[O] .

(6.174)

The ozone concentration can be obtained by simultaneously solving the above equa-tions. A quadratic in [O3] will result. Retaining only the positive square root for thesolution, we obtain after some simplification,

[O3] = [O2](J12J3

)[(1 + 4J3k2[Z]

J1k4

)1/2

− 1

]. (6.175)

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To simplify the above expression, we should have some idea about the relativemagnitudes of the various rate constants. Generally, in the stratosphere [Z] = [N2]∼1024 molecules/m3, J1 ∼ 1.5 × 10−10 min−1, and J3 ∼ 0.019min−1. k2 ∼ 1.2 ×10−43 m6/molecule2 min, k4 ∼ 7.5 × 10−20 m3/moleculemin, [O2] ∼ 1023.5 mole-cules/m3. Hence, [O3] = [O2]((J1/J3)(k2/k4) · [N2])1/2 ≈ 1018.8 molecules/m3 and[O] = 1013.4 molecules/m3.

The steady-state concentration of [O3] given by the equation earlier can berearranged to obtain the expression

[O3] = k2[O2][Z]k4(1 + (J3/k4[O])) (6.176)

with J3/k4[O] ≈ (0.019/7.5 × 10−20 × 1013.4) = 1 × 104. Hence, [O3] ≈ (k2/J3)[O2][Z][O]. Thus we have the ratio

[O3][O] ≈ k2

J3[O2][Z]. (6.177)

The steady-state concentration ratio [O3]/[O] should remain constant (∼2.5 × 105)if [O2] remains constant. This ratio can vary if either [O] or [O3] varies due to otherreactions. Since the production of O atoms by photolysis of O2 is a slow process(J1 is very small), any competing reaction that decomposes O3 faster will reduce theratio rapidly. These competing reactions involve trace species such as OH•, NO, andCFCs. The general catalytic cycle follows the scheme

X + O3 −→ XO + O2,

XO + O −→ O2 + X(6.178)

with the net reaction

O + O3 −→ 2O2. (6.179)

Note that the species X is neither formed nor removed from the system, and thus actsas a catalyst.

The reactionwithOH• is estimated to account for the decomposition of only−15%of the stratospheric ozone. Excited ozonemolecules (λ < 310 nm) give rise to excitedoxygen atoms, which in turn decomposes H2O or CH4 to provide OH•:

O3hν−→ O∗ + O2,

O∗ + CH4 −→ CH3 + HO•,

O∗ + H4O −→ 2HO•.

(6.180)

The OH radicals react with ozone in a self-propagating series of reactions to give twoO2 molecules:

HO• + O3 −→ O2 + HO2,

HO2 + O3 −→ O2 + HO•.(6.181)

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If the last two reactions are fast in comparison to the reaction of ozone with Z, thenthe ratio [O3]/[O] will decrease.

Although a number of species can catalyze the dissociation of ozone in the strato-sphere, a major concern at present is the influence of chlorinated compounds calledCFCs. These have been in widespread use as refrigerants and aerosol propellantsand are purely anthropogenic in origin. They comprise mainly CF2Cl2 (Freon 12,CFC-12) and CFCl3 (Freon 11, CFC-11) and are quite inert in the troposphere. Oncereleased to the atmosphere they are slowly transported to the stratosphere where theyremain for a very long time. Prior to the signing of the Montreal Protocol in 1987,several tons of these compounds were pumped into the atmosphere. For example, theconcentration of CFC-12 continually increased (see Table 6.15) at a rate of 17 pptv/ywith a mean concentration of ∼484 pptv in 1990. Its atmospheric lifetime is ∼130years. Molina and Rowland (1974) first alerted the scientific community to the factthat CFCs can adversely affect the stratospheric ozone layer. Radiations with wave-lengths of 180–220 nm in the stratosphere are absorbed byCFCs to produce the highlyreactive Cl and ClO species (represented generally as ClOx).

CCl2F2hν−→ CF2Cl + Cl•,

CCl2F2 + O• k1−→ CF2Cl + ClO.(6.182)

The ClOx can react with ozone and O atoms as follows:

Cl•O3k2−→ ClO• + O2,

ClO + O• k3−→ Cl + O2.(6.183)

The net effect is O + O3 → 2O2. The Cl atoms are subsequently trapped by hydro-carbons (e.g., CH4) to form HCl, which appears in the troposphere, and are washeddown by rain. From the above scheme, we have

d[ClO]dt

= k1[CCl2F2][O•] + k2[O3][Cl•] − k3[ClO][O•]. (6.184)

Therefore, the steady-state concentration of ClO is given by

[ClO]ss = k1k3

[CCl2F2] + k2k3

[O3][O•] [Cl

•]. (6.185)

Thus, the steady-state concentration of ClO can be maintained in the atmosphereas long as CCl2F2 is added to the stratosphere, even if no CCl2F2 is photolyzed. Ifphotolysis of CCl2F2 does occur, the second term becomes significant. Cl• will befinite and [ClO]ss will increase. Thus, the above chain will sustain high concentrationsof ClO at the expense of O3. In other words, CFCs act as catalysts to remove ozonefrom the stratosphere.

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Depletion of the ozone layer poses a problem. It can affect the amount of UVradiation reaching the earth and can increase the rate of skin cancer and other relatedhealth issues. Indeed, depletions of ozone have been observed over the Antarcticregions and, to a lesser extent in the Arctic regions, depleted ozone has been linkedto the increased ClOx in these regions (Figure 6.46).

With the signing of the Montreal Protocol, the industrialized nations of the worldcurtailed production of CFCs. This action has slowed the rate of increase of CFCsin the atmosphere. It is encouraging to see that industries have now replaced CFCswith alternatives that are environmental friendly. It is possible to use a simple boxmodel for the atmosphere to predict the effect of a voluntary reduction of CFCemissions.

61

020

040

0

CIO

mix

ing

ratio

CIO mixing ratio in ppt

O3 m

ixin

g ra

tio

O3 mixing ratio ppb

3000

2000

1000

0

800

62 63 64 65 66Latitude (degrees South)

67 68 69 70 71 72

FIGURE 6.46 Simultaneously observed ClO and O3 concentrations obtained on September21, 1987, by the ER-2 in the Antarctic, with corrections made for variations in potential tem-perature. (Reprinted from Anderson et al. 1989. Journal of Geophysical Research 94(D9),465–479, American Geophysical Union. With permission.)

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EXAMPLE 6.24 EFFECT OF CFC SOURCE REDUCTION ON THE FUTURE

ATMOSPHERIC CONCENTRATION

Figure 6.47 represents the entire atmosphere as a single box that receives CFCs at aconstant production rate of Stot mol/cm2 y. Even though the rate of CFC productionhas declined since the Montreal Protocol was signed, no significant depletion of CFCsin the stratosphere has yet been noted because of its low reactivity. The long lifetimeof CFCs guarantees a continuous but slow increase in atmospheric concentration evenafter the Montreal Protocol. Let us assume that for several decades the rate of release,Stot , can be assumed to be constant. The only mechanism by which CFCs are removedfrom the system involves photolysis reactions.

Let the overall rate constant for the reaction of a CFC be k(y−1) and its concentrationbe [CFC] (mol/m3 of air). From Section 6.3.1.1, using wv = u = wp = 0 and ri =−k[CFC], we have the following differential equation:

d[CFC]dt

+ k[CFC] = Stot

Zs. (6.186)

Solving the above equation with the initial condition that at t = 0, [CFC] = [CFC]0,we obtain

[CFC] = [CFC]0e−kt + Stot

kZs(1 − e−kt). (6.187)

Note that Stot/kZs = Qs/kVatm , where Qs is the source strength in mol/y and Vatm isthe atmospheric volume (m3).At some time in the future, the CFC concentration shouldreach a steady state, that is, as t → ∞, [CFC]ss → Stot/kZs. Hence,

[CFC][CFC]0 = [CFC]ss

[CFC]0 +(1 − [CFC]ss

[CFC]0

)e−kt . (6.188)

Since [CFC], [CFC]0, and [CFC]ss are concentrations in moles of CFC per m3 of air,they are easily converted to conventional units of ratio of volume of CFC to air (either

Surface sources

Atmosphere

Stot

CFC → Products

Zs

FIGURE 6.47 Box model for the production and dissipation of CFCs in theatmosphere.

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in ppmv or ppbv) by multiplying with 0.0224m3 CFC/mol, which is the molar volumeof CFC.

Let us first calculate the values of [CFC], [CFC]0, and [CFC]ss for the year 1987whenthe Montreal Protocol took effect. In 1987, [CFC]∼450 pptv for CFC-12. If we choosek − 0.0065 y−1, a production rate for CFC-12 of Qs ∼ 3 × 109 mol/y, and Vatm =3.97 × 1018 m3, we obtain [CFC]ss = 2.8 ppbv. If, further, we assume a constant Qs,then

[CFC]

[CFC]0= 6.2 − 5.2e−0.0065t .

Curve 1 of Figure 6.48 shows the resulting profile. This can be considered to be thebase case, that is, a consequence of nonimplementation of the Montreal Protocol. If theMontreal Protocol is to reach a goal of 50% reduction in net CFC emissions, then Qshas to be replaced by 0.5Qs and [CFC]ss = 1.4 ppbv. Hence,

[CFC]

[CFC]0= 3.1 − 2.1e−0.0065t .

Curve 2 of Figure 6.48 shows the expected CFC concentration in air in this case. Notethat even under this scenario, the CFC concentration continues to rise in the atmosphere,albeit at a slower rate. As a consequence, chemical reactions causing the destructionof stratospheric ozone will continue into the next century. Figure 6.48 also considersthe following scenario: What if the Montreal Protocol had attempted the completecessation of CFC production by 1997? The decay of CFC concentration will then begiven by

0

0.5

1

1.5

2

2.5

3

3.5

1980 2000 2020 2040 2060 2080 2100

[CFC

]/[C

FC] 0

Year

G0

0.5 G0

G = 0 in 1997

FIGURE 6.48 Atmospheric concentration of CFCs for various scenarios of theMontreal Protocol.

continued

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[CFC]

[CFC]in1997= e−0.0065t ,

and is shown as curve 3. It is evident that even under these most optimistic conditions,the concentration of CFC in the year 2087 will decrease only to one-half its value in1987. This clearly shows how long the effect of CFCs will linger in the stratosphereeven if a concerted attempt is made in this century to eliminate its production.

In the lower troposphere, ozone performs a different function. It reacts with oxidesof nitrogen (NOx) and atmospheric hydrocarbons and leads to the formation of smog.This process accounts for ∼60–70% of the ozone destroyed in the troposphere.

EXAMPLE 6.25 KINETICS OF SMOG FORMATION IN URBAN AREAS

The chemistry of smog, as we have discussed, is tied to the chemistry of NOx speciesin the atmosphere. The basic reaction is the photochemical dissociation of NO2 (λ =435 nm) that gives rise to NO and O. The O then rapidly reacts with O2 in the presenceof a third body, Z (e.g., N2), to produce ozone. The cycle is completed when O3 reactswith NO to regenerate NO2.

NO2hν−→J1

NO2 + O; J1 ≈ 0.6min−1,

O + O2 + Zk2−→ O3 + Z∗; k2 ≈ 6.1 × 10−34 cm6/(molecule2 s), (6.189)

O3 + NOk3−→ O2 + NO2; k3 ≈ 1.8 × 10−14 cm3/(molecules).

The rate constants given are at 298K.As explained earlier,we distinguish photochemicalrate constants from chemical rate constants by utilizing the symbol J for the former.The rates of formation of the different species are

d[NO2]dt

= −J1[NO2

]+ k3[O3][NO],

d [O]

dt= J1

[NO2

]− k2 [O][O2][Z],

d[O3]

dt= k2 [O]

[O2][Z] − k3 [NO]

[O3].

(6.190)

Applying the PSSA for the intermediate species [O], we obtain

[O]ss = J1k2

[NO2][O2][Z] . (6.191)

If the NO2 formation and dissipation is at steady state, we have

[O3] = J1

k3

[NO2][NO] . (6.192)

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This is an important result, since it states that the concentration of ozone is dependentnot only on the magnitude of NOx , but also on the ratio [NO2]/[NO]. As J1 → 0 (i.e.,nighttime conditions), in the presence of excess of ozone the ratio becomes very large,whereas the lowest ratios are observed in bright sunshine (high J1 ∼20 h−1).

If we now start with a system with initial condition [O3]0 = 0 and [NO]0 = 0, thestoichiometry states that [O3] = [NO2]0 − [NO2] = [NO]. Thus,

J1[NO2][NO] = [NO2]0 − [NO2], [NO2] = k3[NO][NO2]

J1 + k3[NO] .

Now, since [O3] = [NO2]0 − [NO2], we obtain the following quadratic in [O3]:

k3[O3]2 + J1

[O3]− J1

[NO2

]0 = 0. (6.193)

Since [O3] > 0, we keep only the positive root of the above quadratic, and obtain

[O3] = 1

2

⎡⎣{(

J1k3

)2+ 4

J1k3

[NO2

]0

}1/2− J1k3

⎤⎦ . (6.194)

We know that J1 ∼ 0.6min−1. k3, which is in cm3/molecule S, can be converted intoppmv/min by multiplying with 1.47 × 1015. Hence k3 ∼ 26 ppmv/min and J1/k3 =0.02 ppmv. The variation in [O3] with [NO2]0 is shown in Figure 6.49. When[NO2]0 = 0, [O3] = 0 and the ozone concentration increases as more NO is convertedto NO2.

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1 1.2[NO2]0 / ppmv

[O3]

/ ppm

v

[NO2]

[O3]

FIGURE 6.49 Ozone concentration as a function of initial nitrogen dioxideconcentration.

continued

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In urban areas with smog, the concentration of ozone is higher than that predictedabove. This can be interpreted using the expression [O3] = J1[NO2]/k3[NO]. Anyreaction that increases the transformation of NO to NO2 can bring about increasedozone concentrations.What types of materials are responsible for this effect? Table 6.12summarizes the most common constituents of urban smog and their adverse effects.

TABLE 6.12Common Constituents of Urban Smog that React with Ozone

Compound Formula Health Effects

Aldehydes R–CHO Eye irritation, odorHydrocarbons R–CH3 Eye irritation, odorAlkyl nitrates R–ONO2Peroxyacyl nitrates (PAN) R–CO–OONa Toxic to plants, eye irritantAerosols (NH4)2SO4, NH4NO3, etc. Visibility reduction

During daytime automobiles generate NO via combustion:

N2 + O2 −→ 2NO. (6.195)

However, automobiles also generate hydrocarbons (HR) and CO via incompletecombustion, both of which react with atmospheric OH radicals to produce peroxyradicals:

OH + HR + O2 −→ RO2 + H2O. (6.196)

In general, only C3 and higher hydrocarbons are reactive and are called “non-methanehydrocarbons (NMHC).” The peroxy radicals are very efficient in converting NO toNO2:

RO2 + NO + O2 −→ RO + NO2 + HO2. (6.197)

Thus, the ratio [NO2]/[NO] and O3 concentration increase during the daytime.NO2 and HR also react producing aldehydes, ozone, and peroxy acetyl nitrate (PAN).For example,

CH3–CH3 + OH + O2 −→ CH3CH2O2 + H2O,

CH3CH2O2 + NO + O2 −→ CH3CHO + HO2 + NO2,

OH + CH3CHO −→ CH3CO• + H2O,

CH3CO• + O2 −→ CH3COOO

•,

CH3COOO• + NO2 −→ CH3COOONO2.

(6.198)

PAN, so formed, is an eye irritant. Thus to eliminate smog in an urban area, the firstline of defense is to reduce HR emissions andNO from automobiles and power plants.

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0.2

0.4

4 8 12 16 20

NO

NO2 O3

RCHO

PAN

Time of day/h

Mix

ing

ratio

× 1

06

RH

FIGURE 6.50 Diurnal variation in concentrations of various species in a typical urban smog.

In urban areas such as Los Angeles, California, where smog is a common occur-rence, the concentrations of hydrocarbons (saturated and unsaturated) and aldehydesare very large. They are produced from automobile emissions. As an example, agasoline-powered vehicle exhaust consists of∼78% N2, 12% CO2, 5% H2O (vapor),1% unused oxygen,∼2% each of CO and H2,∼0.08% of hydrocarbons, and∼0.06%NO. The remaining several hundred pptv of oxidized hydrocarbons consist of alde-hydes, formaldehyde being the dominant fraction. Typical smog composition in LosAngeles and the diurnal profile are shown in Figure 6.50. Note the peak concentrationsof ozone at noontime when smog is severe. The chemistry of smog in urban areasis incredibly complex due to the presence of a variety of hydrocarbons and aldehy-des that participate in reactions with ozone and nitrogen oxides. Seinfeld (1986) hasdiscussed the salient aspects of these reactions, to which the reader is referred to formore details.

6.4 SOIL AND SEDIMENT ENVIRONMENTS

The land surface contributes about 25% of the earth’s surface area. Compoundsmove between soil and water in the groundwater environment. Sediment is a sinkfor contaminants entering water in lakes, rivers, and estuaries. Similarly, soil and aircompartments exchange chemicals. We discuss three examples of transport models,one for each interface—soil–water, sediment–water, and soil–air.We also discuss soilremediation concepts that use principles from chemical kinetics.

6.4.1 F&T MODELING

We discuss three cases here, namely, groundwater, sediment–water, and soil–airexchange of chemicals.

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6.4.1.1 Transport in Groundwater

Leaking underground storage tanks, old unlined landfills, surface impoundments,and accidental spills are the main causes for groundwater contamination. There aretwo regions in the subsurface that are subject to contamination: the unsaturated(vadose) zone and the saturated zone (Figure 6.51). Liquid spilled on the surfacegradually migrates downward. The light nonaqueous-phase liquid (LNAPL) floats onthe groundwater table. A dense nonaqueous-phase liquid (DNAPL) sinks and poolsat the bottom of the impervious layer. Contaminants slowly dissolve in the ground-water. In the zone of solubilization, globules of the organic solvent slowly dissolvewith time. The transport of the contaminant plume in groundwater, therefore, consistsof two processes: dissolution of globules (ganglia) of pure fluid and movement of thesolubilized fraction.

6.4.1.1.1 Solubilization of Ganglia

An NAPL globule trapped in a soil pore will slowly dissolve in the groundwater.The rate of dissolution will depend on the mass transfer coefficient, kL (m/s),between the NAPL globule and water under specified flow conditions. The flux isgiven by FA = kL

([A]∗ − [A]

). [A]∗ is the aqueous solubility of the NAPL and

[A] is the concentration in water. Consider the NAPL globule to be approximatelyspherical (diameter Dp, m), with a volume ratio θN (m3/m3 of the medium)—seeFigure 6.52. The area per unit volume of the medium a = 6θN/Dp. Note that bothθN and Dp are functions of the distance x from the globule in the flow direc-tion. a(x, t) = 6θN(x, t)/Dp(x, t). Thus the overall concentration change along thex-direction is given by (Hunt et al., 1988)

1

2

3

4

5

Groundwater flow

Groundwater table

Saturatedzone

Less permeable strata

Solute front

Soil surfaceLandfill

Vadose zone

FIGURE 6.51 Groundwater contamination from a leaking source.

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Soil surface

xs

NAPL ganglionGroundwaterflow

Flow velocity, u

FIGURE 6.52 An NAPL ganglion dissolving in groundwater.

d [A]

dt= kL

uD

(6θN

Dp

) ([A]∗ − [A]

), (6.199)

where [A] = f (x, t) and uD is the Darcy velocity, defined later in this section. Theabove equation is applicable for plug-flow conditions. We assume that the reductionin NAPL volume is minimal, and henceDp = D0 and θN = θ0 (the initial values). Thevolume fraction θ0 is the spread over a length Xs. Integrating the resulting equationgives us the concentration of A at the edge of the plume,

[A][A]∗ = 1 − e−[(kL/uD)(6θ0/D0)Xs]. (6.200)

Numerous correlations exist for kL in the chemical engineering literature. Hunt et al.(1988) used the correlation kLεT/1.09uD = Pe−2/3, where Pe is the Peclet numbergiven by uDDp/Dw, and Dw is the molecular diffusivity of the NAPL in water.

EXAMPLE 6.26 NAPL GANGLION SOLUBILIZATION

Let us consider an initial volume fraction θo of 0.05 over a total horizontal extentof 10m. In this case θoXs = 0.5m. For a typical Darcy velocity of 1m/day, and aDo = 0.1m, and for a typical molecular diffusivity in water of 8.6 × 10−5 m2/d, weobtainPe = 1157. If εT = 0.4, kL = 0.024m/d. Hence [A]/[A]∗ = 0.51.As the globuledissolves, both its diameter and volume decrease (Powers et al., 1991). To obtain theactual solubilizedmass in water, the change in volume should be taken into account. Theganglion lifetimes are found to beweak functions of the flowvelocity.As a consequence,very large volumes of water are necessary to reduce the ganglion size. Therefore, oneends up with an expensive above-ground treatment system.

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6.4.1.1.2 Transport of the Dissolved Contaminant in Groundwater

Pollutant that solubilizes in the aqueous phase is transported with the groundwater.As the solubilized pollutant moves through the heterogeneous subterranean soil pore-water, it adsorbs to the soil particles. Hence, we have to consider both soluble andadsorbed contaminant mass in ascertaining the transport in the groundwater environ-ment. Nonideal plug flow is assumed because of the soil heterogeneity. Consider avolume element in the direction of groundwater flow (Figure 6.53). Advective anddispersive terms comprise both feed and effluent rates. The advective term is U[A]and dispersion is given by−D∂[A]/∂x, evaluated at appropriate x co-ordinates.Accu-mulation of mass occurs in both solid and liquid phases. Reaction losses occur in theliquid phase as a result of chemical or microbiological processes. Note that for theporous solid the volume is εAcΔx. The overall material balance is as follows:

rate in = rate out + accumulation + reaction loss.

εAcU [A]|x − εAcD∂[A]∂x

∣∣∣∣x

= εAcU [A]|x+Δx − εAcD∂[A]∂x

∣∣∣∣x+Δx

+ εAcΔx∂[A]tot

∂t+ rAεAcΔx . (6.201)

Note that [A] is the concentration in the mobile phase (water) and [A]tot is the totalconcentration (porewater+ solid). Dividing throughout by εAcΔx and taking the limitas Δx → 0, we obtain after rearrangement,

D∂2[A]∂x2

− U∂[A]∂x

− rA = ∂[A]tot

∂t. (6.202)

The above expression is the general advective–dispersive equation in the x co-ordinate.Note that on the right-hand sideof the equation, [A]tot = ε[A] + (1 − ε)ρsWA,

Reaction loss

Effluent(Advection

+Dispersion)

Feed(Advection

+Dispersion)

Porousmedia

Accumulation on solids

Accumulation in liquid phase

FIGURE 6.53 A volume element and transport kinetics in the groundwater environment.

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the total concentration of i per m3. If transport is in all three directions, we cangeneralize the equation to the following:

∇ · (D∇[A])− U · (∇ [A])− rA = ∂ [A]tot

∂t, (6.203)

where∇ = ∂/∂x + ∂/∂y + ∂/∂z is called the “del operator.” Using the local equilib-rium assumption (LEA) between the solid- and liquid-phase concentrations, WA =Ksw [A], and if rA = 0, we obtain

D∂2[A]∂x2

− U∂[A]∂x

= RF∂[A]∂t

, (6.204)

where RF = ε+ ρbKSW is the retardation factor as defined earlier. ρb = (1 − ε)ρs isthe soil bulk density.

In a plug-flow reactor, if the initial concentration in the fluid is [A]0 and at timet = 0, a step increase in concentration [A]s is applied at the inlet, the followinginitial and boundary conditions apply to the advection–dispersion equation in thex-direction:

[A](x, 0) = [A]0 for x ≥ 0,

[A](0, t) = [A]0 for t ≥ 0,

[A](∞, 0) = [A]0 for t ≥ 0.

(6.205)

The solution is

[A] − [A]0[A]s − [A0] = 1

2

[erfc

(x − (U/RF) t

2√Dt/RF

)+ exp

(Ux

D

)erfc

(x + (U/RF) t

2√Dt/RF

)],

(6.206)

where erfc is the complementary error function (see Appendix 8).It is instructive to consider the simplifications of the advective–dispersion equation

for specific cases:

(i) If dispersion (diffusion) in much larger than advection, that is, D∂2[A]∂x2

�U

∂ [A]

∂xand rA = 0, we have

D∂2[A]∂x2

= RF∂[A]∂t

, (6.207)

which is the well-known Fick’s equation, for which, with the boundaryconditions given earlier, the solution is

[A] − [A]0[A]s − [A0] = 1 − erf

(x√4Dt

). (6.208)

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(ii) If advection is larger than dispersion, that is, D(∂2[A]/∂x2)�

U (∂[A]/∂x), we have

−U ∂[A]∂x

= RF∂[A]∂t

(6.209)

or, alternatively,

− (∂[A]/∂t)(∂[A]/∂x) = U

RF, (6.210)

where U is the Darcy velocity of the groundwater. The left-hand side isinterpreted as the velocity of the pollutant,U ′. ThenU ′/U = 1/RF. In otherwords, the retardation factor scales as the velocity of the groundwater. LargeRF means U ′ < U and the chemical movement is retarded.

A qualitative picture of the effects of dispersion, adsorption, and reaction on themovement of pollutants is shown in Figure 6.54. Dispersion alone will make thepollutant front less sharp. The addition of adsorption will delay the breakthroughtime; however, the concentration at the sampling point will eventually reach the feedconcentration at x = 0. Inclusion of reaction will further decrease the peak maxi-mum and the feed concentration will never be attained at the downstream samplingpoint. The relative magnitudes of dispersion and advection are assessed in terms of adimensionless Peclet number based on dispersivity, Pe = UXs/D.

For soils (porousmedia), the dispersion coefficient has two contributions, one frommolecular diffusion and the other from fluidmovement through the porousmedia. Fora porous medium, the tortuous path that a molecule has to take in traversing the porefluid reduces the effective diffusion coefficient. Dwε

4/3, where ε is the porosity andDw is the molecular diffusivity of the pollutant in water, gives the effective diffusioncoefficient. The dispersion due to fluid motion is related to fluid velocity through asaturated medium, which is given by Darcy’s law:

UD = −κdhdx

, (6.211)

where UD is the Darcy velocity (cm/s), κ is the hydraulic conductivity of the medium(cm/s), and dh/dx is the hydraulic gradient. In terms of volumetric flow, UD = Q/A,where Q is the volumetric flow rate of fluid through a cross-sectional area A. Thenegative sign for UD indicates that the flow of fluid is in the direction of decreasinghydraulic head. Table 6.13 lists typical values of κ for representative media. It canvary over many orders of magnitude depending on the media.

The dispersion constant is given by D = Dwε4/3 + UDαD, where αD is termed

the dispersivity. It is a measure of the media heterogeneity and has units of length.A proposed relationship for αD in terms of the travel distance for a contaminant isαD = 0.017X1.5 (Neumann, 1990), where αD is in m and X is in m.

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No dispersion, adsorption or advectionDispersion only

Time

Dispersion+

Adsorption+

Reaction

Dispersion+

Adsorption

([A] –

[A] 0

)/([A

]*– [A

] 0)

FIGURE 6.54 Effects of dispersion, adsorption, and reaction on the movement of a pollutantin the subsurface.

TABLE 6.13Typical κ Values for Porous Media

Media κ (cm/s)

Gravel 0.03–3Sand (coarse) 9 × 10−5 − 0.6Sand (fine) 2 × 10−5 − 0.02Silt 1 × 10−7 − 0.003Clay 8 × 10−11 − 2 × 10−7

Shale 1 × 10−11 − 2 × 10−7

Source: FromBedient, P.B.,Rifai,H.S., andNewell,C.L. 1994. Groundwater Contamination.Englewood Cliffs, NJ: Prentice-Hall PTR.

EXAMPLE 6.27 TIME TO BREAKTHROUGH FOR A CONTAMINANT PLUME IN

GROUNDWATER

Estimate the concentration of chlorobenzene in the groundwater at a well 1 km froma source after 500 years. The Darcy velocity is 5m/y. The soil has an organic carboncontent of 2%, a porosity of 0.4, and a density of 1.2 g/cm3.

For chlorobenzene, log Kow = 2.91. Hence log Koc = (0.92)(2.91) − 0.23 = 2.45.Ksw = Kocfoc = (102.45)(0.02) = 5.6 L/kg. RF = ε+ (1 − ε)ρsKsw = 4.4. Dw =8.7 × 10−6 cm2/s. αD = 0.017X1.5

s = 530m = 5.3 × 104 cm. UD = 5m/y = 1.6×10−5 cm/s. D = 2.5 × 10−6 + 0.85 = 0.85 cm2/s. Note that Pe = 1.8 and hence bothadvection and dispersion are important.

continued

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Since [A]0 = 0, we have [A]/[A]s on the left-hand side of the equation. Using UDfor U, we have

[A]

[A]s= 1

2

[erfc

(105 − 1.45 × 10−6t

0.51√t

)+ 6.5erfc

(105 − 1.45 × 10−6t

0.51√t

)],

where t is in seconds. If we use t = 500 × 3.17 × 107 s, we obtain [A]/[A]s = 0.336.

EXAMPLE 6.28 TIMEOFTRAVELFORADVECTIVETRANSPORTOFACONTAMINANT

IN GROUNDWATER

Estimate the time taken for a plume of chlorobenzene to reach a groundwater well 100mfrom the source if the Darcy velocity is 1 × 10−4 cm/s for the soil in the last example.Assume only advective transport is significant.

Since advection is dominant over dispersion, U/UD = 1/RF = 0.23. U =(0.23)(1 × 10−4) = 2.3 × 10−5 cm/s.Hence t = Xs/U = 10,000/2.3 × 10−5 = 4.3 ×108s = 13.7 y.

6.4.1.2 Sediment–Water Exchange of Chemicals

Compounds distribute between the various compartments in the environment. One ofthe repositories for chemicals is the sediment. Chemical exchange at the sediment–water interface is, therefore, important in delineating the fate of environmentallysignificant compounds. Sediment contamination arose from the uncontrolled pol-lutant disposal in lakes, rivers, and oceans. As environmental regulations becamestricter, most pollutant discharges to our lakes and waterways became controlled.The contaminants in sediments bioaccumulate in marine species and exposure tohumans becomes likely. Hence the risks posed by contaminated sediments have to beevaluated and sediment remediation strategies determined. The risk-based correctiveaction (RBCA) is predicated upon knowledge of chemical release rate from sedimentand transport through air and water environments (Figure 6.55). The first step in thisprocess requires an understanding of potential release mechanisms.

In the case of the sediment environment, a number of pathways for chemicalexchange between the sediment and water can be identified. These are representedschematically in Figure 6.56. For sediments that rest in quiescent environments, diffu-sion (molecular) retarded by adsorption is the most ubiquitous of transport processes.Advective transport is driven by the nonuniform pressure gradients on the rough sed-iment terrain. Other transport processes include active sediment particle transport. Acomparison of characteristic times for a hypothetical scenario was made by Reible,Valsaraj, and Thibodeaux (1991) and is shown in Table 6.14. Processes with smallhalf-lives are likely to be the most important transport processes.

Diffusion of compounds from sediment in the absence of advection or biodegrada-tion can be represented by the equation derived in the previous section on groundwater

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Contaminatedsediment

Chemical release

Transport/Transformation

Receptor

Risk

Risk-based paradigm

FIGURE 6.55 RBCA paradigm.

transport, where the dispersivity is replaced by molecular diffusion. The geometryconsidered for modeling is shown in Figure 6.57.

[A]w

[A]0w= erf

(z√4D∗

s t

), (6.212)

where D∗s = Dwε

4/3

ε+ ρpKsw.

The release rate (flux) is

FA =(D∗

s

πt

)1/2

W0Aρp, (6.213)

where W0A is the initial sediment contamination.

Air

BedrockGroundwater

Bed sedimentWorms

SuspendedsedimentWater

Dune

“Mobile” bed

“Mud flats”

99

2

10

5

2

168

Evaporation

Deposition

11

7

Bed sediment

4

3 2

FIGURE 6.56 Schematic of the fate and transport processes in bed sediments.

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TABLE 6.14Comparison of the Characteristic Times of SedimentTransport Processes

Mechanism Characteristic Time (Years)

Molecular diffusion (unretarded by sorption) 0.5 (hypothetical)Molecular diffusion (retarded by sorption) 1900Colloidal-enhanced diffusion 1500Sediment erosion (1 cm/y erosion) 10Capped sediment (30 cm effective cap) 21,000Bed load transport (sediment movement) 42 hAdvection (aquifer interactions) 4000Surface roughness (local advection) 69,000Bioturbation 10

Note: Characteristic times are order of magnitude estimates of time required toleach a typical hydrophobic organic compound (e.g., trichlorobiphenyl)from a 10-cm layer of sediment by each of various transport mechanisms.Inmost cases, the characteristic time is 1/e times or half-lives. For advectiveprocesses, the characteristic times represent time for complete removal.The above refer to typical sediment conditions represented in Reible et al.(1991). Although absolute values refer to a particular set of conditions, theranked magnitudes are probably indicative of generic behavior.

6.4.1.3 Soil–Air Exchange of Chemicals

Figure 6.58 represents the chemical transport pathway from soil to the atmosphere.The difference from the saturated case (groundwater) is that in this unsaturated case,

Water

Contaminated sediment

z

Contaminantdiffusion

z = 0x

FIGURE 6.57 Schematic of the diffusive transport of a contaminant from a bed sediment tothe overlying water column.

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(Diffusion + Advection)

Reaction

Soil surface

Vadose zone

Groundwater

Windx

zSoil

FIGURE 6.58 Transport of a contaminant from the soil to the atmosphere.

there exist both pore air and porewater. The mobile phase is air, while both porewaterand solids are immobile with respect to the contaminant. The advective–dispersionequation derived is applicable with the stipulation that [A] represents the pore airconcentration and [A]tot is the total concentration (pore air+ porewater+ solidphase). D should be replaced by the effective molecular diffusivity in partially satu-rated air, Dg, and Ug is the advective gas-phase velocity. RA denotes the biochemicalreaction loss of compound A in pore air and porewater.

Dg∂2[A]g

∂z2− Ug

∂[A]g

∂z− RA = ∂[A]tot

∂t. (6.214)

If the air-filled and water-filled porosities are εg and εw, respectively, then atotal mass conservation gives [A]tot = mass in pore air+mass in porewater+masson soil particles = εg[A]g + εw [A]w + ρs(1 − εg − εw)WA. Since local equilibriumis assumed between the three phases, [A]w = [A]g/Kaw,WA = KSA[A]g. Withthese expressions, we have [A]tot = RF[A]g, whereRF = εg + (εw/Kaw) + (1 − εg −εw)ρsKSA is the retardation factor. If we further consider the case where RA = 0 andadvective velocity is negligible, we have

∂[A]g

∂t= Dg

RF

∂2[A]g

∂z2. (6.215)

Thus molecular diffusion through pore air is the dominant mechanism in this case.The initial condition is [A]g (z, 0) = [A]0 for all z. The following two boundaryconditions are applicable: (i) [A]g(∞, t) = [A]0 for all t and (ii) −Dg

(∂[A]g/∂z

)+ka [A]g (z, t) = 0 at z = 0. The second boundary condition states that at the surfacethere is a reaction ormass transfer loss of chemical, thus contributing to a resistance to

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mass transfer at the soil surface. The solution for the flux from the surface is (Valsarajet al., 1999)

FA = ka[A]0 exp(

k2a t

DgRF

)erfc

(ka

√t

DgRF

)(6.216)

Note that at t = 0,FA = ka[A]0 and as t → ∞,FA = [A]0√DgRF/πt, which is

the solution to Fick’s diffusion equation. Note that FA is a function of both Dg andRF, both of which depend on the soil porewater content. With increasing water in thepore space, Dg and RF will decrease. The decrease in RF will be far more significantsinceKSA has been observed to be a sensitive function of porewater content (Guilhem,1999). Experimental data have verified this prediction (Figure 6.59). Thus flux willbe higher from a wet soil or sediment and lower from a dry sediment. In other words,pesticides and organic compounds will be far more volatile from wet than dry soils.

250

200DBFModel (wet)Model (dry)

humid airdry air

humidair

150

100

50

00 200 400 600

t, h

NA

(ng/

cm2 · h

)

800 1000 1200 0 200 400 600 800 1000 1200

60

50

40

30

20

10

0

NA

(ng/

cm2 · h

)

PHEModel (wet)Model (dry)

0 200 400 600 800 1000 1200

Water

0.010

0.008

0.006

0.004

0.002

0.000

NA

(ng/

cm2 · h

)

0 200 400 600 800 1000 1200

10

8

6

4

2

0

NA

(ng/

cm2 · h

)

PYRModel (wet)Model (dry)

t, h

t, ht, h

FIGURE 6.59 Effects of change in air relative humidity on dibenzofuran, phenanthrene,pyrene, and water flux from a 6.5%moisture University Lake sediment. Model curves for bothhumid air over “wet” sediment and dry air over “dry” sediment are shown. (Data fromValsarajet al. 1999. Environmental Science and Technology 33, 142–148.)

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EXAMPLE 6.29 PESTICIDE VOLATILIZATION RATE FROMA SOIL

Estimate the rate of release (μg/s) of dieldrin (a pesticide, molecular weight 381)applied to a soil at 100μg/g concentration. The soil has a total porosity of 0.5, an foc of1%, and a water saturation of 5%. The soil density is 2 g/cm3 and area of application is1 ha. Assume a surface mass transfer coefficient of 0.1 cm/s.The effective diffusivity Dg in a partially saturated soil is given by DA(ε

10/3a /ε2T),

where DA is the molecular diffusivity in air and εa is the air-filled porosity. Note thatεa = (1 − θw)εT, where θw is the water saturation. For dieldrin, DA = 0.028 cm2/s.Since θw = 0.05, εa = 0.57, εw = θwεT = 0.03. Hence Dg = 0.012 cm2/s.RF = εg + (εw/Kaw) + ρbKSA · Kaw for dieldrin is 8.1 × 10−6. From Chap-

ter 4, we have KSA = Ksw/Kaw = focKoc/Kaw. With log Kow = 5.48, logKoc =(0.92)(5.48) − 0.23 = 4.81. Hence KSA = 7.9 × 107 P/kg and RF = 6.8 × 107.[A]0 = ρbWi/KSA = 1.2 × 10−6 μg/cm3. Hence FA = (0.1)(1.2 × 10−6)[exp(1.88× 10−5). erfc(0.0137)] = 1.12 × 10−7 μg/cm2 s. Hence mass lost, mi = 1.12 ×10−7 × 108 = 11.2μg/s.

6.4.2 SOIL AND GROUNDWATER TREATMENT

As depicted in Figure 6.51, surface spills of solvents can potentially contaminate twozones. The discussion in the previous section reflects the different transport mech-anisms that operate in the two zones. In the vadose zone, pore air and porewatertransport processes are operative, whereas in the saturated zone porewater transportis dominant. The remediation strategy for contaminants in the two zones also dif-fers, the underlying phenomena being driven by the respective operative transportmechanisms.

In both zones, there are three major categories of remedial processes:

(a) Containment, whereby contaminants are prevented from further spreading(b) Removal, whereby contaminants are extracted from the subsurface(c) Treatment, wherein contaminants are separated and treated by appropriate

technologies.

Containment is required in cases where the movement of fluids is to be controlledbefore adverse effects are manifest in nearby communities that depend on drinkingwater supplies from the aquifer. Physical barriers such as slurry walls, grout curtains,and sheet pilings are used for containment. Hydraulic barriers, that is, reversing thehydraulic gradient by a series of pumps and drains, are effective in containing aslow-moving contaminant plume in both zones, especially in the saturated zone.

Removal is the only option in some cases. Highly contaminated surface soils areexcavated and treated before disposal. However, this is often infeasible and expensivefor surface soils and groundwater in the saturated zone.Amore practical solution is theso-called P&Tmethod.As the very name indicates, this entails bringing groundwaterto the surface, separating the contaminants, and discharging thewater to the subsurfaceor to lakes or rivers. A variation of P&T practiced in both the vadose zone and the

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saturated zone is pumping air and/or applying vacuum to strip volatile materials fromsoil pore air or groundwater. In the vadose zone this is called air sparging. Whenwater is used to flush contaminants, it is called in situ soil washing.

Treatment involves a variety of above-ground technologies for both water andair that is recovered from the subsurface. Processes that utilize intrinsic bacterialpopulations are useful in destroying contaminants in-place. These include naturalattenuation and intrinsic bioremediation.

Equilibrium and chemical kinetic aspects of P&T technologies for the vadose zoneand saturated zone in subsurface soil environments are explored below.

6.4.2.1 P&T for NAPL Removal from Groundwater

Figure 6.60 is a schematic for a P&T groundwater remediation. This is a method forremediating contaminated groundwater. Clean water that is brought into the aquiferflushes pollutants from the contaminated region. The water brought to the surfacewill have to be further treated before re-injection or disposal. Although effective incontaining plume migration, P&T is ineffective in removing all of the material fromthe subsurface. Once the free phase NAPL is removed, the residual NAPL is harder toremove by water flushing. It gives rise to the “tailing-off” effect, that is, contaminantsin inaccessible regions dissolve very slowly over extended periods (Figure 6.61).

For a withdrawal rate Q from a well, the maximum removal rate of a contaminantis Q[A], where [A] is the porewater concentration. If the contaminant is present as apure phase, the value of [A] is the same as the aqueous solubility [A]∗. As water ispumped through the aquifer, equilibrium between the solids and porewater dictatesthe maximum mass of contaminant that can be transported through the groundwater.

Off-site

Cone ofdepression

Extractionwell

Remove pollutant

Treat water

Re-injectFrom cleanwater storage

Injectionwell

Injectionwell

Vadosezone

Plume Flow

Groundwatertable

Impervious bedrock

FIGURE 6.60 A basic P&T approach for groundwater remediation.

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00 1 2

0.5

1.0

Theoreticalremoval

Removal withtailing

Efflue

nt co

ncen

trat

ion

Influ

ent c

once

ntra

tion

Water-filled aquifer volume

FIGURE 6.61 Phenomena of tailing-off in conventional P&T process for groundwaterremediation.

The process can be modeled by considering the aquifer to be a completely mixedreactor (CSTR). If we consider a total volume VT of the aquifer, the fluid volume isεVT, where ε is the porosity of solids. A mass balance on the continuous reactor givesthe following:

rate in with water = rate out with water + accumulation rate (solids + pore fluid).

Q[A]in = Q[A] + εVTd[A]dt

+ (1 − ε)VTρsdWA

dt. (6.217)

If local equilibrium between solids and pore fluid is assumed,WA = Ksw[A]. More-over, if pure water is brought into contact from the injection well, [A]in = 0.Hence,

−(Q

RF

) [A]VT

= d[A]dt

, (6.218)

where Q is the volumetric flow rate and RF = ε+ (1 − ε)ρsKsw is the retardationfactor. Integrating the above equation gives

[A][A]0 = exp

(− Qt

RFVT

). (6.219)

Hence, the time taken to flush 50% of the contaminant in porewater will be t1/2 =0.693 (RFVT/Q). Thus, cleanup time will increase with increasing RF and decreasingQ. Note that [A]0 = M0/VTRF, whereM0 is the total initial mass of the contaminant.Note that (Qt/εVT) is the number of pore volumes, NPV flushed in time t. Therefore,

[A][A]0 = exp

(−εNPV

RF

). (6.220)

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EXAMPLE 6.30 NUMBER OF PORE VOLUMES FOR IN SITU FLUSHING OF

RESIDUAL NAPL

Calculate the total number of pore volumes to remove 99% of the following pollutantsin porewater from an aquifer with porosity 0.3 and an organic carbon content of 1%.Soil bulk density is 1.4 g/cm3: (a) benzene, (b) naphthalene, and (c) pyrene.

First calculate Koc from log Koc = 0.937 logKow − 0.006 and Ksw = focKoc. Thencalculate RF = ε+ ρbKsw. For 99% removal [A]/[A]0 = 0.01.

Compound log Ksw Ksw RF NPV = (−RF/ε) ln(0.01)

Benzene 2.13 1.6 2.5 38Naphthalene 3.36 13.9 19.8 304Pyrene 5.13 632.1 885 13,585

Extraction from the saturated zone is incapable of removing most contaminantsthat have low solubility in water. These cases require excessively large volumes ofwater to be pumped (see Example 6.30). As a result, techniques have been sought toenhance P&T. In Chapter 4, we noted that the use of surfactants significantly lowersthe surface tension at oil–water interfaces and also increases the aqueous solubility oforganic compounds through the presence of surfactant micelles. Both of these aspectshave been utilized in enhancing the extent of removal using P&T methods.

Figure 6.62 identifies the different phases into which a pollutant can partitionwhen surfactants are present in the aqueous phase in the form of micelles. If a CSTRis assumed for the aquifer, the overall mass balance with added surfactant in the

WaterMicelle in water

Solid

MonomerPollutant in

micellePollutant in

water

AdsorbedpollutantWater

Adsorbedsurfactant

FIGURE 6.62 Partitioning of a pollutant between various compartments in the groundwaterwhen surfactants are introduced into the subsurface.

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aquifer gives

rate in = rate out (water+micelles) + accumulation (water+ solids+micelles).

0 = Q ([A] − [A]mic) + εVTd[A]dt

+ εVTd[A]mic

dt+ (1 − ε)VTρs

dWA

dt. (6.221)

Since [A]mic = [Surf]Kmic[A] and WA = Ksw[A]i, we find

d[A]dt

= −(Q

VT

)((1 + [Surf]Kmic)

ε+ ε[Surf]Kmic + (1 − ε)ρsKsw

), (6.222)

where Kmic is the micelle–water partition constant for i. [Surf] = [Surf]tot–[CMC],where [Surf]tot is the total added surfactant concentration and [CMC] is the criticalmicellar concentration. Integrating, we obtain

[A][A]0 = exp

(−(Q

VT

)((1 + [Surf]Kmic)

ε+ ε[Surf]Kmic + (1 − ε)ρbKsw

)t

). (6.223)

EXAMPLE 6.31 EXTRACTION OF NAPL RESIDUAL USING A SURFACTANTSOLUTION

For Example 6.30, determine the number of pore volumes required for 99% removalof benzene if a 100mM solution of sodium dodecyl sulfate (SDS) is used for flushing.CMC of SDS is 8mM.

From Chapter 4, logKmic = 1.02 logKow − 0.21 = (1.02)(2.13) − 0.21 = 1.92.[Surf] = 100 − 8 = 92mM = 36 g/L. R′

F = 0.3 + (1.4)(1.6) + (0.3)(101.92)(0.036)

= 3.4. Note that NPV = (Qt)/εVT. Hence,[A][A]0 = exp

(− (1 + [Surf]Kmic) εNPV

R′F

).

For [A]/[A]0 = 0.01,

NPV = (4.605)

(R′

Fε (1 + [Surf]Kmic)

)= 13.

Hence NPV is reduced to 13 from 38 with pure water.

6.4.2.2 In Situ Soil Vapor Stripping in the Vadose Zone

In the vadose zone, applying vacuum using a pump can remove contaminant (Wilsonand Clarke, 1994). Figure 6.63 is a schematic of a soil vapor stripping well. Volatilecontaminants desorb into the pore air that is pulled out of the subsurface. The mecha-nism of removal is similar to that for in situ water flushing in the saturated zone. Justas in the case of the saturated zone, the vadose zone can be considered to be a CSTRand the change in concentration in pore air is obtained from a mass balance between

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all the phases in the subsurface. The contaminant will be distributed between poreair, porewater, solids, and the NAPL phases. The total mass is given by

Mtot = εaVtot [A]g + εwVtot[A]gKAW

+ ρbKSA [A]g Vtot + εNVtotKNA [A]g, (6.224)

or concentration by

[A]tot = Mtot

Vtot=(εa + εw

KAW+ ρbKSA + εNKNA

)[A]g, (6.225)

whereKSA is the soil–air distribution constant for the chemical andKNA is the NAPL–air distribution constant. Note that KSA = KSW/KAW and KNA = (ρN/[A]∗KAW),where ρN is the density of the NAPL and [A]∗ is the aqueous solubility of thecontaminant.

Using the CSTR assumption for the vadose zone between the contaminant sourceand the extraction well, we have the following:

rate in = rate out+ accumulation (air+ solid+water+NAPL).

Qg[A]ing = Qg[A]g + Vtot

d[A]tot

dt. (6.226)

Since [A]ing = 0, the resulting differential equation can be solved with the initialcondition that [A]g = [A]0 at t = 0 to obtain

[A]g

[A]0 = exp

(− Qg

RFVtott

). (6.227)

Activatedcarbon bed AirDemister

NAPL

Zone of saturation

Vadose zone

Flowlines

Blower

FIGURE 6.63 Schematic of a soil vapor stripping well.

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where RF = εa + (εw/Kaw) + ρbKSA + (εN/KNA). The number of pore volumesflushed in time t is

NPV = Qgt

εaVtot= −

(RF

εa

)ln

([A]g[A]0

). (6.228)

The volumetric flow rate of air is given using Darcy’s law (Wilson and Clarke, 1994)

Qg = (Axεaκd)(P2

0 − P21)

2XLP1, (6.229)

where κd is the permeability of soil (Darcy’s constant, m2/atm s), P0 is the inletpressure (1 atm), P1 is the outlet pressure (atm), XL is the distance between the inletand exit (m), and Ax is the area normal to the flow (m2).

Note thatNPV is directly proportional toRF and is logarithmically related to degreeof removal [A]g/[A]0.

EXAMPLE 6.32 EXTRACTION OF VAPORS FROMTHE VADOSE ZONE

Estimate howmuch o-xylene vapors can be removed from 1m3 of the vadose zone of anaquifer that has a total porosity of 0.6 in 1 day. The water content is 10% and the xylenecontent is 20%.Darcy’s constant is 0.01m2/atm s.A vacuum of 0.8 atm is applied acrossa 10-m soil length. The soil organic fraction is 0.02 and bulk density is 1.4 g/cm3. Thedensity of xylene is 0.9 g/cm3 and has an aqueous solubility of 160μg/cm3.To calculate RF, εa = (1 − 0.1 − 0.2)(0.6) = 0.42, εw = 0.06, εN = 0.12. Kaw =

0.215,Koc = 102.3, Ksw = φocKoc = 4.0 P kg. KNA = (0.9)/(160)(0.215)(0.001) =26, KSA = 4/(0.215) = 18. Hence RF = 29. Qg(1)(0.42)(0.01)[(12 − (0.2)2)/(0.2)](1/20) = 0.001m3/s. Hence [A]g/[A]0 = exp

(−((0.001)(8.64 × 104)/(29)(1)

))=

0.05. Thus removal from pore air is 95%.As observed in Chapter 4, air–water partition constant KAW increases with tem-

perature, whereas KSA and KNA decrease with temperature. Hence with increasingT , retardation factor decreases and the efficiency of soil vapor extraction increases.Therefore, heat treatment of soils will accelerate soil–vapor stripping (Wilson andClarke, 1994).

6.4.2.3 Incineration for Ex Situ Treatment of Soils and Solid Waste

Thermal processes are used to destroy organic species in soils and solid waste.There are two types of incinerators employed, namely, rotary kiln and direct flameincinerators. Rotary kiln incinerators are ideal for soil and solid waste combustion.The basic design and operation of a rotary kiln incinerator is described in detail byReynolds, Dupont, and Theodore (1991).

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EXAMPLE 6.33 EFFICIENCY OFA HAZARDOUS WASTE INCINERATOR

A chlorinated hydrocarbon waste is to be incinerated in a rotary kiln at a temperature of1500◦F. The activation energy for the reaction was observed to be 50 kcal/mol and thepre-exponential factor was 3 × 107 P/s. Laboratorywork has shown that the combustiongas flow rate required was 500 SCF/lb at 70 ◦F. If the feed rate is 1000 lb/h at a facevelocity of 30 ft/s and a kiln residence time of 1 s is desired, what should be the lengthand diameter of the incinerator. What is the efficiency of the reactor if it were a PFR?

From the given gas flow rate and feed rate we get feed flow rate at 70 ◦F.Qg = (500 ft3/lb)(1000 lb/h) (1/60 h/min) = 8333 ft3/min. Feed flow rate at 1500 ◦Fis given by Qg = 8333 (TF/Ti) = (8333)(1500 + 460/70 + 460) = 30, 817 ft3/min.We know that Ug = Qg/A = Xs/τR, where Xs is the length of the incinerator andτR is the mean residence time. If the kiln is cylindrical, A = π(D2/4). HenceQg = π(D2/4)(Xs/τR) or D = √4QgτR/πXs. Since UgτR = Xs = (30)(1) = 30 ft,we have D = √

(4)(30817)(1)/(60)(3.14)(30) = 4.7 ft. Thus D = 4.7 ft and Xs =30 ft. Efficiency of a PFR is E = 1 − exp(−kτR). We need k at 1500 ◦F. K =(3 × 107) exp[−50, 000/(1.98)(1960)] = 76 s−1. Hence E = 1 − exp[−(76)(2)] = 1.Complete destruction of the material is possible.

6.5 APPLICATIONS OF CHEMICAL KINETICS INENVIRONMENTAL BIOENGINEERING

Our world is teeming with various biological species. Microorganisms lie at the baseof our evolutionary chain and have played a critical role in changing the earth’senvironment. They possess the ability to transform complex molecules. It has longbeen known that microorganisms are at work in composting waste into manure. Theyare also critical in treating sewage and wastewater. Table 6.15 lists typical exampleswhere microorganisms play a role in environmental engineering.

Soils, sediments, and groundwater contain amilieu of naturally occurringmicroor-ganisms that can break down toxic compounds. By providing sufficient nutrients and

TABLE 6.15Examples of Bioengineering in Environmental Processes

Subsurface bioremediation Addition of microbes to waste materials (groundwater,soils, and sediments) mostly in situ

Wastewater treatment Removal of organic compounds from industrialwastewaters, activated sludge, and trickling filters

Land farming Petroleum exploration and production waste treatment(above ground)

Enhanced NAPL recovery Introduction of bacterial cultures in subsurface soils forsimulation of surfactant production

Oil spill cleanup Microorganisms for oil spill biodegradationEngineered microorganisms To treat special waste materials (highly toxic and

refractory compounds)

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the necessary conditions to initiate their activity, they can act on any given typeof waste.

The ultimate products of biodegradation by microorganisms are inorganic com-pounds (CO2 and H2O). This kind of a complete conversion process is calledmineralization. The following questions regarding mineralization become important:(a) Why do microorganisms transform compounds? (b) What mechanisms do theyemploy in breaking down complex molecules? (c) How do the microorganisms affectchemical transformation rates?. The first question can be answered from thermody-namic principles. The second question needs principles from biochemistry, which wewill not pursue here. The third question concerns chemical kinetics, which is the focusof this section.

Thermodynamics provides clues to the energy of microbial processes. Considerthe oxidation of glucose:

C6H12O6 + 8O2 −→ 6CO2 + 12H2O + 2870 kJ. (6.230)

This is called an exergonic reaction. For every 1mol of glucose consumed, an organ-ism can obtain 2870 kJ of energy. Endergonic reactions, on the other hand, requireorganisms to consume energy from the surroundings for conversion of chemicals.Microorganisms (living cells) couple exergonic and endergonic reactions to accom-plish the goal of lowering the free energy. This is done using intermediate chemicalsthat can store energy released during an exergonic process, and transfer the storedenergy to the site where an endergonic process takes place. This is the basis of liv-ing cell metabolism. Table 6.16 gives the magnitudes of ΔG0 for some commonmicrobial-mediated environmental processes.

To accomplish the types of processes discussed above, living cells make useof intermediates such as adenosine triphosphate (ATP) and guanosine triphosphate(GTP). By far,ATP is the most important and is the universal transfer agent of chemi-cal energy-yielding and energy-requiring reactions. ATP formation will store energy,whereas its hydrolysiswill release energy.Organisms that have adenosine diphosphate(ADP) will first convert it toATP by the addition of a phosphate group, which requiresenergy to be consumed. This is called phosphorylation. Subsequent hydrolysis ofATPtoADP releases the stored energy (ΔG0 = −30 kJ/mol). During a redox process, the

TABLE 6.16A Sampling of Microbially Mediated Environmental Redox Processes

Reaction −ΔG0(w) (kJ/mol)

Fermentation: 14CH2O + 1

4H2O � 14CO2 + 1

2H2 (g) 1.1

Aerobic respiration: 14CH2O + 1

2O2 � 14CO2 (g) + 1

4H2O 119

Nitrogen fixation: 16N2 (g) + 1

3H+ + 1

4CH2O � 13NH

+4 + 1

4CO2 14.3

Carbon fixation: 14CO2 (g) + 1

2H2O � 14CH2O + 1

4O2 (g) −119

Note: CH2O represents 1/6C6H12O6 (glucose). ΔGσ(w) was defined in Section 5.8 as the differencein pe0(w) between the oxidant and reductant.

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electrons that are transferred between compounds by microorganisms also find homein intermediate compounds. Enzymes responsible for the redox transformations cap-ture electrons from electron-rich compounds (dehydrogenases) and store them inintermediate co-enzymes such as nicotine adenine dinucleotide (NAD). A typicalreaction involving the NAD/NADH pair is NAD+ + 2H+ + 2e− � NADH + H+.

The synthesis of pyruvic acid by the partial oxidation of glucose is called gly-colysis. This reaction is a convenient one to show how ATP mediates in a metabolicprocess. Figure 6.64 shows the partial oxidation process. The process starts with theconversion of glucose to glucose-6-phosphate in which one ATP molecule is lost.Subsequent rearrangement to fructose-6-phosphate and loss of another ATP givesfructose-1,6-diphosphate. At this point there is a net energy loss. However, duringfurther transformations, two molecules ofATP are gained and hence overall energy isstored during the process. The conversion of glyceraldehyde-3-phosphate to pyruvicacid is the most energy-yielding reaction for anaerobic organisms. This ATP/ADPcycle is called the fermentative mode of ATP generation and does not involve anyelectron transport.

The complete oxidation of glucose should liberate 38molecules ofATP, equivalentto the free energy available in glucose. If that much has to be accomplished, theelectrons generated during the process should be stored in other compounds thatthen undergo reduction. Electron acceptors generally used by microbes in our naturalenvironment include oxygen, nitrate, Fe(III), SO2−

4 , and CO2. It is to mediate thetransfer of electrons from substrate to electron acceptors that the microorganismsneed intermediate electron transport agents. These agents can also store some ofthe energy released during ATP synthesis. Some examples of these intermediatesare cytochromes and iron–sulfur proteins. The same function can also be performedby compounds that act as H+-carriers (e.g., flavoproteins). The redox potentials ofsome of the electron transport agents commonly encountered in nature are given inTable 6.17. Chappelle (1993) cites the example of E. coli that uses the NADH/NADcycle to initiate redox reactions that eventually releases H+ ions out of its cell. TheNADH oxidation to NAD in the cytoplasm is accompanied by a reduction of theflavoprotein that releases H+ from the cell to give an FeS protein which furtherconverts to flavoprotein via acquisition of 2H+ from the cytoplasm, and in concertwith coenzymeQ releases 2H+ out of the cell.The resulting cytochromeb transfers theelectron tomolecular oxygen formingwater.Thenet result is the use of the energy fromredox reactions to transport hydrogen ions out of the cell. The energy accumulatedin the process is utilized to convert ADP to ATP. The entire sequence of events ispictorially summarized in Figure 6.65 and is sometimes called chemiosmosis, theprocess of harnessing energy from electron transport. More details of these schemesare given in advanced textbooks such as by Schelegel (1992). The main point of thisdiscussion is the part played by electron transport intermediates and ATP synthesisin the metabolic activities of a living cell. Thus, microorganisms provide an efficientroute by which complex molecules can be broken down. The entire process is drivenby the energy storage and release capabilities of microorganisms that are integralparts of their metabolism.

Thermodynamic considerations of the energy of reactions mediated by micro-organisms can provide us information on the feasibility of the transformations. The

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ATPGlucose

Glucose-6-phosphate

Fructose-6-phosphate

Fructose-1,6-diphosphate

ADP

ATP

ADP

Glyceraldehyde-3-phosphate

Diphosphoglyceric acid

3-phospho-glyceric acid

2-phospho-glyceric acid

Phosphoenol pyruvic acid

Pyruvic acid

ATP

ADP

ATP

ADP

FIGURE 6.64 Partial oxidation of glucose to pyruvic acid.

activity of microorganisms will be such that only those with negative overall freeenergies will proceed spontaneously. Those with positive free energies will needmediation byATP andNAD-type compounds.Aswe have seen, biochemical reactionsare thermodynamically feasible if the overall free energy change is negative. However,even ifΔG is negative, not all of the reactions will occur at appreciable rates to be ofany use in a living cell. The answer to this is catalysis caused by the enzymes presentin all living organisms. In Chapter 5, we derived the rate laws for enzyme catalysis.In the following section, we will apply those expressions to study reactor design inenvironmental engineering.

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TABLE 6.17Redox Potentials for ElectronTransport Agents in BiochemicalSystems

Component E0H (V)

NAD +0.1Flavoprotein +0.34Cytochrome c +0.69Oxygen +1.23

Outersolution CytoplasmMembrane

NADH + H+

NAD+

4H+

FlavoproteinFeS protein

Co-enzyme QCytochro me bCytochrome o

0.5O2 + 2H+

2H+

H2O4H+

ADP

H+H+

ATP

FIGURE 6.65 Electron transport and ATP synthesis in a living cell.

6.5.1 ENZYME REACTORS

Biochemical reactors are used in environmental engineering not only for ex situwastetreatment operations, but also in situ waste site remediation processes. Those thatemploy living cells (producing enzymes) are termed fermentors, whereas those thatinvolve only enzymes are called enzyme reactors. We shall first describe the enzymereactor, but the reader is reminded that the same principles are also applicable tofermentors (Bailey and Ollis, 1986; Lee, 1992).

There are three types of enzyme reactors similar to conventional chemical reac-tors such as described in Section 6.1. These are (i) a batch reactor; (ii) a PFR; and(iii) a CSTR. In addition, we will also discuss a convenient operational mode appli-cable to each of the reactors. This is the process of immobilizing (anchoring) theenzymes so that contact between the enzyme and substrate is facilitated.

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6.5.1.1 Batch Reactor

This is the simplestmodeof an enzyme reactor or fermentor.The equation representingthe change in concentration of substrate with time is

−d[S]dt

= Vmax[S]Km + [S] , (6.231)

where Michaelis–Menten kinetics is assumed to hold. Rearranging and integrating,we obtain

1

tln

( [S]0[S])

= Vmax

Km−( [S]0 − [S]

Km

)1

t. (6.232)

This is the integrated rate equation for an enzymatic reaction. For a batch reactor,a plot of (1/t) ln ([S]0/[S]) versus ([S]0−[S])/t should yield −1/Km as the slopeand Vmax/Km as the intercept. From these the Michaelis–Menten parameters can beevaluated. If we define the degree of conversion, χ, such that [h] = [S]0(1 − χ), wecan write

1

tln

(1

1 − χ)

=(Vmax

Km

)−( [S]0χ

Km

)1

t. (6.233)

The above equation gives us the time required for a desired conversion in a batchreactor.

EXAMPLE 6.34 PESTICIDE DISAPPEARANCE FROM SOIL INA BATCH REACTOR

Meikle et al. (1973) performed an experiment designed to estimate the biological decom-position of a pesticide (4-amino-3,5,6-trichloropicolinic acid or picloram) from a soilmatrix. The experiment was conducted in a batch reactor. The following results wereobtained after 423 days of treatment:

[S]0 (ppmw) [S] (ppmw)[S]0 − [S]

(t = 423 days)1t

ln[S]0[S]

3.2 1.56 0.003877 0.0016983.2 1.76 0.003404 0.0014131.6 0.51 0.002577 0.0027031.6 0.69 0.002151 0.0019880.8 0.24 0.001324 0.0028460.8 0.21 0.001395 0.0031620.4 0.12 0.000662 0.002846

continued

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0.4 0.094 0.000723 0.0034240.2 0.029 0.000404 0.0045650.2 0.026 0.000411 0.0048230.1 0.01 0.000213 0.0054430.1 0.013 0.000206 0.0048230.05 0.007 0.000102 0.0046480.05 0.005 0.000106 0.005443

A plot of (1/t) ln([S]0/[S]) versus ([S]0 − [S])/t was obtained as shown in Figure 6.66.The slope of the line was −0.96647, which gives Km = −1/slope = 1.03 ppmw.The intercept was 0.004771, which meant Vmax/Km = 0.004771. Hence Vmax =0.0049 ppmw/day. The correlation coefficient for the straight line was 0.7852. Thisexample should serve to illustrate how a batch reactor can be used to obtain theMichaelis–Menten constants for a biodegradation process.

1 ¥ 10–3

2 ¥ 10–3

3 ¥ 10–3

4 ¥ 10–3

5 ¥ 10–3

6 ¥ 10–3

0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035

(1/t)

ln ([

S]0

/ [S]

)

([S]0 – [S])/t

Y = M0 + M1*X0.0047709M0

–0.96654M1

0.78522R2

FIGURE 6.66 A linearized plot of picloram degradation rate in an enzyme batchreactor.

6.5.1.2 Plug-Flow Enzyme Reactor

The equation describing a PFR for an enzyme reaction is identical to the one describedin Section 6.1.1.3.

rS = −Find [S]

dV, (6.234)

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where Fin is the influent feed rate and V is the volume of the reactor. Thus,

Vmax[S]Km + [S] = −Fin

d [S]

dV. (6.235)

Rearranging and integrating, we obtain

[S]0 − [S]ln ([S] / [S]0)

= Vmax

{τ

ln ([S] / [S]0)

}− Km, (6.236)

where τ = V/Fin is the residence time for the substrate in the reactor. A plot of([S]0 − [S]/ ln ([S] / [S]0)) versus τ/ ln ([S] / [S]0) can be used to arrive at Vmax andKm from the slope and intercept, respectively. The total volume of the reactor requiredfor a given removal is

V =(Fin

Vmax

)[Km ln

( [S]0[S])

+ [S]0 − [S]]. (6.237)

6.5.1.3 Continuous Stirred Tank Enzyme Reactor

If the influent and effluent rates are matched, then from Section 6.1.1.2, the followinggeneral equation should represent the behavior of a CSTR:

Vd[S]dt

= Fin([S]in − [S]) − VrS, (6.238)

where rS = Vmax[S]/Km + [S]. At steady state d[S]/dt is zero, and hence we have

[S] − Km + Vmaxτ[S]([S] − [S]in)

, (6.239)

where τ = V/Fin. From the above equation, one can also obtain the size (volume) ofa reactor required for a specific enzymatic degradation of the substrate.

V =(Fin

Vmax

)(1 − [S]in

[S])

(Km + [S]). (6.240)

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EXAMPLE 6.35 MICROBIAL GROWTHAND SUBSTRATE KINETICS INA CSTR

A microbial culture in a CSTR is a very efficient method for degrading substrates.The technique has been widely employed in the biotechnology field. It has also beenemployed in activated sludge wastewater treatment process. Indigenous bacteria aregrown in a reactor that is fed with the substrate (nutrients) at a specified rate. Theoperating conditions (pH, dissolved oxygen, and temperature) are maintained constantwithin the reactor. The effluent is monitored for both microbial population and substrateconcentration. Figure 6.67 represents the CSTR.The seed culture of microbes is placed within the reactor and nutrients provided to it

in the influent stream that contains the substrate. The submerged bubble generators thatare used to introduce oxygen into the reactor provide mixing. Paddle mixers are alsoused to gently mix the constituents. The biomass in the reactor is uniformly distributedwith no clumping at the vessel wall and the yield factor, yX , is assumed to be constant.A mass balance on the microbial concentration, [X], should include its growth and itsloss via overflow into the effluent and decay within the reactor. We have

Vd[X]dt

= μ [X]V − F [X] − kdecay [X]V . (6.241)

At steady state, d[X]/dt = 0, and hence

μ = F

V+ kdecay. (6.242)

This is called the dilution rate (time−1) of themicrobe. The reciprocalV /F = θ is calledthemean cell residence time. Note that θ is the same as the specific growth rate discussedearlier. Thus, in a CSTR it is an important parameter in understanding microbial growth

Q[X][S]*

[X][S]*

Q[S]in

V

FIGURE 6.67 Schematic of a CSTR for microbial and substrate kinetics.

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kinetics. μ can be large if either V is large or F is small. The maximum value of 1/θ orμ is given by μmax as in the Monod equation.

1

θ+ kdecay = μmax[S]∗

Ks + [S]∗ , (6.243)

[S]∗ = Ks(1 + kdecayθ

)

μmaxθ− (1 + kdecayθ) . (6.244)

We next proceed to write an overall mass balance for the substrate, which gives

Vd[S]dt

= F([S]in − [S]∗) − V

(μ[X]yx

), (6.245)

where yX is the yield factor as defined previously. At steady state, we have

[X] = yx([S]in − [S]∗). (6.246)

Substituting for [S]∗ gives

[X] = yx

{[S]in − Ks

(1 + kdecayθ

)

μmaxθ− (1 + kdecayθ)}. (6.247)

Whenmicroorganisms are grown in the reactor, and their decay proceeds simultaneouslywith constant kdec, the observed yield factor y

obsX has to be corrected to obtain the actual

yX that should be introduced into the above equation.Horan (1990) derived the followingequation for yX :

0

2

4

6

8

10

12

0 0.2 0.4 0.6 0.8

[X] o

r [S]

* /mg

∙ L–1

Time/days

[X]

[S]*

FIGURE 6.68 Microbial growth and substrate decomposition in a CSTR. Ks =15mg/L, [S]in = 15mg/L, Y = 0.6, μmax = 25 d−1.

continued

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yx = yobsX

1 + kdecayθ. (6.248)

As an illustration of the above equation, we shall choose the special case whenkdecay = 0. Other relevant parameters are chosen for a typical bacteria (E. coli) found

in wastewater plants: Ks = 15mg/L, μmax = 25 d−1, yX = 0.6, and [S]in = 15mg/L.Figure 6.68 displays the change in [X] and [S]∗ as a function of the mean residencetime, θ, in the reactor.

It should be mentioned that the above equations only predict the steady-state behav-ior. In actual operation, the unsteady state should be considered whenever the systemexperiences changes in influent concentrations. There will then exist a lag time beforethe substrate consumption and microbial growth approach a steady state. The micro-bial growth lags by several θ values before it adjusts to a new [S]in. This is called thehysteresis effect.

It is also useful to consider here the competition for a substrate S between an organ-ism that utilizes it and other competing complexation processes within the aqueousphase. Consider Figure 6.69. While an enzymatic reaction of species S (an inorganicmetal, for example) occurs via complexation with a cellular enzyme (denoted E), acompeting ligandY in the aqueous phase may bind species S. The cellular concentra-tion of species S is determined by a steady state between cell growth (division) andthe rate of uptake of S. If [S]cell denotes the cellular concentration of S (moles/cell)and μ is the specific growth rate (d−1), then the rate of cell growth is rcell = μ [S]cell.From the reaction scheme in Figure 6.69, the uptake rate of species S is given byruptake = k∗

E [S]tot[E]tot, which is the rate of reaction of species S with the enzymeligand E. This necessarily assumes that the enzyme is in excess of the concentration

E

Aqueous phaseE

E

E

SYSSE

E

E

Cell

kuptake(Fast)

k–Y(Fast)

E

kE(slow)

kY (Fast)

(Quasi-equilibrium)

r = kE [S][E]

FIGURE 6.69 Kinetics of competing biological uptake and complexation in the aqueousphase. (Adapted from Morel, F.M.M. and Herring, J.G. 1993. Principles and Applications ofAquatic Chemistry, NewYork: Wiley.)

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of species S. [S]tot is the total concentration of substrate S in the aqueous phase. k∗E is

the effective complex formation rate constant. For a given ionic strength and pH, k∗E

is fixed. From the steady-state balance of cell growth rate and uptake rate, we obtain

μ = k∗E

( [S]tot

[S]cell

)[E]tot (6.249)

or

ln[S]tot = ln

(1

k∗E

)+ ln

(μ

[S]cell

[S]tot

). (6.250)

If both [S]cell and [E]tot are constant, and k∗E is fixed for a given pH and ionic strength,

then a plot of ln [S]tot versus ln (1/k∗E) should yield a linear plot (Hudson and Morel,

1990). An interesting conclusion of that work is that “. . .complexation kinetics maybe one of the keys to marine ecology. . . .” (Morel and Herring, 1993).

The next two sections will describe some characteristics of bioreaction kineticsfor wastewater treatment and in situ biodegradation of subsurface contaminants. Itis important in these cases to facilitate the contact of pollutant with bioorganisms sothat the reaction is completed in a short time. In the case of wastewater treatment, theenzymes can be isolated from the organisms and used in a completely mixed reactor.This can be made more efficient by attaching the isolated enzymes or biomass ona solid support and using it as a bed reactor. This also facilitates the separation ofbiomass from solution after the reaction. For in situ biodegradation, the organismsshould have easy access to the substrate in the subsurface. In soils and sediments thisis a major impediment. Moreover, nutrient and oxygen limitations in the subsurfaceenvironment will limit the growth and activity of organisms. We shall describe firstthe immobilized enzyme reactor for wastewater treatment. Selected aspects of in situsubsoil bioremediation will follow this.

6.5.1.4 Immobilized Enzyme or Cell Reactor

Enzymes are generally soluble inwater. Hence their reuse after separation from a reac-tor is somewhat difficult. It is, therefore, useful to isolate and graft it onto surfaceswhere they can be immobilized. The surface can then act as a fixed-bed reactor similarto a packed column or ion-exchange column. The enzyme or cell can be easily regen-erated for further use. The reaction can be carried out in a continuous mode where thesubstrate (pollutant) is passed through the reactor bed and the products recovered at theeffluent end. Both chemical and physical methods can be used to immobilize enzymesonto solid substrates (see Table 6.18). The same methods are also useful in immobi-lizing living cells onto solid substrates. Both immobilized enzyme and cell reactorshave been shown to have applications in wastewater treatment (Trujillo et al., 1991).

The use of an immobilized enzyme or cell reactor involves the consideration ofsome factors that are not necessarily addressed in a conventional CSTR. Specifically,we have to consider the different resistances to mass transfer of substrates towardthe reaction site on the immobilized enzyme. The situation is very similar to thatdescribed for sorption and reaction in a natural porous medium (Section 6.1.2.1).There are three sequential steps before a substrate undergoes transformation; these

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TABLE 6.18Immobilization Techniques for Enzymes or Cells

Chemical Physical

Covalent bonding to inactive supports Adsorption onto supportsCopolymerization with support structure Entrapment in cross-linked polymersCrosslinking multifunctional groups Microencapsulation

are indicated in Figure 6.70. The first step, namely, the diffusion in the bulk fluidtoward the liquid boundary layer and the subsequent diffusion through the boundarylayer to the support surface, constitutes the external mass transfer resistance. The laststep, namely, the diffusion to the enzyme reaction site, constitutes the internal masstransfer resistance.

If the enzyme or cell is grafted on the surface of an insoluble support, only thefirst two steps contribute to mass transfer resistance and reaction. In other words, theexternal resistance controls the substrate decomposition. In such a case, the rate oftransfer of solute to the surface of the support is balanced exactly by the substratereaction at the enzyme site:

kLav[S]∞ − [S] = Vmax[S]Km + [S] , (6.251)

where [S] is the substrate concentration at the surface, [S]∞ is the concentration inthe bulk solution, and kL is the overall liquid-phase mass transfer coefficient forsubstrate diffusion across the boundary layer. av is the total surface area per unitvolume of the liquid phase. To simplify further discussion, we define the following

123

Immobilizedenzyme on

solid support

Boundarylayer

Reactionsite

Surface

Bulksolution

Solidsupport

FIGURE 6.70 Steps involved in diffusion and reaction of a substrate at an immobilizedenzyme support.

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dimensionless variables:

[S]∗ = [S][S]∞ ; β = [S]∞

Km; Θ = Vmax

kLav[S]∞ .

The last one, Θ, is called the Dämkohler number. It is defined as the ratio of themaximum rate of enzyme catalysis to the maximum rate of diffusion across theboundary layer. The above equation can now be rewritten as

Θ = (1 − [S]∗)(1 + 1

β[S]∗). (6.252)

IfΘ� 1, the rate is controlled entirely by the reaction at the surface and is given byVmax[S]/Km + [S], whereas if Θ� 1, the rate is controlled by mass transfer acrossthe boundary layer and is given by kLav[S]∞. From the above equation one obtainsthe following quadratic in [S]∗:

([S]∗)2 + α[S]∗ − 1

β= 0, (6.253)

where α = Θ− 1 + (1/β) The solution to the above quadratic is

[S]∗ = α

2

(−1 ±

√1 + 4

βα2

). (6.254)

If α > 0, one chooses the positive root, whereas for α < 0, one has to choose thenegative root for a physically realistic value of [S]∗. As α→ 0, it is easy to show that[S]∗ → (1/β)1/2.

To analyze the lowering of reaction rate as a result of diffusional resistance to masstransfer, we can define an effectiveness factor, ω. The definition is

ω = actual rate of reaction

rate of reaction unaffected by diffusional resistance.

If the reaction is unaffected by diffusional resistance to mass transfer, the rate is givenby the Michaelis–Menten kinetics where the concentration at the surface is the sameas the concentration in the bulk solution. In other words, no gradient in concentrationexists in the boundary layer. Thus,

ω = (Vmax[S]/(Km + [S]))/(Vmax[S]∞/(Km + [S]∞)) = (1 + β)[S]∗/(1 + β[S]∗).(6.255)

ω varies between the limits of 0 and 1.Let us now turn to the case where the resistance to mass transfer is within the

enzyme-grafted solid surface. In this case the diffusion of substrate within the particle

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(a spherical pellet) is balanced by the rate of conversion to products. If we consider aspherical shell of thickness dr, the following mass balance for substrate [S] will hold:

rate of input at (r + dr) − rate of output at r

= rate of consumption in the volume 4πr2 dr.

Ds1

r2∂

∂r

(r2

∂[S]∂r

)= rS, (6.256)

whereDs is the diffusivity of substrate in the fluid within the pellet and r is the radius.rS is given by the Michaelis–Menten kinetic rate law.

Ds1

r2∂

∂r

(r2

∂[S]∂r

)= Vmax[S]Km + [S] . (6.257)

The above second-order differential equation can be cast into a nondimensionalform by using the following dimensionless variables:

[S]∗ = [S][S]∞ ; β = [S]∞

Km; r∗ = r

R,

where [S]∞ is the substrate concentration in the bulk solution and R is the radius ofthe pellet. The resulting equation is

1

(r∗)2∂

∂r∗

((r∗)2 ∂[S]∗

∂r∗

)= R2Vmax

DsKm

( [S]∗1 + β[S]∗

). (6.258)

We now define a Thiele modulus, Φ, as(R2Vmax/9DsKm

)1/2so that the observed

overall rate is expressed in moles per pellet volume per unit time, that is, robsS =

3

RDs

∂[S]∗∂r∗

∣∣∣∣r∗=1

.

The above equation can now be written as

1

(r∗)2∂

∂r∗

((r∗)2 ∂[S]∗

∂r∗

)= (9Φ2)

( [S]∗1 + β[S]∗

). (6.259)

This equation can be solved numerically to get [S]∗ as a function of r∗, and furtherto obtain the effectiveness factor ω as defined by

ω = robsS

rs= (3/R)Ds(∂[S]∗/∂r∗)|r∗=1

(Vmax[S]/Km + [S]) (6.260)

Solutions to ω as a function of Φ are given in sources on biochemical engineering(Lee, 1992; Bailey and Ollis, 1986), to which the reader is referred for further details.

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The above equations are also useful to represent the substrate uptake and utilizationby a microbial cell, if we replace the Michaelis–Menten kinetics by Monod kinetics.This forms the basis of the kinetic analysis of biological floc in activated sludgeprocesses for wastewater treatment. Powell (1967) developed a lumped parametermodel for substrate utilization and kinetics. The rate of substrate utilization was givenby the following equation:

rs =(μmax

ys2σ

)⎡⎣(1 + σ+ [S0]

Ks

)−{(

1 + σ+ [S0]Ks

)2

+ 4[S]0Ks

}1/2⎤⎦, (6.261)

where σ = μmaxδR2/3ysKsDs(R+ δ). δ is the thickness of the stagnant zone of liquidaround a cell of radiusR. [S]0 is the substrate concentration in the bulk fluid.μmax andKs are the Monod kinetics constants. The above equation is identical to that derivedearlier using Michaelis–Menten kinetics.

6.5.1.5 In Situ Subsoil Bioremediation

An important application of environmental bioengineering is the use of specialmicrobes to degrade recalcitrant and persistent chemicals in the subsurface soil andsediment environment. Primarily bacteria, algae, and fungi bring this about. Bacteriaare found in large numbers in the soil; they are small in size comparedwith fungi. Fungiare filamentous in nature and account for a large mass of microorganisms in the soil.

Microbes in the soil can survivewith orwithout oxygen.Accordingly, soilmicrobesare classified as aerobes (requiring oxygen) and anaerobes (require no oxygen).Facultative anaerobes are those that can grow in either environment.Aerobicmicrobesbreak down chemicals into CO2 and H2O. Anaerobes break down chemicals intocompletely oxidized species in addition to some CO2, H2O, and other species (H2S,CH4, SO2, etc.). The biotransformation of chemicals in soil involves several steps,enzymes, and species of many types. Since enzymes are specific to a chemical, thebiodegradation of complex mixtures requires that several species co-exist and exerttheir influence on individual chemicals in the mixture. Microbes act in two ways:(i) they imbibe chemicals and the intracellular enzymes act to degrade the pollutant,and (ii) the organism excretes enzymes (extracellular enzymes) which subsequentlycatalyze the degradation.

The general pathway of degradation of many types of pollutants by specificmicrobes has been elucidated over the past several decades. These aspects aredescribed extensively in a recent book (Alexander, 1994). The specific mechanismsare not the topic of this chapter, whereas we are more interested in the equilibriumand kinetics of biodegradation in the subsoil.

In the earlier section, we noted the kinetics of enzyme catalysis and cell growthin the aqueous environment. In the subsurface soil environment, which is a two-phase system (soil+ porewater), we should expect some similarities with the aqueoussystem. The microbes are attached to the soil surface and they act as a fixed-film(immobilized cell) reactor.

In the subsurface soil environment, the degradation of pollutants and micro-bial growth are most conveniently handled using a Monod-type expression (see

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Section 6.3.1). The substrate degradation rate per unit volume (mol/m3/s) is

rv = 1

yx

μmax[S][X]Ks + [S] , (6.262)

with yX being the yield factor. The term μmax/yX is called the maximum rate ofsubstrate utilization and is represented by a constant, k. The limits of the above rateexpression are of special interest. If [S] � Ks, rv = k [X] and the rate is first order,whereas for [S] � Ks, rv = (k/Ks) [X] [S] and is second order. Further if [X] isconstant, the rate is zero order in [S] at high [S] and first order at low [S].

Other rate expressions that are also based on Monod-type kinetics have beenproposed to describe biotransformations in the subsurface soil environment. Theyare based on the basic premise that as either [X] or [S]→0, rv →0, and that rv shouldincrease linearly with both [X] and [S]. These models comprise two types:

(i) A power law relationship: rv = k[S]n.(ii) A hyperbolic relationship: rv = a[S]

b+ [S] .

Note that the latter resembles Monod kinetics. These expressions do not includemicrobial concentration as a separate parameter and describe the general disappear-ance rates (biotic or abiotic) of chemicals in soils. In most instances, in the subsurfacesoil environment, we have low values of [S] and hence a first-order expression in [S] issufficient. Fortunately, this is the same limiting condition for both the Monod kinet-ics and the empirical rate laws described above. However, the consensus is that atheoretically based approach such as the Michaelis–Menten (or Monod) kinetics ismore appropriate than an empirical rate law whenever empiricism can be avoided. Atheoretical model will offer a greater degree of confidence in the testable conclusionsdrawn from a chemodynamic model.

For a successful biodegradation there are several requirements. These are summa-rized below:

(i) Availability of the chemical to the organism(ii) Presence of the appropriate nutrient(iii) Toxicity and/or inactivation of enzymes(iv) Sufficient acclimation of the microbe to the environment

Whereas the latter three issues are specific to a microbe or a chemical, the first one iscommon to all organisms.

Chemicals can be held in special niches such as micropores and crevices in thesoil structure that are inaccessible to the microbes. Chemicals can adsorb to solidsurfaces and may not readily available in the aqueous phase where the microbes(enzymes) function. In most cases, sorption seems to decrease the rate of degradation.Sorbed molecules may not be in the right conformation to form the requisite enzyme–substrate complex. The sorption site may also be far removed from the microbialcolony. We can think of two separate biodegradation processes: that of the sorbed

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species and that of the truly dissolved species in the aqueous phase. If both are firstorder, rsorbed

v = ρbks [S]ads and rw = kw [S], where ρb is the soil density, [S]ads isthe sorbed mass of S, ks and kw are the respective rate constants. We can write thefollowing expressions for the fraction sorbed and truly dissolved:

f sorbed = kswρs

1 + kswρs,

f w = 1

1 + kswρs,

(6.263)

where [S]ads is the soil sorbent concentration. A composite rate is defined as

r∗ = k∗[S]tot, (6.264)

where [S]tot is the total substrate concentration and

k∗ = f sorbedks + f wkw. (6.265)

The above equation satisfactorily represents the effect of sorption on the overallbiodegradation if it is modeled as a first-order process. Apart from sorption, bio-transformations can also be affected by other environmental factors such as soil pH,temperature, soil moisture, clay content, and nutrients. Valentine and Schnoor (1986)have summarized these issues in an excellent review, to which the reader should referto for more details.

Since bioavailability is a primary limitation in bioremediation, considerable efforthas been expended to facilitate desorption of chemicals from sediment and soil sur-faces. Mobilization via solubilization using surfactant solutions is one of the conceptsunder study. Several workers have suggested that nonionic surfactant solutions used inconjunction with specific microbes can affect the mineralization of several hydropho-bic organics. There are, however, several conflicting reports regarding this approach;in particular the long-term viability of the microbes in the presence of surfactants, thepresence of residual surfactants in the subsurface soil environment, and the degree ofmineralization in the presence of surfactants have been disputed.

6.5.2 KINETICS OF BIOACCUMULATION OF CHEMICALS IN THE AQUATIC

FOOD CHAIN

In Chapter 4, we introduced the bioconcentration factor that defined the equilib-rium partitioning of chemicals between the biota and water. This section is to modelthe kinetics of chemical uptake and dissipation in organisms. Connolly and Thomann(1992) summarized the state-of-the-art in this area.What follows is a concise descrip-tion of their model. There are three important subareas that have to be included insuch a kinetic model, namely, (i) the growth and respiration rates of each organismincluded in the food chain, (ii) the efficiency of chemical transfer across the biologicalmembranes (gills, gut, etc.), and (iii) the rate of dissipation and excretion of chemicalsby the organism.

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To analyze the uptake rate we should first consider the base organisms in the foodchain, namely, phytoplankton and detrital organic material. The uptake by organismsgenerally depends on the ingestion, sorption, metabolism, and growth. For the basein the food chain it is only dependent on sorption and desorption processes. If ks(m3/d/g dry weight) denotes the rate of sorption from the aqueous phase and kd(d−1) the desorption rate, the rate of accumulation within the base organisms isgiven by

d

dt(Worg[A]b

org) = ksWorg[A]w − kdWorg[A]borg, (6.266)

where [A]borg is the concentration of A in the base organism (mol/kg weight of

organism), [A]w is the concentration in the aqueous phase (mol/dm3),Worg is theweight of the organism (kg), ks is a second-order rate constant (dm3/kg d), and kd isa first-order rate constant (d−1). If equilibrium is assumed between the organism andwater, we have

[A]borg

[A]w = ks

kd= KBW (6.267)

as defined in Chapter 4.In this case the change in [A]b

org with t will follow the equation

[A]borg = KBW[A]w(1 − e−kdt). (6.268)

Base of food chain

Excretion

Level 1

Level i

Ingestion

Level nToppredatore.g. human

Sorption/desorption

Pollutantin water

e.g. Fish

e.g. Zooplankton

FIGURE 6.71 Food chain bioaccumulation of a pollutant from water.

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Thus, [A]borg increases in an approximate first-order form to approach the equilibrium

value KBW[A]w.When we move up the food chain (zooplankton → fish → top predator), we have

to consider not only the rate of direct uptake from water by sorption/desorption, butalso by ingestion of the lower species and the portion actually assimilated into thetissues of the animal.Additionally, we also have to take into account themass excretedby the animal (Figure 6.71). Thus, we have three rate processes included in the overalluptake rate: (i) the rate of uptake by sorption, (ii) the rate of uptake by ingestion, and(iii) the rate of loss via desorption and excretion.

EXAMPLE 6.36 APPROACH TO STEADY-STATE UPTAKE IN THE FOOD CHAIN

The earliest attempts at modeling bioaccumulation involved exposing the first level ofaqueous species to organic chemicals and obtaining the rate of uptake. We will nowascertain the approach to steady state in this system. Figure 6.72 treats the organismand water as distinct compartments (CSTRs) between which an organic chemical isexchanged. Neely (1980) described such a model; the following discussion is based onhis approach.As explained earlier, let [A]orgrepresent the moles of chemical A per weight of

organism, and [A]w be the concentration of chemical A in water (mol/dm3 or mol/kg).Expanding the left-hand side of the equation for the rate of uptake

d

dt(Worg[A]org) = Worg

d[A]org

dt+ [A]org

dWorg

dt. (6.269)

If the weight of the organism is constant, dWorg/dt = 0. However, since the organismgrows as it consumes the chemical, its concentration changes in relation to the growth.The overall equation is then

Worgd[A]org

dt= ksWorg[A]w − kdWorg[A]org − [A]org

dWorg

dt. (6.270)

To proceed further, we denote the growth rate of the organisms as

μ = 1

Worg

dWorg

dt(6.271)

Hence,d[A]org

dt= ks[A]w − (kd + μ)[A]org. (6.272)

If the concentration inwater is assumed to change only slightly such that [A]w is constantduring a time interval, τ, we can integrate the above equation to yield

[A]org

[A]w =(

ks

kd + μ)

(1 − e−(kd+μ)t). (6.273)

If μ = 0 and τ is very large, we have [A]org/[A]w = ks/kd = KBW. The approach toequilibrium will be delayed if the organism grows as it feeds on the pollutant A.

continued

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15

10

5

0

Age class (year)

Alewife

Lake trout

PCB

conc

entr

atio

n (u

g/g,

wet

wei

ght)

25

20

15

10

5

0

1 2 3 4 5 6

10 2 3 4 5Age class (year)

6 7 8 9 10

FIGURE 6.72 Comparison of observed PCB concentrations in alewife and laketrout. (Reprinted from Thomann, R.V. and Connolly, J.P. 1984. Environmental ScienceTechnology 18, 68, American Chemical Society. With permission.)

For any species i in the food chain, we have the following rate equation (Connollyand Thomann, 1992):

rate of increase in body burden = rate of sorption + rate of uptake by ingestion

of lower species − rate of excretion.

Let [A]∗org,i be the whole weight body burden in the ith species (mol/organism) andis given as [A]org,iWorg,i. The rate of sorption is given by the product of the uptakerate constant for species i, the organism weight, and the aqueous concentration of thechemical, that is, ksiWorg,i[A]w. This is analogous to oxygen uptake from water. ksihas units of dm3/d kg of organism. An expression for ksi is 10−3WiE/ϕi, where wi isthe total weight of organism i and ϕi is the fraction of lipid content (kg lipid/kg wet

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weight). Ei is the chemical transfer efficiency. A general expression for the rate ofrespiration is (Connolly and Thomann, 1992)

Rel(g/g/d) = βiWγi eρiTeviui , (6.274)

where T is temperature (EC) and ui is the speed of movement of the organism inwater (m/s). βi, γi, ρi, and vi are constants typical to the given species.

The rate of ingestion is given by the product of the chemical assimilation capacityof the organism i on another organism j, denoted as αij, the rate of consumption oforganism i on organism j, denoted as Cij, and the concentration of the pollutant inorganism i, [A]org,i. Cij = pijCi, where pij is the fraction of the consumption of i thatis on j andCi is the weight-specific consumption of i (kg of prey per kg of predator perday). The value of Cij is dependent on the rate of respiration and the rate of growth:C =(R+ μ)/α. The value of α varies from 0.3 to 0.8 and is specific to individualspecies. The rate of uptake via ingestion of lower species is

∑j αijWorg,iCij[A]org,j.

The overall rate of ingestion of a chemical by the species is given by∑n

j αijCij[A]org,j.The summation denotes the uptake of different species j = 1, 2, . . . , n by species i.

The rate of excretion via desorption of the chemical is given by kdiWorg,i[A]org,i,where kdi has units of d−1.

Now, since

d(Worg,i[A]org,j)

dt= Worg,j

d[A]org,j

dt+ [A]org,j

dWorg,j

dt(6.275)

and denoting the growth rate, μi = (1/Worg,i) (dWorg,i/dt

), we have (6.277)

The overall rate equation for bioaccumulation is, therefore, given by

d[A]org,i

dt= ksi[A]w +

n∑j=1

αijCij[A]org,i − kdi[A]org,i − μi[A]org,i. (6.276)

This equation is applied to each age class of the organism. Within each age groupthe various parameters are assumed constant. Since equilibrium is rapidly attainedfor lower levels of the food chain, it may be appropriate to assume no significantchange in concentration with time for such species. This suggests a PSSA for [A]org,i.Therefore, the concentration in the organism i at steady state is given by

[A]∗org,i = kdi[A]w +∑nj=1 αijCij[A]org,i

(kdi + μi) . (6.277)

Connolly and Thomann (1992) employed this equation to obtain the concentrationof PCBs in contaminated fish in Lake Michigan. Figure 6.72 shows the data whereit is assumed that for each age class, constant values of assimilation, ingestion, andrespiration rates are applicable.

If exposure of organism to pollutant A occurs only through the water route, wehave at steady state

[A]∗org,i = ksi

(kdi + μi) [A]w. (6.278)

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Thus, the bioconcentration factor is

KBW = [A]∗org,i

[A]w= ksi

(kdi + μi) (6.279)

Since ksi = 10−3W−γi (Ei/pi), kdi = ksi/Kow (Thomann, 1989) and μi = 0.01W−γ

i ,we can write

KBW = KOW

(1

1 + 10−6(KOW/E)

). (6.280)

The above equation indicates that KBW goes through a maximum at a Kow = 106. AsKow increases, the value of KBW decreases due to decreasing E and increasing μi.

The ideas presented in this section are solely to make the reader aware of theapplications of chemical kinetics principles to obtain the steady-state concentration ofpollutants in a given food chain. Numerous parameters are needed and only averagevalues are justifiable. These limitations are indicative of the state of knowledge inthis area.

6.6 APPLICATIONS IN GREEN ENGINEERING

Manufacturing industries have traditionally relied on what is called “end-of-the-pipe”waste treatment philosophy to take care of the inevitable pollutant releases to theenvironment. This philosophy involves treating the waste stream (air, water, or solidwaste), that is, pollution control, at the end of the manufacturing process so as todecrease the amount discharged to the environment. In many cases, this also ledto a significant portion of valuable materials at trace concentrations being returnedto the environment. As processes became “greener,” more emphasis was placed inrecovering and recycling valuable species from the waste, thus reducing the amountreleased and also reducing the pollution control costs.As pollution regulations becamemore stringent, emphasis was placed on reducing the amount of pollutant produced atthe source itself. Thus was born the concept of “environmentally conscious design”or otherwise called “green engineering” (Allen and Shonnard, 2002). Figure 6.73describes the difference between a “traditional” and “green” process.

To make a process design more environmentally conscious, one has to have meth-ods and tools to properly evaluate the environmental consequences of the variouschemical processes involved and products therein. This means that we have to be ableto quantify environmental impacts and guide product design changes. Figure 6.74shows the various items we need to carry out an environmentally conscious design ofa chemical process.

Green engineering follows a set of core principles laid down in 2003 in whatis known as the “Sandestin Declaration,” which are given in Table 6.19. A def-inition of “Green Engineering” is “the design, commercialization, and use ofprocesses and products that are feasible and economical while minimizing (i) riskto human health and the environment, and (ii) generation of pollutants at the source”(http://www.epa.gov/oppt/greenengineering). Closely aligned with the principles ofgreen engineering are twoother related concepts: (i) sustainability defined as “meeting

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“Traditional” Process

“Green” Process

Energy ChemicalProcess

ChemicalProcess (New)

Pollutioncontrol

Pollutioncontrol

Waste

Products

Release toenvironment

Rawmaterials

Energy ProductsWaste

Release toenvironmentRaw

materialsRecycle

FIGURE 6.73 Differences between a “traditional” and a “green” chemical manufacturingprocess.

Green processdesign

Chemical properties(thermo, kinetics,

transport)

Process designtools

Processoptimization tools

Processsimulation

models

Environmentalimpacts models

Environmentalfate models(fugacity)

FIGURE 6.74 Various steps in an environmentally conscious design of a chemical process.

the needs of the current generation without impacting the needs of future generationsto meet their own needs,” and (ii) Life Cycle Assessment (LCA).

6.6.1 ENVIRONMENTAL IMPACT ANALYSIS

To implement green processes in chemical or environmental engineering processes,we use a hierarchical (three-tiered) approach for Environmental Impact Analysis

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TABLE 6.19The Sandestin Declaration of Green Engineering Principles

Green engineering transforms existing engineering disciplines and practices to those that promote sus-tainability. Green engineering incorporates the development and implementation of technologically andeconomically viable products, processes, and systems to promote human welfare while protecting humanhealth and elevating the protection of the biosphere as a criterion in engineering solutionsTo fully implement green engineering solutions, engineers use the following principles:

1. Engineer processes and products holistically use systems analysis, and integrate environmentalimpact assessment tools

2. Conserve and improve natural ecosystems while protecting human health and well-being3. Use life cycle thinking in all engineering activities4. Ensure that material and energy inputs and outputs are as inherently safe and benign as possible5. Minimize depletion of natural resources6. Strive to prevent waste7. Develop and apply engineering solutions, being cognizant of local geography, aspirations, and

cultures8. Create engineering solutions beyond current or dominant technologies; improve, innovate, and

invent (technologies) to achieve sustainability9. Actively engage communities and stakeholders in the development of engineering solutions

Source: Adapted with permission fromGonzalez, M.A. and Smith, R.L. 2003.Amethodology to evaluateprocess sustainability. Environmental Progress 22, 269–276.

(EIA) (Allen and Shonnard, 2002):

Tier 1: Identify toxicity potential and costs when the problem or process isdefined.

Tier 2: Identify material/energy intensity, emissions, and costs when the recy-cle/separation system is being considered.

Tier 3: Identify emissions, environmental fate, and risk when the overall systemdesign is being considered.

In Tier 1, we make extensive use of the concepts derived from chemical thermody-namics and kinetics. This includes estimating thermodynamics properties of reactantsand products, fugacity models for F&T, kinetic rate constants, and mass transfercoefficients for species. These are then used in Tier 3 relative risk assessments. Thisapproach is a stepwise, hierarchical one and involves the following tasks:

Step 1: Identify reactions and processes.Step 2: Identify inputs/outputs rates of various species—use EPA or other

emission factors, use process flow sheet data.Step 3: Estimate the physicochemical parameters for each species—use thermo-

dynamic correlations, use databases (e.g., EPIWIN).Step 4: Obtain concentrations in various compartments using a multimedia

environmental model (e.g., fugacity model).

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Step 5: Obtain relative risk index (RRI) for various categories such as globalwarming, ozone depletion, smog formation, acid rain, and carcino-genicity.

Step 6: Adjust parameters: (i) go to Step 2 if risks are large and repeat calcula-tions; or (ii) go to Step 1 if reactant substitutions are possible and repeatcalculations.

The RRI for a particular impact category is the sum of contributions for all chemicalsfrom a process weighted by their emission rates (Allen and Shonnard, 2002):

Icategory =N∑i=1

(dimensionless risk index for the category)i × mi, (6.281)

where i = 1, . . . ,N is the number of compounds (species) involved. The risk indexesare generally calculated for the following five categories: global warming poten-tial (GWP), smog formation potential (SFP), acid rain potential (ARP), inhalationtoxicology potential (INHTP), and ingestion toxicology potential (INGTP). The firstthree potentials are available from various resources for specific chemicals and aresummarized inAppendix D inAllen and Shonnard (2002). The last two (noncarcino-genicity toxicity indices) are obtained from the reference dose (RfD) and referenceconcentration (RfC) for ingestion (oral) and inhalation, respectively. Similarly, the car-cinogenicity indices for ingestion (oral) or inhalation pathways are also defined usingthe slope factors instead of the reference doses (see also Chapter 1 for descriptionsof these terms).

To obtain the INHTP or INGTP, one has to first estimate the concentration of achemical in air or water environments using the fugacity model. For examples, letus assume we know the concentration of the chemical in water and air. Once this isobtained, we can obtain the risk index as follows:

INHTPi = (Cia/RfCi)

(Cbm,a/RfCbm), (6.282)

where bm stands for a benchmark chemical.

INGTPi = (Ciw/RfCi)

(Cbm,w/RfCbm). (6.283)

The carcinogenic potentials (both inhalation and ingestion) are determined bysubstituting RfC and RfD with cancer SFs (see Chapter 1 for definition).

To illustrate the above methodology, an example is given below.

EXAMPLE 6.37 ILLUSTRATION OF ENVIRONMENTAL RISK EVALUATION FOR A

CHEMICAL MANUFACTURING PROCESS

Let us consider a plastics manufacturing process that has to treat and recover materialsfrom a gaseous stream.The gaseous stream consists of benzene, toluene, and byproducts

continued

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(CO2 and NOx). Figure 6.75 is the overall process flow sheet. The stream that consistsof 200 kg/h of benzene and toluene is introduced first into an absorption column wherenonvolatile absorption oil is used to strip the benzene and toluene from the stream. Theoil is distilled to separate the benzene and toluene for reuse within the manufacturingprocess. The oil is then recycled into the absorption column. The primary emissionsfrom the process occur through the vents, storage tanks, and other fugitive sources andconsist of traces of benzene, toluene, and utility gases (CO2, NOx). These emissionsare dependent on the oil flow rate through the absorption column.We need to determinethe overall environmental impacts from this process and how they can be minimizedas we change the process parameter, namely, oil flow rate. The methodology describedhere follows closely that given in Allen and Shonnard (2002).

VentVent(recoveryof Bz, Tol) Oil

Products

Benzene,toluene200 kg/h

DistillationAbsorption

FIGURE 6.75 Overall process flow sheet for a typical chemical manufacturingprocess.

Table 6.20 lists the various emission rates resulting from the process schematic givenin Figure 6.75. Ideally, one would use a commercial process simulator such as HYSIS,ASPEN, or SimSci to carry out mass and energy balances on the various system com-ponents and calculate the emission rates of different species generated in the process.In doing so, one will have to use the standard EPA emission factors that are listed forspecific pollutants. We have circumvented that process to generate Table 6.20 usinghypothetical values for the process as an illustration of the calculation of environmen-tal impact indices. The emission rate for benzene rapidly decreases with increasing oilflow rate, whereas that for toluene decreasesmuchmore slowly.With increasing oil flowrates, the utility systems emit more of CO2 and NOx . Typical physicochemical proper-ties required for the calculationwill have to be obtained from thermodynamic argumentspresented in Chapters 2 and 3 or from standard tables such as given in Appendix 1.

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TABLE 6.20Typical Air Emission Rates from the Absorption/Distillation ProcessDescribed in Figure 6.75

Air Emission Rate (kg/h)

Absorption Oil Flow Rate (kg mol/h) Benzene Toluene CO2 NOx

0 200 200 0 050 1 150 150 0.2100 0.05 120 300 0.5250 0.05 10 700 1.0500 0.05 0.2 1500 2.0

Table 6.21 lists the various RRIs for individual chemicals needed for the calculationof overall environmental risk index for each category. Note that the values of GWP,SFP, and ARP for chemicals are listed in various sources (Allen and Shonnard, 2002).Similarly, the RfD and RfC values for benzene and toluene are listed by the Agencyfor Toxic Substance Disease Registry (http://www.atsdr.cdc.gov/). We can now utilizeEquations 6.282 and 6.283 for the calculation of IGWP, ISFP, and IARP using the abovevalues for various values of the oil flow rate given in Table 6.20. For the calculationof IINHTP and IINGTP, using Equations 6.282 and 6.283, the benchmark chemical usedis benzene in Table 6.21. To do so, we have to calculate the concentrations in air andwater compartments for benzene and toluene in the general environment resulting fromreleases from the process. This is done using a fugacity model that is described inChapter 3. Typically, a Level III model is required; however, we use a Level II modelhere as an illustration. We consider a three-compartment model (air, water, and soil) asgiven in Example 3.6. A total emission rate of 0.5mol/h each of benzene and tolueneto air is assumed so that E = 1mol/h. All other values are same as given in Example3.6. Note that the value of E will not affect the calculation of environmental indices inEquations 6.282 and 6.283 since the concentration ratios of benzene and toluene areused. The outputs from Level II fugacity calculation are given in Table 6.22 along withthe calculated values of IINGTPi and IINHTPi for benzene and toluene.

TABLE 6.21Relative Risk Index for Each Chemical in the Process Described inTable 6.20

Category Benzene Toluene CO2 NOx

Global warming potential, IGWPi 3.4 3.3 1.0 40Smog-forming potential, ISFPi 0.42 2.70 — —Acid rain potential, IARPi — — — 1.07RfD for ingestion (mg/kg/d) 0.004 0.2 — —RfC for inhalation (mg/m3) 0.03 0.4 — —

continued

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TABLE 6.22Results from Fugacity Level II Calculations for Air andWater Concentrations

K ′aw Ksw Cia Ciw

Chemical (Pa−1 m−3mol−1) (L−1kg) (mol−1m3) (mol−1m3) IINGTPi IINHTPi

Benzene 557 5 0.11 0.50 1.0 1.0Toluene 676 20 0.10 0.42 42 12

Figure 6.76 shows the results of the calculations of risks for various categories as afunction of the process variable, the absorption oil flow rate. IGWP decreases consid-erably and reaches a minimum at about 50 kgmol/h of oil flow rate. The optimumtherefore with respect to global warming is to work around this value. Initially whenthere is no oil flow, all of the VOCs are released to the air and gradually convertedto CO2 elevating the global warming impact. However, as the oil flow rate increases,more of the VOCs are recovered and recycled reducing the global warming impact.As the oil flow rate increases above 50 kgmol/h, the process utilities increase theemission of CO2 and therefore negate any advantages due to increased recovery ofVOCs. The smog formation potential, ISFP, appears to decrease considerably above100 kgmol/h of oil flow rate. Hence, the maximum reduction in the smog formationpotential can be realized at the highest oil flow rate achievable. This also coincides

0.1

1

10

100

1000

104

0 100 200 300 400 500 600

GWPSFPARPINGTPINHTP

Inde

x

Oil flow rate/kg mol ∙ h–1

FIGURE 6.76 Calculated risks for different categories as a function of the absorptionoil flow rate.

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with the highest recovery of toluene from the gaseous stream. The acid rain potential,IARP, shows a slow steady increase with oil flow rate and relates only to the increasedprocess utilities emission of NOx at high oil flow rates. The optimum value appearsto be about 100 kgmol/h for this case. The noncarcinogenicity indices, IINGTP andIINHTP, show similar behavior and decrease considerably above 100 kgmol/h oil flowrate. The reductions in inhalation and ingestion toxicity are both related to the removalof benzene and toluene from the process waste stream that would otherwise enter thegeneral environment. Thus, a single value of oil flow rate, which minimizes the overallenvironmental index, appears to be not realizable. However, a value between 50 and100 kgmol/h appears to be a good compromise for the process.

Note that in the above illustration of green engineering principles, we have utilizedconcepts from Chapters 2 and 3 in the estimation of physicochemical propertiesof chemicals and the fugacity Level II model to obtain the environmental F&T ofchemicals. Additionally, in the more general case of fugacity Level III model, wewill also have to utilize rate constants in air, water, and soil that were the topics ofChapter 5. This should serve as another application of the basic concepts of chemicalthermodynamics and kinetics in environmental engineering that we have touchedupon in this book.

6.6.2 LIFE CYCLE ASSESSMENT

Another important aspect of green engineering is the so-called LCA. This processinvolves following the so-called “cradle-to-grave” cycle of any component within acomplete manufacturing, usage, and disposal of a product. This is best illustrated inFigure 6.77.A material such as, for example, a metal is used in several manufacturingoperations. One follows themetal from its rawmaterial supply chain to the productionprocess, and further to the usage of the product in several forms. Ultimately, themanufactured products are disposed after use. During the various stages, we have the

Emissions to air, soil or water

Rawmaterialsupply

Production DisposalInputs Products

Recycle Remanufacture Reuse

Life cycle analysis (LCA)

Usage

FIGURE 6.77 Steps in the life cycle assessment for a metal.

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Ore andconcentrate107

Production Ore andconcentrate

APT20

Tungstencarbidepowder

10F & M

220(EUSM)159(FPM) 156

(EUSM)78(FPM)

19Phantom low

Environment

Import/Export

Tungsten: 1975 to 2000All values in Gg W

Secontary production(prompt and obsolute scrap)

Processingwaste

34922

Tailings3

Ore29

51011–5051–100101–125

125+

Ore

All values in Og W

Stock (industrialand governmental)

Use

242130

Other7

Finishedproducts5

Scrap23

Wastemanagement

FIGURE 6.78 Overall cycle and material flows for tungsten. (Reprinted with permissionfrom Harpu, E.M. and Graedel, T.E. 2008. Euvironmental Science and Technology, 42, 3839.Copyright (2008) American Chemical Society.)

opportunity to reuse the product several times if recovered from the disposal process.From the disposedmaterial, some fraction of the metal can be recovered in its originalform and used to remanufacture the product. Finally, the recovered metal may also beused as a raw material in the production chain. During any of the product life cycleprocesses, some fraction of the metal (in any oxidized, reduced, or native form) maybe released to the environment (air, soil, or water).Amaterial flow analysis during theentire life cycle of the metal can be done to ascertain the overall fate of the metal inany manufacturing process. The overall cycle and material flow analysis for tungstenin the United States during the years 1975–2000 is given in Figure 6.78. Once the rateof input into the environment is obtained, a fugacity model can be used to ascertainthe overall environmental fate of the metal in various compartments.

PROBLEMS

6.12 Consider a lake with a total capacity of 1010 m3 with an average volu-metric water flow rate of 103 m3/d. The lake receives a wastewater pulseof pesticide A inadvertently released into it by a local industry due to an

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explosion in a manufacturing unit. The input pulse has a concentration of1000μg/dm3. The pesticide is degradable in water to an innocuous prod-uct B with a rate constant of 0.005 d−1. Determine how long it will takefor 90% of the pollutant to disappear from the lake. Repeat the calcula-tion if in addition to the reaction A → B, the pesticide A also adsorbs tosuspended particles (∼1000mg/L) in the lake and settles out of the waterinto the sediment. Ksw for A is 1000 L/kg.

6.23 Consider the troposphere (total volume V ) to be a well-mixed CSTR.A man-made chemical A is released from surface sources at a constantrate of NA (Tg/y). Assume that the northern and southern hemispheresshow uniform concentrations in A. The initial concentration is [A]0 =0 Tg/m3. If the only removal process in the troposphere is a first-orderreaction with a rate constant kA (y−1), what is the concentration of Ain the troposphere at any given time? (Note: (1/[A])(d[A]/dt) = (1/τ),where τ is the characteristic time of A in the reservoir. kA = 1/τr , whereτr is the characteristic time for reaction ofA and NA/V [A] = 1/τd, whereτd is the characteristic time for inflow ofA. Express [A]/[A]0 as a functionof τr and τd alone.)

6.32 Constructed wetlands have been proposed as wastewater treatment sys-tems. Plants produce enzymes (extracellular and intracellular) that breakdown organic pollutants. The system involves a gentle down-flow of waterover sloping vegetation with a slow accompanying water evaporation rate(20mol/m2 min). Consider a 1 km × 0.1 km × 0.2 km deep wetland. Anonvolatile pesticide run-off (concentration 10−2 mol/m3) is introducedinto the wetland at a flow rate of 0.05m3/min and undergoes a first-orderbiochemical reaction with a rate constant of 10−6 min−1. What is thesteady-state concentration of the pesticide at the outflow from thewetland?Note that the water can be considered to be in plug flow.

6.42 A type of waste combustion incinerator is called a rotary kiln. It is acylindrical reactor inclined to the horizontal and undergoing slow axialrotation to achieve adequate mixing of the reactants. Consider a pesticide(e.g., DDT) containing waste incinerated in the kiln at a temperature of300◦C. If its decomposition is first order with a rate constant, k of 15 s−1

at 300◦C, determine the reactor volume required for 99.99% destructionof DDT from the waste stream flowing into the kiln at a rate of 0.01m3/s.Consider the kiln to be first a CSTR and then repeat the calculation fora PFR.

6.53 Consider a barge containing a tank full of benzene that develops a leakwhile in the Mississippi river near Baton Rouge, Louisiana, and spilled10,000 gallons of its contents into the river. The stretch of the Missis-sippi river between Baton Rouge and New Orleans is 100 miles long.Assume that the pollutant is in plug flow and is completely solubilized inwater. Estimate how long it will take to detect benzene in New Orleans,the final destination of the river. Assume that benzene does not transformvia biodegradation or chemical reactions. Benzene may adsorb to bottomsediments. The following parameters are relevant to the problem: Volu-metric flow rate of the river is 7.6 × 103 m/s, river velocity is 0.6m/s,mean depth of the river is 16m, and its mean width is 1.8 × 103 m. Theaverage organic carbon content of the sediment is 0.2% and the averagesolids concentration in the river water is 20mg/L.

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6.63 On November 1, 1986, a fire in a chemical warehouse near Basel, Switzer-land caused the release of 7000 kg of an organo-phosphate ester (pesticide)into the river Rhine. Assuming that the river is in plug flow, determine theconcentration of the pesticide detected near Strasbourg, which is 150 kmfrom Basel. The relevant chemical properties of the pesticide are as fol-lows: First-order rate constant for reaction in water = 0.2 d−1, first-orderrate constant for reaction on sediment = 0.2 d−1, the sediment–water par-tition constant = 200 L/kg. River Rhine has an average volumetric flowrate of 1400m3/s, a solids concentration of 10mg/L, a resuspension rateof 5 × 10−5 d−1, a total cross-sectional area of 1500m2, and an averagewater depth of 5m.

6.73 A CSTR box model can be useful in estimating pollutant deposition ontoindoor surfaces.

(a) Consider a living room of total volume V and total internal sur-face area A onto which ozone deposition occurs. Let Vd rep-resent the deposition velocity; it is defined as the ratio of theflux to the surface (μg/m2 s) to the average concentration in air(μg/m3). Outside air with ozone concentration Co exchanges withthe indoor air through vents at a known rate Rexc. At steady statewhat will be the value of Vd for 75% deposition of ozone ina room that is 5m × 5m × 2.5m, if Rexc = 5 L/min and Co =75 ppb? Experimental values range from 0.001 to 0.2 cm/s. Whatare the possible sources of such wide distributions in experimentalvalues?

(b) A home near a chemical plant has a total internal volume of 300m3.The normal air exchange between the indoor and outdoor occurs at arate of 0.1 h−1.An episodic release of a chlorinated compound (1,1,1-trichloroethane, TCA) occurred from the plant. The concentrationof TCA was 50 ppbv in the released air. If the normal backgroundconcentration of TCA is 10 ppbv, how long will it take before theconcentration inside the room reaches a steady-state value?

6.83 Consider a beaker of water in equilibrium with CO2 in the air (partialpressure PCO2 ). The pH of the water is made slightly alkaline so that thedissolution of CO2 in water is enhanced. The solution process is controlledby diffusion and reaction in the stagnant boundary layer of thickness δ. CO2undergoes two competing reactions in water:

(a) Dissolution in water: CO2 + H2Ok1�HCO−

3 + H+.

(b) Reaction with OH−: CO2 + OH− k2�HCO−3 .

(i) Obtain first the amount of CO2 in the boundary layer at any timet and the flux to the surface. The ratio of the two will give thetime constant for diffusion in the boundary layer.

(ii) Assume that only the two forward reactions are of importanceand write down the reaction rate equation for the dissolu-tion/reaction of CO2 in water. The reciprocal of the first-orderrate constant (at constant pH) is the time constant for reactionin the boundary layer.

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(iii) Equate the two rate constants and obtain the value of δ if the pHof the solution is 5.5. What does this δ represent?

6.92 A chemical spill near Morganza, Louisiana, occurred at about 1.45 P.M.on Monday 19, 1994, from a barge on the Mississippi river. The barge wascarrying 2000 gallons of a highly flammable product called naphtha. TheU.S. Coast Guard closed the river traffic for about 4 miles from the spillto about 22 miles down river, ostensibly so that naphtha could dissipate.First determine from the literature themajor constituents of naphtha. Usingthis as the key, perform calculations (considering a pulse input) to checkwhether the Coast Guard’s decision was appropriate and prudent. Flowcharacteristics of the Mississippi river are given in problem 6.5.

6.103 Assume a global CO2 increase in air at the rate of 0.41% per year. ThecurrentPCO2 is 358 ppmv. Predict what will be the average pH of rainwaterin the year 2050? Is your estimate realistic? Justify your conclusions.

6.112 Tropospheric CO is an important pollutant. It has a background mix-ing ratio of 45:250 ppbv. In the period between 1950 and 1980, it hasbeen shown that PCO increased at ∼1% per year in the northern hemi-sphere. Recent data (1990–1993) showed a significant decrease in CO inthe atmosphere. Consider the fact that the largest fraction of CO in air is

lost via reaction with the hydroxyl radical: CO + OH• kOH−−→ CO2 + H•.If the CO emission rate from all sources is considered to be repre-sented by a single term Σ Si, the rate of change of PCO is given by(Novelli et al., 1994): dPCO/dt =∑i Si − kOHPCO, what is the con-centration of CO at steady state? Utilize the expression to derive thedifferential expression dPCO/PCO. Assume that CO is mainly emit-ted from automobile exhaust. In the mid-1970s, stringent regulationshave reduced the automobile emissions. Hence assume that no changein CO sources have occurred during 1990–1993. If PCO presently is120 ppb and the [OH] concentration in the atmosphere has increased1 ± 0.8% per year, what is the percent change in PCO during the period1990–1993?

6.122 Using the following data establish the contribution to global warming foreach of the following greenhouse gas. Assume a predicted equilibriumtemperature change of 2◦C for a doubling of the partial pressure of eachgas. (a) Calculate the total temperature increase in the year 3000 due tothe combined emissions of all five gases, if the rate of accumulation isunchecked.

Preindustrial Current Rate ofAtmospheric AccumulationConcentration 1990 in Air

Compound (1750–1800) Concentration (% per year)

CO2 280 ppmv 353 ppmv 0.5CH4 0.8 ppmv 1.72 ppmv 0.9N2O 288 ppbv 310 ppbv 0.25CFC-11 0 280 pptv 4CFC-12 0 484 pptv 4

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(b) If the rate of accumulation of CFC-11 and CFC-12 are reduced to<0.1% through voluntary limits on CFC production, and that of N2Ois reduced to <0.1% via stricter automobile emission checks, whattemperature increase can be anticipated in the year 3000?

6.132 Free radicals formed in the atmosphere from photochemical species suchas aldehydes can increase the concentration of ozone in urban smog. Thefollowing mechanism is suggested for the effect of formaldehyde upon thereaction of ozone in the troposphere:

HCHOhν−→J2

2HO•2 + CO2; J2 = 0.015min−1

HCHO + OH• k4−→HO•2 + CO + H2O;

k4 = 1.1 × 10−11 cm3/molecules

HO•2 + NO

k5−→NO2 + OH•; k5 = 8.3 × 10−12 cm3/molecules

OH• + NO2k6−→HNO3; k6 = 1.1 × 10−11 cm3/molecules.

Along with the Chapman mechanisms, the above reactions can be used toobtain [O3] as a function of t. Derive expressions for the rate of change inconcentrations [NO], [NO2], and [HCHO]. The equations should involveonly terms in [NO2] and [HCHO]. From the resulting equations how doyou propose to obtain [O3] given the initial conditions [NO]0, [NO2]0,and [HCHO]0.

6.142 An enzyme catalysis is carried out in a batch reactor. The enzyme con-centration is held constant at 0.003mol/dm3. The decomposition of thepollutant A−→E P was studied separately in the laboratory. It obeys theMichaelis–Menten kinetics with KM = 5 mM and Vmax = 10mM/min,but at an enzyme concentration of 0.001mol/dm3. How long will it takein the batch reactor to achieve 90% conversion of A to P?

6.152 An organism (Zooglia ramigera) is to be used in a CSTR for waste-water treatment. It obeys Monod kinetics with μmax = 5.5 d−1 and Ks =0.03mg/dm3. A total flow rate of 100mL/min is being envisaged at aninitial pollutant concentration of 100mg/dm3. CX = 10mg/L: (a) If thedesired yield factor is 0.5, what reactor volume is required? (b) Design aCSTR in series with the first one that can reduce the pollutant concentra-tion to 0.1mg/dm3. (c) If the second reactor is a PFR, what size reactorwill be required in part (b)? (d) Which combination do you recommendand why?

6.162 A solution containing a denitrifying bacteriaMicrococcus denitrificans at aconcentration of 1 × 10−6 kg/dm3 is introduced into a semibatch reactorwhere the growth rate is given by r = k([X][O2][S]/Ks + [S]), where[O2] is the oxygen concentration (mol/dm3). An excess of substrate [S]was used. If after 3 h of operation, the concentration of the bacteria is3 × 10−5 kg/dm3, what is the specific growth rate of the bacteria?

6.172 The rate of growth of Escherichia coli obeys Monod kinetics withμmax = 0.9 h−1 and Ks = 0.7 kg/m3. A CSTR is used to grow the bac-teria at 1 kg/m3 with inlet substrate pollutant feed rate of 1m3/h and

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[S]in = 100 kg/m3. If the cell yield is 0.7, calculate (a) the reactor vol-ume required for the maximum rate of production of E. coli? (b) If theeffluent from the reactor is fed to a series reactor, what is the volume ofthe second reactor if the desired pollutant concentration in the exit streamis 0.1 kg/m3?

6.183 Consider the long-term trends of PCBs in Lake Superior. It is an olig-otrophic lake with an average depth of 145m, surface area of 8.2 ×1010 m2, volume of 1.21 × 1013 m3, and a flushing time of 177 years.The DOC content is ∼1 mg/L. KSW = 105 L/kg. It was reported that thePCB content in Lake Superior has been steadily decreasing between 1978and 1992 due to a combination of flushing, sedimentation, biodegradation,and volatilization. A pseudo-first-order rate constant has been suggested.The following equation was said to be applicable:

d[W]/dt = I − k′[W],where [W] is themass of PCBs in the lake (kg), I is the input of PCBs (bothdirect and by gas absorption, kg/y), and k′ is the first-order rate constantfor disappearance of PCBs (y−1). It is related to the overall first-order rateconstant as k′ = +I(t)/W(t):

(a) Justify the equation for d[W]/dt given above. State all necessaryassumptions.

(b) Derive the equation for the overall rate constant k.(c) k′ is given as a sum k′ = kflushing + ksedimentation + kbiodegradation +

kvolatilization. With kbiodegradation = 0, kflushing = 0.006 y−1,

ksedimentation = 0.004 y−1, and kvolatilization = 0.24 y−1 for a com-bination of 82 PCB congeners, calculate the input of PCBs in the year1986 if the 1986 PCB burden is 10,100 kg.

(d) How much PCB (in kg) is lost through volatilization in theyear 1986?

(e) Using an averageHenry’s constant of 1 × 10−4 atmm3/mol and a net1986 PCB flux of 1900 kg/y, determine the air concentration of PCBin 1986. The 1986 water concentration is 2 ng/L (Note: You haveto relate kvolatilization to the mass transfer coefficient Kw throughthe equation kvolatilization = (Kw/h)fw, where h is the average depthof the lake and fw − 0.87 is the dissolved aqueous phase PCBfraction.)

(f) How much PCB will remain in the lake in the year 2020?

6.192 The following data pertain to a local stream in Baton Rouge, Louisiana:

Average stream speed: 0.36m/sAverage stream depth: 3mAverage water temperature: 13CFlow rate of water in the stream: 1m3/sBOD of the stream: 5mg/LCO2,w in the stream: 8mg/L.

At a point along the stream, a treatment plant discharges wastewaterthrough a pipe ∼2 m above the water level with the following charac-teristics:

Average flow rate: 0.1m3/s

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BOD of the wastewater: 20mg/LCO2,w in the wastewater: 2mg/L.

Obtain the following: (a) the dissolved oxygen profile in the stream asa function of distance downstream from the discharge point; (b) themaximum oxygen deficit and its location; and (c) the concentration ofthe waste (BOD) 15miles downstream of the discharge point (outfall).De-oxygenation coefficient, kd is 0.2 d−1.

6.202 Consider the University lake system in Baton Rouge, Louisiana, which hasa potential water quality problem resulting from waste discharged into thenearby Crest Lake. From the following data determine what nutrient level(phosphorus) is to be expected at steady state in the Crest Lake, which isa completely mixed unstratified system.

Mean depth: 1.5mSurface area: 3.4 × 104 m2

Mean settling velocity: 10m/yMean detention time: 561 dPhosphorus concentration in the waste: 10mg/L.

Assume no losses due to volatilization or other reactions of P in the lake.What reduction in discharge level should be accomplished so that the Pconcentration is reduced to <0.2 mg/L at which no fish kills are observedin the lake?

6.213 Consider a two-level system comprising the base food chain phytoplanktonthat forms the food for a fish. They are both exposed to a pollutant (aPCB) in the aqueous phase. Consider the fish population to be in two ageclasses: 0–3 years and 4–6 years. Using the data given below, obtain a plotof PCB concentration in the fish at steady state as a function of the age(year).

Age Class

Parameter 0–3 Years 4–6 Years

Growth rate, μi (d−1) 0.002 0.0005

Respiration rate: R = βWγeρtevu(g/g d) β = 0.05

γ = −0.2ρ = 0◦C−1

ν = 0.01 s/cmSwim speed, u (cm/s): ωWδeφT ω = 1

δ = 0.1φ = 0EC−1

Food assimilation efficiency 0.8PCB assimilation efficiency 0.35KB (μg/kg/μg/L) 4 × 105

Dissolved PCB concentration (ng/L) 5Weight of fish (g) 0.5 3Lipid fraction (kg lipid/kg weight) 0.1 0.1

6.222 Determine the half-life and90%removal time for the following compoundsfrom a well-mixed lake surface water: (a) pyrene, (b) 1,2-dichloroethane,

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(c) p, p′-DDT, (d) chlorpyrifos, and (e) 1,2-dichlorobenzene. The depth ofmixing is 1m.

6.233 The following data were obtained on the oxygen deficit (Δ =C∗

O2–CO2 ) in a pond with a surface aerator. Determine the mass

transfer coefficient for oxygen from the data.

Oxygen Deficitt (min) (% of Saturation)

0 782 624 526 448 3710 3112 2714 23

Note: dΔ/dt = −krΔ, and kr = kLa for a surface aerator.6.242 The absorption cross-section and actinide flux for the photolysis

ClNOhν−→Cl + NO in air are given below:

λi (nm) σλi (cm2/Molecule) Iλi (Photons/cm2 s)

280 10.3 × 10−20 0 × 1014

300 9.5 0.325320 12.1 5.08340 13.7 8.33360 12.2 9.65380 8.3 8.45400 5.1 11.8

Determine the atmospheric half-life for ClNO.6.252 In the surface waters of a natural lake, iron (Fe2+) reacts with

hydroxide in an oxidation–precipitation reaction with a rate −roxdn =kox[Fe2+][O2(aq)][OH−]2 and photochemically dissolves by reductionwith a rate −rphoto = kphoto{FeIIIL}, where {FeIIIL} is the concentration

of a ligand-bound FeIII on surfaces. If the lake volume is V and has avolumetric flow rate of Q, obtain the steady-state concentration of Fe2+in the lake.

6.262 Using a 16-W low-pressureHg lamp photoreactor emitting light at 254 nm,a series of pesticides were subjected to photodegradation in distilled water.The concentrations, absorbances, and quantum yields are given below:

Compound φ Aabs C (mol/L)

Atrazine 0.037 0.08 2.3 × 10−5

Simazine 0.038 0.059 1.8 × 10−5

Metolachlor 0.34 0.005 1.0 × 10−5

The emitted light intensity of the lamp was I0 = 7.1 × 10−8

einstein/L s. Estimate the half-lives of the pesticides in water.

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6.272 The air above the city of Baton Rouge, Louisiana, can be considered tobe well-mixed to a depth of 1 km over an area 10 × 10 km2. CO is beingemitted by both stationary and mobile sources within the city at a rate of10 kg/s. Assume CO is a conservative pollutant. (a) For a wind velocity of5 m/s in the city, what is the steady-state concentration of CO in the city?(b) If after attaining steady state, the wind velocity decreases to 1.5m/s,what is the new steady-state concentration?

6.282 Peroxyacyl nitrate (PAN) is a component of photochemical smog. It isformed from aldehydes by reaction with OH as follows:

CH3CHO + OH• −→ CH3CO + H2O

CH3CO• + O2 −→ CH3CO(O2)

•

CH3CO(O2)• + NO2 + M � CH3C(O)O2NO2 + M

Derive an expression for the production of PAN.6.291 The annual production rate of CH3Cl is 0.3Tg/y and has an averagemixing

ratio of 650 pptv.What is its average residence time in the atmosphere?

6.302 Consider a lake 108 m2 of surface area for which the only source of phos-phorus is the effluent from a wastewater treatment plant. The effluent flowrate is 0.4m3/s and has a phosphorus concentration of 10 g/m3. The lake isalso fed by a stream of 20m3/s flowwith no phosphorus. If the phosphorussettling rate is 10 m/y, estimate the average steady-state concentration ofP in the lake. The depth of the lake is 1m.

6.312 An accident on a highway involving a tanker truck spilled 5000 gallonsof 1,2-dichloroethene (DCA) forming a 2500-m2 pool. DCA has a vaporpressure of 0.1 atm at 298K.The air temperature is 25C and thewind speedaveraged 4m/s. The accident occurred on a sunny afternoon. Estimate thedistance downwind of the spill that would exceed the worker standardexposure limit of 1 ppmv for DCA.

6.322 In a poorly ventilated hut in a third world country, logs are burnt to supplyheat for cooking purposes. The total volume of the hut is 1000m3. Logburning releases CO at the rate of 2mg/h. The air exchange rate is 0.1 h−1.(a) What is the CO concentration after 2 h of cooking inside the hut? (b)Compare the steady-state indoor CO concentration with the U.S. ambientair quality criteria of 0.05 ppmv.

6.332 In 1976, a tragic release of one of the most toxic compounds knownto man (dioxin, i.e., 2,3,7,8-tetrachlorodibenzo-p-dioxin) occurred inSeveso, Italy. The plant was manufacturing 2,4,5-trichlorophenol from1,2,4,5-tetrachlorobenzene when the reaction ran away. The reactor over-pressurized and the relief system opened for 5 min during which time 2 kgof dioxin escaped into the atmosphere through the plant’s roof. The windspeed was 2m/s and leak occurred on an overcast day. What is the likelyconcentration 10 km from the plant site?

6.342 Consider the city of BatonRouge, Louisiana, with a population of 400,000.The morning peak hour traffic is approximately 100,000 vehicles in anarea 300 km2 with an average travel distance of 5 km from 7 a.m. to 10a.m. daily. Assume each vehicle emits 2 g of CO for every 1 km traveled.Determine the CO concentration in the atmosphere at 9 a.m. The initial

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background concentration in the air prior to rush hour traffic is 5mg/m3

and the average wind velocity is 3m/s.6.353 A vent from a hydrocarbon process (process gas) contains appreciable

amounts of pentane (C5), hexane (C6), and heptane (C7), which are tobe recovered using a brine knockout condenser and a carbon adsorptionunit. The knockout condenser already existing and producing a conden-sate has the compositions: 10, 40, and 50 mole percent C5, C6, and C7,respectively. The temperature and pressure of the condensate and tail gasare 10C and 35 psia. Nitrogen is noncondensable under these conditions.As the project engineer, your task is to recover the alkanes in the tail gas.Assume ideal behavior. The Antoine constants for compounds are givenbelow:

Pentane Hexane Heptane

A 6.88 6.91 6.89B 1076 1190 1264C 233 226 217

(a) First calculate the composition of the tail gas that is emitted from theknockout condenser. Express asmole fractions ofC5,C6,C7, and N2.

(b) Express the total concentration of hexane (C6) in kg/m3 and defineit as CF, the inlet feed concentration to the adsorption bed. (Hint: useideal gas law, 14.696 psia = 101352 Pa). Based on laboratory datafor activated carbon, the mass transfer zone length and Freundlichparameters have been calculated for the tail stream: δ = 0.1m. TheFreundlich isotherm is C = KFreunW

n, where KFreun is 10 kg/m3

and n = 2.25. C is the concentration (kg/m3) and W is the adsor-bate concentration (kg/kg). If the adsorbent density is 384 kg/m3, theactual volumetric flow rate through the bed is 1m3/s, what shouldbe the length? The adsorber bed diameter is 2m and breakthroughtime is 1 h. Note that it is not necessary to determine or use the per-cent recovery for the absorber as the given mass transfer zone lengthcorresponds to 99% recovery.

6.362 What is the desired gas flow rate to obtain a 1-h breakthrough in treatingan air stream using a carbon bed of density 0.35 g/cm3. The inlet con-centration of the pollutant is 0.008 kg/m3. The bed length is 2m and hasa cross-sectional area of 6m2. The Freundlich isotherm parameters areKFreun = 500 kg/m3 and n = 2. The mass transfer coefficient is 50 s−1.

6.371 Answer whether the following statements are true or false. Giveexplanations:

1. The mass transfer of a contaminant from a gas stream that is verysoluble in a liquid-absorbing phase is likely to be gas phase controlled.

2. Particulates formed by condensation of gases (i.e., smoke) are likelyto be easily removed from a gas stream by gravity settling.

3. High ozone pollutant levels were responsible for the deaths of asmany as 4000 people in London in the early 1950s.

4. Atmospheric conditions during high winds and overcast skies wouldbe expected to be stable, with relatively little crosswind or verticalpollutant dispersion.

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5. The specification of absorption column diameter is usually done onthe basis of the mechanical design of the column internals and isessentially independent of the compound to be stripped.

6. A crossflow scrubber is more efficient than countercurrent or co-current scrubbers for the removal of particulates from an air stream.

7. A doubling of carbon dioxide partial pressure in the atmosphere effec-tively increases the global average surface temperature by the sameamount.

8. The reaction of ozone with oxides of nitrogen accounts for 60–70%of ozone destruction in the lower troposphere.

6.382 A local industry produces a process air stream that contains vinyl chlo-ride (molecular weight 62.5) at a concentration of 1000 ppmv. The massflow rate of air is 1.9 kg/s at a temperature of 25◦C and a total pressure of1 atm. The environmental group has proposed using an activated carbonbed to remove vinyl chloride. Density of air is 1.2 kg/m3. Given that theadsorption bed is 5m2 in area and 0.35m in thickness, determine the fol-lowing: (a) The lifetime of the bed before regeneration. Answer in hours.The carbon used is Ambersorb XE-347 for which the Freundlich isothermparameters for vinyl chloride are KFreun = 4.16 and n = 2.95, that is,Ce = KFreunW

nsat , where Wsat is the saturation adsorbed concentration

(g/g). The adsorption bed density is 700 kg/m3. The overall mass transfercoefficient is 30 s−1. (b)What is the overall flux of vinyl chloride throughthe bed?Answer in kg/m2 s.What is the mass rate through the bed in kg/h?

6.392 Construct a two-boxmodel for a lake that consists of an epilimnion (awell-stirred upper region) and a hypolimnion (the cold portion of the lake). rp isthe rate of removal (precipitation, sedimentation), rb is the rate of burial,rd is the rate of addition to hypolimnion by dissolution, Qexc is the rateof exchange of water between epilimnion and hypolimnion. Write a massbalance equation for the steady-state concentration of a pollutant in thehypolimnion. Note that at steady state rb = rp − rd. Consider a lake wherethe average river water flow is 100m3/s, an average rainfall deposition of20m3/s, and a mean burial rate of 200mg/s for phosphorus. If the influentto the lake has a P loading of 5μg/L from agricultural pesticide runoff intothe river, is the lake prone to eutrophication (i.e., biological productivity)?Mean P concentration for biological productivity in lakes is 5μg/L.

6.403 Estimate the time to breakthrough for 1,2-dichlorobenzene to reach awell 1 km from a source in the groundwater that has a Darcy velocityof 8m/y. The soil has a 3% organic carbon, porosity of 0.4, and a densityof 1.2 g/cm3. Both advection and dispersion are important in this case.Breakthrough is defined as the time at which the concentration reaches1% of the feed (source).

6.412 The following two contaminants are present in a gasoline mixture: hex-ane and octane. If a gasoline spill has occurred in the subsurface due toan underground storage tank rupture and contaminated the groundwater,how quickly will these chemicals be detected in a monitoring well 100maway?The arrival time of a conservative tracer (chloride ion) was observedto take only 50 days. The soil had an organic carbon fraction of 1% and aporosity of 0.45. The soil density was 1.3 g/cm3.

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6.422 During a routine navigational dredging of a river, a 55-gallon drum con-taining pure benzene was excavated. A small leak was observed througha hole 10 cm2 area on the top of the drum. How much benzene has beenlost to the river water by diffusion during the one year the drum has beenon the bottom sediment surface? The drum has a height of 100 cm.

6.432 Estimate the total volume of water required to remove a 3% residualtrichloroethylene (TCE) spill from 1m3 of an aquifer of porosity 0.3 andorganic carbon content of 1%. The density of TCE is 1.47 g/cm3 and itsaqueous solubility is 1100mg/L. Assume no retention of TCE on aquifersolids.

6.442 Estimate the number of pore volumes required to reduce 99% of theporewater concentration of the following contaminants from an aquiferwith 2% organic carbon content and a porosity of 0.4: 1,2-dichloroethane,1,4-dichlorobenzene, biphenyls, and Aroclor-1242.

6.453 Compare the rates of emission to air of the following pesticides applied to asurface soil at an average concentration of 10μg/g each. The soil porosityis 0.6 with a 3% water saturation and organic carbon content of 3%. Thesoil density is 2 g/cm3 and the area of application is 1 ha.Assume a surfacemass transfer coefficient of 1 cm/s. Repeat your calculation for a saturatedsurface soil of 10%water saturation: (a) lindane, (b) chloropyrifos, and (c)p,p′-DDT.

6.463 The groundwater source in Baton Rouge is high in Fe(II). The CentralTreatment Facility uses aeration to oxidize Fe(II) and precipitate as ironoxy-hydroxide. The reaction rate is given in Section 5.8.1 (Chapter 5). Theobserved rate constant is k = 1.6 × 10−7 L/mol min atm at the water pHof 6.7. The oxygen concentration in water is to be kept at its saturationvalue. If the oxidation reactor is a CSTR, what residence time is requiredto reduce Fe(II) by 90% in the groundwater?

6.473 Emissions from wastewaters in petroleum refineries are categorized intoprimary and secondary. Secondary sources of emissions are generally fromwastewater ponds. Before the water reaches a pond, streams from differ-ent unit operations are mixed in a large vessel where air emissions canbe a problem. There is little or no chemical degradation in these vessels.The only loss mechanism is mass transfer to the vapor space. Althoughno aerators are placed, there is enough turbulence to mix the water. Usea CSTR model to assess the rate of air emissions from a vessel in whichthe inlet concentration of ethyl benzene is 1 × 10−5 g/mL at a flow rateof 2 L/s. The vessel cross-sectional area is 1m2.

6.482 Determine the number of pore volumes of air required for 99% removalof chlorobenzene from the vadose zone of an aquifer. The total volume ofthe aquifer is 1m3. It has a porosity of 0.6, an organic carbon fraction of0.02, and a bulk density of 1.4 g/cm3. The zone is 5% water saturated and5% saturated with chlorobenzene.

6.492 On a clear, calm Monday evening (January 18, 1999), a tanker truck car-rying 22 tons of liquid anhydrous ammonia rolled over as it was makinga sharp turn outside the Farmland Industries chemical plant in Pollock,Louisiana. The liquid ammonia escaped through a valve. Assume that itformed a 2500-m2 pool and turned into gas as it went into the air. Estimate

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the concentration of ammonia vapor 500 m from the accident. Is the con-centration dangerous to the workers in the plant? The wind speed duringthe period was nominal (3m/s).

6.503 Ed Garvey determined that as water flows over hot spots in the Thom-son Island pool of the Hudson River that its PCB concentration increasesrapidly. Data taken from 1993–1995 showed that the “hot-spot” sedi-ments were responsible for about 1.5 pounds of PCBs (3 pounds duringthe summer) moving into the water column. Because the flesh of mostHudson River fish exceeds the FDA limit of 2 ppm for PCBs, various fish-ing bans and advisories are in effect along the river, with children andwomen of child-bearing age told to “eat none” (Rivlin, 1998). Use a com-partmental box model to verify whether or not fish advisories or bansare warranted. Assume: (a) no PCBs in the air, (b) no PCB upstream ofThomson Island pool, and (c) fish in equilibrium with the PCBs in water.The PCB data are evaporation mass transfer coefficient = 5.7 cm/h, logKoc = 4.5. River data: surface area for evaporation = 2.3 × 105 m2, riverflow = 4 × 105 L/s.

6.512 Benzene (=1), chlorobenzene (=2), methyl chloride (=3), and methane(=4) are combusted in a new incinerator designed for a residence time of 1 sand a burner temperature of 1200◦F. However, during the startup the con-versionswere all<20%which is unacceptable.As the contact engineer youare presentedwith two options: (1) extend the burner chamber 10-fold, thusincreasing the residence time to 10 s; or (2) increase the temperature from1200 to 1400◦F.Which option gives the highest conversions?Whichwouldbe cheapest to implement?What are the consequences or concerns of each?

6.522 Heterogeneous reactions on polar stratospheric clouds (PSCs) are respon-sible for the abnormal destruction of ozone in the Antarctic regions in thespringtime. The following series of reactions are shown to be occurring ona surface, M:

2ClO + Mk1−→Cl2O2 + M

Cl2O2J2−→hν

ClOO + Cl

ClOO + Mk3−→Cl + O2 + M

2

(Cl + O3

k4−→ClO + O2

).

Note that the overall scheme is 2O3hν−→ 3O2. The first step in the above

scheme is the slowest reaction. Justify that d[O3]/dt = −k1[ClO]2[M],where k1 = 8 × 10−44 m6/s at 190K (stratospheric temperature).Assumean average [M] = 3 × 1024 m−3 and determine the mean concentration ofClO responsible for the ozone destruction.

6.532 The SO2(g) from power plant smoke stacks dissolves in a surface cloud(e.g., fog) to produceS(IV).The resultingS(IV) then reactswith a dissolvedspecies A (aq) to give S(VI) in a fog droplet. The rate of the reaction isgiven by r = −d[S(IV)]/dt = k[A]aq[S(IV)]aq. Rewrite the above equa-tion to give the rate in terms of the liquid water content of the fog, θL (see

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Section 4.2.1) to given r = 3.6 × 106θLrRT, where the units of r are ppbvSO2 h

−1 and R is the universal gas constant.6.542 Hurricane Katrina flooded several New Orleans homes in 2005. Contam-

inated sediments from Lake Ponchartrain remained as a cake and finallydried out in abandoned homes once the floodwater receded. These sedi-ments are likely to be sources of vapors of organic compounds to the indoorair. Consider a 20m long × 20mwide × 5m high roomwhere the air flowvelocity is only 10m/h.Assume that the emission rate of the organic vaporfrom the sediment to air is 100 ng/m2/h.Assume also that the organic vapordegrades in the indoor air with a first-order rate constant of 10 h−1 andalso that the fresh air entering the room has only negligible organic vaporin it.

(a) Perform a mass balance on the organic vapor in the room and writethe unsteady mass balance differential equation.

(b) Apply the steady-state approximation to obtain the concentration ofthe organic in air.

(c) For the data given above calculate the steady-state concentration.Express your answer in ng/m3.

6.553 A pool of pure benzene of area 0.4 m2 exists at the bottom of a lake ofvolume 1000m3. The volumetric flow rate of water in the lake is 50 L/s.If the average mass transfer coefficient (kmt) for the dissolution of purebenzene into water is 0.1m/s, determine the concentration in the outflowfrom the lake after 12 h of the spill. The solubility of benzene in water is780 mg/L. Assume that the lake is well mixed and initial concentration ofbenzene is zero.

6.562 Estimate the volume of a bioreactor required to achieve a yield of 0.5 of E.coli bacteria using 20mg/L of an activated sludge in the influent streamof acontinuous reactor. The desired flow rate of the sludge is 10m3/d. The bac-terium follows Monod kinetics with Ks = 10mg/L and μmax = 10 d−1.The bacterial decay constant is 2 d−1. It is desired to maintain the bacterialpopulation at 5mg/L through the day. Give the answer in m3.

6.573 Air enters a room (5m × 5m × 8m) at a velocity of 1 m/h and carriesa gaseous pollutant at a concentration of 10μg/m3. An open-hearth fur-nace in the room generates the same pollutant at a rate of 24μg/m2/h. Thepollutant, however, undergoes a zeroth-order reaction in the room with arate constant of 1μg/m3/h.

(a) Write down the appropriate unsteady-state mass balance equation forthe pollutant in the room.

(b) Solve the above equation to obtain concentration as a function of time.(c) What is the pollutant concentration leaving the room 1 h after the start

of the open-hearth furnace?

6.581 State whether the following statements are true or false. Give briefexplanations.

1. In an enzyme reactor we have to continuously provide enzymesexternally for the reaction to proceed.

2. An organic pollutant will travel faster in a water-saturated columnfilled with soil than in a column filled with sand.

3. In the lower atmosphere, smog appears when the sunlight is at itshighest intensity and VOC concentration is lowest.

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4. For the same degree of conversion, a plug-flow reactor requires alarger volume than a CSTR.

6.593 A tanker collided with a tugboat and spilled 9000 gallons of diesel fuel intothe Mississippi River at 1:30 a.m. on July 23, 2008, near the mouth of theriver inNewOrleans, Louisiana.As a result, residents of communities suchas Algiers, Gretna, and the Plaquemines parish living downstream of theriver were asked to conserve water, as water intakes to these communitieswere closed to prevent contamination of the drinking water supply. Con-sider the most soluble component of diesel (conduct a literature survey)and determine how fast it will move with the water to the nearest commu-nity down river.You will have to obtain the relevant flow parameters fromthe USGS website.

6.603 Heavy duty vehicles (hdv) using diesel fuel emits carcinogenic hydrocar-bon, benzo[a]pyrene (BaP) directly to the air. A detailed investigation ofBaP emissions between 1961 and 2004 was recently reported by Beyeaet al. (2008, Environmental Science Technology). Consider the BaltimoreHarbor Tunnel that is 2.31 km long, 6.7m wide, and 4.1m high. The max-imum speed of vehicles within the tunnel is 80 km/h. Wind parallel to thetunnel blows in at 4m/s. It is estimated that 2500 vehicles per hour with12% hdv mix, moving at an average speed of 46 km/h transits through thetunnel. If the estimated emission factor for BaP was 40μg/km in 1961and 5μg/km in 2004, what are the corresponding changes in BaP airconcentrations during this period?

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“78194_A001.tex” — page 425[#1] 10/4/2009 19:33

Appendix1Properties of SelectedChemicals ofEnvironmentalSignificance

Mixing AqueousMolecular Ratio Solution Log(KAW

′/ LogCompound Weight (χi /[−]) (C∗

i /mol/dm3) kPa dm3/mol) (Kow/[−])1. Inorganic Compoundsa−c

CO 28 6–12 × 10−6 9.3 × 10−4 +4.99CO2 44 0.035 3.3 × 10−2 +3.47 0.83N2 28 79 6.2 × 10−4 +5.19 0.67NO 30 ≤5 × 10−8 1.8 × 10−3 +4.73NH3 17 1–10 × 10−8 28.4 +0.22 −1.37NO2 46 ≤5 × 10−7 +4.00H2 2 6 × 10−5 7.7 × 10−4 +5.10 0.45O2 32 21 2.6 × 10−4 +4.90 0.65O3 48 1–10 × 10−6 4.4 × 10−4 +3.91SO2 64 1–10 × 10−9 1.5 +1.92H2S 34 ≤2 × 10−8 0.1 +3.01 0.96H2O 18 3.2 — — −1.15H2O2 34 1 × 10−7 −2.85 −1.08N2O 42 3.0 × 10−5 2.4 × 10−2 +3.60 0.43HNO3 (g) 63 1–10 × 10−8 −3.32HCl (g) 36.5 −1.39

continued

425

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Vapor AqueousMolecular Pressure Solution Log(KAW

′/ LogCompound Weight (P∗

i /kPa) (C∗i /mol/dm3) kPa dm3/mol) (Kow/[−])

2. Organic Compounds(i) Hydrocarbonsd

Methane 16 2.8 × 104 4.1 × 10−1 4.85 1.12Ethane 30 4.0 × 103 8.1 × 10−2 4.90 1.78Propane 44 9.4 × 102 1.3 × 10−2 4.86 2.36n-Butane 58 2.5 × 102 2.6 × 10−3 4.98 2.89n-Pentane 72 7.0 × 101 5.6 × 10−4 5.10 3.62n-Hexane 86 2.0 × 101 1.5 × 10−4 5.26 4.11n-Heptane 100 6.2 × 100 3.1 × 10−5 5.31 4.66n-Octane 114 1.9 × 100 6.3 × 10−6 5.51 5.18n-Decane 142 1.7 × 10−1 2.7 × 10−7 5.85 6.70Benzene 78 1.2 × 101 2.3 × 10−2 2.75 2.13Toluene 92 3.8 × 100 5.6 × 10−3 2.83 2.69Ethylbenzene 106 1.3 × 100 1.6 × 10−3 2.93 3.15Naphthalene 128 3.7 × 10−2 8.7 × 10−4 1.69 3.36Phenanthrene 178 9.0 × 10−5 3.4 × 10−5 0.55 4.57Anthracene 178 7.8 × 10−5 3.3 × 10−5 0.36 4.54Pyrene 202 4.0 × 10−6 4.4 × 10−6 −0.04 5.13Cyclopentane 70 4.3 × 101 2.3 × 10−3 4.27 3.00Cyclohexane 84 1.2 × 101 7.1 × 10−4 4.25 3.44d-Limonene 136 0.26 1.0 × 10−4 1.88 4.57β-Pinene 136 1.7 × 10−4 −1.67 4.37

(ii) Acids and basesd,e

Acetic acid 60 1.6 × 100 1.0 × 102 −1.99 −0.17n-Propionic acid 74 4.5 × 10−1 0.33n-Butanoic 88 1.0 × 10−1 3.2 × 10−2 0.79n-Pentanoic 102 2.0 × 10−2 8.1 × 10−3 0.99n-Hexanoic 116 6.0 × 10−3 9.4 × 10−2 1.90n-Heptanoic 130 2.0 × 10−3 1.8 × 10−2 2.72

(@293K)n-Octanoic 144 1.0 × 10−3 1.7 × 10−2 3.22

(@373K)Benzoic acid 122 2.7 × 10−2 −2.15 1.87

(iii) Alcohols and phenolsd,f ,g

Butanol 74 9.3 × 10−1 1.0 × 100 −0.24 0.88Pentanol 88 3.4 × 10−1 3.0 × 10−1 0.17 1.16Hexanol 102 1.4 × 10−1 1.3 × 10−1 0.25 2.03Heptanol 116 1.5 × 10−2 0.34 2.41Octanol 130 6.8 × 10−3 4.4 × 10−3 0.45 2.84Phenol 94 6.9 × 10−2 1.0 × 100 −1.34 1.482-Chlorophenol 128 1.9 × 10−1 2.2 × 10−1 −1.24 2.172,4-Dichlorophenol 163 2.7 × 10−2 2.752,4,6-Trichlorophenol 197 1.1 × 10−3 4.0 × 10−3 −2.20 3.38Pentachlorophenol 266 1.5 × 10−5 5.2 × 10−5 −1.56 5.04

(@293K)

continued

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Properties of Selected Chemicals of Environmental Significance 427




2-Nitrophenol 139 1.8 × 10−2 1.3 × 10−2 0.13 1.894-Me-2-nitrophenol 153 6.2 × 10−3 3.8 × 10−3 0.21 2.374-Cl-2-nitrophenol 173 4.9 × 10−3 3.9 × 10−3 0.10 2.462,4-Dinitrophenol 230 6.8 × 10−4 1.5 × 10−3 −1.55 1.67

(iv) Halocarbonsh

Methylenechloride 85 5.9 × 101 2.3 × 10−1 2.43 1.15Chloroform 119 2.6 × 101 6.4 × 10−2 2.59 1.93Carbontetrachloride 154 1.5 × 101 6.3 × 10−3 3.33 2.731,2-Dichloroethane 99 9.1 × 100 8.5 × 10−2 2.00 1.471,1,1-Trichloroethane 133 1.6 × 101 8.5 × 10−3 3.49 2.48Hexachloroethane 285 2.4 × 10−2 1.7 × 10−4 2.40Hexachlorobutadiene 261 3.4 × 10−2 1.2 × 10−5 3.44 4.90Chlorobenzene 112 1.6 × 100 4.5 × 10−3 2.54 2.911,2-Dichlorobenzene 147 1.9 × 10−1 9.8 × 10−4 2.28 3.381,3,5-Trichloro-benzene 181 7.8 × 10−2 7.1 × 10−5 3.04 4.02Hexachlorobenzene 285 3.5 × 10−4 2.3 × 10−6 2.18 5.50Endrin 381 4.0 × 10−7 6.6 × 10−10 −0.12 4.56Aldrin 365 8.0 × 10−7 5.5 × 10−7 0.17 6.50Lindane 291 6.4 × 10−5 1.9 × 10−4 −0.49 3.78p,p′-DDT 355 9.5 × 10−8 9.8 × 10−8 −0.02 6.372,3,7,8-Tetrachloro-dibenzo-p-dioxin

322 1.6 × 10−7 3.2 × 10−8 0.70 6.64

Trichlorofluoro-methane 137 1.0 × 102 8.3 × 10−3 4.10 2.16Dichlorodifluoro-methane 121 1.5 × 104 1.6 × 10−2 4.60 2.53Biphenyl 154 1.0 × 10−3 1.3 × 10−4 2.19 4.094-Chlorophenyl 188 2.5 × 10−3 3.1 × 10−5 1.91 4.534,4′-Dichlorobiphenyl 223 8.3 × 10−5 5.1 × 10−6 1.21 5.332,3′,4,4′-Tetrachloro-biphenyl

292 6.7 × 10−6 2.1 × 10−7 1.49 6.31

Chlorocyclohexane 118 1.08 1.8 × 10−3 2.78Bromocyclohexane 163 0.41 3.9 × 10−4 2.97 3.20

(v) Thio compounds and estersh

Dimethyl sulfide 94 3.8 × 100 3.6 × 10−2 2.02 1.77Thiophene 84 1.0 × 101 4.7 × 10−2 2.35 1.81Diethyl phthalate 222 8.3 × 10−4 4.1 × 10−3 −0.70 2.35Di-n-butyl phthalate 278 9.5 × 10−6 3.4 × 10−5 −0.89 4.57

(vi) Mercury and mercury compoundsi

Hg 201 1.6 × 10−4 2.8 × 10−7 +2.86 0.61HgCl2 272 1.3 × 10−5 2.5 × 10−1 −4.14 0.52Hg(OH)2 235 5.1 × 10−4 −2.10 −1.30(CH3)2Hg 230 1.13 × 10−3 1.3 × 10−2 +2.88 2.26

(vii) Ketones and aldehydesb

Acetone 58 2.8 × 101 1.3 × 101 0.47 −0.24Formaldehyde 30 5.2 × 102 4 −1.53 0.35

continued

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Acetaldehyde 44 1.2 × 102 4 0.83 0.52Benzaldehyde 106 1.3 × 10−1 3.1 × 10−2 0.61 1.48Acetophenone −5.53

(viii) Siloxanesj

Octamethyl-cyclotetrasiloxane

297 3.0 × 10−2 1.8 × 10−7 4.39 5.09

(ix) Alkenes and haloalkenesb,h

1-Butene 56 3.6 × 102 4.0 × 10−3 4.40 2.40Styrene 104 6.3 × 10−1 2.4 × 10−3 2.42 3.05Vinyl chloride 62 3.9 × 102 1.7 × 10−1 3.35 0.60Tetrachloroethene 166 2.5 × 100 9.1 × 10−4 3.44 2.88

(x) AminesAniline 93 1.3 × 10−1 3.9 × 10−1 −0.47 0.90Ethylamine 45 1.3 × 102 1.1 × 102 5.03 −0.30

(xi) Perfluoroalkanoic acidsk,l

Perfluorooctane-sulfonic acid(PFOS)

500 3.3 × 10−7 1.4 × 10−3 −3.48 4.13

Perfluorooctanoicacid(PFOA)

383 1.3 × 101 8.9 × 10−3 0.39 4.40

(xii) Polybrominated diphenyl ethers (PBDE)m

PeBDE-tetra 485 4.7 × 10−8 2.7 × 10−8 1.04 6.57PeBDE-penta 564OBDE-hexa 643 6.6 × 10−9 7.7 × 10−10 1.02 6.29OBDE-hepta 722OBDE-octa 801DBDE-deca 959 4.6 × 10−9 <1 × 10−10 >1.6 6.27

(xiii) Alkynitrilesn

Acetonitrile 41 13.4 0.04 0.79 ± 0.04 −0.34Propionitrile 55 5.2 1.08 1.17 ± 0.05 0.16Butyronitrile 69 2.7 0.48 1.48 ± 0.05 0.53Valeronitrile 83 1.5 0.28 1.65 ± 0.07 1.12Isovaleronitrile 83 2.55 ± 0.09 1.07

a To obtain values in conventional units use the following conversion factors: Pressure in kPashould be multiplied by (1/101.325) to convert to atm. Aqueous solubility expressed in mol/dm3

is identical to mol/L. Henry’s constant Kaw′ given in units of kPa dm3/mol should be multiplied

by 4.04 × 10−4 to obtain Kaw, which is a dimensionless molar ratio. All values are at 298 Kunless otherwise indicated. Log Kow values are mostly from Hansch et al. (1995).

b Leo andHansch (1971), Liss and Slater (1974),Mackay and Leinonen (1975), Thibodeaux (1979),Stumm and Morgan (1981), Zhu and Mopper (1990), Morel and Herring (1993), Montgomery(1996), and Ji and Evans (2007).

c For gases the partial pressures are typical of the atmosphere. Aqueous solubility is that at a totalpressure of 101.325 kPa and at 298K.

continued

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Properties of Selected Chemicals of Environmental Significance 429

d Vapor pressure and aqueous solubility at 298K except where noted. For solids and gases the values arefor the subcooled liquid state (Ambrose and Ghiasse, 1987; Cal, 2006; Helburn et al., 2008).

e Verschuren (1983), Schwarzenbach et al. (1993), and Montgomery (1996).f Values for pentachlorophenol are for the solid species.g For nitrophenols the values are for subcooled liquid species at 293K (Schwarzenbach et al., 1988).Values of log Kow are for the neutral species.

h Values for solids and gases are for subcooled liquid state at 298K (Mackay et al., 1992; Schwarzenbachet al., 1993; Sarraute et al., 2008).

i Values from Iverfeldt and Lindqvist (1984), Mason et al. (1996), and Stein et al. (1996).j Values from Mazzoni et al. (1997).k Organization for Economic Co-operation and Development: Joint meeting of the chemical commit-tee and the working party on chemicals, pesticides, and biotechnology: Hazard assessment of PFOS,ENV/JM/RD(2002)17/Final (2002).

l Li et al. (2007).m PBDEs are commercial mixtures and consist of PeBDE (penta-, tetra-, and hexa-BDE), OBDE (hepta-,octa-, and hexa-BDE), DBDE (deca- and nona-BDE). Data from Environment Canada: EcologicalScreening Assessment Report on PBDEs, June 2006.

n Hansch et al. (1995), Lee et al. (1996), Ruelle (2000), Yaffe et al. (2001), and Ji et al. (2008).Note: Excellent compilations of physicochemical properties for a variety of chemicals can be found

at the following sites on the worldwide web: (a) http://webbook.nist.gov chemistry and (b)http://www.chemfinder.camsoft.com.

REFERENCES

Amrose, D. and Ghiassee, N.B. 1987. Vapor pressures and critical temperatures and criticalpressures of some alnoic acids: C1 to C10. Journal of Chemical Thermodynamics 19,505–517.

Cal, K. 2006. Aqueous solubility of liquid monoterpenes at 293K and relationship withcalculated log P value. Yakugaku Zasshi 126, 307–310.

Hansch, C., Leo, A., and Hoekman, D. 1995. Exploring QSAR-Hydrophobic, Electronic andSteric Constants. Washington, DC: American Chemical Society.

Helburn, R., Albritton, J., Howe, G., Michael, L., and Franke, D. 2008. Henry’s law constantsfor fragrance and organic solvent compounds in aqueous industrial surfactants. J Chem.Eng. Data 53, 1071–1079.

Iverfeldt,A. and Lindqvist, O. 1984. The transfer ofmercury at the air-water interface,Proceed-ings of the International Symposium onGas transfer atWater Surfaces,Reidel PublishingCo., NewYork, pp. 533–538.

Ji, C. and Evans, M.E. 2007. Using an internal standard method to determine Henry’s lawconstants. Environmental Toxicology and Chemistry 26, 231–236.

Ji, C., Boisvert, S.M., Arida, A.C., and Day, S.E. 2008. Measurement of Henry’s law constantusing internal standard: A quantitative GC experiment for the instrumental analysis orenvironmental chemistry laboratory. Journal of Chemical Education 65, 369–372.

Leo, A., Hansch, C., and Elkins, D. 1971. Paritition coefficients and their uses. Chem. Rev. 71,525–613.

Li, H., Ellis, D., and Mackay, D. 2007. Measurement of low air-water partition coefficients oforganic acids by evaporation from a water surface, Journal of Chemical and EngineeringData 52, 1580–1584.

Mackay, D., Shiu, W.Y., and Ma, K.C. 1992. Illustrated Handbook of Physical-ChemicalProperties and Environmental Fate for Organic Chemicals. Chelsea, MI: LewisPublishers.

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Mason, R.P., Reinfelder, J.R., and Morel, F.M.M. 1996. Uptake, toxicity and trophic transferof mercury in a coastal diatom. Environmental Science and Technology 30, 1835–1845.

Mazzoni, S.M.,Roy, S., andGrigoras, S. 1997.Eco-relevant properties of selected organosiliconmaterials. In G. Chandra (Ed.), The Handbook of Environmental Chemistry, Part H:Organosilicon Materials,Vol. 3, pp. 53–81. Berlin: Springer-Verlag.

Montgomery, J.H. 1996. Groundwater Chemicals–Desk Reference, 2nd ed. Boca Raton, FL:Lewis Publishers.

Ruelle, P. 2000. The n-octanol and n-hexane/water partition coefficient of environmentally rel-evant chemicals predicted from the mobile order and disorder (MOD) thermodynamics.Chemosphere 40, 457–512.

Sarraute, S., Mokbel, I., Costa Gomes, M.F., Majer, V., and Jose, J. 2008. Atmosphere/waterpartition of halocyclohexanes from vapor pressure and solubility data. AtmosphericEnvironment 42, 4724–4734.

Stein, E.D., Cohen, Y., and Winer, A.M. 1996. Environmental distribution and transformationof mercury compounds. Crit. Rev. Env. Sci. Technol. 26, 1–43.

Thibodeaux, L.J. 1979. Chemodynamics. NewYork, NY: John Wiley & Sons, Inc.Verschueren, K. 1983. Handbook of Environmenta1 Data on Organic Chemicals, 2nd ed.

NewYork, NY: Van Nostrand Reinhold Co.Yaffe, D., Cohen,Y., Espinosa, G., Arenas, A., and Giralt, F. 2001.A fuzzyARTMAP based on

qualitative structure-property relationships (QSPRs) for predicting aqueous solubility oforganic compounds. Journal of Chemical Information and Modeling 41, 1177–1207.

Zhou, X. and Mopper, K. 1990. Apparent partition coefficients of 15 carbonyl compoundsbetween the air and seawater and between air and freshwater; implications for air-seaexchange. Environmental Science and Technology 24, 1864–1869.

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Appendix2Standard Free Energy,Enthalpy, and Entropy ofFormation forCompounds ofEnvironmentalSignificance

Compound (kJ/mol) ΔG◦f (kJ/mol) ΔH◦

f (J/K/mol) S◦f

H+(aq) 0 0 0H2O(l) −237.2 −285.8 69.9H2O(g) −228.6 −241.8 188.7OH−(aq) −157.5 −230.3 −10.7

CO2−3 (aq) −527.9 −677.1 −56.9

HCO−3 (aq) −586.8 −692.0 91.2

H2CO3(aq) −623.2 −699.7 187.0CO2(g) −394.4 −393.5 213.6HOCl(aq) −80.0 −121.1 142.5H2S(g) −33.6 −20.6 205.7H2S(aq) −27.9 −39.8 121.3HS−(aq) 12.0 −17.6 62.8S2− (aq) 85.8 33.0 −14.6

SO2−4 (aq) −744.6 −909.2 20.1

Cl−(aq) −131.6 −167.2 56.5Cl2(aq) 6.9 −23.4 121.0Ca2+(aq) −553.5 −542.8 −53.0CaCO3(s) −1127.8 −1207.4 88.0

continued

431

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Compound (kJ/mol) ΔG◦f (kJ/mol) ΔH◦

f (J/K/mol) S◦f

Fe(metal) 0 0 27.3Fe2+(aq) −78.8 −89.1 −138.0FeS(s) −100.6 −100.1 60.4FeS2(s) −167.2 −178.5 53.0Fe(OH)3(s) −697.6 −824.2 106.8Na+(aq) −262.2 −240.5 59.1NH3(g) −16.5 −46.1 192.0NH3(aq) −26.6 −80.3 111.0NH+

4 (aq) −79.4 −132.5 113.4CH4(g) −50.8 −74.8 186.0

Note: All values are at 298 K.Source: Obtained fromWagman, D.D. et al. SelectedValues of Chemical

Thermodynamics Properties. US National Bureau of Standards,Technical Notes, pp. 270–273 (1968), pp. 270–274 (1969),pp. 270–275 (1971).

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Appendix3Selected Fragment (bj)and Structural Factors(Bk) for Octanol–WaterPartition ConstantEstimation

Fragment bj bφ

j bφφ

j bj , Other Types

Fragment Constanta

(a) Hydrocarbon increments−H 0.23 0.23>C< 0.20 0.20=C<aromatic 0.13b

=CH-aromatic 0.355−CH3 0.89 0.89−C6H5 1.90

(b) Oxygen increments−O− −1.82 −0.61 0.53 Vinylc −1.21−O-aromatic −0.08−OH −1.64 −0.44 Benzylc −1.34

(c) Carbonyl increments−C(O)− −1.90 −1.19 −0.50−C(O)-aromatic −0.59−C(O)H −1.10 −0.42−C(O)O− −1.49 −0.56 −0.09 Vinyl −1.18,

benzyl −1.38−C(O)O-aromatic −1.40−C(O)OH −1.11 −0.03 Benzyl −1.03−C(O)NH2 −2.18 −1.26 Benzyl −1.99

continued

433

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Fragment bj bφ

j bφφ

j bj , Other Types

(d) Nitrogen increments−N< −2.18 −0.93 −0.50−N<aromatic −1.12−NH− −2.15 −1.03 −0.09−NH-aromatic 0.65−NH2− −1.54 −1.00 Benzylc −1.35−NO2 −1.16 −0.03−CN −1.27 −0.34 Benzyl −0.88

(e) Halogen increments−Cl 0.06 0.94 Vinyl 0.50−Br 0.20 1.09 Vinyl 0.64

Feature BkStructural Factors

(a) Geometric featuresMultiple bond (unsaturation)Double bond −0.09d

Triple bond −0.50d

Skeletal flexingHydrocarbon chains −(N − 1)(0.12)e

Alicyclic rings −(N − 1)(0.19)Chain branchingNonpolar chain −0.13Polar chain −0.22

(b) Electronic featuresPolyhalogenation2 on the same C 0.603 on the same C 1.594 on the same C 2.882 on adjacent sp3 C 0.283 on adjacent sp3 C 0.564 on adjacent sp3 C 0.84

Polar Fragments In Chain In Aromatic Ring

On same C −0.42(b1 + b2)On adjacent C −0.26(b1 + b2) −0.16(b1 + b2)On C separated by one C −0.10(b1 + b2) −0.08(b1 + b2)Intramolecular hydrogen bondingWith −OH 1.0With −NH 0.6

a The superscript φ denotes attachment to an aromatic ring (e.g., C6H5−O−). φφ denotes fragmentto two aromatic rings (e.g., C6H5−CO−C6H5). Values are adapted from Lyman et al. (1990) andBaum (1998).

b Contribution of C shared by aromatic rings is 0.225, for C shared by aromatic rings and bonded to ahetero atom or a nonisolating C is 0.44.

c Vinyl means an isolated double bond >C=C<, benzyl means C6H5CH2−group.d Value includes deductions for removing hydrogen atoms to produce unsaturation.e N denotes the number of bonds in chain or ring.

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Appendix4Concentration Unitsfor Compartments inEnvironmentalEngineering

Concentrations of pollutants can be expressed in several units. The following is asummary of some of the common ones in air, water, and soil matrices that also appearin the text:

COMPARTMENT: AIR

The following are the common units to represent air concentration of a pollutant:

(i) A volumetric ratio (ppmv)

(ppmv) =(ViVT

)× 106,

where Vi is the volume of pollutant i and VT is the total volume(air+ pollutant). The advantage of the volume unit is that gaseous con-centrations reported in these units do not change upon gas compression orexpansion.

(ii) Amixed unitCia (μg of i/m3 of air), which is related to ppmv (at temperatureT degrees K and PT total pressure),

(ppmv) =(RT

PT

)(1

Mi

)(μg/m3)

103,

where Mi is the molecular weight of pollutant i, R is the gas constant0.08205P atm/K/mol, and1000 is a conversion factor (1000 L= 1m3).Notethat at 1 atm total pressure and standard ambient temperature of 298K, theconversion factor RT /1000PT is 0.0245.

435

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(iii) Atmospheric “mixing ratio”Note that ppmv is also equal to (Pi/PT)106 since volumes are related topartial pressures through the ideal gas law. Pi is the partial pressure of thepollutant in the gas. Analogously ppmv is also equal to (ni/nT)106, wheren represents the number of moles. Note that (Pi/PT)106 is also called the“mixing ratio χi.”

COMPARTMENT: WATER

The three most common units of a pollutant concentration in water are

(i) Molarity, Ciw = moles of i per liter (or per dm3) of solution.(ii) Molality, mi = moles of i per kilogram of water.(iii) Mole fraction of i, xi = ni/(ni + nT). Note that for most “dilute solutions,”

mole fraction and molarity are related through xi = CiwVw, where Vw is themolar volume of water (0.018 L/mol).

A common, but less precise unit of pollutant concentration in water is:(iv) Parts per million by mass, ppmm= (mg of i per kg of solution). Note that

for water since the density is 1 kg/L, ppmv is equivalent to mg of i per literof solution, ppmv = mg/L.

For acids and bases, the preferred unit of concentration is “normality,” which isthe number of equivalents per liter of solution. The number of equivalents per moleof acid equals the number of moles of H+ the acid can potentially produce.

COMPARTMENT: SOIL/SEDIMENT

Concentration of a pollutant on soil/sediment is expressed as

(i) ppmm = (Wi/Wsoil)106, where Wi is the grams of pollutant i and Wsoil isthe total amount of soil in grams. Note that this is equivalent to milligramsof i per kilogram of soil or sediment. Hence ppmm = mg/kg.

(ii) Moles of i per square meter of surface area of soil/sediment= (mg of iper kilogram of soil) (10−3Sa/Mi), where Sa is the specific area of the soil(m2 per kilogram of soil) and Mi is the molecular weight of i.

(iii) Percent by weight of i = (g of i/g of soil)100.

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Appendix5Dissociation Constantsfor EnvironmentallySignificant Acids andBases

Compound Ka1 Ka2 Ka3

Acids: HA + H2O � H3O+ + A−; Ka =[H3O+] [

A−]

[HA]Formic acid 2.1 × 10−4

Acetic acid 1.7 × 10−5

Propionic acid 1.4 × 10−5

Butyric acid 1.5 × 10−5

Valeric acid 1.6 × 10−5

Phenol 1.0 × 10−10

2-Nitrophenol 6.0 × 10−8

2-Chlorophenol 3.0 × 10−9

2,4-Dinitrophenol 6.0 × 10−6

2,4-Dichlorophenol 1.4 × 10−8

2,4,6-Trichlorophenol 1.0 × 10−6

Pentachlorophenol 1.8 × 10−5

H2CO3 4.3 × 10−7 4.7 × 10−11

H2SO3 1.3 × 10−2 6.0 × 10−8

H2S 9.1 × 10−8 1.3 × 10−13

H3PO4 7.5 × 10−3 6.2 × 10−8 4.8 × 10−13

H3BO3 7.3 × 10−10 1.8 × 10−13 1.6 × 10−14

H2SO4 1.2 × 10−2

HF 3.5 × 10−4

continued

437

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Compound Kb1 Kb2 Kb3

Bases: B + H2O � BH+ + OH−; Kb =[BH+] [OH−]

[B]Acetate 5.5 × 10−10

Ammonia 1.8 × 10−5

N2H4 8.9 × 10−5

Aniline 2.5 × 10−5

Methylamine 2.2 × 10−11

Tributylamine 1.3 × 10−11

Glycine 4.4 × 10−3 1.7 × 10−10

Urea 7.9 × 10−1

Fe(OH)3 3.1 × 10−12 5.0 × 10−12

Al(OH)3 5.0 × 10−9 2.0 × 10−10

CaOH+ 3.5 × 10−2

MgOH+ 2.5 × 10−3

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Appendix6Bond Contributions toLog Kaw for the Meylanand Howard Model

(A) Bond Contributions

Bond Contribution, qi Bond Contribution, qi

C–H −0.1197 Car–H −0.1543C–C 0.1153 Car–Car 0.2638 (intraring C to C)C–Car 0.1619 0.1490 (external C to C)C–Col 0.0635 Car–Cl −0.0241C–Ctr 0.5375 Car–O 0.3473C–N 1.3001 Car–OH 0.5967C–O 1.0855 Car–N 0.7304C–Cl 0.3335 CO–H 1.2101C–Br 0.8187 CO–O 0.0714Col–H −0.1005 CO–CO 2.4000Col=Col 0.0000 O–H 3.2318Col–Col 0.0997 O–O −0.4036Col–Cl 0.0426 N–N 1.0956Col–O 0.2051 N–H 1.2835Ctr–H 0.0040Ctr /Ctr 0.0000

(B) Correction Factors

Feature Correction Factor, Qj

Linear or branched alkane −0.75Cyclic alkane −0.28Monoolefin −0.20

continued

439

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(B) Correction Factors

Feature Correction Factor, Qj

Linear or branched alkane −0.75Linear or branched aliphatic alcohol −0.20Additional alcohol functional group above one −3.00A chloroalkane with only one chlorine +0.50A totally halogenated halofluoroalkane −0.90

Source: Adapted from Baum, E. 1998. Chemical Property Estimation:Theory and Application, Boca Raton, FL: CRC Press.

Note: Col, olefinic C; Car , aromatic C; and Ctr , bonded to a triple bond.

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Appendix7Regression Analysis(the Linear Least-SquaresMethodology)

Regression analysis is a powerful tool for establishing a relationship between two ormore variables. There are a number of cases in environmental engineering where adependent variable y and an independent variable x are measured to give an arrayof data points (xi, yi). Examples are linear free energy relationships, and numerousexamples where adsorption data, partitioning data, reaction kinetic data, etc. are fittedto a straight-line equation. In each case, a plot of xi versus yi is made and the best-fitcurve to the data is determined. The method is called linear least squares. Considerthe figure given below:

y

x

a

b

y = a + bx

The best fit through the data points is a straight line with a slope of b and anintercept of a. The objective is to estimate (i) the slope b and intercept a from the datapoints, and (ii) the degree of fit, that is, how well do the data fit the straight line?

We make the primary assumption that there is negligible error in the independentvariable x. For any given value of xi, there is an observed value yi and a correspondingcalculated value yi, calcd = a+ bxi. Thus, the residual sum of errors in the estimation

441

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is given by

W =∑i

(yi − yi, calcd

)2 =∑i

(yi − a− bxi)2.

To obtain the value of a and b, one minimizes the function W with respect to both aand b, that is, differentiate the function with respect to a and b and set them equal tozero. Thus,

∂W

∂a= −2

∑i

( yi − a− bxi) = 0

and

∂W

∂b= −2

∑i

( yi − a− bxi) xi = 0.

The above equations give us the following two equations:

na+ b∑i

xi =∑i

yi,

a∑i

xi + b∑i

x2i =∑i

xiyi,

where n is the total number of paired (x, y) data points.Solving the above equations gives us the values of a and b as follows:

a =(∑

i x2i

) (∑i yi)− (∑i xi

) (∑i xiyi

)(n∑

i x2i

)− (∑i xi)2 ,

b =(n∑

i xiyi)− (∑i xi

∑i yi)

(n∑

i x2i

)− (∑i xi)2 .

The respective standard deviations in the slope and intercept are then given by thefollowing equations:

σa = σy[ ∑

i x2i

n∑

i x2i − (∑i xi

)2]1/2

,

σb = σy[

n

n∑

i x2i − (∑i xi

)2]1/2

,

where

σ2y =∑

i (yi − a− bxi)2

(n− 2).

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Regression Analysis (the Linear Least-Squares Methodology) 443

The denominator in the above equation (n− 2) represents the total number of degreesof freedom excluding the two degrees of freedom already used up, namely the slopeand the intercept. σy is called the standard error of estimate.

The goodness of fit is ascertained in terms of the regression coefficient, which isgiven by the following equation:

r2 =∑

i

(yi, calcd − y

)2∑

i (yi − y)2,

where yi, calcd is the value estimated from the regression equation, yi is the measuredvalue, and y is the mean of the measured values. In the case of an exact fit, r2 willbe unity and will decrease in magnitude as the quality of the fit of the model to thedata diminishes. An r2 of near zero indicates that the best estimate of the dependentvariable is not any better than the overall mean of the independent variable estimatedfrom the data. It is possible that a poor model can give a high r2 and, alternatively, alow r2 can be displayed for a good model. Hence, the correlation coefficient shouldonly be taken as a general indicator of the goodness of the fit.

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Appendix8Error Function andComplementary ErrorFunction Definitions

The “error function” is defined as

erf(x) =(

2√π

) ∫ x0e−y2 dy.

The “complementary error function” is defined as

erfc(x) = 1 − erf(x).

Some characteristics of the error functions are given below:

erf(0) = 0 erfc(0) = 1erf(∞) = 1 erfc(∞) = 0erf(−∞) = −1erfc(−x) = −erf(x) erfc(−x) = 1 − erf(x) = 1 + erf(x) = 2 − erfc(x)

The following is a partial table of error function values. For a complete tabulation seeAbramovitz, M. and Stegun, I.A. (1965)Handbook of Mathematical Functions. NewYork: Dover Publications.

x erf(x)

0.00 0.00000000000.02 0.02256457470.04 0.04511110610.06 0.06762159440.08 0.09007812580.10 0.11246291600.20 0.22270258920.30 0.3286267595

445

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x erf(x)

0.40 0.42839235500.50 0.52049987780.60 0.60385609080.70 0.67780119380.80 0.74210096470.90 0.79690821241.00 0.84270079292.00 0.99532226503.00 1.00000

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Appendix9Cancer Slope Factor andInhalation Unit Risk forSelected Carcinogens

ReferenceCancer Oral Oral Reference InhalationSlope Factor Dose (RfD) Inhalation unit Risk Concentration

Chemical (mg/kg/d) (mg/kg/d) (per μg/m3) (RfC) (mg/m3)

Arsenic 1.5 3.4 × 10−4 4.3 × 10−3 —Benzene 4.0 × 10−3 4.0 × 10−3 2.2 × 10−6 3 × 10−2

Cadmium — 5 × 10−4 (water) 1.8 × 10−3 —1 × 10−3 (food)

Carbon tetrachloride 0.13 7.4 × 10−4 1.5 × 10−5 —Chloroform — 1 × 10−2 2.3 × 10−5 —Methyl chloride 0.0075 — — 9 × 10−2

p,p′-DDT 0.34 5 × 10−4 — —Dieldrin 0.16 5 × 10−5 — —Hexachloroethane 0.014 1 × 10−3 4.0 × 10−6 —PCBs 0.04–2.0 — 1.0 × 10−4 —Naphthalene — 2.0 × 10−2 — 3 × 10−3

Vinyl chloride 0.72–1.5 3.0 × 10−3 (4.4–8.8) × 10−6 1.0 × 10−1

Source: U.S. EPA IRIS database (1989). Available at http://www.epa.gov/iriswebp/iris/subst/index.html.

RfD: An estimate of a daily oral exposure to human population that is likely to be without anappreciable risk of deleterious effects during a lifetime.

RfC: An estimate of a continuous inhalation exposure to the human population that is likely to bewithout an appreciable risk of deleterious effects during a lifetime.

Oral slope factor: An upper bound, approximating a 95% confidence limit, on the increased cancerrisk froma lifetimeoral exposure to an agent.This estimate, usually expressed in units of a populationaffected per mg/kg d, is generally for use in the low-dose region of the dose–exposure relationship,that is, for exposures corresponding to risks < 1 in 100.

Inhalation unit risk: The upper bound excess lifetime cancer risk estimated to result from continuousexposure to an agent at a concentration of 1μg/L in water or 1μg/m3 in air. The interpretation ofinhalation unit risk would be as follows: If unit risk = 2 × 10−6 μg/L, two excess cancer cases areexpected to develop per 1,000,000 people if exposed daily for a lifetime to 1 μg of the chemical in1 L of drinking water.

447

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Appendix10U.S. National AmbientAir Quality Standards

Primary Standards Secondary Standards

Pollutant Level Averaging Time Level Averaging Time

Carbon monoxide 9 ppm (10mg/m3) 8 h None35 ppm(40mg/m3)

1 h

Lead 1.5μg/m3 Quarterly average Same as primaryNitrogen dioxide 0.053 ppm

(100μg/m3)

Annual (arithmeticmean)

Same as primary

Particulate matter(PM10)

150μg/m3 24 h Same as primary

Particulate matter(PM2.5)

15.0μg/m3 Annual (arithmeticmean)

Same as primary

35μg/m3 24 h Same as primaryOzone 0.075 ppm

(2008 Standard)8 h Same as primary

0.08 ppm(1997 Standard)

8 h Same as primary

0.12 ppm 1 h (applies only inlimited areas)

Same as primary

Sulfur dioxide 0.03 ppm Annual (arithmeticmean)

0.5 ppm(1300μg/m3)

3 h

0.14 ppm 24 h

Source: Available at http://www.epa.gov/air/criteria.html#1.

449

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“78194_C007.tex” — page 451[#1] 22/4/2009 19:27

Index

A

Absorption, 153, 285oxygen, 152, 299

Accelerated growth phase, 252Accidental spill scenario, 323Acid-base

catalyzed reaction, 225enthalpy, 24–25reaction, 24–25solubility and octanol–water partition

constant, 84surface exchange reaction, 156

Acid–base hydrolysismechanisms, 227rate for organic compounds, 226

Acid rain, 132, 330potential, 401

Acridines, 167Acrolein, 329Activated carbon adsorption

breakthrough time, 327continuous mode fixed-bed operation, 173treating wastewater stream, 174–175

Activated complex theorychemical reaction kinetics, 210–212reaction, 210

Activated sludge process, 273Activation energy, 208–215Active zone, 324, 325Activity, 43, 46, 47Activity coefficients, 47

binary liquid mixture, 87correlation equations, 87nonideal solutions, 69–71variation, 65water, 82

Adenosine diphosphate (ADP), 377Adenosine triphosphate (ATP), 377

living cell synthesis, 380Adiabatic system, 14

dry and moist, 16expansion, 19, 21lapse rate, 16, 320

ADP. SeeAdenosine diphosphate (ADP)Adsorption, 95–96, 153

air environment, 324–326equilibria examples, 90isotherm, 89, 90, 93–95

wave analysis, 325zone, 324, 325

Advection, 281and dispersion equation, 360, 361

Aeration basins air stripping, 299–302Aeration lagoon, 302Aerobes, 391Aerosols, 136

dry deposition velocities, 143formula and health effects, 356fraction wet deposition, 138–142pollutants dry deposition, 143

Afterburners, 327Air–aerosol partition constant, 167–170Air bubble in water

cross-section, 35fugacity capacities definition, 52

Air environment, 313–356adsorption, 324–326air pollution control, 324–329aqueous droplets, 330–336atmospheric processes, 330–356box models, 313–318dispersion models, 319–323F&T models, 313–323global warming and greenhouse effect,

337–346photolytic rate constants, 247stratosphere and troposphere ozone, 347–356thermal destruction, 327–329

Air pollutantconcentration, 435–436control, 324–329plume emission, 321

Air relative humidity, 368Air sparging, 370Air stripping, 303Air-vegetation partition constant, 171–174Air–water

aerosol-bound fraction wet deposition,138–142

aerosol-bound pollutants drydeposition, 143

atmospheric aqueous dropletsthermodynamics, 147–149

chemical exchange, 120contact and equilibrium, 121equilibrium constant, 119equilibrium in atmospheric chemistry,

136–151451

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452 Index

Air–water (Continued)gases dry deposition flux from atmosphere,

144–146mass transfer coefficients, 145separation processes, 121vapor species wet deposition, 137waste treatment systems air/water equilibrium,

150–151Air–water partition constant, 120, 123, 126,

128–129, 145ionic strength, 128–129measured, 128measurements methods, 125temperature, 127temperature dependence, 127value comparison, 126

Air–water phase equilibrium, 119–135environmental variables effects, 126–135Henry’s constants estimation from groups

contributions, 123–124Henry’s law of constants experimental

determination, 125Aitken particles, 149Alcohols, 84Aldehydes, 356Aldrin

Freundlich isotherm, 174liquid phase mass transfer coefficient, 293

AlewifePCB concentration, 396

Aliphatic alcohols, 84Aliphatic hydrocarbons, 84Alkanes

solubility and octanol–water partitionconstant, 84

washout ratios range, 140Alkenes, 84Alkyl benzoates, 84Alkyl halides, 84Alkyl nitrate, 356Alkylbenzenes, 84Alkynes, 84Alumina surface properties, 158Ambient air organics, 169Ammonia

air–water partition constants, 134, 145Antoine constants, 60pH effect, 134rainwater concentration, 331

Anaerobes, 391Aniline, 62Anionic surfactants effects, 131Anisotropic forces, 32

surface water molecule, 33Anthracene

air–water partition constant and DOC, 131experimental and predicted vapor pressure, 62

photolytic rate constants, 247vegetation atmosphere partition coefficient vs.

temperature correlations, 172Antoine constants, 60Antoine equation, 60Aquatic food chain bioaccumulation kinetics,

393–397Aquatic stream

box models, 313transport and fate processes, 313

Aqueous concentration, 37Aqueous droplets

air environment, 330–336Kohler curves, 149oxidation times, 337

Aqueous solubilityKohler curves, 149octanol–water partition constant

calculations, 83using SAR correlations, 84

Arbitrary plane, 35ARP. SeeAcid rainArrhenius, Svänte, 208Arrhenius complex, 224Arrhenius equation, 208, 209, 212Arrhenius expression, 230Atmosphere

aerosols, 138aqueous droplets thermodynamics, 147–149carbon dioxide flux, 147carbon dioxide increase, 347CFC concentration, 353chemical reactions, 206detergent, 218global temperature change, 347greenhouse gases, 340moisture, 136particles and droplets properties, 137photolysis rate constant, 246–247pollutants, 138pressure and temperature profiles, 18–20processes, 330–356residence time, 319solar energy adsorbed and reflected, 338

ATP. SeeAdenosine triphosphate (ATP)Autocatalysis in natural waters manganese

oxidation, 235–236Autocatalytic reaction concentration

profile, 235Automobile emissions, 316Autooxidation kinetics, 236

B

Barrier, 8Batch operation, 299Batch process, 151

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Index 453

Batch reactors, 267, 268ideal reactors, 269pesticide disappearance, 381

Bed sedimentsdiffusive transport, 366fate and transport process, 365

Beer–Lambert law, 244Beer’s law, 244Belousov–Zhabotinsky reaction, 234Benzanthracene, 62Benzene

accidental spill scenario, 323air–water partition constants and air–water

mass transfer coefficients, 145Antoine constants, 60concentration in wastewater lagoon, 136degrees of freedom, 76excess Gibbs energy, 70experimental and predicted vapor pressure, 62flux, 151Freundlich isotherm, 174hydrophobicity, 67liquid phase mass transfer coefficient, 293soil bulk density, 372surface impoundment, 151thermal oxidation reaction rate parameters, 329

Benzo[a]anthracene, 172Benzo[a]pyrene, 172BET. See Brauner-Emmett-Teller (BET) equationBinary liquid mixture, 87Bioaccumulation, 165Biochemical oxygen demand (BOD), 203

curve, 296defined, 295

Biochemical systems, 380Bioconcentration factor, 165, 393

equilibrium thermodynamics applications,164–167

Biological uptake, 386Biota

equilibrium thermodynamics applications,164–167

fugacity capacities definition, 52Biphenyl

Antoine constants, 60bioconcentration factor, 167hydrophobicity, 67liquid phase mass transfer coefficient, 293molar volume, 50

BOD. See Biochemical oxygen demand (BOD)Boltzmann constant, 211Boltzmann equation, 101Bond contribution scheme, 124Box models. See also specific type

air environment, 313–318aquatic stream, 313material balance, 314

oceans, 343–345urban atmosphere, 314well-mixed, 316

Branching ratio, 208Brauner-Emmett-Teller (BET) equation, 97,

99–100Brauner-Emmett-Teller (BET) isotherm,

97, 98shapes, 99

Bromoform, 164Brönsted environmental engineering, 243Bulk phases, 32Bulk water molecule, 33Butane

free energy values, 82thermodynamic function, 77

Butanol adsorption, 96Butene, 329

C

CalTOX, 6Canadian Environmental Protection Act (CEPA), 7Capillary condensation, 64Capillary phenomenon, 34Carbon dioxide

air–water partition constant pH effect, 134atmosphere, 147atmosphere global temperature change, 347equilibrium between air and water, 135–136flux, 147monthly average concentration, 339percent contributions emissions, 340seawater, 147water, 205

Carbon monoxideair–water partition constants, 145concentration, 316standards, 449

Carbon tetrachlorideair–water mass transfer coefficients, 145air water partition constant, 126Freundlich isotherm, 174hydrophobicity, 67

Carcinogen cancer slope factor, 447–448Carnot engine, 17, 21Catalysis. See also Heterogeneous catalysis;

Homogeneous catalysisautocatalysis, 235–236defined, 221environmental, 221–236enzyme, 247–253surface, 228–233

Catalyzed reactions mechanisms and rates,222–223

Cationic surfactant effects, 131Cell kinetics, 252Celsius thermometers, 15

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454 Index

Centimeter-gram-second (CGS) system ofunits, 11

CEPA. See Canadian Environmental ProtectionAct (CEPA)

CFC. See Chlorofluorocarbons (CFC)CGS. See Centimeter-gram-second (CGS) system

of unitsChain reactions, 216–218Chapman mechanism, 348ChemCAN, 6Chemical equilibrium, 190

and Gibbs Free Energy, 26–27Chemical kinetics. See Chemical reaction kinetics;

Environmental systems applications ofchemical kinetics

Chemical manufacturing processenvironmental risk evaluation, 401–405traditional vs. green, 399

Chemical potential, 29, 43, 45calculation, 47–48defined, 30equilibrium criteria, 52significance, 30

Chemical process, 399Chemical process industry (CPI), 86Chemical reaction kinetics, 189–266

activated complex theory, 210–212activation energy, 208–215catalyzed reactions mechanisms and rates,

222–223chain reactions, 216–218chemical reaction progress toward

equilibrium, 190–192definition, 3, 189environmental catalysis, 221–236environmental photochemical reactions,

244–246environmental reactions autocatalysis,

234–236enzyme catalysis, 247–253heterogeneous catalysis, 228homogeneous catalysis, 224–226initial rate method, 195integrated rate laws, 195–201ionic strength effects on rate constants,

219–220isolation method, 195kinetic rate laws, 194–207linear free energy relationships, 215Michaelis–Menten kinetics, 247–253Monod kinetics, 247–253rates, 240–243reaction mechanisms, 216–218reaction rate, order, and rate constants, 193redox reactions, 237–243reversible reactions, 198–200solutions reactions, 219–220

solvent effect on reaction rates, 213–214steady-state approaches series reactions,

201–206surface catalysis, 228–233

Chemical sediment–water exchange, 364–365Chemical soil–air exchange, 366–368Chemical thermodynamics, 13–42

chemical equilibrium and Gibbs Free Energy,26–27

chemical potential standard states, 31curved interfaces andYoung–Laplace

equation, 33–34definition, 3enthalpy and heat capacity, 22equilibrium, 13first and second laws combination, 25first law of thermodynamics, 16free energy variation with temperature

and pressure, 26–27Gibbs adsorption equation, 37Gibbs–Duhem relationship for single

phase, 30Gibbs Free Energy and chemical potential,

29–30maximum work, 28reaction, formation, and combustion

enthalpies, 23–24role, 10second law of thermodynamics, 17surfaces and colloidal systems

thermodynamics, 32–37surface tension, 32surface thermodynamics and Gibbs

equation, 36surface thickness and Gibbs dividing

surface, 35thermodynamics laws, 13–25third law of thermodynamics, 18–21zeroth law of thermodynamics, 15

Chemiosmosis, 378Chlordane

adsorption isotherm, 94air to particulate partition constant and

temperature relationship, 169Chloride, 164Chlorinated pesticides, 170Chlorobenzene

Freundlich isotherm, 174octanol–water partition constant, 69sequential breakthrough, 164thermal oxidation reaction rate parameters, 329

Chlorobiphenyl, 69Chlorofluorocarbons (CFC), 128, 144, 317,

350–352atmospheric concentration, 353production and dissipation, 352source reduction effect, 352–354

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Index 455

Chloroformair water partition constant, 126, 128air–water partition constants and air–water

mass transfer coefficients, 145Antoine constants, 60Freundlich isotherm, 174hydrophobicity, 67sequential breakthrough, 164streams, 295

Chloromethanes, 127Chromophores, 245Chrysene, 247Clausius–Clapeyron equation, 59, 61, 126

multicomponent equilibrium thermodynamics,58–62

Closed system, 13CMC. See Critical micellar concentration (CMC)Coexistence point, 58Cohesive energy density, 215Colloids, 129

system thermodynamics, 32–37Combustion incinerator, 275Co-metabolite compound, 254Composite rate, 393Compounds. See also specific name

Antoine constants, 60standard free energy, enthalpy, and entropy,

431–432vs. temperature, 65

Concentrationdriving force, 121SI and CGS unit relationships, 11

Conservation of mass, 267Contaminant transport, 367Continuous-flow stirred tank reactors (CSTR),

267, 268, 273, 371approach, 301box models, 313, 314, 315–316enzyme reactors, 383–386epilimnion of lake, 292ideal reactors, 269microbial growth, 384model, 272, 292one-box model, 291reactor volumes, 272substrate kinetics, 384surface impoundment, 272

Continuous mode fixed-bed operation, 173Continuous process, 151Convection term, 281CPI. See Chemical process industry (CPI)Cradle-to-grave cycle, 405Cristobolite, 234Critical micellar concentration (CMC), 131Critical oxygen deficit, 297Critical point, 59Critical radius, 148

CSTR. See Continuous-flow stirred tank reactors(CSTR)

Cyclohexane, 82Cyclopentane

free energy values, 82octanol–water partition constant, 69

D

Damköhler number, 281, 389Darcy velocity, 362Davies equation, 49DCE. See Dichloroethane (DCE)Deacon’s power law, 320Death phase, 252Debye–Huckel equation, 49Debye–Huckel limiting law, 80, 220Debye–Huckel theory, 49, 79–80Decane, 62Decelerated growth phase, 252Dehydrogenases, 378Del operator, 361Dense nonaqueous-phase liquid

(DNAPL), 358Deoxygenation, 297Departure from equilibrium, 123Deposition

dry, 137, 143flux, 143fog pesticide, 142

Detergents, 131Dibenzofuran, 368Dibromosuccinic acid decomposition, 209Dichlorobenzene

air/water partition constants and air/watermass transfer coefficients, 145

Antoine constants, 60hydrophobicity, 67

Dichloroethane (DCE)adsorption, 95–96Freundlich isotherm, 174linear adsorption isotherms, 97

Diffuse-double-layer model, 101Diffuse region, 32Diffusion-limited region, 284Diffusive transport, 366Dilute solutions, 55–68Dilution rate, 384Direct reaction, 307Dispersion model, 281

air environment, 319–323environmental systems chemical kinetics

applications, 277number, 278plug-flow schematic, 277soil column, 279

Dispersivity, 362

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456 Index

Dissociated ionic species in water, 89Dissolution

ammonia in water, 134carbon dioxide in water, 133water, 132

Dissolved air flotation, 151Dissolved organic colloids (DOC),

129–131, 159DNAPL. See Dense nonaqueous-phase liquid

(DNAPL)DOC. See Dissolved organic colloids (DOC)Dose-response model, 6Double-layer thickness, 103Dry adiabatic lapse rate, 19, 20Dry and moist adiabatic lapse rates, 16Dry deposition, 137

velocities in aerosols, 143Dynamic equilibrium, 58

E

Eadie-Hofstee plot, 250–251Earth’s atmosphere. See alsoAtmosphere

solar energy adsorbed and reflected, 338Effective double-layer thickness, 102Electrochemistry, 237Electron acceptors, 378Electron activity in rainwater, 238Electron-deficient silicon atom, 155Electron transport, 380Emission, 321

automobile, 316Endergonic reaction, 377End-of-the-pipe waste treatment philosophy, 398Energy. See also Free energy; Gibbs free energy;

Linear free energy relationships(LFER)

activation, 208–215cohesive density, 215consumption, 1enthalpy, and entropy, 431–432Hammett linear free energy relationships, 215Helmholtz free energy, 14, 25Marcus free energy relationships, 243modes of transfer, 14population and pollution, 1solar, 338work and heat unit relationships, 11

Enthalpy, 14acid–base reaction, 24–25activation, 212chemical thermodynamics, 22

Entropically unfavorable, 75Entropy, 18

activation, 212nonpolar compounds in water

solution, 76

Environmentactivity coefficients variation, 65awareness, 2catalysis, 221–236chemical concentration units, 44chemical kinetics, 201chemicals properties, 425–430compartments, 5, 52compounds standard free energy, enthalpy, and

entropy, 431–432compounds vs. temperature, 65exposure pathways risks, 6impact, 1integrated rate laws, 197lapse rate, 320matrix concentrations calculation, 45mercury flux, 146models and fugacity, 51–54partitioning, 53–54phases, 14processes, 376processes chemical thermodynamics and

kinetic applications, 9reactions, 240separation processes, 8significant acids and bases dissociation

constants, 437–438standards and criteria, 2stress, 1sulfur dioxide flux, 145system, 13, 197

Environmental bioengineering chemical kineticsapplications, 375–397

aquatic food chain bioaccumulation kinetics,393–397

batch reactors, 381enzyme reactors, 380–392

Environmental engineeringchemical thermodynamics and kinetics, 4–10compartments concentration units, 435–436definition, 3discipline, 2–3equilibrium partition constants, 243equilibrium partitioning, 4fate and transport modeling, 5–7pillars, 3separation processes design, 8–10thermodynamics and kinetics applications,

4–10Environmental Protection Agency (EPA), 2Environmental systems applications of chemical

kinetics, 267–419adsorption, 324–326aeration basins air stripping, 299–302air environment, 313–356air pollution control, 324–329aquatic food chain bioaccumulation kinetics,

393–397

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Index 457

aqueous droplets, 330–336atmospheric processes, 330–356batch reactors, 269, 381box models, 313–318chemical sediment–water exchange, 364–365chemical soil–air exchange, 366–368CSTR, 269, 276, 383–386dispersion and reaction, 280dispersion models, 277, 319–323environmental bioengineering chemical

kinetics applications, 375–397environmental impact analysis, 399–405enzyme reactors, 380–392ex situ soil and solid waste incineration, 375F&T modeling, 313–323, 357–368fate and transport, 290–294fluid–fluid interfaces kinetics and transport,

284–285global warming and greenhouse effect,

337–346green engineering applications, 398–405groundwater transport, 358–363heterogeneous medium reaction, 281–285ideal reactors, 267–276immobilized enzyme or cell reactors, 387–390in situ soil vapor stripping in Vadose Zone,

373–374in situ subsoil bioremediation, 391–392lakes and oceans chemicals, 290–292life cycle assessment, 405NAPL groundwater removal via P&T,

370–372natural streams biochemical oxygen demand,

295–298natural waters photochemical reactions,

311–312nonideal reactors, 277–289oxidation reactor, 303–308plug-flow enzyme reactors, 382plug-flow reactor (PFR) or tubular

reactor, 270porous medium diffusion and reaction,

286–289reactor types, 267–289soil and groundwater treatment, 369–375soil and sediment environments, 357–375stratosphere and troposphere ozone, 347–356surface waters chemicals, 293–294tanks-in-series model, 278–279thermal destruction, 327–329wastewater treatment photochemical reactions,

309–310water environment, 290–298water pollution control, 299–312

Enzyme(s), 248immobilization techniques, 388substrate interactions, 248

Enzyme catalysis, 248chemical reaction kinetics, 247–253reaction rate, 249

Enzyme reactors, 380–392batch reactors, 381continuous stirred tank enzyme reactors,

383–386environmental bioengineering chemical

kinetics applications, 380–392immobilized enzyme or cell reactors, 387–390in situ subsoil bioremediation, 391–392plug-flow enzyme reactors, 382

EPA. See Environmental Protection Agency (EPA)EPICS method, 125Epilimnion of lake

CSTR model, 292CSTR one-box model, 291

Equilibrium. See also Henry’s constantchemical potential criteria, 52chemical thermodynamics, 13defined, 14environmental compartments, 5partitioning compounds, 144state, 14

Equilibrium constants, 190, 192environmental engineering, 243environmental engineering chemical

thermodynamics and kinetics, 4intrinsic and conditional, 156standard thermodynamic data, 201

Equilibrium thermodynamics applications,119–188

aerosol-bound fraction wet deposition,138–142

aerosol-bound pollutants drydeposition, 143

air-to-aerosol partition constant, 167–170air–water equilibrium in atmospheric

chemistry, 136–151air–water phase equilibrium, 119–135atmospheric aqueous droplets

thermodynamics, 147–149bioconcentration factor, 164–167biota/water partition constant, 164–167environmental variables effects on air water

partition constant, 126–135gases dry deposition flux from atmosphere,

144–146Henry’s constants estimation from groups

contributions, 123–124Henry’s law of constants experimental

determination, 125metal ions adsorption on soils and sediments,

155–158organic molecules on soils and sediments

adsorption, 159–163soils and sediments partitioning from water,

153–154

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458 Index

Equilibrium thermodynamics applications(Continued)

soil–water sediment–water equilibrium,152–163

vapor species wet deposition, 137vegetation atmosphere partition coefficient,

171–174waste treatment systems air/water equilibrium,

150–151wastewater treatment activated carbon

adsorption, 172–174Error function, 445–446Esters, 84Ethane

free energy values, 82thermal oxidation reaction rate

parameters, 329thermodynamic function, 77

Ethanol, 329Ethers

bioconcentration factor and octanol–waterpartition constant correlation, 167

solubility and octanol-water partitionconstant, 84

Ethylbenzene, 86Freundlich isotherm, 174

EU. See European Union (EU) 7th amendmentEuropean Union (EU) 7th amendment, 7Evaporation

benzene from water, 122mass transfer coefficient, 122well-stirred surface, 293

Excess partial molar Gibbs energy, 70Exchangeable fraction, 138Exergonic reaction, 377Exponential growth phase, 252Extended Debye–Huckel equation, 49Extensive properties, 14Extracellular enzymes, 391

F

F&T. See Fate and transport (F&T) modelingFacultative anaerobes, 391Fate and transport (F&T) modeling, 291

air environment, 313–323aquatic stream, 313bed sediments, 365environmental engineering chemical

thermodynamics and kinetics, 5–7lake, 292soil and sediment environments, 357–368water environment, 290–294

Fermentative mode, 378Fermentors, 380Fick’s equation for molecular diffusion,

277, 361

Field gradient, 8Film theory, 122First and second laws combination, 25First law of thermodynamics, 16Flory–Huggins equation, 87Fluid–fluid interfaces kinetics and transport,

284–285Fluoranthene

air to particulate partition constant andtemperature relationship, 169

vegetation atmosphere partition coefficient vs.temperature correlations, 172

Flushing, 292Flux, 120, 137, 343

air relative humidity, 368gases dry deposition, 144–146mass transfer, 123mercury, 146water, 368

Fog pesticide deposition, 142Food chain

bioaccumulation of water pollutants, 394steady-state uptake, 395

Forceper unit length, 33SI and CGS unit relationships, 11

Fractional conversion, 193Fragment constant, 67

octanol-water partition constant, 68–69Free energy

change of reaction, 191per unit surface area, 33solution Henry’s constant, 81and temperature example, 28variation with temperature and pressure,

26–27Free radicals chain reaction, 216Freon, 224Freundlich isotherm, 94, 172, 326

adsorption, 92hexachlorobutadiene adsorption, 174–175

Fröessling correlation, 283Fugacity, 43, 45–54

capacity, 51coefficient, 46concept, 46condenses phases, 46derived, 45environmental models, 51–54environmental partitioning model, 53–54equilibrium criteria, 52gases, 46ionic strength and activity coefficients, 49–50level II calculation, 54liquid-gas interface, 91solutes and activity coefficients, 47–48

Fulvic acid, 130

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Index 459

G

GAC. See Granular activated carbon (GAC)Gaia hypothesis, 337Ganglion solubilization, 359Gases

dry deposition from atmosphere, 144–146dry deposition velocities, 143phase on to solid, 98phase reaction activation parameters, 213schematic adsorption, 98

Gaussian air dispersion model, 323Geothite surface properties, 158GHG. See Greenhouse gases (GHG)Gibbs, J. Willard, 26Gibbs activation energy, 208Gibbs adsorption equation, 37Gibbs convention, 36Gibbs dividing surface, 35Gibbs–Duhem equation, 37Gibbs–Duhem relationship for

single phase, 30Gibbs equation, 36

nonionic and ionic systems, 89Gibbs free energy, 14, 25, 80, 191

activation, 212chemical equilibrium, 26–27chemical thermodynamics, 29–30equilibrium criteria, 52liquid dissolution, 72nonideal solutions, 86partial molar, 30

Gibbs function variables, 192Gibbs–Helmholtz equation, 27Gibbs–Helmholtz relation, 72Gibbsite surface properties, 158Global carbon reservoirs and fluxes, 343Global mixing model, 317Global surface warning, 346Global warming potential (GWP), 401

air environment, 337–346Glucose, 379Glycolysis, 378Granular activated carbon (GAC), 93

adsorption isotherm, 94Green engineering applications, 398–405

environmental impact analysis, 399–405life cycle assessment, 405Sandestin Declaration principles, 400

Greenhouse gases (GHG), 338air environment effect, 337–346atmosphere, 340radiative forcing effects, 341

Grotthus–Draper law, 244Groundwater

contaminant plume, 363–364contaminant transport, 364contamination, 358

dissolving NAPL, 359environment, 360P&T approach, 370P&T process, 371pollutant travel time, 162–163remediation, 370, 371transport of soil and sediment, 358–363treatment, 369–375

Group contribution scheme, 124Growth kinetics, 252GTP. See Guanosine triphosphate (GTP)Guanosine triphosphate (GTP), 377Guggenheim quasi-chemical approximation, 87Guntelberg equation, 49Guoy–Chapman theory, 101, 103GWP. See Global warming potential (GWP)

H

Half-cell reactions, 238, 239Half-life of reactant, 197Halobenzenes, 84Halogenated benzenes, 84Halogenated hydrocarbons, 167Hammet environmental engineering, 243Hammett linear free energy relationships, 215Hazardous waste incinerator efficiency, 376HAZCHEM, 6Heat, 14

capacity, 22defined, 23engine, 20Hess’s law of summation, 24

Helmholtz free energy, 14, 25Henry, William, 55Henry’s constant, 213

estimation from groups contributions,123–124

free energy of solution, 81standard deviation, 125vapor-liquid equilibrium, 55

Henry’s Law, 55, 71different units, 57experimental determination, 125illustration, 57

Heptachlor degrees of freedom, 76Hess’s law of heat summation, 24Heterogeneous catalysis, 221

chemical reaction kinetics, 228environmental engineering, 222Langmuir isotherm, 250water ester hydrolysis, 231–232

Heterogeneous reaction, 281–285diffusion and reaction-limited

regions, 283potential energy surface, 231solid surface, 229

“78194_C007.tex” — page 460[#10] 22/4/2009 19:27

460 Index

Hexacane molar volume, 50Hexachlorobenzene, 86

air-to-particulate partition constant andtemperature relationship, 169

air–water partition constant, 128Freundlich isotherm, 174

Hexachlorobutadiene adsorption, 174–175Hexadecane molar volume, 50Hexane

octanol–water partition constant, 69thermal oxidation reaction rate

parameters, 329Hoff complex, 224Homogeneous catalysis, 221

chemical reaction kinetics, 224–226environmental engineering, 222

Horizontal dispersion coefficients, 323Humic acid, 130

hematite and kaolinite sorption, 159Hurricane heat engine, 20Hydration forces, 75Hydraulic conductivity, 362Hydraulic gradient, 362Hydrocarbons

dissolution in water, 82formula and health effects, 356

Hydrogen bonds, 74Hydrogen halide formation, 216Hydrogen peroxide, 60Hydrometeors, 136Hydrophobic hydration, 76Hydrophobicity, 33, 67

effect, 77molecule misnomer, 81molecules in water, 77–78nonideal solutions correlations, 72–81octanol–water partition constant, 129pollutants, 129

Hydrosols defined, 136Hygroscopic, 79Hysteresis effect, 386

I

Ideal fluids, 43–44concentration units, 44dilute solution definition, 44

Ideal gas, 31, 43law, 43

Ideal reactors, 267–276batch reactors, 269CSTR, 269, 276environmental systems chemical kinetics

applications, 267–276PFR, 270types, 268

Ideal solution, 31

Ideal solutions, 55–68curved surfaces vapor pressure, 63–65Henry’s Law, 55linear free energy relationships, 69liquid–liquid equilibrium, 66–69octanol–water partition constant, 66–68organic compounds vapor pressure, 58–62Raoult’s Law, 56–57vapor-liquid equilibrium, 55–56

ILCR. See Incremental lifetime cancer risk (ILCR)Illite surface properties, 158Immobilization techniques, 388Immobilized enzyme, 388Incinerators

air pollution control, 330industrial effluents, 327

Incremental lifetime cancer risk (ILCR), 7Indoor air pollution model, 317Induced air flotation, 151Induced fit model, 248Induction period, 202, 234Inert, 14Infinite dilution, 72Ingestion toxicology potential (INGTP), 401INGTP. See Ingestion toxicology potential

(INGTP)Inhalation toxicology potential (INHTP), 401Inhalation unit risk, 447–448INHTP. See Inhalation toxicology potential

(INHTP)Initial rate method, 195Inner sphere (IS) complex, 241Inorganic compounds, 60

solubility in water, 88Inorganic gases, 132

water, 204Inorganic pollutants, 145. See also specific nameInsecticide, 94. See also specific chemicalsIntegrated rate laws, 195–201Intensive properties, 14Interface, 32

adsorption and surface, 88–102tension, 33

Internal pressure, 215International Panel on Climate Change

(IPCC), 338International System of Units (SI), 11International Union of Pure and Applied

Chemistry (IUPAC), 11, 15Intracellular enzymes, 391Intrinsic bioremediation, 370Ionic strength

effects on rate constants, 219–220, 221fugacity, 49–50vs. mean ionic activity coefficient, 49

IPCC. See International Panel on Climate Change(IPCC)

“78194_C007.tex” — page 461[#11] 22/4/2009 19:27

Index 461

Iron oxide surface properties, 158IS. See Inner sphere (IS) complexIsolated system, 13Isolation method, 195Isotherm constants, 100Isotropic forces, 32

bulk water molecule, 33IUPAC. See International Union of Pure and

Applied Chemistry (IUPAC)

J

Junge’s equation, 170

K

Kelvin equation, 15, 63, 64, 147application, 64pure water droplets, 149

Ketones, 84Kilojoules, 26Kinetics. See also Chemical reaction kinetics;

Environmental systems applications ofchemical kinetics; Michaelis-Mentenkinetics

applications, 4–10aquatic food chain bioaccumulation, 393–397autooxidation, 236biological uptake, 386carbon dioxide in water, 205cell, 252growth, 252Monod, 247–253, 391rate constants, 243rate laws, 194–207salt effects, 220smog formation urban area, 354

Kohler curves, 148aqueous solution droplets, 149pure water droplets, 149

Kurbatov plot, 158

L

LAB. See Linear alkyl benzene sulfonates (LAB)Lakes and oceans chemicals, 290–292Lake trout, 396Lambert’s law, 244Langmuir adsorption

constant, 230constant for water on aerosols, 171heterogeneous catalysis, 250isotherm, 92, 94, 97, 159, 229–230plot, 250–251

Langmuir–Hinshelwood mechanism, 229Langmuir–Rideal mechanism, 229, 230

LAS. See Linear alkyl sulfates (LAS)LCA. See Life Cycle Assessment (LCA)LEA. See Local equilibrium assumption (LEA)Lead standards, 449Lewis–Randall rule, 48LFER. See Linear free energy relationships

(LFER)Life Cycle Assessment (LCA), 399, 405Light nonaqueous-phase liquid

(LNAPL), 358Lindane

air–water partition constant, 126air–water partition constants and air–water

mass transfer coefficients, 145experimental and predicted vapor pressure, 62

Linear adsorptionconstant, 91equation, 91isotherms, 97isotherm schematic, 92

Linear alkyl benzene sulfonates (LAB), 131Linear alkyl sulfates (LAS), 131Linear free energy relationships (LFER), 160, 161,

216, 242between bioconcentration factor and

octanol–water partition constant, 166chemical reaction kinetics, 215defined, 69environmental engineering, 243

Linear least-squares methodology, 441–444Lineweaver–Burke plot, 250–251Liquid dissolution, 72Liquid–gas interface, 91Liquid–gas system, 95Liquid–phase mass transfer coefficient, 293Liquid phase on to solid, 98LNAPL. See Light nonaqueous-phase liquid

(LNAPL)Local equilibrium assumption (LEA), 361Lock-and-key model, 248Louisiana sediment/soil composition, 146

M

Malathion, 247Manganese, 236Marcus theory, 241, 242, 243Margules equation, 87Mass balance, 5, 267Mass transfer coefficient, 122Mass transfer flux, 123Maximum useful work calculation, 28–29McDevit–Long theory, 49Mean cell residence time, 384Mean ionic activity coefficient

calculation, 51vs. ionic strength, 49

“78194_C007.tex” — page 462[#12] 22/4/2009 19:27

462 Index

Meniscus, 65Mercury

environment, 146liquid phase mass transfer coefficient, 293washout ratios range, 140

Metal ions adsorption on soils and sediments,155–158

Metal oxide, 154semiconductor, 310

Methaneair–water mass transfer coefficients, 145behavior, 75free energy values, 82thermal oxidation reaction rate parameters, 329thermodynamic function, 77thermodynamic transfer functions, 61

Methyl chloride, 329Meylan and Howard model, 439–440Micelle–water partition constant, 373Michaelis–Menten kinetics, 381, 391

chemical reaction kinetics, 247–253equation, 249, 253–254parameter estimation, 250–251rate law, 390

Microbially mediated environmental redoxprocesses, 377

Microorganisms growth curve, 252Microscopic reversibility principle, 199Millington–Quirk approximation, 286Mineralization, 377Mirex, 247Molar enthalpy, 73Molar extinction coefficient, 244Molar heat capacity equations, 23Mole fraction of component, 32Molecularity, 194

aqueous solubility parameters, 83estimating activity coefficient, 82–83theories of solubility, 80–81

Monod kinetics, 391chemical reaction kinetics, 247–253equation, 252, 253–254

Monophenyl terephthalate (MPT), 231heterogeneous catalysis, 232

Montreal Protocol, 338, 351, 353MPT. SeeMonophenyl terephthalate (MPT)Multicomponent equilibrium thermodynamics,

43–118Clausius–Clapeyron equation, 58–62concentration units, 44condenses phases, 46curved surfaces vapor pressure, 63–65dilute solution definition, 44environmental models, 51–54fugacity, 45–54gases, 46Henry’s Law, 55

ideal and nonideal fluids, 43–44ideal solutions and dilute solutions, 55–68ionic strength and activity coefficients, 49–50LFER, 69liquid–liquid equilibrium, 66–69nonideal solutions, 69–87octanol-water partition constant, 66–68organic compounds vapor pressure, 58–62Raoult’s Law, 56–57solutes and activity coefficients, 47–48surfaces and interfaces adsorption, 88–102vapor–liquid equilibrium, 55–56

Multicomponent heterogeneous systems, 43Multimedia approach, 5Multimedia fate and transport model, 5, 6, 7Muscovite surface properties, 158Mutarotation, 226–227

N

Naphthalene, 83, 86air–water mass transfer coefficients, 145Antoine constants, 60experimental and predicted vapor pressure, 62hydrophobicity, 67molar volume, 50photolytic rate constants, 247soil bulk density, 372

NAPL. See Nonaqueous-phase liquid (NAPL)Natural attenuation, 370Natural environment, 13

composition, 4Natural environmental systems, 267Natural gas, 329Natural water

organic compounds, 218photochemical reactions, 311–312photochemical transient species, 312

Negatively charged surface, 102Nernst equation for electrochemical

cells, 239Net depletion, 89Net surface excess, 89Nitrate, 331Nitrogen, 60Nitrogen dioxide, 449

ozone concentration, 355Nitrous oxide, 145Nonaqueous-phase liquid (NAPL), 358, 370

extraction, 373flushing, 372ganglion solubilization, 359groundwater dissolving, 359groundwater removal via P&T, 370–372removal, 370–375

Nonideal fluids, 43–44

“78194_C007.tex” — page 463[#13] 22/4/2009 19:27

Index 463

Nonideal reactors, 277–289dispersion model, 277tanks-in-series model, 278–279

Nonideal solutions, 69–87activity coefficient, 69–70activity coefficient and solubility, 70–71aqueous solubility prediction theoretical and

semi-empirical approaches, 83–87excess functions, 70excess Gibbs Free Energy models, 86first-generation group contribution methods,

83–85hydrophobicity correlations, 72–81inorganic compounds solubility

in water, 88molecular theories of solubility, 80–81nonpolar solutes hydrophobic hydration,

75–77second-generation group contribution

methods, 86–87solute hydrophilic interactions in water, 79solute hydrophobic interactions, 78structure activity relationships and activity

coefficients in water, 82UNIFAC method, 86–87water structural features, 73–74

Nonpolar solutes hydrophobic hydration, 75–77NRTL, 70

equation, 87

O

OC. See Organic matter (OC)Oceans two-box model, 343–345Octadecane molar volume, 50Octanol–water partition constant, 161, 433–434Open system, 13

sulfur dioxide oxidation, 333–334Organic adsorption rate and hydrolysis rate, 228Organic and inorganic particulate sizes, 129Organic carbon, 159Organic colloids characterized, 129Organic compounds

acid–base hydrolysis rate, 226Antoine constants, 60Freundlich isotherm, 174homogeneously catalyzed, 227–228hydrophobicity, 67photodegradation, 311seawater molar volume, 50vapor pressure, 58–62

Organic matter (OC), 160partition constant, 161

Organic moleculessoils and sediments adsorption, 159–163water, 209

Organic pollutants, 145

OS. See Outer sphere (OS) complexOscillatory reactions, 234Outer sphere (OS) complex, 241Overall effectiveness factor, 288Oxidation, 237

chain reaction, 218Marcus free energy relationships, 243states, 237sulfur dioxide to sulfate, 332times aqueous droplet, 337

Oxidation reactor, 303–308Oxide minerals, 158Oxygen

air/water partition constants and air/watermass transfer coefficients, 145

deficit, 296, 297–299demand, 295–296uptake, 152, 299

Ozoneair-water partition constants, 145concentrations, 355continuous reactor, 308decomposition in impure water, 306decomposition in pure water, 305depletion in water, 308reactor, 309standards, 449

P

P&T. See Pump-and-treat (P&T) technologyPAH. See Polyaromatic hydrocarbon (PAH)PAN. See Peroxyacyl nitrates (PAN)Parathion, 247Partial molar Gibbs free energy, 30Partial pressure, 31Particulate, 129Partition coefficients, 52. See also specific typePartition constant, 66. See also specific typePartitioning pollutant, 372Pasquill stability categories, 320, 322PCB. See Polychlorinated biphenyls (PCB)PCDD washout ratios range, 140Peclet number, 281Pentachlorophenol, 145Pentafluorobenzene, 60Pentane, 82Permissible process, 28Peroxyacyl nitrates (PAN), 356Pesticides

bioconcentration factor and octanol-waterpartition constant correlation, 167

photolytic rate constants, 247volatilization rate, 369

PFR. See Plug flow reactors (PFR)Phase addition, 8Phase creation, 8Phase diagram, 58

“78194_C007.tex” — page 464[#14] 22/4/2009 19:27

464 Index

Phase equilibrium, 13Phenanthrene

air relative humidity, 368air-to-particulate partition constant and

temperature relationship, 169air–water mass transfer coefficients, 145air–water partition constant, 128–129, 145experimental and predicted vapor pressure, 62molar volume, 50vegetation atmosphere partition coefficient vs.

temperature correlations, 172washout ratios range, 140

Phenol, 62pH of rainwater concentration, 331Phosphorylation, 377Photochemical transient species, 312Photochemistry, 244Photodegradation, 311Photolytic rate constants, 247Phthalate esters, 231Physical chemistry, 237Picloram degradation rate, 382Plug flow, 301Plug flow reactors (PFR), 268, 278, 330

combustion incinerator, 275environmental systems chemical kinetics

applications, 382ideal reactors, 270reactor volumes, 272schematic, 271, 277

Poisson–Boltzmann equation, 155, 232Poisson equation, 101Polar compounds, 75Pollutants

air concentration, 435–436atmosphere, 138bioconcentration factor, 167dispersion, adsorption, and reaction, 363between dissolved organic colloids and

water, 130gaseous plume reflection, 322partitioning, 372soil/sediment concentration, 436standards, 449washout ratios range, 140water concentration, 436

Pollution, 1Polyaromatic hydrocarbon (PAH)

air-to-particulate partition constant andpressure relationship, 170

air-to-particulate partitioning, 169–170photolytic rate constants, 247vegetation atmosphere partition coefficient vs.


Polychlorinated biphenyls (PCB)

air-to-particulate partition constant andpressure relationship, 170

air–water partition constants and air–watermass transfer coefficients, 145

concentration, 396molecular volume, 84octanol–water partition constant, 84solubility, 84vegetation atmosphere partition coefficient vs.


Polycyclic aromaticssolubility and molecular volume, 84

Polynuclear aromaticssolubility and octanol–water partition

constant, 84Population and pollution, 1Porewater solute partitioning, 154Porous medium, 363

diffusion and reaction, 287Porous medium diffusion and reaction

environmental systems chemical kineticsapplications, 286–289

Potential determining ions, 155Potential energy surface, 210

heterogeneous reaction, 231Potential radius, 148Poynting correction, 46Pre-exponential factor, 208Pressure, 11Pressure-volume-temperature (P-V-T)

relationship, 43Process, 14Propagators, 307Propane

free energy values, 82thermal oxidation reaction rate

parameters, 329thermodynamic function, 77

PSDF washout ratios range, 140Pseudo-first-order rate, 195Pseudo-second-order rate, 195Pseudo-steady-state approximate

(PSSA), 202, 306Pulse input, 280Pump-and-treat (P&T) technology, 174

groundwater remediation, 370, 371Pure water, 132

activated, 148Kelvin equation, 149Kohler curves, 149

PVT. See Pressure-volume-temperature (P-V-T)relationship

Pyreneair relative humidity, 368air to particulate partition constant and

temperature relationship, 169

“78194_C007.tex” — page 465[#15] 22/4/2009 19:27

Index 465

Antoine constants, 60hydrophobicity, 67photolytic rate constants, 247soil bulk density, 372washout ratios range, 140

Pyruvic acid oxidation, 379

Q

Quantum efficiency, 245Quartz

equilibrium and rate constants, 234surface properties, 158

Quasi-steady state, 190

R

Radiative forcing, 345Radionuclides, 140Rain or fog pollutants scavenging, 335Rainwater

constituents, 331electron activity, 238

Raoult’s Law, 147, 165activity coefficient, 57ideal solutions and dilute solutions, 56–57illustration, 57

Rate constant, 194ionic strength effect, 221

Rate-determining step, 204RBCA. See Risk-based corrective action (RBCA)Reaction, 281

acid catalyzed and base catalyzed, 222activated complex, 210coordinate, 210defined, 193equilibrium, 132extent, 191limited regions, 283, 284loss, 292quotient, 192rate constant, 198

Reactors. See also specific typemass balance, 269material balance with dispersion and

reaction, 280Reaeration, 297Real gas, 43

mixture, 48Real solutions, 48Redox potential discontinuity (RPD), 237Redox reactions, 237–243Reductive dechlorination, 237Reflection, 322Regression analysis, 441–444Regression coefficient, 443Relative risk index (RRI), 401

Remedial process categories, 369Residence time, 315Retardation factor, 163, 361, 367Revelle buffer factor, 341Reversible reactions, 198–200Reynolds number, 283Risk-based corrective action (RBCA), 364

paradigm, 365River pulse input, 280Rotary kiln incinerator, 327, 328RPD. See Redox potential discontinuity (RPD)RRI. See Relative risk index (RRI)

S

Salting in, 50Salting-out

behavior, 128process, 50

Sandestin Declaration, 398, 400Saturated zone, 324Scatchard–Hildebrand equation, 214Scavenging

rain or fog pollutants, 335ratio, 336

Schmidt number, 283Seawater

activity of water, 48carbon dioxide flux, 147

Second law of combination, 25Second law of thermodynamics, 17Sedimentation

rate constant and expression, 292transport processes, 366

Sediment–water exchange, 364–365Self-exchange reaction, 242Separation factor, 301Separation processes, 8, 10

air–water contact and equilibrium, 121design, 8–10

Series reaction, 203Setschenow equation, 50Sevin, 247SFP. See Smog formation potential (SFP)Sherwood number, 283SI. See International System of Units (SI)Silanol, 154Silica

dissolution in water, 233–234equilibrium and rate constants, 234

SimpleBOX, 6Smog formation potential (SFP), 401

urban area, 354Sodium montmorillonite surface properties, 158Soil–air exchange, 366–368Soil and sediment environments, 357–375

bulk density, 372chemical sediment-water exchange, 364–365

“78194_C007.tex” — page 466[#16] 22/4/2009 19:27

466 Index

Soil and sediment environments (Continued)chemical soil–air exchange, 366–368concentration units, 44ex situ soil and solid waste incineration, 375F&T modeling, 357–368fugacity capacities definition, 52groundwater transport, 358–363in situ soil vapor stripping in Vadose Zone,

373–374NAPL groundwater removal via P&T,

370–372partitioning from water, 153–154pesticide volatilization rate, 369pollutant concentration, 436soil and groundwater treatment, 369–375solute partitioning, 154treatment, 369–375

Soil vapor stripping well, 374Soil water, 162Soil–water sediment–water equilibrium, 152–163

metal ions adsorption on soils and sediments,155–158

organic molecules on soils and sedimentsadsorption, 159–163

soils and sediments partitioning from water,153–154

Solar energy adsorbed and reflected, 338Solid agent, 8Solid–air interface, 89Solid mixture, 48Solid surface, 229Solid–vapor equilibrium, 59, 60Solid-water interface, 89Solubility

alcohols, 84group contributions, 85–86nonpolar solute in water, 81octanol–water partition constant, 84

Soluteactivity coefficients, 47–48fugacity, 47–48hydrophilic interactions, 79hydrophobic interactions, 78molecule, 74partitioning soil and porewater, 154sequential breakthrough, 164

Solventinternal pressure, 215sublation, 303

Sorption, 153Sparingly soluble compound, 71Specific growth rate, 252Specific rate, 194Stable, 14Standard deviations, 442Standard enthalpy, 24Standard error of estimate, 443

Standard states, 23Stanton number, 326Stark–Einstein law, 244, 245State of equilibrium, 14, 43Stationary phase, 252Steady state, 190, 276

approaches series reactions, 201–206food chain uptake, 395

Stern layer, 103Steroids, 84Stoichiometric equation, 190Stratosphere–troposphere, 318Streams

chloroform loss, 295material balance, 294oxygen deficit, 297–299

Streeter–Phelps oxygen-sagcurve, 298equation, 297

Stripping, 285Structural factor, 67Sub-cooled liquid states, 59, 71Submerged aeration, 152Submerged bubble aeration, 300Sulfate

rainwater concentration, 331washout ratios range, 140

Sulfur dioxideair–water mass transfer coefficients, 145air–water partition constant pH effect, 134, 145Antoine constants, 60oxidation and open system effect, 333–334standards, 449water, 206

Supercritical states, 59Suppressors, 307Surface

catalysis, 228–233changed surfaces adsorption, 101–102chemical thermodynamics, 32–37colloidal systems thermodynamics, 32–37complexation model, 101, 157concentration, 37curved interfaces andYoung–Laplace

equation, 33–34deficiency, 35defined, 32, 35equilibrium adsorption isotherms at interfaces,

90–100excess, 35Gibbs adsorption equation, 37Gibbs equation for nonionic and ionic

systems, 89and interfaces adsorption, 88–102surface tension, 32tension, 33thermodynamics and Gibbs equation, 36

“78194_C007.tex” — page 467[#17] 22/4/2009 19:27

Index 467

thickness, 35thickness and Gibbs dividing surface, 35water environment, 293–294water molecule anisotropic forces, 33waters chemicals, 293–294

Surface impoundmentbenzene flux, 151CSTR model, 272exit vs. inlet concentration ratio, 274primary loss mechanisms, 273VOC transport, 151

Surface tension, 37chemical thermodynamics, 32substances, 34surfaces and colloidal systems

thermodynamics, 32System variables, 14

T

Taft linear free energy relationships. See Linearfree energy relationships (LFER)

Tanks-in-series model, 277, 278–279TCE. See Trichloroethane (TCE)Terminators, 307Tetrachloroethane, 292Tetrachloroethylene

Antoine constants, 60experimental and predicted vapor pressure, 62Freundlich isotherm, 174

Tetradecane, 50Thermal destruction of air environment, 327–329Thermal incinerator (afterburner), 328Thermal oxidizers, 327Thermodynamics, 4–10

defined, 14developments, 15equilibrium partitioning, 4fate and transport modeling, 5–7laws, 13–25separation processes design, 8–10transfer functions, 75

Thermometer, 15Thiele modulus, 288, 289, 390Third law of thermodynamics

chemical thermodynamics, 18–21Three dimensional ice lattice structure, 73Time-invariant state, 276Titanium, 222, 309–311Toluene

air water partition constant, 128Freundlich isotherm, 174hydrophobicity, 67thermal oxidation reaction rate parameters, 329

Toxic Substances Control Act (TSCA), 7Trace gases, 337Transition state, 210

Triaryl phosphate estershydrolysis, 215

Trichlorobenzene, 83Trichlorobiphenyl, 128Trichloroethane (TCE)

adsorption, 95–96linear adsorption isotherms, 97

Trichloroethylene, 173Trichlorofluoromethane, 145Trifluralin, 247Trimethylpentane, 82Triple point, 58Trout, 396Trouton’s rule, 61, 71TSCA. See Toxic Substances Control Act (TSCA)Tubular reactor. See Plug flow reactors (PFR)Tungsten, 406Two-film theory of mass transfer, 284

U

Uncatalyzed reaction, 225UNIFAC. See Universal Functional group Activity

Coefficient (UNIFAC) methodUNIQUAC. See Universal Quasi Chemical

(UNIQUAC) equationUnited States National Ambient Air Quality

Standards, 449–450Units and dimensions, 11Universal Functional group Activity Coefficient

(UNIFAC) method, 87–89nonideal solutions, 86–87

Universal Quasi Chemical (UNIQUAC)equation, 87

Universal solvent, 70Unsteady-state, 190Urban area

atmosphere, 314carbon monoxide concentration, 316smog constituents, 356smog diurnal variations, 357smog formation kinetics, 354well-mixed box model, 316

V

Vadose zone, 375van der Waals forces, 33van Laar equation, 87Vapor–liquid equilibrium

dilute solutions, 55–56Henry’s constant definitions, 55ideal solutions, 55–56

Vapor pressure, 62above aqueous droplets, 149–150curved surfaces, 63–65estimation, 62

Vapors extraction, 375

“78194_C007.tex” — page 468[#18] 22/4/2009 19:27

468 Index

Vapor species wet deposition, 137Vegetation. see alsoAir-vegetation partition

constantatmosphere partition coefficient, 171–174atmospheric pollutants uptake, 172

Vermiculite surface properties, 158Vertical and horizontal dispersion coefficients, 323Vicinal water, 154Vinyl chloride, 329Viscosity, 11Volatile organic compounds (VOC), 98, 144

desorption, 299reaction rate constant, 198transport and surface impoundment, 151

Volatilizationcompound from water, 122rate constant and expression, 292

W

Washout ratio, 138calculation, 141

Wasted work, 28Waste treatment

separation processes, 120systems air/water equilibrium, 150–151

Wastewater lagoonbenzene concentration, 136submerged bubble aeration, 300

Wastewater oxidationozone in continuous reactor, 308ozone reactor, 309

Wastewater stream, 174–175Wastewater tank, 152Wastewater treatment, 173Water, 70

air relative humidity, 368and atmosphere chemicals exchange, 120confined in capillary, 65dissolution, 132

droplets in air, 64ester hydrolysis, 231–232fugacity capacities definition, 52heterogeneous catalysis, 231–232and organic solvents concentration units, 44partition constants and mass transfer

coefficients, 145photochemical reactions, 311–312photochemical transient species, 312pollutant concentration, 436tetrahedral cage molecule, 74

Water environment, 290–298aeration basins air stripping, 299–302fate and transport, 290–294lakes and oceans chemicals, 290–292natural streams biochemical oxygen demand,

295–298natural waters photochemical reactions,

311–312oxidation reactor, 303–308photolytic rate constants, 247surface waters chemicals, 293–294wastewater treatment photochemical reactions,

309–310water pollution control, 299–312

Water pollutantscontrol, 299–312food chain bioaccumulation, 394

Well-mixed box model, 316Well-mixed troposphere and stratosphere, 318Well-stirred surface evaporation, 293Wet deposition, 137Wilson equation, 87Work, 14, 16, 28

Y

Young–Laplace equation, 34, 63chemical thermodynamics, 33–34

elements of environmental engineering: thermodynamics and kinetics, third edition

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