Lecture series 7 - Civil and Environmental Engineering | SIU

Transcript Lecture series 7 - Civil and Environmental Engineering | SIU

CE 510 Hazardous Waste Engineering

Department of Civil Engineering Southern Illinois University Carbondale Instructors: Jemil Yesuf Dr. L.R. Chevalier Lecture Series 7: Biotic and Abiotic Transformations

Course Goals

       Review the history and impact of environmental laws in the United States Understand the terminology, nomenclature, and significance of properties of hazardous wastes and hazardous materials Develop strategies to find information of nomenclature, transport and behavior, and toxicity for hazardous compounds Elucidate procedures for describing, assessing, and sampling hazardous wastes at industrial facilities and contaminated sites Predict the behavior of hazardous chemicals in surface impoundments, soils, groundwater and treatment systems Assess the toxicity and risk associated with exposure to hazardous chemicals Apply scientific principles and process designs of hazardous wastes management, remediation and treatment

Abiotic and Biotic Transformations

 Abiotic  Chemical and physical transformations  Hydrolysis, Redox reactions, Photolysis,…  Biotic   Transformation of contaminants through biological processes Results in mineralization of both natural and engineered organic compounds

BIOLOGICAL TREATMENT OF HAZARDOUS WASTE DEGRADATION OF ORGANIC WASTE BY THE ACTION OF MICROORGANISMS This degradation alters the molecular structure of the organic compound

TWO DEGREES OF DEGRADATION BIOTRANSFORMATION Breakdown of organic compound to daughter compound MINERALIZATION Complete breakdown of organic compound into cellular mass, carbon dioxide, water and inert inorganic residuals

Schematic diagram of biodegradation process

A A A bacterial cell A A A A A An organic reactant A is bound to an extracellular enzyme

Schematic diagram of biodegradation process

A A bacterial cell A A A The enzyme transports the organic reactant A into the cell.

Schematic diagram of biodegradation process

The organic reactant provides the energy to synthesize new cellular material, repair damage, and transport nutrients across the cell boundary B C A O 2 CO 2 H 2 0

Schematic diagram of biodegradation process

A Enzyme bound chemicals A A A bacterial cell Transport of chemicals across the cell boundary A A bacterial cell A A A A A A A A CO 2 B Breakdown of chemicals C H 2 0 O 2

Definitions

 Microbes need carbon and energy source (electron donors)     Light – phototrophs – carry out photosynthesis Chemical sources – chemotrophs   Inorganic source – lithotroph   Ammonia, NH 3, Ferrous iron, Fe 2+, Sulfide, HS Manganese, Mn 2+ NH 3 + O 2  NO 2 - + H 2 O + Energy Organic source – organotrophs  Examples include the food you eat  C 8 H 10 + 10.5O

2  8CO 2 + 5H 2 O + Energy Autotrophs – obtain carbon from carbon dioxide  6CO 2 + Energy + 6H 2 O  C 6 H 12 O 6 + 6O 2 Heterotrophs – obtain carbon from organic matter  C 8 H 10 + 10.5O

2  8CO 2 + 5H 2 O + Biomass

Definitions

 Microbes also need electron acceptor Source: Newell et al., 1995 The biochemical energy associated with alternative degradation pathways can be represented by the redox potential of the alternative electron acceptors The more positive the redox potential, the more energetically favorable is the reaction utilizing that electron acceptor.

See Textbook example 7.7

Governing Variables

 Chemical structure and Oxidation state     Persistent hazardous wastes – some halogenated solvents, pesticides, PCBs  xenobiotics Branching, hydrophobicity, HC saturation and increased halogenation are reported to decrease rates of biodegradation and reactivity Oxidation state of a contaminant is an important predictor of abiotic and biotic transformation This number changes when an oxidant acts on a substrate.  Redox reactions occur when oxidation states of the reactants change

Class Example

What is the average oxidation state of carbon in a) Methane b) c) d) TCA TCE PCE

Solution

a) b) c) d) Methane TCA TCE PCE (-IV) (0) (I) (+II)

Governing Variables

 Presence of reactive species  Abiotic and biotic transformations require the presence of   Oxidant Hydrolyzing agent (nucleophile)   Microorganisms Appropriate transforming species  Availability  Sorption  NAPLs

Other Variables

   Dissolved oxygen  Aerobic and anerobic biodegradations Temperature  Two fold increase in reaction rate for each rise of 10ºC  pH Empirical equation in biological treatment engineering: k 2 = k 1 Θ (T2-T1)  Optimal pH for growth varies

Oxidation-Reduction (Redox) Reactions   Living organisms utilize chemical energy through redox reactions This is a coupled reaction  Transfer of electrons from one molecule to another  Electron acceptor - Oxidizing agents  Electron donor - Reducing agents

Redox Reactions

The tendency of a substance to donate electrons or accept electrons is expressed as the reduction potential E o (measured in volts) Negative E o Positive E o – donors - acceptors e e -

Redox Reactions

Oxidation Process in which an atom or molecule loses an electron Reduction Process in which an atom or molecule gains an electron e e -

Redox Reactions

Oxidation Process in which an atom or molecule loses an electron Na (s)  Na + + e Reduction Process in which an atom or molecule gains an electron Cl 2(g) + 2e  2Cl e e -

Redox Reactions

These “half reactions” occur in pairs.

Together they make a complete reaction.

2Na (s)  2Na + + 2e Cl 2(g) + 2e  2Cl Na (s) + Cl 2(g)   Na + + 2Cl -

Tables for Half Reactions

Reduction Half-Reaction Li + (aq) + e -> Li(s) Ca 2+ (aq) + 2e -> Ca(s) Na + (aq) + e -> Na(s) Mg 2+ (aq) + 2e -> Mg(s) 2H + (aq) + 2e Fe 3+ (aq) + e -> H 2 (g) -> Fe 2+ (aq) Ag + (aq) + e -> Ag(s) Hg 2+ (aq) + 2e -> Hg(l) 2Hg 2+ (aq) + 2e -> Hg 2 2+ (aq) NO 3 (aq) + 4H + (aq) + 3e -> NO(g) + 2H 2 O(l) O O 2 3 (g) + 4H (g) + 2H F 2 (g) + 2e + + (aq) + 4e (aq) + 2e -> 2F (aq) -> 2H 2 O(l) -> O 2 (g) + H 2 O(l) Standard Potential E ° (volts) -3.04

-2.76

-2.71

-2.38

0 0.77

0.8

0.85

0.9

0.96

1.23

2.07

2.87

These equations are written as reductions.

For oxidation, the equation would be in reverse.

E o would also change signs.

Tables for Half Reactions

-2.76

-2.71

-2.38

0 0.77

0.8

0.85

0.9

0.96

1.23

2.07

2.87

A full redox reaction is a combination of a reduction equation and an oxidation equation

Redox Equations

Redox pairs (O/R) are expressed such that the oxidizing agent (electron acceptor) is written on the left, while the reducing agent (electron donor) is written on the right.

To pair two reactions as redox, one of the pairs are written as a reduction, the other as oxidation.

CO 2 /C 6 H 12 O 6 and O 2 /H 2 O

Redox Equations

To determine whether a chemical is oxidized or reduced, consider E o from the standard reduction table. For the pairs below: CO 2 /C 6 H 12 O 6 and O 2 /H 2 O 6CO 2 + 24H O 2 (g) + 4H + + +24e + 4e = C 6 H 12 O 6 = 2H 2 O E o E o = -0.43 V = 0.82 V The negative E 0 value indicates that this reaction should be written in reverse (oxidation)

Balancing Redox Equations

Consider the metabolism of glucose by aerobic microorganisms. Write the balanced reaction that combines the redox pairs CO 2 /C 6 H 12 O 6 and O 2 /H 2 O.

(work as class example)

Solution

Glucose is the energy source, and the electron donor. It will be oxidized. Oxygen, on the other hand, is the electron acceptor, it will be reduced.

1. Write the two half reactions

C

H

O

2 

H

 2

O CO

2    (

oxidation

)    (

reduction

)

Solution

2. Balance the main elements other than oxygen and hydrogen

2 

 2



no change



3. Balance oxygen by adding H 2 0 and hydrogen by adding H+

2  4

  6

 6

2  2

 24



Solution

4. Balance the charge by adding electrons

2  4

  6

 6

2  4

  2

 24

  24

 5. Multiply each half reaction by the appropriate integer that will result in the same number of electrons in each. Then add the two half reactions to come up with the balanced reaction.

Solution

2  4

  6

 6

2  4

  2

 24

  24



6 6

2  6

 24

  2

 6

2   12

 24

  24



6  6

2  6

2  6

Example

Balance the redox reaction of sodium dicromate (Na 2 Cr 2 O 7 ) with ethyl alcohol (C 2 H 5 OH) if the products of the reaction are Cr +3 and CO 2 strategy

Strategy

       Balance the principal atoms Balance the non-essential ions Balance oxygen with H 2 O Balance hydrogen with H + Balance charges with electrons Balance the number of electrons in each half reaction and add together Subtract common items from both sides of the equation.

Solution

Free Energy of Formation, G

f o  Energy released or energy required to form a molecule from its elements   By convention, G f 0 of the elements (O 2 , C, N 2 ) in their standard state is zero.

Some representative values G given on the next slide f 0 are

Free Energy of Formation, G

f o Compound C 6 H 12 O 6 CO 2 O 2 H 2 0 CH 4 N 2 0 G f o , kJ/mole -917.22

-394.4

0 -237.17

-50.75

104.18

Using G aA + bB f 0 you can calculate whether a reaction will occur. For the reaction  cC + dD D G o = cG f o (C)+dG f o (D) – aG f 0 (A) – bG f o (B)

Class Example

One mole of methane (CH D G o .

4 ) and two moles of oxgyen are in a closed container. Determine if the reaction below will proceed as written based on CH 4 + 2O 2  CO 2 + 2H 2 0

Solution

Compound CO 2 O 2 H 2 0 CH 4 G f o , kJ/mole -394.4

0 -237.17

-50.75

CH 4 + 2O 2  CO 2 + 2H 2 0 D G o = cG f o (C)+dG f o (D) – aG f 0 (A) – bG f o (B) =(-394.4)+2(-237.17) -(-50.75)-2(0) = -817.99 kJ/mole This is a large negative value, the reaction will proceed as written.

Relationship between

G

o The electromotive force, E o is related to ΔG o

and

E

o D

G o

 

nFE

0 Where ΔG o = the Gibbs energy of reaction at 1 atm and 25 o C n = number of electrons in the reaction F = caloric equivalent of the faraday = 23.06 kcal/volt-mole E o is related to the equilibrium constant, K, by:

E o



ln(

)

Where: R=universal gas constant=0.00199 kcal/mol o K T=temperature( o K)

Binary Fission

 1 2 4 8 16 32   P = P o (2) n P o is the initial population at the end of the accelerated growth phase  P is the population after n generations

Microbial Growth

Time

Microbial Growth

Lag Phase Adjustment to new environment, unlimited source of nutrient and substrate Time

Microbial Growth

Lag Phase Accelerated growth phase bacteria begin to divide at various rates Time

Microbial Growth

Lag Phase Exponential growth phase differences in growth rates not as significant because of population increase Accelerated growth phase Time

Microbial Growth

Stationary phase substrate becomes exhausted or toxic by products build up resulting in a balance between the death and reproduction rates Lag Phase Exponential growth phase Accelerated growth phase Time

Microbial Growth

Lag Phase Stationary phase Death phase Exponential growth phase Accelerated growth phase Time

Rates of Transformation

   Kinetics of transformations are difficult to quantify Furthermore, soil, groundwater and hazardous waste treatment systems are so complex that the exact transformation pathway cannot be elucidated However, the prediction of rates is necessary in order to  Perform site characterization   Perform facilities assessment Design treatment systems

Rates of Transformation

Generalized equation 

   



C = Contaminant concentration k = proportionality constant (units dependent on reaction order) n = reaction order

Zero Order Kinetics

 

C t d

   



n d

   

0 

C o



First Order Kinetics

 

C t d

   

d dt

   



1 

C o e



Second Order Kinetics



   



enzyme





but d





k dt C OH dt d



enzyme



 

d Therefore



d dt



'  0 (

steady state

)

where k

' 

[

enzyme

Text Problem 7.4

The biodegradation rate of benzo[a]pyrene has been described by the expression 

    

Where, k=3X10 -15 L/cell-h [C] = conc. of benzo[a]pyrene [X] = biomass conc.

During a bioremediation project of a contaminated groundwater, the biomass concentration reached a steady state at 7.1X10

11 cell/L during treatment and remained at approximately that concentration through out the project. If Co is 25 ug/L and the hydraulic detention time of the groundwater as it passes through the control volume is 10 days, determine the effluent concentration of benzo[a]pyrene as the water exits the system.

Solution

[X] = 7.1X10

11 t = 240 days cell/L C o = 25 ug/L 

    

k’ = k[X] = (3x10 -15 L/cell-hr)(7.1x10

= 0.00213 hr -1 11 cell/L) Therefore, C = C o e -k’t = (25 ug/L) e -(0.00213 hr-1 x 240 hr) = 15 ug/L

…end of solution

Michaelis-Menton Kinetics

It is a saturation phenomena described by:



max



C K m

where V = rate of transformation (mg/L  h) V max = maximum rate of transformation (mg/L C = contaminant concentration (mg/L) K m = half-saturation constant (mg/L)  h)

Michaelis-Menton Kinetics

V max 0.5 V max ..

Contaminant Concentration (mg/L)

Class Example

Describe how you would get Km and Vmax from the following data.

Initial Conc. (mg/L) 8 14 23 32 47 55 65 Initial Rate (mg/(L-min)) 1.2

1.6

2.4

2.7

2.8

Summary of Important Points and Concepts

   Biotransformation refers to the breakdown of a chemical into daughter compounds whereas mineralization is the complete breakdown of a compound Redox reactions can be used to determine the biological or chemical oxidation/reduction of waste Estimates of the kinetics of waste reduction are necessary to assess and design treatment of hazardous waste.