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
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
A
B
C
CO2
O2
H2 0
Schematic diagram of
biodegradation process
Enzyme bound chemicals
A
A
Transport of chemicals across
the cell boundary
A
bacterial
cell
A
A
A
A
A bacterial
cell A
A
A A
A
A
Breakdown of chemicals
A
B
C
CO2
O2
H2 0
Definitions
 Microbes need carbon and energy source (electron
donors)


Light – phototrophs – carry out photosynthesis
Chemical sources – chemotrophs
 Inorganic source – lithotroph


Ammonia, NH3, Ferrous iron, Fe2+, Sulfide, HS-Manganese, Mn2+
NH3 + O2  NO2- + H2O + Energy
 Organic source – organotrophs


Examples include the food you eat

C8H10 + 10.5O2  8CO2 + 5H2O + Energy
Autotrophs – obtain carbon from carbon dioxide
 6CO2 + Energy + 6H2O  C6H12O6 + 6O2

Heterotrophs – obtain carbon from organic matter
 C8H10 + 10.5O2  8CO2 + 5H2O + 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) TCA
c) TCE
d) 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
Empirical equation in biological treatment
engineering: k2 = k1 Θ(T2-T1)
 pH

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 Eo
(measured in volts)
e-
Negative Eo – donors
Positive Eo - acceptors
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
e-
Na(s) Na+ + eReduction
Process in which an atom
or molecule gains an
electron
Cl2(g) + 2e-  2Cl-
e-
Redox Reactions
These “half reactions” occur in pairs.
Together they make a complete reaction.
2Na(s) 2Na+ + 2eCl2(g) + 2e-  2ClNa(s) + Cl
2(g)
 Na+ + 2Cl-
Tables for Half Reactions
Reduction
Standard
Potential
Half-Reaction
E° (volts)
Li+(aq) + e- -> Li(s)
-3.04
Ca2+(aq) + 2e- -> Ca(s)
-2.76
Na+(aq) + e- -> Na(s)
-2.71
Mg2+(aq) + 2e- -> Mg(s)
-2.38
2H+(aq) + 2e- -> H2(g)
0
Fe3+(aq) + e- -> Fe2+(aq)
0.77
Ag+(aq) + e- -> Ag(s)
0.8
Hg2+(aq) + 2e- -> Hg(l)
0.85
2Hg2+(aq) + 2e- -> Hg22+(aq)
0.9
NO3-(aq) + 4H+(aq) + 3e- -> NO(g) + 2H2O(l)
0.96
O2(g) + 4H+(aq) + 4e- -> 2H2O(l)
1.23
O3(g) + 2H+(aq) + 2e- -> O2(g) + H2O(l)
2.07
F2(g) + 2e- -> 2F-(aq)
2.87
These
equations are
written as
reductions.
For oxidation,
the equation
would be in
reverse.
Eo would also
change signs.
Tables for Half Reactions
Reduction
Standard
Potential
Half-Reaction
E° (volts)
Li+(aq) + e- -> Li(s)
-3.04
Ca2+(aq) + 2e- -> Ca(s)
-2.76
Na+(aq) + e- -> Na(s)
-2.71
Mg2+(aq) + 2e- -> Mg(s)
-2.38
2H+(aq) + 2e- -> H2(g)
0
Fe3+(aq) + e- -> Fe2+(aq)
0.77
Ag+(aq) + e- -> Ag(s)
0.8
Hg2+(aq) + 2e- -> Hg(l)
0.85
2Hg2+(aq) + 2e- -> Hg22+(aq)
0.9
NO3-(aq) + 4H+(aq) + 3e- -> NO(g) + 2H2O(l)
0.96
O2(g) + 4H+(aq) + 4e- -> 2H2O(l)
1.23
O3(g) + 2H+(aq) + 2e- -> O2(g) + H2O(l)
2.07
F2(g) + 2e- -> 2F-(aq)
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.
CO2/C6H12O6 and O2/H2O
Redox Equations
To determine whether a chemical is oxidized or reduced,
consider Eo from the standard reduction table. For the
pairs below:
CO2/C6H12O6 and O2/H2O
6CO2 + 24H+ +24e- = C6H12O6 Eo = -0.43 V
O2(g) + 4H+ + 4e- = 2H2O Eo = 0.82 V
The negative E0 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 CO2/C6H12O6 and O2/H2O.
(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 6 H 12 O 6  CO 2    ( oxidation )
O 2  H 2O
   ( reduction )
Solution
2. Balance the main elements other than
oxygen and hydrogen
C 6 H 12 O 6  6 CO 2
O 2  H 2O
 no
change

3. Balance oxygen by adding H20 and hydrogen by adding
H+
C 6 H 12 O 6  6 H 2 O  6 CO 2  24 H
O2  4 H

 2 H 2O

Solution
4. Balance the charge by adding electrons
C 6 H 12 O 6  6 H 2 O  6 CO 2  24 H
O2  4 H


 24 e


 4e  2 H 2O
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
C 6 H 12 O 6  6 H 2 O  6 CO 2  24 H
O2  4H


 24 e

 24 e


 4e  2 H 2O
C 6 H 12 O 6  6 H 2 O  6 CO 2  24 H
6 O 2  24 H


 24 e  12 H 2 O
C 6 H 12 O 6  6 O 2  6 CO 2  6 H 2 O

Example
Balance the redox reaction of sodium dicromate
(Na2Cr2O7) with ethyl alcohol (C2H5OH) if the products of
the reaction are Cr+3 and CO2
strategy
Strategy
 Balance the principal atoms
 Balance the non-essential ions
 Balance oxygen with H2O
 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
Solution
Free Energy of Formation,
o
Gf
 Energy released or energy required to
form a molecule from its elements
 By convention, Gf0 of the elements (O2,
C, N2) in their standard state is zero.
 Some representative values Gf0 are
given on the next slide
Free Energy of Formation,
Compound
Gfo, kJ/mole
C6H12O6
-917.22
CO2
-394.4
O2
0
H20
-237.17
CH4
-50.75
N20
104.18
o
Gf
Using Gf0 you can
calculate whether a
reaction will occur. For
the reaction
aA + bB  cC + dD
DGo = cGfo(C)+dGfo(D) –
aGf0(A) – bGfo(B)
Class Example
One mole of methane (CH4) and two moles of
oxgyen are in a closed container. Determine if the
reaction below will proceed as written based on
DGo.
CH4 + 2O2  CO2 + 2H20
Solution
Compound
Gfo, kJ/mole
CO2
-394.4
O2
0
H20
-237.17
CH4
-50.75
CH4 + 2O2  CO2 + 2H20
DGo = cGfo(C)+dGfo(D) –
aGf0(A) – bGfo(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 DGo and DEo
o
The electromotive force, E is related to ΔG
D G   nFE
o
o
0
Where
o
o
ΔG = the Gibbs energy of reaction at 1 atm and 25 C
n = number of electrons in the reaction
F = caloric equivalent of the faraday = 23.06 kcal/volt-mole
o
E is related to the equilibrium constant, K, by:
E 
o
RT
nF
ln( K )
Where:
o
R=universal gas constant=0.00199 kcal/mol- K
o
T=temperature( K)
Binary Fission
1
2
4
8
16
32
 P = Po(2)n
 Po is the initial population at the end
of the accelerated growth phase
 P is the population after n generations
Bacterial numbers
(log)
Microbial Growth
Time
Bacterial numbers
(log)
Microbial Growth
Lag
Phase
Adjustment to new
environment, unlimited
source of nutrient and
substrate
Time
Bacterial numbers
(log)
Microbial Growth
Lag
Phase
Accelerated growth phase
bacteria begin to divide at various rates
Time
Bacterial numbers
(log)
Microbial Growth
Exponential growth phase
Lag
Phase
differences in growth rates not
as significant because of
population increase
Accelerated growth phase
Time
Microbial Growth
Bacterial numbers
(log)
Stationary phase
Lag
Phase
Exponential
growth phase
substrate becomes
exhausted or toxic byproducts build up
resulting in a balance
between the death and
reproduction rates
Accelerated growth phase
Time
Microbial Growth
Bacterial numbers
(log)
Stationary phase
Death phase
Lag
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

d C 
dt
 k C 
n
C = Contaminant concentration
k = proportionality constant (units
dependent on reaction order)
n = reaction order
Zero Order Kinetics

d C 
dt

d C 
dt
 k C 
n
 k C 
C t  C o  kt
0
First Order Kinetics

d C 
dt

d C 
dt
 k C 
n
 k C 
Ct  C oe
1
 kt
Second Order Kinetics

d C 
dt

d C 
dt
but 
 k C enzyme
 or
 k C OH .
d enzyme
dt


d OH .
 0 ( steady state )
dt
Therefore ,

d C 
dt
 k C  where k '  k [ enzyme / OH .]
'
Text Problem 7.4
The biodegradation rate of benzo[a]pyrene has been described by
the expression

d C 
dt
 k C  X

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 7.1X1011
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] = 7.1X1011 cell/L
t = 240 days
Co = 25 ug/L

d C 
dt
 k C  X

k’ = k[X] = (3x10-15 L/cell-hr)(7.1x1011 cell/L)
= 0.00213 hr-1
Therefore,
C = Coe-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:
V  V max
C
C  Km
where
V = rate of transformation (mg/Lh)
Vmax = maximum rate of transformation (mg/Lh)
C = contaminant concentration (mg/L)
Km = half-saturation constant (mg/L)
Michaelis-Menton Kinetics
Rate (mg/L-min)
Vmax
0.5 Vmax
..
Km
Contaminant Concentration (mg/L)
Class Example
Describe how you would get Km and Vmax from the
following data.
Initial Conc. (mg/L)
Initial Rate
(mg/(L-min))
8
1.2
14
1.6
23
2.4
32
2.7
47
2.8
55
2.8
65
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.