Title: Ch3 Transformation Processes
1Ch3 Transformation Processes
- 3.A Governing Concepts?
- Stoichiometry
- Equilibrium
- Kinetics
- 3.B Phase Changes and Partitioning
- 3.C Acid-Base Reactions
- 3.D Oxidation-Reduction Reactions
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3- Stoichiometry, is the application of material
balance to transformation processes. - Chemical equilibrium, describes how species
partition between phases and how elements
partition among chemical species if some specific
restrictions are met. - Kinetics, deals with the rate of reactions and
provides information on how species
concentrations evolve.
4Transformation Processes
- All phenomena that alter the chemical or physical
state of environmental impurities - Phase change, aicd-base reactions,
oxidation-reduction reactions
5Stoichiometry
- Material balance
- Atoms and Charges conservation
- NOT molecules conservation
- A set of algebraic equations
6 aCO2 bNO3- cHPO42- dH eH2O ?
C106H263O110N16P1 fO2
Ca 106
O2a 3b 4c e 110
2f Nb 16
Hc d 2e 263
Pc 1
/b2c d 0
7Chemical Equilibrium
- Base on theory of Thermodynamics
- A steady-state condition NOT a static condition
- A time dependence reaction
- Equilibrium equations
- Equilibrium constant, K
8- EXAMPLE 3.A.3 Water Vapor Concentration at
Equilibrium - Consider a sealed jar that is partially
filled with pure liquid water and otherwise
contains nitrogen gas. Assume that the
temperature of the system is fixed at 293 K
(20?). What is the steady-state molar
concentration of H2O(g) in the gas phase above
the liquid water? - SOLUTION
- The water vapor pressure in the jar will
reach the saturation vapor pressure (that is, the
relative humidity will be 100 percent). The
saturation vapor pressure of water at T 293 K
is 2338 Pa (see 3.B.1). - From the ideal gas law and the rule of
partial pressures
9Kinetics
- Time for equilibrium in a system
- Rate-related problems
- Reaction rates
- Chemical descript to mathematical description
- Rate laws
- Link reaction rate to concentration of reactants
- Reaction order
- Zeroth, first, second order or enzyme reactions
- Determined empirically from experiments
10- Rate Laws
- For reaction ?, the expected form of the rate law
is - R?k?AaBß
- To be more explicit, rate laws for ? that are
zeroth, first, or second order are shown below - Zeroth-order reaction aß 0 R?
k? - First-order reaction a 1, ß 0
R? k? A -
or a 0, ß 1 R? k? B - Second-order reaction aß 1 R?
k? AB - or a
2, ß 0 R? k?A2 -
or a 0, ß 2 R? k?B2
11- Radon-222 is a naturally occurring radioactive
gas formed by the decay of radium-226, a trace
element in soil and rock. The radioactive decay
of radon can be described by the elementary
reaction. - radon-222 gt polonium-218 alpha
particle - The rate constant for this reaction is k 2.1 X
10-6 s-1, independent of temperature.At time t
0, a batch reactor is filled with air containing
radon at concentration C0.How does the radon
concentration in the reactor change over time?
12- For t?0, only one reaction influences the radon
concentration, CRn. The change in radon
concentration is related to the reaction rate by
the expression - Since radon decay is an elementary reaction, we
expect the rate law to be first-order - R kCRn
13- The following expressions give the time rate of
change in radon concentration and the initial
condition -
-
CRn (0)C0 - The differential equation can be solved by direct
integration following rearrangement (see Appendix
D, D.1 for details) to obtain - CRn (t)C0exp-kt
-
-
14- The radon concentration decays exponentially
toward zero with a characteristic time tk-1
5.5 d (Figure 3.A.1)
15- Empirical Determination of Reaction Order and
Rate Constant - The differential equations and solutions
describing the change in concentration over time
for zeroth-, first-, and second-order reactions
are summarized below. The functions, A(t), were
obtained by direct integration in each case. - Zeroth-order reaction
- First-order reaction
- Second-order reaction
16- (a) The data conform to a zeroth-order reaction
with a rate constant K0 0.097 mM min-1.
17- (b) The compound decays by a first-order reaction
with k1 0.03 min-1.
18- (c) The compound decays by a second-order
reaction with k2 0.0095 mM-1 min-1.
19- Characteristic Time of Kinetic Processes and the
Equilibrium Assumption - Fast kinetics t r ltltt
- Kinetics may be ignored reaction proceeds
rapidly to completion, or equilibrium conditions
are effectively instantaneously achieved
(e.g.,.acid-base.reactions in a water treatment
process) - Slow kinetics t r gtgtt
- Chemical transformations (and the associated
kinetics) are slow enough to be neglected
altogether (e.g., carbon monoxide oxidation in
urban air) - Intermediate kinetics t r t
- We cannot make any useful simplifying
approximations we must consider in detail the
effects of chemical kinetics on the system (e.g.,
ozone in urban air)
20- The system we consider is a glass jar that can be
sealed and maintained at a constant temperature
(Figure3.A.3). Half of the jars volume is
initially filled with pure liquid water the
other half is filled with dry nitrogen (N,) gas
at a pressure of 1 atm . - Eventually, the rates of evaporation and
condensation become balanced, and the state of
chemical equilibrium is attained. - An exploration of the rates of evaporation and
condensation in this system will help us to
better understand the relationship between
kinetics an equilibrium.
21we can say that for a fixed temperature, the
rate of evaporation,?evaporation (with units of
molecules per time) is proportional to the
gas-water interface area
where k1 is a constant that depends on liquid
water temperature, Tw.A key factor is the rate at
which the gaseous water molecules strike the
surface. This rate will be proportional to the
inter- facial surface area, S
22water molecules, Cg (units mol/m3)?condensation
is proportional to both surface area and
gas-phase water molecule concentration
where Tg is the gas temperature. At equilibrium,
the liquid and gas temperatures are equal, Tw Tg
T, and the rates of evaporation and
condensation are equal. Therefore.
K' is a temperature-dependent constant
23fixed temperature
The rate of change in this number is given by the
difference between the rates of evaporation and
condensation
with initial condition
24The solution method is disctissed in detail in
Appendix D (D.1) The solution is given by
equation 3.A.33 and is plotted in Figure 3.A.5.
Note that the steady-state condition from
equation 3.A.31 is d(Cg)/dt 0, which leads to
the equilibrium result Cg k1/k2. This same
condition is obtained from equation 3.A.33 as t ?
8. The characteristic time required for the
water vapor concentration to approach equilibrium
can be determined by dividing the stock of water
vapor molecules in the system, VgCg, by the rate
of flow out due to condensation, k2SCg . The
result is t VgCg (k2SCg)-1 H/k2. This is also
seen to be the reciprocal of the argument of fin
the exponential term in equation 3.A.33.
25Figure 3.A.5 Time-dependent behavior of the
concentration of watervapor molecules in the gas
phase of the system depicted in Figure 3.A.3. The
steady-state concentration, k1/k2, is approached
with a characteristic time of H/k2,. The initial
rate of increase in vapor concentration is k1/H
and reflects the effects of evaporation alone,
since no condensation occurs when Cg 0.
26Phase Changes and Partitioning
- Vapor Pressure
- Liquid-gas
- Dissolution of species in water
- Gas-liquid, solid-liquid
- Sorption
- Gas-solid, liquid-solid
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28Vapor Pressure
- Saturation, subsaturation, supersaturation
- Chemical structure is more strong factor than
molecular weight - Nonvolatile impurity reduces its vapor pressure
29Dissolution of species in water
- Partitioning between gas and water
- Henrys Law (CKP)
- Solubility of Nonaqueous-phase liquids
- CK
- Dissolution and precipitation of solid
- AaBb K
30(No Transcript)
31Sorption
- Adsorption and Absorption
- Sorption isotherms - Equilibrium partitioning
- Linear, Langmuir, Freundlich
- Theory develop and Empirical data
32Acid-Base Reactions
- Hydrogen ion H H3O
- Chemical Equilibrium more important, Kinetics
less important - KwKw(T)HOH-, Ex3.c.1
- pH and pKA
- Carbonate systems
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34Carbonate systems
35Carbonate systems
36Oxidation-Reduction reactions
- Electrons transfer
- Kinetics more important, Chemical equilibrium
less important - Oxidation state
- Corrosion (Metal is oxidized)
37Corrosion control methods
- Physical isolate (paint in air)
- Physical isolate (deposition a layer of CaCO3 on
inner surface of pipes ) - Eliminate corrosive compound (O2)
- Cathodic protection (sacrificial anode)
- Corrosion and Scale
38- Combusion
- Atmospheric Oxidation Processes
- Microbial Reactions
- Microbial Redox processes
- Photosynthetic
- Aerobic respiration
- Nitrogen fixation
- Nitrification
- Nitrate reduction and Denitrification
- Methane Formation
39- Combustion Stoichiometry
- If exactly enough air is provided to fully
oxidize the fuel without any excess oxygen, than
the complete combustion of a pure hydrocarbon
fuel using air as the as oxidizer can be
represented by this overall reaction
40- Combustion Stoichiometry
- The relative amounts of fuel and air for
nonstoichiometric combustion may be expressed in
terms of the equivalence ratio, (?), defined by
the expression
41- Combustion Stoichiometry
- The overall reaction for complete combustion of a
pure hydrocarbon fuel under fuel-lean conditions
can be expressed as follows
42- Photolytic Reactions
- In photolytic reactions, the energy to break the
chemical bond of a reactant is supplied by
absorption of a photon of light. - In the troposphere, photolysis is caused by
light of wavelengths 280 nm lt ? lt 730 nm (410 nm
lt ? lt 650 nm defines the visible range).
Shorter-wavelength light is absorbed by
stratospheric ozone and oxygen molecules and does
not penetrate to the troposphere.
43Microbial Reactions
- Photosynthetic Production of Biomass
- Photosynthetic microorganisms (algae and some
bacteria) carry out photosynthesis reactions. In
these reactions, energy-rich carbohydrate
molecules are produced by combining carbon
dioxide and water, using energy derived from
sunlight. Overall, these reactions can be written
in the form given below.
44Microbial Reactions
- Aerobic Respiration
- In the presence of oxygen, microorganisms degrade
biomass to from carbon dioxide and water.
Chemical energy that is released can be used by
the organism. This process is the reverse of
photosynthesis Carbon is oxidized and oxygen is
reduced.
45Microbial Reactions
- Nitrogen Fixation
- We refer to compounds such as ammonia and nitrate
that contain a single nitrogen atom as fixed
nitrogen species. Certain groups of bacteria are
capable of converting gaseous nitrogen to fixed
nitrogen, in the form of the ammonium ion.
46Microbial Reactions
- Nitrate Reduction or Denitrification
- When oxygen is not available as the oxidizer to
degrade biomass, microorganisms can use nitrate
as the oxidizer (electron acceptor) instead.
47Microbial Reactions
- Sulfate Reduction
- Some environments that contain biodegradable
materials lack both oxygen and nitrate to serve
as the oxidizing agent. In such cases, sulfate
may serve that role.
48Microbial Reactions
- Methane Formation (Methanogenesis)
- In the absence of oxygen, nitrate, and sulfate,
biomass can still be converted to carbon dioxide
as shown in the following reaction.
49Microbial Reactions
- Microbial Kinetics
- To predict the rate of contaminant degradation
by microorganisms, we must also predict changes
in the microbial population. - the rates of microbial growth and contaminant
degradation are described by kinetic rate
equations of a general form.
50Microbial Reactions
- Microbial Kinetics
- The model equation for the rate of change of
microbial cell concentration (X) in a batch
reactor has this form
51Microbial Reactions
- Microbial Kinetics
- This equation contains three parameters µ is the
net specific growth rate of cells, rg is the cell
growth rate coefficient, and kd is the cell death
rate coefficient. - The most widely accepted form for describing the
dependence of rg on S is
52Microbial Reactions
- Microbial Kinetics
- The net specific growth rate of cells can then be
written in a form known as the Monod equation
(sometimes called a saturation reaction)
53Microbial Kinetics
54Microbial Reactions
- Microbial Kinetics
- In fact, the cell yield coefficient represents
the mass of cells produced per mass of substrate
consumed. - Therefore, the rate of change of substrate
concentration due to microbial degradation is
modeled as
55Microbial Reactions
- Biochemical Oxygen Demand
- Two factors are relevant in assessing the
oxygen-depleting significance of BOD. - The first factor is stoichiometric. We want to
know the total amount of oxygen that is required
for biodegradation of oxidizable compounds. - The second factor is kinetic. We want to know how
rapidly oxygen will be consumed in the oxidation
process.
56Microbial Reactions
- Measuring BOD
- The basic procedure for measuring stoichiometric
BOD is simple, consisting of the following steps. - 1. Measure the initial dissolved oxygen content
of water to be analyzed. Call this D0(0). - 2. Fill a 300 mL glass bottle with a sample of
the water. Seal the bottle with a stopper. - 3. Incubate the water in the dark at 20ºC for 5
days. - 4. Measure the dissolved oxygen content of the
incubated water. Call this DO5. - 5. Compute the five-day BOD as BOD5 D0(0)
DO5, where all have units of mg L-1.
57EXAMPLE 3.D.3
- A BOD test is run using 100 mL of treated
wastewater mixed with 200 mL of pure water. The
initial DO of the mix is 9.0 mg/L. After 5 days,
the DO is 4.0 mg/L. After a long period of time,
the DO is 2.0 mg/L and no longer seems to be
decreasing. Assume that nitrification has been
inhibited so that the only BOD being measured is
carbonaceous.
58EXAMPLE 3.D.3
- (a) What is the 5-day BOD of the wastewater
(mg/L)? - (b) Estimate the ultimate carbonaceous BOD
(mg/L). - (c) What is the remaining BOD after 5 days
(mg/L)? - (d) Estimate the reaction rate constant, kBOD (d
-1).
59EXAMPLE 3.D.3
- SOLUTION
- (a) The change in dissolved oxygen content in the
test sample during the first five days is 9.0 -
4.0 5.0 mg/L. Wastewater comprises only
one-third of the test sample. To correct for
dilution, we multiply by 3, so the BOD5 content
of the wastewater is 15 mg/L.
60EXAMPLE 3.D.3
- SOLUTION
- (b) The ultimate carbonaceous BOD is the
difference between the initial and final DO
levels, corrected for dilution, so BODu (9.0 -
2.0) 3 21 mg/L. - (c) BODu BOD5 BOD(5), so the BOD remaining at
5 days is BOD(5) 21 - 15 6 mg/L.
61EXAMPLE 3.D.3
- SOLUTION
- (d) The rate constant is estimated from the
first-order model, given BODu and BOD(5)