Ch3 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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• 化學反應程序
• 觀念：
– 化學計量(反應質量平衡、物質不滅)
– 化學平衡(動態平衡、平衡方程式)
– 化學動力(反應階次、酵素反應)
• 反應類型：
– 相轉換(分配)、
– 酸鹼反應、
– 氧化還原反應
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•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.
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Transformation Processes
• All phenomena that alter the chemical or
physical state of environmental impurities
– Phase change, aicd-base reactions,
oxidation-reduction reactions
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Stoichiometry
•
•
•
•
Material balance
Atoms and Charges conservation
NOT molecules conservation
A set of algebraic equations
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aCO2 + bNO3- + cHPO42- +dH+ + eH2O → C106H263O110N16P1 + fO2
C：a = 106
O：2a + 3b + 4c + e = 110 + 2f
N：b = 16
H：c + d + 2e = 263
P：c = 1
+/–：–b–2c + d = 0
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Chemical Equilibrium
• Base on theory of Thermodynamics
• A steady-state condition NOT a static
condition
• A time dependence reaction
• Equilibrium equations
• Equilibrium constant, K
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• 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:
nH 2O
V
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PH 2O
2338Pa
mol


 0.96 3
1
1
RT
8.314 Jmol K  293K
m
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Kinetics
• 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
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 Rate Laws
• For reaction γ, the expected form of the rate law is
Rγ=kγ[A]α[B]β
• To be more explicit, rate laws for γ that are zeroth, first, or second
order are shown below:
– Zeroth-order reaction :
– First-order reaction :
α=β= 0
α= 1, β= 0
or α= 0, β= 1
– Second-order reaction :
α=β= 1
or α= 2, β= 0
or α= 0, β= 2
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Rγ = k γ
Rγ= kγ [A]
Rγ= kγ [B]
Rγ= kγ [A][B]
Rγ= kγ[A]2
Rγ= kγ[B]2
10
• 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 —> 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?
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• 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
dC Rn
R
dt
• Since radon decay is an elementary reaction, we expect the rate law to
be first-order
R = kCRn
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• The following expressions give the time rate of change in radon
concentration and the initial condition:
dC Rn
 kCRn
dt
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]
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The radon concentration decays exponentially toward zero with a
characteristic time τ~k-1 = 5.5 d (Figure 3.A.1)
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 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:
A
d[ A]
 k 0
[ A]0  A0
 [ A](t )  A0  k 0 t
for0  t  0
dt
k0
– First-order reaction:
d [ A]
 k 0 [ A]
dt
[ A]0  A0

[ A](t )  A0 exp( k1t )
– Second-order reaction:
d[ A]
 2k 2 [ A]2
dt
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[ A]0  A0

[ A](t ) 
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A0
1  2k 2 tA0
15
• (a) The data conform to a zeroth-order reaction with a rate constant
K0 = 0.097 mM min-1.
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• (b) The compound decays by a first-order reaction with k1= 0.03
min-1.
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•
(c) The compound decays by a second-order reaction with k2=
0.0095 mM-1 min-1.
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• Characteristic Time of Kinetic Processes and the Equilibrium
Assumption
– Fast kinetics
τ r <<τ
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
τ r >>τ
Chemical transformations (and the associated kinetics) are slow
enough to be neglected altogether (e.g., carbon monoxide oxidation
in urban air)
– Intermediate kinetics τ r ～τ
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)
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• The system we consider is a glass jar that can be sealed and
maintained at a constant temperature (Figure3.A.3). Half of the jar’s
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.
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we can say that for a fixed temperature, the rate of
evaporation,γevaporation (with units of molecules per time) is
proportional to the gas-water interface area:
3.A.24
 evaporation  S
 evaporation  k1 Tw S
3.A.25
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
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water molecules, Cg (units: mol/m3)γcondensation is
proportional to both surface area and gas-phase water
molecule concentration:
 condensation  SC g
 condensation  k 2 Tg SC g
3. A.26
3. A.27 
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.
k1 T 
3.A.28
C g T  
 K T 
k 2 T 
K' is a temperature-dependent constant
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fixed temperature
Pg T  
ng
V
RT  C g T RT  K T RT  K T 
3. A.29
The rate of change in this number is given by the difference between
the rates of evaporation and condensation:
d C gVg 
dt
dC g
dt

  evaporation   condensation
k1
k
 2 Cg
H
H
3. A.30
3. A.31
with initial condition
C g 0  0
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3. A.32
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The 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.
C g t  
k1
k2

 k 2 
1

exp
t 


H



3. A.33
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 → ∞.
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 τ~ 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.
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Figure 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.
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Phase 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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• 化學反應程序
• 反應類型：
– 相轉換(溶解)：蒸汽壓(揮發性溶劑)、
氣液相(Henry’s Law, KH )、NAPL(水溶
解度, KWS)、固液相(溶解度積, Ksp)
– 相轉換(吸附):固液相、氣固相(吸附數
學模式, Kads, qmax, kf)
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Vapor Pressure
• Saturation, subsaturation, supersaturation
• Chemical structure is more strong factor
than molecular weight
• Nonvolatile impurity reduces its vapor
pressure
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Dissolution of species in water
• Partitioning between gas and water
– Henry’s Law (C=KP)
• Solubility of Nonaqueous-phase liquids
– C=K
• Dissolution and precipitation of solid
– [A]a[B]b = K
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Sorption
• Adsorption and Absorption
• Sorption isotherms - Equilibrium partitioning
– Linear, Langmuir, Freundlich
– Theory develop and Empirical data
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Acid-Base Reactions
• Hydrogen ion [H+] [H3O+]
• Chemical Equilibrium more important,
Kinetics less important
• Kw=Kw(T)=[H+][OH-], Ex3.c.1
• pH and pKA
• Carbonate systems
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• 化學反應程序
• 反應類型：
– 酸鹼反應: pH 值、強酸弱酸(碳酸鹽系
統、酸雨)
– 氧化還原反應：氧化態、腐蝕(陰極防
蝕法)、燃燒(過剩空氣量)、大氣氧化
反應(自由基)、微生物反應(有機物氧
化、硝化、脫硝、微生物動力學、
BOD特性)
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Carbonate systems
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Carbonate systems
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Oxidation-Reduction reactions
• Electrons transfer
• Kinetics more important, Chemical
equilibrium less important
• Oxidation state
• Corrosion (Metal is oxidized)
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Corrosion 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
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• Combusion
• Atmospheric Oxidation Processes
• Microbial Reactions
–
–
–
–
–
–
–
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Microbial Redox processes
Photosynthetic
Aerobic respiration
Nitrogen fixation
Nitrification
Nitrate reduction and Denitrification
Methane Formation
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• 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:
m
m
m


C m H n   n  O2  3.78 N 2  
 nCO2  H 2 O  3.78 n   N 2
4
2
4


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• Combustion Stoichiometry
– The relative amounts of fuel and air for
nonstoichiometric combustion may be expressed in
terms of the equivalence ratio, (ψ), defined by the
expression
m

m
f
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f
ma 
ma s
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• Combustion Stoichiometry
– The overall reaction for complete combustion of a
pure hydrocarbon fuel under fuel-lean conditions
can be expressed as follows:
1 m
m
3.78  m 
C n H m   n  O2  3.78 N 2  
 nCO2  H 2 O 
 n   N 2  aO2
 4
2
  4
where
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 1  m 
a    1 n  
   4 
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• 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 < λ < 730 nm
(410 nm < λ < 650 nm defines the visible
range). Shorter-wavelength light is absorbed
by stratospheric ozone and oxygen molecules
and does not penetrate to the troposphere.
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Microbial 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.
hv
CO2  H 2 O CH 2 O  O2
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Microbial 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.
CH 2O O2  CO2  H 2O
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Microbial 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.
3CH 2 O  2 N 2  3H 2 O  4 H 
 3CO2  4 NH 4

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
Microbial 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.
5CH 2O 4NO3  4H 
5CO2  7H 2O  2N 2

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
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Microbial 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.
2CH 2 O  2 H  SO4 
 2CO2  2 H 2 O  H 2 S

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2
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Microbial 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.
2CH 2 O 
 2CO2  CH 4
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Microbial 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.
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Microbial Reactions
• Microbial Kinetics
– The model equation for the rate of change of
microbial cell concentration (X) in a batch
reactor has this form:
dX
 X  cell growth rate  cell decay rate  rg X  k d X
dt
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Microbial 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
km S
rg  Y
Ks  S
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Microbial 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):
km S
1 dX

Y
 kd
X dt
Ks  S
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Microbial Kinetics
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Microbial 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
rg
km S
dS

X 
X
dt
Y
Ks  S
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Microbial 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.
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Microbial Reactions
•
Measuring BOD
•
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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.
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EXAMPLE 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.
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EXAMPLE 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).
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EXAMPLE 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.
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EXAMPLE 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.
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EXAMPLE 3.D.3
• SOLUTION
– (d) The rate constant is estimated from the firstorder model, given BODu and BOD(5):
BOD 5  BOD u exp  k BOD  5days 
Substituting and solving for k BOD yields the result :
kBOD  
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1
 6
 ln    0.25d 1
5days
 21 
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