2015/9/18Environmental Engineering Science1 Ch3 Transformation Processes 3.A Governing Concepts? –Stoichiometry –Equilibrium –Kinetics 3.B Phase Changes

112/04/21 Environmental Engineering Science 1

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

• 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



•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.


Transformation Processes

• All phenomena that alter the chemical or physical state of environmental impurities

– Phase change, aicd-base reactions, oxidation-reduction reactions


Stoichiometry

• Material balance

• Atoms and Charges conservation

• NOT molecules conservation

• A set of algebraic equations


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


Chemical Equilibrium

• Base on theory of Thermodynamics

• A steady-state condition NOT a static condition

• A time dependence reaction

• Equilibrium equations

• Equilibrium constant, K


• 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:

31196.0

293314.8

233822

m

mol

KKJmol

Pa

RT

P

V

n OHOH


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


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 : α=β= 0 Rγ = kγ

– First-order reaction : α= 1, β= 0 Rγ= kγ [A]

or α= 0, β= 1 Rγ= kγ [B]

– Second-order reaction : α=β= 1 Rγ= kγ [A][B]

or α= 2, β= 0 Rγ= kγ[A]2

or α= 0, β= 2 Rγ= kγ[B]2


• 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?


• 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

dt

dCR Rn


• 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]

RnRn kC

dt

dC


The radon concentration decays exponentially toward zero with a

characteristic time τ~k-1 = 5.5 d (Figure 3.A.1)


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:

0

00000 0)]([0][

][

k

AtfortkAtAAAk

dt

Ad

)exp()]([0][][][

1000 tkAtAAAAkdt

Ad

02

00

22 21

)]([0][][2][

tAk

AtAAAAk

dt

Ad


• (a) The data conform to a zeroth-order reaction with a rate constant K0 = 0.097 mM min-1.


• (b) The compound decays by a first-order reaction with k1= 0.03 min-1.


• (c) The compound decays by a second-order reaction with k2= 0.0095 mM-1 min-1.


• 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)


• 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.


24..3 ASnevaporatio

25..31 ASTk wnevaporatio

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:

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


water molecules, Cg (units: mol/m3)γcondensation is proportional to both surface area and gas-phase water molecule concentration:

26..3 ASCgoncondensati

27..32 ASCTk ggoncondensati

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.

28..3

2

1 ATKTk

TkTCg

K' is a temperature-dependent constant


fixed temperature 29..3 ATKRTTKRTTCRT

V

nTP g

gg

30..3 Adt

VCdoncondensatinevaporatio

gg

31..321 ACH

k

H

k

dt

dCg

g

The rate of change in this number is given by the difference between the rates of evaporation and condensation:

with initial condition

32..300 ACg


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.

33..3exp1 2

2

1 AtH

k

k

ktCg

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.


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.


Phase Changes and Partitioning

• Vapor Pressure– Liquid-gas

• Dissolution of species in water– Gas-liquid, solid-liquid

• Sorption– Gas-solid, liquid-solid

• 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)



Vapor Pressure

• Saturation, subsaturation, supersaturation

• Chemical structure is more strong factor than molecular weight

• Nonvolatile impurity reduces its vapor pressure


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



Sorption

• Adsorption and Absorption• Sorption isotherms - Equilibrium partitioning

– Linear, Langmuir, Freundlich– Theory develop and Empirical data


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

• 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 )



Carbonate systems


Carbonate systems


Oxidation-Reduction reactions

• Electrons transfer

• Kinetics more important, Chemical equilibrium less important

• Oxidation state

• Corrosion (Metal is oxidized)


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


• Combusion

• Atmospheric Oxidation Processes

• Microbial Reactions– Microbial Redox processes– Photosynthetic– Aerobic respiration– Nitrogen fixation– Nitrification– Nitrate reduction and Denitrification– Methane Formation


• 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:

22222 478.3

278.3

4N

mnOH

mnCONO

mnHC nm


• 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

saf

af

mm

mm


• Combustion Stoichiometry– The overall reaction for complete combustion of a

pure hydrocarbon fuel under fuel-lean conditions can be expressed as follows:

41

1

4

78.3

278.3

4

1222222

mnawhere

aONm

nOHm

nCONOm

nHC mn


• 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.


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.

2222 OOCHOHCO hv


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.

OHCOOOCH 2222


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.

42222 434323 NHCOHOHNOCH


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.

22232 275445 NOHCOHNOOCH


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.

SHOHCOSOHOCH 2222

42 2222


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.

422 22 CHCOOCH


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.


Microbial Reactions

• Microbial Kinetics– The model equation for the rate of change of

microbial cell concentration (X) in a batch reactor has this form:

XkXrratedecaycellrategrowthcellXdt

dXdg


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

SK

SkYr

s

mg


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):

ds

m kSK

SkY

dt

dX

X

1


Microbial Kinetics


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

XSK

SkX

Y

r

dt

dS

s

mg


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.


Microbial 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.


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.


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).


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.


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.


EXAMPLE 3.D.3

• SOLUTION – (d) The rate constant is estimated from the first-

order model, given BODu and BOD(5):

125.021

6ln

5

1

:

5exp5

ddays

kBOD

resulttheyieldskforsolvingandngSubstituti

dayskBODBOD

BOD

BODu

Documents

2015/9/18Environmental Engineering Science1 Ch3 Transformation Processes 3.A Governing Concepts? –Stoichiometry –Equilibrium –Kinetics 3.B Phase Changes