lecture 8 aosc 434 air pollution russell r. dickerson

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LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

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Third Order A + B + C → Products O + O ₂ + M → O ₃ + M(3) Rate equation: d[A]/dt = d[B]/dt = d[C]/dt = -k[A][B][C] Units of the rate of a reaction are: conc./time. Rate constants for reactions of order n must have units:

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Page 1: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

LECTURE 8AOSC 434AIR POLLUTIONRUSSELL R. DICKERSON

Page 2: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

RECAP: KINETICSFirst Order

A → ProductsN₂O₅ → NO₂ + NO₃ (1)

Rate equation:

Second OrderA + B → Products

NO + O₃ → NO₂ + O₂ (2)Rate equation:

d[A]/dt = d[B]/dt = - k[A][B]

)(0

0

][][

)]/[]ln([][/][

tkt

t

eAA

tkAAAkdtAd

Page 3: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

Third OrderA + B + C → ProductsO + O₂ + M → O₃ + M (3)

Rate equation:d[A]/dt = d[B]/dt = d[C]/dt = -k[A][B][C]

Units of the rate of a reaction are: conc./time. Rate constants for reactions of order n must have units:

1)1( timeconc n

Page 4: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

LifetimeIf t = 1/k then , then the time corresponding to 1/k = τ, the lifetime. This is similar to “half life” concept:

Pseudo first order rate constantsIf [B] is approximately constant then k’ ≈ k[B].

37.0]/[][ 10 eAA t

)2ln(/)2ln(2/1 kt

'/1][

1 kBk

Page 5: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

III KINETICS b) Activation EnergyThe energy hill that reactants must climb in order to produce products; a barrier to thermodynamic equilibrium

ENERGY DIAGRAM

C + DA + B

AB*

ΔH

Ea

A + B → AB*

AB* + M → AB + M*

AB* → C + D

(Activate complex)

(Quenching)

(Reaction)

Page 6: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

c) Arrhenius Expressions

The Arrhenius or pre-exponential factor, “A” is related to the number of collisions the reactants make, and has a maximum value of about 2x10⁻¹º cm³ s⁻¹ for a second order reaction. If the orientation of the reactants is important, and it usually is, then A << 10⁻¹º. is the activation energy in units of J/mole. For an endothermic reaction if is less than or equal to ΔHºrxn. If is positive then the reaction will go faster at higher temperature; if is negative, then the reaction will go slower at high temperature. For example, for Reactions (1) to (3):

RTaE

Aek

aE aE

aEaE

16353

13122

1141

)/510exp(106.6

)/1450exp(103.2

14.0)/10317exp(1024.1

scmTk

scmTk

sTk

Page 7: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

The rate constant for Reaction (3), k₃, holds only if the reaction proceeds in an argon or helium atmosphere. If the third body (M) is a diatomic or triatomic, the rate constant goes up. For a N₂O atmosphere k is 1.6 time faster, and water vapor it is fifteen times faster than in air.

A recent evaluation (NASA, 1997) of the rate constant k₃ for the Earth’s atmosphere is given in a different form:

GENERAL EXAMPLES(2)

What happen if dilute car exhaust mixes with “clean” air? Let [NO]0 = 1.0 ppm and [O₃]0 = 50 ppb. Assume P = 1.0 atm and T = 25º C. What is the rate of loss of ozone? Note that we have to match the units of the concentrations to the rate constant.

163.2343 )300/(100.6 scmTk

13122

223

)/1500exp(100.3

scmTk

ONOONO

Page 8: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

Ozone disappears very quickly, but this is only the initial rate of loss. At the end of one second the concentration of the reactants, ozone especially, will be less than the initial concentration. To calculate a realistic rate of loss, we can assume that d[NO]/dt = 0. This is a pretty good assumption because [NO]>>[O₃]. Let

k[NO] = k’k’ = 0.45 s⁻¹

τ ≈ 2 sIn 2.0 s the ozone concentration is down to e⁻¹ or to about 37% of the

initial concentration. In 10 s the concentration is down to e ⁵ or 0.67%.⁻

sppbscmdtOd

cmMO

cmMNO

scmk

ONOkdtOd

/5.22105.5/][

1025.11050][

105.2100.1][

108.1)298(

]][[/][

13113

31293

3136

1314

33

Page 9: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

Equilibrium and Rate Constants

At equilibrium production equals loss.

]][[/][)(]][[/][)(

BAkdtAdALOSSDCkdtAdAPROD

DCBA

f

r

eqr

f KDCBA

kk

1

]][[]][[

]][[]][[0/][

DCkBAkdtAd

rf

Page 10: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

Equilibrium and Rate Constants

For a reaction of arbitrary order:

aA + bB R cC + dD

eqba

dc

r

f KBADC

kk

][][][][

Page 11: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

Photolysis ReactionsSunlight drives photochemical smog. How do we deal with this special

class of reactions?AB + hv → AB* Photoexitation AB* → A + B Photodissociation or photolysisM + AB* → AB + M* Quenching AB* → AB + h Reemission

The rate depends on the intensity of sunlight and chemical characteristics of AB. The rate constant is represented as j(AB) to distinguish it from ordinary first order rate constants.

0

)()()()( dIABj

Page 12: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

Where:σ = absorption cross-sectionΦ = quantum yieldI = solar intensity

EXAMPLENO₂ + h → NO + O

j(NO₂) ≈ 1.0 x 10⁻² s⁻¹ (at noon)The lifetime of NO₂ with respect to photolysis is about 100 s.

Steady StateWhen a molecule is in “steady state” it’s concentration is constant. This

means that the production and destruction rates are matched, it does not necessarily mean that the system is in thermodynamic equilibrium, but a system in equilibrium is also in steady state.

Page 13: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

EXAMPLEA + B = C + D (1)

A + B + M → AB + M (2)A + AB → 2A + B (3)

Rate Equations

)(]][[]][][[/][

]][[]][][[]][[]][[]][[]][[2]][][[]][[]][[][/[

]][[]][][[]][[]][[/][

32

32

3211

33211

3211

notationhandshortRRABAkMBAkdtABd

ABAkMBAkBAkDCkABAkBAkMBAkBAkDCkdtAd

ABAkMBAkBAkDCkdtBd

Page 14: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

At steady state d[AB]/dt = 0.0 and R2 = R3.

k₂[A][B][M] = k₃[A][AB]Which means

The concentration of AB depends only on the ratio of the rate constant for reactions 2 and 3, and on the concentration of B. It is independent of [A]. What is the steady state concentration of B?

Any new reaction that are discovered can be added to the scheme in a similar manner.

]][[][]][[]][[][

]][][[]][[]][[]][[

21

13

2131

MAkAkDCkBAkB

MBAkBAkBAkDCk

SS

3

2

3

2 ]][[])[(

]][][[][kkMB

AkMBAkAB

Page 15: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

Photostationary State

Here is a real-world example.NO₂ + hv → NO + O (1)NO + O₃ → NO₂ + O₂ (2)O + O₂ + M → O₃ + M (3)

If we assume ozone is in steady state then production equals loss.

But what is ?R1 = R3

j(NO₂)[NO₂] = k₃[O][O₂][M]

][]][][[][

]][[]][][[

2

233

3223

23

NOkMOOkO

ONOkMOOkRR

SS

SSO][

Page 16: LECTURE 8 AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

Substituting

What happens when the sun comes up on a polluted air mass? Let [NO] = 50 ppb and [NO₂] = 500 ppb. At noon j(NO₂) = 10⁻² s⁻¹, but near sunrise

Later in the day j(NO₂) = 1.0x10⁻², if the ratio of NO:NO₂ remains the same:

]][[])[(][

23

22

MOkNONOjO SS

][])[(][

2

223 NOk

NONOjO SS

ppbcmO

scmk

sNOj

SS 40100.1)50(100.1)500(100.1][

100.1

10)(

31214

3

3

13142

132

SMOGEXTREMEppb400][O SS3