Transcript LECTURE 8 AOSC 434 RUSSELL R. DICKERSON AIR POLLUTION

LECTURE 8
AOSC 434
AIR POLLUTION
RUSSELL R. DICKERSON
RECAP: KINETICS
First Order
A → Products
N₂O₅ → NO₂ + NO₃
Rate equation:
(1)
d [ A] / dt  k[ A]
ln([ A]t /[ A]0 )  k  t
[ A]t  [ A]0 e (  k t )
Second Order
A + B → Products
NO + O₃ → NO₂ + O₂
Rate equation:
d[A]/dt = d[B]/dt = - k[A][B]
(2)
Third Order
A + B + C → Products
O + 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:
conc  ( n 1)time1
Lifetime
1
If t = 1/k then [ A]t /[ A]0  e  0.37 , then the time corresponding to 1/k = τ,
the lifetime. This is similar to “half life” concept:
Pseudo first order rate constants
If [B] is approximately constant then k’ ≈ k[B].
t1/ 2  ln( 2) / k  ln( 2) 

1
 1/ k'
k  [B ]
III KINETICS b) Activation Energy
The energy hill that reactants must climb in order to produce products; a
barrier to thermodynamic equilibrium
ENERGY DIAGRAM
AB*
Ea
A+B
A + B → AB*
AB* + M → AB + M*
AB* → C + D
ΔH
C+D
(Activate complex)
(Quenching)
(Reaction)
c) Arrhenius Expressions
k  Ae
  Ea 


 RT 


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.
Ea is the activation energy in units of J/mole. For an endothermic reaction if Ea
is less than or equal to ΔHºrxn. If Ea is positive then the reaction will go faster
at higher temperature; if Ea is negative, then the reaction will go slower at
high temperature. For example, for Reactions (1) to (3):
k1  1.24 1014 exp( 10317 / T )  0.14 s 1
k 2  2.3 10 12 exp( 1450 / T ) cm 3 s 1
k3  6.6 10 35 exp( 510 / T ) cm 6 s 1
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:
k3  6.0 10 34 (T / 300) 2.3 cm 6 s 1
GENERAL EXAMPLES
NO  O3  NO2  O2
(2)
k 2  3.0 10 12 exp( 1500 / T ) cm3 s 1
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.
d [O3 ] / dt  k[ NO][O3 ]
k (298)  1.8 10 14 cm3 s 1
[ NO]  1.0 10 6  M  2.5 1013 cm 3
[O3 ]  50 10 9  M  1.25 1012 cm 3
d [O3 ] / dt  5.5 1011 cm 3 s 1  22.5 ppb / s
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≈2s
In 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%.
Equilibrium and Rate Constants
A B  C  D
PROD ( A)  d [ A] / dt  kr [C ][ D]
LOSS ( A)  d [ A] / dt  k f [ A][ B]
At equilibrium production equals loss.
d [ A] / dt  0 
k f [ A][ B]  kr [C ][ D]
1
 [ A][ B] 
  K eq
 
kr  [C ][ D] 
kf
Equilibrium and Rate Constants
For a reaction of arbitrary order:
aA + bB R cC + dD
 [C ]c [ D]d 
  a b   Keq
kr  [ A] [ B] 
kf
Photolysis Reactions
Sunlight drives photochemical smog. How do we deal with this special
class of reactions?
AB + hv → AB*
Photoexitation
AB* → A + B
Photodissociation or photolysis
M + 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.

j ( AB)    ( )   ( )  I ( )d
0
Where:
σ = absorption cross-section
Φ = quantum yield
I = solar intensity
EXAMPLE
NO₂ + 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 State
When 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.
EXAMPLE
A+B=C+D
A + B + M → AB + M
A + AB → 2A + B
(1)
(2)
(3)
Rate Equations
d [ B ] / dt  k1[C ][ D ]  k1[ A][ B ]  k2 [ A][ B ][ M ]  k3[ A][ AB]
d [ A][/ dt  k1[C ][ D ]  k1[ A][ B ]  k2 [ A][ B ][ M ]  2k3[ A][ B ]  k3[ A][ AB]
 k1[C ][ D ]  k1[ A][ B ]  k2 [ A][ B ][ M ]  k3[ A][ AB]
d [ AB] / dt  k2 [ A][ B ][ M ]  k3[ A][ AB]
 R2  R3 ( short hand notation)
At steady state d[AB]/dt = 0.0 and R2 = R3.
k₂[A][B][M] = k₃[A][AB]
Which means
[ AB] 
k2 [ A][ B ][ M ]
k
 [ B ][ M ] 2
( k3 [ A])
k3
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?
k 1[C ][ D ]  k3 [ A][ B ]  k1[ A][ B ]  k 2 [ A][ B ][ M ]
[ B ]SS 
k3 [ A][ B ]  k 1[C ][ D ]
k1[ A]  k 2 [ A][ M ]
Any new reaction that are discovered can be added to the scheme in a
similar manner.
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.
R3  R2
k3[O ][O2 ][ M ]  k2 [ NO][O3 ]
[O3 ]SS 
But what is [O]SS ?
k3[O ][O2 ][ M ]
k2 [ NO]
R1 = R3
j(NO₂)[NO₂] = k₃[O][O₂][M]
j ( NO2 )[ NO2 ]
[O]SS 
k3 [O2 ][ M ]
Substituting
[O3 ]SS 
j ( NO2 )[ NO2 ]
k 2 [ NO]
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
j ( NO2 )  10 3 s 1
k 2  1.0 10 14 cm3 s 1
1.0 10 3 (500)
12
3
[O3 ]SS 

1
.
0

10
cm
 40 ppb
1.0 10 14 (50)
Later in the day j(NO₂) = 1.0x10⁻², if the ratio of NO:NO₂ remains the same:
[O3 ]SS  400 ppb
EXTREME
SMOG