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KINETICS AND ATMOSPHERIC
CHEMISTRY
NC A&T Lecture
February 1, 2011
John Orlando
[email protected]
What is KINETICS?
From Wikipedia, the free encyclopedia
Boulder Kinetics
“… a race from the banks of Boulder Reservoir and back by human-powered
vehicles timed on speed and judged for style.”
The Kinetics, a rock band
Kinetics (physics), the study of motion and its causes
What is KINETICS?
Chemical kinetics - the study of chemical reaction rates
Oxford English Dictionary – “A field of study concerned with the
mechanisms and rates of chemical reactions or other kinds of process;
But first, let’s review a little !!
What do we know so far?
Macroscopic properties of gases:
For atmospheric T and P, the atmosphere is an ideal gas.
Scale Height:
Lapse Rate:
This is the Dry Lapse Rate, ≈10 K/km
In practice, air contains humidity → 7 K/km
PV = nRT
What do we know so far?
Structure of the atmosphere
From Lutgens and Tarbuck, 2001
What do we know so far?
General Motions of Air:
What do we know so far?
Atmospheric Composition
Mostly ‘inert’ species – N2, O2, H2O, CO2, Ar
Not much chemistry?
What do we know so far?
Atmospheric Composition
Mostly ‘inert’ species – N2, O2, H2O, CO2, Ar
LOTS OF REACTIVE TRACE GASES !!
So, Actually, Lots of Chemistry!
Natural sources – NO from soil, many hydrocarbons (isoprene) from plants
Anthropogenic sources – hydrocarbons, NO, …
What do we know so far?
Atmospheric Composition (besides the basic N2, O2, H2O, etc.)
Emissions – Lots of “stuff” out there.
Urbanski et al.,
Wildland Fires and Air
Pollution, 2009
What do we know so far?
Atmospheric Composition (besides the basic N2, O2, H2O, etc.)
Emissions – Lots of “stuff” out there.
Urbanski et al.,
Wildland Fires and Air
Pollution, 2009
The atmosphere needs a way to remove
these species.
GENERAL DESCRIPTION:
The atmosphere (particularly the troposphere) acts as a low-temperature,
slow-burning combustion engine.
Takes all the emissions (reduced compounds) and ‘burns’ (oxidizes) them:
CH4
CO2 + H2O
Isoprene
Other by-products, such
as O3, particles, acids,
nitrates, etc.
(2ry POLLUTANTS)
DMS, NH3
GENERAL DESCRIPTION:
The atmosphere (particularly the troposphere) acts as a low-temperature,
slow-burning combustion engine.
Takes all the emissions (reduced compounds) and ‘burns’ (oxidizes) them:
CH4
CO2 + H2O
Isoprene
Other by-products, such
as O3, particles, acids,
nitrates, etc.
(2ry POLLUTANTS)
DMS, NH3
GENERAL DESCRIPTION:
The atmosphere (particularly the troposphere) acts as a low-temperature,
slow-burning combustion engine.
Takes all the emissions (reduced compounds) and ‘burns’ (oxidizes) them:
OH
HO2
CH4
CO2 + H2O
Isoprene
Other by-products, such
as O3, particles, acids,
nitrates, etc.
(2ry POLLUTANTS)
DMS, NH3
NO
NO2
MACROSCOPIC : (1044 molecules in the atmosphere)
MICROSCOPIC : (about 25 molecules in a 10 nm cube)
KINETIC THEORY OF GASES
KINETIC THEORY OF GASES:
(most any Physical Chemistry text, including Adamson, 1979; Atkins, 1986).
1) Molecules Move !! (they have kinetic energy):
Average Velocity:
For N2, can show that c is about 4 x 104 cm/sec at 298 K
KINETIC THEORY OF GASES:
(most any Physical Chemistry text, including Adamson, 1979; Atkins, 1986).
2) Molecules bump into walls!! (pressure on wall of a vessel)
Pressure:
KINETIC THEORY OF GASES:
(most any Physical Chemistry text, including Adamson, 1979; Atkins, 1986).
3) Molecules collide with each other!!
Frequency of collisions is about 5 x 109 sec-1 / atm at 298 K.
With velocity of 4 x 104 cm/s, Mean Free Path = 7 x 10-6 cm at atmospheric P.
KINETIC THEORY OF GASES:
(most any Physical Chemistry text, including Adamson, 1979; Atkins, 1986).
3) Molecules collide with each other!!
Frequency of collisions is about 5 x 109 sec-1 / atm at 298 K.
With velocity of 4 x 104 cm/s, Mean Free Path = 7 x 10-6 cm at atmospheric P.
4) Molecules can react with each other when they collide !
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
First, a couple of definitions.
ELEMENTARY REACTION
From Wikipedia - An elementary reaction is a chemical reaction in which one or
more of chemical species react directly to form products in a single reaction step.
Usually involves 1-3 molecules, with bimolecular most common:
e.g., OH + CH4  CH3 + H2O
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
COMPLEX REACTION SCHEME OR MECHANISM
- Made up of a bunch of elementary reactions
- e.g., the oxidation of CH4 to CH2O in the polluted troposphere leads to the
following net effect:
CH4 + 4 O2  CH2O + H2O + 2 O3
OH + CH4  CH3 + H2O
CH3 + O2  CH3O2
CH3O2 + NO  CH3O + NO2
CH3O + O2  CH2O + HO2
HO2 + NO  OH + NO2
NO2 + hn  NO + O
NO2 + hn  NO + O
O + O2  O3
O + O2  O3
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
COMPLEX REACTION SCHEME OR MECHANISM
- Explicitly describing the chemical mechanism occurring in the troposphere
would probably require > millions of reactions.
- e.g., Aumont et al., ACP 2005.
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
BACK TO ELEMENTARY REACTIONS (BIMOLECULAR)
Bimolecular reactions are the most common type of elementary reaction in
the atmosphere
Typically are of the form
A-B + C  A + B-C
CH4 + OH  CH3 + HOH
Rate of the chemical reaction (disappearance of reactants or appearance of
products):
k is the rate constant, units of (time)-1 (concentration)-1
[AB] and [C] are concentrations
Then rate in units of (concentration) (time)-1
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
ELEMENTARY REACTIONS (BIMOLECULAR)
What determines the value of the rate constant???
Recall:
Frequency of collisions is about
5 x 109 sec-1 / atm at 298 K.
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
ELEMENTARY REACTIONS (BIMOLECULAR)
Frequency of collisions is about 5 x 109 sec-1 / atm at 298 K.
So, let’s say that OH and CH4 are the two reactants:
OH + CH4  H2O + CH3
In the troposphere, typically daytime
[OH] = 4.1 x 10-14 atm
[CH4] = 1.86 x 10-6 atm
So, IF REACTION OCCURRED ON EVERY COLLISION,
Rate = k [OH] [CH4] =
(5e9 sec-1/atm-1) * (4.14e-14 atm) * (1.86e-6 atm) = 3.8e-10 atm sec-1
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
ELEMENTARY REACTIONS (BIMOLECULAR)
So, IF REACTION OCCURRED ON EVERY COLLISION,
Rate = (5 x 109 sec-1/atm-1) * (4.1 x 10-14 atm) * (1.86 x 10-6 atm) =
3.8x 10-10 atm sec-1
Usually, we work in molecules cm-3, rather than in atmospheres
So, given that there are 2.45 x 1019 molecule cm-3 in 1 atm of gas at 298 K:
Rate = (2 x 10-10 cm3 molecule-1 s-1) * (1 x 106 molecule cm-3) * (4.56 x 1013
molecule cm-3) = 9.1 x 109 molecule cm-3 s-1
BUT IN REALITY, REACTION DOES NOT OCCUR ON EVERY COLLISION!!!
Why NOT? – any ideas?
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
ELEMENTARY REACTIONS (BIMOLECULAR)
BUT IN REALITY, REACTION DOES NOT OCCUR ON EVERY COLLISION!!!
Two main reasons:
1) There are energetic limitations. Colliding molecules must possess
sufficient energy to overcome an ‘activation energy’ that typically exists.
2) Also, there may be ‘geometrical limitations’. Molecules must approach
each other in such a way that the appropriate bonds can break / form.
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
ELEMENTARY REACTIONS (BIMOLECULAR)
BUT IN REALITY, REACTION DOES NOT OCCUR ON EVERY COLLISION!!!
Ea
From Wikipedia
k = A * exp(-Ea/RT)
A is the pre-exponential factor, and accounts for the geometry limitations.
Ea is activation energy.
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
ELEMENTARY REACTIONS (BIMOLECULAR)
k = A * exp(-Ea/RT)
So, Let’s go back to the OH / CH4 reaction.
IF REACTION OCCURRED ON EVERY COLLISION,
k = 2 x 10-10 cm3 molecule-1 s-1
Turns out that k = 2.45 x 10-12 * exp(- 3525 cal / RT)
k = 6.3 x 10-15 cm3 molecule-1 s-1 at 298 K
k = 5.2 x 10-16 cm3 molecule-1 s-1 at 210 K
Only about 1 in 30000 OH/CH4 collisions results in reaction at 298 K.
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
ELEMENTARY REACTIONS (BIMOLECULAR)
Another example
HO2 + NO  OH + NO2
– two radical species; no barrier to reaction (attractive forces).
HO2 + NO
HOO-NO
OH + NO2
Reaction turns out to have a “negative activation energy”.
k = 3.5 x 10-12 exp(500 cal / RT) cm3 molecule-1 s-1
(Colder molecules more likely to react – less able to overcome attraction).
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
ELEMENTARY REACTIONS (BIMOLECULAR)
REACTION
A-Factor
(cm3 molecule-1 s-1)
Activation
Energy (cal.)
Rate constant k at 298 K
(cm3 molecule-1 s-1)
•OH + CH4  •CH3 + H2O
•OH + H2  H• + H2O
•OH + CH3OH  •CH2OH + H2O
2.45e-12
5.5e-12
6.7e-12
3525
3975
1200
6.3e-15
6.7e-15
8.9e-13
NO3• + CH3CHO  HNO3 + CH3CO•
1.4e-12
3800
2.4e-15
HO2• + CH3O2•  O2 + CH3OOH
NO + CH3O2•  NO2 + CH3O•
CH3C(O)OO• + CH3C(O)OO• 
CH3C(O)O• + CH3C(O)O• + O2
3.8e-13
4.2e-12
2.9e-12
-1600
-360
-1000
5.6e-12
7.7e-12
1.5e-11
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
EQUILIBRIUM – All elementary reactions are reversible. At equilibrium, rate
of forward and reverse reactions must be equal.
[ HO…H-CH3 ]
Ea = 3525 calories
Ea = 17450 calories
OH + CH4
kf [OH] [CH4] = kr [CH3] [H2O]
HOH + CH3
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
EQUILIBRIUM – All elementary reactions are reversible. At equilibrium, rate
of forward and reverse reactions must be equal.
[ HO…H-CH3 ]
Ea = 3525 calories
kf = 6.3e-15 cm3 molec-1 s-1
Ea = 17450 calories
kr = 1.2e-25 cm3 molec-1 s-1
OH + CH4
kf [OH] [CH4] = kr [CH3] [H2O]
HOH + CH3
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
“Equilibrium” and “Steady-State” are different:
Equilibrium is a very precise, physical concept - established when forward
and reverse rates of all reactions in a system are equal.
Steady-State is more conceptual and approximate
- A (short-lived) species, often an intermediate in a chemical
scheme, is being produced and destroyed at roughly the same rate.
Production Rate = Loss Rate
O(1D) + H2O
OH
Reaction with CH4
HO2 + NO
Reaction with CO
Reaction with Isoprene
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
Let’s look at this in more detail.
Consider a reaction scheme like this:
OH + CH4  CH3 + H2O
CH3 + O2  CH3O2
CH3O2 + NO  CH3O + NO2
CH3O + O2  CH2O + HO2
HO2 + NO  OH + NO2
NO2 + hn  NO + O
NO2 + hn  NO + O
O + O2  O3
O + O2  O3
Can assume CH3 to be in ‘steady-state’. (CH3O, CH3O2, OH, HO2, NO, NO2 too).
What is the steady-state [CH3]? (What do we need to know?)
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
OH + CH4  CH3 + H2O
CH3 + O2  CH3O2
CH3O2 + NO  CH3O + NO2
CH3O + O2  CH2O + HO2
HO2 + NO  OH + NO2
NO2 + hn  NO + O
NO2 + hn  NO + O
O + O2  O3
O + O2  O3
Can assume CH3 to be in ‘steady-state’. (CH3O, CH3O2, OH, HO2, NO, NO2 too).
Assume [OH] = 106, [CH4] = 4.6 x 1013, [O2] = 5 x 1018, all in molecule cm-3
Appropriate rate constants:
k1 = 6.3 x 10-15 cm3 molecule-1 s-1
k2 = 1 x 10-12 cm3 molecule-1 s-1
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
OH + CH4  CH3 + H2O
CH3 + O2  CH3O2
CH3O2 + NO  CH3O + NO2
CH3O + O2  CH2O + HO2
HO2 + NO  OH + NO2
NO2 + hn  NO + O
NO2 + hn  NO + O
O + O2  O3
O + O2  O3
Can assume CH3 to be in ‘steady-state’. (CH3O, CH3O2, OH, HO2, NO, NO2 too).
Assume [OH] = 106, [CH4] = 4.6 x 1013, [O2] = 5 x 1018, all in molecule cm-3
Appropriate rate constants:
k1 = 6.3 x 10-15 cm3 molecule-1 s-1
k1[OH][CH4] = k2[O2][CH3]ss
k2 = 1 x 10-12 cm3 molecule-1 s-1
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
OH + CH4  CH3 + H2O
CH3 + O2  CH3O2
CH3O2 + NO  CH3O + NO2
CH3O + O2  CH2O + HO2
HO2 + NO  OH + NO2
NO2 + hn  NO + O
NO2 + hn  NO + O
O + O2  O3
O + O2  O3
Can assume CH3 to be in ‘steady-state’. (CH3O, CH3O2, OH, HO2, NO, NO2 too).
Assume [OH] = 106, [CH4] = 4.6 x 1013, [O2] = 5 x 1018, all in molecule cm-3
Appropriate rate constants:
k1 = 6.3 x 10-15 cm3 molecule-1 s-1
k1[OH][CH4] = k2[O2][CH3]ss
k2 = 1 x 10-12 cm3 molecule-1 s-1
[CH3]ss = 0.06 molecule cm-3 (Very small !!)
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
Termolecular reactions (three-body reactions).
(Strangely enough, closely related to “unimolecular reactions” as we will
see in a minute or two).
e.g.,
NO3 + NO2  N2O5
N2O5  NO3 + NO2
So, what is actually going on?
- not as simple as they look (not
elementary reactions)
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
Termolecular reactions (three-body reactions).
e.g.,
NO3 + NO2  N2O5
NO3 + NO2
(N2O5) *
(N2O5) * has a choice –
forwards or backwards
N2O5
NO3 + NO2 = (N2O5) *
(N2O5) * = NO3 + NO2
(N2O5) * + M = N2O5 + M
where “M” = N2, or to a lesser extent, O2
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
Termolecular reactions (three-body reactions).
NO3 + NO2 = (N2O5) *
(N2O5) * = NO3 + NO2
(N2O5) * + M = N2O5 + M
ka
kb
kc,
where “M” = N2 or O2
Interested in the rate of formation of (stabilized) N2O5:
Assume steady-state for (N2O5)*
: ka [NO3] [NO2] = { kb + kc [M] } [(N2O5)*]
Or [(N2O5)*] = { ka [NO3] [NO2] } / { kb + kc [M] }
Substitution yields
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
Termolecular reactions (three-body reactions).
NO3 + NO2 = (N2O5) *
(N2O5) * = NO3 + NO2
(N2O5) * + M = N2O5 + M
ka
kb
kc,
where “M” = N2 or O2
Substitution yields:
Consider low-pressure limit,
[M]  0
(termolecular)
And high-pressure limit
[M]  
(bimolecular)
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
Termolecular reactions (three-body reactions).
NO3 + NO2 = (N2O5) *
(N2O5) * = NO3 + NO2
(N2O5) * + M = N2O5 + M
ka
kb
kc,
where “M” = N2 or O2
Unfortunately, life is even more complicated. AHHH!!!
Jurgen Troe:
where Fc is the “broadening factor”.
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
Termolecular reactions (three-body reactions).
NO3 + NO2 = (N2O5) *
(N2O5) * = NO3 + NO2
(N2O5) * + M = N2O5 + M
ka
kb
kc,
where “M” = N2 or O2
RATE COEFFICIENT (cm3 molecule-1 s-1)
1.0e-12
Fc = 0.6
Fc = 1
8.0e-13
6.0e-13
4.0e-13
2.0e-13
0.0
0.1
1
10
100
PRESSURE (Torr)
1000
10000
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
Termolecular reactions (three-body reactions).
(Strangely enough, closely related to “unimolecular reactions”)
e.g.,
NO3 + NO2  N2O5
N2O5  NO3 + NO2
- not as simple as they look (not
elementary reactions)
N2O5 + M  (N2O5) * + M
(N2O5) * + M  N2O5 + M
(N2O5) *  NO2 + NO3 + M
Analogous treatment:
Consider low-pressure limit,
[M]  0
(bimolecular)
And high-pressure limit
[M]  
(unimolecular)
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
Termolecular reactions (three-body reactions) / Unimolecular reactions:
Main “Take-home” Message: Set of chemical reactions that lead to
recombination of reactive species (radicals), and formation of reservoirs:
NO3 + NO2 + M  N2O5 + M
OH + NO2 + M  HNO3 + M
HO2 + NO2 + M  HO2NO2 + M
CH3C(O)OO + NO2 + M  CH3C(O)OONO2 + M
ClO + NO2 + M  ClONO2 + M
ClO + ClO + M  ClOOCl + M
Often reversible (equilbrium).
Even though the formation of the reservoir is exothermic, the reverse
reaction results in a gain in entropy to compensate.
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
Photolysis reactions: (brief introduction to next week).
Some generalities:
- Sunlight provides energy across the electromagnetic spectrum,
which can be absorbed by molecules.
- Energies of chemical bonds typically correspond to UV photons.
- Thus, absorption of UV sunlight can lead to photolytic destruction
of certain molecules.
e.g.,
NO2 + hn  NO + O(3P)
“j-value” – unimolecular ‘rate constant’. Vary with spectral properties of the
molecule of interest, but also with solar intensity (as a fn of wavelength)
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
Photolysis reactions: (brief introduction to next week).
e.g.,
NO2 + hn  NO + O(3P)
Assume constant solar intensity (j-value is constant), and assume no
production of NO2:
Then, rearrangement and integration leads to:
[NO2]t = [NO2]o exp -(j*t) (exponential decay of NO2)
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
Photolysis reactions: (brief introduction to next week).
e.g.,
NO2 + hn  NO + O(3P)
O2 + hn  O(1D) + O(3P), upper stratosphere and above
O2 + hn  O(3P) + O(3P), stratosphere and above
O3 + hn  O(1D) + O2
O3 + hn  O(3P) + O2
NO3 + hn  NO2 + O(3P)
NO3 + hn  NO + O2
CH2O + hn  HCO + H
CH2O + hn  CO + H2
HONO + hn  OH + NO
Important at all altitudes
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
HOW DO WE MEASURE RATE CONSTANTS IN THE LAB???
TWO BASIC METHODS:
TIME–RESOLVED METHODS (includes “Flash Photolysis” and “Flow Tube”)
INDIRECT METHODS (“Relative Rate” method)
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
HOW DO WE MEASURE RATE CONSTANTS IN THE LAB???
“Flash Photolysis” – e.g., Bryukov et al., J. Phys. Chem. A., 2004, v. 108, 10464-10472.
OH + CH4  H2O + CH3
Basic requirements:
A method for getting CH4 in the vessel,
and knowing its concentration.
A method of generating reactive
radicals (in this case, OH) ‘instantaneously’.
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
HOW DO WE MEASURE RATE CONSTANTS IN THE LAB???
“Flash Photolysis” – e.g., Bryukov et al., J. Phys. Chem. A., 2004, v. 108, 10464-10472.
OH + CH4  H2O + CH3
Basic requirements:
A fast (ms to ms) method for detecting and
quantifying [OH], as it decays via reaction
with CH4.
Generally, detection is via absorption or
fluorescence-based methodologies. In this
case, Laser-Induced Fluorescence (LIF)
is used.
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
HOW DO WE MEASURE RATE CONSTANTS IN THE LAB???
Looking at an elementary bimolecular reaction, so:
To simply measurement, often use “Pseudo-First-Order” conditions. Keep
[OH] produced in the flash small compared to amount of CH4 present.
Then, [CH4] remains essentially constant (change due to reaction with OH very
small), and :
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
HOW DO WE MEASURE RATE CONSTANTS IN THE LAB???
IDEALIZED OBSERVED OH TEMPORAL PROFILE
LIF SIGNAL (proportional to [OH])
1.2
1.0
0.8
0.6
0.4
0.2
0.0
-4
-2
0
2
4
TIME (ms)
6
8
10
12
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
HOW DO WE MEASURE RATE CONSTANTS IN THE LAB???
IDEALIZED OBSERVED OH TEMPORAL PROFILE
1.0
0.0
0.8
Ln { [OH]t/[OH]o}
LIF SIGNAL (proportional to [OH])
1.2
0.6
0.4
0.2
-0.5
-1.0
Col 1 vs Col 2
-1.5
-2.0
0.0
-2.5
-4
-2
0
2
4
TIME (ms)
6
8
10
12
0
2
4
6
TIME (ms)
8
10
12
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
HOW DO WE MEASURE RATE CONSTANTS IN THE LAB???
IDEALIZED OBSERVED OH TEMPORAL PROFILE
1.0
0
0.8
Ln { [OH]t/[OH]o}
LIF SIGNAL (proportional to [OH])
1.2
0.6
0.4
0.2
-1
-2
Col 1 vs Col 2
-3
-4
0.0
-4
-2
0
2
4
TIME (ms)
6
8
10
12
-5
0
2
4
6
TIME (ms)
8
10
12
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
HOW DO WE MEASURE RATE CONSTANTS IN THE LAB???
-1
-2
Col 1 vs Col 2
-3
PLOT OF k' VERSUS METHANE
(Slope give the value for k)
-4
0.5
-5
0
2
4
6
8
10
12
0.4
TIME (ms)
0.3
k'
Ln { [OH]t/[OH]o}
0
0.2
0.1
k = k [CH4]
0.0
0
1
2
3
[CH4]
4
5
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
HOW DO WE MEASURE RATE CONSTANTS IN THE LAB???
“Flow Tube”
e.g., Vaghjiani, J. Chem. Phys., 104, 5479 (1996).
O(3P) + N2H4 
Basic concept:
Gases flow through a tube at a precisely-known velocity.
Concentration of reactive radical is made at the bottom
of the tube.
Because velocity known, exact reaction time is known.
Make measurements of [O(3P)] as a function of reaction
distance, which can be converted to reaction times.
Again use pseudo-first order conditions, so that decay is
logarithmic with time.
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
HOW DO WE MEASURE RATE CONSTANTS IN THE LAB???
RELATIVE RATE METHOD:
-Basic concept – measure an unknown rate coefficient relative to a known one.
- e.g., if want to measure k for reaction of OH with “X” relative to “Y”, expose X
and Y simultaneously to OH, and measure relative rate of decay of the two
species.
and
Because both X and Y exposed to same
OH ‘field’,
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
HOW DO WE MEASURE RATE CONSTANTS IN THE LAB???
RELATIVE RATE METHOD:
From Orlando et al., J. Phys. Chem. 105, 3564,
2001.
k(OH + allyl alcohol) = 4.5 x 10-11 cm3 molec-1 s-1
k(OH + 1-penten-3-ol) = 6.7 x 10-11 cm3 molec-1 s-1
k(OH + 2-penten-1-ol) = 10.6 x 10-11 cm3 molec-1 s-1
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
OK, WHAT ABOUT REACTION MECHANISMS?
e.g., - What are products of
the oxidation of a particular
hydrocarbon by OH?
Environmental chambers
typically used.
e.g., OH + unsaturated alcohols
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
OK, WHAT ABOUT REACTION MECHANISMS?
From Orlando et al., J. Phys. Chem. 105, 3564, 2001.
Reaction of OH with cis-2-penten-1-ol leads
to the products shown.
(glycolaldehyde, propanal, formaldehyde).
HO-CH2-CH=CHCH2CH3
HO-CH2-CH=O, O=CHCH2CH3, HCHO
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
One More Concept – Atmospheric lifetime.
If a compound is emitted into the atmosphere, how long (on average) will it
take to remove it (assuming no other production or emission).
Consider CH4 and its reaction with OH:
OH + CH4  CH3 + H2O
Let’s assume that, on average, [OH] = 106 molecule cm-3. Then k [OH] = k
And
The atmospheric lifetime is then defined as the inverse of this k value.
In this particular case, with k = 3.7 x 10-15 cm3 molecule-1 s-1 at 273 K (roughly the average
tropospheric temperature), k = 3.7 x 10-9 s-1.
The inverse of this quantity is the lifetime, t = 2.7 x 108 s, or about 8.5 years.
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
One More Concept – Atmospheric lifetime.
One more example – Isoprene, C5H8.
Reacts with OH (daytime only), O3 (day and night) and NO3 (nighttime only).
Calculate isoprene lifetime during daylight hours and during nighttime hours,
assuming the following rate constants and concentrations:
[OH] = 106 molecule cm-3, 12-hr daylight; k(OH+Isoprene) = 1 x 10-10 cm3 molecule-1 s-1
[NO3] = 108 molecule cm-3, 12-hr nighttime; k(NO3+Isoprene) = 7 x 10-13 cm3 molecule-1 s-1
[O3] = 1012 molecule cm-3, 24-hr average; k(O3+Isoprene) = 1.3 x 10-17 cm3 molecule-1 s-1
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
One More Concept – Atmospheric lifetime.
One more example – Isoprene, C5H8.
Reacts with OH (daytime only), O3 (day and night) and NO3 (nighttime only).
Daytime loss via OH and O3 reactions.
[OH] = 2 x 106 molecule cm-3, 12-hr daylight; k(OH+Isoprene) = 1 x 10-10 cm3 molecule-1 s-1
[O3] = 1012 molecule cm-3, 24-hr average; k(O3+Isoprene) = 1.3 x 10-17 cm3 molecule-1 s-1
Then Isoprene Loss Rate = kOH [OH] + kO3[O3] = 2 x 10-4 s-1 + 1.3 x 10-5 s-1 = 2.13 x 10-4 s-1.
Then, lifetime t = 1/ (Loss Rate) = 4700 sec. = 1.3 hrs.
REACTION KINETICS: (follows Brasseur, Orlando and Tyndall, pp. 95-114.)
One More Concept – Atmospheric lifetime.
One more example – Isoprene, C5H8.
Reacts with OH (daytime only), O3 (day and night) and NO3 (nighttime only).
Nighttime loss via NO3 and O3 reactions.
[NO3] = 108 molecule cm-3, 12-hr nighttime; k(NO3+Isoprene) = 7 x 10-13 cm3 molecule-1 s-1
[O3] = 1012 molecule cm-3, 24-hr average; k(O3+Isoprene) = 1.3 x 10-17 cm3 molecule-1 s-1
Then Isoprene Loss Rate = kNO3 [NO3] + kO3[O3] = 7 x 10-5 s-1 + 1.3 x 10-5 s-1 = 8.3 x 10-5 s-1.
Then, lifetime t = 1/ (Loss Rate) = 12000 sec. = 3.3 hrs.
A LITTLE BIT OF REVIEW:
Spent the last while talking about kinetic theory of gases, and elementary
reactions.
Molecules move, collide with each other, and (when possible) react with each other!
Talked about how to look at individual elementary reactions
– bimolecular, termolecular/unimolecular, photolysis reactions.
What is the rate of these reactions? Rate constant? How do we measure rate constants,
etc?
A LITTLE BIT OF REVIEW:
The atmosphere (particularly the troposphere) acts as a low-temperature,
slow-burning combustion engine.
Takes all the emissions (reduced compounds) and ‘burns’ (oxidizes) them:
OH
HO2
CH4
CO2
Isoprene
Other by-products, such
as O3, particles, acids,
nitrates, etc.
(2ry POLLUTANTS)
DMS, NH3
NO
NO2
OH
HO2
CH4
Isoprene
NMHC
DMS, NH3
etc.
CO2, Ozone, other
secondary pollutants
NO
NO2
So, how do these elementary reactions
come together to create the engine?
Consider a very simplified atmosphere, for illustrative purposes.
Use CO to represent all the ‘fuel’, combine with NO, NO2, OH, HO2, and see
what happens.
THE TROPOSPHERIC “ENGINE”:
Start with “Odd Nitrogen” Family:
“Active Species”
“Reservoir Species”
:
:
NO and NO2, collectively called NOx
HNO3, (HO2NO2, HONO, many others)
Consider first NO and NO2: Present in the atmosphere mainly due to emissions from
anthropogenic activity, but also from soils and lightning.
NO + O3  NO2 + O2,
NO2 + hn  NO + O,
O + O2 + M  O3 + M,
k = 2 x 10-14 cm3 molecule-1 s-1
j = 8 x 10-3 s-1
(very fast, assume instantaneous)
Assume [O3] = 1 x 1012 molecule cm-3, and assume ‘photo-stationary state’
Then:
k [NO] [O3] = j [NO2]
Rearrange: [NO2] / [NO] = k [O3] / j = 2.5
Loss is often via reaction of OH with NO2, OH + NO2 + M HNO3 + M
THE TROPOSPHERIC “ENGINE”:
Now the “Odd Hydrogen” Family:
“Active Species”
:
OH and HO2, (CH3O2, …), called HOx
“Reservoir Species”
:
H2O2, HNO3, (HO2NO2, HONO, many others)
THE TROPOSPHERIC “ENGINE”:
Now the “Odd Hydrogen” Family:
Production:
Consider first OH and HO2:
O3 + hn  O(1D) + O2
O(1D) + H2O  OH + OH
Conversion of OH to HO2:
OH + CO (+O2)  HO2 + CO2
OH + O3  HO2 + O2,
dominant (when all ‘fuel’ considered)
usually minor
Conversion of HO2 back to OH:
HO2 + O3  OH + 2 O2
HO2 + NO  OH + NO2,
(followed by NO2 + hn  NO + O, O + O2 + M  O3 + M,
which generates O3 !!)
Losses of HOx via two processes:
HO2 + HO2 + M  HOOH + O2 + M
OH + NO2 + M = HNO3 + M