Vietnam part 4 CARL p1 50

This section explores some of the important interactions of HOx (OH + HO2) in the
atmosphere. Due to several rapid oxidation and reduction reactions OH and HO2 rapidly
reach a quasi steady state. As will be demonstrated later, the characteristic time to reach
quasi-steady state is dependent on the average lifetime of the species.

1


It is worthwhile looking again at the most abundant species in the Earth’s lower
atmosphere. Just as with the interaction of the Sun’s radiation with the Atmosphere, one
needs to approach reactivity by considering first any possible reactions between a given
radical and the most abundant species since, even if the reaction is quite inefficient (having
a rate bi-molecular rate constant below, say, 1 x 10-16 cm3 s-1) , the actual rate of reaction –
rate constant multiplied by the co-reactant concentration – could still the larger than for
other more efficient reactions.

2


As far as the atmosphere is concerned, essentially all energy received from the Sun is in the
form of electromagnetic radiation. Other sources of energy from, for example, fossil fuel
combustion or transfer from the warm Earth's core, are so small in comparison that they can
be neglected. The average amount of electromagnetic flux reaching the Earth is quite
precisely known at (1366  3) W m-2, (this is equivalent to about 14 100% efficient, 100 W
household light bulbs per square metre) perpendicular to the direction of the photons. Over
the spherical surface of the Earth, this energy averages to 342 W per square meter of the
Earth’s surface. Solar radiation, more precisely, ultraviolet radiation (wavelengths less than
400 nm), is, of course, hugely important to the chemistry of the atmosphere. Without it, the
atmosphere would be inert and any substances released from the Earth's surface would not
be removed by either chemical reaction or photo-dissociation, leading to their increasing
atmospheric concentrations and the related environmental impact. The value 1366 Wm-2 is

known as the Solar constant, S.

3


The Sun behaves very much like a black-body radiator, which is an object that is able to
absorb and emit photons of all wavelengths. Its electromagnetic spectrum follows closely
that of a black body of temperature 5800 K, with the greatest deviations occurring at very
short wavelengths. 5800 K is essentially the average surface temperature of the Sun. The
peak intensity of the emission is found in the visible region close to 500 nm, which happens
to be close to the visual response peak of the human eye at 555 nm. As can be seen in this
graph, both the intensity and the shape of the spectrum of the radiation reaching the Earth’s
surface is modified by absorption of (and scattering by) several atmospheric species.
Amongst these absorbers, both O2 and O3 are prominent in the ultraviolet (UV) and visible
regions of the spectrum. The hashed area shows the total photon flux that would reach the
Earth's surface if atmospheric species did not absorb at all. The difference between the
average incoming flux of 342 W m-2 and the flux arriving at the Earth's surface is accounted
for by reflection, mostly from water clouds and other aerosols.

4


Most of the atmosphere is composed of N2, O2, and H2O vapour. The ratio of the former
two is essentially constant in the troposphere, stratosphere, and mesosphere, while the
concentration of water vapour changes spatially in three dimensions ranging from a fraction
of one percent to about four percent by number of molecules per unit volume. Although
most of the interesting chemistry of the atmosphere occurs between those minor species that
make up only a small fraction of the atmosphere (orange block, above), the macroscopic
structure of the lower and middle atmosphere is governed by the interaction of O2 with
sunlight, as will be discussed shortly.

In order to consider the fate of molecules when subjected to UV and visible radiation, one
may begin by simply looking at the dissociation energies of typical molecular bonds. In the
case of diatomic molecules, the bond dissociation energy is equal to the difference in
enthalpy of formation of XY and of X + Y. N2 has one of the strongest molecular bonds
encountered in nature. The bond dissociation energy can be related to the minimum photon
energy (and hence its associated wavelength) necessary to produce N atoms: for N2, 127 nm
or less is required to dissociate N2 to N + N, assuming that there is no barrier to the
dissociation process that would mean even shorter wavelengths were required for
dissociation. According to the Sun’s emission spectrum as observed from space, relatively
few photons of wavelength less than 127 nm reach the Earth's atmosphere. O2, on the other
hand, may dissociate a longer wavelengths of up to 240 nm. In this spectral region, many
more photons are available.

5


The change in light intensity as it passes through a gas can be easily described by the BeerLambert expression, which predicts an exponential decrease in intensity with distance if the
concentration of the absorbing species remains constant. Please note the different units used
for absorption and also to the fact that the product (, k or) cL is dimensionless. In this
course we will tend to use absorption-cross section with units of cm2 (per molecule) as we
will normally use concentrations with units of (molecules) cm-3.

6


Absorption of light in the atmosphere is interesting for three reasons. (1) It shields animals
and plants from harmful UV wavelengths (2) it heats the atmosphere (3) it produces highly
reactive species.
In order to quantify the absorption process, one needs to know the absorption cross-section
(or equivalently, the molar extinction coefficient, or molar absorption coefficient) of each

molecule and what the resulting dissociation products are, if any. Shown here is the
absorption cross-section for O2 and H2O. That of N2 is not shown, but it becomes significant
compared to the other two only below 150 nm, though it does not dissociate until 127 nm or
less, as already noted. Note, the values (and units) of the absorption cross section. The
highest is of the order of 10-17 cm2. This is considered to be a very high value for a molecule
in the atmosphere, although atoms can have much greater peak absorption cross-sections
than this. The essential physical interpretation of absorption cross-section of 10-17 cm2 is
that, according to a photon corresponding to a particular wavelength, the molecule appears
to have a surface area in the direction of the photons approach, of 10-17 cm2. Such a surface
area corresponds to a diameter of (4 x 10-17/)0.5 = 35 Ǻ, which is several times greater than
the collision diameter of O2. To put this into perspective, remember that Iabs/I0 = cL, where
 is the molecular absorption cross-section (cm-2 per molecule), c is the concentration
(molecules per cm-3), and L is the path length over which absorption occurs (cm). At ground
level, O2 has a concentration of about 5 x 1018 (molecules) cm-3. Thus for 20% absorption, L
= 0.2/(10-17x 5 x 1018) = 0.004 cm. Therefore, a layer of air of only 0.1 cm thick would
appear entirely "black" if the absorption cross section of air in the visible was 10-17 cm2
(only a fraction light of exp[-(10-17 cm2 x 5 x 1018 cm-3 x 0.1 cm)] = 0.007 would penetrate
such a layer. At the other extreme, in order to view an object through air that is 1 km away,
we can say the light absorption needs to be minimal (no more than 10 %, for example).
What should be the absorption cross section in this case? Here  = 0.9/(5 x 1018 cm2 x 1000
m x 100 cm/m) = 2 x 10-24 cm2. These figures are good to bear in mind. Naturally, one has
to consider that the concentrations of most absorbing species in the atmosphere are orders of
magnitude less than that of O2, and one must also take into account of the logarithmic

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change in pressure with altitude, which means that any given absorption path length in the
atmosphere in the vertical direction will not usually have a constant concentration.
Also shown on this figure, is the actinic flux: the spectral irradiance of the Sun directly

above the atmosphere (say at an altitude of 200 km) in linear units. Since most of the light
reaching the atmosphere is lies at wavelengths longer than the main absorption bands of O2
and H2O, a better representation of the influence of O2 and H2O on the solar spectrum is
that of a log/linear plot. This is given on the next page.

7


On the previous graph, it was difficult to see the very small absorption cross-sections
associated with O2 and H2O at longer wavelengths where the Sun's radiation flux begins to
increase rapidly. A logarithmic scale for the y-axis shows these more clearly. Shown also
are two examples of the range of the effectiveness of O2 in reducing the Sun's light
intensity. It can be clearly seen that at 150 nm the absorption cross-section is so large that a
path of O2 (at ground level) of only 0.1 mm is necessary to reduce the incident intensity by
a factor e (that is, a factor of 2.72). At 240 nm, 9 km of O2 is required to achieve the same
reduction factor.
It is quite clear then that due to absorption of O2 alone, light at 150 nm cannot reach the
Earth’s surface. Would the atmosphere be transparent at 240 nm due to absorption of O2
only? In order to answer this question one would need to take into account the exponential
changing concentration of O2 with altitude, as already mentioned. An example calculation is
given later on the page relating air pressure to altitude.

8


In order to work out the amount of light (or number of photons in this case) absorbed by the
atmosphere, three item of information are required. (1) the absorption cross-section as a
function of wavelength (2) the initial light intensity as a function of wavelength (3) the
concentration of the absorption species as a function of distance (for this we also assume a
constant T of 250 K). It is also more convenient to simplify the absorption cross section and

the photon flux data as indicated by the red lines. In practice this means having a table of
absorption cross section values for say each nm and the same for the photon flux.

9


The rate of absorption of photons at any given altitude can be calculated by the integration
given above. Note that there is also a factor that takes into account that not all absorbed
photons lead to photo-dissociation, this is called the quantum yield. To perform an
integration one needs to known the various functions and then be able to integrate them.
This is normally not possible and numerical integration is performed as described in the
lower box. Here F(), (), and () are taken from tables. The units of photolysis rate is
(molecules) s-1. For the numerical integration one only has to consider a relevant
wavelength range that correspond to the range over which photo-dissociation may occur
and the range over which there is a reasonably high actinic flux.

10


Molecular orbitals considered as a perturbation of atomic orbitals are used to find the
ordering of molecular states and the correlation as we consider heavier molecules. Don’t
take the shapes of the above orbitals literally, they just represent the symmetry. The shaded
areas represent positive portions of the wave-function, the white areas represent negative
portions. Note the ‘u’ and ‘g’ assignments. If the wavefunction at position x,y,z is the same
as the wavefunction at –x,-y,-z (where 0,0,0 is the centre of the molecule) then the
wavefunction is said to be even and is given the assignment ‘g’ (meaning even).
Alternatively, if the wavefunction becomes negative (but is otherwise the same) then the ‘u’
assignment is given (meaning odd). Mathematical operations on even and odd functions can
be treated the same as mathematical operations on +1 and -1. Thus g x g = g and u x u = g
and g x u = u and u x u x g x u x g x u = g etc. ‘u’ and ‘g’ assignments are given only to

homonuclear diatomic molecules.
Two important things to note from the above correlation diagram. (1) the higher the nuclear
charge, the lower the orbital energy. This is why the energy of the 1s orbital goes down
from right to left. (2) Diatomic molecular states with the same symmetry cannot cross, thus
g cannot cross with another g, etc.

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The production of OH in the atmosphere, especially in the troposphere, is somewhat associated with
O2 as seen shortly. It is instructive to consider in a little more detail the spectroscopy of O2, since
there exists two low-lying electronically-excited states that are metastable (i.e., their radiative
lifetime is sufficiently long that they are more likely to react or be quenched under tropospheric and
stratospheric conditions than to emit a photon. The O2 system also serves as a brief reminder of
some basic electronic spectroscopy.
In deriving molecular terms, one needs to consider only those electrons outside a filled molecular
sub-shell, since filled shells have zero total spin angular momentum and zero total orbital angular
momentum. In this case they are the two (2g*) electrons, each having ml = 1 or -1. Note, for atoms,
the projection of angular momentum along an axis was given by the quantum number ml which took
all integer values from +l to -l. For molecular orbitals, this is different and ml can take only the
values +1 AND -1, not any in between. The reason for this is can be seen with 2p molecular
orbitals (given on the following page). The angular momentum projection corresponding to ml = 0 is
derived from the pz atomic orbitals that are used to construct the 2p molecular sub-shell, not the
2p sub-shell. This type of reasoning is the same for molecular orbitals of higher angular
momentum.
We can write down all the ways in which the two electrons can be distributed.
Each electron has a orbital angular momentum projection on the inter nuclear axis according to ml =
+1 or -1 (classically, rotation in opposite directions) and a spin projection of +1/2 or -1/2. How can
these electrons then be arranged
------------------------------------------------------------ml=1

ml=-1




















-------------------------------------------------------------

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Ml

2

-2

0

0

0

0

Ms

0

0

0

-1

1

0

------------------------------------------------------------Take the highest value of ML (= 2). This value corresponds to L = 2 and, for MOLECULAR orbitals
will be accompanied by ML = -2. The orbital term sign is defined as the positive value of ML
(ML) thus we have a  state ( = 2). There are two  states, one corresponding to  = +2, the
other to  = -2. Thus, all molecular states with a non-zero angular momentum have an ORBITAL
degeneracy of 2. The total spin of the  state is S = 0, thus the state should be termed 1 (no spin
degeneracy). We are left with four states having ML = 0. For these we then first consider the

highest MS value. Recall the vector summation of s for atoms. We use the same procedure here.
Thus an Ms of 1 implies an S of 1, which is associated with MS values of +1, 0, -1. Thus these
microstates give a 3 term. The remaining microstate gives a 1 term. Two further considerations
are whether the wave function changes sign when reflecting through centre on the molecule
(inversion) (g or u). Both electrons here have g symmetry and g x g = g. The second consideration
is whether or not the wave function is symmetric, or anti-symmetric with respect reflection in the xz
plane (+ or -). The Pauli principle states that the overall wave function should be anti symmetric w.
r. t exchange of the electrons. I will just state here that singlet spin functions are anti-symmetric in
this respect, triplet spin functions are symmetric in this respect. Thus +g state has a spatial part that
is symmetric w.r.t. inversion (g) and symmetric w.r.t. reflection on the xz plane (+). To make the
overall wave function anti-symmetric overall one must couple this symmetric spatial function with
an anti-symmetric spin function, i.e. with a singlet spin function. The converse holds for -g spatial
function. Thus the allowed states for O2 from the (2p)2 valence shell are
X3-g, a1g, b1+g
Again Hund’s rules are applied for the energy ordering for states derived from the same sub-shell.
The capital 'X' denotes the ground state. Small 'a' and 'b' denote excited state of different spin than
the ground state. If the spins of the excited states were the same then one would use 'A' and 'B'.
Because these states arise from the same sub-shell it is expected that the inter-nuclear bond lengths
are about the same.
The y-axis above has units of kJ mol-1. Are we concerned about the electronic levels of O2? Yes,
because firstly, 193 nm radiation dissociates O2 to give O(3P). If you look at the absorption cross
section for O2 the strongest absorption band peaks at 150 nm (800 kJ mol-1) and decreases very
sharply so that at 200 nm (600 kJ mol-1) the absorption cross section is about one million times
smaller. Which transition is responsible for this?
Note that O(1D) then can only be produced at wavelengths shorter than 175 nm in the photodissociation of O2. Thus no O(1D) is produced in this manner in the stratosphere or troposphere.
Note that the threshold for O atom production in the photodissociation of O2 is 242.2 nm, however
O3 production has been observed in pure O2 irradiated with 248 nm. The mechanism might involve
(though this is not clear) absorption of light by a small fraction of O2 in its =1 state. The O atom
formed then forms O3 which photodisociates to O + O2( <=26). The vibrationally-excited O2 can
then readily photo-dissociate at wavelengths longer than 242.2 nm. This mechanism is important for

O3 production in the upper stratosphere and mesosphere, though it is only a small contribution to the
whole.

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The origin of the “Northern Lights” is the generation of electronically excited species
(though not by UV radiation). Some prominent emissions arise from electronically excited
O atoms. O(1S) to O(1D) is spin allowed and therefore quite intense. Transition though from
O(1D) to ground electronic state O(3P) is spin forbidden. The long radiative lifetime of
O(1D) makes it able to undergo collisions in the atmosphere. Most of these collisions lead to
de-excitation to O(3P) without emitting a photon. One then only see this red emission at
high altitudes (low pressures).

13


This graph represents the penetration of UV light through the atmosphere. In fact the graph
actually displays absorbed photons for several wavelengths as a function of altitude. Here,
as expected, the longer wavelengths, corresponding to lower absorption cross-section of O2,
penetrate the atmosphere much further than do the shorter wavelengths. Also note that the
peak absorptions are much greater too. This is because the initial photon flux increases with
wavelength.
Since one photon produces two O atoms, the rate of production of O atoms by this process
is double the photolysis rate of O2.

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It is known that O-atoms react very rapidly with O2 throughout the atmosphere. A collisions

between two mutually reactive species do not always lead to chemical reaction. This can
occur for several reasons that will be explained later. At 20 km, about one in ten thousand
collisions result in reaction. Since O2 has a concentration that is more than ten thousand
times greater than any other species that reacts rapidly with O atoms it is immediately
apparent that this must be the major O-atom loss route in the atmosphere.
The depiction of the reaction above appears at first sight to be rather complicated. It is my
own version of representing the chemical reactions for this course, but it contains much
information that should make each process easier to appreciate. The simplified version
found in nearly all texts is given in the orange box. You should use this latter version when
writing out formulas.

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This is a reminder of the temperature profile of the atmosphere. As this discussion has
already pointed out. The troposphere, stratosphere and mesosphere are formed only as a
consequence of the interaction of sunlight with O2. The ozone that results contributes to the
heating of the atmosphere causing a large-scale temperature inversion that defines the
stratosphere.

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Above is a plot of the absorption spectrum of O3 (note the log scale). Based purely on
thermo-chemical considerations (change in enthalpy), one may dissociate O3 on absorption
of light of wavelength 1130 nm (this is in the infrared). In practice though significant
dissociation of O3 does not occur until the uv spectral region, at which point various other
dissociation products are possible (these are given in the next page). What is quite clear
from the above graph is that the absorption cross-section of O3 increases extremely rapidly
in the region 350 nm to 290 nm. Thus, all other things being equal, the dissociation

probability of O3 at 290 nm will be of the order 100 000 times greater than at 350 nm. Of
course, the actual dissociation frequency at a particular wavelength depends on the
availability of light at that wavelength – this is where the calculation of the actual
dissociation rate of O3 becomes complicated.
You will notice that the amount of light reaching the Earth's surface varies according to the blue line
given above. It is actual no surprise that it has a trend that is opposite to the absorption cross-section
of O3. This is because O3 actually plays the largest role in absorption of the Sun’s light. Since the
frequency of dissociation of O3 at any wavelength is the product of the absorption cross-section, the
intensity of light, and the quantum yield for the particular process involved (see later), the
wavelength range that is responsible for most O3 dissociation in the lower atmosphere lies in the
290-nm to 315-nm region, depending on the angle of the sun. This also means that the peak
wavelength for average O3 dissociation also depends on latitude – at the equator the peak lies toward
shorter wavelengths.

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The ozone absorption cross section can be rationalized by looking at the various potentialenergy surfaces (PESs) involved in photon absorption. Note that the PES given here is not
accurate for all geometries of O3: it is merely an indicative slice through a multidimensional surface. According to the PES significant absorption will take place vertically
from the ground vibrational state. The vertical transition here is an expression of the BornOppehiemer separation of electronic and nuclear motion dictating that the internuclear
distances of the molecule do not change on a time scale associated with photon absorption.
The peak in absorption cross-section for any given band corresponds to the maximum
overlap in wavefunctions. Thus the Hartely band has a peak where the vertical line
intersects the PES leading to O(1D) + O2(a1g). A second peak occurs in the visible region.
This is associated with the green PESs (note these are singlet surfaces so that transition are
allowed, whereas transitions to the grey surfaces are nearly completely forbidden).
Curiously, absorption in the Chappuis and Wulf bands of ozone do not lead to photodissociation; rather fluorescence is observed. This fact is not obvious from the given PESs.

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One of the important considerations when calculating photo-dissociation rates is the
quantum yield for the process. According to the potential energy surfaces there are two
routes for the production of electronically-excited O-atoms, O(1D). The overall contribution
has three parts.
(1) Direct photo-dissociation to O(1D) + O2(a1g) – this process is spin allowed and has a
zero-Kelvin cut-off of 306 nm. At higher temperature a greater fraction of vibrationally
excited O3 is present, which enables photo-dissociation at longer wavelengths than the
306-nm cut-off.
(2) Vibrationally-enhanced photo-dissociation to O(1D) + O2(a1g) – not only does
absorption of vibrationally-excited O3 produce absorption beyond 306 nm, the
wavefunction of the second vibrational level of O3 overlaps the electronic excited state
more than does the ground vibrating level, resulting in a greater quantum yield per unit
population. The 0 K and 400 K quantum yields differ due to vibrationally-excited O3
(3) Spin-forbidden absorption leading to O(1D) + O2(X3-g) – this has a constant quantum
yield of around 0.07 and contributes at all wavelengths up to a cut-off of around 420
nm.

19


In a separate section of this course a detailed analysis of the formation route of OH was
undertaken. Here it is summarized. Once being formed via photo-dissociation of O3
metastable, electronically-excited O(1D) will react rapidly with many species in the
atmosphere at nearly collision frequency (that is nearly every collision undergone by O(1D)
results either in reaction or in physical quenching). Since then the rate constants for reaction
or quenching of O(1D) are similar (ranging by a facto of about 10) for all species one need
first to look at the concentrations of potential co-reactants in order to find the main O(1D)
removal routes (remember rate of removal is a product of bimolecular rate constant
multiplied by co-reactant concentration).

By far the three most abundant species in the atmosphere are N2 (having a constant mixing
ratio), O2 (having a constant mixing ratio), and H2O (which is highly variable both spatially
and temporally). Other reaction of O(1D) are not important for the overall O(1D) removal
rate BUT might be an important source of other species (e.g. stratospheric NO in the
reaction of O(1D) with NO2).

20


Here are two dataset for experimental determination of the rate constants for the reaction of
OH + O(1D).

21


In order to calculate then the average fraction of [O(1D)] (square brackets indicate
concentration) one must calculate the relative rate of removal of O(1D) due to N2, O2, and
H2O. Above are two example calculations: one at 1 km altitude and the other a 27 km
altitude. Concentrations of N2 and O2 are calculated on the basis of an exponentially
decreasing pressure with altitude given by P = P0exp(-H/7400 m), where P0 is the pressure
at ground level (1 Bar), H is altitude in m. Given a value of P one then calculates the
concentration (normally in molecules per cm3, since the rate constants are normally given in
cm3 s-1 molecule-1) of species by first calculating the total concentration (n/V) – referred to
more accurately as number density – using the ideal-gas law, n/V = P/kBT. Where kB here
(not to be confused with rate constant) has a value of 1.38 x 10-23 JK-1 and P is in Pa (Nm-2).
Once total n/V is found (remembering the factor 1000 in the conversion of 1 m3 to 1 cm3 for
V) then the concentration of the species can be found if its mixing ratio (i.e., fraction) is
known at that particular altitude. At atmospheric pressure and 298 K n/V = 2.5 x 1019
molecules cm-3 (note that the unit “molecules” can be, and often is, omitted since it isn’t an
SI unit).

The factor 2 on the O(1D) + H2O calculations is required since two OH are produced per
reaction. The small fraction of O(1D) that produces OH at 27 km is a reflection of the very
dry stratosphere, since most H2O is condensed out below the Tropopause.

22


At a given location there is a strong correlation between measured OH concentration and
the production rate of O(1D) that is derived from flux measurements and local ozone
concentration. The scatter around the correlation line mainly reflects the uncertainty in the
combination of [OH] measurements and J(O1D) determinations. However there is a
relatively large deviation at low light levels from the observed value which indicates
another source of OH and further, that this source is not operative at higher light levels.
Such alternative sources include photo-dissociation of HONO and HCHO during the early
mornings (see later); later in the day [HONO] and [HCHO] has diminished due to photodissociation and relatively slow rates of re-formation.

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