Vietnam part 4 CARL p1 50

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

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 HO2rapidly

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

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

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.

4

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

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

6

The change in light intensity as it passes through a gas can be easily described by the

Beer-Lambert 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

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 I abs /I 0 = 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

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 O2only? 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

9

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

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

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

11

The production of OH in the atmosphere, especially in the troposphere, is somewhat associated with

O2as 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 m l = 1 or -1 Note, for atoms,

the projection of angular momentum along an axis was given by the quantum number m l which took

all integer values from +l to -l For molecular orbitals, this is different and m l 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 m l = 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 m l =

+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

-

m l=1     

m l=-1     

-

M l 2 -2 0 0 0 0

M s 0 0 0 -1 1 0

-

Take the highest value of M L (= 2) This value corresponds to L = 2 and, for MOLECULAR orbitals

will be accompanied by M L = -2 The orbital term sign is defined as the positive value of M L

( M L ) 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 M L = 0 For these we then first consider the

highest M S 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 M S 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

X 3 

-g , a 1 g, b 1  +

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( 3 P) 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( 1 D) then can only be produced at wavelengths shorter than 175 nm in the

photo-dissociation of O2 Thus no O( 1 D) is produced in this manner in the stratosphere or troposphere

Note that the threshold for O atom production in the photodissociation of O2is 242.2 nm, however

O3production has been observed in pure O2irradiated 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 O3which photodisociates to O + O2(  <=26) The vibrationally-excited O2can

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

12

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)

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

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

16

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

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.

The ozone absorption cross section can be rationalized by looking at the various

potential-energy 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

multi-dimensional 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

Born-Oppehiemer 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

photo-dissociation; rather fluorescence is observed This fact is not obvious from the given PESs

18

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 O3overlaps 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(X

3-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

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)

In order to calculate then the average fraction of [O(D)] (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 = P 0 exp(-H/7400 m), where P 0 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/k B T Where k B 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

photo-dissociation and relatively slow rates of re-formation

On a global scale, the dominating reactions involving the OH radical are those with CH4and CO, with the latter being more important (as a removal route for OH) owing to its greater rate constant The latter reaction is also the major source of HO2 in the troposphere Using the measured background levels of CO and CH4 one can see that the average lifetime

of OH is less than two seconds Since nearly all OH production has its origin in

photo-dissociation (of O3, HCHO, and HONO) it is expected that OH concentrations are extremely low during the night Under these conditions removal of alkanes and alkenes by reactions of NO3 and Cl with alkanes become significant; the latter mainly in the marine boundary layer where Cl is generated by heterogeneous reaction between NaCl(s) and ,for example, ClONO2 to give Cl2 which is readily photo-dissociated in the near uv region The chemistry on the following pages is essentially the main HOx radical chemistry in the, so-called, unpolluted (or remote) troposphere atmosphere, the main characteristic being that the concentration of NOx is very low indeed The remote troposphere is distinguished from the free troposphere in that the former is in unpolluted regions at ground level and the latter refers to regions above the planetary boundary layer where species are considered to be relatively well-mixed Gape Grim is a could place to measure remote troposphere chemistry since air masses their have generally travelled a few thousand kilometres over ocean