Spectroscopy & Photochemistry I -...

1

Required Reading: FP Chapter 3B, 3C, 4

Required Reading: Jacob Chapter 7

Atmospheric ChemistryCHEM-5151 / ATOC-5151

Spring 2013Jose-Luis Jimenez

Spectroscopy & Photochemistry I

Importance of Spectroscopy and Photochemistry I

• Most chemical processes in the atmosphere are initiated by photons

– Photolysis of O3 generates OH – the most important atmospheric oxidizer:O3 + hv O2 + O(1D)

O(1D) + H2O 2 OH

– Solar photodissociation of many atmospheric molecules is often much faster than any other chemical reactions involving them:

NO2 + hv O + NO (source of O3 in the troposphere)

CF2Cl2 + hv CF2Cl + Cl (photolysis of CFCs in the stratosphere)

HONO + hv OH + NO (source of OH in the troposphere)

NO3 + hv O2 + NO or O + NO2 (removal of NO3 generated at night)

Cl2 + hv Cl + Cl (source of Cl radicals)

H2CO + hv H2 + CO or H + HCO (important in hydrocarbon oxidation)

etc.

2

Importance of Spectroscopy and Photochemistry II

• Absorption of solar and earth radiation by atmospheric molecules directly influences the energy balance of the planet

– Greenhouse effect (CO2, H2O, N2O, CFCs)

– Stratospheric temperature inversion (O3 photochemistry)

• Spectroscopy of atmospheric molecules is used to detect them in situ

– OH is detected via its electronic transition at 310 nm

– NH3 is detected via its fundamental vibrational transition at 1065 cm-1, etc.

Blackbody Radiation

Linear ScaleLog Scale

From R.P. Turco, Earth Under Siege: From Air Pollution to Global Change, Oxford UP, 2002.

Interactive demo: http://phet.colorado.edu/en/simulation/blackbody-spectrum

P/A=T4

T = b

3

Solar & Earth Radiation Spectra• Sun is a radiation source with

an effective blackbody temperature of about 5800 K

• Earth receives circa 1368 W/m2 of energy from solar radiation

• Clicker Question: are relative vertical scales correct in right plot?

A. YesB. Somewhat offC. Completely wrongD. I don’t know

From Turco

Adapted from S. Nidkorodov

Solar Radiation Spectrum

From Turco

UVC B A

Photon Energy

IR

4

Solar Radiation: Initiator of Atmos. ReactionsAverage thermal energy of collisions:

Each degree of freedom ~ ½ kT (per molecule)Collision energy ~ RT = 8.3 J mol-1 K-1 x T (per mole)RT = 2.5 kJ mol-1 @ 300 K

Energy of photons (E = hv):300 nm photon = 380 kJ mol-1

600 nm photon = 190 kJ mol-1

Typical bond strengths:D0(O2) = 495 kJ mol-1

D0(Cl2) = 243 kJ mol-1

C-H, O-H, C-O ~ 400 kJ mol-1

Atmospheric chemistry on Earth is driven by photolysis, not by thermal excitation!!!

Adapted from S. Nidkorodov

What is light?• Dual nature

– Photon: as particle• Energy but no

mass

– As wave: electric and magnetic fields oscillating in space and time

• Wavelength, frequency

• c ~ 3 x 109 m/s

From F-P&P

Discuss in class: at a fundamental physical level, why are molecules capable of absorbing light?

5

The Electromagnetic Spectrum

• Units used for photon energies and wavelengths:– 1 eV = 8065.54 cm-1 = 96.4853 kJ/mol = 23.0605 kcal/mol

– 1 Å = 0.1 nm = 10-10 m;

– 1 micron (m) = 10-6 m = 1000 nm

– Planck's constant = 6.626068 × 10-34 m2 kg / s

• Clicker Q: the energy of a green photon ( = 530 nm) is:

A. 1 kJ/mol B. 230 kJ/mol C. 230 kcal/mol

D. 6 x 10-7 kJ/mol E. I don’t know

1

c

v

hE

(wavenumber)

Reminder of EM Spectrum

6

Types of radiation important in lower atmosphere

• Ultraviolet and visible radiation ( = 100-800 nm)– Excites bonding electrons in molecules– Capable of breaking bonds in molecules photodissociation)– Ultraviolet photons ( = 100-300 nm) have most energy, can

break more and stronger bonds. We will pay special attention to them.

• Infrared radiation ( = 0.8 - 300 m)– Excites vibrational motions in molecules– With very few exceptions, infrared radiation is not energetic

enough to break molecules or initiate photochemical processes

• Microwave radiation ( = 0.5 - 300 mm)– Excites rotational motions in molecules

http://phet.colorado.edu/en/simulation/molecules-and-light



Atmospheric ChemistryCHEM-5151 / ATOC-5151

Spring 2013Jose-Luis Jimenez

Spectroscopy & Photochemistry II

7

v', J', …

v", J", …

Eph

oton

Fundamentals of Spectroscopy• Molecules have energy in translation,

vibration, rotation, and electronic state– Translation (= T) cannot be changed directly

with light

– We will focus on the other 3 energy types

• Molecule can absorb radiation efficiently if:– The photon energy matches the energy

spacing between molecule’s quantum levels

– Optical transition between these quantum levels is allowed by “selection rules”

– “Forbidden” transitions can occur but are weaker

Vibrational Energy & Transitions• Bonds can be

viewed as “springs”

• Energy levels are quantized, – Ev = hvvib(v+1/2)

– vvib is constant dependent on molecule

– v = 0, 1, 2… is vibrational quantum number

From F-P&P

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Vibrational Energy Levels• Ideally: Harmonic Oscillator

– Restoration force of “spring” follows Hooke’s law: F= k x

– Ev = hvvib(v+1/2), v = 0, 1, 2…

– Energy levels are equally spaced

• Really: Anharmonic oscillator– Restauration force rises sharply at small

r, bond breaks at large r

– Vibrational quantum levels are more closely spaced as v increases

...yhxhhνE eevib 32

21

21

21 vvv

From F-P&P

http://phet.colorado.edu/en/simulation/energy-skate-park

Rotational level manifolds for different vibrational quanta

overlap with each other

Example: Ground Electronic State of HF

HF molecular constants

Bv=0 = 20.557 cm-1 (rotational constant)

= 4138.32 cm-1 (harmonic frequency)

xe = 89.88 cm-1 (anharmonicity)

v

)1(

hνE

JBJE

EEE

vib

rot

vibrottotal

Possible rovibrational transition:v=0 v=1

J=14 J=15

From S. Nidkorodov

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Vibration-rotation of HCl• Molecules vibrate and rotate simultaneously

From F-P&P

Electronic Transitions (ETs)• Molecules can undergo an

ET upon absorption of an appropriate photon– Simultaneous vibrational

and rotational transitions– No restriction on v, many

vib. trans. can occur– J = -1, 0, +1

• P, Q, and R branches

• Frank-Condon principle– Time for ET so short (10-15

s) that internuclear distance cannot change

– “vertical” transitions

From F-P&P

10

From Wayne

Pathways for Loss of e- Excitation

• Photophysicalprocesses– Lead to emission

of radiation– Energy converted

to heat

• Photochemical processes– Dissociation,

ionization, reaction, isomerization

Business Items• Piazza details

– Student’s answer, instructor answer, comments

– Email alerts digest

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Repulsive States• No minima in PE

vs r curves

• Dissociation occurs immediately after absorption of light

From F-P&P

Molecules & Light Simulation

http://phet.colorado.edu/en/simulation/molecules-and-light

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Quantum Yields ()• Relative efficiency of various photophysical and

photochemical processes:

• E.g.: NO3 + hv NO3* (3)

NO3* NO2 + O (4a)

NO + O2 (4b)

NO3 + hv (4c)

and so on

• i Are wavelength dependent, above all important at different

absorbed photons ofnumber Total

i processby proceeding molecules excited ofNumber i

absorbed photons ofnumber Total

formed molecules NO ofNumber 24 a

Quantum Yields IIFrom F-P&P

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Atmospheric Chemistry

ATOC-5151 / CHEM-5151

Spring 2013Prof. Jose-Luis Jimenez

Spectroscopy and Photochemistry III

On Beer’s Law

http://phet.colorado.edu/en/simulation/beers-law-lab

“The taller the glass, the darker the brew,The less the amount of light that comes through”

Clicker prediction:If I set up(1) 200 Mconcentration & 1 cm path(2) 100 Mconcentration and 2 cm pathThe transmittance will be:A. SameB. MoreC. LessD. A lot lessE. I don’t know

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Gas Absorption: Beer-Lambert Law I

• Allows the calculation of the decay in intensity of a light beam due to absorption by the molecules in a medium

Definitions:• A = ln(I0/I) = Absorbance = LN(also “optical depth”)• absorption cross section

[cm2/molec]• L absorption path length [cm]• n density of the absorber

[molec/cm3]

)exp(0 NLII

Solve in class: Show that in the small absorption limit the relative change in light

intensity is approximately equal to absorbance.

From F-P&P & S. Nidkorodov

Strength of the bands

Quantify w/ absorption cross-section for one molecule: cm2

a. 1017 cm2

b. 10 cm2

c. 10-12 cm2

d. 10-17 cm2

e. 10-20 cm2

What is a large (but still realistic) value of for molecules?

From Tolbert

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Physical interpretation of • , absorption cross section (cm2 / molecule)

– Effective area of the molecule that photon needs to traverse in order to be absorbed.

– The larger the absorption cross section, the easier it is to photoexcite the molecule.

– E.g., pernitric acid HNO4

Collisions 10-15 cm2/molec

Light absorption 10-18 cm2/molec From S. Nidkorodov

Measurement of absorption cross sections is, in principle, trivial. We need a light source, such as a lamp (UV), a cell to contain the molecule of interest, a spectral filter (such as a monochromator) and a detector that is sensitive and responds linearly to the frequency of radiation of interest:

Measurements are repeated for a number of concentrations at each wavelength of interest.

Gas cell

n [#/cm3]DetectorFilter

L

I0

L

I

Measurement of Absorption Cross Sections

From S. Nidkorodov

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A sample contains 1 mbar of molecules of interest (A) with =2x10-21 cm2/molec and 1 bar of impurity (I) with =2x10-18 cm2/molec,

both at =530 nm. Which contributes the most to the total absorbance in a 50 cm cell with 1012 photons cm-2 s-1 at the wavelength

of interest?

A) Molecules A contribute mostB) Impurity I contributes mostC) They contribute equally D) They are both negligibleE) I don’t know

Clicker Q

Oxygen Spectrum• O2 photolysis in the 200-

240 nm range is the major source of O3 in the stratosphere

• O2 can absorb nearly all radiation with = 10-200 nm high up in the atmosphere

From Brasseur and Solomon

Clicker Q: Estimate the length of air column at P = 0.01 mbar and T= 200 K (characteristic of 80

km altitude) required to reduce the radiation flux at 150 nm by 10 orders of magnitude. Neglect

the fact that a substantial portion of oxygen is atomized at

this altitude.

Hole in the spectrum

coincident with Lyman line of

H-atom

A. 2.3 cm B. 23 mC. 23 km D. 230 mE. I don’t know

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Calculation of Photolysis Rates I

• “First-order process”

• What does JA for a given molecule depend on?A. Light intensity from above

B. Light intensity from all directions

C. Absorption cross section () & Path length (L)

D. Quantum yield for fluorescence (f)

E. I don’t know

][][

AJdt

AdA

Generic reaction: A + hν B + C

Calculation of Photolysis Rates II

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

AdFAJdt

AdAAA

Generic reaction: A + hν B + C

JA – first order photolysis rate of A (s-1)

σA() – wavelength dependent cross section of A (cm2/#)

A() – wavelength dependent quantum yield for photolysis

F() – spectral actinic flux density (#/cm2/s)

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Group ProblemThe graphs below show the approximate solar flux at the Earth surface, absorption cross section of NO2 molecule, and photodissociation quantum yield of NO2. What is the photodissociation lifetime of NO2(NO2)?

A. 16 sB. 6 s-1

C. 6000 sD. 116 s-1

E. I don’t know

Par

t of

2005

Fin

al e

xam

Corollary: What are the smallest cross sections that matter in the atmosphere?

Group ProblemThe graphs below show the approximate solar flux at the Earth surface, absorption cross section of NO2 molecule, and photodissociation quantum yield of NO2. What is the photodissociation lifetime of NO2(NO2)?

A. 16 sB. 6 s-1

C. 6000 sD. 116 s-1

E. I don’t know

Par

t of

2005

Fin

al e

xam

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TUV Model from NCAR (as run for previous slides)

http://cprm.acd.ucar.edu/Models/TUV/Interactive_TUV/

Actinic Flux @ surface & 20 kmO

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Nitrogen Dioxide (NO2)• NO2 is one of a very few atmospheric molecules that absorb & photolyze in the visible range• Photolysis of NO2 generates ozone in the troposphere:

NO2 + hv NO + O(3P)O(3P) + O2 + M NO2

• Absorption cross sections are structured, and have a non-trivial dependence on T,P.• NO2 contributes to the brown color of air in very polluted cities (but most due to aerosols!).

8x10-19

6

4

2

600550500450400350300Wavelength, nm

Cro

ss S

ecti

on(c

m2

mol

ec-1

, bas

e e)

From S. Nidkorodov

Photochemistry of NO2• Photolysis occurs with nearly 100% yield below 397.8 nm. O-atom immediately makes O3

NO2 + hv NO + O(3P)O(3P) + O2 (+ M) O3

• Between 398 nm and 415 nm, room temperature NO2 still partially photodissociates because of contributions of internal energy to the process, but the quantum yield declines rapidly with wavelength

• Above 410 nm, electronic excitation of NO2 can result in the following processes:NO2

* NO2 + hv FluorescenceNO2

* + O2 NO2 + O2(a1g) Electronic energy transferNO2

* + N2 NO2 + N2(v) Electronic-to-vibrational energy transferNO2

* + NO2 NO + NO3 Disproportionation (lab conditions only)

From F-P&P

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Clicker Q• What is the order of magnitude of the shortest

possible species lifetime due to photolysis at the 210 nm radiation peak @ 20 km altitude?A. 50 years

B. 50 months

C. 50 days

D. 50 hrs

E. 50 min

Corollary: what about the surface?

20 km

Surface

Examples of Photolysis Rates

From F-P&P

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General Remarks

• In order to photodissociate a molecule it must be excited above its dissociation energy (D0).

• In the lower troposphere, only molecules with D0 corresponding to > 290 nm are photochemically active. Most common atmospheric molecules, including N2, CO, O2, CO2, CH4, NO, etc. are stable against photodissociation in the troposphere.

• In addition, the molecule should have bright electronic transitions above D0. For example, HNO3 has a low dissociation energy (D0 = 2.15 eV) but it needs UV for its photolysis because it does not have appropriate electronic transitions in the visible.

• In general, both the absorption cross sections and photodissociation quantum yields are wavelength dependent.

• Photoionization processes are generally not important in the lower atmosphere (ionization potentials for most regular molecules > 9 eV).

Photodissociation is the most important class of photochemical process in the atmosphere:

AB + hv A + B

From S. Nidkorodov

Chappuis bands

Hartley bands

O3 Absorption Spectrum

Huggins bands

From Yung & DeMore

Infraredabsorption

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O2 Photochemistry• Schumann continuum very

efficient screening radiation below 200 nm

• Solar radiation more intense towards longer

• Dissociation of O2 in Herzberg continuum (200-240 nm) is very important for O3 in the stratosphereO2 + hv O(3P) + O(3P) O(3P) + O2 (+ M) O3

• Troposphere > 290 nm– Not enough energyfor O2 dissociation– O3 from NO2 + hv


From F-P&P

Importance of O3

• Central role in atmospheric chemistry• Highly reactive• Highly toxic => health effects in humans• Crop degradation => billions of $ in losses• Absorbs UV

– Shield surface from hard UV– Its photolysis produces O(1D), which yields OH

• OH is most important tropospheric oxidant

– Photolysis to O(3P) regenerates O3, not important!

• Absorbs IR– Greenhouse gas

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O3 Photochemistry• Most important aspect is production of O(1D) (and thus OH)

O3 + hv O2 + O(1D) (1D) 90% below 305 nm

O3 + hv O2 + O(3P) (3P) 10% below 305 nm

O(1D) + H2O 2 OH OH yield 10% (at the surface)

O(1D) + M O(3P) the rest of O(1D) atoms are quenched

Energy Threshold

From F-P&P

UV absorption by O2 and O3

http://www.ccpo.odu.edu/SEES/ozone/oz_class.htm

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O3

O2

Solar Radiation Spectrum IV• Solar spectrum is strongly

modulated by atmospheric absorptions

• Remember that UV photons have most energy– O2 absorbs extreme UV in

mesosphere; O3 absorbs most UV in stratosphere

– Chemistry of those regions partially driven by those absorptions

– Only light with >290 nm penetrates into the lower troposphere

– Biomolecules have bonds that can break with UV absorption => damage to life

• Importance of protection provided by O3 layerFrom F-P&P

Solar Radiation Spectrum vs. altitude

• Very high energy photons are depleted high up in the atmosphere• Some photochemistry is possible in stratosphere but not in troposphere

• Only > 290 nm in trop.

From F-P&P

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Photoionization

Clicker Q: is photoionization important in the troposphere?A. Yes for many

speciesB. Yes for NaC. Yes for Na & MgD. NoE. I don’t know

Chlorofluorocarbons (CFCs)

• No other CFC sinks than photolysis

• Known that Cl would destroy O3

• 1995 Nobel Prize (M&R) is the idea in this slide: CFCs will provide large source of Cl in stratosphere and lead to O3destruction

Absorption Spectra

Photolysis Rates


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Photolysis Rates for O2 , NO2, and O3Typical values for photodissociation coefficients, J, for O3, O2, & NO2 as a function of altitude. Photolysis rate for O2 is stronglyaltitude dependent because the lower you go the less UV radiation capable of breaking O2 is available (self-shielding). Photolysis rate for O3becomes altitude dependent below 40 km for similar self-shielding reasons. On the contrary, visible NO2 photolysis occurs with about the same rate throughout the atmosphere because there is not enough of it for self-shielding.

Solve in class: Based on the figures shown here, estimate the lifetime of NO2, O2 and O3 against

photodissociation at 20 km and 50 km.

From Yung & DeMore

From Warneck

Solar Radiation Spectrum II

From Turco

From F-P&P

•Solar spectrum is strongly modulated by atmospheric scattering and absorption

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Scattering by Gases• Purely physical

process, not absorption

• Approximation:

• Strongly increases as decreases

• Reason why “sky is blue” during the day

From Turco

420

5 /)1(10044.1 ntsg




ATOC-5151 / CHEM-5151


Spectroscopy and Photochemistry IV

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Solar Radiation IntensityTo calculate solar

spectral distribution in any given volume of air at any given time and location one must know the following:– Solar spectral

distribution outside the atmosphere

– Path length through the atmosphere

– Wavelength dependent attenuation by atmospheric molecules

– Amount of radiation indirectly scattered by the earth surface, clouds, aerosols, and other volumes of air

From F-P&P

Solar Radiation: Further DetailsO

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Solar Zenith Angle

sech

L

pathlength Vertical

pathlength Actualm Mass"Air "

S o la r F lu x

W a v e len g th [n m ]1 0 0 0

Sol

ar F

lux

[Ph

oton

s cm

-2 s

-1 n

m-1

]

1 0 1 2

1 0 1 3

1 0 1 4

1 0 1 5

S Z A = 0 ºS Z A = 8 6 º

3 0 0 4 0 0 5 0 0 7 0 0

At large SZA very little UV-B radiation

reaches the troposphere

• Aside form the altitude, the path length through the atmosphere critically depends on the time of day and geographical location.

• Path length can be calculated using the flat atmosphere approximation for zenith angles under 80º. Beyond that, Earth curvature and atmospheric refraction start to matter.

Radiation vs. time of day, location, season

From F-P&P

Q: summer/winter solstices intensity at 445 nm: @ 8 am? @ noon?

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Surface Albedo

• Wavelength dependent!

• Question: for the same incident solar flux, will you tan faster over snow or over a desert?

)Radiation(Incident

)Radiation( Reflected)(

Albedo

From F-P&P

Direct Attenuation of Radiation

I radiation intensity (e.g., F)I0 radiation intensity above atmospherem air masst attenuation coefficient due to

– absorption by gases (ag)– scattering by gases (sg)– scattering by particles (sp)– absorption by particles (ap)

apspagsg

m-t

ttttt

eI I

0

Rayleigh scatteringtsg -4

Deep UV – O, N2, O2

Mid UV & visible – O3

Near IR – H2OInfrared – CO2, H2O, others

tag

tsp -n

much more

complex

From F-P&P & S. Nidkorodov

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Scattering & Absorption by Particles• Particles can scatter

and absorb radiation• Scattering efficiency is

very strong function of particle size– For a given wavelength

• Visible: ~ 0.5 m – Particles 0.5-2 m are

most efficient scatterers!

• Will discuss in more detail in aerosol lectures

From Jacob




ATOC-5151 / CHEM-5151


Spectroscopy and Photochemistry V (and last!)

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Review Clicker Q

• Jose said during last lecture that: – ozone is a greenhouse gas

– the absorption spectrum of ozone to the right is centered around 1000 cm-1 = 1 m

• Is that correct?A. Yes

B. Only approximately

C. No

D. I don’t know

O3 Infraredabsorption

Database of Absorption Spectraht

tp://

ww

w.a

tmos

pher

e.m

pg.d

e/en

id/2

295

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Example of Spectra from Database

Limitations of Available Information

Spectra in database:

Quantum yields in database:

Much less data for quantum yields, much harder to measure than absorption spectra. (Why?)

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Structure of Important N-Species

• From Jacobson (1999) Table B– http://cires.colorado.edu/jimenez/AtmChem/Jacobson_Table_B.pdf– Many other species there, useful when you don’t know detailed

structure

Note sensitivity of photolysis rates to shape of absorption spectra

Solve in class: Estimate the photodissociation lifetimes of N2O and

HONO at 20 km. Compare those with

characteristic times for vertical transport in the

stratosphere ( 2 years).

Absorption Spectra and Photolysis Rates of Selected N-Species

From S. Nidkorodov

From Warneck

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Nitrous Acid (HONO)• HONO + hv OH + NO 1 below 400 nm

• HONO, like NO2, has strong absorption in visible and a highly structured spectrum. Its photochemical lifetime in the atmosphere is ~ a few minutes.

• Very important as source of OH radicals in the morning

From F-P&P

All aldehydes have relatively weak transitions at around 300 nm. Photoexcitation of aldehydes normally results in a release of HCO, which is quickly converted into HO2 and CO in the atmosphere:

RCHO + hν R + HCO (slow)

R + O2 RO2 (very fast)

HCO + O2 HO2 + CO (very fast)

Formaldehyde (HCHO)• Two competing photodissociation channels:

HCHO + hν H + HCO (a)

HCHO + hν H2 + CO (b)

• Channel (a) leads to HO2

radical production via:

H + O2 + M HO2 (19)

HCO + O2 HO2 + CO (20)

• Sources of HO2 are effectively sources of OH because:

HO2 + NO OH + NO2 (17)

HCHO can be a dominant Hox source (e.g. Mexico City)

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Sources of HOx = OH + HO2 in Mexico City

• HOx drive smog and secondary aerosol chemistry

• HONO photolyzes at long , very important in early morning

• HCHO (formaldehyde) is dominant source

• O3 source needs to wait for O3 to be produced! (depends on others)

From R. Volkamer & W. Brune (MIT & PSU)

Nitric Acid (HNO3)HNO3 + hv OH + NO2 1 between 200 and 315 nm

HNO3 + hv O + HONO requires vacuum UV; is only 0.03 at 266 nm

HNO3 + hv H + NO3 requires vacuum UV; is only 0.002 at 266 nm

• Overall relatively long

lifetime against photolysis

From F-P&P

38

Nitrate Radical (NO3)• NO3 is important in nighttime chemistry. It has unusually large cross sections in the red photodissociates in seconds in the morning.

NO3+ hv NO2 + O(3P) is dependent; important towards the blue

NO3+ hv NO + O2

is dependent; important in the red; competes with fluorescence; this process is nearly thermoneutral but it is inhibited by a tall energy barrier.

From F-P&P

Nitrous Oxide (N2O)• N2O is extremely long-lived because it is unreactive, and it does not absorb much above 200 nm. Below 200 nm:

N2O + hv N2 + O(1D)- Note: this is a popular laboratory method of generating O(1D); 1

• Subsequent reactions of O(1D) with N2O lead to production of other nitrogen oxides in the stratosphere:

N2O + O(1D) NO + NOmajor source of NO in the stratosphere

N2 + O2

competing step

• Because of its stability, N2O is used a "tracer". Concentrations of other molecules are often compared to that of N2O to see is they are well mixed.

From Calvert & Pitts

Solve in class: Find a conversion between absorption coefficient in cm-1atm-1 an

cross sections in cm2/# at room temperature. Evaluate N2O cross sections

at 193 nm using the data shown here.

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Atm. Profiles of Important N-SpeciesImportant N-containing molecules in lower atmosphere: – extremely photo-stable

N2O – easily-degradable NO2,

NO3, and N2O5, HO2NO2

– multitude of poorly quantified organic nitrates, RONO2 (most nitrates are relatively stable towards UV)

Generally, concentrations are inversely proportional to photodissociation / reaction rates in the atmosphere

Alt

itud

e (k

m)

Tota

l Den

sity

From NASA

HW 5 discussion• How to load data from databases into Igor

• Interpolation in HW5.2

• Diagram for HW5.3– NO2 + hv → NO + O (1)

O + O2 + M → O3 (2)NO + O3 → NO2 + O2 (3)

– O3 + hv → O(1D) + O2 (4)

– O(1D) + H2O → 2 OH (5)

– O(1D) + O2 → O(3P) (6)

– O(1D) + N2 → O(3P) (7)

– OH + NO2 → HNO3 (8)

– H + O2 → HO2 (9)

– HO2 + NO → OH + NO2 (10)

– HO2 + HO2 → H2O2 + O2 (11)

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Space for Diagram for HW5.3

Spectroscopy & Photochemistry I -...

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