Chemistry · Lesson
Atmospheric reactions, aerosols and photochemistry
Connect photon energy, gas amount fractions and explicitly bounded reaction networks while distinguishing local chemical transformation from transport.
Back to the Chemistry mapBy the end, you can
- Convert a specified vacuum wavelength to photon and molar photon energy.
- Keep absorbed photons distinct from expected chemical events.
- Convert an amount fraction with the stated gas temperature and pressure.
- Balance a photochemical cycle and distinguish chemical-loss time from local removal time.
Air chemistry involves gases, particles and surfaces
A gaseous species, a particle-associated constituent and an adsorbed species occupy different environments. Transport and phase exchange can change their distribution. Aerosols are populations of particles, with size, composition and state affecting their behaviour.
A mixing ratio and an amount concentration have different meanings and units. Conversion needs the specified temperature, pressure and gas model. Particle number, mass and surface-area concentrations describe different aspects of a population.
Energy, event count and gas concentration have different denominators
For an original symbolic single-photon teaching case, prescribe vacuum wavelength 410 nm, absorbed-photon amount 4.20 μmol and reaction quantum yield 0.600. The incident photon amount and any real molecular absorption spectrum are unspecified. Use exact SI Planck constant h=6.62607015×10⁻³⁴ J s, speed of light c=299,792,458 m s⁻¹ and Avogadro constant N_A=6.02214076×10²³ mol⁻¹.
With E=hc/λ, photon energy is 4.84498990×10⁻¹⁹ J; multiplying by N_A and converting joules to kilojoules gives 291.772111 kJ mol⁻¹ of photons. Expected event amount is separately 4.20×0.600=2.52 μmol. Photon energy alone does not establish absorption, bond cleavage or a reaction rate.
Changing only the specified wavelength to 315 nm gives 6.30617732×10⁻¹⁹ J per photon and 379.766875 kJ mol⁻¹, an energy factor 410/315=1.30158730. Holding the teaching photon amount and yield fixed leaves expected events at 2.52 μmol. An actual molecule need not retain that yield or absorption at a different wavelength.
For a separate hypothetical ideal gas mixture, prescribe ozone amount fraction y=42.0 nmol mol⁻¹=4.20×10⁻⁸, total pressure 93,000 Pa and temperature 283 K. The ideal relation gives ozone amount concentration c_O3=yp/(RT), using molar gas constant R=8.31446261815324 J mol⁻¹ K⁻¹.
The result is 1.66001349×10⁻⁶ mol m⁻³. With a supplied rounded ozone molar mass 48.0 g mol⁻¹, mass concentration is 79.6806477 μg m⁻³. At 311 K with the same amount fraction and pressure, these become 1.51055890×10⁻⁶ mol m⁻³ and 72.5068273 μg m⁻³. The amount fraction stays fixed; concentration per volume changes. No actual air measurement or threshold is supplied.
Photochemistry needs both absorbed light and a reaction pathway
A photon’s energy depends on wavelength. A molecule’s absorption and the fate of its excited state determine whether absorbed light leads to the particular reaction of interest. Incident light is different from absorbed photons and from the resulting chemical events.
A photochemical rate can change with the light field, absorption, quantum yield and other conditions. A reaction equation alone does not provide all those quantities. A change in concentration can also include transport or reactions that occur without light.
A closed matter cycle and a local removal model answer different questions
Specify this three-step atmospheric matter model: NO₂+hν→NO+O(³P); O(³P)+O₂+M→O₃+M; NO+O₃→NO₂+O₂. Here hν denotes absorbed photon energy and M is a nonconsumed bath partner such as N₂ or O₂. O(³P) specifies the oxygen-atom electronic state; alternative states and reactions are outside this model.
For a restricted photostationary relation, prescribe NO₂ photolysis coefficient J=0.00500 s⁻¹, NO₂ and NO number densities 3.60×10¹⁰ and 1.20×10¹⁰ molecules cm⁻³, and NO/O₃ coefficient k=2.00×10⁻¹⁴ cm³ molecule⁻¹ s⁻¹. Assume the named oxygen-atom step is rapid and the listed NO reaction dominates the stated ozone-loss balance.
Equating J[NO₂]=k[NO][O₃] gives [O₃]=7.50×10¹¹ molecules cm⁻³ and cycle rate 1.80×10⁸ molecules cm⁻³ s⁻¹. Doubling only J to 0.0100 s⁻¹ doubles those model values to 1.50×10¹² and 3.60×10⁸. Summing all three matter equations cancels the chemical species, including the third body; the completed closed matter cycle produces no net ozone. Photon energy input still drives the cycle. Other chemical routes or transport require another network.
For a separate homogeneous fixed-volume symbolic parent X, assign bimolecular coefficient 1.80×10⁻¹² cm³ molecule⁻¹ s⁻¹ and maintained radical density 5.50×10⁵ molecules cm⁻³. Their product is pseudo-first-order chemical coefficient k_chem=9.90×10⁻⁷ s⁻¹. Also assign transport-out coefficient k_out=7.00×10⁻⁷ s⁻¹ and no input, so total local parent-removal coefficient is 1.69×10⁻⁶ s⁻¹.
Start with 2.80×10⁹ parent CoreUnits cm⁻³; each symbolic product Z retains one tracked CoreUnit. At 4.50 days, parent=1.45142730×10⁹, chemically converted=7.89992292×10⁸ and transported-out=5.58580408×10⁸ CoreUnits cm⁻³. The three inventories sum to the initial pool. These are symbolic-unit balances, not complete atoms or charges of an unknown real product.
Chemical-only characteristic time 1/k_chem is 11.6909839 days; local parent-removal time 1/(k_chem+k_out) is 6.84856454 days. Doubling maintained radical density gives k_chem=1.98×10⁻⁶ s⁻¹, total 2.68×10⁻⁶ s⁻¹, local time 4.31868436 days and remaining parent at 4.50 days 9.87711545×10⁸ CoreUnits cm⁻³. Transport remains fixed and does not destroy the conserved unit. These times are neither global persistence estimates nor the ages of each molecule.
Parent / 10⁹ CoreUnits cm⁻³
Time / days
- Assigned k_chem=9.90×10⁻⁷ s⁻¹
- Doubled radical density
- — First local-removal model
Underlying data
| Time / days | Parent / 10⁹ CoreUnits cm⁻³ |
|---|---|
| 0 | 2.8 · Assigned k_chem=9.90×10⁻⁷ s⁻¹ |
| 4.5 | 1.4514273001027196 · Assigned k_chem=9.90×10⁻⁷ s⁻¹ |
| 6 | 1.1659350897156509 · Assigned k_chem=9.90×10⁻⁷ s⁻¹ |
| 0 | 2.8 · Doubled radical density |
| 4.5 | 0.987711544551216 · Doubled radical density |
| 6 | 0.6978908367309945 · Doubled radical density |
Original symbolic fixed-volume model X(t)=2.80×10⁹exp[−(k_chem+k_out)t], with t converted from days to seconds. Circles use maintained radical density 5.50×10⁵ molecules cm⁻³; squares double that density, retaining the same transport coefficient. Dashed curve shows only the first model. Parent decline does not mean the tracked CoreUnit is destroyed or determine global atmospheric persistence.
A reactive intermediate belongs to a network
Radicals can participate in initiation, propagation, termination and branching steps. A species can be regenerated while the net reaction converts other constituents. Cancelled intermediates help derive a net equation but do not disappear from the kinetic explanation.
A short sequence can illustrate atom balance without predicting a complete atmosphere. Competing reactions, changing concentrations, illumination and surfaces can alter the network. State the molecular or electronic state when it changes the reaction account.
A lifetime estimate needs its removal model
Under a stipulated constant removal coefficient, a species can show exponential decay and a characteristic lifetime. Several independent first-order removal processes add their coefficients within that model. A local chemical-loss time and a global atmospheric persistence estimate need different evidence.
Formation and transport can sustain a concentration even while removal occurs. The original rate examples below are invented conditions, not current pollution forecasts or instructions for generating or handling reactive atmospheric species.
Worked examples
Calculate energy and expected events separately
Use 410 nm vacuum wavelength, 4.20 μmol absorbed photons and assigned yield 0.600.
- E=hc/λ=4.84498990×10⁻¹⁹ J per photon.
- Molar energy is 291.772111 kJ mol⁻¹ of photons.
- Expected events are 2.52 μmol.
Absorbed-photon energy and the event expectation have different bases.
Convert an amount fraction at stated pressure and temperature
Use ozone y=4.20×10⁻⁸, 93,000 Pa, 283 K and supplied molar mass 48.0 g mol⁻¹.
- c_O3=yp/(RT)=1.66001349×10⁻⁶ mol m⁻³.
- Multiplying by 48.0 g mol⁻¹ gives 79.6806477 μg m⁻³.
- The inverse conversion recovers the specified amount fraction.
Gas conditions belong in a concentration-per-volume conversion.
Close the specified photostationary rates
Use J=0.00500 s⁻¹, NO₂ 3.60×10¹⁰ and NO 1.20×10¹⁰ molecules cm⁻³, and k=2.00×10⁻¹⁴ cm³ molecule⁻¹ s⁻¹.
- The photolysis flux is 1.80×10⁸ molecules cm⁻³ s⁻¹.
- Equating it with the NO/O₃ flux gives O₃ 7.50×10¹¹ molecules cm⁻³.
- Summing the three stated matter steps gives zero net ozone production.
A circulating photochemical matter cycle differs from net ozone production.
Include chemical conversion and outward transport
Use the specified symbolic parent, maintained radical and transport model for 4.50 days.
- Chemical and total local coefficients are 9.90×10⁻⁷ and 1.69×10⁻⁶ s⁻¹.
- Parent, converted and transported pools are 1.45142730×10⁹, 7.89992292×10⁸ and 5.58580408×10⁸ CoreUnits cm⁻³.
- Chemical-only and local characteristic times are 11.6909839 and 6.84856454 days.
Transport removes parent from the boundary without destroying its tracked unit.
Try it yourself
Change only wavelength to 315 nm, deliberately retaining the same absorbed photon amount and assigned yield. Find energy and expected events.
Show the answer and reasoning
6.30617732×10⁻¹⁹ J per photon; 379.766875 kJ mol⁻¹; 2.52 μmol expected events.
Energy increases by 410/315. Holding a toy yield fixed does not establish the actual yield at the new wavelength.
Change only gas temperature to 311 K. Find ozone amount and mass concentration.
Show the answer and reasoning
1.51055890×10⁻⁶ mol m⁻³ and 72.5068273 μg m⁻³.
At fixed pressure and amount fraction the concentration scales by 283/311.
Double J to 0.0100 s⁻¹ while retaining the other restricted photostationary inputs. Find O₃ and cycle flux.
Show the answer and reasoning
O₃ 1.50×10¹² molecules cm⁻³; cycle flux 3.60×10⁸ molecules cm⁻³ s⁻¹.
Both double, while the summed three-step matter cycle still produces no net ozone.
Double maintained radical density while keeping transport unchanged. Find chemical coefficient, local time and parent after 4.50 days.
Show the answer and reasoning
k_chem=1.98×10⁻⁶ s⁻¹; local time 4.31868436 days; parent 9.87711545×10⁸ CoreUnits cm⁻³.
Total coefficient is 2.68×10⁻⁶ s⁻¹. It is not twice the original total because transport stays fixed.
What to carry forward
- Photon energy alone does not establish an absorbed-photon inventory or reaction yield.
- Gas amount fractions and amount concentrations require different denominators.
- Summing a reaction cycle can reveal zero net formation despite continuing reactions.
- A local removal time can include outward transport as well as chemical transformation.
References
Teaching resources and reference material supporting this lesson’s explanations and examples.
- AR6 WGI Annex VII glossary: lifetimeIPCC Working GroupI international assessment; individual glossary editors not extractedSource · PDF
- Glossary of Atmospheric Chemistry Terms: photostationary stateJack G. Calvert; IUPAC international recommendationsSource · PDF
- NOx24: NO+O3→NO2+O2 evaluated rateR. Atkinson; D.L. Baulch; R.A. Cox; J.N. Crowley; R.F. Hampson; R.G. Hynes; M.E. Jenkin; M.J. Rossi; J. Troe; international IUPAC atmospheric evaluationSource · PDF
- The International System of Units: defining constantsNIST institutional SP330, adopting international SI definitions; individual editors not extractedSource
- Quantum yield, Q04991IUPAC Compendium of Chemical Terminology; original photochemistry recommendationsSource
- Aerosol, A00176IUPAC original Recommendations2007 on sols, gels, networks and inorganic-organic hybrid materialsSource