The Chapman Cycle and Stratospheric Ozone

Introduction

In 1930 Sydney Chapman proposed that the stratospheric ozone layer is made and destroyed by oxygen alone, driven by sunlight. His four-reaction mechanism correctly explains why there is an ozone layer, why it lies in the stratosphere and why it heats the air around it. It is also a model case of applying kinetics and the steady-state approximation to a real natural system. Understanding the Chapman cycle, and where it fails, is the starting point for understanding ozone depletion.

Core explanation

The four reactions.

1. O₂ + hν → O + O (λ < 242 nm), rate constant J₁ 2. O + O₂ + M → O₃ + M, rate constant k₂ 3. O₃ + hν → O₂ + O (λ < about 1100 nm, strongest 200–310 nm), rate constant J₃ 4. O + O₃ → 2O₂, rate constant k₄

Reaction 1 is the only true source of ozone: it creates oxygen atoms from stable O₂. Reaction 2 is a termolecular association; the third body M carries away the energy released, otherwise the new O₃ would fall apart. Reaction 3 is where ozone absorbs harmful UV-B and UV-C and converts it into heat. Reaction 4 is the slow chemical sink that finally returns ozone to O₂.

Odd oxygen. Reactions 2 and 3 are fast and simply swap O and O₃ back and forth; neither changes the total O + O₃. Chemists therefore group them as the odd-oxygen family , Oₓ = O + O₃. Oₓ is produced only by reaction 1 (two per photon) and destroyed only by reaction 4 (two per reaction).

Partitioning within the family. Because reactions 2 and 3 are much faster than 1 and 4, O and O₃ reach a rapid steady state:

k₂[O][O₂][M] = J₃[O₃], so [O]/[O₃] = J₃ / (k₂[O₂][M])

In the lower and middle stratosphere this ratio is very small (well below 10⁻⁴), so almost all odd oxygen is ozone. Higher up, [M] falls and the ratio grows.

Steady-state ozone. Setting Oₓ production equal to loss, 2J₁[O₂] = 2k₄[O][O₃], and substituting for [O] gives:

[O₃] = [O₂] × √( J₁ k₂ [M] / (J₃ k₄) )

Why a layer? J₁ needs short-wavelength UV, which is plentiful at high altitude but absorbed as it penetrates deeper. [O₂] and [M] increase downwards. The product of a source increasing with height and air density increasing with depth creates a maximum. The ozone number density peaks at about 20–25 km, while the ozone mixing ratio peaks higher, near 30–35 km.

Where Chapman fails. When measured rate constants are used, the Chapman expression predicts roughly twice the observed ozone. Something else must be destroying odd oxygen: catalytic cycles involving HOx, NOx, ClOx and BrOx radicals, which accelerate the net reaction O + O₃ → 2O₂.

Formulae

[O]/[O₃] = J₃ / (k₂[O₂][M]). Steady-state ozone: [O₃] = [O₂] √(J₁k₂[M] / (J₃k₄)). 1 Dobson unit = 2.69 × 10¹⁶ O₃ molecules cm⁻²; a typical global column is about 300 DU.

Step-by-step reasoning

To derive the Chapman steady state:

1. Identify the odd-oxygen family, Oₓ = O + O₃. 2. Write d[Oₓ]/dt = 2J₁[O₂] − 2k₄[O][O₃] and set it to zero. 3. Apply the fast-partition steady state to express [O] in terms of [O₃]. 4. Substitute and rearrange to get [O₃]² and hence [O₃]. 5. Interpret the dependence on J₁, J₃ and [M] at different altitudes.

Visual explanation

Draw a triangle of species: O₂ at the base, O and O₃ at the top joined by a thick double arrow for the fast reactions 2 and 3. A thin arrow up from O₂ (reaction 1) feeds the family, and a thin arrow back down (reaction 4) drains it. The thick fast loop spins many times for every trip through the thin arrows.

Real-world analogy

Think of money moving between a current account and a savings account (O and O₃). Transfers between them are frequent but do not change your total wealth. Only your salary (reaction 1) increases it, and only spending (reaction 4) reduces it. Total wealth settles where salary balances spending.

Real-world example

Ozone columns are measured worldwide in Dobson units by ground spectrometers and satellites. The global average is about 300 DU — equivalent to a layer of pure ozone just 3 mm thick at the surface — yet this thin shield absorbs almost all UV-C and most UV-B, protecting DNA in living organisms.

Why?

Why is ozone photolysis (reaction 3) not a true loss of ozone? The oxygen atom it produces almost always recombines with O₂ to reform O₃ within seconds. The absorbed UV energy ends up as heat, but odd oxygen is conserved; only reaction 4 permanently removes it.

Common misconception

"Ozone is densest where its mixing ratio is highest." The mixing ratio peaks near 30–35 km, but air is thinner there. The number density — molecules per cm³ — peaks lower, near 20–25 km, because the total air density is much greater.

Worked example

Question: At 30 km, J₃ = 1 × 10⁻³ s⁻¹, k₂ = 1 × 10⁻³³ cm⁶ molecule⁻² s⁻¹ (approximately), [O₂] = 8 × 10¹⁶ cm⁻³ and [M] = 4 × 10¹⁷ cm⁻³. Estimate [O]/[O₃].

Reasoning: [O]/[O₃] = J₃ / (k₂[O₂][M]) = 1 × 10⁻³ ÷ (1 × 10⁻³³ × 8 × 10¹⁶ × 4 × 10¹⁷) = 1 × 10⁻³ ÷ 32 ≈ 3 × 10⁻⁵.

Answer: About 3 × 10⁻⁵, so odd oxygen at 30 km is almost entirely ozone.

Quick check

1. Which Chapman reaction is the only one that creates odd oxygen, and why does it occur only at high altitude? Answer: O₂ photolysis; it needs photons below 242 nm, which are absorbed before they reach lower altitudes.

Exam focus

Be able to write all four reactions with correct wavelength conditions and label source, interconversion and sink. The derivation of steady-state ozone using odd oxygen is a favourite long-answer question; state each assumption explicitly.

Advanced insight

Reaction 2 is pressure dependent, like other termolecular associations, and has a negative temperature dependence. The Chapman ozone expression therefore rises as [M] increases, but the supply of short-wavelength photons falls with depth. Modern models embed the Chapman reactions alongside dozens of radical reactions and transport; transport from the tropical source region explains why the thickest columns occur at high latitudes in spring.

Summary

The Chapman cycle forms ozone by O₂ photolysis followed by O + O₂ + M, interconverts O and O₃ by ozone photolysis, and removes odd oxygen by O + O₃. Treating Oₓ as a family and applying the steady-state approximation gives [O₃] = [O₂]√(J₁k₂[M]/(J₃k₄)). It explains the ozone layer's existence and position but overestimates ozone by about a factor of two.

Practice questions

1. Why is a third body M needed in the reaction O + O₂ → O₃? Answer: M removes the excess energy of the newly formed O₃, which would otherwise dissociate back to O + O₂. 2. Show that reactions 2 and 3 together produce no net chemical change. Answer: Adding O + O₂ + M → O₃ + M and O₃ + hν → O₂ + O cancels all species, leaving only the conversion of light into heat. 3. How would the Chapman steady-state ozone change if J₁ were doubled, other factors fixed? Answer: [O₃] is proportional to √J₁, so it increases by a factor of √2, about 1.4. 4. Give the main reason the Chapman mechanism overpredicts ozone. Answer: It ignores catalytic destruction of odd oxygen by radical families such as HOx, NOx, ClOx and BrOx.