The Hydroxyl Radical: Detergent of the Troposphere

Introduction

Every year the atmosphere receives hundreds of millions of tonnes of methane, carbon monoxide, hydrocarbons and sulfur gases. Most of them do not accumulate indefinitely, because they are oxidised to products that dissolve in rain or are taken up by surfaces. The main agent of this cleansing is not O₂, which is too unreactive, but the hydroxyl radical, OH. Present at well under one part per trillion, OH nonetheless acts as the "detergent" of the troposphere.

Core explanation

Why not O₂? Molecular oxygen has a triplet ground state and strong bond, so it reacts only very slowly with closed-shell molecules such as CH₄ or CO. A radical is needed to start oxidation by abstracting an atom or adding to a double bond.

Primary production. The main source of OH is ozone photolysis at wavelengths below about 330 nm:

O₃ + hν → O₂ + O(¹D)

Most O(¹D) atoms collide with N₂ or O₂ and are quenched to ground-state O(³P), which simply reforms ozone. A small fraction, typically about 10% near the surface in humid air, react with water vapour:

O(¹D) + H₂O → 2OH

The fraction depends on water vapour, which is why OH production is highest in the humid, sunny tropics. Other primary sources include photolysis of nitrous acid (HONO), formaldehyde and peroxides.

Reaction with trace gases. OH reacts with almost everything reduced:

OH + CO → H + CO₂ (followed by H + O₂ + M → HO₂ + M) OH + CH₄ → CH₃ + H₂O

Many of these reactions convert OH into HO₂. OH and HO₂ form the HOx family .

Recycling. HO₂ can be converted back to OH, most importantly by NO:

HO₂ + NO → OH + NO₂

In air with moderate NOx, OH is recycled many times before HOx is lost, greatly amplifying the effect of each primary radical. HOx is terminated by radical–radical reactions such as HO₂ + HO₂ → H₂O₂ + O₂, or by OH + NO₂ + M → HNO₃ + M.

Steady-state concentration. OH has a lifetime of only about one second, so its concentration is set by a local balance of production and loss:

[OH]ss ≈ P(OH) / Σ kᵢ[Xᵢ]

where the sum runs over all species X that react with OH — its OH reactivity . Typical midday values are 10⁶ to 10⁷ molecules cm⁻³; the global, 24-hour mean is about 1 × 10⁶ molecules cm⁻³, a mixing ratio of roughly 0.04 ppt.

Oxidising capacity. Because nearly all removal of reduced gases passes through OH, global OH sets their lifetimes. It is inferred from the decay of methyl chloroform (CH₃CCl₃), a gas with well-known emissions and removed almost entirely by OH. Such studies show that global OH has been fairly stable, varying by only a few per cent between years.

Formulae

P(OH) = 2 J(O¹D)[O₃] × f, where f = k(O¹D + H₂O)[H₂O] / (k(O¹D + H₂O)[H₂O] + k q[M]). Steady state: [OH] ≈ P(OH) / Σkᵢ[Xᵢ]. OH reactivity = Σkᵢ[Xᵢ], in s⁻¹.

Step-by-step reasoning

To estimate a steady-state OH concentration:

1. Calculate the primary OH production rate from ozone photolysis and the fraction of O(¹D) reacting with water. 2. Add any recycling source, such as HO₂ + NO. 3. Sum the pseudo-first-order loss rates kᵢ[Xᵢ] for CO, CH₄, VOCs and NO₂. 4. Divide production by total loss. 5. Check that the answer lies in the expected range of 10⁶–10⁷ cm⁻³.

Visual explanation

Draw OH and HO₂ as two circles joined by arrows: CO, CH₄ and VOCs push OH to HO₂; NO pushes HO₂ back to OH. Feeding in from above is O₃ + hν + H₂O. Leaking out below are HNO₃ and H₂O₂, the termination products.

Real-world analogy

OH is like a small team of cleaners in a vast building. Each cleaner works for only a moment before needing replacement, but new cleaners arrive constantly and the old ones are often sent back to work (recycling). The total cleaning done depends on how fast cleaners arrive and how much mess there is.

Real-world example

Carbon monoxide from fires and traffic is converted to CO₂ over about two months, and methane over about a decade, almost entirely by OH. If OH were reduced — for example by greatly increased CO emissions consuming it — methane would last longer and its climate effect would grow.

Why?

Why is OH concentration so much higher by day than at night? Its main source requires ultraviolet sunlight to photolyse ozone and other precursors. With a lifetime of about a second, OH vanishes within moments of sunset once production stops.

Common misconception

"Because OH is present in such tiny amounts it cannot matter." What matters is the rate, k[OH][X]. OH reacts very rapidly with many gases and is continuously regenerated, so its total oxidising flux is enormous despite its tiny concentration.

Worked example

Question: In clean air, OH is lost mainly to CO (100 ppb, k = 2.3 × 10⁻¹³ cm³ molecule⁻¹ s⁻¹) and CH₄ (1900 ppb, k = 6.4 × 10⁻¹⁵). Take air density 2.5 × 10¹⁹ cm⁻³. Calculate the OH reactivity and lifetime.

Reasoning: [CO] = 2.5 × 10¹² cm⁻³, so k[CO] = 0.58 s⁻¹. [CH₄] = 4.75 × 10¹³, so k[CH₄] = 0.30 s⁻¹. Total ≈ 0.88 s⁻¹.

Answer: OH reactivity ≈ 0.9 s⁻¹; OH lifetime ≈ 1/0.9 ≈ 1.1 s.

Quick check

1. Why is only a small fraction of O(¹D) converted into OH? Answer: Most O(¹D) atoms are quenched by collisions with the far more abundant N₂ and O₂ before they meet a water molecule.

Exam focus

Write the two-step primary source of OH and at least two loss reactions. Be able to apply the steady-state approximation and to explain why NO recycling increases OH. Quote typical concentrations with sensible units.

Advanced insight

Field campaigns in forests with high isoprene emissions have measured OH at levels well above those predicted by standard mechanisms. This led to the discovery of new recycling pathways, such as unimolecular isomerisation of isoprene peroxy radicals that regenerates HOx without NO — an area still being refined.

Summary

OH is the main oxidant of the troposphere. It is produced when O(¹D), from ozone photolysis, reacts with water vapour. OH converts CO, CH₄ and VOCs into oxidised products and is recycled from HO₂ by NO. Its short lifetime means its concentration is a local steady state, around 10⁶ cm⁻³ on global average, which sets the lifetimes of many trace gases.

Practice questions

1. Write the two reactions by which ozone photolysis produces OH. Answer: O₃ + hν → O₂ + O(¹D), then O(¹D) + H₂O → 2OH. 2. Explain why OH production is greater in the tropics than at high latitudes. Answer: The tropics have stronger UV sunlight and much more water vapour, so both ozone photolysis and the O(¹D) + H₂O reaction are faster. 3. How does NO increase OH concentrations in moderately polluted air? Answer: HO₂ + NO → OH + NO₂ recycles HO₂ back into OH, so each primary radical takes part in more oxidation steps. 4. Why is methyl chloroform used to estimate global mean OH? Answer: Its emissions are well known and it is removed almost entirely by OH, so its observed decay rate reveals the average OH concentration.