Termolecular and Pressure-Dependent Reactions

Introduction

Two reactive fragments can join, but the newly formed molecule may contain enough excess internal energy to separate again. A collision with a third particle can remove energy and stabilise the product. This gives a net equation often written A + B + M → AB + M and a pressure-dependent rate. The compact equation should not automatically be read as one simultaneous three-particle impact: a short-lived energised complex and a later stabilising collision are often the more useful microscopic picture.

Core explanation

Consider A + B ⇌ AB followed by AB + M → AB + M. Formation of energised AB can be fast, but without stabilisation it can dissociate back to A+B. The bath-gas molecule M carries away some excess energy in a collision and reappears chemically unchanged. At low pressure, a newly formed AB may rarely meet M before it dissociates, so more M raises the stabilisation probability. The observed rate can take the limiting form r = k₃[A][B][M]. If concentrations are mol L⁻¹ and r is mol L⁻¹ s⁻¹, k₃ has units L² mol⁻² s⁻¹. IUPAC defines a third body as a species that can help energise a molecule or bring about combination between atoms or radicals.

At a sufficiently high bath-gas concentration, stabilisation may become fast enough that increasing [M] has little further effect on the association rate. A high-pressure limiting expression is r = k₂,∞[A][B], where k₂,∞ has units L mol⁻¹ s⁻¹. Between these limits lies a fall-off region in which an effective second-order rate constant k₂,eff(T,p) depends on temperature and pressure. One often writes r = k₂,eff[A][B], with k₂,eff ≈ k₃[M] at low pressure and k₂,eff → k₂,∞ at high pressure. The same reaction thus appears third order in concentration at one limit and second order at another, without changing its net chemical formula.

The phrase termolecular reaction is used in different contexts. Molecularity strictly counts reactant entities in an elementary microscopic event. A genuinely simultaneous collision of three free particles is less probable than a two-particle collision under ordinary dilute conditions, so many observed third-order laws arise from sequential association and stabilisation. The rate-law order alone cannot prove a single three-body elementary collision. This is another example of why measured order and elementary molecularity should be distinguished.

The identity of M matters. At equal concentrations, helium, argon and a polyatomic bath gas can differ in how effectively each removes internal energy from AB . A gas mixture therefore may be described using an effective bath-gas concentration that weights partners by collision efficiency, but those weights can depend on temperature and reaction channel. Pressure dependence should be measured with composition controlled; changing total pressure by adding a different gas is not necessarily equivalent to compressing an existing mixture. A primary experimental and theoretical study of methyl-radical recombination measured fall-off and bath-gas effects across temperature and pressure.

Some recombination rate constants decrease as temperature rises over a stated range. With the local Arrhenius definition Ea = −R d ln k/d(1/T), that behaviour corresponds to a negative apparent Ea. It does not mean a molecule climbs a negative physical barrier. The observed coefficient combines capture, energised-complex lifetimes, redissociation and collisional stabilisation. Higher temperature can make the complex more likely to redissociate before stabilisation, or change energy-transfer efficiency. A negative apparent Ea is therefore a property of a fitted composite rate law over certain conditions, not a universal statement about the potential-energy surface. The methyl-radical recombination study reports negative temperature dependence in its measured range.

Pressure-dependent chemistry appears in combustion, atmospheric radical networks and plasma processing. In each, a model that uses one fixed rate constant across all pressures can mispredict products. A more complete approach computes energy-specific AB formation and dissociation, plus collision-driven energy transfer, often with RRKM and a master equation. The simple limits remain essential checks on a sophisticated numerical model: the low-pressure result should scale with [M] under suitable conditions, and the high-pressure result should approach a finite association limit.

Step-by-step reasoning

1. Write a plausible energised-complex scheme and identify how M transfers energy. 2. Determine whether stabilisation collisions are scarce or abundant compared with redissociation. 3. At low pressure, use r = k₃[A][B][M] as a limiting expression when supported by the mechanism. 4. At high pressure, use r = k₂,∞[A][B] if stabilisation has reached saturation. 5. Check units and state which concentrations or partial pressures are held fixed in comparisons. 6. Interpret unusual temperature slopes in terms of the complete mechanism and measurement range.

Visual explanation

Draw A and B meeting to form a high-energy AB above the stable AB energy level. One arrow from AB returns to A+B; another arrow marked “collision with M” descends to AB + M. Beneath, plot effective second-order k₂,eff against [M]. The line rises approximately linearly near zero and bends toward a high-pressure plateau. Add a small note beside the plot: “Changing M identity may shift the curve.”

Real-world analogy

Two people can briefly link hands on an unstable moving platform. A third person can steady them before they separate. More helpers initially make successful pairing more likely, but once help arrives almost every time, adding helpers gives little benefit. This illustrates stabilisation probability, not a literal human-scale mechanism for energy transfer or quantum-state populations.

Real-world example

In a radical recombination experiment, two methyl radicals form an energised ethane molecule. At low pressure, some energised molecules redissociate before collisions remove enough energy. By varying helium or argon bath-gas pressure, researchers measure how the effective association rate approaches a high-pressure limit. Temperature scans can reveal negative apparent activation energies in a range where stabilisation becomes less favorable as temperature rises. Product analysis and pressure control help distinguish recombination from competing radical reactions.

Why?

Why does M appear in the low-pressure rate law even though it does not appear in the net product formula? At low pressure, forming AB is not sufficient for lasting product. The probability that AB meets a collision partner before falling apart grows roughly with [M] when such collisions are rare. M controls the yield of stable AB but is regenerated in the stabilising collision, so it affects kinetics without being consumed stoichiometrically.

Common misconception

“Third-order rate law means three particles hit simultaneously.” Sequential formation of AB and later stabilisation can produce the same low-pressure concentration dependence. Another misconception is that a negative fitted Ea means a negative energy barrier. Apparent slopes of composite pressure-dependent rate constants include competing processes and can have either sign. Always give temperature and pressure conditions when comparing these coefficients.

Worked example

At a stated low-pressure condition, let k₃ = 2.0 L² mol⁻² s⁻¹, [A] = 0.010 mol L⁻¹, [B] = 0.020 mol L⁻¹ and [M] = 0.10 mol L⁻¹. Then r = k₃[A][B][M] = 2.0 × 0.010 × 0.020 × 0.10 = 4.0 × 10⁻⁵ mol L⁻¹ s⁻¹. If only [M] doubles while the system remains in the low-pressure limit, the model predicts 8.0 × 10⁻⁵ mol L⁻¹ s⁻¹. It does not predict indefinite doubling: at higher [M] the system eventually bends toward its high-pressure limit.

Quick check

1. What units must k₃ have if r = k₃[A][B][M] with molar concentrations? Answer: L² mol⁻² s⁻¹, so multiplication by three mol L⁻¹ factors gives mol L⁻¹ s⁻¹.

Exam focus

Distinguish net association A+B→AB from the mechanism A+B⇌AB followed by AB +M→AB+M. Give both limiting rate laws and the units of their coefficients. Explain why low-pressure order in M disappears at the high-pressure limit. State that bath-gas identity affects collisional energy transfer and that a negative apparent Ea is a fitted temperature-slope parameter, not a physical barrier height.

Advanced insight

In energy-resolved master-equation calculations, AB occupies many internal-energy grains. Collisions move it among grains, and each grain can dissociate with an energy-specific RRKM rate. The apparent rate coefficient can depend on total angular momentum, multiple product channels and the time window of observation. A simple one-parameter fall-off curve may fit one dataset while failing at another temperature or with a different bath gas. Such failures are useful evidence about energy-transfer models and potential-energy surfaces.

Summary

Third-body recombination stabilises an energised association complex by transferring internal energy to a bath gas. At low pressure, r often scales as [A][B][M]; at high pressure, it approaches a form proportional to [A][B] alone. Observed order can change across fall-off without a new net reaction. Negative apparent activation energies can result from the combined temperature response of capture, redissociation and stabilisation, so they must be interpreted within a mechanism and stated conditions.

Practice questions

1. If low-pressure k₃ is fixed and [M] triples while [A] and [B] stay fixed, what happens to r in that limit? Answer: It triples because r = k₃[A][B][M].

2. What is the high-pressure limiting rate law for the association discussed here? Answer: r = k₂,∞[A][B], where further increases in [M] have little effect under the limiting model.

3. Why does a third body not need to appear in the net chemical equation? Answer: It exchanges energy in a collision and emerges chemically unchanged, so it is not consumed overall.

4. Give one reason a rate coefficient for recombination might decrease when temperature rises. Answer: The energised complex may redissociate more readily before it can be collisionally stabilised, lowering the net association yield.

5. Does a negative apparent Ea require a negative activation free energy for every elementary step? Answer: No. It can arise from temperature-dependent competition among capture, redissociation and stabilisation in a composite rate.