Chain Reaction Kinetics

Introduction

An overall reaction can proceed through a small number of reactive intermediates that are regenerated repeatedly. One newly made radical may drive many product-forming cycles before it is lost. That amplification is the defining feature of a chain reaction. To understand its measured rate, separate the steps that start chains, keep them going and stop them. The chain length then connects a microscopic radical lifetime with macroscopic product formation.

Core explanation

A chain mechanism has three functional parts. Initiation generates a reactive carrier, often a radical, from a stable precursor. Propagation consumes a carrier in one step but forms a carrier in the same or a subsequent step, allowing the sequence to repeat. Termination removes carriers by combination, disproportionation, scavenging or a wall reaction. The carrier can change identity across a propagation cycle; it need not be the same radical at every step. The IUPAC chain-reaction definition emphasises continuous regeneration of reactive intermediates through repetitive propagation steps, while its termination definition concerns their destruction or inactivation.

For a simple symbolic example, let an initiator make R·. Propagation can be represented as R· + S → P + R·, meaning one net product P forms while the carrier is regenerated. This compressed one-step notation stands for a cycle, possibly containing multiple elementary reactions. Termination might be 2R· → inactive product. If the radical production rate is R i in mol L⁻¹ s⁻¹ and bimolecular termination events have rate k t[R·]², then radical loss is 2k t[R·]² because each event consumes two radicals. Under a steady-state approximation, R i ≈ 2k t[R·]² and [R·] ≈ √[R i/(2k t)]. The factor of two follows from how k t and R i were defined; different textbook conventions can absorb it into a termination coefficient, so definitions must be stated.

If the propagation cycle produces one P per event and its rate is R p = k p[S][R·], substitution gives R p ≈ k p[S]√[R i/(2k t)]. This predicts a square-root dependence on the initiation rate under these particular assumptions. Doubling the radical source does not necessarily double product formation, because a higher radical population also increases bimolecular termination. Other termination routes change the algebra: a first-order scavenger loss can give a different dependence, and branching chains can grow much faster than a simple steady state permits.

Chain length δ is the average number of propagation cycles per initiated chain. In a simple steady mechanism with one product per propagation cycle and a clearly defined initiation event, δ ≈ R p/R i after aligning the rate definitions. IUPAC defines chain length as the average number of repetitions of the closed propagation cycle and relates it to overall reaction rate divided by initiation rate. If each initiation event produces two carriers, or one cycle produces more than one counted product, define the denominator and numerator accordingly before calculating. A large δ means a small initiation flux can sustain substantial overall conversion; it does not mean the radical concentration itself is large.

Radical chain chemistry appears in halogenation, oxidation, combustion and polymerisation. In photochemical initiation, light can create radicals at a controllable rate. Oxygen or an added inhibitor may trap radicals and slow propagation, sometimes producing an induction period before rapid product formation. Surfaces and reactor dimensions can matter if walls terminate chains. These observations are mechanistic tests: changing light intensity, inhibitor concentration or vessel surface-to-volume ratio can alter a chain reaction in characteristic ways.

The simple radical balance is not a universal chain model. A realistic mechanism may contain several carriers and cycles, branching steps that multiply carriers, chain transfer that moves a reactive site, and temperature-dependent competition. Deriving a rate law then requires writing every carrier balance and applying either steady-state approximations or numerical integration. A primary ACS study of radical polymerisation kinetics uses the same initiation–termination balance to predict propagation rates in its idealised limit. The next pages examine a classic multi-carrier gas reaction and branching-chain behaviour.

Step-by-step reasoning

1. List reactive carriers and mark the steps that first create them. 2. Identify each closed propagation cycle and count product formed per cycle. 3. Identify all carrier-loss routes, including bimolecular termination and trapping by impurities or walls. 4. Write a differential balance for each carrier and apply steady state only when justified. 5. Derive the product-rate expression with all stoichiometric factors retained. 6. Define chain length consistently and test predicted dependencies by changing initiation or termination conditions.

Visual explanation

Draw an arrow from a stable initiator to a dot labeled R·. From R·, draw a circular arrow through S and P back to R·, marked “propagation.” Draw a separate arrow from two R· dots to a stable product, marked “termination.” Beside the diagram plot radical production as a horizontal line and bimolecular radical loss as an upward-curving function of [R·]. Their intersection illustrates the steady-state radical concentration; moving the initiation line upward shifts the intersection by a square-root relation.

Real-world analogy

A reusable tool enables many assembly steps after someone first supplies it, until the tool breaks or is removed. Initiation supplies the tool, propagation reuses it and termination removes it. The analogy captures repeated cycles, but radicals are chemical species consumed and regenerated in elementary reactions rather than durable physical tools.

Real-world example

In a light-initiated polymerisation, a photoinitiator makes radicals that add monomer repeatedly. The reactive chain end is regenerated after each monomer addition, so one initiation event can yield many repeat units. Termination by radical combination stops growth. Varying light intensity changes radical creation; the simple steady-state model predicts monomer-consumption rate may grow with the square root of initiation rate when bimolecular termination dominates. Real polymer systems may also show viscosity and diffusion effects.

Why?

Why can a reaction continue rapidly after only a small initiation rate? Propagation regenerates the carrier, so one carrier can pass through many cycles before termination. The overall product rate counts all cycles, whereas initiation counts only new chains. If δ is large, their ratio is large. This amplification explains why inhibitors that intercept a small radical population can strongly affect the macroscopic reaction.

Common misconception

“A chain reaction consumes one radical for every product molecule.” Propagation may regenerate a carrier, so the same chain can make many product molecules. Another error is omitting the factor of two in radical loss for 2R· termination while using k t[R·]² as the event rate. A third is assuming every chain reaction is explosively branching; ordinary nonbranching chains can reach a steady radical population and proceed controllably.

Worked example

Suppose radicals form at R i = 2.0 × 10⁻⁸ mol L⁻¹ s⁻¹ and terminate by 2R· → inactive product with k t = 1.0 × 10⁸ L mol⁻¹ s⁻¹. The stated convention gives [R·] = √[R i/(2k t)] = √[(2.0 × 10⁻⁸)/(2.0 × 10⁸)] = 1.0 × 10⁻⁸ mol L⁻¹. If k p = 1.0 × 10³ L mol⁻¹ s⁻¹ and [S] = 0.10 mol L⁻¹, R p = k p[S][R·] = 1.0 × 10⁻⁶ mol L⁻¹ s⁻¹. With one product per cycle and the stated initiation-rate convention, δ ≈ R p/R i = 50 propagation cycles per initiated carrier on average.

Quick check

1. If bimolecular termination events occur at k t[R·]², what is the radical disappearance rate from that step? Answer: 2k t[R·]², because each termination event destroys two radicals.

Exam focus

Classify steps by what they do to reactive carrier number, not by their position in a written list. Write radical balances with stoichiometric factors and state the steady-state approximation. Distinguish chain length from polymer degree of polymerisation unless a particular model connects them. Explain why square-root initiation dependence arises only with bimolecular termination and the other assumptions used here.

Advanced insight

In a network with two radical carriers, one propagation step may convert R· to Q· and another convert Q· back to R·. A single-radical balance would miss this cycle. An inhibitor can produce a time-dependent induction period before a steady chain rate emerges, so early-time data may not fit steady-state algebra. If branching creates more than one carrier per cycle, initiation and termination no longer necessarily balance at a small stable radical concentration; nonlinear feedback can cause ignition or explosion limits. Those cases require coupled balances and heat-transfer analysis.

Summary

Chain reactions repeatedly regenerate reactive carriers through propagation. Initiation creates carriers, and termination removes or inactivates them. In a simple steady model with bimolecular radical termination, carrier concentration scales as the square root of initiation rate, while product formation can greatly exceed initiation. Chain length counts average propagation cycles per initiated chain and must be defined using consistent rate conventions.

Practice questions

1. If R i rises by a factor of nine with k t and [S] fixed, how does R p change in the simple model? Answer: It rises by a factor of three because [R·] and hence R p scale as √R i.

2. What distinguishes a propagation step from a termination step? Answer: Propagation sustains or regenerates a reactive carrier; termination destroys or inactivates carriers and ends that chain.

3. If R p = 4.0 × 10⁻⁵ and R i = 2.0 × 10⁻⁷ mol L⁻¹ s⁻¹ with one product per cycle, estimate chain length. Answer: δ ≈ R p/R i = 200 cycles per initiated chain under the stated convention.

4. Name a condition that could invalidate the square-root radical result. Answer: Significant first-order radical trapping, multiple carrier types, branching or failure of steady state would change the balance.

5. Under the simple steady-state model, how does [R·] change if R i becomes four times larger? Answer: It doubles, because [R·] is proportional to the square root of R i.