Chain-Reaction Kinetics

Introduction

Some reactions proceed far faster than their slowest elementary step would suggest. The explanation is a chain mechanism : a single reactive intermediate, once formed, passes through a loop of steps that consumes reactant, makes product and regenerates itself. Each carrier can therefore turn over hundreds or thousands of reactant molecules before it is destroyed. Chain kinetics underlies combustion, atmospheric ozone chemistry, polymerisation and many industrial halogenations, and it is the clearest application of the steady-state approximation to a real reaction network.

Core explanation

A chain mechanism is organised into three classes of step.

Initiation produces carriers from stable molecules, often by homolysis driven by heat or light. For the hydrogen–bromine reaction, H₂ + Br₂ → 2HBr, initiation is Br₂ → 2Br•. Initiation is usually slow because it breaks a bond without forming one.

Propagation steps each consume one carrier and produce another, so the number of carriers is conserved. In the H₂/Br₂ system: Br• + H₂ → HBr + H• and H• + Br₂ → HBr + Br•. Adding these two steps gives the overall stoichiometry, H₂ + Br₂ → 2HBr, with Br• returned at the end of the cycle. Propagation steps are where the product is actually made.

Termination removes carriers, usually when two radicals combine: 2Br• → Br₂. Because termination is second order in carrier concentration, it becomes more important as carriers accumulate.

Some chains also contain inhibition or retardation steps, in which product destroys a carrier without ending the chain completely. In H₂/Br₂ the step H• + HBr → H₂ + Br• slows the reaction as HBr builds up.

The rate law comes from applying the steady-state approximation to every carrier. Setting d[H•]/dt = 0 and d[Br•]/dt = 0, and solving, gives the celebrated result found by Bodenstein and Lind:

rate = k[H₂][Br₂]^(1/2) ÷ (1 + k′[HBr]/[Br₂])

The half-order dependence on Br₂ is a fingerprint of initiation by homolysis balanced by second-order termination: at steady state, initiation rate = termination rate, so k i[Br₂] = k t[Br•]², giving [Br•] ∝ [Br₂]^(1/2). The denominator reflects inhibition by HBr. A rate law with fractional orders and a denominator is a strong sign of a chain.

The kinetic chain length is the rate of propagation divided by the rate of initiation. Long chains mean that a tiny initiation rate can support a large overall rate, which is why trace impurities that destroy carriers can dramatically slow a chain reaction.

Formulae

Steady state for carriers: rate of initiation = rate of termination (for a chain with no branching).

For initiation X₂ → 2X• (rate constant k i) and termination 2X• → X₂ (k t): [X•] = (k i[X₂]/k t)^(1/2).

Chain length ν = rate of propagation ÷ rate of initiation.

Step-by-step reasoning

To derive a chain rate law:

1. Write every elementary step and label it initiation, propagation, inhibition or termination. 2. Identify all carriers (the species that appear and disappear within the cycle). 3. Write a net rate expression for each carrier and set it to zero. 4. Add the carrier equations; propagation terms often cancel, leaving initiation = termination. 5. Solve for carrier concentrations and substitute into the rate of product formation.

Visual explanation

Draw the propagation steps as a closed loop: Br• enters at the top, reacts with H₂ to release HBr, becomes H•, reacts with Br₂ to release a second HBr and returns as Br•. Arrows entering the loop from outside show initiation; arrows leaving it show termination. The rate is set by how many carriers circulate and how fast each lap is.

Real-world analogy

A chain resembles a relay team that never tires. Recruiting a new runner (initiation) is slow, but each runner completes many laps (propagation), handing the baton on each time. The team only shrinks when two runners collide and both drop out (termination).

Real-world example

Free-radical chlorination of methane in light is a chain reaction: Cl• abstracts H from CH₄, and the methyl radical then abstracts Cl from Cl₂. Chain lengths can reach thousands, which is why small traces of oxygen, a radical scavenger, noticeably slow the process.

Why?

Why does the Br₂ order come out as one half rather than one? Carriers are made in pairs by a first-order process but destroyed by a second-order recombination. Balancing a first-order supply against a second-order loss makes the carrier concentration depend on the square root of [Br₂].

Common misconception

"The slowest step of a chain is rate-determining, so the rate equals the initiation rate." In fact the overall rate equals the initiation rate multiplied by the chain length; product is formed in the propagation steps, many times per initiation event.

Worked example

Question: In a chain reaction, carriers are initiated at 2.0 × 10⁻⁹ mol dm⁻³ s⁻¹ and product forms at 4.0 × 10⁻⁵ mol dm⁻³ s⁻¹. Each propagation cycle forms one product molecule. Estimate the chain length.

Reasoning: ν = rate of propagation ÷ rate of initiation = (4.0 × 10⁻⁵) ÷ (2.0 × 10⁻⁹) = 2.0 × 10⁴.

Answer: Each initiation event leads to about 20 000 propagation cycles.

Quick check

1. Which type of step in a chain mechanism actually produces most of the product molecules? Answer: The propagation steps, because each carrier cycles through them many times before termination.

Exam focus

Be ready to classify steps, apply the steady-state approximation to each carrier and show that initiation equals termination. Explain a half-order term as the signature of pairwise initiation and recombination termination.

Advanced insight

If a propagation step produces more carriers than it consumes, the chain branches , as in H• + O₂ → OH• + O•. Branching can make carrier numbers grow exponentially, leading to explosion limits that depend sharply on pressure and temperature. The steady-state treatment then fails, because no steady state exists above the branching threshold.

Summary

Chain reactions proceed through initiation, propagation and termination. Carriers are regenerated in propagation, so one initiation event yields many product molecules. Applying the steady-state approximation to each carrier gives rate laws with fractional orders and inhibition denominators, as in H₂ + Br₂. Chain length connects the slow initiation rate to the fast observed rate.

Practice questions

1. Classify Cl₂ → 2Cl•, Cl• + CH₄ → HCl + CH₃•, and 2CH₃• → C₂H₆. Answer: Initiation, propagation and termination respectively. 2. Why does HBr appear in the denominator of the H₂/Br₂ rate law? Answer: H• reacts with HBr to regenerate H₂, removing H• that would otherwise make product, so HBr inhibits the reaction. 3. Show that pairwise initiation with rate constant k i and termination 2X• → X₂ with k t gives [X•] proportional to [X₂]^(1/2). Answer: At steady state k i[X₂] = k t[X•]², so [X•] = (k i[X₂]/k t)^(1/2). 4. Explain why traces of a radical scavenger can greatly reduce the rate of a long chain reaction. Answer: Each scavenged carrier would have completed many propagation cycles, so removing a few carriers removes a large amount of product formation.