The Hydrogen-Bromine Chain Reaction

Introduction

The overall equation H₂ + Br₂ → 2HBr looks simple, yet its rate does not follow directly from that stoichiometry. A radical-chain mechanism produces fractional bromine dependence and inhibition by the HBr product. Deriving the rate law is a useful test of mechanistic reasoning: one must identify carriers, write both steady-state balances and count product formation and consumption correctly. The result shows why a measured rate law can reveal more about a reaction than its balanced equation.

Core explanation

Use this simplified thermal mechanism with rate constants k₁ through k₅. Initiation: Br₂ → 2Br·, rate k₁[Br₂]. Propagation step 1: Br· + H₂ → HBr + H·, rate k₂[Br·][H₂]. Propagation step 2: H· + Br₂ → HBr + Br·, rate k₃[H·][Br₂]. Inhibition: H· + HBr → H₂ + Br·, rate k₄[H·][HBr]. Termination: 2Br· → Br₂, event rate k₅[Br·]². The inhibition step consumes a product HBr and diverts H· away from the product-forming propagation step. A University of Michigan reaction-engineering treatment uses this classical chain example to connect observed rate behaviour with a proposed mechanism.

Begin with the hydrogen-radical balance. H· is produced only by propagation step 1 in this simplified scheme and lost by propagation step 2 and the inhibition step. Under steady state, k₂[Br·][H₂] ≈ k₃[H·][Br₂] + k₄[H·][HBr]. Solve for [H·]: [H·] ≈ k₂[Br·][H₂]/(k₃[Br₂] + k₄[HBr]). The denominator shows competition for H·. More Br₂ sends H· through productive propagation; more HBr sends it through the reverse-like inhibiting route.

The bromine-radical balance initially seems longer. Initiation makes two Br· per Br₂ event. Propagation step 1 consumes Br·; propagation step 2 and inhibition each regenerate Br·. Termination destroys two Br· per event. Thus d[Br·]/dt = 2k₁[Br₂] − k₂[Br·][H₂] + k₃[H·][Br₂] + k₄[H·][HBr] − 2k₅[Br·]². The middle three terms cancel because the H· steady-state balance says k₂[Br·][H₂] = k₃[H·][Br₂] + k₄[H·][HBr]. Consequently 0 ≈ 2k₁[Br₂] − 2k₅[Br·]² and [Br·] ≈ √[(k₁/k₅)[Br₂]]. The square root reflects the balance between first-order radical initiation and second-order radical termination.

Now count HBr carefully. Propagation step 1 forms one HBr; propagation step 2 forms one more; inhibition consumes one. Therefore d[HBr]/dt = k₂[Br·][H₂] + k₃[H·][Br₂] − k₄[H·][HBr]. Because k₂[Br·][H₂] equals the sum of the other two H· loss terms, this expression simplifies to d[HBr]/dt = 2k₃[H·][Br₂]. Substitution gives d[HBr]/dt = 2k₂k₃[Br·][H₂][Br₂]/(k₃[Br₂]+k₄[HBr]). Finally substitute the Br· expression: d[HBr]/dt = 2k₂√(k₁/k₅)[H₂][Br₂]^(1/2)/{1+(k₄/k₃)[HBr]/[Br₂]}. The same law can be written with [Br₂]^(3/2) in the numerator and [Br₂]+(k₄/k₃)[HBr] in the denominator.

At the start of a run, [HBr] may be small. Then the denominator is near one and HBr production is approximately proportional to [H₂][Br₂]^(1/2). As product accumulates, HBr inhibits its own formation through the H·+HBr step. This is not merely a generic “product slows reaction” rule; it follows from a specified elementary competition and appears in the rate-law denominator. A measured fractional order does not require a half molecule of bromine colliding with hydrogen. It arises mathematically from eliminating a radical whose concentration is set by initiation and termination.

This derivation is conditional. It assumes the listed steps dominate, radical concentrations are low and approximately steady, initiation and termination rates follow the stated forms, and concentrations are meaningful in a well-mixed gas regime. Photochemical initiation can replace the simple thermal k₁[Br₂] expression and change the bromine dependence. Wall termination or other radical losses can also alter it. Agreement between the derived rate law and data supports the mechanism but does not by itself prove every elementary step uniquely; independent radical, light-intensity and pressure measurements strengthen the case.

Step-by-step reasoning

1. Write every elementary step and its mass-action rate, including the number of radicals created or destroyed per event. 2. Set the H· formation and loss rates approximately equal and solve for [H·]. 3. Write the full Br· balance, then use the H· steady-state relation to cancel the propagation and inhibition terms. 4. Solve the remaining initiation–termination balance for [Br·]. 5. Form the net HBr production rate, subtracting the inhibition step that consumes HBr. 6. Substitute both radical concentrations and examine the early-time and product-rich limits.

Visual explanation

Draw a loop with Br· at the left, H· at the right and two arrows between them. The upper arrow, Br·+H₂→H·+HBr, makes the first HBr. The lower arrow, H·+Br₂→Br·+HBr, makes the second. Add a competing arrow from H· through HBr back to Br·, marked “inhibition: consumes HBr.” Above the loop show Br₂ splitting into two Br·; below it show two Br· combining. Beside the loop, box [Br·] ∝ √[Br₂] and the HBr-containing denominator of the final rate law.

Real-world analogy

A relay team passes a baton between two runners; each successful handoff can complete part of a repeated circuit. A side route returns the baton but cancels some progress, while a separate event removes batons entirely. The analogy helps distinguish propagation, inhibition and termination, but radical reactions are not coordinated choices and their rates follow molecular encounters.

Real-world example

In a carefully controlled gas experiment, an investigator measures initial HBr formation at several H₂ and Br₂ concentrations, then repeats the measurement with added HBr. A square-root-like initial dependence on Br₂ and a reduced rate with added product are consistent with the chain model. Changing the light exposure can test whether initiation is thermal or photochemical. The observed behaviour must be compared with actual temperature, reactor walls and potential side reactions before a complete mechanism is assigned.

Why?

Why does product HBr enter a denominator rather than only the numerator as a product of the overall reaction? The elementary H·+HBr→H₂+Br· step competes with H·+Br₂→HBr+Br·. More HBr draws H· into an unproductive route that consumes a product molecule. Solving the H· steady-state balance puts both competing loss rates in a denominator, so the rate of productive H· consumption falls as [HBr] rises.

Common misconception

“The rate must be k[H₂][Br₂] because the balanced equation uses one of each.” Overall coefficients do not determine a composite mechanism's rate law. Another error is counting only positive HBr-producing steps and forgetting H·+HBr consumes HBr. A third is dropping the factor of two in initiation or termination radical balances, which can corrupt intermediate concentrations even if some factors later cancel.

Worked example

Use the dimensionless inhibition ratio q = (k₄/k₃)[HBr]/[Br₂]. Suppose fixed [H₂] and [Br₂] give an uninhibited initial production rate of 0.020 mol L⁻¹ s⁻¹ when [HBr] is negligible. Later, if q = 1.0 while the reactant concentrations are momentarily treated as the same for comparison, the model gives rate 0.020/(1+1.0) = 0.010 mol L⁻¹ s⁻¹. When q = 3.0, the rate is 0.020/4 = 0.0050 mol L⁻¹ s⁻¹. This isolates the inhibition effect; in a real closed reactor, [H₂] and [Br₂] also decline as reaction proceeds.

Quick check

1. Why does [Br·] vary as √[Br₂] in the stated thermal model? Answer: Initiation creates Br· at a rate proportional to [Br₂], while bimolecular termination destroys radicals at a rate proportional to [Br·]²; steady balance gives the square root.

Exam focus

Show both radical balances and retain stoichiometric factors for two radicals formed or lost. Derive the net HBr rate by adding both productive propagation contributions and subtracting HBr consumption. State the final denominator clearly, then discuss the early-time [HBr]≈0 limit. Explain that fractional order is a consequence of an intermediate steady state, not a fractional elementary collision.

Advanced insight

If initiation is photochemical and its rate depends on absorbed photon flux rather than simply k₁[Br₂], the radical steady state can produce a different dependence on bromine concentration. Similarly, walls may terminate Br· with an approximately first-order loss instead of the bimolecular k₅[Br·]² term, changing the square-root law. Experimental changes in light intensity, vessel geometry and added radical scavengers can distinguish these alternatives. The apparent rate law is therefore a prediction of a complete mechanism plus conditions, not an immutable property of the overall H₂+Br₂ equation.

Summary

The hydrogen–bromine reaction is a radical chain with Br· and H· carriers. Steady state gives [Br·] proportional to √[Br₂], and competition between H·+Br₂ and H·+HBr places product HBr in a rate-law denominator. Correct net HBr production counts two productive propagation steps and subtracts the inhibiting consumption step. The derived fractional order and product inhibition illustrate how a composite mechanism determines observed kinetics.

Practice questions

1. Write the steady-state expression for [H·] in the stated mechanism. Answer: [H·] = k₂[Br·][H₂]/(k₃[Br₂]+k₄[HBr]).

2. At negligible [HBr], what are the apparent orders in H₂ and Br₂ under this model? Answer: First order in H₂ and one-half order in Br₂ for the HBr production rate.

3. If (k₄/k₃)[HBr]/[Br₂] = 2, what fraction of the corresponding uninhibited rate remains at fixed reactant concentrations? Answer: One-third, because the denominator is 1+2 = 3.

4. Why must the net HBr rate subtract the inhibition-step rate? Answer: The inhibition step uses one HBr molecule as a reactant, so it removes product while redirecting the H radical.

5. Which step directly causes HBr inhibition? Answer: H· + HBr → H₂ + Br·, because it consumes HBr and diverts H· from productive reaction with Br₂.