Chain Reaction Kinetics: The Hydrogen-Bromine Rate Law Derived

Introduction to Non-Elementary Chain Kinetics

In elementary reactions, the order of reaction matches its molecularity because the stoichiometric equation reflects the single molecular event occurring at the microscopic level. However, a vast majority of gas-phase chemical transformations are complex or non-elementary. For such systems, the experimentally observed rate expression often deviates drastically from stoichiometric coefficients. Students who have mastered fundamental rate laws and reaction orders quickly encounter reactions whose rate laws contain non-integer exponents or concentration terms in the denominator.

The thermal synthesis of hydrogen bromide from molecular hydrogen and molecular bromine is the classic textbook prototype of a non-elementary chain reaction (श्रृंखला अभिक्रिया):

H₂(g) + Br₂(g) → 2 HBr(g)

When this reaction was investigated experimentally under thermal conditions between 200 °C and 300 °C, measurements revealed an empirical rate law that puzzled chemists for over a decade:

d[HBr]/dt = (k [H₂] [Br₂]^1/2) / (1 + m [HBr] / [Br₂])

Here, k and m are empirical constants. This experimental equation exhibits two peculiar characteristics:

• Fractional Order: The rate depends on the square root of the bromine concentration ([Br₂]^1/2), giving an apparent order of 0.5 with respect to bromine at early stages.
• Product Retardation: The product concentration [HBr] appears in the denominator, meaning that as hydrogen bromide accumulates, it actively slows down the overall reaction velocity.

Explaining this empirical formula requires postulating a closed sequence of elementary free-radical steps and solving for the concentrations of transient intermediates using the steady-state approximation. For students preparing for higher-tier competitive examinations such as IIT-JAM, CSIR-NET, GATE, and CUET-PG, deriving this rate law is an indispensable exercise that tests mechanistic reasoning, algebraic precision, and radical kinetics. At ABC Chemistry, founder Mr. Aniket Bhardwaj (20+ years teaching chemistry) emphasizes mastering this derivation from first principles in our classroom-centre and online classes (call 9212142427 for academic guidance).

The Five Elementary Steps of the Hydrogen-Bromine Mechanism

A chain reaction proceeds through reactive intermediates (chain carriers or श्रृंखला वाहक) that are continually generated, consumed, and regenerated. In the hydrogen-bromine reaction, the chain carriers are atomic bromine radicals (Br•) and atomic hydrogen radicals (H•). The complete mechanism comprises five elementary steps: initiation, two propagation steps, inhibition (retardation), and termination.

1. Chain Initiation (श्रृंखला प्रारंभन)

Br₂ + M → 2 Br• + M   (Rate constant: k₁)

Thermal energy dissociates molecular bromine into two bromine free radicals upon collision with an inert third body M (which can be any gas molecule present, such as H₂, Br₂, or an inert gas). Why does Br₂ dissociate rather than H₂? The bond dissociation enthalpy of the Br–Br bond is approximately 193 kJ/mol, whereas that of the H–H bond is roughly 436 kJ/mol. Under moderate thermal conditions, only bromine molecules possess a significant Boltzmann fraction of collisions energetic enough to undergo homolytic cleavage. The forward rate of initiation can be expressed as:

r_initiation = k₁ [Br₂] [M]

Because the third body M participates symmetrically in both initiation and termination, its concentration cancels out in the final rate expression, allowing us to absorb [M] into the effective initiation rate constant k₁ for clarity: r_init = k₁ [Br₂].

2. First Chain Propagation Step (श्रृंखला संचरण)

Br• + H₂ → HBr + H•   (Rate constant: k₂)

An atomic bromine radical attacks a molecular hydrogen molecule to form the first molecule of hydrogen bromide and generates a highly reactive atomic hydrogen radical. Thermochemically, this step is endothermic by approximately 70 kJ/mol, carrying a substantial activation energy of around 75 kJ/mol. Because of this high activation barrier, step 2 is the slowest step in the forward chain cycle and represents the primary kinetic bottleneck of the reaction.

3. Second Chain Propagation Step

H• + Br₂ → HBr + Br•   (Rate constant: k₃)

The atomic hydrogen radical produced in step 2 rapidly attacks a bromine molecule. This reaction generates a second molecule of HBr and regenerates the bromine radical Br•, which can re-enter step 2. Thermochemically, this step is strongly exothermic (around −170 kJ/mol) with an extremely low activation energy of roughly 4 kJ/mol. Consequently, step 3 proceeds at a very high velocity.

4. Chain Inhibition or Retardation (श्रृंखला मंदन / संदमन)

H• + HBr → H₂ + Br•   (Rate constant: k₄)

As the reaction progresses, product molecules of HBr accumulate in the reaction vessel. A reactive hydrogen atom (H•) now has two competing reaction pathways: it can collide with a reactant molecule Br₂ (step 3, leading to forward chain propagation), or it can collide with a product molecule HBr (step 4). Step 4 is the exact microscopic reverse of step 2. Because this step consumes the product HBr and destroys the reactive H• intermediate while regenerating H₂, it acts as a chain-retarding or inhibiting step. This competition between Br₂ and HBr for the intermediate H• radical is the microscopic origin of the [HBr] term in the denominator of the rate law.

5. Chain Termination (श्रृंखला समापन)

2 Br• + M → Br₂ + M   (Rate constant: k₅)

Two bromine radicals recombine via a three-body collision to reform molecular bromine, terminating the radical chain. A common conceptual question among postgraduate aspirants is: why does termination occur via recombination of two Br• radicals, rather than two H• radicals (2 H• → H₂) or cross-recombination (H• + Br• → HBr)?

The answer lies in the relative concentrations of the two radical intermediates. Because step 2 (consumption of Br•) is endothermic and slow, while step 3 (consumption of H•) is exothermic and fast, the steady-state concentration of bromine atoms is several orders of magnitude higher than that of hydrogen atoms ([Br•] >> [H•]). Therefore, the collision probability between two Br• radicals vastly exceeds that of any collision involving the vanishingly scarce H• radicals. Bimolecular recombination of Br• is the sole significant termination path.

Steady-State Treatment: Deriving the Rate Law

To determine the overall rate of formation of product HBr, we apply the steady-state approximation (स्थिर अवस्था सन्निकटन). The SSA posits that because reactive radical intermediates (H• and Br•) are consumed almost as rapidly as they are produced, their net rates of change over time are virtually zero throughout the main course of the reaction.

Step 1: Write the Rate of Formation of HBr

Hydrogen bromide is produced in step 2 and step 3, and consumed in step 4. Thus, the net rate of formation of HBr is:

d[HBr]/dt = k₂ [Br•] [H₂] + k₃ [H•] [Br₂] − k₄ [H•] [HBr]   (Equation 1)

Step 2: Apply the Steady-State Approximation to [H•]

The hydrogen radical is formed in step 2 and consumed in steps 3 and 4:

d[H•]/dt = k₂ [Br•] [H₂] − k₃ [H•] [Br₂] − k₄ [H•] [HBr] = 0   (Equation 2)

Rearranging Equation 2 to group terms containing [H•]:

k₂ [Br•] [H₂] = [H•] (k₃ [Br₂] + k₄ [HBr])

Solving for the steady-state concentration of [H•]:

[H•] = (k₂ [Br•] [H₂]) / (k₃ [Br₂] + k₄ [HBr])   (Equation 3)

Step 3: Simplify the Rate Expression for d[HBr]/dt

Notice a convenient algebraic relationship from Equation 2:

k₃ [H•] [Br₂] − k₄ [H•] [HBr] = k₂ [Br•] [H₂] − 2 k₄ [H•] [HBr]

Alternatively, rewrite Equation 1 by grouping [H•]:

d[HBr]/dt = k₂ [Br•] [H₂] + [H•] (k₃ [Br₂] − k₄ [HBr])

Now substitute Equation 3 for [H•] into Equation 1:

d[HBr]/dt = k₂ [Br•] [H₂] + [ (k₂ [Br•] [H₂]) / (k₃ [Br₂] + k₄ [HBr]) ] × (k₃ [Br₂] − k₄ [HBr])

Factor out k₂ [Br•] [H₂]:

d[HBr]/dt = k₂ [Br•] [H₂] × [ 1 + (k₃ [Br₂] − k₄ [HBr]) / (k₃ [Br₂] + k₄ [HBr]) ]

Combine the terms over a common denominator:

d[HBr]/dt = k₂ [Br•] [H₂] × [ (k₃ [Br₂] + k₄ [HBr] + k₃ [Br₂] − k₄ [HBr]) / (k₃ [Br₂] + k₄ [HBr]) ]

The product inhibition terms (+k₄ [HBr] and −k₄ [HBr]) cancel out cleanly in the numerator:

d[HBr]/dt = (2 k₂ k₃ [Br•] [H₂] [Br₂]) / (k₃ [Br₂] + k₄ [HBr])

Divide both the numerator and denominator by k₃ [Br₂]:

d[HBr]/dt = (2 k₂ [Br•] [H₂]) / [ 1 + (k₄ / k₃) ([HBr] / [Br₂]) ]   (Equation 4)

Step 4: Apply the Steady-State Approximation to [Br•]

Now we express the unknown intermediate concentration [Br•] in terms of stable reactant concentrations. Bromine atoms are formed in step 1 (initiation), step 3 (propagation), and step 4 (inhibition), and consumed in step 2 (propagation) and step 5 (termination):

d[Br•]/dt = 2 k₁ [Br₂] − k₂ [Br•] [H₂] + k₃ [H•] [Br₂] + k₄ [H•] [HBr] − 2 k₅ [Br•]² = 0   (Equation 5)

Add Equation 2 (d[H•]/dt = 0) and Equation 5 (d[Br•]/dt = 0) together. Notice what happens to the propagation and inhibition terms:

(−k₂ [Br•] [H₂] + k₂ [Br•] [H₂]) + (k₃ [H•] [Br₂] − k₃ [H•] [Br₂]) + (k₄ [H•] [HBr] − k₄ [H•] [HBr]) = 0

All propagation and retardation steps cancel out entirely! This leaves only the initiation and termination terms:

2 k₁ [Br₂] − 2 k₅ [Br•]² = 0

2 k₁ [Br₂] = 2 k₅ [Br•]²

[Br•]² = (k₁ / k₅) [Br₂]

[Br•] = (k₁ / k₅)^1/2 [Br₂]^1/2   (Equation 6)

Step 5: Substitute [Br•] into the Rate Expression

Substitute Equation 6 into Equation 4:

d[HBr]/dt = [ 2 k₂ (k₁ / k₅)^1/2 [H₂] [Br₂]^1/2 ] / [ 1 + (k₄ / k₃) ([HBr] / [Br₂]) ]

This theoretical result matches the empirical rate law discovered by Bodenstein:

d[HBr]/dt = (k [H₂] [Br₂]^1/2) / (1 + m [HBr] / [Br₂])

where the empirical parameters are identified as explicit combinations of elementary rate constants:

• Overall forward rate constant: k = 2 k₂ (k₁ / k₅)^1/2
• Inhibition constant: m = k₄ / k₃

Why the Rate Law is Fractional and Dynamic

The derived rate equation provides deep physical insight into how microscopic chain processes govern macroscopic reaction orders. Two major features demand careful evaluation:

1. Origin of the Fractional Order (1/2 with Respect to Br₂)

Why does [Br₂] appear to the power of 1/2? The fractional exponent arises directly from the mismatch in radical molecularity between chain initiation and chain termination:

• Initiation produces two radicals per dissociated molecule: Br₂ → 2 Br•.
• Termination consumes two radicals bimolecularly: 2 Br• → Br₂.

Setting the initiation rate equal to the termination rate yields [Br•]² ∝ [Br₂], which dictates that the steady-state radical pool is proportional to [Br₂]^1/2. Because the rate-limiting propagation step requires a single Br• radical (Br• + H₂ → HBr + H•), the rate of product formation directly inherits this half-power dependence. Whenever a chain reaction features thermal quadratic termination of radicals originating from a diatomic precursor, a half-order dependence inevitably emerges.

2. Dynamic Variation of the Overall Reaction Order

Because the denominator contains a concentration ratio, the apparent reaction order is not constant; it evolves as the reaction progresses through different concentration regimes:

• Case I: Initial Rate Regime ([HBr] ≈ 0): At the beginning of the reaction, virtually no product has formed. Therefore, [HBr] ≈ 0, which makes m [HBr] / [Br₂] << 1. The denominator approaches 1, and the rate law simplifies to:
  d[HBr]/dt ≈ k [H₂] [Br₂]^1/2
  In this initial regime, the reaction is first-order in H₂, half-order in Br₂, and has an overall reaction order of 1 + 0.5 = 1.5 (three-halves order).
• Case II: Late Stage or Added Product ([HBr] >> [Br₂]): When significant HBr accumulates, or if HBr is added to the initial mixture, the inhibition term dominates: m [HBr] / [Br₂] >> 1. The denominator approximates m [HBr] / [Br₂], giving:
  d[HBr]/dt ≈ [ k [H₂] [Br₂]^1/2 ] / [ m [HBr] / [Br₂] ] = (k / m) [H₂] [Br₂]^3/2 [HBr]⁻¹
  Here, the apparent order with respect to Br₂ rises to 1.5, the order with respect to H₂ remains 1, and the reaction exhibits a negative first-order (−1) with respect to product HBr. The overall order remains 1 + 1.5 − 1 = 1.5, but the reaction rate is severely suppressed.
• Case III: Excess Bromine ([Br₂] >> [HBr]): Maintaining a high partial pressure of bromine suppresses the retardation ratio ([HBr] / [Br₂]), driving the denominator back toward unity and preserving fast synthesis rates.

Comparison: Hydrogen-Bromine vs. Hydrogen-Chlorine Systems

Both the H₂–Br₂ reaction and the H₂–Cl₂ reaction are halogenation chain reactions, but their kinetic profiles, quantum yields, and empirical rate laws differ profoundly. Understanding these differences is a favourite testing ground in advanced chemical science examinations.

The dramatic contrast in quantum yield (Φ) and reaction rates between the two systems stems directly from the thermochemistry of the first propagation step:

• In the hydrogen-chlorine system, the attack of Cl• on H₂ has a modest activation energy (23 kJ/mol). Once a single chlorine atom is generated by absorbing a photon, it triggers a chain that cycles hundreds of thousands of times before a radical collides with the wall or an impurity. This generates a massive quantum yield (10⁴ to 10⁶) and can produce violent photochemical explosions. Furthermore, the reverse inhibition step (H• + HCl → H₂ + Cl•) is endothermic by 188 kJ/mol and carries an activation energy exceeding 200 kJ/mol, rendering product inhibition completely negligible under standard conditions.
• In the hydrogen-bromine system, the attack of Br• on H₂ is heavily endothermic (ΔH ≈ +70 kJ/mol) with an activation barrier of 75 kJ/mol. At room temperature, a bromine radical collides millions of times with H₂ molecules without possessing sufficient energy to react; instead, it recombines with another Br• radical. Even at elevated temperatures, the chain length remains very short, keeping the quantum yield around 0.01 to 2. Moreover, because the activation energy for H• attacking HBr is low (only ~5 kJ/mol), product inhibition sets in aggressively as HBr accumulates.

Strategic Preparation for CSIR-NET, GATE, and IIT-JAM

Chemical kinetics constitutes a foundational, regularly examined pillar of physical chemistry across Indian competitive examinations. To excel in questions related to non-elementary reactions and chain kinetics, aspirants should adopt a structured preparation methodology:

1. Derive from Scratch: Avoid memorizing the final Bodenstein rate law in isolation. Practice setting up the net differential equations for [H•] and [Br•] on blank paper. Examiners frequently alter the mechanism (for example, introducing a photochemical initiation step, a wall termination step, or omitting the inhibition step) and ask you to determine the revised rate law.
2. Master Cancellation Patterns: Notice how adding the steady-state conditions d[H•]/dt = 0 and d[Br•]/dt = 0 immediately eliminates all propagation and inhibition steps, leaving only initiation and termination. Recognizing this pattern saves critical minutes in timed exams.
3. Analyze Boundary Conditions: Be prepared to state the apparent reaction order in limiting conditions: initial rates ([HBr] → 0), late stages ([HBr] >> [Br₂]), or high halogen concentrations. Questions often ask for the apparent order with respect to Br₂ under specific limiting cases.
4. Examine Official Papers: Candidates should build their own personal topic tally by reviewing official previous year papers. Rather than relying on secondary hearsay or unverified claims, verify the steady presence of complex reaction kinetics directly in official brochures and past test papers.
5. Utilize Quality Resources: Supplement your textbook study with verified structured physical chemistry study material that includes step-by-step mathematical proofs and practice question sets.

For students seeking systematic classroom-centre or online mentoring, ABC Chemistry provides comprehensive courses for IIT-JAM, CSIR-NET, and GATE under the guidance of Mr. Aniket Bhardwaj. Inquiries regarding admissions, batches, and academic guidance can be directed to 9212142427.

Frequently Asked Questions

Why is the reaction order with respect to Br₂ equal to 1/2 rather than 1?

The fractional order of 1/2 arises because bromine enters the active radical pool through unimolecular dissociation (Br₂ → 2 Br•) but leaves the pool through bimolecular recombination of two radicals (2 Br• → Br₂). Setting the initiation rate equal to the termination rate yields [Br•]² ∝ [Br₂], meaning the concentration of chain-carrying bromine atoms is proportional to [Br₂]^1/2. Because the rate-determining propagation step involves one Br• radical, the overall rate inherits this square-root dependence.

Why does HBr appear in the denominator of the rate law?

The product HBr appears in the denominator because it acts as a kinetic inhibitor (retarder). As HBr accumulates, it competes with Br₂ for reactive atomic hydrogen radicals via the reverse reaction: H• + HBr → H₂ + Br•. This step consumes the product and diverts H• away from the forward propagation step (H• + Br₂ → HBr + Br•). Mathematically, solving the steady-state equation for [H•] places the sum (k₃ [Br₂] + k₄ [HBr]) in the denominator, which normalizes to 1 + m ([HBr] / [Br₂]).

Why do we neglect radical recombination between two H• atoms in the termination step?

Recombination between two hydrogen atoms (2 H• → H₂) is neglected because the steady-state concentration of H• is orders of magnitude lower than that of Br• ([Br•] >> [H•]). The first propagation step (Br• + H₂ → HBr + H•) is endothermic and slow, whereas the second step (H• + Br₂ → HBr + Br•) is exothermic and extremely fast. Consequently, hydrogen atoms are consumed virtually instantaneously upon formation, making collisions between two H• atoms statistically negligible compared to collisions between abundant Br• atoms.

What is the initial order of the reaction before any HBr has formed?

At the start of the reaction, [HBr] ≈ 0, which makes the inhibition term m [HBr] / [Br₂] negligible compared to 1. The rate law reduces to d[HBr]/dt = k [H₂] [Br₂]^1/2. In this initial regime, the reaction is first-order with respect to H₂ and half-order with respect to Br₂, giving an overall initial reaction order of 1.5 (or 3/2).

Why does the hydrogen-chlorine reaction have a much higher quantum yield than the hydrogen-bromine reaction?

The quantum yield of the H₂–Cl₂ reaction is enormous (10⁴ to 10⁶) because its first propagation step (Cl• + H₂ → HCl + H•) has a low activation energy (~23 kJ/mol) and is nearly thermoneutral, allowing chains to cycle thousands of times before termination. In contrast, the corresponding step in the H₂–Br₂ system (Br• + H₂ → HBr + H•) is heavily endothermic (ΔH ≈ +70 kJ/mol) with a high activation barrier (~75 kJ/mol). Most Br• radicals recombine before acquiring enough energy to react with H₂, keeping the chain length short and the quantum yield very low (~0.01 at room temperature).

How does adding an inert gas affect the thermal hydrogen-bromine reaction rate?

In a purely thermal gas-phase system, adding an inert gas increases the concentration of collision partners M. However, because M participates equally as a third body in both the initiation step (Br₂ + M → 2 Br• + M) and the bimolecular termination step (2 Br• + M → Br₂ + M), its concentration cancels out in the steady-state radical balance ([Br•] = (k₁/k₅)^1/2 [Br₂]^1/2). Thus, an inert gas does not alter the fundamental rate law, provided it does not alter thermal conductivity or reactor wall collisions significantly.