Mechanism and Rate Law Determination
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Direct answer
The rate law comes from experiment, never from the balanced equation: only for an elementary step do the stoichiometric coefficients enter the rate expression directly. In a multi-step mechanism, the slowest step — the rate-determining step — dictates the rate law, and if an intermediate appears in it, that intermediate must be eliminated using the equilibrium of a fast preceding step or the steady-state approximation. The classic cautionary example is H2 + Br2 → 2HBr, whose chain mechanism yields rate = k[H2][Br2]^1/2, a half-order that no balanced equation could predict. Conversely, H2 + I2 does follow rate = k[H2][I2] despite a two-step mechanism — matching stoichiometry never proves a reaction is elementary.
What you must remember
- Slow step writes the law: for NO2 + CO → NO + CO2 below 500 K, rate = k[NO2]^2 because the slow step is NO2 + NO2 → NO3 + NO, with NO3 + CO fast after.
- Eliminate intermediates: replace [intermediate] via the fast pre-equilibrium before it (K = k(forward)/k(backward)) or via steady state d[intermediate]/dt = 0.
- Famous complex laws: rate = k[H2][Br2]^1/2 for the HBr chain reaction; fractional and zero orders are mechanistic fingerprints, not anomalies.
- Molecularity versus order: molecularity (1, 2, rarely 3) applies to elementary steps only; overall order can be fractional, zero, or negative.
- Units of k by order: zero order mol L^-1 s^-1, first order s^-1, second order L mol^-1 s^-1 — a standalone JEE Main question type.
- Zero-order classics: ammonia decomposition on hot tungsten and photochemical H2 + Cl2 — surface or light saturation, not [reactant], controls the rate.
From data to law to mechanism
An initial-rates table does the heaviest lifting. Suppose doubling [A] quadruples the rate while doubling [B] leaves the rate untouched. The exponent of A is 2 (rate scales as its square), the exponent of B is 0, so rate = k[A]^2. Insert any single run — rate 4.0 × 10^-3 mol L^-1 s^-1 at [A] = 0.1 M — and k = 4.0 × 10^-3/(0.1)^2 = 0.4 L mol^-1 s^-1. The method is always divide-and-compare: one experiment over another, one concentration at a time.
Then test a proposed mechanism: if the slow step is 2A → product, the law rate = k[A]^2 is reproduced and the mechanism stands; if the slow step were A + B → product, the law would demand a [B] dependence the data deny. This two-way check — data to law, law to mechanism — is exactly how JEE Advanced frames its comprehension passages, and the half-order in [Br2] is the standard exhibit that intermediate I atoms were correctly eliminated through the pre-equilibrium Br2 ⇌ 2I.
How the exam frames it
Three question formats dominate. First, the direct derivation: given a mechanism with one slow step and one intermediate, write the rate law — the error to avoid is leaving the intermediate in your answer. Second, matching units of k to the order, where students forget that k's dimensions change with order. Third, the conceptual trap: from the equation 2N2O5 → 4NO2 + O2, students happily write rate = k[N2O5]^2 and call it stoichiometric common sense; the observed law is indeed first order in N2O5, but for mechanistic reasons (slow unimolecular decomposition of N2O5), not because the equation said so. Whenever an option claims "overall order equals the sum of stoichiometric coefficients", that option is bait.
Frequently asked questions
Why can't the balanced equation give the rate law?
The balanced equation only conserves atoms; it hides the sequence of elementary collisions, and only elementary steps have rate laws written directly from their coefficients.
How is an intermediate removed from a rate law?
Express its concentration from the fast equilibrium that precedes the slow step, or apply the steady-state approximation setting its formation and consumption rates equal.
What makes H2 + Br2 a textbook example?
Its chain mechanism gives rate = k[H2][Br2]^1/2, a fractional order impossible to guess from stoichiometry and a permanent warning against coefficient-based rate laws.
What are the units of a second-order rate constant?
L mol^-1 s^-1, from rate (mol L^-1 s^-1) divided by concentration squared (mol^2 L^-2).
Which reaction is zero order and why?
Ammonia decomposing on a hot tungsten surface — the catalyst surface saturates, so the rate no longer depends on NH3 concentration and stays constant with time.