Elementary Steps and the Rate-Determining Step

Unimolecular means one particle rearranges or breaks apart, giving rate = k[A]. Bimolecular means two particles collide, giving rate = k[A][B] or k[A]². Termolecular steps require three particles to meet simultaneously and are rare because such a triple collision is very improbable.

Molecularity counts reacting particles in one elementary forward event: A → products is unimolecular; 2 A → products and A + B → products are bimolecular. The elementary step's rate is k[A]² or k[A][B], respectively. It is never fractional or zero molecularity. Fractional and zero experimental orders can describe the combined mechanism, so order and molecularity are different ideas. For a reversible elementary step, write separate forward and reverse rate laws.

If one step is much slower than the others, it acts as a bottleneck. Under a valid bottleneck approximation, begin with the slow step’s elementary rate law and eliminate any intermediate before comparing with the observed law. When the slow step is first, the prediction is immediate because only reactants appear in it.

If the slow elementary step is 2 NO₂ → NO₃ + NO, its event rate is k₁[NO₂]². A subsequent rapid NO₃ + CO → NO₂ + CO₂ cancels NO₃ and one NO₂, giving net NO₂ + CO → NO + CO₂. The resulting prediction is second order in NO₂ and zero order in CO under this approximation, even though CO appears in the net equation. Net stoichiometry alone cannot produce that prediction.

If the slow step contains an intermediate, its elementary law is only a starting point. You must eliminate the intermediate through a justified earlier equilibrium or another kinetic approximation. Merely circling the slow step does not finish the derivation. On a multipeak energy diagram, compare a barrier from its preceding well, rather than declaring the highest absolute peak automatically rate controlling under every condition.

Two tests must both pass. First, the steps must sum to the overall balanced equation with intermediates cancelling. Second, the rate law predicted by the rate-determining step must match the experimental rate law. Passing both makes a mechanism plausible, never proven, because another mechanism might make the same predictions.

Test a mechanism in this order: balance each proposed step; add and cancel to recover the observed reaction; classify intermediates and catalysts; derive the rate law using the stated approximations; compare every predicted order with experiment. A scheme that sums correctly but predicts the wrong concentration dependence is inconsistent. A scheme passing both tests is a candidate, since additional observations such as an isotope effect or detection of an intermediate can distinguish mechanisms with identical rate laws.

Add proposed elementary steps and cancel species that appear on both sides to recover the overall reaction. An intermediate is formed then consumed; a catalyst is consumed then regenerated. Neither appears in the net equation, but their roles differ by the order of appearance. This bookkeeping checks a mechanism before attempting a rate-law derivation.

In A + C → AC followed by AC + B → AB + C, C is supplied and regenerated, whereas AC is formed and later consumed. Both disappear from the net A + B → AB equation, but C may still affect rate because it participates in the pathway. A catalyst is not required to appear unchanged in every step; regeneration across the full catalytic cycle is what matters. A transition state is distinct from AC: it is a barrier arrangement rather than a chemical species residing in a local energy minimum.