Distinguish between order and molecularity. Explain, using the decomposition of H₂O₂ catalysed by I⁻, how the slow step decides the rate law.

1. 1 · Define both in one line each. Order = sum of powers in the rate law (experimental). Molecularity = number of species colliding in an elementary step.
2. 2 · Give two differences.
   • Order can be 0 or fractional; molecularity only 1, 2 or 3.
   • Order applies to complex reactions too; molecularity only to elementary steps.
3. 3 · Write the two steps. H₂O₂ + I⁻ → H₂O + IO⁻ (slow); H₂O₂ + IO⁻ → H₂O + I⁻ + O₂ (fast).
4. 4 · Link to the rate law. The slow step involves one H₂O₂ and one I⁻, so rate = k[H₂O₂][I⁻]: second order.
5. 5 · Close in one line. So the rate-determining step fixes the order of a complex reaction.

Scoring tip: Mark the slow step clearly with the word "slow" and name IO⁻ as the intermediate.