Which of the following could not be rate determining step?

The correct answer is Option (2) → O₃ ⇌ O₂ + O

A complex reaction takes place in a series of elementary steps. The rate of the whole reaction is controlled by the slowest step, the rate determining step. A step written as a fast reversible equilibrium is not the slow step: it reaches equilibrium quickly before the slow step that follows it.

• Option (1) NO₂ + NO₂ → NO₃ + NO — Can be a rate determining step. It is the slow first step in the reaction of \( \ce{NO2} \) with CO; the fast step \( \ce{NO3 + CO -> NO2 + CO2} \) follows. Hence rate = k[NO₂]².

• Option (2) O₃ ⇌ O₂ + O — Cannot be a rate determining step. In the decomposition of ozone this step is a fast equilibrium; the slow step comes after it:

      \( \ce{O3 <=> O2 + O} \) (fast)
      \( \ce{O + O3 -> 2O2} \) (slow)

• Option (3) H₂O₂ + I⁻ → H₂O + IO⁻ — Can be a rate determining step. It is the slow step in the iodide-catalysed decomposition of \( \ce{H2O2} \); \( \ce{IO^- + H2O2 -> H2O + I^- + O2} \) is fast. Hence rate = k[H₂O₂][I⁻].

• Option (4) NO₂ + F₂ → NO₂F + F — Can be a rate determining step. It is the slow first step in \( \ce{2NO2 + F2 -> 2NO2F} \); the F atom then reacts quickly with a second \( \ce{NO2} \). Hence rate = k[NO₂][F₂].

NCERT (2026-27 edition):
Molecularity of a Reaction – “… the overall rate of the reaction is controlled by the slowest step in a reaction called the rate determining step. Consider the decomposition of hydrogen peroxide which is catalysed by iodide ion in an alkaline medium.
\( \ce{2H2O2} \xrightarrow[\text{Alkaline medium}]{\ce{I^-}} \ce{2H2O + O2} \)
The rate equation for this reaction is found to be
\( \text{Rate} = \frac{-d[\ce{H2O2}]}{dt} = k[\ce{H2O2}][\ce{I^-}] \)
This reaction is first order with respect to both H₂O₂ and I^–. Evidences suggest that this reaction takes place in two steps
(1) \( \ce{H2O2 + I^- -> H2O + IO^-} \)
(2) \( \ce{H2O2 + IO^- -> H2O + I^- + O2} \)
Both the steps are bimolecular elementary reactions.”