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The decomposition of H2O2 catalysed by I^- has rate = k[H2O2][I^-] and proceeds via a slow first step. What is the molecularity of the rate determining step?
AUnimolecular
BZero
CTrimolecular
DBimolecular
Answer & Solution
Correct answer: D. Bimolecular
1. The slow step is $H_2O_2 + I^- \rightarrow H_2O + IO^-$.
2. Two species collide simultaneously in this step.
3. So its molecularity is two, i.e. bimolecular.
4. This matches the observed overall second order rate law.
_Source: NCERT Class 12 Chemistry Ch 3 "Chemical Kinetics", p.9_
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