2 H2O2(aq) > 2 H2O(l) + O2(g) in the presence of I-(aq) is proposed to be: Step 1 (slow): H2O2 + I- > H2O + OI- Step 2 (fast): H2O2 + OI- > H2O + O2 + I- What is the molecularity of the rate determining step?

Molecularity of the rate determining step = 2

Explanation:

Step 1 (slow): H₂O₂ + I⁻ > H₂O + OI⁻

Step 2 (fast): H₂O₂ + OI⁻ > H₂O + O₂ + I⁻

The rate determining step in a reaction mechanism is also considered as slowest step.

Slowest step is also considered its highest activation energy in energy profile diagram.

In this case intermediate  (IO⁻) is formed.

Step 1 considered as a slowest step.

So,  Rate = K [H₂O₂][I⁻]

  Molecularity = 2

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