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

i just learnt this recently but i need to look through my notes again

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error