Question

For the reaction:
2H$_2$O$_2$ $\xrightarrow{I^-}$ 2H$_2$O + O$_2$
Given mechanism:
(I) H$_2$O$_2$ + I$^-$ → H$_2$O + IO$^-$ (slow)
(II) H$_2$O$_2$ + IO$^-$ → H$_2$O + I$^-$ + O$_2$ (fast)
(1) Write rate law for the reaction.
(2) Write the overall order and molecularity of the reaction.

Hint:

Rate law is determined from the rate-determining (slowest) step only.

Answer 1

(1) The rate is determined by the slowest step: \[ \text{Rate} = k[\text{H}_2\text{O}_2][\text{I}^-] \] (2) The overall reaction order is 2, with first-order dependence on H$_2$O$_2$ and first-order dependence on I$^-$
The slow step has a molecularity of 2, involving both H$_2$O$_2$ and I$^-$ in a bimolecular process.