Question

For decomposition of $H_2O_2$ by $I^-$: Step I: $H_2O_2 + I^- \rightarrow H_2O + IO^-$ (slow). Step II: $H_2O_2 + IO^- \rightarrow H_2O + I^- + O_2$ (fast). (a) Write rate law. (b) Determine order w.r.t. $H_2O_2$ and $I^-$ and overall order. (c) Molecularity of Step II.

Hint:

Rate law comes from the slow (rate-determining) step only. Molecularity = number of species colliding in a single step.

Answer 1

Step 1: Rate-determining step.
Step I is slow; Step II is fast. The overall rate is determined by the slowest step (Step I).
Step 2: Rate law from slow step.
Step I is elementary: $H_2O_2 + I^- \rightarrow H_2O + IO^-$. Rate law: \[ \text{Rate} = k[H_2O_2][I^-] \]
Step 3: Order of reaction.
Order w.r.t. $H_2O_2 = 1$; order w.r.t. $I^- = 1$; overall order $= 1 + 1 = 2$ (second order).
Step 4: Molecularity of Step II.
Step II: $H_2O_2 + IO^- \rightarrow H_2O + I^- + O_2$. Two molecular species collide in this elementary step. Molecularity $= 2$ (bimolecular).