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.
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Solution and Explanation
For a multi-step reaction, the overall rate law is determined by the slowest (rate-determining) step only.
Step 2: Rate Law
The slow step is Step I: $H_2O_2 + I^- \rightarrow H_2O + IO^-$. The rate law is written from this step directly (since it is elementary): $$\text{Rate} = k[H_2O_2][I^-]$$
Step 3: Order Analysis
Order with respect to $H_2O_2$ = 1 (first order). Order with respect to $I^-$ = 1 (first order). Overall order $= 1 + 1 = 2$ (second order).
Step 4: Molecularity of Step II
In Step II: $H_2O_2 + IO^- \rightarrow H_2O + I^- + O_2$. Two species ($H_2O_2$ and $IO^-$) are involved in the collision. Molecularity =
2 (bimolecular).
Final Answer:
(a) Rate $= k[H_2O_2][I^-]$
(b) Order w.r.t. $H_2O_2$ = 1; Order w.r.t. $I^-$ = 1; Overall order = 2
(c) Molecularity of Step II = 2 (bimolecular)
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