Question

Consider the decomposition of hydrogen peroxide (\( \mathrm{H_2O_2} \)) as per the equation given below :

\[ 2\mathrm{H_2O_2} \xrightarrow[\text{Alkaline medium}]{\mathrm{I^-}} 2\mathrm{H_2O} + \mathrm{O_2} \]

The mechanism for this reaction was found to be :

Step I : \( \mathrm{H_2O_2 + I^-} \rightarrow \mathrm{H_2O + IO^-} \) (slow)

Step II : \( \mathrm{H_2O_2 + IO^-} \rightarrow \mathrm{H_2O + I^- + O_2} \) (fast)

Determine the order of reaction with respect to \( \mathrm{H_2O_2} \), \( \mathrm{I^-} \) and overall order of reaction.

Hint:

The slow step I fixes the rate law, rate = k[H2O2][I-]. Each reactant is first order and the overall order is 2.

Answer 1

Step 1: Find the rate-determining step:
In a multi-step mechanism, the slowest step controls the overall rate.
Here Step I is slow and Step II is fast. So Step I is the rate-determining step.

Step 2: Write the rate law from the slow step:
Step I is an elementary step, so its rate law follows from its molecularity. One molecule of \(\mathrm{H_2O_2}\) and one ion of \(\mathrm{I^-}\) collide.
\[ \text{Rate} = k[\mathrm{H_2O_2}][\mathrm{I^-}] \]
The intermediate \(\mathrm{IO^-}\) appears only in the fast step, so it does not enter the rate law.

Step 3: Read the orders from the rate law:
The power of \([\mathrm{H_2O_2}]\) is 1, so the order with respect to \(\mathrm{H_2O_2}\) is 1.
The power of \([\mathrm{I^-}]\) is 1, so the order with respect to \(\mathrm{I^-}\) is 1.
The overall order is the sum of the powers: \(1 + 1 = 2\).
Note that \(\mathrm{I^-}\) is used in Step I and made again in Step II. It is a catalyst, yet it still appears in the rate law.

Final Answer:
Order with respect to \(\mathrm{H_2O_2}\) is 1, with respect to \(\mathrm{I^-}\) is 1, and the overall order is 2.
\[ \boxed{\text{Rate} = k[\mathrm{H_2O_2}][\mathrm{I^-}],\ \text{overall order} = 2} \]