The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q)(\text { slow }) \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) \quad(\text { fast }) \end{aligned} $$ (a) Write the chemical equation for the overall process. (b) Identify the intermediate, if any, in the mechanism. (c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.

Step by step solution

01

1. Writing the chemical equation for the overall process

To find the overall chemical equation, we can add the two given equations and then cancel out any species that appear on both sides of the equation. First, let's write down the two equations provided: (i) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q)\) (ii) \(\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q)\)

02

2. Identifying the intermediate

An intermediate is a species that is produced in one step of a mechanism and consumed in another step. In this case, \(\mathrm{IO}^{-}(a q)\) is produced in step (i) and consumed in step (ii), so it is the intermediate.

03

3. Predicting the rate law for the overall process

Since the first step of the mechanism is rate determining (slow), the rate law for the overall process will be determined by the rate law of the first step. In the first step, we have: \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q)\) The rate law for this elementary reaction is given by: Rate \(= k[\mathrm{H}_{2} \mathrm{O}_{2}][\mathrm{I}^{-}]\) where k is the rate constant, and the concentration of each reactant is raised to the power of its stoichiometric coefficient (which is 1 for both reactants in this case). So, the rate law for the overall process is given by: Rate \(= k[\mathrm{H}_{2} \mathrm{O}_{2}][\mathrm{I}^{-}]\).

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