The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q) \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) \end{aligned} $$ (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.

To determine the overall reaction, add the two given reactions together, cancel out any species that appear on both sides of the equation, and combine the remaining species into a single equation. Given reactions: 1) \(\mathrm{H}_2\mathrm{O}_2(aq) + \mathrm{I}^-(aq) \longrightarrow \mathrm{H}_2\mathrm{O}(l) + \mathrm{IO}^-(aq)\) 2) \(\mathrm{IO}^-(aq) + \mathrm{H}_2\mathrm{O}_2(aq) \longrightarrow \mathrm{H}_2\mathrm{O}(l) + \mathrm{O}_2(g) + \mathrm{I}^-(aq)\) Adding both reactions: \(\mathrm{H}_2\mathrm{O}_2(aq) + \mathrm{I}^-(aq) + \mathrm{IO}^-(aq) + \mathrm{H}_2\mathrm{O}_2(aq) \longrightarrow \mathrm{H}_2\mathrm{O}(l)+\mathrm{IO}^-(aq) + \mathrm{H}_2\mathrm{O}(l) + \mathrm{O}_2(g) + \mathrm{I}^-(aq)\) Now, we can cancel out the intermediate species \(\mathrm{IO}^-\) and simplify the equation: \(\mathrm{H}_2\mathrm{O}_2(aq) + \mathrm{H}_2\mathrm{O}_2(aq) \longrightarrow 2\mathrm{H}_2\mathrm{O}(l) + \mathrm{O}_2(g)\) The overall reaction is: \(\mathrm{2H}_2\mathrm{O}_2(aq) \longrightarrow 2\mathrm{H}_2\mathrm{O}(l) + \mathrm{O}_2(g)\)

The intermediate species is a species that is produced in one step and consumed in another step. In this case, \(\mathrm{IO}^-(aq)\) is produced in the first step and consumed in the second step, so it is the intermediate species.

Assuming that the first step is rate-determining, we can write the rate law for the overall reaction based on the first step: 1) \(\mathrm{H}_2\mathrm{O}_2(aq) + \mathrm{I}^-(aq) \longrightarrow \mathrm{H}_2\mathrm{O}(l) + \mathrm{IO}^-(aq)\) The rate law based on this step is: \(Rate = k[\mathrm{H}_2\mathrm{O}_2][\mathrm{I}^-]\) where \(k\) is the rate constant, and \([\mathrm{H}_2\mathrm{O}_2]\) and \([\mathrm{I}^-]\) are the concentrations of \(\mathrm{H}_2\mathrm{O}_2\) and \(\mathrm{I}^-\), respectively.