Manganese(IV) oxide, $\text{MnO}_2$, speeds up the decomposition of hydrogen peroxide, $\text{H}_2\text{O}_2$, as shown: $\text{2H}_2\text{O}_2(aq) \rightarrow \text{2H}_2\text{O}(l) + \text{O}_2(g)$. The pathway includes formation of the intermediate species, $\text{Mn}^{2+}$, in the first stage, which is then consumed in the second stage.
(a)[3]
State and use the appropriate electrode potentials, $E^\ominus$, to form two equations that illustrate how $\text{MnO}_2$ catalyses this reaction.
(b)[1]
At these conditions, the reaction is first order with respect to hydrogen peroxide, and the rate constant, $k$, is $2.0 \times 10^{-6}\,\text{s}^{-1}$ at $298\,\text{K}$. Calculate the initial decomposition rate for a $0.75\,\text{mol dm}^{-3}$ hydrogen peroxide solution at $298\,\text{K}$.
(c(i))[1]
State the meaning of the term rate-determining step.
(c(ii))[1]
Use this mechanism to derive a balanced equation for this reaction.
(c(iii))[1]
Deduce the reaction order with respect to each of these: $\text{H}_2\text{O}_2$, $\text{I}^-$, $\text{H}^+$.
Worked solution & mark scheme
This 7-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Accurate electrode potentials and $E_{cell}$ values” …