Chemistry 9701 · AS & A Level · Reaction kinetics

Reaction kinetics — practice question

Nitrogen monoxide, $\text{NO}$, combines with oxygen to make nitrogen dioxide, $\text{NO}_2$. The equation is $\text{2NO}(g) + \text{O}_2(g) \rightleftharpoons \text{2NO}_2(g)$. The rate equation for the forward reaction is given as: $\text{rate} = k[\text{NO}]^2[\text{O}_2]$.
(a)[1]

Complete the table below by giving: the order of reaction with respect to $[\text{NO}]$; the order of reaction with respect to $[\text{O}_2]$; the overall order of reaction.

(b(i))[2]

Using the data from experiment 1, Calculate the value of the rate constant, $k$. State the units of $k$.

(b(ii))[1]

Calculate the value of $[\text{NO}]$ for experiment 2.

(c)[1]

Define what is meant by the term rate-determining step.

(d(i))[2]

Peroxodisulfate ions, $\text{S}_2\text{O}_8^{2-}$, react with iodide ions, $\text{I}^-$. The equation is $\text{S}_2\text{O}_8^{2-} + 2\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_2$. The rate equation for the reaction without any catalyst is $\text{rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]$. Suggest equations for a two-step mechanism for this reaction, and state which step is the rate-determining step.

(d(ii))[1]

A large excess of peroxodisulfate ions is mixed with iodide ions. At the instant of mixing, $[\text{I}^-] = 0.00780\,\text{mol dm}^{-3}$. Under the conditions used, the half-life of $[\text{I}^-]$ is $48\,\text{s}$. Calculate the iodide ion concentration $192\,\text{s}$ after the peroxodisulfate and iodide ions are mixed.

Worked solution & mark scheme

This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: The order of reaction with respect to $[NO]$ is 2 and the order with respect to $[O_2]$ is 1, so the overall order is 3

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