Chemistry 9701 · AS & A Level · Reaction kinetics

Reaction kinetics — practice question

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

Complete the table below: - 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]

Use the experiment 1 data to calculate the rate constant, $k$. State the units of $k$. $k =$ ................. , units $=$ .................

(b(ii))[1]

Calculate the value of $[\text{NO}]$ in experiment 2. $[\text{NO}] =$ ................. $\text{mol dm}^{-3}$

(c)[1]

Define rate-determining step.

(d(i))[2]

Suggest equations for a two-step mechanism for this reaction, stating which of the two steps is the rate-determining step. Peroxodisulfate ions, $\text{S}_2\text{O}_8^{2-}$, react with iodide ions, $\text{I}^-.$ $\text{S}_2\text{O}_8^{2-} + 2\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_2$ The rate equation for the reaction in the absence of any catalyst is shown. $\text{rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]$ step 1 ........................................ step 2 ........................................ rate-determining step = ................................

(d(ii))[1]

An excess of peroxodisulfate ions is mixed with iodide ions. Right after 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. iodide ion concentration = ................. $\text{mol dm}^{-3}$

Worked solution & mark scheme

This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: with respect to $[NO]$, the order is 2

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