Chemistry 9701 · AS & A Level · Reaction kinetics

Reaction kinetics — practice question

In aqueous solution, persulfate ions, $\text{S}_2\text{O}_8^{2-}$, undergo reaction with iodide ions, as illustrated by reaction 1. reaction 1: $\text{S}_2\text{O}_8^{2-} + 2\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_2$ The speed of reaction 1 is being studied. A sample of $\text{S}_2\text{O}_8^{2-}$ is combined with a large excess of iodide ions at a known concentration. The graph in Fig. 5.1 presents the results obtained.
(a(i))[1]

From Fig. 5.1, find the initial rate of reaction 1. Show your working.

(a(ii))[1]

The rate equation for reaction 1 is rate $= k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]$. Suggest why a large excess of iodide ions makes it possible to determine the rate constant from the half-life in this investigation.

(b)[2]

The reaction of persulfate ions, $\text{S}_2\text{O}_8^{2-}$, with iodide ions is catalysed by $\text{Fe}^{2+}$ ions. Write two equations to show how $\text{Fe}^{2+}$ catalyses reaction 1.

(c)[1]

Describe the effect of an increase in temperature on the rate constant and the rate of reaction 1.

(d)[1]

In aqueous solution, thiosulfate ions, $\text{S}_2\text{O}_3^{2-}$, react with hydrogen ions, as shown in reaction 2. reaction 2: $\text{S}_2\text{O}_3^{2-} + 2\text{H}^+ \rightarrow \text{SO}_2 + \text{S} + \text{H}_2\text{O}$ Under certain conditions, the reaction is first order with respect to $[\text{S}_2\text{O}_3^{2-}]$ and zero order with respect to $[\text{H}^+]$. The rate constant, $k$, for this reaction is $1.58 \times 10^{-2}\,\text{s}^{-1}$. Calculate the half-life, $t_{\frac{1}{2}}$, for reaction 2.

(e)[2]

The compound nitrosyl bromide, NOBr, can be formed as shown in reaction 3. reaction 3: $2\text{NO}(g) + \text{Br}_2(g) \rightarrow 2\text{NOBr}(g)$ The rate is first order with respect to $[\text{NO}]$ and first order with respect to $[\text{Br}_2]$. The reaction mechanism consists of two steps. Suggest equations for the two steps in this mechanism. State which step is the rate-determining step.

Worked solution & mark scheme

This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: Tangent drawn at $t=0$ and gradient worked out at $t=0$ (between 0.016 and 0.040)

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