Home / AS & A Level Chemistry 25.2 Partition coefficients: Exam Style Questions Paper 4

AS & A Level Chemistry 25.2 Partition coefficients: Exam Style Questions Paper 4

A Level Chemistry Paper 4 - All Chapters
Question

In aqueous solution, persulfate ions, \(S_2O_8^{2–}\), react with iodide ions, as shown in reaction 1.

reaction 1   \(S_2O_8^{2–} + 2I^– \to 2SO_4^{2–} + I_2\)
The rate of reaction 1 is investigated. A sample of \(S_2O_8^{2–}\) is mixed with a large excess of iodide ions of known concentration. The graph in Fig. 5.1 shows the results obtained.

Concentration vs. time graph

(i) Use Fig. 5.1 to determine the initial rate of reaction 1. Show your working.

(ii) The rate equation for reaction 1 is rate = \(k [S_2O_8^{2–}] [I^–]\). Suggest why a large excess of iodide ions allows the rate constant to be determined from the half-life in this investigation.

(b) The reaction of persulfate ions, \(S_2O_8^{2–}\), with iodide ions is catalysed by \(Fe^{2+}\) ions. Write two equations to show how \(Fe^{2+}\) catalyses reaction 1.

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

(d) In aqueous solution, thiosulfate ions, \(S_2O_3^{2–}\), react with hydrogen ions, as shown in reaction 2.
reaction 2    \(S_2O_3^{2–} + 2H^+ \to SO_2 + S + H_2O\)
The rate of reaction is first order with respect to \([S_2O_3^{2–}]\) and zero order with respect to \([H^+]\) under certain conditions. The rate constant, k, for this reaction is \(1.58 \times 10^{–2} s^{–1}\). Calculate the half-life, \(t_{\frac{1}{2}}\), for reaction 2.

(e) The compound nitrosyl bromide, NOBr, can be formed as shown in reaction 3.
reaction 3    \(2NO(g) + Br_2(g) \to 2NOBr(g)\)
The rate is first order with respect to [NO] and first order with respect to \([Br_2]\). The reaction mechanism has two steps. Suggest equations for the two steps of this mechanism. State which is the rate-determining step.

▶️ Answer/Explanation
Solution

(a)(i) The initial rate is determined by drawing a tangent at \(t = 0\) on the graph and calculating its gradient. From Fig. 5.1, the gradient is approximately 0.028 mol dm⁻³ s⁻¹.

Explanation: The tangent’s slope at \(t = 0\) gives the initial rate, as rate = \(-\frac{d[S_2O_8^{2–}]}{dt}\). The calculated value falls within the range 0.016–0.040 mol dm⁻³ s⁻¹.

(a)(ii) With a large excess of \(I^–\), its concentration remains effectively constant, simplifying the rate equation to pseudo-first order: rate = \(k’ [S_2O_8^{2–}]\), where \(k’ = k[I^–]\).

Explanation: The half-life of a first-order reaction depends only on \(k’\), allowing \(k\) to be derived if \([I^–]\) is known.

(b)
Equation 1: \(2Fe^{2+} + S_2O_8^{2–} \to 2Fe^{3+} + 2SO_4^{2–}\)
Equation 2: \(2Fe^{3+} + 2I^– \to 2Fe^{2+} + I_2\)

Explanation: \(Fe^{2+}\) is oxidised by \(S_2O_8^{2–}\) in step 1 and reduced back by \(I^–\) in step 2, regenerating the catalyst.

(c) Increasing temperature raises the rate constant (\(k\)) and thus the reaction rate, as more molecules possess energy ≥ activation energy.

Explanation: The Arrhenius equation \(k = Ae^{-E_a/RT}\) shows \(k\) increases exponentially with temperature.

(d) For a first-order reaction, \(t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0158} \approx 43.9 \text{ s}\).

Explanation: The half-life is independent of concentration for first-order reactions, calculated directly from \(k\).

(e)
Step 1 (slow/RDS): \(NO + Br_2 \to NOBr_2\)
Step 2 (fast): \(NOBr_2 + NO \to 2NOBr\)
Rate-determining step: Step 1

Explanation: The rate law (first order in both NO and Br₂) matches the stoichiometry of the slow step, confirming it as the RDS.

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