In aqueous solution, persulfate ions, S₂O₈²⁻, react with iodide ions, as shown in reaction 1. reaction 1 S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂ The rate of reaction 1 is investigated. A sample of S₂O₈²⁻ is mixed with a large excess of iodide ions of known concentration. The graph in Fig. 5.1 shows the results obtained. [Figure 5.1]

1. (a)(i)Use Fig. 5.1 to determine the initial rate of reaction 1. Show your working. rate = mol dm⁻³ min⁻¹[1]
2. (a)(ii)The rate equation for reaction 1 is rate = k [S₂O₈²⁻] [I⁻]. Suggest why a large excess of iodide ions allows the rate constant to be determined from the half-life in this investigation.[1]
3. (b)The reaction of persulfate ions, S₂O₈²⁻, with iodide ions is catalysed by Fe²⁺ ions. Write two equations to show how Fe²⁺ catalyses reaction 1. equation 1 equation 2[2]
4. (c)Describe the effect of an increase in temperature on the rate constant and the rate of reaction 1.[1]
5. (d)In aqueous solution, thiosulfate ions, S₂O₃²⁻, react with hydrogen ions, as shown in reaction 2. reaction 2 S₂O₃²⁻ + 2H⁺ → SO₂ + S + H₂O The rate of reaction is first order with respect to [S₂O₃²⁻] and zero order with respect to [H⁺] under certain conditions. The rate constant, k, for this reaction is 1.58 × 10⁻² s⁻¹. Calculate the half-life, t½, for reaction 2. t½ = s[1]
6. (e)The compound nitrosyl bromide, NOBr, can be formed as shown in reaction 3. reaction 3 2NO(g) + Br₂(g) → 2NOBr(g) The rate is first order with respect to [NO] and first order with respect to [Br₂]. The reaction mechanism has two steps. Suggest equations for the two steps of this mechanism. State which is the rate-determining step. step 1 step 2 rate-determining step =[2]

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