Peroxodisulfate ions, S₂O₈^2–, react with iodide ions in aqueous solution to form iodine and sulfate ions.

S₂O₈^2–(aq) + 2I^–(aq) → 2SO₄^2–(aq) + I₂(aq)                 equation 1

The initial rates of reaction are compared by timing how long it takes to produce a particular amount of iodine using four different initial concentrations of S₂O₈^2–. The results are shown in the table.

a. Plot a suitable graph to calculate rate of reaction.

b. Deduce the order of reaction with respect to peroxodisulfate ions. Explain your answer.

c. The reaction is first order with respect to iodide ions. Use this information and your answer to part b to write the overall rate equation for the reaction.

d. The reaction between peroxodisulfate ions and iodide ions is slow. The reaction can be speeded up by adding a few drops of Fe3+ (aq) ions. The following reactions then take place:

2I^–(aq) + 2Fe³⁺(aq) → I₂(aq) + 2Fe²⁺(aq)                                 equation 2

2Fe²⁺(aq) + S₂O₈^2–(aq) → 2Fe³⁺(aq) + 2SO₄^2–(aq)             equation 3

i. What type of catalysis is occurring here? Explain your answer.

ii. By referring to equations 1, 2 and 3 above, suggest why Fe³⁺(aq) ions catalyse the reaction between peroxodisulfate ions and iodide ions.

Transcribed Image Text:

[s,0,21/moldm3 Initial rate of reaction/s1 0.0200 4.16 x 10-3 0.0150 3.12 x 10-3 0.0120 2.50 x 10-3 0.0080 1.70 x 10-3