1.1. Peroxodisulafte(VI) ions, S₂O₈^2-, reacts with iodide ions, I^–, in solution to form sulfate ions, SO₄^2-. and iodine, I_2. 

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

If the reaction is carried out in the presence of a fixed amount of aqueous thiosulfate ions, S₂O₃^2- (aq), the iodine produced is reduced back to iodide ions.

2S₂O₃^2- (aq) + I₂ (aq) → S₄O₆^2- (aq) + 2I^– (aq)

When all of the thiosulfate ions have been used up, the iodine remaining will react with starch solution to produce a blue black colour.

Using appropriate apparatus and the chemicals listed below, plan an experiment to find the order of this reaction with respect to iodide, I^– (aq). Your answer should quote suitable volumes of solutions for the experiment

Chemicals

• 1.00 mol dm^-3 aqueous potassium iodide, KI (aq)

• 0.040 mol dm^-3 aqueous dipotassium peroxodisulfate(VI), K₂S₂O₈ (aq)

• 0.010 mol dm^-3 aqueous sodium thiosulfate, Na₂S₂O₃ (aq)

• 1% aqueous starch solution

• Distilled water

1.2. Explain how the results can be used to show that the reaction is first order with respect to iodide ions, I^– (aq).

1.3. In a separate experiment, a student reacts together:

• 7.2 x 10^-3 mol dm^-3 I^– (aq)

• 3.8 x 10^-3 mol dm^-3 S₂O₈^2- (aq)

The initial rate is 8.95 x 10^-3 mol dm^-3 s^-1. The reaction is first order with respect to both I^– and S₂O₈^2-.

Calculate the rate constant k for this reaction. State the units, if any.