Rates of reaction - kinetics (A2)
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I am doing the iodine clock reaction using potassium iodide solution, potassium peroxodisulphate sodium thiosulphate and starch. What is the point of the sodium thiosulphate as it has order 0 and so doesn't affect the reaction?? Also what is the mechanism of the reaction. Really confused. Thanks v much.
Corrie writes ....
There have been several question about this reaction over the years - try putting peroxodisulphate into the FAQ Search on React. Here's part of one of the earlier answers, which I have adapted:
One of the best known clock reactions involves the chemicals you have mentioned in your question. When all these ingredients are mixed together the following reactions occur.
First the potassium iodide is oxidised by the potassium peroxodisulphate to form iodine:
2KI(aq) + K2S2O8(aq) -> 2K2SO4 (aq) + I2(aq)
The iodine then immediately reacts with the sodium thiosulphate, so no colour appears:
I2(aq) + 2Na2S2O3(aq) ----> Na2S4O6(aq) + 2NaI(aq)
However, when the supply of sodium thiosulphate is exhausted, the iodine can no longer be removed, so its colour becomes immediately evident. If starch has also been added to the mixture, this event is heralded by the sudden appearance of a deep blue-black colour.
The thiosulphate does not affect the rate of the reaction between iodide and peroxodisulphate. Adding a fixed amount of thiosulphate means that the time for the clock to 'tick' (go black) changes as the rate of this reaction changes. This time can than be used as a measure of the rate, as the rate is proportional to 1/time - the longer the time, the lower the rate, and vice versa.
Several previous FAQs have asked about the mechanism. You'll find them on the React site.
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updated: 12 February 2007
