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Calculating rate constants - peroxodisulfate iodide clock reaction

Hi all, have this TL4.2 homework to do.

It's on the iodine-peroxodisulfate clock reaction:

S2O82- + 2I- ---> I2 + 2SO42-
I2 + 2S2O32- ---> 2I- + S4O62-
I2 + starch indicator ---> blue-black ppt

The iodide reacts with peroxodisulfate to produce iodine and sulfate ions. The iodine is then instantly reconverted back to iodide by thiosulfate, as seen in the second equation.

I used 0.04M peroxodisulfate, 1M iodide and 0.01M thiosulfate.

Problem is I can't calculate the amount of iodine produced when the blue-black precipitate appears. In my case, there are 4x10-5 mol of thiosulfate in each solution, 1.6x10-4 mol of peroxodisulfate, and excess iodide.

I thought that the amount of peroxodisulfate used would be 1.4x10-4 mol, because 0.2x10-4 is used to exhaust thiosulfate, and there is no net yield in iodine during this time. Therefore by the time the precipitate appears, that 1.4x10-4 mol of iodine is produced.

But apparently not all iodine forms when the blue black precipitate appears. And if I don't know how many moles of iodine are produced, I can't calculate the rate constant. The one I got, which is possibly wrong, was 0.11 mol-2 dm6 s-1.

I was using the formula

rate = difference in iodine concentration / time

Is there any way to do this with reactants? That's the only way I can think of getting around this.

Cheers
Reply 1
There is something wrong with the numbers you listed, care to elaborate?

I thought that the amount of peroxodisulfate used would be 1.4x10-4 mol


Used, or left?

But apparently not all iodine forms when the blue black precipitate appears.


No idea what you mean. All iodine that was produced up to this moment was consumed by thiosulfate oxidation, so you can easily calculate how much iodine was produced up to this moment. Color means some excess iodine was made, but its amount doesn't matter, what matters is that the thiosulfate was consumed - so from the stoichiometry you can easily calculate how much peroxodisulfate reacted up to this moment - and amount over time is the average rate.

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