In the iodine clock reaction with S2O3 ion I have found that one of the exponents in the reaction rate is 1.55. Can I round it to 2? Or shoud I write it as a fruction 3/2?

I know that reaction rate can be a fraction. But it is a fraction in this case?

Thank you! :)

For sure it can be 1.55 you must leave it as it is.

Thank you but iodine doesn't react with oxygen

This is in the wrong thread. Here is the original thread.

https://www.askmehelpdesk.com/chemistry/why-after-titration-s2o3-2-using-starh-indicator-blue-reappears-351397.html

Iodine certainly won't react with oxygen, but will it react with iodide? It certainly doesn't do it very often, but it can happen (the Winkler test requires a Manganese catalyst). Oxygen is a sufficiently strong oxidizer.

Winkler test for dissolved oxygen - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Winkler_test_for_dissolved_oxygen)

Time-resolved reaction of oxygen(1.DELTA.) with iodide in aqueous solution - The Journal of Physical Chemistry (ACS Publications) (http://pubs.acs.org/doi/abs/10.1021/j100334a026)

2I^- \rightleftharpoons I_2 + 2 e^- E = - 0.54 volts

O_2 + 4H^+ + 4e^- \rightleftharpoons 2H_2O E = 1.23 V

ΔE = +.69 volts

You can see that the electrochemical potential is there (which says nothing about kinetics). In addition, it only takes a tiny amount of iodine to turn starch blue. The last thing is that it's the only thing I can think of that's present that could oxidize iodide to iodine. Iodine has to be present or the starch wouldn't turn blue. I'm open to other suggestions.