I have a question in regards to the iodine clock reaction experiment. I was carr

Question

I have a question in regards to the iodine clock reaction experiment. I was carrying out the experiment over different temperatures received an unusual time for one of the runs in an ice bath solution. I'm just wondering if there would be a reason as to why the ice bath would run faster than a run done at room temperature? (below are the results and solution used)

_______________________
RESULTS

0°     1:42:03
25°   7:07:22
40°   3:30:75
65°   3:06:25

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Solutions used:
• 20ml Buffer A (100ml 0.75M NaOAC Soln + 100ml 0.22M HOAc Soln + Vitex diluted in a 500ml volumetric flask)
• 5 ml H3AsO3 Arsenous acid
• 5 ml IO3- Potassium Iodate
• 25 ml I- Potassium Iodide (added when timer starts)

Explanation / Answer

The rate of the reaction depends upon its reaction mechanism. The slowest step is always rate determining step in any reaction mechanism.

IO3- + 3 H2SO3 ----> I- + 3 SO42- + 6 H+

At low temperatures the NaHSO3 does not decompose, but at higher temperatures it will. Chemical reactions are thermodynamically “favorable” if the overall energy the the products is lower than the overall energy of the reactants.

at higher temperatures NaHSO3 is unstable and decompose, hence HSO3- is no longer sufficiently available to proceed the reaction in forward direction.

Hope this helped you!

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