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Topic: Collision Theory and Orientation Help  (Read 3510 times)

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Offline getfirefox

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Collision Theory and Orientation Help
« on: March 02, 2010, 06:35:30 PM »
Hey guys.

I just did Persulfate variation of the iodine clock experiment.

http://en.wikipedia.org/wiki/Iodine_clock_reaction

According to collision theory explained on this page, molecules need to hit each other in certain positions to react.

Which molecule needs to hit which bond and in what direction/orientation in the iodine clock reaction?

2I(aq) + S2O82−(aq) → I2 (aq) + 2SO42−(aq)

and

I2 (aq) + 2S2O32−(aq) → 2I(aq) + S4O62−(aq)

Thanks guys!

Offline renge ishyo

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Re: Collision Theory and Orientation Help
« Reply #1 on: March 02, 2010, 07:23:47 PM »
Although I cannot provide a detailed account of the reaction mechansim for you (no atoms are directly transferred in these reactions between the two species, only electrons, which makes things like collision orientations different to determine...also species like I- and I2 can engage in side reactions to form complex iodides like I3- etc. which might actually play a role for all we know), but a few things can be learned by examining the structures of the sulfur containing species in this reaction:

For the first step, the bond between the two oxygens in the center of [S2O82-] must be broken by the transfer of two electrons so the action in some way will occur around here (see the image at: http://www.3dchem.com/inorganicmolecule.asp?id=644). Since the oxygens are protected by a "shield" of surrounding electrons from the surrounding oxygens, and the iodide ions are rather large and negatively charged, the first step is probably going to be pretty slow kinetically as it will be hard to get these two species to get close to one another to react. On the other hand,the thermodynamics of the reaction appear very favorable at first glance due to the formation of the two sulfate molecules in the end and the relief of bond strain from the oxygen oxygen bond.

On the other hand the second step of the reaction should be pretty fast as the reducing sulfurs are clearly exposed (see the image at: http://www.3dchem.com/inorganicmolecule.asp?id=1200). Two of these molecules will come together and form a new bond to make S4O6 (shown here: http://www.chemthes.com/entity_datapage.php?id=2463). This forms another "oxygen cage" protecting the central sulfur atoms from attack. In the process electrons are transferred to the neutral iodine molecule to make more iodide ions. This reaction is expected to be very fast due to the exposed sulfur atom in S2O32-.

It should be noted that neither one of these reactions has a direct influence on the color you eventually observe. That is a reaction between iodine and starch and is a whole subject in and of itself. The purpose of the second reaction above is merely to delay the color change reaction between I2 and starch from taking place (the color change reaction would take place upon production of I2 normally, but the fast nature of the reaction of S2O32- with I2 delays the starch reaction until all the S2O32- is used up).

To see an example of the sort of structure that forms when iodine reacts with starch look here: http://en.wikipedia.org/wiki/Iodine_test (this is the species at the end that ultimately produces the color...note the presence of the complex I3- ion in the center of the helix in the image...)

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