[Big-Daddy]

This website, http://www.napavalley.edu/people/fquinlan/Documents/Chem%20120/Microsoft%20Word%20-%20Cobalt%20Titration%20_2_.pdf, gives equilibrium constant for I₂ + I^-  I₃^- to be 710, but according to Wikipedia (the page on "Triiodide") this equilibrium is endergonic - which should mean that ΔG°>0 so K<1? Which piece of information is wrong?

[dunowen.com]

The equilibrium constant of endergonic reaction where ΔG° > 0 is less than 1: K < 1.

hope that helps!

[Big-Daddy]

I'm afraid that doesn't help at all - I noticed exactly the same facts in my OP. My question is about the discrepancy between the reported fact that the equilibrium in study is endergonic and that Wikipedia's equilibrium constant for it is greater than 1 (710), whereas if it is endergonic the equilibrium constant should be less than 1.

[Corribus]

Unfortunately I don't have any data to definitively back this up (not from lack of effort, mind you), but one possibility is that the reaction is exergonic in aqueous phase but endergonic in gas phase. I do have some data to suggest this might be the case.

Here is what we can say for sure:

The reaction is almost certainly exergonic in aqueous phase, supporting the link you provided in your first post. The equilibrium constant does also appear to be approximately 750 or thereabouts - there is plenty of primary literature to support this (see, e.g., http://pubs.acs.org/doi/pdf/10.1021/ja00950a005). There does seem to be some variation in reported values because there are some other possible competing equilibria going on. The standard Gibbs energy change is in the vicinity of -16.4 kJ/mol at 298 K, clearly exergonic.  The reaction appears to be primarily enthalpy driven in the aqueous phase (ΔH° ~ -17 kJ/mol), almost entropy neutral (ΔS° ~ -2 J/mol K; slightly unfavorable, but not nearly enough to make up for the heat liberation at room temperature).

Wikipedia does not specify the phase of the reaction for context when talking about endergonicity. However, we may speculate that the reaction is endergonic in the gas phase. Here's my hand-wavy argument. In the gas phase, the enthalpic terms will be very different. In dilute gas phase, the biggest contributer to the enthalpy change of a reaction is bond breaking and bond forming. Tri-iodide is a linear molecule/ion, so to an approximation, all you are doing is forming an I-I bond, which costs about 150 kJ/mol. This isn't completely accurate because there's an overall negative charge, but either way - it's pretty clear the gas phase reaction is going to be endothermic.  This is in agreement with the NIST chemistry webbook's reaction data, ΔH° = +136 kJ/mol (http://webbook.nist.gov/cgi/cbook.cgi?ID=B227&Units=SI&Mask=8#Thermo-React).  You'll notice this is very different from the aqueous system. I'm sure you can think of a good reason why.

As for entropy, well, the aqueous reaction is dominated by the enthalpy change. There's no reason to assume that the gas phase reaction will behave similarly. BUT given that the gas phase reaction will be resulting in a reduction in the # of moles of gas, it seems very likely the gas phase reaction will be pretty entropically unfavorable.  So, in the gas phase you'll have a reaction that is pretty endothermic AND entropically unfavorable. The sum result would be an endergonic gas phase reaction.

[Big-Daddy]

Unfortunately I don't have any data to definitively back this up (not from lack of effort, mind you), but one possibility is that the reaction is exergonic in aqueous phase but endergonic in gas phase. I do have some data to suggest this might be the case.

Good idea - the difference probably comes down to phases. The difference is, I thought the reaction was only possible in aqueous conditions ... clearly that was wrong. But how do these ions exist in the first place in the gaseous phase? Surely there would have to be a cation to counter the presence of I^- or I₃^-, but most cations prefer to form solid lattices than go into the gaseous phase, so how do we get the charges to balance in order to have ions in the gaseous phase?

Wikipedia does not specify the phase of the reaction for context when talking about endergonicity. However, we may speculate that the reaction is endergonic in the gas phase. Here's my hand-wavy argument. In the gas phase, the enthalpic terms will be very different. In dilute gas phase, the biggest contributer to the enthalpy change of a reaction is bond breaking and bond forming. Tri-iodide is a linear molecule/ion, so to an approximation, all you are doing is forming an I-I bond, which costs about 150 kJ/mol. This isn't completely accurate because there's an overall negative charge, but either way - it's pretty clear the gas phase reaction is going to be endothermic.  This is in agreement with the NIST chemistry webbook's reaction data, ΔH° = +136 kJ/mol (http://webbook.nist.gov/cgi/cbook.cgi?ID=B227&Units=SI&Mask=8#Thermo-React).  You'll notice this is very different from the aqueous system. I'm sure you can think of a good reason why.

I'm not sure I follow - you are forming a new I-I bond, and the bond dissociation energy of I-I is +150 kJ/mol, so wouldn't you get 150 kJ/mol out of the formation of this bond, so that ΔH° = -150 kJ/mol for this reaction?

[Corribus]

Whether or not a reaction happens in the gas phase does not change the fact that many reactions are calculated that way. It is easier to simulate reaction thermochemistry in dilue gas phase than in condensed aqueous phase. This is particularly the case in earlier decades, where taking into account intermolecular interactions was simply not possible given computation limitations at the time. If you go to the NIST chemistry webbook, you'll see many reactions are/were calculated this way, even if it isn't particularly realistic. That said, there is some potential relevance to having thermochemical values in the gas phase. Photochemical reactions like I^- + I₂ could occur in the gas phase. (I'm not saying that they do, just that they might.) Anyway, important thing is that you know what phase you're in when you're throwing around thermochemical information, because it does make a difference.

I'm not sure I follow - you are forming a new I-I bond, and the bond dissociation energy of I-I is +150 kJ/mol, so wouldn't you get 150 kJ/mol out of the formation of this bond, so that ΔH° = -150 kJ/mol for this reaction?

Yeah, you're right, my apologies, I accidentally had it reversed. It would be easier to discuss this quantitatively if I could find enough actual data for the standard molar entropies and standard heats of formation for the three species involved using a cursory search of the internet, but I couldn't.  Either way, though, it's a bad approximation, because I₃ is ionic and the two bonds in I₃^- aren't equivalent to the I-I bond in I₂. The paper I quoted earlier gives an overall reaction enthalpy of 136 kJ/mol but I can't access that paper at the moment to check it out and see the details.  From just consideration of # of moles of gas consumed/producedm, it is pretty clear that the entropy change will be negative, which may by itself be enough to make the reaction unfavorable.

At the end of the day, it's just speculation on my part that this is why wikipedia calls the reaction endergonic. It's possible that whoever wrote the article made a simple error and the reaction is exergonic in both aqueous and gas phases at room temperature. Easy enough to do (as you saw above). What is definitely clear is that the reaction is exergonic in aqueous phase a room temperature.

[Big-Daddy]

Ok, thanks for the help.