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I just came across this reaction

$\ce{Na2S2O3 + 2HCl -> 2NaCl + SO2 + S + H2O}$

I was wondering if it was possible to predict the products of this reaction if I didn't know it (or any inorganic reaction for that matter).

Can we use arrow-pushing (as we do in organic chemistry) to get the products of an inorganic reaction? Can we devise a mechanism to find the products? It seems unrealistic to me how anyone could learn the many reaction of inorganic chemistry.

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  • $\begingroup$ The arrow pushing formulism is expandable to any reaction of the main group elements involving the forming or breaking of covalent bonds. Organic chemistry is just the largest subset of these reactions. The remaining "inorganic only" reactions are a much smaller set. $\endgroup$
    – Ben Norris
    Commented Dec 3, 2012 at 11:32
  • $\begingroup$ If that is so, then can you please explain how I can obtain the product from the above reaction using arrow pushing. Can you give me a link to the image or something, I would like to know. $\endgroup$
    – carboncopy
    Commented Dec 3, 2012 at 14:09
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    $\begingroup$ See amazon.com/Arrow-Pushing-Inorganic-Chemistry-Approach/dp/…. $\endgroup$
    – Leponzo
    Commented Feb 26, 2015 at 12:36

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The general answer to this question is no, one cannot simply predict the products of any reaction. How do we know that one of the products is not $\ce{S2O6^{2-}}$ or perhaps even $\ce{S2O5Cl2^{2-}}$?

The best that can be done without any guidance from experiment is to compute the thermodynamically favored products. One does this (at standard temperature and pressure) by writing down all plausible products and then computing the Gibbs free energy ($\Delta G$) for the reactions. The one with the smallest $\Delta G$ is the thermodynamically preferred.

But be warned: there are mechanistic and kinetic factors to take into account. The thermodynamically favored reaction may, as a practical matter, not take place. The standard example is diamond --> graphite, which is thermodynamically allowed, but don't hold your breath waiting for it to happen.

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  • $\begingroup$ So how do chemists possibly make sense of all the reactions ? $\endgroup$
    – carboncopy
    Commented Dec 5, 2012 at 12:26

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