6
$\begingroup$

I am a high school student studying basic chemistry and I am confused about the process of writing decomposition reactions.

The question I have encountered asks me to decompose $\ce{BaCO3}$.

Online, I have found that this decomposes to $\ce{BaO + CO2}$.

My question is, essentially, is there a method to determine which elements go where. I'm having trouble finding answers.

$\endgroup$
  • 3
    $\begingroup$ A hint is usually in the name. Carbonates usually turn into $\ce{CO_2}$, hydrates become $\ce{H_2O}$ - nitrates become $\ce{NO_x}$ - nitrides becomes $\ce{N_2}$... and so forth and so on. These substances were named before we had spectroscopy - many of them were named after the gas that they decompose to. $\endgroup$ – Stian Yttervik Sep 10 '19 at 16:41
  • 3
    $\begingroup$ I disagree with the question being labelled as too broad. The answer winds up being too broad, in that there aren't a set of rules that could be condensed into a stackexchange post, but the OP's question is clear and M. Farooq's answer gives a clear explanation of why the question is not as simple as it might seem. $\endgroup$ – Tyberius Sep 10 '19 at 21:34
11
$\begingroup$

Chemistry, despite being several hundred year old science, it is still very much empirical. In the beginning you will have to learn and accept some facts. Later you will start to see patterns (hence the periodic table). If you learn that the carbonate of barium decomposes into carbon dioxide and barium oxide, it will be safe to say that Mg, Ca, Ba, Sr, will do the same. Here you have learnt four more chemical reactions. If you recall the periodic functions in mathematics, which repeat themselves periodically, in the same way, elements repeat their properties in the periodic table.

When you strongly heat inorganic salts, usually you get smaller products (e.g., carbon dioxide and barium oxide) which are energetically more stable with respect to the starting material.

| improve this answer | |
$\endgroup$
2
$\begingroup$

The recipe isnt really complicated:

  • (I guess you already know that) Barium metal and oxygen are very reactive with each other. So you would have to go to very high temperatures to reverse that reaction.
  • Carbonates contain $\ce{CO3^{2-}}$, which is $\ce{CO2}$ with an extra Oxygen anion $\ce{O^{2-}}$. Such decomposition reactions usually make as little rearrangements as necessary, so it is likely that the $\ce{O=C=O}$ motive is retained.
  • You might also know that $\ce{CO2}$ is very stable, and only decomposes to $\ce{CO2 -> CO + C}$ at rather high temperatures (Bouduard equillibrium).
  • More complicated structures (molecules) are usually unstable at higher temperatures, so we can just forget about them.
  • You probably already know other carbonates give off carbon dioxide upon heating.
  • Gas evolution is usually favoured, because gas has high entropy.

Voilà, thats it. The thing was more or less clear after the second of my bullet points, but the rest fits in as well. If the solution was sth else, that would be quite unusual and you would either have heard about it or your teacher wouldnt expect you to know. ;)

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.