8
$\begingroup$

Many metal oxides decompose into the free metal and oxygen gas at high temperatures, but why doesn't $\ce{CaO}$ do that? What happens to $\ce{CaO}$ at high temperatures?

This is from problem 2 of the 2011 USNCO local exam:

Oxygen gas can be produced by the decomposition of all of the following substances EXCEPT

(A) calcium oxide. (B) hydrogen peroxide. (C) mercury(II) oxide. (D) ozone.

The correct answer to this question was (A).

$\endgroup$
  • 1
    $\begingroup$ Nothing happens to $\ce{CaO}$ at high temperatures. $\endgroup$ – Jan Dec 25 '15 at 0:42
8
$\begingroup$

Stability, the thermodynamic property, is defined as whether a decomposition reaction is exergonic or endergonic at a certain temperature. The reaction $\ce{CaO -> Ca + 1/2 O2}$ is endergonic across a very large temperature range. But that doesn’t necessarily help you.

In that list of four, we have to clearly molecular compounds and two mostly ionic ones. We need different discussions for both. Basically, the molecules are unstable towards decomposition into $\ce{O2}$ and something ($\ce{H2O}$/$\ce{O2}$) because they contain energetic bonds whose replacement by an $\ce{O=O}$ bond is favourable.

For the ionic ones, we have a number of different discussions we can do. For one, the metallic partner is calcium versus mercury; while the latter is known since ancient times and pretty stable in liquid form the former is not: It will readily react to form salts. For another, the $\ce{Ca^2+}$ ion is rather small and thus a hard Lewis acid according to Pearson’s concept. $\ce{Hg^2+}$ is much larger and thus a much softer acid. Therefore, the adduct of $\ce{Ca^2+}$ and $\ce{O^2-}$ is predicted to be more stable (hard–hard interaction) than that of $\ce{Hg^2+}$ with $\ce{O^2-}$. Finally, we can also consider heats of formation which should (I unfortunately don’t have values) be much more negative in the case of $\ce{CaO}$ than for $\ce{HgO}$ — therefore, the latter can be decomposed with enough heat due to $\Delta G = \Delta H - T \Delta S$ while the former cannot.

$\endgroup$
  • 1
    $\begingroup$ I think the chemistry olympiad question was getting at hard-soft acid-base theory mainly. $\endgroup$ – Dissenter Dec 25 '15 at 5:28
  • 1
    $\begingroup$ @Dissenter - if you think you can explain it with HSAB (which looks like it would work), go ahead and add an answer. $\endgroup$ – Ben Norris Dec 25 '15 at 19:04
  • 1
    $\begingroup$ @BenNorris HSAB is one of the three explanations given in my answer … $\endgroup$ – Jan Dec 25 '15 at 20:56
  • $\begingroup$ CaO can thermally decompose at the right temperature and partial pressure of oxygen, it just so happens that everything becomes a gas at such temperatures. $\endgroup$ – A.K. Dec 26 '15 at 19:07
2
$\begingroup$

In short ... before the decomposition of CaO, it start to melts. That means, melting point of CaO is less than the theoretical decomposition temperature of CaO. So it do not decompose upon high temperature. It melts. Compounds like Na2O , K2O, MgO .... etc also do not decompose due to this reason.

In fact these compounds are thermodynamicaly more stable.

Compound like Silver oxide, oxides of mercury readily decompose as their decompose temperature is less than their melting point. Oxides of elements below Hydrogen in electro-chemical series act like this. That is these compounds are thermodynamically less stable with respect to above compounds. (that doesn't mean that these compounds are thermodynamically unstable. Only less stable with respect to Na2O , K2O, MgO .... etc.)

$\endgroup$
  • 4
    $\begingroup$ Okay, so it melts. What happens if you keep heating it? Do you get oxygen gas eventually? $\endgroup$ – user2357112 Dec 25 '15 at 9:50
  • 5
    $\begingroup$ Yes, I don't see why the physical state would matter. Solid, liquid, gas, so what? The question is whether decomposition is endothermic or exothermic at any given temperature, isn't it? $\endgroup$ – Tom Zych Dec 25 '15 at 11:48
  • $\begingroup$ I'm tempted to downvote you - all compounds decompose given high enough temperature. $\endgroup$ – Mithoron Dec 26 '15 at 0:46
1
$\begingroup$

A comparison of the heats of formation of the 4 molecules yield the correct result. At -630 kJ/mol /\ Hf, CaO requires the least energy to form - and therefore the most energy to decompose - compared to HgO (-90), MgO (-600), and H2O2 (-188). In other words, CaO is the happiest of the 4 molecules at 'high temperatures'.

CaO will eventually decompose if the temperature is increased sufficiently, but the other 3 molecules will do it first. I think the term 'high temperatures' is a source of confusion. :)

$\endgroup$
  • $\begingroup$ Welcome to Chemistry.SE! Please note that formulas can be better expressed with \$\ce{}\$ for chemical formulas/equations, \$\mathrm{}\$ for math term/equations, and \$\pu\$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies. $\endgroup$ – A.K. Aug 30 '18 at 13:46

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.