Why do carbon dioxide and sodium hydroxide not form sodium oxide and carbonic acid?

$\ce{CO2 + 2NaOH -> Na2O + H2CO3}$

as opposed to:

$\ce{CO2 + 2NaOH -> Na2CO3 + H2O}$

This is in the context of fractional distillation of liquified air, before which carbon dioxide must be removed using sodium hydroxide.

  • 2
    $\begingroup$ Take a look at Standard enthalpy of reaction and you can find the answer by yourself. $\endgroup$ Mar 12, 2015 at 8:01
  • $\begingroup$ Could you elaborate, so we can say what's wrong with reasoning behind your first reacion? You know that Na2O is very basic? $\endgroup$
    – Mithoron
    Mar 12, 2015 at 14:54
  • $\begingroup$ I've updated my question. I do know that sodium oxide is basic, and I was expecting a conjugate acid and a conjugate base to be formed. $\endgroup$ Mar 12, 2015 at 14:59

1 Answer 1


Sodium oxide is not a base in either the Bronsted-Lowry or Arrhenius senses. It is technically a base anhydride, meaning that it can be hydrated to yield a base:

  1. $$\ce{Na2O + H2O -> 2NaOH}$$

Carbon dioxide can similarly be viewed as an acid anhydride, meaning it can be hydrated to form an acid.

  1. $$\ce{CO2 + H2O <=> H2CO3}$$

The difference is that the hydration of sodium oxide is essentially irreversible. One program I used estimated $K_{eq}=10^{39}$ for sodium oxide hydration. In contrast the hydration of carbon dioxide is reversible, with $K_{eq}\approx1.5\times10^{-3}$.

The reaction that you are wondering about, $\ce{CO2 +2NaOH⟶Na2O +H2CO3}$, is essentially the dehydration of sodium hydroxide by $\ce{CO2}$ to form sodium oxide and carbonic acid, that is, it is reaction 2 minus reaction 1. But the equilibrium constants I just mentioned show that sodium oxide has an "affinty" for water that is about 42 orders of magnitude greater than carbon dioxide's affinity for water. So that reaction is very unfavorable. The reverse reaction would be extremely favored: Sodium oxide would be a powerful reagent for $\ce{CO2}$ removal prior to air liquefaction, at least as powerful as sodium hydroxide.

If you are curious, here is the R code I used to estimate the equilibrium constant of sodium oxide hydration, using the package CHNOSZ:

> subcrt(c("Na2O","H2O","NaOH"),c(-1,-1,2),T=25)
info.character: found H2O(liq), also available in gas
subcrt: 3 species at 298.15 K and 1 bar (wet) 
     coeff         name formula state ispecies
2031    -1 sodium oxide    Na2O    cr     2031
1       -1        water     H2O   liq        1
1157     2         NaOH    NaOH    aq     1157

      logK         G         H         S         V       Cp
1 39.01742 -53229.29 -57143.24 -13.24728 -36.04979 -40.8745
  • $\begingroup$ On a related note, on of the commenters asked me to look up the standard enthalpy of my proposed reaction. Any ideas why? $\endgroup$ Mar 12, 2015 at 16:34
  • 1
    $\begingroup$ It would be very positive, a hint that the reaction is likely to be thermodynamically unfavorable. (I say "hint" because favorability is determined by $\Delta G=\Delta H - T \Delta S$, so in theory it would be possible for an extremely large positive entropy change to counteract the very large enthalpy change, thus making $\Delta G$ negative and the reaction thermodynamically feasible, but since the reaction consumes a gas, it's entropy change is probably negative, which would only make even more disfavored.) $\endgroup$
    – Curt F.
    Mar 12, 2015 at 16:38
  • 1
    $\begingroup$ $\ce{O^2-}$ is a base according to Brønsted and Lowry’s definition as it takes up $\ce{H+}$ ions to form $\ce{OH-}$. $-1$ $\endgroup$
    – Jan
    Oct 26, 2015 at 14:49
  • $\begingroup$ Good point Jan. I was going off of Wikipedia which says A base anhydride is neither an Arrhenius base, nor a Brønsted–Lowry base, since it does not accept protons and do not increase the hydroxide ion concentration of water. However, a base anhydride is a Lewis base, since it will share an electron pair with some Lewis acids, most notably acidic oxides. It sounds like that is not completely correct. If someone edits Wikipedia I'll edit my answer. $\endgroup$
    – Curt F.
    Oct 26, 2015 at 18:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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