Why doesn't $\ce{NaOH(g)}$ react with $\ce{CO2(g)}$ to produce gaseous sodium carbonate $\ce{(Na2CO3(g))}$:

$$ \ce{2 NaOH(g) + CO2(g) → Na2CO3(g) + H2O(g)} \tag{3} $$

Why don't we see any gaseous $\ce{Na2CO3}$ at high temperatures (above 800)?

  • 3
    $\begingroup$ Check their boiling points. You will probably have difficulties to find it for the carbonate. $\endgroup$
    – Poutnik
    Mar 5 at 7:05
  • $\begingroup$ Most carbonates decompose, to some extent at least, at high T. A quick look at MPs BPs and guessing temperatures indicates that there could be some gas phase reaction. As temperatures cool rapidly gas phase reactions should minimize. $\endgroup$
    – jimchmst
    Mar 5 at 18:09
  • $\begingroup$ @Poutnik curiously, Wikipedia reports neither a boiling point nor a decomposition point for sodium carbonate. $\endgroup$ Mar 6 at 17:21
  • $\begingroup$ @OscarLanzi I assume it is not a surprise. $\endgroup$
    – Poutnik
    Mar 6 at 17:26

2 Answers 2


You are looking at a high entropy situation (high temperature, gas phase, and likely low vapor density as sodium hydroxide does not achieve one atmosphere vapor pressure until 1388°C). Under these conditions the proposed equilibrium

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

tends to favor more gas molecules with fewer atoms apiece, which is seen to be the left side (the sodium compounds, even if ionic, exist in the gas phase as neutral clusters of said ions).

Upon condensation of the sodium compounds the gas phase entropies are more nearly balanced, with only one three-atom molecule remaining gaseous on either side, and enthalpy effects that favor forming the carbonate come to the fore.

Incidentally, sodium hydroxide is one of only a few metal hydroxides that actually make it to the gas phase with significant vapor pressure. Most metal hydroxides, even with the lighter alkali metal lithium, decompose by emitting water vapor, due to those sane entropy effects, before they would become volatile.


Your equation implies bot $\ce{NaOH}$ and $\ce{Na2CO3}$ have to be the gas state. Well! It is hard to obtain. $\ce{NaOH}$ is gaseous at temperatures above $1500$°C. And $\ce{Na2CO3}$ does not exist in the gaseous phase. It melts at 851°C, and is decomposed at higher temperatures.


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