# Why NaOH(s,l) + CO2(g) → Na2CO3(s)? Why not NaOH(g) + CO2(g) → Na2CO3(g)?

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)?

• Check their boiling points. You will probably have difficulties to find it for the carbonate. Mar 5 at 7:05
• 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. Mar 5 at 18:09
• @Poutnik curiously, Wikipedia reports neither a boiling point nor a decomposition point for sodium carbonate. Mar 6 at 17:21
• @OscarLanzi I assume it is not a surprise. Mar 6 at 17:26

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