When gases leave the catalytic reactor in the contact process of sulfuric acid generation, both $\ce{SO2}$ and $\ce{SO3}$ will be present. According to descriptions I have read, the trioxide reacts with the water spray, but the dioxide does not, and instead rises as a gas and is recirculated into the system.

I thought that sulfur dioxide was highly soluble in water. If so, should it not dissolve into the water spray and thereby contaminate the oleum?

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    $\begingroup$ $\ce{SO3}$ reacts rapidly and exothermically with $\ce{H2O}$, forming the liquid $\ce{H2SO4}$, sulfuric acid. This removes the $\ce{SO3}$ from the mixed gases, allowing yet more to react. In fact, the resulting acid can contain more $\ce{SO3}$ than $\ce{H2O}$. This is called oleum or fuming sulfuric acid. $\endgroup$ Commented Aug 5, 2016 at 2:10
  • $\begingroup$ Nobody knows why $\ce{SO2}$ is not soluble in oleum. The prediction of the solubilities is one of the yet unknown questions to be solved in chemistry. There are plenty of theories for explaining the solubilities of a group of substances in a group of solvents, using different parameters like polarity, dipole moments, electronegativities, entropy, etc. But no theories are general enough to explain the solubility of any solute in any solvent. $\endgroup$
    – Maurice
    Commented Mar 1, 2022 at 9:57

2 Answers 2


You are true that $\ce{SO2}$ is very soluble in water but it depends on the temperature. Also, $\ce{SO3}$ reacts with water then dissolves in it in a very exothermic way. My understanding is that, if the quantity of water is well adjusted, you will actually get hot concentrated sulfuric acid and sulfur dioxide is not quite soluble in it (ca. 8 g/kg at 120 °C). So $\ce{SO3}$ reacts with water and sulfur dioxide gas can be recirculated.



When gases leave the catalytic reactor The hot gas mixture is then passed through concentrated sulfuric acid , Sulfur trioxide readily reacts with concentrated sulfuric acid to form oleum but other gases like O2, N2 don't react and pass through
The reaction is as follows:

SO3(g) + H2SO4(l) → H2S2O7(l) (Oleum)

In this step, the SO2 and unreacted O2 pass through the absorption system relatively unaffected, while the SO3 is captured and converted into oleum.


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