I'd like to know the physical mechanism of oleum "fuming" on its surface. Dissolved SO3 reacts very exothermically with some air humidity. Why does the mist occur instead of dissolving created H2SO4 in the solution?
You've pretty much answered your own question here.
No, that is too simplified.
The reaction takes place on the liquid's surface. The heat of reaction heats up surface's surroundings.
What does there vaporize? Air humidity? Then why does it condensate 1 cm above the surface? Or sulfuric acid with (~300°C, p(atm)) boiling point? Generally, why the sulfuric acid is being carried away if it can dissolve?
My bet is that the concentration (partial pressure) of SO3 above oleum is high enough to start condensation in the presence of water vapor. That will happen above the surface, in the gas phase.
Or sulfuric acid with (~300°C, p(atm)) boiling point?
Cp(H2SO4, liq) ~ 60 J/molK, and the enthalpy of the reaction between SO3 and water vapor at the surface is around 160 kJ/mol. Fume? You betcha --- that's the old "acid to water, never water to acid" rule from HS chemistry.
My bet is that the concentration (partial pressure) of SO3 above oleum is high enough to start condensation in the presence of water vapor. That will happen above the surface, in the gas phase.
Fume? You betcha --- that's the old "acid to water, never water to acid" rule from HS chemistry.
When water putted into acid or conversely acid putted into water, the liquid may spit around due to gas expansion, as I see. Also that is different physical process than oleum fuming.
