Before the specific temperature, even if the temperature doesn't reach the boiling point of azeotrope or ethanol or water, we know that there are still some ethanol evaporate. In this time(before the boiling point of azeotrope(because the boiling point of azeotrope is the smallest)), will there be some water come out with ethanol BECAUSE OF hydrogen bond? I'm confused that if the hydrogen bond will count before the boiling point.
Azeotropes exist for miscible liquids with mixing enthalpy sufficiently different to zero, what leads to existence of either minimum or maximum of total vapour pressure for particular molar fractions.
It also leads to partial pressure not being linear function of the molar fraction, being a deviation from the Raoult's law.
Hydrogen bonds themselves do not lead to the azeotrope. The important thing is the change in the average enthalpy of intermolecular bonding after mixing.
Mixture of liquids forming an azeotrope acts as if the azeotrope mixture were the 3rd liquid and if there were evaporating the azeotrope and just one ( for binary azeotropes ) of the pure solvents, depending on the composition.
If the mixture is more ethanol rich than the ethanol-water azeotrope, there is evaporating azeotrope plus ethanol. And vice versa.
So at any temperature, there is always some partial vapour pressure of all volatile components, no matter if they form an azeotrope or not, or if there are hydrogen bonds or not.