Here's why that is the case.
As you know, the standard enthalpy of vaporization of a substance tells you how much heat is needed in order for a specific quantity, usually one mole, of that substance to undergo a liquid
In other words, if you want one mole of a substance to go from liquid at its boiling point to vapor at its boiling point, you must ensure a change in enthalpy equal to its standard enthalpy of vaporization.
Now, an important clue as to why ethane,
As you know, in order for the molecules of a substance to go from liquid to vapor, you need to provide them with enough energy so that they overcome the intermolecular forces of attraction that they exhibit in the liquid state.
This means that the stronger the intermolecular forces of attraction are, the higher the boiling point of a given substance, since you'd need to supply it with more heat in order to allow its molecules to go from liquid to vapor.
Now, a liquid boils when its vapor pressure is equal to the external pressure. Take a look at the boiling points of ethane and propane at
So, according to these values, ethane boils at a lower temperature than propane. This tells you that ethane molecules need less heat than propane molecules in order to overcome their intermolecular forces of attraction and go into vapor state.
This is why the standard enthalpies of vaporization are different for these two compounds.
So, based on this explanation, which standard enthalpy of vaporization would you predict is smaller?
Well, since you need less heat to boil one mole of ethane at its boiling point, you can say that
#DeltaH_"vap ethane" < DeltaH_"vap propane"#
SIDE NOTE As you know, both molecules are nonpolar, which means that the difference in the strength of their intermolecular forces comes from the magnitude of the London dispersion forces they exhibit.