Is the activation energy for vaporization equal to the enthalpy of vaporization?

When ethanol, say, evaporates, the intermolecular forces between molecules need to be broken. As a vapour, there are no new forces being created, so it is a purely endothermic process, with no "hump" in the enthalpy profile, i.e. the potential energy maximum *is* the final state. Is this logic correct?


If so, then wouldn't the reverse process have no energy barrier? Presumably the only reason this wouldn't be spontaneous at any temperature would be because of entropy changes.


