Explain why the enthalpies of vaporization of the following...

Explain why the enthalpies of vaporization of the following substances increase in the order $\mathrm{CH}_{4} < \mathrm{NH}_{3} < $ $\mathrm{H}_{2} \mathrm{O}$ , even though all three substances have approximately the same molar mass.

Key Concepts

Hydrogen Bonding

Hydrogen bonding is a particularly strong type of dipole–dipole interaction that occurs when hydrogen is bonded to highly electronegative atoms such as oxygen, nitrogen, or fluorine. The extensive hydrogen bonding networks that can form significantly increase the energy required to vaporize a substance, markedly raising its enthalpy of vaporization.

Dipole–Dipole Interactions

Dipole–dipole interactions occur between polar molecules due to the attraction between permanent dipoles. These interactions are stronger than London dispersion forces and contribute to higher enthalpies of vaporization in molecules where these forces are significant.

London Dispersion Forces

London dispersion forces are weak intermolecular attractions that arise due to temporary fluctuations in electron densities. They are present in all molecules but are the dominant force in nonpolar molecules, leading to lower enthalpies of vaporization compared to substances with stronger interactions.

Intermolecular Forces

Intermolecular forces are the attractions between molecules that hold them together in the liquid phase. The primary types include London dispersion forces, dipole–dipole interactions, and hydrogen bonding. Differences in these forces directly affect the energy needed to break the liquid into vapor.

Enthalpy of Vaporization

Enthalpy of vaporization is the energy required to convert a substance from a liquid phase to a gas phase at constant pressure. This property is strongly influenced by the strength of the interactions between molecules in the liquid state, with stronger intermolecular forces requiring more energy for vaporization.

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