Question

The hydrogen fluoride molecule, HF, is more polar than a water molecule, \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) (for example, HF has a greater dipole moment). Yet the molar enthalpy of vaporization for liquid hydrogen fluoride is lesser than that of water. Explain.

Short answer

Less heat is required to overcome hydrogen bonding in case of HF than for \({{\rm{H}}_{\rm{2}}}{\rm{O}}\)_.

Step-by-step solution

Definition of liquid

One of the states of matter is the liquid state. Because the particles in a liquid are free to move, it lacks a defined form even while having a defined volume.

Explanation of why molar enthalpy of vaporization is less in HF

As an HF molecule is capable of forming hydrogen bonds with other HF molecules. Therefore, less heat is required to overcome hydrogen bonding in case of \({\rm{HF}}\)than in \({{\rm{H}}_{\rm{2}}}{\rm{O}}\)_.

As a result, molar enthalpy of vaporization of HF is less than that of water.

Also, the water molecule has two hydrogen atoms that can form hydrogen bonds with other water molecules. In comparison, hydrogen fluoride contains only one hydrogen that is capable of forming hydrogen bonds with other \({\rm{HF}}\)molecules.

Therefore, less heat is required to overcome hydrogen bonding in case of HF than in\({{\rm{H}}_{\rm{2}}}{\rm{O}}\)_.