Question

Explain why the molar enthalpies of vaporization of the following substances increase in the order \({\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{ < }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{\rm{ < }}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}\), even though the type of IMF (dispersion) is the same.

Short answer

\({\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{ < }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{\rm{ < }}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}\) is the order of the molar enthalpies of vaporization.

Step-by-step solution

Explaining molar enthalpies of vaporization

As we know, dispersion forces or boiling points increase with increase in molecular mass or size. Therefore, as the number of atoms comprising the molecules in this homologous series increases, so does the extent of intermolecular attraction via dispersion forces.

Calculating molar mass

The mass (in grams) of one mole of a substance is referred to as its molar mass. You may compute an element's molar mass by multiplying the element's atomic mass by the conversion factor in grams per mole (g/mol).

As the number of carbons in the compound increases, the molar mass of the compound also increases.

Comparison between given compounds

Now, as we go from left to right in the series \({\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{,}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{\rm{,}}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}\)the molecular mass increases, and therefore the energy required to overcome these forces also increases.

Therefore, the order of molar enthalpies of vaporization is as follows:

\({\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{ < }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{\rm{ < }}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}\)_.