Question

Which substance in each pair would be expected to be more volatile at a particular temperature? Explain your reasoning. a. \(\mathrm{H}_{2} \mathrm{O}(l)\) or \(\mathrm{H}_{2} \mathrm{~S}(l)\) b. \(\mathrm{H}_{2} \mathrm{O}(l)\) or \(\mathrm{CH}_{3} \mathrm{OH}(l)\) c. \(\mathrm{CH}_{3} \mathrm{OH}(l)\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)\)

Short answer

In summary: a. H₂S is more volatile than H₂O due to weaker dipole-dipole interactions. b. CH₃OH is more volatile than H₂O due to weaker hydrogen bonding. c. CH₃OH is more volatile than CH₃CH₂OH due to weaker van der Waals forces.

Step-by-step solution

Pair a: \(\mathrm{H}_{2} \mathrm{O}(l)\) or \(\mathrm{H}_{2} \mathrm{S}(l)\)

In this pair, we have water and hydrogen sulfide. Water has very strong hydrogen bonding due to the highly electronegative oxygen and the hydrogen atoms bonded to it. Hydrogen sulfide, on the other hand, has weaker dipole-dipole interactions due to the lower electronegativity of sulfur compared to oxygen. Therefore, water has stronger intermolecular forces than hydrogen sulfide. Since the volatility of a substance is inversely related to the strength of its intermolecular forces, we can conclude that hydrogen sulfide is expected to be more volatile than water at a particular temperature.

Pair b: \(\mathrm{H}_{2} \mathrm{O}(l)\) or \(\mathrm{CH}_{3} \mathrm{OH}(l)\)

In this pair, we have water and methanol. As mentioned earlier, water has very strong hydrogen bonding due to the highly electronegative oxygen and the hydrogen atoms bonded to it. Methanol, too, exhibits hydrogen bonding due to the presence of an oxygen-hydrogen bond. However, the hydrogen bonding in water is stronger because of the presence of two hydrogen atoms bonded to the electronegative oxygen, compared to only one hydrogen atom in methanol. Therefore, the intermolecular forces in water are stronger than those in methanol. Since volatility is inversely related to the strength of intermolecular forces, we can conclude that methanol is expected to be more volatile than water at a particular temperature.

Pair c: \(\mathrm{CH}_{3} \mathrm{OH}(l)\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)\)

In this pair, we have methanol and ethanol. Both substances have hydrogen bonding due to the presence of an oxygen-hydrogen bond. However, ethanol has a larger molar mass and a longer carbon chain compared to methanol, which results in stronger van der Waals forces. Since the hydrogen bonding strength in both substances is similar, we can compare their volatilities based on their van der Waals forces. As ethanol has stronger van der Waals forces, it has stronger overall intermolecular forces than methanol. Thus, we can conclude that methanol is expected to be more volatile than ethanol at a particular temperature.

Key concepts

Intermolecular Forces

Intermolecular forces are the attractions between molecules that influence the physical properties of substances, such as boiling and melting points, as well as volatility. Volatility refers to how easily a substance vaporizes; substances with weaker intermolecular forces are generally more volatile because they require less energy to transition into a gaseous state. There are several types of intermolecular forces:

• Van der Waals Forces – These include London dispersion forces and dipole-dipole interactions, and are generally weaker than other types of intermolecular forces.
• Hydrogen Bonding – A particularly strong type of intermolecular force that occurs when hydrogen is covalently bonded to a highly electronegative element, such as oxygen, nitrogen, or fluorine.

By assessing the types of intermolecular forces present in a substance, we can predict its volatility. Substances with strong hydrogen bonds, for example, exhibit lower volatility due to stronger molecular attractions.

Hydrogen Bonding

Hydrogen bonding is a strong type of dipole-dipole interaction found in molecules where hydrogen is bonded to highly electronegative atoms like oxygen, nitrogen, or fluorine. This results in significant partial charges, causing strong attractions between molecules. An example is water (\(\mathrm{H}_{2}\mathrm{O}\)), which forms hydrogen bonds between the oxygen atom of one molecule and the hydrogen atoms of neighboring molecules.
A high number of hydrogen bonds leads to higher boiling points and lower volatility, as seen with water compared to hydrogen sulfide (\(\mathrm{H}_{2}\mathrm{S}\)), which has weaker hydrogen bonding due to sulfur's lower electronegativity. The presence and strength of hydrogen bonds are critical to understanding the molecular behavior and physical properties of substances.

Van der Waals Forces

Van der Waals forces are weaker intermolecular forces that include attractions like London dispersion forces and dipole-dipole interactions. These forces occur due to temporary or permanent dipoles in molecules.
- **London Dispersion Forces:** Present in all molecules, particularly significant in nonpolar compounds. They arise from momentary charges created by electron movement. - **Dipole-Dipole Interactions:** Occur between polar molecules with permanent dipole moments due to differences in electronegativity.
Van der Waals forces are generally weaker than hydrogen bonds, but play a significant role in contributing to the overall intermolecular forces within a substance. They become more pronounced as the molar mass increases or in longer hydrocarbon chains, influencing physical properties like boiling points and volatility.

Molar Mass

Molar mass is the mass of one mole of a substance's molecules and is crucial in assessing the strength of van der Waals forces. Higher molar mass often means stronger dispersion forces due to increased electron cloud sizes and polarizability.
This is evident when comparing methanol (\(\mathrm{CH}_{3}\mathrm{OH}\)) and ethanol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\)). Despite both being alcohols, ethanol has a higher molar mass with an added carbon and hydrogen, enhancing van der Waals forces compared to methanol.
Thus, a higher molar mass usually leads to stronger intermolecular forces, reducing volatility and increasing boiling points, though it is always essential to consider other forces like hydrogen bonding when predicting physical properties.

Dipole-Dipole Interactions

Dipole-dipole interactions occur between polar molecules where partial positive and negative charges attract each other. These interactions are stronger than dispersion forces but weaker than hydrogen bonds.
Consider water (\(\mathrm{H}_{2}\mathrm{O}\)) and hydrogen sulfide (\(\mathrm{H}_{2}\mathrm{S}\)). Water has strong hydrogen bonds due to dipole-dipole interactions enhanced by its polar O-H bonds, leading to less volatility. Hydrogen sulfide has weaker dipole-dipole interactions because sulfur is less electronegative than oxygen, decreasing the overall intermolecular attraction and making \(\mathrm{H}_{2}\mathrm{S}\) more volatile.
Dipole-dipole interactions are critical in the assessment of molecular volatility and boiling points, especially in polar substances where such forces are more significant.