Question

In each of the following groups of substances, pick the one that has the given property. Justify your answer. a. highest boiling point: $\mathrm{HBr},\mathrm{Kr}$, or ${\mathrm{Cl}}_2$ b. highest freezing point: ${\mathrm{H}}_2\mathrm{O},\mathrm{NaCl}$, or HF c. lowest vapor pressure at ${25}^{\circ }\mathrm{C}:{\mathrm{Cl}}_2,{\mathrm{Br}}_2$, or ${\mathrm{I}}_2$ d. lowest freezing point: ${\mathrm{N}}_2,\mathrm{CO}$, or ${\mathrm{CO}}_2$ e. lowest boiling point: ${\mathrm{CH}}_4,{\mathrm{CH}}_3{\mathrm{CH}}_3$, or ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_3$ f. highest boiling point: HF, HCl, or HBr CC=O g. lowest vapor pressure at ${25}^{\circ }\mathrm{C}:{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_3,{\mathrm{CH}}_3{\mathrm{CCH}}_3$, or ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}$

Short answer

a. $\mathrm{HBr}$ has the highest boiling point due to the presence of hydrogen bonds which are stronger than the dispersion forces in $\mathrm{Kr}$ and ${\mathrm{Cl}}_2$. b. $\mathrm{NaCl}$ has the highest freezing point because its ionic bonds are stronger than the hydrogen bonds in ${\mathrm{H}}_2\mathrm{O}$ and HF. c. ${\mathrm{I}}_2$ has the lowest vapor pressure at ${25}^{\circ }\mathrm{C}$ due to its stronger dispersion forces as a result of its larger molecular size. d. ${\mathrm{N}}_2$ has the lowest freezing point because it has weaker dispersion forces compared to the other substances in the group. e. ${\mathrm{CH}}_4$ has the lowest boiling point due to weaker dispersion forces resulting from its smaller molecular size. f. HF has the highest boiling point because it has the strongest hydrogen bonds, as fluorine is the most electronegative element. g. ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}$ has the lowest vapor pressure at ${25}^{\circ }\mathrm{C}$ because of the stronger hydrogen bonds in comparison to the dispersion forces in the other substances in the group.

Step-by-step solution

Problem a - Highest Boiling Point

For the group ($\mathrm{HBr},\mathrm{Kr}$, or ${\mathrm{Cl}}_2$), we need to determine the substance with the highest boiling point. Boiling points depend on the type and strength of intermolecular forces present in the given substances. The stronger the intermolecular forces, the higher the boiling point. $\mathrm{HBr}$ has a hydrogen bond between hydrogen and bromine atoms, $\mathrm{Kr}$ exhibits London dispersion forces, and ${\mathrm{Cl}}_2$ exhibits London dispersion forces as well. Hydrogen bonds are stronger than dispersion forces, therefore $\mathrm{HBr}$ has the highest boiling point among these substances.

Problem b - Highest Freezing Point

For the group (${\mathrm{H}}_2\mathrm{O},\mathrm{NaCl}$, or HF), we need to determine the substance with the highest freezing point. Freezing points also depend on the strength of intermolecular forces. ${\mathrm{H}}_2\mathrm{O}$ has hydrogen bonds, $\mathrm{NaCl}$ is an ionic compound with electrostatic attraction, and HF has hydrogen bonds. Among hydrogen bonds and ionic bonds, ionic bonds are stronger. Therefore, $\mathrm{NaCl}$ has the highest freezing point among these substances.

Problem c - Lowest Vapor Pressure at ${25}^{\circ }\mathrm{C}$

For the group (${\mathrm{Cl}}_2,{\mathrm{Br}}_2$, or ${\mathrm{I}}_2$), we need to determine the substance with the lowest vapor pressure at ${25}^{\circ }\mathrm{C}$. Vapor pressure depends on the strength of intermolecular forces, with stronger forces resulting in lower vapor pressure. All three substances, ${\mathrm{Cl}}_2$, ${\mathrm{Br}}_2$, and ${\mathrm{I}}_2$ exhibit London dispersion forces. These forces increase with increasing molecular size. Since ${\mathrm{I}}_2$ has the largest molecular size, it has the strongest dispersion forces and therefore the lowest vapor pressure at ${25}^{\circ }\mathrm{C}$.

Problem d - Lowest Freezing Point

For the group (${\mathrm{N}}_2,\mathrm{CO}$, or ${\mathrm{CO}}_2$), we need to determine the substance with the lowest freezing point. ${\mathrm{N}}_2$ exhibits London dispersion forces, $\mathrm{CO}$ has dipole-dipole interactions, and ${\mathrm{CO}}_2$ has London dispersion forces. Dispersion forces are weaker than dipole-dipole interactions. Among the two substances with dispersion forces, ${\mathrm{N}}_2$ has a smaller molecular size which results in weaker dispersion forces. Therefore, ${\mathrm{N}}_2$ has the lowest freezing point.

Problem e - Lowest Boiling Point

For the group (${\mathrm{CH}}_4,{\mathrm{CH}}_3{\mathrm{CH}}_3$, or ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_3$), we need to determine the substance with the lowest boiling point. All three substances exhibit London dispersion forces. These forces increase with increasing molecular size. Since ${\mathrm{CH}}_4$ has the smallest molecular size, it has the weakest dispersion forces and therefore the lowest boiling point.

Problem f - Highest Boiling Point

For the group (HF, HCl, or HBr), we need to determine the substance with the highest boiling point. All three substances exhibit hydrogen bonds, with the strength of hydrogen bonds increasing with increasing electronegativity. Since fluorine is the most electronegative element, HF has the strongest hydrogen bonds and therefore the highest boiling point.

Problem g - Lowest Vapor Pressure at ${25}^{\circ }\mathrm{C}$

For the group (${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_3,{\mathrm{CH}}_3{\mathrm{CCH}}_3$, or ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}$), we need to determine the substance with the lowest vapor pressure at ${25}^{\circ }\mathrm{C}$. ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_3$ and ${\mathrm{CH}}_3{\mathrm{CCH}}_3$ exhibit London dispersion forces, while ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}$ exhibits hydrogen bonds due to the -OH group. Hydrogen bonds are stronger than dispersion forces, resulting in lower vapor pressure. Therefore, ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}$ has the lowest vapor pressure at ${25}^{\circ }\mathrm{C}$.

Key concepts

Understanding Boiling Point

The boiling point of a substance is the temperature at which it changes from a liquid to a gas, which occurs when vapor pressure equals atmospheric pressure. Substances with stronger intermolecular forces typically have higher boiling points because more energy is required to overcome these forces.

For instance, in the exercise, the boiling point of substances like HBr, Kr, and Cl₂ are compared. The presence of hydrogen bonding in HBr increases its boiling point compared to Kr and Cl₂, which rely on weaker London dispersion forces. London dispersion forces are present in all molecules but are the only type of intermolecular force in nonpolar molecules like Kr and Cl₂.

Role of Molecular Size and Electronegativity

Furthermore, the size of the molecules and electronegativity play a role in the strength of intermolecular forces. Larger molecules generally have greater dispersion forces, and elements with higher electronegativity form stronger hydrogen bonds, influencing the boiling point.

Deciphering Freezing Point

The freezing point is the temperature at which a liquid turns into a solid. The principles dictating freezing points are similar to those for boiling points, where intermolecular forces are key players. A substance with stronger intermolecular forces will have a higher freezing point because more energy must be removed for the molecules to form a rigid structure.

As seen in the original exercise, NaCl has a higher freezing point than H₂O and HF because ionic bonds in NaCl, which arise from electrostatic attractions between oppositely charged ions, are generally much stronger than hydrogen bonds. While H₂O and HF do form hydrogen bonds, the strength of ionic forces in NaCl outweighs them, leading to its higher freezing point.

Impact of Compound Type

It's important to note that the type of compound, whether ionic or molecular, also affects the freezing point. Ionic compounds typically have higher melting and freezing points due to the strong electrostatic forces between ions in the crystalline lattice structure.

Vapor Pressure's Role in Phase Changes

The vapor pressure of a substance indicates how much the substance tends to evaporate. At higher temperatures, a substance's vapor pressure increases as more molecules have enough energy to escape into the gas phase. A lower vapor pressure at a given temperature suggests that fewer molecules are escaping, which is usually due to stronger intermolecular forces within the substance.

In comparing Cl₂, Br₂, and I₂, I₂, with larger molecules, has stronger London dispersion forces and thus a lower vapor pressure as observed in the exercise. This phenomenon happens because it is more difficult for molecules to leave the liquid phase when they are held together by stronger forces.

Molecules in Action

For substances with similar molecular structures, differences in molecular mass can affect vapor pressure. Heavier molecules often have lower vapor pressures, due primarily to increased dispersion forces resulting from more electrons and larger surface areas. This is significant in understanding how various substances will behave under the same conditions.