Question

In each of the following groups of substances, pick the one that has the given property. Justify each answer. a. highest boiling point: \(\mathrm{CCl}_{4}, \mathrm{CF}_{4}, \mathrm{CBr}_{4}\) b. lowest freezing point: \(\mathrm{LiF}, \mathrm{F}_{2}, \mathrm{HCl}\) c. smallest vapor pressure at \(25^{\circ} \mathrm{C}: \mathrm{CH}_{3} \mathrm{OCH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) d. greatest viscosity: \(\mathrm{H}_{2} \mathrm{S}, \mathrm{HF}, \mathrm{H}_{2} \mathrm{O}_{2}\) e. greatest heat of vaporization: \(\mathrm{H}_{2} \mathrm{CO}, \mathrm{CH}_{3} \mathrm{CH}_{3}, \mathrm{CH}_{4}\) f. smallest enthalpy of fusion: \(\mathrm{I}_{2}, \mathrm{CsBr}, \mathrm{CaO}\)

Short answer

a) \(\mathrm{CBr}_{4}\) has the highest boiling point due to its larger size and stronger dispersion forces compared to \(\mathrm{CCl}_{4}\) and \(\mathrm{CF}_{4}\). b) \(\mathrm{F}_{2}\) has the lowest freezing point due to its weak intermolecular forces (dispersion forces) compared to \(\mathrm{LiF}\) with ionic bonds and \(\mathrm{HCl}\) with hydrogen bonds. c) \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\) has the smallest vapor pressure at 25°C due to its strong intermolecular forces (hydrogen bonding) compared to \(\mathrm{CH}_{3}\mathrm{OCH}_{3}\) with dipole-dipole forces and \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{3}\) with dispersion forces. d) \(\mathrm{H}_{2}\mathrm{O}_{2}\) has the greatest viscosity due to its hydrogen bonding compared to \(\mathrm{HF}\) with dipole-dipole forces and \(\mathrm{H}_{2}\mathrm{S}\) with dispersion forces. e) \(\mathrm{H}_{2}\mathrm{CO}\) has the greatest heat of vaporization due to its strong dipole-dipole forces compared to \(\mathrm{CH}_{3}\mathrm{CH}_{3}\) and \(\mathrm{CH}_{4}\), which have weak dispersion forces. f) \(\mathrm{I}_{2}\) has the smallest enthalpy of fusion due to its weak dispersion forces compared to \(\mathrm{CsBr}\) and \(\mathrm{CaO}\) with strong ionic bonds.

Step-by-step solution

a) Highest boiling point

For the given group of substances, we have to compare the boiling points of \(\mathrm{CCl}_{4}, \mathrm{CF}_{4}\), and \(\mathrm{CBr}_{4}\). The boiling point mostly depends on the strength of intermolecular forces. Since all of these are nonpolar molecules, the strength of dispersion forces acts. Dispersion forces are an increase in size and mass. In these substances, we have the order of atomic masses: Br > Cl > F. So, we have the order of boiling points as: \(\mathrm{CBr}_{4} > \mathrm{CCl}_{4} > \mathrm{CF}_{4}\). Therefore, \(\mathrm{CBr}_{4}\) has the highest boiling point.

b) Lowest freezing point

For the given group of substances, we have to compare the freezing points of \(\mathrm{LiF}, \mathrm{F}_{2}\), and \(\mathrm{HCl}\). Freezing points depend on the intermolecular forces between molecules. Lithium fluoride (LiF) is an ionic substance with strong ionic bonds, fluorine gas (F2) consists of non-polar molecules with weak dispersion forces, and hydrogen chloride (HCl) is a polar molecule with hydrogen bonds. The strength of these forces in order is: ionic bonds > hydrogen bonds > dispersion forces. So, we have the order of freezing points as: \(\mathrm{LiF} > \mathrm{HCl} > \mathrm{F}_{2}\). Therefore, \(\mathrm{F}_{2}\) has the lowest freezing point.

c) Smallest vapor pressure at 25°C

For the given group of substances, we have to compare the vapor pressures of \(\mathrm{CH}_{3} \mathrm{OCH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\), and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) at \[25^{\circ}C\]. Vapor pressure depends on the intermolecular forces between molecules. Dimethyl ether (\(\mathrm{CH}_{3} \mathrm{OCH}_{3}\)) is a polar molecule, ethanol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\)) has hydrogen bonding, and propane (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{3}\)) is nonpolar with weak dispersion forces. The smaller the intermolecular forces, the higher the vapor pressure. So, we have the order of vapor pressures as: \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{3} > \mathrm{CH}_{3}\mathrm{OCH}_{3} > \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\). Therefore, ethanol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\)) has the smallest vapor pressure at 25°C.

d) Greatest viscosity

For the given group of substances, we have to compare the viscosities of \(\mathrm{H}_{2} \mathrm{S}, \mathrm{HF}\), and \(\mathrm{H}_{2}\mathrm{O}_{2}\). Viscosity depends on the molecular weight and hydrogen bonding. In these substances, hydrogen peroxide (\(\mathrm{H}_{2}\mathrm{O}_{2}\)) forms hydrogen bonds. The strength of these forces in order is: hydrogen bonds > dipole-dipole forces > dispersion forces. So, we have the order of viscosities as: \(\mathrm{H}_{2}\mathrm{O}_{2} > \mathrm{HF} > \mathrm{H}_{2}\mathrm{S}\). Therefore, hydrogen peroxide (\(\mathrm{H}_{2}\mathrm{O}_{2}\)) has the greatest viscosity.

e) Greatest heat of vaporization

For the given group of substances, we have to compare the heats of vaporization of \(\mathrm{H}_{2} \mathrm{CO}, \mathrm{CH}_{3}\mathrm{CH}_{3}\), and \(\mathrm{CH}_{4}\). Heat of vaporization depends on the strength of intermolecular forces. Methane (\(\mathrm{CH}_{4}\)) and ethane (\(\mathrm{CH}_{3}\mathrm{CH}_{3}\)) are both nonpolar molecules with only dispersion forces acting, whereas formaldehyde (\(\mathrm{H}_{2}\mathrm{CO}\)) is a polar molecule with dipole-dipole forces. The stronger the intermolecular forces, the higher the heat of vaporization. So, we have the order of heat of vaporizations as: \(\mathrm{H}_{2}\mathrm{CO} > \mathrm{CH}_{3}\mathrm{CH}_{3} > \mathrm{CH}_{4}\). Therefore, formaldehyde (\(\mathrm{H}_{2}\mathrm{CO}\)) has the greatest heat of vaporization.

f) Smallest enthalpy of fusion

For the given group of substances, we have to compare the enthalpies of fusion of \(\mathrm{I}_{2}, \mathrm{CsBr}\), and \(\mathrm{CaO}\). Enthalpy of fusion depends on the strength of intermolecular forces. Iodine (\(\mathrm{I}_{2}\)) has weak dispersion forces, cesium bromide (\(\mathrm{CsBr}\)) is an ionic compound with strong ionic bonds, and calcium oxide (\(\mathrm{CaO}\)) also has ionic bonds. The stronger the forces, the higher the enthalpy of fusion. So, we have the order of enthalpies of fusion as: \(\mathrm{CaO} > \mathrm{CsBr} > \mathrm{I}_{2}\). Therefore, iodine (\(\mathrm{I}_{2}\)) has the smallest enthalpy of fusion.

Key concepts

Boiling Point

The boiling point is the temperature at which a liquid turns into a vapor. It is deeply influenced by the intermolecular forces present within the substance. Stronger intermolecular forces require more energy (heat) to break, thus raising the boiling point.

• Nonpolar molecules like \(\mathrm{CBr}_{4}\\), \(\mathrm{CCl}_{4}\\), and \(\mathrm{CF}_{4}\\) primarily experience dispersion forces, which increase as molecular size and mass increase.
• As a result, \(\mathrm{CBr}_{4}\\) has the highest boiling point in the series due to its larger atomic mass and stronger dispersion forces compared to the other compounds.

Understanding boiling points in relation to intermolecular forces helps predict which substances will transition to vapor more readily under specific conditions.

Freezing Point

The freezing point is the temperature at which a liquid becomes solid. Like boiling points, freezing points are heavily affected by the nature and strength of intermolecular forces.

• Ionic bonds, as seen in \(\mathrm{LiF}\\), are very strong, leading to high freezing points.
• In contrast, \(\mathrm{F}_{2}\\) is composed of weak dispersion forces, hence retains the lowest freezing point among the given substances.
• Polar molecules like \(\mathrm{HCl}\\) fall in between, having intermediate freezing points due to hydrogen bonding.

By analyzing the types of bonds and molecular interactions, we can understand why certain substances freeze at lower temperatures than others.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid. Substances with weaker intermolecular forces tend to have higher vapor pressures, as molecules escape the liquid phase more easily.

• For instance, \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_3\\) exhibits weak dispersion forces, granting it the highest vapor pressure.
• Ethanol, \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}\\), capable of strong hydrogen bonding, results in a lower vapor pressure.
• Dimethyl ether falls in between due to its polar nature and the presence of dipole-dipole interactions.

Understanding vapor pressure assists in predicting how easily a substance will evaporate at a given temperature.

Viscosity

Viscosity is a measure of a liquid's resistance to flow. Substances with stronger intermolecular forces generally have higher viscosities.

• \(\mathrm{H}_2\mathrm{O}_2\\) exhibits significant hydrogen bonding, leading to a higher viscosity compared to the more moderately interacting \(\mathrm{HF}\) and weakly interacting \(\mathrm{H}_2\mathrm{S}\\).
• In simple terms, stronger attractions among molecules slow their movement, resulting in thicker or more viscous liquids.

Viscosity is an important property to consider in processes like lubrication and fluid transport.

Heat of Vaporization

Heat of vaporization is the energy required to convert a liquid into a gas at its boiling point. It is directly proportional to the strength of intermolecular forces.

• \(\mathrm{H}_2\mathrm{CO}\) requires more energy due to its dipole-dipole interactions, giving it the greatest heat of vaporization compared to nonpolar \(\mathrm{CH}_3\mathrm{CH}_3}\) and \(\mathrm{CH}_4\).
• Nonpolar compounds, with their weaker dispersion forces, have relatively low heats of vaporization.

Knowing a substance's heat of vaporization helps in understanding the energy costs associated with phase changes.

Enthalpy of Fusion

Enthalpy of fusion is the energy required to change a solid to a liquid at its melting point. It signifies how strong the internal forces are within a solid.

• Substances like \(\mathrm{CaO}\) and \(\mathrm{CsBr}\), which have strong ionic bonds, require significant energy, leading to high enthalpies of fusion.
• Conversely, weak dispersion forces in \(\mathrm{I}_2\) make it easier to melt, resulting in a low enthalpy of fusion.

Understanding this concept is essential in areas such as metallurgy and material science, where melting plays a crucial role.