Problem 51 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) Place the following substances in order of increasing volatility: \(\mathrm{CH}_{4}, \mathrm{CBr}_{4}, \mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CH}_{3} \mathrm{Cl}, \mathrm{CHBr}_{3},\) and \(\mathrm{CH}_{2} \mathrm{Br}_{2}\). (b) How do the boiling points vary through this series? (c) Explain your answer to part (b) in terms of intermolecular forces.

Step-by-Step Solution

Answer

The order of increasing volatility for the given substances is: \(\mathrm{CBr}_{4} < \mathrm{CHBr}_{3} < \mathrm{CH}_{2} \mathrm{Br}_{2} < \mathrm{CH}_{2} \mathrm{Cl}_{2} < \mathrm{CH}_{3} \mathrm{Cl} < \mathrm{CH}_{4}\) The boiling points decrease as the volatility increases: \(\mathrm{CBr}_{4} > \mathrm{CHBr}_{3} > \mathrm{CH}_{2} \mathrm{Br}_{2} > \mathrm{CH}_{2} \mathrm{Cl}_{2} > \mathrm{CH}_{3} \mathrm{Cl} > \mathrm{CH}_{4}\) This trend can be explained by the presence and strength of intermolecular forces (London dispersion forces and dipole-dipole forces). Stronger intermolecular forces lead to higher boiling points. The varying boiling points are due to the differences in molecular mass and the type of intermolecular forces present.

Step 1: Determine molecular mass

First, we need to determine the molecular mass for each of the substances. Since volatility is affected by molecular mass, this will help us determine the substances' volatility order. - \(\mathrm{CH}_{4}\): \(12.01 + (4 \times 1.01) = 16.05 \, g/mol\) - \(\mathrm{CBr}_{4}\): \(12.01+(4 \times 79.904) = 331.73 \, g/mol\) - \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\): \(12.01+ (2 \times 1.01) + (2 \times 35.453) = 84.93 \, g/mol\) - \(\mathrm{CH}_{3} \mathrm{Cl}\): (\(12.01+(3 \times 1.01)+35.453 = 50.48 \, g/mol\) - \(\mathrm{CHBr}_{3}\): \(12.01+(3 \times 79.904) + 1.01= 252.74 \, g/mol\) - \(\mathrm{CH}_{2} \mathrm{Br}_{2}\): \(12.01+(2 \times 1.01)+(2 \times 79.904) = 173.84 \, g/mol\)

Step 2: Determine the type of intermolecular forces

For each substance, we will determine the type of intermolecular force present. There are mainly three types of intermolecular forces: London dispersion forces (LDF), dipole-dipole forces, and hydrogen bonding. All substances have LDF, but the strength might vary. We will also consider other types of forces if applicable and use them to estimate boiling points and volatility. - \(\mathrm{CH}_{4}\): London dispersion forces - \(\mathrm{CBr}_{4}\): London dispersion forces - \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\): London dispersion forces and dipole-dipole forces - \(\mathrm{CH}_{3} \mathrm{Cl}\): London dispersion forces and dipole-dipole forces - \(\mathrm{CHBr}_{3}\): London dispersion forces and dipole-dipole forces - \(\mathrm{CH}_{2} \mathrm{Br}_{2}\): London dispersion forces and dipole-dipole forces

Step 3: Place the substances in order of increasing volatility

Since larger molecules with higher molecular mass have stronger LDFs, substances with greater molecular mass will have lower volatility. Similarly, dipole-dipole forces will also lead to lower volatility as they are stronger than LDFs. Considering these factors, we can place the given substances in the following order of increasing volatility: \(\mathrm{CBr}_{4} < \mathrm{CHBr}_{3} < \mathrm{CH}_{2} \mathrm{Br}_{2} < \mathrm{CH}_{2} \mathrm{Cl}_{2} < \mathrm{CH}_{3} \mathrm{Cl} < \mathrm{CH}_{4}\)

Step 4: Discuss how the boiling points vary through this series

As the volatility of the substances increases, their boiling points are expected to decrease. The reason being, more volatile substances escape to the gas phase more easily, requiring less energy (i.e., lower boiling point) to boil. With this in mind, the boiling points would decrease in the following order: \(\mathrm{CBr}_{4} > \mathrm{CHBr}_{3} > \mathrm{CH}_{2} \mathrm{Br}_{2} > \mathrm{CH}_{2} \mathrm{Cl}_{2} > \mathrm{CH}_{3} \mathrm{Cl} > \mathrm{CH}_{4}\)

Step 5: Explain the answer in terms of intermolecular forces

The varying boiling points in this series can be explained by the presence and strength of their intermolecular forces. London dispersion forces are present in all of the substances, but their strength increases with increasing molecular mass. Additionally, the substances that also possess dipole-dipole forces result in stronger intermolecular forces leading to higher boiling points. In this series, the strongest intermolecular forces are found in \(\mathrm{CBr}_{4}\) due to its high molecular mass, causing the highest boiling point. On the other hand, \(\mathrm{CH}_{4}\) has the weakest intermolecular forces (only LDF present) and the lowest molecular mass, hence it has the lowest boiling point. The boiling points of other substances can be similarly explained based on the molecular mass and the type of intermolecular forces present.

Key Concepts

Volatility

Volatility refers to how easily a substance can convert from a liquid to a vapor, or in other words, how readily a substance evaporates. A volatile substance will quickly become a gas, requiring less energy to do so.

• The weaker the intermolecular forces, the higher the volatility.
• Substances with low molecular masses tend to be more volatile.
• Dipole-dipole interactions and hydrogen bonding can significantly reduce volatility.

In the given exercise, methane (\(\mathrm{CH}_{4}\)) is the most volatile due to its small size and only having weaker London dispersion forces. On the other hand, \(\mathrm{CBr}_{4}\) is the least volatile as it has a larger molecular mass and stronger London dispersion forces.

Boiling Points

The boiling point of a substance is the temperature at which it changes from a liquid to a gas. This phase change requires the breaking of intermolecular forces. Substances with strong intermolecular forces require more energy (higher temperature) to boil, resulting in higher boiling points. As volatility increases, the boiling point typically decreases for the following reasons:

• Increased volatility means a substance already seeks to vaporize easily, thus lowering the energy needed to overcome intermolecular forces.
• The presence and strength of intermolecular forces directly influence boiling points.

In this exercise, \(\mathrm{CBr}_{4}\) has the highest boiling point due to its strong London dispersion forces, whereas \(\mathrm{CH}_{4}\) has the lowest boiling point because it has weak London dispersion forces.

Molecular Mass

Molecular mass plays a critical role in determining both volatility and boiling points. It is the total mass of all atoms in a molecule and is usually expressed in grams per mole (g/mol). Factors to consider:

• Higher molecular mass typically implies stronger London dispersion forces.
• Strong intermolecular attractions due to higher molecular mass can result in higher boiling points and lower volatility.

From the original exercise, substances with higher molecular masses like \(\mathrm{CBr}_{4}\) have greater London dispersion forces, leading to higher boiling points and lower volatility. Conversely, \(\mathrm{CH}_{4}\), with the smallest molecular mass in this series, exhibits the weakest forces and the highest volatility.