Problem 52 - Chemistry: The Central Science in SI Units, Global Edition

Question

True or false: \((\mathbf{a}) \mathrm{CBr}_{4}\) is more volatile than \(\mathrm{CCl}_{4}\). (b) \(\mathrm{CBr}_{4}\) has a higher boiling point than \(\mathrm{CCl}_{4}\). (c) \(\mathrm{CBr}_{4}\) has weaker intermolecular forces than \(\mathrm{CCl}_{4}\). (d) \(\mathrm{CBr}_{4}\) has a higher vapor pressure at the same temperature than \(\mathrm{CCl}_{4}\).

Step-by-Step Solution

Answer

(a) False, (b) True, (c) False, and (d) False.

Step 1: Analyzing Intermolecular Forces

Both \(\mathrm{CCl}_{4}\) and \(\mathrm{CBr}_{4}\) are nonpolar molecules due to their tetrahedral structure and balanced distribution of electronegativity. Therefore, their main intermolecular forces are London dispersion forces. London dispersion forces depend on polarizability: the ease with which the electron cloud of an atom or molecule can be distorted. Generally, larger atoms or molecules have a more easily distorted electron cloud, leading to stronger London dispersion forces. Bromine atoms are larger and more polarizable than chlorine atoms, resulting in stronger London dispersion forces for \(\mathrm{CBr}_{4}\) compared to \(\mathrm{CCl}_{4}\). Now, let's evaluate each statement:

Step 2: Statement (a): Volatility

Volatility refers to the tendency of a substance to vaporize. A substance with stronger intermolecular forces will require more energy to break those forces and vaporize, making it less volatile. Since \(\mathrm{CBr}_{4}\) has stronger London dispersion forces than \(\mathrm{CCl}_{4}\), it is less volatile. This statement is False.

Step 3: Statement (b): Boiling Point

The boiling point is the temperature at which the vapor pressure of a liquid becomes equal to the atmospheric pressure. A higher boiling point corresponds to stronger intermolecular forces. As we have discussed, \(\mathrm{CBr}_{4}\) has stronger London dispersion forces than \(\mathrm{CCl}_{4}\). Thus, it has a higher boiling point, making this statement True.

Step 4: Statement (c): Intermolecular Forces

We have already established that \(\mathrm{CBr}_{4}\) has stronger London dispersion forces than \(\mathrm{CCl}_{4}\). This statement claims that \(\mathrm{CBr}_{4}\) has weaker intermolecular forces, which contradicts our analysis. Therefore, this statement is False.

Step 5: Statement (d): Vapor Pressure at the Same Temperature

At a given temperature, a substance with weaker intermolecular forces will have a higher vapor pressure. Since \(\mathrm{CBr}_{4}\) has stronger intermolecular forces than \(\mathrm{CCl}_{4}\), its vapor pressure at the same temperature will be lower. This statement is False. In summary, the solutions to each statement are: (a) False, (b) True, (c) False, and (d) False.

Key Concepts

London Dispersion Forces

London dispersion forces are fascinating yet often overlooked die-hard components of chemistry. They are the weakest type of intermolecular force but are present in all molecules, whether polar or nonpolar. These forces arise due to temporary fluctuations in electron distribution around an atom or molecule, creating an instantaneous dipole. These dipoles induce similar dipoles in neighboring molecules, resulting in a weak attraction.

What's important to understand about these forces is that they are stronger in larger atoms or molecules. This is because larger atoms have more electrons, and their electron clouds are more easily distorted or polarizable. This means that the bigger the electron cloud, the more pronounced the London dispersion forces are. An example is bromine, which is larger than chlorine, meaning that compounds with bromine atoms like \(\mathrm{CBr}_{4}\) have stronger London dispersion forces compared to those with chlorine atoms like \(\mathrm{CCl}_{4}\). Understanding the basics of this interaction helps in predicting other properties like boiling points and volatility.

Volatility

Volatility is a measure of how readily a substance vaporizes. Essentially, it tells us how likely a liquid is to escape into the gas phase. Substances with high volatility evaporate quickly. This characteristic is intimately connected to the strength of intermolecular forces. If the forces holding molecules together in a liquid are weak, the molecules can more easily escape into the vapor, meaning the substance is more volatile.

Considering \(\mathrm{CCl}_{4}\) and \(\mathrm{CBr}_{4}\), we notice that \(\mathrm{CBr}_{4}\) has stronger intermolecular forces due to its bigger bromine atoms which enhance the London dispersion forces. Therefore, \(\mathrm{CBr}_{4}\) is less volatile than \(\mathrm{CCl}_{4}\), as more energy is needed to overcome these stronger attractions and facilitate vaporization.

Boiling Point

The boiling point of a liquid is a crucial concept and is defined as the temperature at which the vapor pressure of the liquid matches the atmospheric pressure, causing the liquid to transform into vapor. Higher boiling points indicate stronger intermolecular forces, as more energy is required to separate the molecules enough to go from liquid to gas.

In our example, \(\mathrm{CBr}_{4}\) has a higher boiling point than \(\mathrm{CCl}_{4}\). This is due to its stronger London dispersion forces, stemming from the larger bromine atoms as compared to the smaller chlorine atoms. The greater energy requirement for \(\mathrm{CBr}_{4}\) signifies that its molecules are held together more strongly, reflecting a higher boiling point.

Vapor Pressure

Vapor pressure is a property that tells us how much a liquid is tending to become a gas. When a liquid is in a closed container, molecules at the surface can escape into the vapor phase. The vapor pressure is the pressure created by these vaporized molecules in equilibrium with their liquid form.

An essential detail regarding vapor pressure is that it tends to be lower for substances with strong intermolecular forces, because fewer molecules have the energy to overcome these forces to escape into the vapor phase. For instance, given that \(\mathrm{CBr}_{4}\) has stronger London dispersion forces, its vapor pressure will be lower than that of \(\mathrm{CCl}_{4}\) at the same temperature. This means that \(\mathrm{CBr}_{4}\) is less inclined to evaporate than \(\mathrm{CCl}_{4}\) under the same conditions.