Question

The table shown here lists the molar heats of vaporization for several organic compounds. Use specific examples from this list to illustrate how the heat of vaporization varies with (a) molar mass, (b) molecular shape, (c) molecular polarity, (d) hydrogen-bonding interactions. Explain these comparisons in terms of the nature of the intermolecular forces at work. (You may find it helpful to draw out the structural formula for each compound.)

Short answer

In summary, the heat of vaporization is affected by various factors such as molar mass, molecular shape, molecular polarity, and hydrogen-bonding interactions. For example, Octane has a higher heat of vaporization than Methane due to its larger molar mass, Propane has a higher heat of vaporization than Cyclopropane due to its linear shape, Dichloromethane has a higher heat of vaporization than Carbon Tetrachloride due to its molecular polarity, and Water has a higher heat of vaporization than Hydrogen Sulfide due to its hydrogen-bonding interactions. These differences can be explained by the nature of intermolecular forces present in each compound.

Step-by-step solution

To illustrate the effect of molar mass on the heat of vaporization, let's consider two compounds, one with a smaller molar mass and another with a larger molar mass. For instance, Methane (CH₄) and Octane (C₈H₁₈) can be considered. The molar mass of Methane is around 16 g/mol, and for Octane, it's around 114 g/mol. Generally, as the molar mass increases, the heat of vaporization should also increase. This is because larger molecules have more electrons, leading to stronger London dispersion forces. These forces must be overcome for a substance to vaporize, requiring more energy. Therefore, the heat of vaporization for Octane should be higher than Methane. #Step 2: Molecular Shape #

To illustrate the effect of molecular shape on the heat of vaporization, let's consider two compounds with different shapes: Propane (C₃H₈) and Cyclopropane (C₃H₆). Propane is linear, while Cyclopropane has a ring structure. Molecules with a long linear structure tend to have more extensive contact with neighboring molecules. This increased contact leads to stronger London dispersion forces requiring more energy to overcome those forces. Thus, linear Propane should have a higher heat of vaporization than the ring-shaped Cyclopropane. #Step 3: Molecular Polarity #

To illustrate the effect of molecular polarity on the heat of vaporization, let's consider two compounds: Carbon Tetrachloride (CCl₄) and Dichloromethane (CH₂Cl₂). Carbon Tetrachloride is a nonpolar molecule, while Dichloromethane is polar. Polar molecules have a positive and a negative side, causing electric forces (dipole-dipole interactions) between these molecules. These interactions are stronger than the London dispersion forces present in the nonpolar molecules. As a result, it requires more energy to break these stronger forces, making the heat of vaporization higher for polar Dichloromethane than for nonpolar Carbon Tetrachloride. #Step 4: Hydrogen-Bonding Interactions #

To illustrate the effect of hydrogen-bonding interactions on the heat of vaporization, let's consider two compounds: Water (H₂O) and Hydrogen Sulfide (H₂S). Water forms hydrogen bonds as it has highly electronegative oxygen atoms, while Hydrogen Sulfide doesn't form hydrogen bonds due to its less electronegative sulfur atoms. Hydrogen bonding is a strong intermolecular force compared to others like London dispersion or dipole-dipole interactions. This stronger force requires more energy to be broken, so the heat of vaporization will be higher for water than for hydrogen sulfide, even though water has a smaller molar mass than Hydrogen Sulfide.

Key concepts

Molar Mass

When diving into the concept of molar mass and its effect on the heat of vaporization, it's fundamental to understand that molar mass is essentially the mass of one mole of a substance, measured in grams per mole (g/mol). It embodies the collective mass of the molecule's atoms.

For a tangible comparison, consider Methane (CH₄) with a low molar mass versus Octane (C₈H₁₈) with a relatively high molar mass. The molar mass of Methane is about 16 g/mol, whereas Octane has a molar mass of approximately 114 g/mol. This significant difference illustrates a general trend; the greater the molar mass, the higher the heat of vaporization.

Why does this happen? Larger molecules, like Octane, have a higher electron count, leading to stronger London dispersion forces between molecules. These forces are a type of intermolecular force that increases with the number of electrons. To vaporize a substance, these dispensation forces must be overcome, which requires more heat for substances with a higher molar mass.

Molecular Shape

Molecular shape also plays a pivotal role in the heat of vaporization. The shape affects how molecules pack together and, ultimately, the strength of the intermolecular forces between them.

Consider Propane (C₃H₈), with its extended linear structure, versus Cyclopropane (C₃H₆), which has a cyclic shape. Because of its straight-chain structure, Propane molecules can align closely, maximizing the area of contact and leading to stronger London dispersion forces. Cyclopropane, on the other hand, being ring-shaped, cannot pack as efficiently.

As a result, the linearly shaped Propane exhibits a higher heat of vaporization than Cyclopropane. Despite having similar molecular formulas, their differing shapes result in varying strengths of intermolecular attractions, affecting the required heat for vaporization.

Molecular Polarity

Molecular polarity implies that the molecule has an uneven distribution of electron density, creating distinct positive and negative regions. This leads to dipole-dipole interactions, where the positive side of one polar molecule attracts the negative side of another.

For instance, Carbon Tetrachloride (CCl₄) is a nonpolar molecule with evenly distributed electron density. Contrastingly, Dichloromethane (CH₂Cl₂) is polar. The polarity of Dichloromethane introduces dipole-dipole interactions which are absent in nonpolar molecules like Carbon Tetrachloride. These dipole-dipole forces add to the intermolecular forces' strength and require more energy—in the form of heat—to overcome. Therefore, Dichloromethane demonstrates a greater heat of vaporization than Carbon Tetrachloride.

This example shows why polar substances often have higher heats of vaporization compared to their nonpolar counterparts, directly attributing to the strength of their intermolecular attractions.

Hydrogen-Bonding Interactions

Hydrogen bonds are a particularly strong type of dipole-dipole interaction observed when hydrogen is bonded to a highly electronegative atom such as nitrogen, oxygen, or fluorine. This interaction greatly affects the heat required for a substance to vaporize.

Take the comparison of Water (H₂O) and Hydrogen Sulfide (H₂S); water is capable of hydrogen bonding due to its oxygen atoms, which are highly electronegative. On the other hand, Hydrogen Sulfide does not form hydrogen bonds as its sulfur atoms are less electronegative. Despite having a lower molar mass, water's ability to form hydrogen bonds means it has a significantly higher heat of vaporization than Hydrogen Sulfide.

This example reflects how hydrogen bonds, even in small quantities, substantially increase a molecule's intermolecular forces, leading to correspondingly higher heat requirements for vaporization.

Intermolecular Forces

Intertwined with the previous concepts is the overarching category of intermolecular forces. These forces are what keep molecules together in a liquid state, and they must be overcome to convert the liquid into a vapor. There are several types of intermolecular forces, such as London dispersion, dipole-dipole, and hydrogen bonding, each progressively stronger than the last.

Substances with primarily London dispersion forces, typically nonpolar molecules, usually require less heat to vaporize than those with dipole-dipole interactions. In turn, molecules with hydrogen bonds require even more heat because hydrogen bonds are significantly stronger.

The complexity and strength of these forces determine the heat of vaporization and explain why substances with seemingly similar molar masses can have drastically different heat requirements to transition from liquid to gas. Understanding these forces and the factors that affect them is essential for students studying physical properties of substances.