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.) $$ \begin{array}{ll} \text { Compound } & \begin{array}{l} \text { Heat of Vaporization } \\ \mathbf{( k J / m o l )} \end{array} \\ \hline \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3} & 19.0 \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} & 27.6 \\ \mathrm{CH}_{3} \mathrm{CHBrCH}_{3} & 31.8 \\ \mathrm{CH}_{3} \mathrm{COCH}_{3} & 32.0 \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br} & 33.6 \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 47.3 \end{array} $$

Short answer

The heat of vaporization increases as molar mass increases: CH3CH2CH3 (19.0 kJ/mol) < CH3CH2CH2CH2CH3 (27.6 kJ/mol). No clear trend is seen with molecular shape. Polar compounds have higher heat of vaporization: CH3CHBrCH3 (31.8 kJ/mol) < CH3COCH3 (32.0 kJ/mol) < CH3CH2CH2Br (33.6 kJ/mol). Heat of vaporization is highest for compounds with hydrogen bonding: CH3CH2CH2OH (47.3 kJ/mol). Larger molecule or stronger intermolecular forces (polarity, hydrogen bonding) lead to higher heat of vaporization due to increased resistance to phase change from liquid to vapor.

Step-by-step solution

(Step 1: Identify trends in heat of vaporization with molar mass)

Compare the molar mass of each compound and its corresponding heat of vaporization. Arrange the molecules in the increasing order of their molar mass and look for any clear patterns or trends.

(Step 2: Identify trends in heat of vaporization with molecular shape)

Examine the structure of each compound and identify its molecular shape. Compare it with other molecules in the list and check if there is any pattern in heat of vaporization based on molecular shape.

(Step 3: Identify trends in heat of vaporization with molecular polarity)

Determine the molecular polarity of each compound using their molecular structure. Molecules with higher polarities tend to have stronger intermolecular forces. Compare the heat of vaporization for polar and non-polar compounds.

(Step 4: Identify trends in heat of vaporization with hydrogen-bonding interactions)

Identify whether each compound can form hydrogen bonds or not. Strong hydrogen bonds can increase intermolecular forces and ultimately the heat of vaporization. Compare the heat of vaporization for compounds with hydrogen bonding and those without hydrogen bonding.

(Step 5: Explain the comparisons in terms of intermolecular forces)

Analyze the trends observed in the heat of vaporization with respect to molar mass, molecular shape, molecular polarity, and hydrogen-bonding interactions, and provide a detailed explanation for these trends in terms of the intermolecular forces at work.

Key concepts

Heat of Vaporization

The heat of vaporization is an important concept when discussing the phase change from liquid to gas. It represents the amount of energy required to convert one mole of a liquid into a gas at constant temperature and pressure. This process involves breaking the intermolecular forces holding the liquid molecules together. Several factors can affect the heat of vaporization:

• Molar Mass: Heavier molecules may require more energy to vaporize because of stronger dispersion forces.
• Molecular Polarity: Polar molecules exhibit dipole interactions, increasing the heat of vaporization.
• Hydrogen Bonding: This strong intermolecular force significantly raises the heat of vaporization for compounds that can form hydrogen bonds.

Understanding the interplay of these forces helps explain the differences in the heat of vaporization among various substances.

Molar Mass

Molar mass is directly related to the heat of vaporization in many substances. Generally, as the molar mass of a compound increases, its heat of vaporization also increases. This occurs because heavier molecules tend to have greater London dispersion forces. Consider the compounds given in the exercise:

• Propane (\( ext{CH}_3 ext{CH}_2 ext{CH}_3 \)) has a lower molar mass and a heat of vaporization of 19.0 kJ/mol.
• Pentane (\( ext{CH}_3 ext{CH}_2 ext{CH}_2 ext{CH}_2 ext{CH}_3 \)), with a higher molar mass, requires more energy at 27.6 kJ/mol.

This trend highlights how larger molecules increase intermolecular attractions, demanding more energy to break apart during vaporization.

Molecular Polarity

Molecular polarity refers to the distribution of electrical charge over the atoms in a molecule. Polar molecules, like acetone (\( ext{CH}_3 ext{COCH}_3 \)), have unequal sharing of electrons, resulting in a dipole moment. Such molecules have stronger intermolecular forces like dipole-dipole interactions. Compare the following compounds to see this effect:

• Acetone (\( ext{CH}_3 ext{COCH}_3 \)) has a higher heat of vaporization (32.0 kJ/mol) due to its polarity.
• Nonpolar propane (\( ext{CH}_3 ext{CH}_2 ext{CH}_3 \)), displays weaker intermolecular forces and thus lower energy requirements at 19.0 kJ/mol.

This illustrates that increased molecular polarity enhances intermolecular attractions, demanding more energy for phase changes.

Hydrogen Bonding

Hydrogen bonding is one of the strongest types of intermolecular forces and occurs when hydrogen is bonded to more electronegative atoms like oxygen, nitrogen, or fluorine. For example, look at:

• 1-Propanol (\( ext{CH}_3 ext{CH}_2 ext{CH}_2 ext{OH} \)) has a high heat of vaporization (47.3 kJ/mol) due to hydrogen bonds formed by the hydroxyl group.
• In contrast, 1-bromopropane (\( ext{CH}_3 ext{CH}_2 ext{CH}_2 ext{Br} \)) is unable to hydrogen bond, resulting in a lower heat of vaporization at 33.6 kJ/mol.

Hydrogen-bonding substantially increases the energy needed for vaporization, showcasing its impact on molecular interactions.