Problem 92 - Chemistry The Central Science In Si Units

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, \((\mathbf{c})\) 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{tabular}{lc} \hline Compound & Heat of Vaporization (kJ/mol) \\ \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 \\ \hline \end{tabular}

Step-by-Step Solution

Answer

Heat of vaporization increases with molar mass, linear shape, polarity, and is highest with hydrogen bonds.

Step 1: Analyze Molar Mass Effect

The molar heat of vaporization generally increases with increasing molar mass, given similar types of intermolecular interactions. For example, comparing \( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{3} \) and \( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3} \), which are both alkanes with no significant polar interactions, the larger pentane has a higher heat of vaporization (27.6 \( \text{kJ/mol} \)) than propane (19.0 \( \text{kJ/mol} \)), which correlates with its larger molar mass and greater London dispersion forces.

Step 2: Explore Molecular Shape Influence

Molecular shape can affect the heat of vaporization by influencing how tightly molecules pack together. Linear structures like \( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3} \) (pentane) have more surface area for intermolecular interactions compared to more compact or branched structures, thus exhibiting stronger dispersion forces and a higher heat of vaporization than more compact molecules of similar mass.

Step 3: Consider Polarity Effects

Polarity increases a compound's heat of vaporization due to stronger intermolecular forces. Take \( \mathrm{CH}_{3} \mathrm{CHBrCH}_{3} \) (31.8 \( \text{kJ/mol} \)) versus \( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{3} \) (19.0 \( \text{kJ/mol} \)); the former's polar nature due to the bromine atom results in stronger dipole-dipole interactions, raising its vaporization heat compared to the nonpolar propane.

Step 4: Identify Hydrogen Bonding Contributions

Hydrogen bonding significantly boosts the heat of vaporization. For instance, \( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH} \) (47.3 \( \text{kJ/mol} \)) can form hydrogen bonds due to its hydroxyl group, creating much stronger intermolecular attractions compared to \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br} \) (33.6 \( \text{kJ/mol} \)), which has weaker dipole interactions as a haloalkane.

Key Concepts

Molar Mass

Molar mass is a crucial factor when it comes to understanding the strengths of intermolecular forces and the subsequent heat of vaporization for organic compounds. In general, as the molar mass of a molecule increases, so too does its heat of vaporization. This trend arises from the enhanced London dispersion forces in larger molecules. These forces are temporary and occur due to the movement of electrons within atoms, leading to momentary charges that attract neighboring molecules.

Take, for example, propane (\( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3} \)), with a molar mass of about 44 g/mol compared to pentane (\( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} \)); whose molar mass is about 72 g/mol. Despite being alkanes with similar intermolecular forces, pentane exhibits a higher heat of vaporization (27.6 kJ/mol) over propane's 19.0 kJ/mol.

This increase reflects the greater molar mass of pentane, as larger and heavier molecules can exhibit stronger dispersion forces due to their increased surface area, allowing them to "stick" together more effectively during phase changes.

Molecular Shape

The shape of a molecule can dramatically affect its physical properties, especially its intermolecular interactions and resulting heat of vaporization. This is because molecular shape determines how molecules pack in space and affect the total surface area available for intermolecular forces like dispersion forces.

Consider linear structures such as pentane (\( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} \)) compared to more compact or branched structures. Linear molecules can closely align with each other, maximizing contact and enhancing dispersion forces, which increases the heat of vaporization. This is why pentane, with its long-chain structure, showcases a higher heat of vaporization than more compact molecules, even if they have a similar molar mass.

In essence, the more extended the molecular structure, the greater the opportunities for interaction between molecular surfaces, leading to higher energy requirements for phase changes from liquid to vapor.

Molecular Polarity

Molecular polarity is a key player in determining the strength of intermolecular forces, influencing a compound’s heat of vaporization. Polarity arises from the differences in electronegativity between atoms in a molecule, leading to an unequal sharing of electrons and resulting in dipoles. These polar molecules experience dipole-dipole interactions in addition to dispersion forces, leading to stronger intermolecular attractions.

For example, consider the polar molecule \( \mathrm{CH}_{3} \mathrm{CHBrCH}_{3} \) which has a heat of vaporization of 31.8 kJ/mol. This molecule compares to nonpolar propane (\( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3} \)) with a significantly lower heat of vaporization (19.0 kJ/mol). The presence of the electronegative bromine atom in \( \mathrm{CH}_{3} \mathrm{CHBrCH}_{3} \) creates a dipole, which results in stronger dipole-dipole interactions, thereby increasing the energy necessary for vaporization.

Ultimately, greater polarity translates into stronger intermolecular attractions, increasing the energy needed to overcome these forces during phase transitions.

Hydrogen Bonding

Hydrogen bonding is one of the strongest types of intermolecular forces and plays a significant role in boosting the heat of vaporization for certain compounds. This special type of dipole-dipole interaction occurs when hydrogen atoms are bonded to highly electronegative atoms like oxygen, nitrogen, or fluorine.

Consider ethyl alcohol (\( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} \)), which has a substantial heat of vaporization of 47.3 kJ/mol. The hydroxyl group (\( \mathrm{-OH} \)) forms strong hydrogen bonds not present in propyl bromide (\( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br} \)), which has a lower heat of vaporization (33.6 kJ/mol).

These hydrogen bonds significantly boost intermolecular attractions, requiring more energy to break them apart during the vaporization process. Thus, hydrogen bonding not only influences physical properties but also highlights the crucial role of molecular structure in determining phase transition energies.