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Rationalize the difference in boiling points in each pair: (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}\left(-23^{\circ} \mathrm{C}\right)\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \quad\left(78^{\circ} \mathrm{C}\right)\), (b) \(\mathrm{CO}_{2}\left(-78.5^{\circ} \mathrm{C}\right)\) and \(\mathrm{CS}_{2}\left(46.2^{\circ} \mathrm{C}\right),(\mathbf{c}) \mathrm{CH}_{3} \mathrm{COCH}_{3}\left(50.5^{\circ} \mathrm{C}\right)\) and \(\mathrm{CH}_{3} \mathrm{COOH}\left(101^{\circ} \mathrm{C}\right)\).

Short Answer

Expert verified
Boiling points vary due to different intermolecular forces like hydrogen bonds, dipole-dipole, and Van der Waals interactions, which affect energy needed to separate molecules.

Step by step solution

01

Understand the concept of boiling point differences

Boiling point differences can be understood by analyzing the strength of intermolecular forces within each compound. Stronger forces result in higher boiling points as more energy is required to separate molecules.
02

Compare intermolecular forces in (a)

For (a), \((\mathrm{CH}_{3})_{2} \mathrm{O}\) is an ether with only dipole-dipole interactions and weak van der Waals forces, leading to a lower boiling point of -23°C. In contrast, \({\mathrm{CH}_{3}{CH}_{2}{OH}}\) is an alcohol containing hydrogen bonds, which are stronger, resulting in a higher boiling point of 78°C.
03

Compare intermolecular forces in (b)

For (b), \({\mathrm{CO}_{2}}\) is a linear molecule with Van der Waals interactions, leading to a lower boiling point of -78.5°C. \({\mathrm{CS}_{2}}\) has similar interactions but owing to its higher molar mass, it has stronger dispersion forces, resulting in a higher boiling point of 46.2°C.
04

Compare intermolecular forces in (c)

For (c), \({\mathrm{CH}_{3}COCH}_{3}}\) (acetone) mainly exhibits dipole-dipole interactions and weak Van der Waals forces, resulting in a boiling point of 50.5°C. \({\mathrm{CH}_{3} \mathrm{COOH}}\) (acetic acid) forms dimers through hydrogen bonding, significantly increasing its boiling point to 101°C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intermolecular Forces
Intermolecular forces are the attractions between molecules in a substance. These forces determine the physical properties of substances such as boiling points. The stronger the intermolecular forces, the more energy is required to break these forces, leading to a higher boiling point. There are several types of intermolecular forces:
  • Van der Waals forces: These are weak forces that arise from temporary dipoles in molecules.
  • Dipole-dipole interactions: These occur between polar molecules, where positive and negative charges attract.
  • Hydrogen bonds: These are the strongest type of intermolecular forces. They occur in molecules where hydrogen is bonded to a highly electronegative atom like oxygen, nitrogen, or fluorine.
Understanding these forces is key to explaining why different substances have different boiling points.
Hydrogen Bonds
Hydrogen bonds are a specific, strong type of dipole-dipole interaction. They occur when a hydrogen atom, covalently bonded to a small, highly electronegative atom such as oxygen or nitrogen, experiences strong attractions to another electronegative atom with a lone pair of electrons. These bonds are crucial in determining the boiling points of molecules. For example, alcohols like ethanol (CH₃CH₂OH) can form hydrogen bonds due to the presence of an OH group. This results in much higher boiling points compared to molecules only capable of dipole-dipole interactions, such as ethers.

Hydrogen bonds significantly increase the boiling point because more energy is needed to disrupt these strong interactions during the phase change from liquid to gas. A substance with hydrogen bonds would require more heat to boil compared to substances with only weak forces, like Van der Waals forces.
Dipole-Dipole Interactions
Dipole-dipole interactions occur between molecules that have permanent dipoles, meaning they have a constant separation of charge due to differences in electronegativity in their atoms. These interactions are stronger than Van der Waals forces but weaker than hydrogen bonds.

In molecules such as acetone (CH₃COCH₃), the carbonyl group creates a dipole due to the difference in electronegativity between carbon and oxygen. This results in moderate-strength dipole-dipole interactions, leading to a boiling point higher than non-polar molecules but lower than hydrogen-bonded molecules. Although not as strong as hydrogen bonds, these interactions can still significantly affect the boiling point of a compound.
Van der Waals Forces
Van der Waals forces are the weakest of the intermolecular forces. They arise from fluctuations in the electron distribution within molecules, leading to temporary dipoles that can induce dipoles in neighboring molecules. These forces are more pronounced in larger molecules with greater electron clouds, as they are more easily polarizable. These forces are particularly relevant for non-polar molecules such as carbon dioxide (COâ‚‚) and carbon disulfide (CSâ‚‚). While both molecules primarily exhibit Van der Waals forces, CSâ‚‚ has a higher boiling point due to its higher molar mass, enhancing its dispersion forces compared to COâ‚‚.

Despite being the weakest, Van der Waals forces can still contribute to the physical properties of substances, especially in the absence of stronger forces like hydrogen bonds or dipole-dipole interactions.

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Most popular questions from this chapter

Butane and 2 -methylpropane, whose space-filling models are shown here, are both nonpolar and have the same molecular formula, \(\mathrm{C}_{4} \mathrm{H}_{10}\), yet butane has the higher boiling point \(\left(-0.5^{\circ} \mathrm{C}\right.\) compared to \(\left.-11.7{ }^{\circ} \mathrm{C}\right)\). Explain.

The table below shows some physical properties of compounds containing O-H groups. \begin{tabular}{lccc} \hline Liquid & Molecular Weight & Experimental Dipole Moment & Boiling Point \\\ \hline \(\mathrm{CH}_{3} \mathrm{OH}\) & 32.04 & 1.7 & \(64.7^{\circ} \mathrm{C}\) \\\ \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) & 74.12 & 1.66 & \(117.7^{\circ} \mathrm{C}\) \\ \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) & 62.07 & 1.5 & \(197.3^{\circ} \mathrm{C}\) \\ \hline \end{tabular} Which of the following statements best explains these data? (a) The larger the dipole moment, the stronger the intermolecular forces, and therefore the boiling point is lowest for the molecule with the largest dipole moment. (b) The dispersion forces increase from \(\mathrm{CH}_{3} \mathrm{OH} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\); since the boiling point also increases in this order, the dispersion forces must be the major contributing factor for the boiling point trend; \((\mathbf{c}) \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) has two groups capable of hydrogen bonding per molecule, whereas \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) have only one; therefore, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) has the highest boiling point.

If \(42.0 \mathrm{~kJ}\) of heat is added to a \(32.0-\mathrm{g}\) sample of liquid methane under \(101.3 \mathrm{kPa}\) of pressure at a temperature of \(-170^{\circ} \mathrm{C}\), what are the final state and temperature of the methane once the system equilibrates? Assume no heat is lost to the surroundings. The normal boiling point of methane is \(-161.5^{\circ} \mathrm{C}\). The specific heats of liquid and gaseous methane are 3.48 and \(2.22 \mathrm{~J} / \mathrm{g}-\mathrm{K}\), respectively. [Section 11.4\(]\)

A watch with a liquid crystal display (LCD) does not function properly when it is exposed to low temperatures during a trip to Antarctica. Explain why the LCD might not function well at low temperature.

Which member in each pair has the stronger intermolecular dispersion forces? (a) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{CH}_{3} \mathrm{OH},\) (b) \(\mathrm{CBr}_{3} \mathrm{CBr}_{3}\) or \(\mathrm{CCl}_{3} \mathrm{CCl}_{3}\) (c) \(\mathrm{C}\left(\mathrm{CH}_{3}\right)_{4}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\).

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