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Chapter 14: Problem 35

Which substance in each pair would be expected to have a lower boiling point? Explain your reasoning. a. \(\mathrm{CH}_{3} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) b. \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) c. \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{CH}_{4}\)

Short Answer

Expert verified
In summary, the substances with lower boiling points in each pair are: a. \(\mathrm{CH}_{3} \mathrm{OH}\) (methanol) b. \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) (ethane) c. \(\mathrm{CH}_{4}\) (methane) This is because \(\mathrm{CH}_{3} \mathrm{OH}\) has weaker London dispersion forces than 1-propanol, \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) lacks hydrogen bonding present in ethanol, and \(\mathrm{CH}_{4}\) only has London dispersion forces, unlike the strong hydrogen bonding in water.

Step by step solution

01

Pair a: \(\mathrm{CH}_{3} \mathrm{OH}\) vs \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)

\(\mathrm{CH}_{3} \mathrm{OH}\) (methanol) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (1-propanol) are both alcohols, so they can form hydrogen bonds. However, 1-propanol has a longer carbon chain than methanol, which increases the strength of London dispersion forces. The hydrogen bonding in both substances somewhat offsets this difference, but overall, the stronger dispersion forces in 1-propanol will make its boiling point higher than methanol's. Therefore, \(\mathrm{CH}_{3} \mathrm{OH}\) (methanol) has a lower boiling point.
02

Pair b: \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) vs \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)

Both \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) (ethane) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (ethanol) have London dispersion forces since they are both non-polar. However, ethanol can also form hydrogen bonds due to its -OH group. Hydrogen bonding is a much stronger force than dispersion forces, and thus, ethanol will have a higher boiling point than ethane. Therefore, \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) (ethane) has a lower boiling point.
03

Pair c: \(\mathrm{H}_{2} \mathrm{O}\) vs \(\mathrm{CH}_{4}\)

Water (\(\mathrm{H}_{2} \mathrm{O}\)) has a unique molecular structure that allows for extensive hydrogen bonding, which makes its boiling point significantly higher than expected. On the other hand, \(\mathrm{CH}_{4}\) (methane) only has London dispersion forces acting between its molecules as it is a nonpolar molecule. The hydrogen bonding in water is much stronger than the dispersion forces in methane, so water has a higher boiling point. Therefore, \(\mathrm{CH}_{4}\) (methane) has a lower boiling point.

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Key Concepts

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

Hydrogen Bonding
Hydrogen bonding is a type of intermolecular force that occurs when a hydrogen atom is bonded to a highly electronegative atom, like oxygen, nitrogen, or fluorine. This force is particularly strong because the hydrogen becomes slightly positive and is strongly attracted to the lone pairs on the electronegative atom in another molecule.

In alcohols like methanol (\(\mathrm{CH}_{3}\mathrm{OH}\)) and ethanol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\)), the -OH group allows for significant hydrogen bonding. While both methanol and ethanol can engage in this type of bonding, their boiling points can vary depending on the presence of other forces such as London dispersion forces.

Key Points:
  • Hydrogen bonds are much stronger than most other intermolecular forces.
  • They significantly impact boiling points, leading to higher boiling points when compared to substances that can't form hydrogen bonds.
  • Ethanol has a higher boiling point than nonpolar molecules like ethane due to its ability to form hydrogen bonds.
London Dispersion Forces
London dispersion forces, also known as van der Waals forces, are the weakest type of intermolecular forces. They occur in all molecular substances due to the temporary fluctuations in electron distribution within molecules that create instantaneous dipoles.

These forces are predominant in nonpolar substances where no permanent dipoles exist, like methane (\(\mathrm{CH}_{4}\)) or ethane (\(\mathrm{CH}_{3}\mathrm{CH}_{3}\)). Although weaker than hydrogen bonds, London dispersion forces can grow stronger in larger molecules due to an increased number of electrons. For instance, in comparing methanol and 1-propanol, the latter's larger size makes its dispersion forces stronger.

Important Insights:
  • These forces increase with greater molecular size and heavier molecular weight.
  • Larger molecules have higher boiling points due to stronger dispersion forces.
  • While weaker than hydrogen bonds, these forces still play a critical role in intermolecular interactions.
Molecular Structure Analysis
The molecular structure plays a crucial role in determining boiling points because it influences the type and strength of intermolecular forces present.

In smaller molecules like methane (\(\mathrm{CH}_{4}\)), only London dispersion forces are at play, leading to a relatively low boiling point. In contrast, water (\(\mathrm{H}_{2}\mathrm{O}\)) has extensive hydrogen bonding due to its bent shape and the presence of -OH groups, resulting in a significantly higher boiling point.

Factors Affecting Boiling Point:
  • The polarity of a molecule: Polar molecules engage in stronger dipole-dipole interactions, including hydrogen bonding, compared to nonpolar ones.
  • The molecular size and shape: Larger and more complex molecules tend to have stronger dispersion forces.
  • The presence of functional groups: Groups such as -OH or -NH are capable of hydrogen bonding, elevating the boiling point.
Understanding these elements allows us to predict and rationalize the boiling points of different substances based on their molecular characteristics.

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Most popular questions from this chapter

Consider the following molecules: \(\mathrm{CH}_{3} \mathrm{OH}, \mathrm{CH}_{4}, \mathrm{H}_{2} \mathrm{O}, \mathrm{C}_{2} \mathrm{H}_{6}\). a. Draw the Lewis structure for each molecule, and indicate whether each is polar or nonpolar. b. At room temperature, two of these compounds exist as a liquid, and two of these compounds exist as a gas. State which two compounds are liquids at room temperature and which two are gases. Be sure to justify your answer completely. Include discussions of intermolecular forces in your response. c. Rank the compounds from lowest to highest boiling point. Justify your answer.

The molar heat of fusion of aluminum metal is \(10.79 \mathrm{~kJ} / \mathrm{mol}\), whereas its heat of vaporization is \(293.4 \mathrm{~kJ} / \mathrm{mol}\). a. Why is the heat of fusion of aluminum so much smaller than the heat of vaporization? b. What quantity of heat would be required to vaporize \(1.00 \mathrm{~g}\) of aluminum at its normal boiling point? c. What quantity of heat would be evolved if \(5.00 \mathrm{~g}\) of liquid aluminum freezes at its normal freezing point? d. What quantity of heat would be required to melt 0.105 mole of aluminum at its normal melting point?

The molar heat of fusion of benzene is \(9.92 \mathrm{~kJ} / \mathrm{mol}\). Its molar heat of vaporization is \(30.7 \mathrm{~kJ} / \mathrm{mol}\). Calculate the heat required to melt \(8.25 \mathrm{~g}\) of benzene at its normal melting point. Calculate the heat required to vaporize \(8.25 \mathrm{~g}\) of benzene at its normal boiling point. Why is the heat of vaporization more than three times the heat of fusion?

Consider a sample of ice being heated from -5 \({ }^{\circ} \mathrm{C}\) to 5 \({ }^{\circ} \mathrm{C}\). Describe on both a macroscopic and a microscopic basis what happens to the ice as the temperature reaches 0 \({ }^{\circ} \mathrm{C}\).

Although the noble gas elements are monatomic and could not give rise to dipole-dipole forces or hydrogen bonding, these elements still can be liquefied and solidified. Explain.
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