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Chapter 11: Problem 63

Predict the order of increasing vapor pressure at a given temperature for the following compounds: a. \(\mathrm{FCH}_{2} \mathrm{CH}_{2} \mathrm{~F}\) b. \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) c. \(\mathrm{FCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) Explain why you chose this order.

Short Answer

Expert verified
(b) < (c) < (a), due to decreasing hydrogen bonding strengths.

Step by step solution

01

Understanding Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase. The vapor pressure of a compound at a given temperature is an indicator of its volatility. Generally, the weaker the intermolecular forces in a compound, the higher its vapor pressure.
02

Analyzing Intermolecular Forces

Let's examine each compound:- (a) \( \text{FCH}_2 \text{CH}_2 \text{F} \) has dipole-dipole interactions and hydrogen atoms substituted with fluorine, making its intermolecular forces relatively weak compared to hydrogen bonding.- (b) \( \text{HOCH}_2 \text{CH}_2 \text{OH} \) contains two hydroxyl (\( \text{-OH} \)) groups, allowing strong hydrogen bonding, which is a strong intermolecular force.- (c) \( \text{FCH}_2 \text{CH}_2 \text{OH} \) has one \( \text{-OH} \) group enabling hydrogen bonding, but also possesses a fluorine which can contribute to dipole interactions.
03

Determining the Order of Vapor Pressure

Based on the strength of intermolecular forces, we determine:- (b) \( \text{HOCH}_2 \text{CH}_2 \text{OH} \) will have the lowest vapor pressure due to strong intermolecular hydrogen bonds from its two \( \text{-OH} \) groups.- (c) \( \text{FCH}_2 \text{CH}_2 \text{OH} \) has a moderate vapor pressure because it has only one \( \text{-OH} \) group but also polar interactions due to fluorine.- (a) \( \text{FCH}_2 \text{CH}_2 \text{F} \) will have the highest vapor pressure as it lacks \( \text{-OH} \) groups and relies more on dipole-dipole interactions which are weaker.
04

Conclusion

Thus, the order of increasing vapor pressure is:(b) \( \text{HOCH}_2 \text{CH}_2 \text{OH} \) < (c) \( \text{FCH}_2 \text{CH}_2 \text{OH} \) < (a) \( \text{FCH}_2 \text{CH}_2 \text{F} \).

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Key Concepts

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

Intermolecular Forces
Intermolecular forces (IMFs) are the attractive forces that occur between molecules. They are crucial because they influence various physical properties of a substance, such as boiling point, melting point, and vapor pressure.
When it comes to vapor pressure, compounds with stronger intermolecular forces tend to have lower vapor pressures as their molecules are more strongly attracted to one another, making it harder for them to escape into the vapor phase.
Comparatively, substances with weaker intermolecular forces will evaporate more easily, resulting in higher vapor pressures.
  • Types of Intermolecular Forces:
    • London Dispersion Forces: These are the weakest forces arising from temporary dipoles formed when electrons move around the nucleus.
    • Dipole-Dipole Interactions: These occur between polar molecules with a permanent dipole.
    • Hydrogen Bonding: These bonds are strong special dipole-dipole interactions that occur when hydrogen is directly bonded to highly electronegative atoms like nitrogen, oxygen, or fluorine.
Hydrogen Bonding
Hydrogen bonding is a type of strong dipole-dipole interaction that is particularly significant in determining the properties of water and other substances with molecules containing hydrogen attached to highly electronegative atoms like oxygen, nitrogen, or fluorine.
These bonds have notable strength compared to other dipole-dipole interactions due to the small size of the hydrogen atom and the large electronegativity difference with the atom it bonds to.
This leads to strong attractions and impacts several physical properties such as higher boiling points and lower vapor pressures.
For compounds with hydrogen bonding,
  • Vapor Pressure: Molecules with strong hydrogen bonding tend to stay as a liquid at a given temperature, resulting in lower vapor pressure.
  • Examples: In the exercise, the compound HOCH鈧侰H鈧侽H features hydrogen bonding due to its two hydroxyl groups, resulting in a lower vapor pressure compared to its counterparts without hydrogen bonding or with fewer such interactions.
Dipole-Dipole Interactions
Dipole-dipole interactions occur between polar molecules where oppositely charged ends attract. These are stronger than London dispersion forces but weaker than hydrogen bonds.
These interactions contribute to the stability of a liquid phase by aligning polar molecules in a manner that decreases their kinetic energy, thereby impacting boiling points and vapor pressures.
The presence of dipole-dipole interactions can lead to higher boiling points compared to nonpolar molecules of similar size.
  • Influence on Vapor Pressure:
    • Molecules with only dipole-dipole interactions generally have a moderate vapor pressure, unless they form hydrogen bonds.
    • The compound FCH鈧侰H鈧侳, for instance, relies on dipole-dipole interactions without additional hydrogen bonding, giving it a relatively higher vapor pressure compared to those that do.
  • Comparison:
    • In the original exercise solution, FCH鈧侰H鈧侳 has the highest vapor pressure due to its weaker dipole-dipole interactions, while FCH鈧侰H鈧侽H, with both an OH group and fluorine, has stronger interactions and, thus, a lower vapor pressure than FCH鈧侰H鈧侳.

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Most popular questions from this chapter

An electric heater coil provided heat to a \(15.5-\mathrm{g}\) sample of iodine, \(\mathrm{I}_{2}\), at the rate of \(3.48 \mathrm{~J} / \mathrm{s}\). It took \(4.54 \mathrm{~min}\) from the time the iodine began to melt until the iodine was completely melted. What is the heat of fusion per mole of iodine?

What is the coordination number of \(\mathrm{Cs}^{+}\) in \(\mathrm{CsCl}\) ? of \(\mathrm{Na}^{+}\) in \(\mathrm{NaCl}\) ? of \(\mathrm{Zn}^{2+}\) in \(\mathrm{ZnS} ?\)

For the carbon and nitrogen family hydrides, we have the following boiling points: Carbon \(\quad\) Nitrogen Family, \({ }^{\circ} \mathbf{C} \quad\) Family, \({ }^{\circ} \mathbf{C}\) \(\mathrm{CH}_{4},-164 \quad \mathrm{NH}_{3},-33\) \(\mathrm{SiH}_{4},-112 \quad \mathrm{PH}_{3},-88\) \(\begin{array}{ll}\mathrm{GeH}_{4},-88 & \mathrm{AsH}_{3},-55\end{array}\) \(\mathrm{SnH}_{4},-52 \quad \mathrm{SbH}_{3},-17\) Account for the following: a. The general trend in the boiling points of the binary hydrides. b. The unusual boiling point of ammonia. c. The observation that the nitrogen family hydrides have boiling points that are notably higher than those of the carbon family.

Carbon disulfide, \(\mathrm{CS}_{2}\), is a volatile, flammable liquid. It has a vapor pressure of \(400.0 \mathrm{mmHg}\) at \(28.0^{\circ} \mathrm{C}\) and \(760.0\) \(\mathrm{mmHg}\) at \(46.5^{\circ} \mathrm{C}\). What is the heat of vaporization of this substance?

Steam at \(100^{\circ} \mathrm{C}\) was passed into a flask containing \(275 \mathrm{~g}\) of water at \(21^{\circ} \mathrm{C}\), where the steam condensed. How many grams of steam must have condensed if the temperature of the water in the flask was raised to \(83{ }^{\circ} \mathrm{C} ?\) The heat of vaporization of water at \(100^{\circ} \mathrm{C}\) is \(40.7 \mathrm{~kJ} / \mathrm{mol}\) and the specific heat is \(4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\)
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