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Why are the dipole-dipole interactions between polar molecules not important in the vapor phase?

Short Answer

Expert verified
Dipole-dipole interactions between polar molecules are not important in the vapor phase because the increased distance between molecules and their rapid motion significantly reduce the strength and occurrence of these interactions, rendering them negligible in the vapor phase.

Step by step solution

01

Introduction to dipole-dipole interactions

Dipole-dipole interactions are a type of intermolecular force that occurs between polar molecules. Polar molecules have a partial positive charge on one side and a partial negative charge on the other side, due to differences in electronegativity between the bonded atoms. Dipole-dipole interactions are the attractive forces between the positive side of one polar molecule and the negative side of another polar molecule.
02

Phases of matter

There are three main phases of matter: solid, liquid, and gas (vapor). The main difference between these phases lies in the arrangement, movement, and interaction of molecules. In the solid phase, molecules are closely packed together with strong intermolecular forces. In the liquid phase, molecules are still in close proximity but can move more freely, resulting in weaker intermolecular forces. In the vapor phase, molecules are widely separated and move rapidly, leading to minimal intermolecular forces.
03

Effect of phase on dipole-dipole interactions

As we move from the solid to vapor phase, the intermolecular forces become weaker due to the increased distance between molecules. This is particularly important for dipole-dipole interactions because these forces depend on the proximity and orientation of the polar molecules. The strength of the dipole-dipole interaction decreases rapidly when molecules are further apart.
04

Importance of dipole-dipole interactions in the vapor phase

In the vapor phase, molecules are further apart due to the increase in energy and motion, resulting in a significant reduction in the strength of the dipole-dipole interactions. Since the strength of the interaction decreases rapidly with distance, these forces become negligible in the vapor phase. Furthermore, the rapid motion of molecules in the vapor phase makes it difficult for the dipoles to align and interact, reducing the significance of dipole-dipole forces even more. In conclusion, dipole-dipole interactions between polar molecules are not important in the vapor phase because the increased distance between molecules and rapid motion reduce the strength and occurrence of these interactions, making them negligible in the vapor phase.

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Most popular questions from this chapter

Which substance in each pair would be expected to be more volatile at a particular temperature? Explain your reasoning. a. \(\mathrm{H}_{2} \mathrm{O}(l)\) or \(\mathrm{H}_{2} \mathrm{~S}(l)\) b. \(\mathrm{H}_{2} \mathrm{O}(l)\) or \(\mathrm{CH}_{3} \mathrm{OH}(l)\) c. \(\mathrm{CH}_{3} \mathrm{OH}(l)\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)\)

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.

What are London dispersion forces and how do they arise in a nonpolar molecule? Are London forces typically stronger or weaker than dipole-dipole attractions between polar molecules? Are London forces stronger or weaker than covalent bonds? Explain.

Two molecules that contain the same number of each kind of atom but that have different molecular structures are said to be isomers of each other. For example, both ethyl alcohol and dimethyl ether (shown below) have the formula \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\) and are isomers. Based on considerations of intermolecular forces, which substance would you expect to be more volatile? Which would you expect to have the higher boiling point? Explain. \(\mathrm{CH}_{3} \stackrel{\text { dimethyl ether }}{\mathrm{CH}}_{3} \quad \mathrm{CH}_{3} \stackrel{\text { ethyl alcohol }}{\mathrm{CH}}_{2}-\mathrm{OH}\)

Identify the most important type of forces (ionic, hydrogen bonding, dipole- dipole, or London dispersion forces) among atoms or molecules present in the solids of each of the following substances.

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