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Chapter 11: Problem 12

At room temperature, \(\mathrm{CO}_{2}\) is a gas, \(\mathrm{CCl}_{4}\) is a liquid, and \(\mathrm{C}_{60}\) (fullerene) is a solid. List these substances in order of (a) increasing intermolecular energy of attraction and (b) increasing boiling point.

Short Answer

Expert verified
(a) CO2, CCl4, C60 (b) CO2, CCl4, C60

Step by step solution

01

Understanding Intermolecular Forces

First, analyze the types of intermolecular forces present in each substance. CO2 is nonpolar and experiences London dispersion forces; CCl4 is also nonpolar with London dispersion forces but has larger molecular size, resulting in stronger London forces; C60, being a large molecule, has even stronger London dispersion forces due to increased surface area.
02

Order by Intermolecular Forces

Based on the strength of the intermolecular forces, we order the substances as follows, from weakest to strongest force: 1. CO2 (gas) 2. CCl4 (liquid) 3. C60 (solid)
03

Understanding Boiling Points

Boiling point increases with the strength of intermolecular forces. A weaker intermolecular attraction corresponds to a lower boiling point, and a stronger intermolecular attraction leads to a higher boiling point.
04

Order by Boiling Point

Based on the strength of intermolecular forces determined in Step 2, order the substances from lowest to highest boiling point: 1. CO2 (gas) 2. CCl4 (liquid) 3. C60 (solid)

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Key Concepts

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

London dispersion forces
London dispersion forces, often known as van der Waals forces, are the weakest type of intermolecular forces that occur between molecules. They arise due to the momentary fluctuations in electron density within a molecule, creating temporary dipoles. These temporary dipoles induce similar dipoles in adjacent molecules, leading to an attraction between them. The strength of London dispersion forces is influenced by several factors:
  • Molecular Size: Larger molecules have more electrons, which means a greater chance of creating temporary dipoles, resulting in stronger dispersion forces.
  • Surface Area: Molecules with larger surface areas have more space to interact with neighboring molecules, increasing the strength of these forces.
The substances in our exercise,
  • CO鈧�, being the smallest, has the weakest London dispersion forces, which is why it exists as a gas at room temperature.
  • CCl鈧�, being larger, experiences stronger London forces and is found as a liquid.
  • C鈧嗏個, with a very large surface area, has the strongest London forces, maintaining its solid state at room temperature.
boiling point
The boiling point of a substance is the temperature at which its vapor pressure equals the atmospheric pressure. At this temperature, the substance transitions from a liquid to a gas. The boiling point is closely tied to the strength of intermolecular forces present within the substance. Stronger forces mean more energy is required to separate the molecules, leading to a higher boiling point. Consider the examples from the exercise:
  • 颁翱鈧�: Contains weak London dispersion forces, so less energy is required to separate the molecules, resulting in a low boiling point and gaseous state at room temperature.
  • 颁颁濒鈧�: With stronger dispersion forces than CO鈧�, its boiling point is higher, thus it exists as a liquid under the same conditions.
  • 颁鈧嗏个: Due to its substantial molecular size and strong London forces, it has the highest boiling point. This requires significant energy to overcome the attractive forces, making it a solid at room temperature.
molecular polarity
Molecular polarity refers to the distribution of electrical charge over the atoms joined by the bond in a molecule. The overall polarity of a molecule is determined by the shape of the molecule and the polarity of its individual bonds. A molecule is considered nonpolar if:
  • It has a symmetrical shape such that polar bonds cancel out.
  • It does not have polar bonds, to begin with.
In our example:
  • 颁翱鈧�: The linear and symmetrical structure of CO鈧� means that any polar bonds effectively cancel out, making it nonpolar.
  • 颁颁濒鈧�: The tetrahedral symmetry results in nonpolarity as the polarities of the carbon-chlorine bonds cancel out.
  • 颁鈧嗏个: This molecule, although comprising nonpolar bonds, has a large structure which primarily influences the strength of dispersion forces it experiences.
While molecular polarity affects intermolecular interactions such as dipole-dipole forces, the nonpolar nature of these molecules means London dispersion forces are their primary intermolecular attraction.

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Most popular questions from this chapter

Use the normal boiling points propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) \quad-42.1^{\circ} \mathrm{C}\) \(\begin{array}{lc}\text { propane }\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) & -42.1^{\circ} \mathrm{C} \\ \text { butane }\left(\mathrm{C}_{4} \mathrm{H}_{10}\right) & -0.5^{\circ} \mathrm{C} \\ \text { pentane }\left(\mathrm{C}_{5} \mathrm{H}_{12}\right) & 36.1^{\circ} \mathrm{C} \\\ \text { hexane }\left(\mathrm{C}_{6} \mathrm{H}_{14}\right) & 68.7^{\circ} \mathrm{C}\end{array}\) heptane \(\left(\mathrm{C}_{7} \mathrm{H}_{16}\right) \quad 98.4{ }^{\circ} \mathrm{C}\) to estimate the normal boiling point of octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\). Explain the trend in the boiling points.

Describe how a cholesteric liquid crystalline phase differs from a smectic A liquid crystalline phase.

For many years drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags of porous clay pots. How many grams of water can be cooled from 35 to \(20^{\circ} \mathrm{C}\) by the evaporation of \(60 \mathrm{~g}\) of water? (The heat of vaporization of water in this temperature range is \(2.4 \mathrm{~kJ} / \mathrm{g}\). The specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K}\).)

Based on their composition and structure, list \(\mathrm{CH}_{3} \mathrm{COOH}\), \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}\), and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) in order of (a) increasing intermolecular forces, (b) increasing viscosity, (c) increasing surface tension.

As the intermolecular attractive forces between molecules increase in magnitude, do you expect each of the following to increase or decrease in magnitude? (a) Vapor pressure, (b) heat of vaporization, (c) boiling point, (d) freezing point, (e) viscosity, (f) surface tension, \((\mathrm{g})\) critical temperature.
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