/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 What types of intermolecular for... [FREE SOLUTION] | 91Ó°ÊÓ

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What types of intermolecular forces are important in the liquid phase of (a) \(\mathrm{C}_{2} \mathrm{H}_{6}\) and (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH} ?\)

Short Answer

Expert verified
Ethane primarily has London dispersion forces; isopropanol has hydrogen bonding and dispersion forces.

Step by step solution

01

Identify the Molecule

First, we'll identify the structural formula and properties of the given molecules: \(\mathrm{C}_{2}\mathrm{H}_{6}\) (ethane) and \((\mathrm{CH}_{3})_{2}\mathrm{CHOH}\) (isopropanol). Ethane is a small, non-polar hydrocarbon, while isopropanol contains an OH group, indicating the presence of a polar functionality.
02

Consider Non-Polar Molecule Forces for Ethane

Since \(\mathrm{C}_{2}\mathrm{H}_{6}\) is a non-polar molecule, it will largely exhibit London dispersion forces (induced dipole-induced dipole interactions) between its molecules. London dispersion forces occur due to temporary changes in electron density that induce a weak attraction between molecules.
03

Consider Polar Molecule Forces for Isopropanol

\((\mathrm{CH}_{3})_{2}\mathrm{CHOH}\), isopropanol, includes a hydroxyl group (\(\mathrm{OH}\)), making it capable of hydrogen bonding, a strong dipole-dipole interaction. Additionally, isopropanol will also exhibit London dispersion forces due to its overall molecular structure.
04

Review the Dominant Intermolecular Forces

Summarizing, \(\mathrm{C}_{2}\mathrm{H}_{6}\) will primarily exhibit London dispersion forces due to its non-polarity. \((\mathrm{CH}_{3})_{2}\mathrm{CHOH}\) will experience both hydrogen bonding, due to the hydroxyl (\(\mathrm{OH}\)) group, and London dispersion forces.

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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 are a type of weak intermolecular force that occur in all molecules, whether they are polar or non-polar. These forces are due to temporary fluctuations in electron distribution within an atom or molecule, which create a temporary dipole. This, in turn, can induce a temporary dipole in neighboring atoms or molecules, leading to an attractive interaction.

These are often the only forces acting between non-polar molecules. For example, in ethane \( (\mathrm{C}_{2}\mathrm{H}_{6}) \), which is non-polar, London dispersion forces are the main type of intermolecular attraction. Despite their weak nature, they become significant when the molecules have a large mass or many atoms, because the larger the electron cloud, the more polarizable it is. More polarizable electron clouds will lead to stronger temporary dipoles and, consequently, stronger London dispersion forces during interactions.
Hydrogen Bonding
Hydrogen bonding is a special type of dipole-dipole interaction that occurs when hydrogen is bonded to electronegative atoms like nitrogen, oxygen, or fluorine. This results in a large dipole moment due to the high electronegativity difference.

For instance, in isopropanol \( (\mathrm{CH}_{3})_{2}\mathrm{CHOH} \), the hydrogen atoms are bonded to an oxygen atom in the hydroxyl \( (\mathrm{OH}) \) group. The oxygen atom is more electronegative than hydrogen, creating a strong dipole. This dipole can then form hydrogen bonds with other molecules, which is a stronger interaction than ordinary dipole-dipole or London dispersion forces. The presence of hydrogen bonding explains the higher boiling and melting points of substances like isopropanol when compared to their non-polar counterparts.
Polar and Non-Polar Molecules
Molecules can generally be classified as polar or non-polar based on their symmetry and the distribution of electron density. In polar molecules, the electron distribution is uneven, resulting in partial charges within the molecule. This causes polar molecules, like isopropanol \( (\mathrm{CH}_{3})_{2}\mathrm{CHOH} \), to exhibit dipole-dipole interactions or hydrogen bonding if specific conditions are met.

On the other hand, non-polar molecules, such as ethane \( \mathrm{C}_{2}\mathrm{H}_{6} \), have a symmetrical distribution of electron density. Since their electron density is evenly distributed, they do not have permanent dipoles and primarily rely on London dispersion forces for intermolecular attractions. These differences largely influence the physical properties of substances, such as boiling points and solubilities.

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Most popular questions from this chapter

If you place \(1.0 \mathrm{L}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in a room that is \(3.0 \mathrm{m}\) long, \(2.5 \mathrm{m}\) wide, and \(2.5 \mathrm{m}\) high, will all of the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethanol at \(25^{\circ} \mathrm{C}\) is \(59 \mathrm{mm} \mathrm{Hg},\) and the density of the liquid at this temperature is \(0.785 \mathrm{g} / \mathrm{cm}^{3}\)

Answer each of the following questions with increases, decreases, or does not change. (a) If the intermolecular forces in a liquid increase, the normal boiling point of the liquid ____. (b) If the intermolecular forces in a liquid decrease, the vapor pressure of the liquid ____. (c) If the surface area of a liquid decreases, the vapor pressure ____. (d) If the temperature of a liquid increases, the equilibrium vapor pressure ____.

The critical temperature and pressure of chloromethane are \(416 \mathrm{K}\) and 66.1 atm, respectively. (Chloromethane's triple point is at \(175.4 \mathrm{K} \text { and } 0.0086 \text { atm. })\) Can \(\mathrm{CH}_{3} \mathrm{Cl}\) be liquefied at or above room temperature? Explain briefly.

The density of copper metal is \(8.95 \mathrm{g} / \mathrm{cm}^{3} .\) If the radius of a copper atom is \(127.8 \mathrm{pm},\) is the copper unit cell simple cubic, body-centered cubic, or face-centered cubic?

Assuming that in a simple cubic unit cell the spherical atoms or ions just touch along the cube's edges, calculate the percentage of empty space within the unit cell. (Recall that the volume of a sphere is \((4 / 3) \pi r^{3},\) where \(r\) is the radius of the sphere.) (IMAGE NOT COPY)

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