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(a) Place the following substances in order of increasing volatility: \(\mathrm{CH}_{4}, \mathrm{CBr}_{4}, \mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CH}_{3} \mathrm{Cl}, \mathrm{CHBr}_{3},\) and \(\mathrm{CH}_{2} \mathrm{Br}_{2}\). (b) How do the boiling points vary through this series? (c) Explain your answer to part (b) in terms of intermolecular forces.

Short Answer

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(a) Order of volatility: \(\mathrm{CBr}_4, \mathrm{CHBr}_3, \mathrm{CH}_2 \mathrm{Br}_2, \mathrm{CH}_2 \mathrm{Cl}_2, \mathrm{CH}_3 \mathrm{Cl}, \mathrm{CH}_4\). (b) Boiling points increase in the opposite order. (c) Larger molecular weights offer stronger London dispersion forces, raising boiling points.

Step by step solution

01

Understand Volatility

Volatility refers to a substance's tendency to vaporize. It is inversely related to the boiling point; substances that are more volatile have lower boiling points. Thus, to order the substances by volatility, we need to consider their boiling points.
02

Identify The Intermolecular Forces

Substances have different intermolecular forces such as London dispersion forces, dipole-dipole interactions, and hydrogen bonding. Among the given molecules: \(\mathrm{CH}_4, \mathrm{CBr}_4, \mathrm{CH}_2 \mathrm{Cl}_2, \mathrm{CH}_3 \mathrm{Cl}, \mathrm{CHBr}_3,\) and \(\mathrm{CH}_2 \mathrm{Br}_2\), we will consider the dominant force: London dispersion, since none have hydrogen bonding.
03

Assess Molecular Weights

For non-polar or weakly polar compounds, heavier molecules generally have stronger London dispersion forces, leading to higher boiling points and lower volatility. Arranging the given compounds by increasing molecular weight: \(\mathrm{CH}_4 \lt \mathrm{CH}_3 \mathrm{Cl} \lt \mathrm{CH}_2 \mathrm{Cl}_2 \lt \mathrm{CH}_2 \mathrm{Br}_2 \lt \mathrm{CHBr}_3 \lt \mathrm{CBr}_4\).
04

List in Order of Volatility

Based on the information and understanding that higher molecular weight implies higher boiling point and lower volatility, the order of increasing volatility (or increasing boiling points) is: \(\mathrm{CBr}_4, \mathrm{CHBr}_3, \mathrm{CH}_2 \mathrm{Br}_2, \mathrm{CH}_2 \mathrm{Cl}_2, \mathrm{CH}_3 \mathrm{Cl}, \mathrm{CH}_4\).
05

Boiling Point Trend

Volatility is inversely related to boiling points, hence the order by increasing boiling point should be the reverse of the volatility order: \(\mathrm{CH}_4, \mathrm{CH}_3 \mathrm{Cl}, \mathrm{CH}_2 \mathrm{Cl}_2, \mathrm{CH}_2 \mathrm{Br}_2, \mathrm{CHBr}_3, \mathrm{CBr}_4\).
06

Explain Using Intermolecular Forces

The boiling points follow the trend primarily due to London dispersion forces, which are stronger in molecules with greater molecular weights. More massive molecules have more electron cloud distortions providing stronger dispersion forces and requiring more energy (higher temperature) to break, thus having higher boiling points.

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Key Concepts

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

Volatility
Volatility describes how easily a substance turns into vapor. When a substance has high volatility, it means it can change from a liquid to a gas at a lower temperature. This is tied directly to the concept of boiling points because more volatile substances have lower boiling points. They require less heat energy to move their molecules from a liquid to a gaseous state. This makes the relationship between volatility and boiling point an inverse one.

To understand volatility better, think of a pot of water. When you heat it, you can see steam rising. A highly volatile substance would start producing steam at a lower temperature compared to something less volatile. In chemistry, substances with lighter molecular weights and weaker intermolecular forces tend to be more volatile.

For instance, in a group of compounds like
  • Methane (\( \mathrm{CH}_{4}\)), a very light molecule, would have higher volatility.
  • On the other extreme, Carbon Tetrabromide (\( \mathrm{CBr}_{4}\)), being heavier, is less volatile with a higher boiling point.
Boiling Points
Boiling points are the temperature at which a liquid becomes a gas. This happens when molecules have enough energy to overcome the intermolecular forces holding them in the liquid state. Therefore, compounds with stronger intermolecular forces will naturally have higher boiling points.

By understanding boiling points, one can predict how substances behave under heating and their relative energy requirements for evaporation. A substance with a lower boiling point is more volatile as these require less energy to escape into the air.
  • The sequence of boiling points for compounds like \( \mathrm{CH}_4\), \( \mathrm{CH}_3 \mathrm{Cl}\), and \( \mathrm{CBr}_4\) is directly tied to their molecular weights and the type of intermolecular forces they experience.
Larger and heavier molecules attract each other more, requiring more heat energy to separate them.
  • Thus, they boil at higher temperatures indicating a lower volatility.
London Dispersion Forces
London dispersion forces are a type of intermolecular force that arises from the random movement of electrons. They are present in all molecules but play a more significant role in non-polar molecules where no permanent dipoles exist. These forces get stronger with increasing molecular size and mass because larger electron clouds are more easily distorted, creating temporary dipoles.

The larger the molecule, the more significant these forces are, and the more energy is needed to break them apart, raising the boiling point.
  • For example, heavier compounds like \( \mathrm{CBr}_4\), which is non-polar, have substantial London dispersion forces due to a significant electron cloud.
This means they tend to have a lower volatility and higher boiling point as compared to lighter molecules like \( \mathrm{CH}_4\).

In this way, London dispersion forces are one of the key factors that determine the physical properties of compounds, especially when you're considering volatility and boiling points.

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Most popular questions from this chapter

(a) How does the average kinetic energy of molecules compare with the average energy of attraction between molecules in solids, liquids, and gases? (b) Why does increasing the temperature cause a solid substance to change in succession from a solid to a liquid to a gas? (c) What happens to a eas if you put it under extremelv high pressuret

Indicate whether each statement is true or false: (a) The liquid crystal state is another phase of matter, just like solid, liquid, and gas. (b) Liquid crystalline molecules are generally spherical in shape, (c) Molecules that exhibit a liquid crystalline phase do so at well-defined temperatures and pressures. (d) Molecules that exhibit a liquid crystalline phase show weaker-than- expected intermolecular forces. (e) Molecules containing only carbon and hydrogen are likely to form liquid crystalline phases. (f) Molecules can exhibit more than one liquid crystalline phase.

(a) Which is generally stronger, intermolecular interactions or intramolecular interactions? (b) Which of these kinds of interactions are broken when a liquid is converted to a gas?

Due to the environmental concern of fluorocarbons as refrigerants, a refrigerant based on a mixture of hydrocarbons was used as a replacement. It is a patented blend of ethane, propane, butane, and isobutane. Isobutane has a normal boiling point of \(-12^{\circ} \mathrm{C}\). The molar specific heat of liquid phase and gas phase isobutane are \(129.7 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) and \(95.2 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) respectively. The heat of vaporization for this compound is \(21.3 \mathrm{~kJ} / \mathrm{mol}\). Calculate the heat required to convert \(25.0 \mathrm{~g}\) of isobutane from a liquid at \(-50^{\circ} \mathrm{C}\) to a gas at \(40^{\circ} \mathrm{C}\).

True or false: (a) Molecules containing polar bonds must be polar molecules and have dipole-dipole forces. (b) For the halogen gases, the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table. (c) In terms of the total attractive forces for a given substance, the more polar bonds there are in a molecule, the stronger the dipole-dipole interaction. \((\mathbf{d})\) All other factors being the same, total attractive forces between linear molecules are greater than those between molecules whose shapes are nearly spherical. (e) The more electronegative the atom, the more polarizable it is.

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