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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

Expert verified
The given substances can be ordered by increasing volatility as follows: \(\mathrm{CH}_{4}<\mathrm{CH}_{3} \mathrm{Cl}<\mathrm{CH}_{2} \mathrm{Cl}_{2}<\mathrm{CH}_{2} \mathrm{Br}_{2}<\mathrm{CHBr}_{3}<\mathrm{CBr}_{4}\). The boiling points increase through the series due to stronger London dispersion forces and the presence of dipole-dipole interactions in some molecules.

Step by step solution

01

Understanding volatility, boiling points, and intermolecular forces

Volatility refers to the ability of a substance to vaporize easily. A higher volatility means that a substance changes into a vapour state at lower temperatures. Boiling point is the temperature at which a substance changes from liquid to gas. Lower boiling points imply greater volatility. Intermolecular forces are attractive forces between molecules. The stronger the intermolecular forces, the higher the boiling point and the lower the volatility. In the given substances, the primary intermolecular forces are London dispersion forces and dipole-dipole interactions.
02

Assessing the London dispersion forces

In simple terms, London dispersion forces (LDF) increase with the size and complexity of a molecule. A larger, more complex molecule usually has stronger LDF and higher boiling points. Comparing the molecules given: \(\mathrm{CH}_{4}, \mathrm{CBr}_{4}, \mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CH}_{3} \mathrm{Cl}, \mathrm{CHBr}_{3}\), and \(\mathrm{CH}_{2}\mathrm{Br}_{2}\), we observe that the presence of halogens will increase LDF due to an increase in molecular size. Thus, LDF will generally increase in the order \(\mathrm{CH}_{4} < \mathrm{CH}_{3} \mathrm{Cl} < \mathrm{CH}_{2} \mathrm{Cl}_{2} < \mathrm{CH}_{2}\mathrm{Br}_{2} < \mathrm{CHBr}_{3} < \mathrm{CBr}_{4}\).
03

Assessing the dipole-dipole interactions

We need to determine whether any of these molecules possess dipole moments that will lead to dipole-dipole interactions. Considering the molecular symmetry in \(\mathrm{CH}_{2}\mathrm{Cl}_{2}\) and \(\mathrm{CH}_{2}\mathrm{Br}_{2}\), both have no net dipole moment, despite having polar C-Cl and C-Br bonds. However, \(\mathrm{CH}_{3}\mathrm{Cl}\) and \(\mathrm{CHBr}_{3}\) have asymmetric structures and therefore net dipole moments, which will lead to dipole-dipole interactions.
04

Ordering the substances based on volatility

Considering the information above, let's order the substances by increasing volatility (decreasing boiling points). As LDF increase in the order \(\mathrm{CH}_{4} < \mathrm{CH}_{3} \mathrm{Cl} < \mathrm{CH}_{2} \mathrm{Cl}_{2} < \mathrm{CH}_{2}\mathrm{Br}_{2} < \mathrm{CHBr}_{3} < \mathrm{CBr}_{4}\) and dipole-dipole interactions make \(\mathrm{CH}_{3}\mathrm{Cl}\) and \(\mathrm{CHBr}_{3}\) less volatile, a reasonable order from the most volatile to the least volatile would be: 1. \(\mathrm{CH}_{4}\) 2. \(\mathrm{CH}_{3} \mathrm{Cl}\) 3. \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) 4. \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) 5. \(\mathrm{CHBr}_{3}\) 6. \(\mathrm{CBr}_{4}\)
05

Evaluating the boiling points

Looking at the order of volatility we determined, we can easily observe how the boiling points vary through this series. Boiling points increase as we move from \(\mathrm{CH}_{4}\) to \(\mathrm{CBr}_{4}\) since the substances become less volatile.
06

Explaining the variation of boiling points in terms of intermolecular forces

The variation in boiling points can be explained due to the differences in London dispersion forces and dipole-dipole interactions between the molecules. As the size and complexity of molecules increase, the London dispersion forces become stronger, resulting in higher boiling points. With the addition of dipole-dipole interactions in molecules like \(\mathrm{CHBr}_{3}\) and \(\mathrm{CH}_{3} \mathrm{Cl}\), the boiling points are further increased.

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Key Concepts

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

Intermolecular Forces
Intermolecular forces are the invisible forces that occur between molecules. They are crucial in determining how molecules interact, their state of matter, and properties like boiling points and volatility.
These forces include three main types:
  • London dispersion forces (or van der Waals forces): Present in all molecules, and the only forces at work in nonpolar molecules.
  • Dipole-dipole interactions: Occur between molecules with permanent dipole moments due to asymmetry in electron distribution.
  • Hydrogen bonds: A special case of dipole-dipole interactions, which are stronger due to the presence of hydrogen atoms.
In the context of our exercise, it's essential to understand how these intermolecular forces impact the physical properties of substances. The stronger the intermolecular forces, the less volatile a substance is, meaning it evaporates less easily. Conversely, these strong forces result in higher boiling points as more energy (heat) is required to separate the molecules into a gaseous form.
London Dispersion Forces
London dispersion forces (LDF) are a type of intermolecular force arising from temporary shifts in electron density in atoms or molecules. Even in nonpolar molecules, where electrons are evenly distributed, these shifts can create a temporary dipole. Other nearby molecules are influenced by this temporary dipole, resulting in an attraction between them.
In many molecules, LDF dominates because it is present in all substances. The strength of LDF increases with:
  • Molecular size: Larger molecules have a greater area over which these forces can interact, meaning stronger dispersion forces.
  • Molecular complexity or mass: More complex or heavier molecules have more electrons, which can shift more easily to create temporary dipoles.
Taking the molecules listed in the exercise, such as \( \mathrm{CH}_4, \mathrm{CBr}_4, \mathrm{CH}_2 \mathrm{Cl}_2 \), and so forth, we can predict that \( \mathrm{CBr}_4 \), a larger molecule, will experience stronger London dispersion forces compared to \( \mathrm{CH}_4 \), a much smaller molecule.
Boiling Points
The boiling point of a substance is the temperature at which it transitions from a liquid to a gas. This change occurs when the molecules have enough energy to overcome the intermolecular forces holding them together.
Boiling points are influenced by:
  • The strength of the intermolecular forces: Stronger forces result in higher boiling points.
  • The molecular weight and structure: More massive and complex molecules tend to have higher boiling points due to increased London dispersion forces.
For example, in the series \( \mathrm{CH}_4 \), \( \mathrm{CH}_2\mathrm{Cl}_2 \), and \( \mathrm{CBr}_4 \), the boiling points will generally increase from \( \mathrm{CH}_4 \) to \( \mathrm{CBr}_4 \). This is because \( \mathrm{CBr}_4 \) has the largest molecular size, experiencing the strongest London dispersion forces, thus needing more energy (higher temperature) to enter the gaseous state.

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Most popular questions from this chapter

The fluorocarbon compound \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) has a normal boiling point of \(47.6^{\circ} \mathrm{C}\). The specific heats of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(l)\) and \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(g)\) are \(0.91\) and \(0.67 \mathrm{~J} / \mathrm{g}-\mathrm{K}\), respectively. The heat of vaporization for the compound is \(27.49 \mathrm{~kJ} / \mathrm{mol}\). Calculate the heat required to convert \(35.0 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) from a liquid at \(10.00{ }^{\circ} \mathrm{C}\) to a gas at \(105.00{ }^{\circ} \mathrm{C}\).

(a) Two pans of water are on different burners of a stove. One pan of water is boiling vigorously, while the other is boiling gently. What can be said about the temperature of the water in the two pans? (b) A large container of water and a small one are at the same temperature. What can be said about the relative vapor pressures of the water in the two containers?

(a) What is the significance of the triple point in a phase diagram? (b) Could you measure the triple point of water by measuring the temperature in a vessel in which water vapor, liquid water, and ice are in equilibrium under 1 atm of air? Explain.

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) melts at \(-114^{\circ} \mathrm{C}\) and boils at \(78^{\circ} \mathrm{C}\). The enthalpy of fusion of ethanol is \(5.02 \mathrm{~kJ} / \mathrm{mol}\), and its enthalpy of vaporization is \(38.56 \mathrm{~kJ} / \mathrm{mol}\). The specific heats of solid and liquid ethanol are \(0.97\) and \(2.3 \mathrm{~J} / \mathrm{g}-\mathrm{K}\), respectively. (a) How much heat is required to convert \(42.0 \mathrm{~g}\) of ethanol at \(35^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C}\) ? (b) How much heat is required to convert the same amount of ethanol at \(-155^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C}\) ?

Suppose you have two colorless molecular liquids, one boiling at \(-84^{\circ} \mathrm{C}\), the other at \(34^{\circ} \mathrm{C}\), and both at atmospheric pressure. Which of the following statements is correct? For each statement that is not correct, modify the statement so that it is correct. (a) The higher-boiling liquid has greater total intermolecular forces than the lower- boiling liquid. (b) The lowerboiling liquid must consist of nonpolar molecules. (c) The lower-boiling liquid has a lower molecular weight than the higher-boiling liquid. (d) The two liquids have identical vapor pressures at their normal boiling points. (e) At \(-84^{\circ} \mathrm{C}\) both liquids have vapor pressures of \(760 \mathrm{~mm} \mathrm{Hg}\).

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