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True or false: (a) \(\mathrm{CBr}_{4}\) is more volatile than \(\mathrm{CCl}_{4}\). (b) \(\mathrm{CBr}_{4}\) has a higher boiling point than \(\mathrm{CCl}_{4}\). (c) \(\mathrm{CBr}_{4}\) has weaker intermolecular forces than \(\mathrm{CCl}_{4}\). (d) \(\mathrm{CBr}_{4}\) has a higher yapor pressure at the same temperature than \(C O\)

Short Answer

Expert verified
(a) False, (b) True, (c) False, (d) False.

Step by step solution

01

Volatility Comparison

Substances with higher volatility tend to have weaker intermolecular forces, resulting in easier evaporation. Volatility is inversely related to boiling point. In comparing the boiling points of carbon tetrahalides, \( \mathrm{CCl}_{4} \) has a boiling point of approximately 76.7°C, while \( \mathrm{CBr}_{4} \) boils at about 189.5°C. Since \( \mathrm{CBr}_{4} \) has a higher boiling point, it is less volatile than \( \mathrm{CCl}_{4} \). Thus, statement (a) is false.
02

Boiling Point Analysis

The boiling point is influenced by the strength of intermolecular forces. Stronger forces result in higher boiling points. Given that the boiling point of \( \mathrm{CBr}_{4} \) is 189.5°C as compared to 76.7°C for \( \mathrm{CCl}_{4} \), \( \mathrm{CBr}_{4} \) indeed has a higher boiling point. Thus, statement (b) is true.
03

Intermolecular Forces Evaluation

Intermolecular forces such as London dispersion forces increase with molar mass and size of the electron cloud.\( \mathrm{CBr}_{4} \) has a higher molar mass and a larger electron cloud compared to \( \mathrm{CCl}_{4} \), thus, has stronger intermolecular forces than \( \mathrm{CCl}_{4} \). Hence, statement (c) is false.
04

Vapor Pressure Comparison

At a given temperature, substances with higher vapor pressures evaporate more easily. \( \mathrm{CO} \), being smaller and lighter with weaker intermolecular forces, is more volatile than \( \mathrm{CBr}_{4} \), resulting in a higher vapor pressure compared to \( \mathrm{CBr}_{4} \).Therefore, statement (d) is false.

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Key Concepts

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

Understanding Volatility
Volatility is a measure of how easily a substance evaporates. In essence, it tells us how readily molecules in a liquid escape into the gas phase. Imagine a liquid that quickly turns into vapor at room temperature; such substances are considered to be highly volatile. Volatility is primarily influenced by the strength of intermolecular forces within a substance. These forces act as a barrier against evaporation. When intermolecular forces are weak, molecules can escape the liquid phase easily, resulting in higher volatility. Conversely, strong intermolecular forces mean that more energy is needed for molecules to evaporate, leading to lower volatility.A key indicator of volatility is the boiling point. The lower the boiling point, the higher the volatility, because it implies that less heat is needed for evaporation to occur. This inverse relationship helps in comparing the volatility of substances like carbon tetrahalides. For example, since \( \mathrm{CCl}_{4} \) boils at a lower temperature than \( \mathrm{CBr}_{4} \), it is therefore more volatile.
All About Boiling Point
The boiling point of a substance is the temperature at which its liquid phase converts to gas as rapidly as the pressure allows. At this temperature, the vapor pressure of the liquid becomes equal to the pressure exerted on the liquid by the surrounding environment, causing the liquid molecules to break free and form gas. Boiling points serve as an indicator of the strength of a substance's intermolecular forces. Specifically, the stronger these forces are, the more energy (and hence higher temperature) is necessary to transition into the gas phase. Hence, substances with higher boiling points tend to have stronger intermolecular forces.Several factors influence the boiling point:
  • Molar Mass: Generally, an increase in molar mass results in higher boiling points due to greater London dispersion forces.
  • Intermolecular Force Types: Forces like hydrogen bonding and dipole-dipole interactions can significantly raise boiling points.
Thus, in comparing \( \mathrm{CCl}_{4} \) and \( \mathrm{CBr}_{4} \), the higher boiling point of \( \mathrm{CBr}_{4} \) indicates its stronger intermolecular forces.
Explaining Vapor Pressure
Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid form. It reflects the tendency of molecules to escape from the liquid or solid into the gas phase. Generally, higher vapor pressures correspond to more volatile substances, as more molecules are in the vapor phase at a given time.Factors Influencing Vapor Pressure:
  • Temperature: As temperature increases, vapor pressure rises because more molecules have the kinetic energy needed to escape into vapor.
  • Intermolecular Forces: Weaker forces within a liquid allow more molecules to escape, resulting in higher vapor pressure.
To illustrate, consider carbon monoxide (\( \mathrm{CO} \)) and \( \mathrm{CBr}_{4} \). \( \mathrm{CO} \), having weaker intermolecular forces, exhibits higher vapor pressure than \( \mathrm{CBr}_{4} \). This makes \( \mathrm{CO} \) more conducive to evaporation while \( \mathrm{CBr}_{4} \) remains mostly in its liquid state at the same temperature.

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Most popular questions from this chapter

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

(a) List the following molecules in order of increasing polarizability: \(\mathrm{GeCl}_{4}, \mathrm{CH}_{4}, \mathrm{SiCl}_{4}, \mathrm{SiH}_{4}\), and \(\mathrm{GeBr}_{4}\) (b) Predict the order of boiling points of the substances in part (a).

(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 \(40^{\circ} \mathrm{C}\).

The molecules Propanol \(\quad\) Ethyl methyl ether have the same molecular formula \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\right)\) but different chemical structures, (a) Which molecule(s), if any, can engage in hydrogen bonding? (b) Which molecule do you expect to have a larger dipole moment? \((\mathbf{c})\) One of these \(\mathrm{mol}\) ecules has a normal boiling point of \(97,2^{\circ} \mathrm{C},\) while the other one has a normal boiling point of \(10.8^{\circ} \mathrm{C}\). Assign each molecule to its normal boiling point. [Sections 11.2 and 11.5\(]\)

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