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One of the attractive features of ionic liquids is their low vapor pressure, which in turn tends to make them nonflammable. Why do you think ionic liquids have lower vapor pressures than most room-temperature molecular liquids?

Short Answer

Expert verified
Ionic liquids have lower vapor pressures than most room-temperature molecular liquids because their particles are held together by strong ionic bonds, which require more energy to break compared to the weaker intermolecular forces in molecular liquids. This leads to low vapor pressure and makes ionic liquids nonflammable.

Step by step solution

01

Introduction to Vapor Pressure

Vapor pressure is the pressure exerted by the vapor in equilibrium with its liquid at a given temperature. It is an important factor when studying the boiling points, evaporation, and condensation of liquids. The magnitude of vapor pressure depends on the attractive forces between the particles in the liquid and the amount of energy required to break these forces and allow those particles to escape to the vapor phase. If stronger forces are present, the particles need more energy to escape, and the vapor pressure will be low. Therefore, we need to look at the intermolecular forces in ionic and molecular liquids to understand the difference in their vapor pressures.
02

Ionic Liquids and Molecular Liquids

Ionic liquids are composed of ions (e.g., positive and negative ions) held together by strong electrostatic forces, called ionic bonds. In contrast, molecular liquids are composed of neutral molecules held together by weaker intermolecular forces, such as van der Waals forces, dipole-dipole interactions, and hydrogen bonds. To understand the difference in vapor pressures, we need to compare the strength of these forces in ionic and molecular liquids.
03

Comparing Intermolecular Forces

At a basic level, we can compare the forces by noting that ionic bonds are usually stronger than the intermolecular forces present in molecular liquids. Ionic bonds are the result of the strong electrostatic attraction between positive and negative ions, while the forces between neutral molecules, such as van der Waals forces and dipole-dipole interactions, are much weaker. Even hydrogen bonds, which are relatively stronger than other intermolecular forces, are still weaker than ionic bonds.
04

Vapor Pressure Relationship to Intermolecular Forces

As we established earlier, the strength of the forces between particles determines the vapor pressure of a liquid. Since ionic bonds are stronger than intermolecular forces in molecular liquids, ionic liquids require more energy to break these bonds and allow particles to escape into the vapor phase. This leads to a lower vapor pressure in ionic liquids compared to molecular liquids.
05

Conclusion

Ionic liquids have lower vapor pressures than most room-temperature molecular liquids because they have stronger attractive forces (ionic bonds) between their particles compared to the weaker intermolecular forces found in molecular liquids. This means that ionic liquids require more energy to break these bonds and evaporate, resulting in a lower vapor pressure. Hence, ionic liquids tend to be nonflammable due to their low vapor pressure.

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Most popular questions from this chapter

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the ClausiusClapeyron equation (Equation 11.1) derive the following relationship between the vapor pressures, \(P_{1}\) and \(P_{2}\), and the absolute temperatures at which they were measured, \(T_{1}\) and \(T_{2}:\) $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\text {vap }}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (b) Gasoline is a mixture of hydrocarbons, a major component of which is octane \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)\). Octane has a vapor pressure of \(13.95\) torr at \(25^{\circ} \mathrm{C}\) and a vapor pressure of \(144.78\) torr at \(75^{\circ} \mathrm{C}\). Use these data and the equation in part (a) to calculate the heat of vaporization of octane. (c) By using the equation in part (a) and the data given in part (b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise 11.80. (d) Calculate the vapor pressure of octane at \(-30^{\circ} \mathrm{C}\).

If \(42.0 \mathrm{~kJ}\) of heat is added to a \(32.0\) - \(\mathrm{g}\) sample of liquid methane under 1 atm of pressure at a temperature of \(-170^{\circ} \mathrm{C}\), what are the final state and temperature of the methane once the system equilibrates? Assume no heat is lost to the surroundings. The normal boiling point of methane is \(-161.5^{\circ} \mathrm{C}\). The specific heats of liquid and gaseous methane are \(3.48\) and \(2.22 \mathrm{~J} / \mathrm{g}-\mathrm{K}\), respectively. [Section 11.4]

For a given substance, the liquid crystalline phase tends to be more viscous than the liquid phase. Why?

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 vapor pressure at the same temperature than \(\mathrm{CCl}_{4}\).

Which member in each pair has the greater dispersion forces? (a) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{H}_{2} \mathrm{~S}\), (b) \(\mathrm{CO}_{2}\) or \(\mathrm{CO}\), (c) \(\mathrm{SiH}_{4}\) or \(\mathrm{GeH}_{4}\).

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