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Chapter 11: Problem 17

Describe the intermolecular forces that must be overcome to convert these substances from a liquid to a gas: (a) SO \(_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{COOH},(\mathbf{c}) \mathrm{H}_{2} \mathrm{S}\) .

Short Answer

Expert verified
To convert (a) SO\(_{2}\) from a liquid to a gas, we must overcome London dispersion forces and dipole-dipole interactions. For (b) CH\(_{3}\)COOH, we must overcome London dispersion forces, dipole-dipole interactions, and hydrogen bonding. And for (c) H\(_{2}\)S, we must overcome London dispersion forces and dipole-dipole interactions. Increasing the temperature provides the kinetic energy needed for the molecules to overcome these intermolecular forces and enter the gaseous phase.

Step by step solution

01

(a) Intermolecular forces in SO\(_{2}\)

(Sulfur dioxide, or SO\(_{2}\), is a polar molecule due to the differences in electronegativity between sulfur and oxygen atoms. This results in two types of intermolecular forces: London dispersion forces, which are present in all molecules, and dipole-dipole interactions, which are specific to polar molecules.)
02

(a) Overcoming these forces

(To convert SO\(_{2}\) from a liquid to a gas, we must provide energy to overcome both the London dispersion forces and the dipole-dipole interactions. As the temperature increases, the kinetic energy of SO\(_{2}\) molecules increases, allowing them to overcome these forces and enter the gaseous phase.)
03

(b) Intermolecular forces in CH\(_{3}\)COOH

(Acetic acid, or CH\(_{3}\)COOH, is a polar molecule with an additional feature, an -OH group. This results in three types of intermolecular forces being present: London dispersion forces, dipole-dipole interactions, and hydrogen bonding, the latter of which occurs due to the presence of hydrogen atoms bonded to highly electronegative oxygen atoms.)
04

(b) Overcoming these forces

(To convert CH\(_{3}\)COOH from a liquid to a gas, we must provide energy to overcome all three intermolecular forces: London dispersion forces, dipole-dipole interactions, and hydrogen bonding. As the temperature increases, the kinetic energy of CH\(_{3}\)COOH molecules increases, allowing them to overcome these forces and enter the gaseous phase.)
05

(c) Intermolecular forces in H\(_{2}\)S

(Hydrogen sulfide, or H\(_{2}\)S, is also a polar molecule due to the difference in electronegativity between hydrogen and sulfur atoms. This results in two types of intermolecular forces being present: London dispersion forces and dipole-dipole interactions.)
06

(c) Overcoming these forces

(To convert H\(_{2}\)S from a liquid to a gas, we must provide energy to overcome both the London dispersion forces and the dipole-dipole interactions. As the temperature increases, the kinetic energy of H\(_{2}\)S molecules increases, allowing them to overcome these forces and enter the gaseous phase.)

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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 the weakest type of intermolecular forces and are present in all molecules. They arise from temporary fluctuations in electron density within an atom or molecule. At any given moment, the electrons may be distributed unevenly, creating a temporary dipole. This temporary dipole can induce a similar dipole in a neighboring molecule, leading to a weak attraction between them.

Despite their weak nature, London dispersion forces play a significant role in the behavior of molecules, especially nonpolar ones. The strength of these forces increases with greater molecular size and mass, as larger molecules have more electrons that can contribute to temporary dipole formation.
  • Present in all molecules
  • Affect weaker and nonpolar molecules more significantly
  • Increase with greater molecular size and mass
Dipole-Dipole Interactions
Dipole-dipole interactions occur between polar molecules, where there is an uneven distribution of electron density resulting in partial charges, or dipoles, within the molecule. The positive end of one polar molecule is attracted to the negative end of another, leading to an attractive force between them.

These interactions are generally stronger than London dispersion forces because the dipoles are permanent rather than temporary. They are especially important in the properties of polar substances, influencing boiling points, melting points, and solubility.
  • Occur in polar molecules
  • Strong due to permanent dipoles
  • Influence physical properties like boiling point
Hydrogen Bonding
Hydrogen bonding is a specific, stronger type of dipole-dipole interaction that occurs when hydrogen is covalently bonded to a highly electronegative atom, such as oxygen, nitrogen, or fluorine. This interaction is due to the large difference in electronegativity, which leaves hydrogen with a partial positive charge and the other element with a partial negative charge.

These bonds have a significant effect on the properties of compounds, including raising boiling and melting points, increasing viscosity, and influencing solubility. They are responsible for many unique properties of substances, such as the high boiling point of water.
  • Occurs with hydrogen bonded to O, N, or F
  • Stronger than typical dipole-dipole interactions
  • Significantly affects physical properties

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Most popular questions from this chapter

In terms of the arrangement and freedom of motion of the molecules, how are the nematic liquid crystalline phase and an ordinary liquid phase similar? How are they different?

Hydrazine \(\left(\mathrm{H}_{2} \mathrm{NNH}_{2}\right),\) hydrogen peroxide \((\mathrm{HOOH}),\) and water \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) all have exceptionally high surface tensions compared with other substances of comparable molecular weights. (a) Draw the Lewis structures for these three compounds. (b) What structural property do these substances have in common, and how might that account for the high surface tensions?

Liquid butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) is stored in cylinders to be used as a fuel. The normal boiling point of butane is listed as \(-0.5^{\circ} \mathrm{C}\) . (a) Suppose the tank is standing in the sun and reaches a temperature of \(35^{\circ} \mathrm{C}\) . Would you expect the pressure in the tank to be greater or less than atmospheric pressure? How does the pressure within the tank depend on how much liquid butane is in it? (b) Suppose the valve to the tank is opened and a few liters of butane are allowed to escape rapidly. What do you expect would happen to the temperature of the remaining liquid butane in the tank? Explain. (c) How much heat must be added to vaporize 250 \(\mathrm{g}\) of butane if its heat of vaporization is 21.3 \(\mathrm{kJ} / \mathrm{mol}\) ? What volume does this much butane occupy at 755 torr and \(35^{\circ} \mathrm{C} ?\)

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 \(78^{\circ} \mathrm{C} ?(\mathbf{b})\) How much heat is required to convert the same amount of ethanol at \(-155^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C} ?\)

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