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Chapter 9: Problem 31

Explain in terms of forces between structural units why (a) HI has a higher boiling point than \(\mathrm{HBr}\). (b) \(\mathrm{GeH}_{4}\) has a higher boiling point than \(\mathrm{SiH}_{4}\). (c) \(\mathrm{H}_{2} \mathrm{O}_{2}\) has a higher melting point than \(\mathrm{C}_{3} \mathrm{H}_{8}\). (d) \(\mathrm{NaCl}\) has a higher boiling point than \(\mathrm{CH}_{3} \mathrm{OH}\).

Short Answer

Expert verified
Question: Explain the differences in boiling points and melting points of the following compounds: (a) HI vs HBr Boiling points (b) GeH4 vs SiH4 Boiling points (c) H2O2 vs C3H8 Melting points (d) NaCl vs CH3OH Boiling points Answer: (a) HI has a higher boiling point than HBr due to stronger London dispersion forces, as HI is larger in size and has more electrons than HBr. (b) GeH4 has a higher boiling point than SiH4 due to stronger London dispersion forces, as it is larger in size and has more electrons than SiH4. (c) H2O2 has a higher melting point than C3H8 due to the presence of stronger hydrogen bonding compared to London dispersion forces in C3H8. (d) NaCl has a higher boiling point than CH3OH because ion-ion interactions present in NaCl are stronger compared to the hydrogen bonding present in CH3OH.

Step by step solution

01

(a) HI vs HBr: Boiling Points

To compare the boiling points of HI and HBr, we first need to analyze the intermolecular forces between the molecules. Both molecules exhibit London dispersion forces. In general, the strength of London dispersion forces increases with the size of the molecules. Molecules with more electrons have more polarizability, which leads to greater attraction between molecules. HI has a larger molecular size and more electrons than HBr. Hence, London dispersion forces are stronger in HI, resulting in a higher boiling point than HBr.
02

(b) GeH4 vs SiH4: Boiling Points

To compare the boiling points of GeH4 and SiH4, we again need to analyze the intermolecular forces between the molecules. Both molecules consist of covalent bonds, and neither exhibits hydrogen bonding or dipole-dipole interactions. Thus, the primary intermolecular forces present are London dispersion forces. GeH4 has more electrons than SiH4, and it is a bigger molecule, leading to stronger London dispersion forces between GeH4 molecules. As a result, GeH4 has a higher boiling point than SiH4.
03

(c) H2O2 vs C3H8: Melting Points

To compare the melting points of H2O2 and C3H8, we need to focus on the intermolecular forces present in their structural units. H2O2 (hydrogen peroxide) exhibits hydrogen bonding due to the presence of O-H bonds within the molecule. C3H8 (propane) is a nonpolar molecule and does not possess any hydrogen bonding or dipole-dipole interactions. Therefore, its primary intermolecular forces are London dispersion forces. Hydrogen bonding is a much stronger intermolecular force compared to London dispersion forces. Hence, H2O2 has a higher melting point than C3H8 due to the presence of stronger intermolecular forces.
04

(d) NaCl vs CH3OH: Boiling Points

Comparing the boiling points of NaCl and CH3OH requires understanding the predominant intermolecular forces present in both compounds. NaCl is an ionic compound consisting of Na+ and Cl- ions, which interact through ion-ion interactions. These forces are quite strong compared to other intermolecular forces. CH3OH (methanol) is a polar molecule that exhibits hydrogen bonding due to the presence of O-H bonds. Although hydrogen bonding is strong compared to dispersion forces, it is weaker compared to ion-ion interactions present in NaCl. Therefore, NaCl has a higher boiling point than CH3OH due to the presence of stronger ion-ion interactions.

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Most popular questions from this chapter

Mt. McKinley in Alaska has an altitude of \(20,320 \mathrm{ft} .\) Water \(\left(\Delta H_{\mathrm{vap}}=\right.\) \(40.7 \mathrm{~kJ} / \mathrm{mol}\) ) boils at \(77^{\circ} \mathrm{C}\) atop Mt. McKinley. What is the normal atmospheric pressure at the summit?

A liquid has a vapor pressure of \(159 \mathrm{~mm} \mathrm{Hg}\) at \(20^{\circ} \mathrm{C}\) and \(165 \mathrm{~mm} \mathrm{Hg}\) at \(30^{\circ} \mathrm{C}\). Different amounts of the liquid are added to three identical evacuated steel tanks kept at \(20^{\circ} \mathrm{C}\). The tanks are all fitted with pressure gauges. For each part, write \(\mathbf{L} / \mathbf{G}\) if both liquid and gas are present. G if only gas is present. I if the situation is impossible. (a) The pressure gauge in Flask I registers a pressure of \(256 \mathrm{~mm} \mathrm{Hg}\). (b) The pressure gauge in Flask II registers a pressure of \(135 \mathrm{~mm} \mathrm{Hg}\). (c) The pressure gauge in Flask III registers a pressure of \(165 \mathrm{~mm} \mathrm{Hg}\) at \(30^{\circ} \mathrm{C}\). The temperature is lowered to \(20^{\circ} \mathrm{C}\), and the gauge registers a pressure of \(159 \mathrm{~mm} \mathrm{Hg}\).

Methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH}\), has a normal boiling point of \(64.7^{\circ} \mathrm{C}\) and has a vapor pressure of \(203 \mathrm{~mm} \mathrm{Hg}\) at \(35^{\circ} \mathrm{C}\). Estimate (a) its heat of vaporization \(\left(\Delta H_{\text {vap }}\right)\). (b) its vapor pressure at \(40.0^{\circ} \mathrm{C}\).

Which of the following statements are true? (a) The critical temperature must be reached to change liquid to gas. (b) To melt a solid at constant pressure, the temperature must be above the triple point. (c) \(\mathrm{CHF}_{3}\) can be expected to have a higher boiling point than \(\mathrm{CHCl}_{3}\) because CHF \(_{3}\) has hydrogen bonding. (d) One metal crystallizes in a body-centered cubic cell and another in a face-centered cubic cell of the same volume. The two atomic radii are related by the factor \(\sqrt{1.5}\)

Of the four general types of solids, which one(s) (a) are generally insoluble in water? (b) have very high melting points? (c) conduct electricity as solids?
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