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Chapter 11: Problem 18

Which type of intermolecular force accounts for each of these differences: (a) \(\mathrm{CH}_{3} \mathrm{OH}\) boils at \(65^{\circ} \mathrm{C} ; \mathrm{CH}_{3} \mathrm{SH}\) boils at \(6^{\circ} \mathrm{C}\). (b) Xe is liquid at atmospheric pressure and \(120 \mathrm{~K}\), whereas \(\mathrm{Ar}\) is a gas under the same conditions. (c) \(\mathrm{Kr}\), atomic weight 84 , boils at \(120.9 \mathrm{~K},\) whereas \(\mathrm{Cl}_{2},\) molecular weight about \(71,\) boils at \(238 \mathrm{~K}\). (d) Acetone boils at \(56^{\circ} \mathrm{C}\), whereas 2 -methylpropane boils at \(-12^{\circ} \mathrm{C}\)

Short Answer

Expert verified
(a) The difference in boiling points between CH₃OH and CH₃SH can be attributed to hydrogen bonding in methanol. (b) The difference between Xe and Ar can be explained by the stronger London dispersion forces in Xenon. (c) The higher boiling point of Cl₂ compared to Kr is due to stronger dipole-dipole interactions in Cl₂. (d) The difference in boiling points between acetone and 2-methylpropane can be attributed to the stronger dipole-dipole interactions in acetone.

Step by step solution

01

(a) Comparing CH3OH and CH3SH boiling points

We need to compare methanol (CH3OH) with methanethiol (CH3SH). Methanol is an alcohol with an OH group that can participate in hydrogen bonding, a strong intermolecular force. Methanethiol, on the other hand, has an SH group which lacks the hydrogen bonding capability. Both molecules have similar sizes, so the difference in boiling points (65°C for CH3OH and 6°C for CH3SH) can be attributed to the presence of hydrogen bonding in methanol.
02

(b) Comparing Xe and Ar properties at 120 K

Xenon (Xe) and Argon (Ar) are both noble gases, which means that they do not form molecules and have complete electron shells. The only intermolecular force present in these gases is the London dispersion force. As Xe has a larger atomic mass and more electrons than Ar, the London dispersion forces are stronger in Xenon, causing it to be in a liquid state at 120 K and atmospheric pressure. In contrast, Ar is still in the gaseous state under the same conditions.
03

(c) Comparing boiling points of Kr and Cl2

Krypton (Kr) is a noble gas with an atomic weight of 84, and Cl2 (chlorine) is a diatomic molecule with a molecular weight of about 71. In the case of Kr, the only intermolecular force present is the London dispersion force. For Cl2, there is a dipole-dipole interaction between the molecules due to the presence of polar covalent bonds between the two chlorine atoms. The dipole-dipole interaction in Cl2 is stronger than the dispersion forces in Kr, which explains the higher boiling point of Cl2 (238 K) compared to Kr (120.9 K).
04

(d) Comparing boiling points of acetone and 2-methylpropane

Acetone is a polar molecule with a ketone functional group (C=O), whereas 2-methylpropane (also known as isobutane) is a nonpolar hydrocarbon. The primary intermolecular forces in acetone are dipole-dipole interactions between the polar C=O groups, while the primary intermolecular forces in 2-methylpropane are the weaker London dispersion forces. This difference in intermolecular forces accounts for the difference in boiling points, with acetone boiling at 56°C and 2-methylpropane boiling at -12°C.

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Most popular questions from this chapter

Propyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) and isopropyl alcohol \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH}\right],\) whose space- filling models are shown, have boiling points of \(97.2^{\circ} \mathrm{C}\) and \(82.5^{\circ} \mathrm{C}\), respectively. Explain why the boiling point of propyl alcohol is higher, even though both have the molecular formula \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\).

The relative humidity of air equals the ratio of the partial pressure of water in the air to the equilibrium vapor pressure of water at the same temperature times \(100 \% .\) If the relative humidity of the air is \(58 \%\) and its temperature is \(68^{\circ} \mathrm{F}\), how many molecules of water are present in a room measuring \(12 \mathrm{ft} \times 10 \mathrm{ft} \times 8 \mathrm{ft} ?\)

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?

Rationalize the difference in boiling points in each pair: (a) \(\mathrm{HF}\left(20^{\circ} \mathrm{C}\right)\) and \(\mathrm{HCl}\left(-85^{\circ} \mathrm{C}\right),(\mathbf{b}) \mathrm{CHCl}_{3}\left(61{ }^{\circ} \mathrm{C}\right)\) and \(\mathrm{CHBr}_{3}\) \(\left(150^{\circ} \mathrm{C}\right),(\mathrm{c}) \mathrm{Br}_{2}\left(59^{\circ} \mathrm{C}\right)\) and \(\mathrm{ICl}\left(97^{\circ} \mathrm{C}\right)\)

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 \(0.97 \mathrm{~J} / \mathrm{g}-\mathrm{K}\) 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 \(78^{\circ} \mathrm{C}\) ?
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