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Chapter 11: Problem 23

Butane and 2 -methylpropane, whose space-filling models are shown here, are both nonpolar and have the same molecular formula, \(\mathrm{C}_{4} \mathrm{H}_{10},\) yet butane has the higher boiling point \(\left(-0.5^{\circ} \mathrm{C}\) compared to \(-11.7^{\circ} \mathrm{C}\right) .\) Explain.

Short Answer

Expert verified
Although both butane and 2-methylpropane have the same molecular formula \(\mathrm{C}_{4}\mathrm{H}_{10}\) and are nonpolar, butane has a higher boiling point due to its linear structure, which allows for a larger surface area contact between molecules and leads to stronger London dispersion forces. In contrast, 2-methylpropane's branched structure results in weaker London dispersion forces, leading to a lower boiling point.

Step by step solution

01

Understand the molecular structures of butane and 2-methylpropane

First, let's analyze the structures of butane and 2-methylpropane. Both molecules have the molecular formula \(\mathrm{C}_{4}\mathrm{H}_{10}\) and are hydrocarbons. The molecule butane has a linear (or more precisely zigzag) structure, while 2-methylpropane has a branched structure with three carbon atoms in the main chain and one methyl group attached to the second carbon atom.
02

Analyze the polarity of butane and 2-methylpropane

Both butane and 2-methylpropane are nonpolar molecules. In each case, the carbon-hydrogen bonds within the molecules are not polar enough to give the overall molecule any significant polarity. Additionally, due to the symmetry of both molecules, any small dipoles that may be present will cancel each other out, resulting in nonpolar molecules.
03

Understand boiling points

Boiling points are determined by the strength of the intermolecular forces between molecules. The stronger the intermolecular forces, the higher the boiling point. The primary types of intermolecular forces are hydrogen bonding, dipole-dipole interactions, and London dispersion forces. Since both butane and 2-methylpropane are nonpolar, their intermolecular forces are primarily due to London dispersion forces. Such forces are temporary, weak attractive forces due to temporary fluctuations in the electron distribution of the molecules.
04

Relate molecular structures to London dispersion forces

London dispersion forces are influenced by the molecular shape and surface area. In general, larger surface areas tend to create stronger London dispersion forces. Butane's linear structure has a larger surface area in contact with neighboring molecules, which allows for more points of interaction and thus stronger London dispersion forces. In contrast, 2-methylpropane’s branched structure has a smaller surface area of contact due to its more spherical nature, which results in weaker London dispersion forces compared to butane.
05

Compare boiling points

Due to the stronger London dispersion forces between butane molecules as a result of its larger surface area, butane has a higher boiling point (-0.5°C) than 2-methylpropane (-11.7°C). Therefore, the difference in boiling points between the two molecules can be explained by their differing molecular structures.

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Most popular questions from this chapter

Look up and compare the normal boiling points and normal melting points of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{H}_{2} \mathrm{S}\) . Based on these physical properties, which substance has stronger intermolecular forces? What kinds of intermolecular forces exist for each molecule?

The fluorocarbon compound \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{F}_{3}\) has a normal boiling point of \(47.6^{\circ} \mathrm{C}\) . The specific heats of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{F}_{3}(l)\) and \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{F}_{3}(g)\) are 0.91 and \(0.67 \mathrm{J} / \mathrm{g}-\mathrm{K}\) , respectively. The heat of vaporization for the compound is 27.49 \(\mathrm{kJ} / \mathrm{mol}\) . Calculate the heat required to convert 35.0 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{F}_{3}\) from a liquid at \(10.00^{\circ} \mathrm{C}\) to a gas at \(105.00^{\circ} \mathrm{C}\) .

(a) What atoms must a molecule contain to participate in hydrogen bonding with other molecules of the same kind? (b) Which of the following molecules can form hydrogen bonds with other molecules of the same kind: \(\mathrm{CH}_{3} \mathrm{F}, \mathrm{CH}_{3} \mathrm{NH}_{2}, \mathrm{CH}_{3} \mathrm{OH}, \mathrm{CH}_{3} \mathrm{B}\) ?

Indicate whether each statement is true or false: (a) The critical pressure of a substance is the pressure at which it turns into a solid at room temperature. (b) The critical temperature of a substance is the highest temperature at which the liquid phase can form. (c) Generally speaking, the higher the critical temperature of a substance, the lower its critical pressure. (\boldsymbol{d} ) In general the more intermolecular forces there are in a substance, the higher its critical temperature and pressure.

(a) What phase change is represented by the "heat of fusion" of a substance? (b) Is the heat of fusion endothermic or exothermic? (c) If you compare a substance's heat of fusion to its heat of vaporization, which one is generally larger?
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