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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 \(\left(-0.5^{\circ} \mathrm{C}\right.\) compared to \(\left.-11.7{ }^{\circ} \mathrm{C}\right)\). Explain.

Short Answer

Expert verified
Butane has a higher boiling point because its straight-chain structure results in stronger dispersion forces, requiring more energy to change state.

Step by step solution

01

Understanding Molecular Structure

Both butane and 2-methylpropane have the same molecular formula, \(\mathrm{C}_4\mathrm{H}_{10}\), but their structures differ. Butane is a straight-chain alkane, while 2-methylpropane is a branched structure.
02

Interaction Between Molecules

Because butane is a straight-chain molecule, it has a larger surface area in contact with other butane molecules, leading to slightly stronger dispersion forces compared to the more compact, branched structure of 2-methylpropane.
03

Boiling Point Correlation

Stronger intermolecular forces in butane require more energy (in the form of heat) to overcome when changing from a liquid to a gas, resulting in a higher boiling point compared to 2-methylpropane.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Structure
When we talk about molecular structure, we're referring to the arrangement of atoms within a molecule. Even if two substances have the same molecular formula, their structure can make a significant difference. In the case of butane and 2-methylpropane, both share the same molecular formula \( ext{C}_4 ext{H}_{10} \). But it's their structural geometry that distinguishes them. Butane is a linear molecule, which means it forms a line-like shape, while 2-methylpropane is branched.

This structural difference affects how closely these molecules pack together. Straight-chain molecules like butane often have more contact with each other compared to branched molecules. This is important when we consider how these molecules interact at the molecular level.
Dispersion Forces
Dispersion forces, also known as London dispersion forces, are a type of intermolecular force. These are weak attractive forces that arise due to temporary dipoles induced when atoms shift in electron density. This effect is universal and occurs in all atoms and molecules, but it is particularly relevant in nonpolar substances like butane and 2-methylpropane.

In butane, the linear configuration offers a more extended surface area for these temporary dipoles to interact. This results in increased dispersion forces between butane molecules compared to the more compact and less accessible structure of 2-methylpropane. Supporting stronger interactions among the molecules, these forces increase the energy needed to separate them, leading to higher boiling points.
Intermolecular Forces
Intermolecular forces are forces of attraction between molecules, which can greatly affect a substance's physical properties. Boiling point is one of these properties influenced by intermolecular forces. The main types of intermolecular forces are London dispersion forces, dipole-dipole interactions, and hydrogen bonds.

For butane and 2-methylpropane, which are both nonpolar, the most significant intermolecular force at play is the London dispersion force. Since butane provides an elongated shape, it affords stronger dispersion forces compared to the more compact 2-methylpropane. This means that as butane molecules attract each other more strongly, it takes more heat energy to convert them from a liquid state to a gaseous state. Therefore, butane has a higher boiling point than 2-methylpropane due to stronger intermolecular attractions.

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Most popular questions from this chapter

The table shown here lists the molar heats of vaporization for several organic compounds. Use specific examples from this list to illustrate how the heat of vaporization varies with (a) molar mass, (b) molecular shape, \((\mathbf{c})\) molecular polarity, (d) hydrogen-bonding interactions. Explain these comparisons in terms of the nature of the intermolecular forces at work. (You may find it helpful to draw out the structural formula for each compound.) \begin{tabular}{lc} \hline Compound & Heat of Vaporization (kJ/mol) \\ \hline \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) & 19.0 \\ \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) & 27.6 \\ \(\mathrm{CH}_{3} \mathrm{CHBrCH}_{3}\) & 31.8 \\ \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) & 32.0 \\ \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\) & 33.6 \\ \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) & 47.3 \\ \hline \end{tabular}

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) is pressurized into liquid and stored in cylinders to be used as a fuel. The normal boiling point of propane is listed as \(-42^{\circ} \mathrm{C}\). (a) When converting propane into liquid at room temperature of \(25^{\circ} \mathrm{C}\), 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 propane is in it? (b) Suppose the fuel tank leaks and a few liters of propane escape rapidly. What do you expect would happen to the temperature of the remaining liquid propane in the tank? Explain. (c) How much heat must be added to vaporize \(20 \mathrm{~g}\) of propane if its heat of vaporization is \(18.8 \mathrm{~kJ} / \mathrm{mol} ?\) What volume does this amount of propane occupy at \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C} ?\)

(a) Do you expect the viscosity of glycerol, \(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\), to be larger or smaller than that of 1 -propanol, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH} ?\) (b) Explain. [Section 11.3\(]\)

(a) Two pans of water are on different burners of a stove. One pan of water is boiling vigorously, while the other is boiling gently. What can be said about the temperature of the water in the two pans? (b) A large container of water and a small one are at the same temperature. What can be said about the relative vapor pressures of the water in the two containers?

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the Clausius-Clapeyron 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 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 \(1.86 \mathrm{kPa}\) at \(25^{\circ} \mathrm{C}\) and a vapor pressure of \(19.3 \mathrm{kPa}\) at \(75^{\circ} \mathrm{C}\). Use these data and the equation in part (a) to calculate the heat of vaporization of octane. \((\mathbf{c})\) 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 11.81 . (d) Calculate the vapor pressure of octane at \(-30^{\circ} \mathrm{C}\).

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