Chem 101 Chapter 13- Intermolecular Forces, Liquids, and Solids

Chem 101 Chapter 13- Intermolecular Forces, Liquids, and Solids
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Unformatted text preview: States of Matter 13— Intermolecular Forces, Liquids, and Solids The Mystery of the Disappearing Fingerprints The events of September 11, 2001, are etched in everyone’s memory. The specter of domestic terrorism, however, was first raised almost two years earlier. That’s when a man was apprehended in December 1999 at the U.S.–Canadian border with bomb materials and a map of Los Angeles International Airport. Although he claimed innocence, his fingerprints were on the bomb materials, and he was convicted of an attempt to bomb the airport. Each of us has a unique fingerprint pattern, as first described by John Purkinji in 1823. Not long after his discovery, English colonists in India began using fingerprints on contracts because they believed it made the contract appear more binding. Not until late in the 19th century, however, was fingerprinting used as an identifier. Sir Francis Galton, a British anthropologist and a cousin of Charles Darwin, established that a person’s fingerprints do not Taking an official fingerprint at the local police station. Charles D. Winters Dusting for fingerprints on a glass coffee mug. 588 Charles D. Winters Chapter Goals See Chapter Goals Revisited (page 633). Test your knowledge of these goals by taking the exam-prep quiz on the General ChemistryNow CD-ROM or website. 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9 Chapter Outline States of Matter and the Kinetic-Molecular Theory Intermolecular Forces Hydrogen Bonding Summary of Intermolecular Forces Properties of Liquids The Solid State: Metals The Solid State: Structures and Formulas of Ionic Solids Other Kinds of Solid Materials The Physical Properties of Solids • • • • • • • Describe intermolecular forces and their effects. Understand the importance of hydrogen bonding. Understand the properties of liquids. Understand cubic unit cells. Relate unit cells for ionic compounds to formulas. Describe the properties of solids. Understand the nature of phase diagrams. 13.10 Phase Diagrams change over the course of a lifetime and that no two prints are exactly the same. Fingerprinting has since become an accepted tool in forensic science. In 1993 in Knoxville, Tennessee, Art Bohanan thought he could use it to solve the case of the kidnapping of a young girl. The girl had been taken from her home and driven away in a green car. The girl soon managed to escape from her attacker and was able to describe the car to the police. Close-up of a fingerprint. After four days the police found the car and arrested its owner. But had the girl been in that car? Art Bohanan inspected the car for her fingerprints and even used the latest technique, fuming with superglue. No prints were found. The abductor of the girl was eventually convicted on the basis of other evidence, but Bohanan wondered why he had never found her prints in the car. He decided to test the permanence of children’s fingerprints compared with adults’ fingerprints. To his amazement, he found that children’s prints disappear in a few hours, whereas an adult’s prints can last for days. Bohanan said, “It sounded like the compounds in children’s fingerprints might simply be evaporating faster than adult’s.” The residue deposited by fingerprints is 99% water. The other 1% contains oils, fatty acids, esters, salts, urea, and amino acids. Scientists at Oak Ridge National Laboratory studied the fingerprints of 50 child and adult volunteers, identifying the compounds present by such techniques as mass spectrometry [ page 127]. What they found clarified the mystery of the disappearing fingerprints. Children’s fingerprints contain more low-molecular-weight fatty acids than do adult fingerprints. (Fatty acids consist of a carbon– hydrogen chain with a carboxylic acid group, ¬ CO2H, at one end. See page 505.) Due to the relatively low molecular weight and low polarity of these acids, their intermolecular forces are weak and the compounds are volatile. As a consequence, children’s fingerprints simply evaporate. In contrast, adult fingerprints contain esters of long-chain fatty acids with long-chain alcohols. These are waxes, semisolid or solid organic compounds with high molecular weights and low volatility. Examples of waxes are lanolin, a component of wool, and the carnuba wax used in furniture polish. Carnuba wax Charles D. Winters O CH3(CH2)30 C O (CH2)33CH3 Portion from fatty acid Portion from longchain alcohol Before puberty, children do not produce waxy compounds in their skin. However, sebaceous glands in adult skin produce sebum, a complex mixture of organic compounds (triglycerides, fatty acids, cholesterol, and waxes). Only a few of these glands are found on the hands; most are located on the mid-back, forehead, and chin. When you touch your face, this mixture of compounds is transferred to your fingers, and you can leave a fingerprint that is unique to you. 589 590 Chapter 13 Intermolecular Forces, Liquids, and Solids To Review Before You Begin • Review ion–ion attraction in ionic compounds (page 377) • Know how to use electronegativity to determine the polarity of covalent bonds (Section 9.8) • Be able to determine the polarity of molecules (Section 9.9). • • • • Throughout the chapter this icon introduces a list of resources on the General ChemistryNow CD-ROM or website (http://now that will: help you evaluate your knowledge of the material provide homework problems allow you to take an exam-prep quiz provide a personalized Learning Plan targeting resources that address areas you should study he vast majority of the known chemical elements are solids at 25 °C and 1 atm of pressure. Only 11 elements occur as gases under these conditions (H2, N2, O2, F2, Cl2, and the six noble gases), and only two elements occur as liquids (Hg and Br2). Many common compounds are gases (such as CO2 and CH4) or liquids (H2O) at standard temperature and pressure, but, as is the case with the elements, the largest number of compounds are solids. The primary objective in this chapter is to elucidate the macroscopic properties of the liquid and solid states by looking at the particulate level—that is, the level of atoms, molecules, and ions. You will find this a useful chapter because it explains, among other things, why your body is cooled when you sweat, how bodies of water can influence local climate, why one form of pure carbon (diamond) is hard and another (graphite) is slippery, and why many solid compounds form beautiful crystals. T 13.1—States of Matter and the Kinetic-Molecular Theory The kinetic-molecular theory of gases [ Section 12.6] assumes that gas molecules or atoms are widely separated and that these particles can be considered to be independent of one another. Consequently, we can relate the properties of gases under most conditions by a simple mathematical equation, PV nRT, known as the ideal gas law equation (Equation 12.4). Liquids and solids present a more complicated picture, however. In these states, the particles are close enough together that attractive forces between them can have a considerable effect. The ideal gas law is valid because, at the temperatures and pressures at which gases exist, we can usually ignore these forces. In contrast, when these forces are introduced in liquids and solids, it is not possible to create a simple “ideal liquid equation” or “ideal solid equation.” How different are the states of matter at the particulate level? We can get a sense of this by comparing volumes occupied by equal numbers of molecules of a material in different states. Figure 13.1a shows a flask containing about 300 mL of liquid nitrogen. If all of the liquid were allowed to evaporate, the gaseous nitrogen would fill a large balloon (more than 200 L volume) to a pressure of 1 atm at room temperature. A large amount of space exists between molecules in a gas, whereas in liquids the molecules are close together. The increase in volume when converting liquids to gases is strikingly large. In contrast, no dramatic change in volume occurs when a solid is converted to a liquid. Figure 13.1b shows the same amount of liquid and solid benzene side by side. As you see, they are not appreciably different in volume. This means that the atoms in the liquid are packed together about as tightly as the atoms in the solid phase. We know that gases can be compressed easily, a process that involves forcing the gas molecules closer together. The air–fuel mixture in your car’s engine, for example, is routinely compressed by a factor of about 10 before it is ignited. In contrast, the molecules, ions, or atoms in liquid or solid phases strongly resist forces that 13.2 Intermolecular Forces 591 Nitrogen gas Figure 13.1 Contrasting gases, liquids, and solids. (a) When a 300-mL sample of liquid nitrogen evaporates, it will produce more than 200 L of gas at 25 °C and 1.0 atm. In the liquid phase, the molecules of N2 are close together; in the gas phase, they are far apart. (b) The same volume of liquid benzene, C6H6, is placed in two test tubes, and one tube (right) is cooled, freezing the liquid. The solid and liquid states have almost the same volume, showing that the molecules are packed together almost as tightly in the liquid state as they are in the solid state. Photos: Charles. D. Winters Liquid nitrogen (a) (b) Liquid benzene Solid benzene would push them closer together. Thus, a characteristic of liquids and solids is a lack of compressibility. For example, the volume of liquid water changes only by 0.005% per atmosphere of pressure applied to it. In the gaseous state, atoms or molecules are relatively far apart because the forces between the particles are not strong enough to pull them together and overcome their kinetic energy. In liquids and solids, much stronger forces pull the particles together and limit their motion. In solid ionic compounds, the positively and negatively charged ions are held together by electrostatic attraction [ page 112]. In molecular solids and liquids, the forces between molecules, called intermolecular forces, are based on various electrostatic attractions that are weaker than the forces between oppositely charged ions. By comparison, the attractive forces between the ions in ionic compounds are usually in the range of 700 to 1100 kJ/mol, and most covalent bond energies are in the range of 100 to 400 kJ/mol (Table 9.10). As a rough guideline, intermolecular forces are generally 15% (or less) of the values of bond energies. See the General ChemistryNow CD-ROM or website: • Screen 13.2 States of Matter, to view an animation of gases, liquids, and solids at the molecular level 13.2—Intermolecular Forces Intermolecular forces influence chemistry in many ways: • They are directly related to properties such as melting point, boiling point, and the energy needed to convert a solid to a liquid or a liquid to a vapor. • They are important in determining the solubility of gases, liquids, and solids in various solvents. • They are crucial in determining the structures of biologically important molecules such as DNA and proteins. You will encounter examples of these relationships in this and subsequent chapters. 592 Chapter 13 Intermolecular Forces, Liquids, and Solids Bonding in ionic compounds depends on the electrostatic forces of attraction between oppositely charged ions. Similarly, the intermolecular forces attracting one molecule to another are electrostatic. Recall that molecules can have polar bonds owing to the differences in electronegativity of the bonded atoms [ Section 9.8]. Depending on the orientation of these polar bonds, an entire molecule can be polar, with one portion of the molecule being negatively charged and another portion being positively charged. Interactions between the polar molecules can have a profound effect on molecular properties and are the subject of this section. See the General ChemistryNow CD-ROM or website: • Screen 13.3 Intermolecular Forces, for an outline of the important intermolecular forces ■ Coulomb’s Law The force of attraction between oppositely charged ions depends directly on the product of the ion charges and inversely on the square of the distance between the ions (Equation 3.1, page 112). The energy of the attraction is also proportional to the ion charge product, but it is inversely proportional to the distance between them. Interactions Between Ions and Molecules with a Permanent Dipole The distribution of bonding electrons in a molecule often results in a permanent dipole moment (Section 9.9). Because polar molecules have positive and negative ends, if a polar molecule and an ionic compound are mixed, the negative end of the dipole will be attracted to a positive cation (Figure 13.2). Similarly, the positive end of the dipole will be attracted to a negative anion. Forces involved in the attraction between a positive or negative ion and polar molecules are less than those for ion–ion attractions, but they are greater than other forces between molecules, whether polar or nonpolar. Ion–dipole attraction can be evaluated based on the equation describing the attraction between opposite charges, Coulomb’s law (Equation 3.1). It informs us that the force of attraction between two charged objects depends on the product of their charges divided by the square of the distance between them (see Section 3.3). Therefore, when a polar molecule encounters an ion, the attractive forces depend on three factors: • The distance between the ion and the dipole. The closer the ion and dipole, the stronger the attraction. • The charge on the ion. The higher the ion charge, the stronger the attraction. • The magnitude of the dipole. The greater the magnitude of the dipole, the stronger the attraction. The formation of hydrated ions in aqueous solution is one of the most important examples of the interaction between an ion and a polar molecule. (See Figure 13.2 and “A Closer Look: Hydrated Salts.”) The energy associated with the hydration of ions—which is generally called the solvation energy or, for ions in water, the enthalpy of hydration—can be substantial. The solvation energy or enthalpy for an individual ion cannot be measured directly, but values can be estimated. For example, the solvation or hydration of sodium ions is described by the following reaction: Na (g) x H2O(/) ¡ [Na(H2O)x] (aq)(x probably 6) ¢Hrxn 405 kJ Water surrounding a cation Water surrounding an anion Active Figure 13.2 Ion–dipole interactions. When an ionic compound such as NaCl is placed in water, the polar water molecules surround the cations and anions. See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. The energy of attraction depends on 1/d, where d is the distance between the center of the ion and the oppositely charged “pole” of the dipole. As the ion radius becomes larger, d increases and the enthalpy of hydration becomes less exothermic. The trend in the enthalpy of hydration of the alkali metal 13.2 d d K d K ,r H d 133 pm 321 kJ/mol Li , r H Li d d d Intermolecular Forces d d Mg2 d d Mg2 H ,r 79 pm 1922 kJ/mol 593 d 78 pm 515 kJ/mol Increasing force of attraction; more exothermic enthalpy of hydration Figure 13.3 Enthalpy of hydration. The energy evolved when an ion is hydrated depends on the ion charge and the distance d between the ion and the polar water molecule. The distance d increases as ion size increases. cations illustrates this property, as do the hydration enthalpy values for Mg2 , Li , and K in Figure 13.3. Ion Radius (pm) 78 98 133 149 165 Enthalpy of Hydration (kJ/mol) 515 405 321 296 263 Cation Li Na K Rb Cs It is interesting to compare these values with the enthalpy of hydration of the H ion, estimated to be 1090 kJ/mol. This extraordinarily large value is due to the tiny size of the H ion. See the General ChemistryNow CD-ROM or website: • Screen 13.4 Intermolecular Forces (2), to view an animation of ion–dipole forces Example 13.1—Hydration Energy Problem Explain why the enthalpy of hydration of Na ( 405 kJ/mol) is somewhat more exothermic than that of Cs ( 263 kJ/mol), whereas that of Mg2 is much more exothermic ( 1922 kJ/mol) than that of either Na or Cs . Strategy The strength of ion–dipole attractions depends directly on the size of the ion charge and the magnitude of the dipole, and inversely on the distance between them. To judge the ion–dipole distance, we need ion sizes from Figure 8.15. Solution The relevant ion sizes are Na 98 pm, Cs 165 pm, and Mg2 79 pm. From these values we can predict that the distances between the center of the positive charge on the metal ion and the negative side of the water dipole will vary in this order: Mg2 Na Cs . The hydration energy varies in the reverse order (with the hydration energy of Mg2 being 594 Strong attraction d Mg2 d d Chapter 13 Intermolecular Forces, Liquids, and Solids the most negative value). Notice also that Mg2 has a 2 charge, whereas the other ions are 1 . The greater charge on Mg2 leads to a greater force of ion–dipole attraction than for the other two ions, which have only a 1 charge. As a result, the hydration energy for MG2 is much more negative than for the other two ions. Exercise 13.1—Hydration Energy d Na d d Which should have the more negative hydration energy, F or Cl ? Explain briefly. Interactions Between Molecules with Permanent Dipoles d Cs d d When a polar molecule encounters another polar molecule, of the same or a different kind, the positive end of one molecule is attracted to the negative end of the other polar molecule. d d Weak attraction d d Many molecules have dipoles, and their interactions occur by dipole–dipole attraction. For polar molecules, dipole–dipole attractions influence the evaporation of a liquid and the condensation of a gas (Figure 13.4). An energy change occurs in both processes. Evaporation requires the addition of heat, specifically the enthalpy of vaporization (¢H °ap) [ Section 6.3; see also Section 13.5]. The value for the env thalpy of vaporization has a positive sign, indicating that evaporation is an endothermic process. The enthalpy change for the condensation process—the reverse of evaporation—has a negative value, because heat is transferred out of the system. The greater the forces of attraction between molecules in a liquid, the greater the energy that must be supplied to separate them. Thus, we expect polar compounds to have a higher value for their enthalpy of vaporization than nonpolar compounds with similar molar masses. Comparisons between a few polar and nonpolar molecules that illustrate this trend appear in Table 13.1. For example, notice that ¢H °ap for polar molecules is greater than for nonpolar molecules of approxiv mately the same size and mass. Figure 13.4 Evaporation at the molecular level. Energy must be supplied to separate molecules in the liquid state against intermolecular forces of attraction. Vapor H vaporization (endothermic) Photos: Charles D. Winters H condensation (exothermic) Liquid 13.2 Intermolecular Forces 595 A Closer Look Hydrated Salts Solid salts with waters of hydration are common. The formulas of these compounds are given by appending a specific number of water molecules to the end of the for- mula, as in BaCl2 2 H20. Sometimes the water molecules simply fill in empty spaces in a crystalline lattice, but often the cation in these salts is directly associated with water molecules. For example, the compound CrCl3 6 H20 is better written as [Cr(H2O)4Cl2]Cl 2 H2O. Four of the six water molecules are associated with the Cr3+ ion by ion–dipole attractive forces; the remaining two water molecules are in the lattice. Common examples of hydrated salts are listed in the table. Compound Na2CO3 10 H2O Na2S2O3 5 H2O MgSO4 7 H2O CaSO4 2 H2O CuCO3 5 H2O Common Name Washing soda Hypo Epsom salt Gypsum Blue vitriol Users Water softener Photography Cathartic, dyeing and tanning Wallboard Biocide Hydrated cobalt(II) chloride, CoCl2 6 H2O. In the solid state the compound is best described by the formula [Co(H2O)4Cl2] 2 H2O. The cobalt(II) ion is surrounded by four water molecules and two chloride ions in an octahedral arrangement. In water, the ion is completely hydrated, now being surrounded by six water molecules. Cobalt(II) ions and water molecules interact by ion–dipole forces. This is an example of a coordination compound, a class of compounds discussed in detail in Chapter 22. The boiling point of a liquid is also related to intermolecular forces of attraction. As the temperature of a substance is raised, its molecules gain kinetic energy. Eventually, when the boiling point is reached, the molecules have sufficient kinetic energy to escape the forces of attraction of their neighbors. The higher the forces of attraction, the higher the boiling point. In Table 13.1 you see that the boiling point for polar ICl is greater than that for nonpolar Br2, for example. Intermolecular forces also influence solubility. A qualitative observation on solubility is that “like dissolves like.” In other words, polar molecules are likely to dissolve in a polar solvent, and nonpolar molecules are likely to dissolve in a nonpolar solvent (Figure 13.5) [ Chapter 14]. The converse is also true; that is, it is unlikely that polar molecules will dissolve in nonpolar solvents or that nonpolar molecules will dissolve in polar solvents. Charles D. Winters ■ Dissolving Substances Other factors besides energy are also important in controlling the mixing of substances. See Section 14.2. Table 13.1 Molar Masses and Boiling Points of Nonpolar and Polar Substances Nonpolar M (g/mol) N2 SiH4 GeH4 Br2 28 32 77 160 BP (°C) 196 112 90 59 H° vap (kJ/mol) 5.57 12.10 14.06 29.96 CO PH3 AsH3 ICl M (g/mol) 28 34 78 162 Polar BP (°C) 192 88 62 97 H° vap (kJ/mol) 6.04 14.06 16.69 — 596 Chapter 13 Intermolecular Forces, Liquids, and Solids Ethylene glycol Photos: Charles D. Winters Hydrocarbon (a) Ethylene glycol (HOCH2CH2OH), a polar compound used as antifreeze in automobiles, dissolves in water. (b) Nonpolar motor oil (a hydrocarbon) dissolves in nonpolar solvents such as gasoline or CCl4. It will not dissolve in a polar solvent such as water, however. Commercial spot removers use nonpolar solvents to dissolve oil and grease from fabrics. Figure 13.5 “Like dissolves like.” For example, water and ethanol (C2H5OH) can be mixed in any ratio to give a homogeneous mixture. In contrast, water does not dissolve in gasoline to an appreciable extent. The difference in these two situations is that ethanol and water are polar molecules, whereas the hydrocarbon molecules in gasoline (e.g., octane, C8H18) are nonpolar. The water–ethanol interactions are strong enough that the energy expended in pushing water molecules apart to make room for ethanol molecules is compensated for by the energy of attraction between the two kinds of polar molecules. In contrast, water–hydrocarbon attractions are weak. The hydrocarbon molecules cannot disrupt the stronger water–water attractions. See the General ChemistryNow CD-ROM or website: • Screen 13.4 Intermolecular Forces (2), to view an animation of dipole–dipole forces Interactions Involving Nonpolar Molecules Many important molecules such as O2, N2, and the halogens are not polar. Why, then, does O2 dissolve in polar water? Perhaps even more difficult to imagine is how the N2 of the atmosphere can be liquefied (see Figure 13.1). Some intermolecular forces must be acting between O2 and water and between N2 molecules, but what is their nature? Dipole/Induced Dipole Forces Polar molecules such as water can induce, or create, a dipole in molecules that do not have a permanent dipole. To see how this situation can occur, picture a polar water molecule approaching a nonpolar molecule such as O2 (Figure 13.6). The electron cloud of an isolated (gaseous) O2 molecule is symmetrically distributed be- 13.2 (a) d d d (b) The dipole of water induces a dipole in O2 by distorting the O2 electron cloud. d d d Intermolecular Forces 597 d d Polar ethanol (C2H5OH) induces a dipole in nonpolar I2 Photos: Charles D. Winters Figure 13.6 Dipole/induced dipole interaction. (a) A polar molecule such as water can induce a dipole in nonpolar O2 by distorting the molecule’s electron cloud. (b) Nonpolar I2 dissolves in polar ethanol (C2H5OH). The intermolecular force involved is a dipole/induced dipole force. tween the two oxygen atoms. As the negative end of the polar H2O molecule approaches, however, the O2 electron cloud becomes distorted. In this process, the O2 molecule itself becomes polar; that is, a dipole is induced in the otherwise nonpolar O2 molecule. The result is that H2O and O2 molecules are now attracted to one another, albeit only weakly. Oxygen can dissolve in water because a force of attraction exists between water’s permanent dipole and the induced dipole in O2. Chemists refer to such interactions as dipole/induced dipole interactions. The process of inducing a dipole is called polarization, and the degree to which the electron cloud of an atom or a molecule can be distorted depends on the polarizability of that atom or molecule. This property is difficult to measure experimentally. It makes sense, however, that the electron cloud of an atom or molecule with a large, extended electron cloud, such as I2, can be polarized more readily than the electron cloud in a much smaller atom or molecule, such as He or H2, in which the valence electrons are close to the nucleus and more tightly held. In general, for an analogous series of compounds, say the halogens or alkanes (such as CH4, C2H6, C3H8, and so on), the higher the molar mass, the greater the polarizability of the molecule. The solubilities of common gases in water illustrate the effect of interactions between a dipole and an induced dipole. Table 13.2 reveals a trend toward higher solubility with increasing mass of the nonpolar gas. As the molar mass of the gas increases, the polarizability of the electron cloud increases, and the strength of the dipole/induced dipole interaction increases. London Dispersion Forces: Induced Dipole/Induced Dipole Forces Iodine, I2, is a solid and not a gas at room temperatures and pressures, proving that nonpolar molecules must also experience intermolecular forces. An estimate of these forces is provided by the enthalpy of vaporization of the substance at its boiling point. The data in the following table suggest that the forces in this case can range from very weak (N2, O2, and CH4 have low enthalpies of vaporization and very low boiling points) to more substantial (I2 and benzene). Table 13.2 The Solubility of Some Gases in Water* Molar Mass (g/mol) 2.01 28.0 32.0 Solubility at 20 °C (g gas/100 g water)† 0.000160 0.00190 0.00434 Gas H2 N2 O2 *Data taken from J. A. Dean: Lange’s Handbook of Chemistry, 14th ed., pp. 5.3–5.8. New York, McGraw-Hill, 1992. † Measured under conditions where pressure of gas pressure of water vapor 760 mm Hg. ■ Dissolving O2 In Water Oxygen dissolves in water to the extent of about 43 ppm. This property is important because microorganisms use dissolved oxygen to convert the organic substances dissolved in water to simpler compounds. The quantity of oxygen required to oxidize a given quantity of organic material is called the biological oxygen demand (BOD). Highly polluted water often has a high concentration of organic matter and so has a high BOD. 598 Figure 13.7 Induced dipole interactions. Momentary attractions and repulsions between nuclei and electrons create induced dipoles and lead to a net stabilization due to attractive forces. Chapter 13 Intermolecular Forces, Liquids, and Solids Two nonpolar atoms or molecules (depicted as having an electron cloud that has a time-averaged spherical shape). Momentary attractions and repulsions between nuclei and electrons in neighboring molecules lead to induced dipoles. Correlation of the electron motions between the two atoms or molecules (which are now dipolar) leads to a lower energy and stabilizes the system. ¢Hvap (kJ/mol) N2 O2 CH4 (methane) Br2 C6H6 (benzene) I2 5.57 6.82 8.2 29.96 30.7 41.95 Element/Compound BP (°C) 196 183 161.5 58.8 80.1 185 ■ Van der Waals Forces The name “van der Waals forces” is a general term applied to intermolecular interactions. P. W. Atkins: Quanta: A Handbook of Concepts, 2nd ed., p. 187, Oxford, Oxford University Press, 2000. To understand how two nonpolar molecules can attract each other, recall that the electrons in atoms or molecules are in a state of constant motion (Figure 13.7). When two atoms or nonpolar molecules approach each other, attractions or repulsions between their electrons and nuclei can lead to distortions in these electron clouds. That is, dipoles can be induced momentarily in neighboring atoms or molecules, and these induced dipoles lead to intermolecular attractions. Thus, the intermolecular force of attraction in liquids and solids composed of nonpolar molecules is an induced dipole/induced dipole force. Chemists often call them London dispersion forces. See the General ChemistryNow CD-ROM or website: • Screen 13.5 Intermolecular Forces (3), to view an animation of induced dipole forces and for an exercise and tutorial on intermolecular forces Example 13.2—Intermolecular Forces Problem Suppose you have a mixture of solid iodine, I2, and the liquids water and carbon tetrachloride (CCl4). What intermolecular forces exist between each possible pair of compounds? Describe what you might see when these compounds are mixed. Strategy First decide whether each substance is polar or nonpolar. Next, use the “like dissolves like” guideline (Figure 13.5) to decide whether iodine will dissolve in water or CCl4 and whether CCl4 will dissolve in water. Solution Iodine, I2, is nonpolar. As a molecule composed of large iodine atoms, it has an extensive electron cloud. Thus, the molecule is easily polarized. Iodine could interact with water, a polar molecule, by induced dipole/dipole forces. 13.3 Hydrogen Bonding 599 Carbon tetrachloride, a tetrahedral molecule, is not polar [ Section 9.9]. As a consequence, it can interact with iodine only by dispersion forces. Water and CCl4 could interact by dipole/induced dipole forces, but the interaction is expected to be weak. The photo here shows the result of mixing these three compounds. Iodine does dissolve to a small extent in water to give a brown solution. When this brown solution is added to a test tube containing CCl4, the liquid layers do not mix. (Polar water does not dissolve in nonpolar CCl4.) When the test tube is shaken, however, nonpolar I2 dissolves preferentially in nonpolar CCl4, as evidenced by the disappearance of the color of I2 in the water layer (top) and the appearance of the purple I2 color in the CCl4 layer (bottom). Charles D. Winters Nonpolar I2 Polar H2O Shake the test tube Polar H2O Br2 I2 Nonpolar CCl4 Charles D. Winters Nonpolar CCl4 and I2 Induced dipole/induced dipole forces. The molecules Br2 (left) and I2 (right) are both nonpolar. They are a liquid and a solid, respectively, implying that there are forces between the molecules sufficient to cause them to be in a condensed phase. Forces between nonpolar substances are known as London dispersion forces or induced dipole/induced dipole forces. Exercise 13.2—Intermolecular Forces You mix water, CCl4, and hexane (CH3CH2CH2CH2CH2CH3). What type of intermolecular forces can exist between each pair of these compounds? If you mix the three liquids, describe what observations you might make. 13.3—Hydrogen Bonding Hydrogen fluoride and many other compounds with O ¬ H and N ¬ H bonds have exceptional properties. We can see this by examining the boiling points for hydrogen compounds of elements in Groups 4A through 7A (Figure 13.8). Generally, the boiling points of related compounds increase with molar mass because of increasing dispersion forces. This trend is seen in the boiling points of the hydrogen compounds of Group 4A elements, for example (CH4 SiH4 GeH4 SnH4). The same effect is also operating for the heavier molecules of the hydrogen compounds of elements of Groups 5A, 6A, and 7A. The boiling points of NH3, H2O, and HF, however, are greatly out of line with what might be expected based on molar mass alone. If we were to extrapolate the curve for the boiling points of H2Te, H2Se, and H2S to the expected boiling point of water, water should boil around 90 °C. The boiling point of water is almost 200 °C higher than the expected value! Similarly, the boiling points of NH3 and HF are much higher than would be expected based on molar mass. Why do the properties of water, ammonia, and hydrogen fluoride differ so from the extrapolated values? Because the temperature at which a substance boils depends on the attractive forces between molecules, the boiling points of H2O, HF, and NH3 clearly indicate strong intermolecular attractions. The unusually high boiling points in these compounds are due to hydrogen bonding. A hydrogen bond is an attraction between the hydrogen atom of an X ¬ H bond and Y, where X and Y are atoms of highly electronegative elements and Y has a lone pair of electrons. Hydrogen bonds are an 600 Chapter 13 Intermolecular Forces, Liquids, and Solids 100 H2O HF Temperature (°C) 0 NH3 H2S HCl 100 PH3 SiH4 H2Se AsH3 HBr GeH4 H2Te SbH3 HI SnH4 CH4 0 2 3 Period 4 5 Active Figure 13.8 The boiling points of some simple hydrogen compounds. The effect of hydrogen bonding is apparent in the unusually high boiling points of H2O, HF, and NH3. See the General ChemistryNow CD-Rom or website to explore an interactive version of this figure accompanied by an exercise. extreme form of dipole–dipole interaction where one atom involved is always H and the other atom is most often O, N, or F. A bond dipole arises as a result of a difference in electronegativity between bonded atoms [ Section 9.8]. The electronegativities of N (3.0), O (3.5), and F (4.0) are among the highest of all the elements, whereas the electronegativity of hydrogen is much lower (2.2). The large difference in electronegativity means that N ¬ H, O ¬ H, and F ¬ H bonds are very polar. In bonds between H and N, O, or F, the more electronegative element takes on a significant negative charge and the hydrogen atom acquires a significant positive charge. In hydrogen bonding, there is an unusually strong attraction between an electronegative atom with a lone pair of electrons (an N, O, or F atom in another molecule or even in the same molecule) and the hydrogen atom of the N ¬ H, O ¬ H, or F ¬ H bond. A hydrogen bond can be represented as d d d d X H Y H The hydrogen atom becomes a bridge between the two electronegative atoms X and Y, and the dashed line represents the hydrogen bond. The most pronounced effects of hydrogen bonding occur where X and Y are N, O, or F. Energies associated with most hydrogen bonds involving these elements are in the range of 5 to 30 kJ/mol. 13.3 Hydrogen Bonding 601 Types of Hydrogen Bonds [X—H - - - :Y] N—H - - - :N— N—H - - - :O— N—H - - - :F— O—H - - - :N— O—H - - - :O— O—H - - - :F— F—H - - - :N— F—H - - - :O— F—H - - - :F— Hydrogen bonding has important implications for any property of a compound that is influenced by intermolecular forces of attraction. For example, it is important in determining structures of molecular solids, one example of which is acetic acid. In the solid state, two molecules of CH3CO2H are joined to one another by hydrogen bonding (Figure 13.9). Hydrogen bonding is also an important factor in the structure of some synthetic polymers. In nylon, for example, the N ¬ H unit of the amide interacts with a carbonyl oxygen on an adjacent polymer chain [ Figure 11.19]. Hydrogen bonding in Kevlar, also a polyamide, gives this material the exceptional strength-to-weight ratio needed for its use in making canoes, ski equipment, and bullet-proof vests. Photo: Charles D. Winters Figure 13.9 Hydrogen bonding. Two See the General ChemistryNow CD-ROM or website: • Screen 13.6 Hydrogen Bonding, for a description of hydrogen bonding acetic acid molecules can interact through hydrogen bonds. This photo shows partly solid glacial acetic acid. Notice that the solid is denser than the liquid, a property shared by virtually all substances. The notable exception is water. Example 13.3—The Effect of Hydrogen Bonding Problem Ethanol, CH3CH2OH, and dimethyl ether, CH3OCH3, have the same formula but a different arrangement of atoms (they are isomers). Predict which of these compounds has the higher boiling point. Ethanol, CH3CH2OH Dimethyl ether, CH3OCH3 Strategy Inspect the structure of each molecule to decide whether each is polar and, if polar, whether hydrogen bonding is possible. Charles D. Winters Solution Although these two compounds have identical masses, they have different structures. Ethanol possesses an O ¬ H group, so hydrogen bonding between ethanol molecules makes an important contribution to its intermolecular forces. CH3CH2 O H H O CH2CH3 hydrogen bonding in ethanol, CH3CH2OH In contrast, dimethyl ether, although a polar molecule, presents no opportunity for hydrogen bonding because there is no O ¬ H bond. We can predict, therefore, that intermolecular forces Tooth whiteners and hydrogen bonding. Most tooth-whitening products contain urea, (NH2)2CO, and hydrogen peroxide, H2O2 (a mixture sometimes referred to as carbamide peroxide). Hydrogen peroxide, the active ingredient, is stabilized by hydrogen bonding with urea. 602 Chapter 13 Intermolecular Forces, Liquids, and Solids will be larger in ethanol than in dimethyl ether and that ethanol will have the higher boiling point . Indeed, ethanol boils at 78.3 °C, whereas dimethyl ether has a boiling point of 24.8 °C, more than 100 °C lower. Dimethyl ether is a gas, whereas ethanol is a liquid under standard conditions. Exercise 13.3—Hydrogen Bonding Using structural formulas, describe the hydrogen bonding between methanol (CH3OH) molecules. What physical properties of methanol are likely to be affected by hydrogen bonding? Hydrogen Bonding and the Unusual Properties of Water One of the most striking differences between our planet and others in our solar system is the presence of large amounts of water on earth. Three fourths of the planet is covered by oceans, the polar regions are vast ice fields, and even soil and rocks hold large amounts of water. Although we tend to take water for granted, almost no other substance behaves in a similar manner. Water’s unique features reflect the ability of H2O molecules to cling tenaciously to one another by hydrogen bonding. One reason for ice’s unusual structure, and water’s unusual properties, is that each hydrogen atom of a water molecule can form a hydrogen bond to a lone pair of electrons on the oxygen atom of an adjacent water molecule. In addition, because the oxygen atom in water has two lone pairs of electrons, it can form two more hydrogen bonds with hydrogen atoms from adjacent molecules (Figure 13.10a). The result is a tetrahedral arrangement for the hydrogen atoms around each oxygen, involving two covalently bonded hydrogen atoms and two hydrogen-bonded hydrogen atoms. To achieve the regular arrangement of hydrogen-bonded water molecules linked by hydrogen bonding, ice has an open-cage structure with lots of empty space (Figure 13.10). The result is that ice has a density about 10% less than that of liquid water, which explains why ice floats. (In contrast, virtually all other solids sink in their liquid phase.) We can also see in this structure that the oxygen atoms are arranged at the corners of puckered, six-sided rings, or hexagons. Snowflakes are always based on six-sided figures [ page 101], a reflection of the internal molecular structure of ice. ■ Energy of Hydrogen Bonding A hydrogen bond between water molecules has an estimated energy of 22 kJ/mol. For comparison, the O ¬ H covalent bond energy is 463 kJ/mol. Figure 13.10 The structure of ice. (a) The oxygen atom of a water molecule attaches itself to four other water molecules by hydrogen bonds. Notice that the four groups that surround an oxygen atom are arranged tetrahedrally. Each oxygen atom is covalently bonded to two hydrogen atoms and hydrogen-bonded to hydrogen atoms from two other molecules. The hydrogen bonds are longer than the covalent bonds. (b) In ice, the structural unit shown in part (a) is repeated in the crystalline lattice. This computer-generated structure shows a small portion of the extensive lattice. Notice the six-member, hexagonal rings. The corners of each hexagon are O atoms, and each side is composed of a normal O ¬ H bond and a longer hydrogen bond. (This structure is found on the General ChemistryNow CD-ROM or website.) O H Hydrogen bond S. M. Young (a) (b) 13.3 Hydrogen Bonding 603 1.0000 0.9999 Density (g/ml) 0.9998 0.9997 When ice melts at 0 °C, a relatively large increase in density occurs (Figure 13.11), as a result of the breakdown of the regular structure imposed on the solid state by hydrogen bonding. When the temperature of liquid water is raised from 0 °C to 4 °C, another surprising thing occurs: The density of water increases. For almost every other substance known, density decreases as its temperature is raised. Once again, hydrogen bonding is the reason for water’s seemingly odd behavior. At a temperature just above the melting point, some of the water molecules continue to cluster in ice-like arrangements, which require extra space. As the temperature is raised from 0 °C to 4 °C, the final vestiges of the ice structure disappear and the volume contracts further, giving rise to the increase in density. Water’s density reaches a maximum at about 4 °C. From this point, the density declines with increasing temperature in the normal fashion. Because of the way that water’s density changes as the temperature approaches the freezing point, lakes do not freeze solidly from the bottom up in the winter. When lake water cools with the approach of winter, its density increases, the cooler water sinks, and the warmer water rises. This “turnover” process continues until all of the water reaches 4 °C, the maximum density. (This is the way oxygen-rich water moves to the lake bottom to restore the oxygen used during the summer and nutrients are brought to the top layers of the lake.) As the temperature decreases further, the colder water stays on the top of the lake, because water cooler than 4 °C is less dense than water at 4 °C. With further heat loss, ice can then begin to form on the surface, floating there and protecting the underlying water and aquatic life from additional heat loss. Extensive hydrogen bonding is also the origin of the extraordinarily high heat capacity of water. Although liquid water does not have the regular structure of ice, hydrogen bonding still occurs. With a rise in temperature, the extent of hydrogen bonding diminishes. Disrupting hydrogen bonds requires heat. The high heat capacity of water explains, in large part, why oceans and lakes have such an enormous effect on weather. In autumn, when the temperature of the air is lower than the temperature of the ocean or lake, the ocean or lake gives up heat to the atmosphere, moderating the drop in air temperature. So much heat is available to be given off for each one-degree drop in temperature that the decline in water temperature is gradual. For this reason the temperature of the ocean or of a large lake generally remains higher than the average air temperature until late in the autumn. Hydrogen bonds involving water are also responsible for the structure and properties of one of the strangest substances on earth: methane hydrate (Figure 13.12). When methane is mixed with water at high pressures and low temperatures, solid methane hydrate forms. Although the substance has been known for years, vast deposits of methane hydrate were recently discovered deep within sediments on the floor of the world’s oceans. How they formed remains a mystery, but what is important is their size. It is estimated that the global methane hydrate deposits contain approximately 1013 tons of carbon, or about twice the combined amount in all known reserves of coal, oil, and conventional gas. Water 0.9180 0.9170 Ice 8 6 4 2 0 2 4 6 8 10 Temperature (°C) Active Figure 13.11 The temperature dependence of the densities of ice and water. See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. ■ Hydrogen Bonding in DNA Hydrogen bonding occurs extensively in biochemical systems. For example, hydrogen bonding takes place between the two bases thymine and adenine in opposing chains of the DNA molecule. Thymine H To chain CH3 C C C O H N N H C N C N C C H Adenine N C H N C O N H Screen 13.7 The Weird Properties of Water, to view an animation of the transformation of ice to water and for a table listing all of the unusual properties of water See “The Chemistry of Life: Biochemistry,” pages 530–545. To c h ain See the General ChemistryNow CD-ROM or website: 604 Chapter 13 Intermolecular Forces, Liquids, and Solids Figure 13.12 Methane hydrate. (a) This interesting substance is found in Photo: John Pinkston and Laura Stern/U.S. Geological Survey/Science News, 11-9-96 huge deposits hundreds of feet down on the floor of the ocean. When a sample is brought to the surface, the methane oozes out of the solid, and the gas readily burns as it escapes from the solid hydrate. (b) The structure of solid methane hydrate consists of methane molecules trapped in a lattice of water molecules. Each point of the lattice shown here is an O atom of an H2O molecule. The edges are O ¬ H ¬ O hydrogen bonds. Such structures are often called “clathrates.” See E. Suess, G. Bohrmann, J. Greinert, and E. Lausch: Scientific American, pp. 76–83, November 1999. See also “The Chemistry of Energy and Fuels,” page 287. (a) Methane hydrate burns as methane gas escapes from the solid hydrate. (b) Methane hydrate consists of a lattice of water molecules with methane molecules trapped in the cavity. 13.4—Summary of Intermolecular Forces Intermolecular forces involve molecules that are polar or those in which polarity can be induced (Table 13.3). London dispersion forces are found in all molecules, both nonpolar and polar, but dispersion forces are the only intermolecular forces that allow nonpolar molecules to interact. Furthermore, several types of intermolecular forces can be at work in a single type of molecule (Figure 13.13). A very large molecule, for example, can have polar or nonpolar regions. In general, the strength of intermolecular forces is in the order dipole–dipole (including H-bonding) dipole/induced dipole induced dipole/induced dipole Table 13.3 Summary of Intermolecular Forces Type of Interaction Ion–dipole Dipole–dipole Factors Responsible for Interaction Ion charge, magnitude of dipole Dipole moment (depends on atom electronegativities and molecular structure) Very polar X ¬ H bond (where X F, N, O) and atom Y with lone pair of electrons An extreme form of dipole–dipole interaction. Dipole–induced dipole Induced dipole–induced dipole (London dispersion forces) Dipole moment of polar molecule and polarizability of nonpolar molecule Polarizability 2–10 0.05–40 H2O . . . I2 I2 . . . I2 Approximate Energy (kJ/mol) 40–600 20–30 Example Na . . . H2O H2O, HCl Hydrogen bonding, X ¬ H . . . :Y 5–30 H2O . . . H2O 13.4 Summary of Intermolecular Forces 605 50 45 40 Intermolecular Force (kJ/mol) 35 30 25 20 15 10 5 0 Ar CO HI HBr HCl NH3 H2O Dipole–dipole force Induced dipole/induced dipole force Dipole/induced dipole force Total force Figure 13.13 Intermolecular forces for various molecules. (Forces are reported in terms of energies in kJ/mol.) The total intermolecular force for atomic argon and weakly polar CO is small (8–9 kJ/mol) and consists entirely of dispersion forces. The polar molecules HI, HBr, and HCl have larger intermolecular forces (21–26 kJ/mol), but dispersion forces dominate in every case. For HCl, dipole–dipole forces contribute 3.3 kJ/mol to the total force of 21.1 kJ/mol. Highly polar water molecules have the largest intermolecular forces (47.2 kJ/mol). Water molecules interact primarily through dipole forces, but induced forces are also present. Example 13.4—Intermolecular Forces Problem Decide which are the most important intermolecular forces involved in each of the following and place them in order of increasing strength of interaction: (a) liquid methane, CH4; (b) a mixture of water and methanol (CH3OH); and (c) a solution of bromine in water. Strategy For each molecule we consider its structure and then decide whether it is polar. If polar, consider the possibility of hydrogen bonding. Solution (a) Methane is a covalently bonded molecule. Based on the Lewis structure we can conclude that it must be a tetrahedral molecule and that it cannot be polar. The only way methane molecules can interact with one another is through induced dipole/induced dipole forces. (b) Both water and methanol are covalently bonded molecules, both are polar, and both have an O ¬ H bond. They therefore interact through the special dipole–dipole force called hydrogen bonding. d H H d d O d H d H O CH3 and H3C d O d H d O Hd d 606 Chapter 13 Intermolecular Forces, Liquids, and Solids (c) Nonpolar molecules of bromine, Br2, interact by induced dipole forces, whereas water is a polar molecule. Therefore, dipole/induced dipole forces are involved when Br2 molecules interact with water. (This is similar to the I2–ethanol interaction in Figure 13.6.) In order of increasing strength, the likely order of interactions is liquid CH4 6 H2O and Br2 6 H2O and CH3OH Exercise 13.4—Intermolecular Forces Decide which type of intermolecular force is involved in (a) liquid O2; (b) liquid CH3OH; and (c) O2 dissolved in H2O. Place the interactions in order of increasing strength. 13.5—Properties of Liquids Of the three states of matter, liquids are the most difficult to describe precisely. The molecules in a gas under normal conditions are far apart and may be considered more or less independent of one another. The structures of solids can be described more readily because the particles that make up solids—atoms, molecules, or ions— are close together and are in an orderly arrangement. The particles of a liquid interact with their neighbors, like the particles in a solid, but, unlike in solids, there is little order in their arrangement. In spite of a lack of precision in describing liquids, we can still consider the behavior of liquids at the molecular level. In the following sections we will look further at the process of vaporization, at the vapor pressure of liquids, at their boiling points and critical properties, and at the behavior that results in their surface tension, capillary action, and viscosity. Vaporization Vaporization or evaporation is the process in which a substance in the liquid state becomes a gas. In this process, molecules escape from the liquid surface and enter the gaseous state. To understand evaporation, we have to look at molecular energies. Molecules in a liquid have a range of energies (Figure 13.14) that closely resembles the distribution of energies for molecules of a gas (see Figure 12.14). As with gases, the average energy for molecules in a liquid depends only on temperature: The higher the temperature, the higher the average energy and the greater the relative number of molecules with high kinetic energy. In a sample of a liquid, at least a few molecules Figure 13.14 The distribution of energy among molecules in a liquid sample. T2 is a higher temperature than T1, and more molecules have an energy greater than the value marked E in the diagram at T2 than at T1. Relative number of molecules T1 T2 T1 Number of molecules having enough energy to evaporate at lower temperature, T1 E Number of molecules having enough energy to evaporate at higher temperature, T2 T2 Energy 13.5 Properties of Liquids 607 Figure 13.15 Evaporation. Some molecules at the surface of a Vapor liquid have enough energy to escape the attractions of their neighbors and enter the gaseous state. At the same time, some molecules in the gaseous state can reenter the liquid. Liquid have very high energy; that is, they may have more kinetic energy than the potential energy of the intermolecular attractive forces holding the liquid molecules to one another. If these high-energy molecules find themselves at the surface of the liquid, and if they are moving in the right direction, they can break free of their neighbors and enter the gas phase (Figure 13.15). Vaporization is an endothermic process because energy must be added to the system to break the intermolecular forces of attraction holding the molecules together. The heat energy required to vaporize a sample is often given as the standard molar enthalpy of vaporization, H vap (in units of kilojoules per mole; see Tables 13.1 and 13.4 and Figure 13.4). Liquid vaporization heat energy absorbed by liquid Vapor Hvap molar heat of vaporization A molecule in the gas phase will eventually transfer some of its kinetic energy by colliding with slower gaseous molecules and solid objects. If this molecule comes in contact with the surface of the liquid again, it can reenter the liquid phase in the process called condensation. Vapor condensation heat energy released by vapor Liquid Condensation is the opposite of vaporization. Condensation is exothermic, so energy is transferred to the surroundings. The enthalpy change for condensation is equal but opposite in sign to the enthalpy of vaporization. For example, the enthalpy change for the vaporization of 1.00 mol of water at 100 °C is 40.7 kJ. On condensing 1.00 mol of water vapor to liquid water at 100 °C, the enthalpy change is 40.7 kJ. In the discussion of intermolecular forces for polar and nonpolar molecules, we pointed out the relationship between the ¢H °ap values for various substances and v the temperatures at which they boil (Table 13.4). Both properties reflect the attractive forces between particles in the liquid. The boiling points of nonpolar liquids (e.g., the hydrocarbons, atmospheric gases, and the halogens) increase with increasing atomic or molecular mass, a reflection of increased intermolecular dispersion forces. The alkanes listed in Table 13.4 show this trend clearly. Similarly, the boiling points and enthalpies of vaporization of the heavier hydrogen halides (HX, where X Cl, Br, and I) increase with increasing molecular mass. For these molecules, hydrogen bonding is not as important as it is in HF, so dispersion forces and ordinary dipole–dipole forces account for their intermolecular attractions (see 608 Chapter 13 Intermolecular Forces, Liquids, and Solids Table 13.4 Molar Enthalpy of Vaporization and Boiling Points for Common Substances* Compound Polar Compounds HF HCl HBr HI NH3 H2O SO2 Nonpolar Compounds CH4 (methane) C2H6 (ethane) C3H8 (propane) C4H10 (butane) Monatomic Elements He Ne Ar Xe Diatomic Elements H2 N2 O2 F2 Cl2 Br2 Molar Mass (g/mol) 20.0 36.5 80.9 127.9 17.0 18.0 64.1 16.0 30.1 44.1 58.1 4.0 20.2 39.9 131.3 2.0 28.0 32.0 38.0 70.9 159.8 ¢H° vap (kJ/mol)† 25.2 16.2 19.3 19.8 23.3 40.7 24.9 8.2 14.7 19.0 22.4 0.08 1.7 6.4 12.6 0.90 5.6 6.8 6.6 20.4 30.0 Boiling Point (°C) (Vapor pressure 760 mm Hg) 19.7 84.8 66.4 35.6 33.3 100.0 10.0 161.5 88.6 42.1 0.5 268.9 246.1 185.9 108.0 252.9 195.8 183.0 188.1 34.0 58.8 *Data taken from D. R. Lide: Basic Laboratory and Industrial Chemicals, Boca Raton, FL, CRC Press, 1993. † ¢H °ap is measured at the normal boiling point of the liquid. v Figure 13.13). Because dispersion forces become increasingly important with increasing mass (Figure 13.13), the boiling points are in the order HCl HBr HI. Also notice in Table 13.4 the very high heats of vaporization of water and hydrogen fluoride that result from extensive hydrogen bonding. See the General ChemistryNow CD-ROM or website: • Screen 13.8 Properties of Liquids (1): Enthalpy of Vaporization, to view an animation of the vaporization process and for a table of ¢H° values vap Example 13.5—Enthalpy of Vaporization Problem You put 1.00 L of water (about 4 cups) in a pan at 100 °C, and the water slowly evaporates. How much heat must have been supplied to vaporize the water? Strategy Three pieces of information are needed to solve this problem: 1. ¢H°ap for water v 40.7 kJ/mol at 100 °C 13.5 Properties of Liquids 609 2. 3. The density of water at 100 °C 0.958 g/cm3 (This is needed because ¢H°ap has units of v kilojoules per mole, so you first must find the mass of water and then the amount.) Molar mass of water 18.02 g/mol 103 cm3) is equivalent to 958 g, and this mass is in Solution A volume of 1.00 L (or 1.00 turn equivalent to 53.2 mol of water. 1.00 L a ˇ 1000 mL 0.958 g 1 mol H2O ba ba b 1L 1 mL 18.02 g 53.2 mol H2O Therefore, the amount of energy required is 53.2 mol H2O a 40.7 kJ b mol 2.16 103 kJ 2160 kJ is equivalent to about one quarter of the energy in your daily food intake. The Image Bank/Getty Images Exercise 13.5—Enthalpy of Vaporization The molar enthalpy of vaporization of methanol, CH3OH, is 35.2 kJ/mol at 64.6 °C. How much energy is required to evaporate 1.00 kg of this alcohol at 64.6 °C? Water is exceptional among the liquids listed in Table 13.4 in that an enormous amount of heat is required to convert liquid water to water vapor. This fact is important to your environment and your own physical well-being. When you exercise vigorously, your body responds by sweating to rid itself of the excess heat. Heat from your body is consumed in the process of evaporation, and your body is cooled. Heats of vaporization and condensation of water also play an important role in weather (Figure 13.16). For example, if enough water condenses from the air to fall as an inch of rain on an acre of ground, the heat released exceeds 2.0 108 kJ! This is equivalent to about 50 tons of exploded dynamite, or the energy released by a small bomb. Figure 13.16 Rainstorms release an enormous quantity of energy. When water vapor condenses, energy is evolved to the surroundings. The enthalpy of condensation of water is large, so a large quantity of heat is released in a rainstorm. Vapor Pressure If you put some water in an open beaker, it will eventually evaporate completely. If you put water in a sealed flask (Figure 13.17), however, the liquid will evaporate only until the rate of vaporization equals the rate of condensation. At this point, no further change will be observed in the system. The situation is an example of what chemists call a dynamic equilibrium. Liquid VJ Vapor Molecules move continuously from the liquid phase to the vapor phase, and from the vapor phase back to the liquid phase. Even though these changes occur on the molecular level, no change can be detected on the macroscopic level. The rate at which molecules move from liquid to vapor is the same as the rate at which they move from vapor to liquid; thus, there is no net change in the masses of the two phases. In contrast to a closed flask, water in an open beaker does not reach an equilibrium with gas-phase water molecules. Instead, air movement and gas diffusion remove the water vapor from the vicinity of the liquid surface, so many water molecules are not able to return to the liquid. ■ Equilibrium Equilibrium is a concept used throughout chemistry and one to which we shall return often. This situation is signaled by connecting the two states or the reactants and products by a set of double arrows ( VJ ). See Chapters 16–18 in particular. 610 Chapter 13 Intermolecular Forces, Liquids, and Solids I NI TI AL E QUI L I B R I UM Time Volatile liquid Ptotal = Pvapor Vapor pressure at temperature of measurement Hg in tube open to flask Active Figure 13.17 Vapor pressure. A volatile liquid is placed in an evacuated flask (left). At the beginning, no molecules of the liquid are in the vapor phase. After a short time, however, some of the liquid evaporates, and the molecules now in the vapor phase exert a pressure. The pressure of the vapor measured when the liquid and the vapor are in equilibrium is called the equilibrium vapor pressure (right). See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. ■ Equilibrium Vapor Pressure At the conditions of T and P given by any point on a curve in Figure 13.18, the pure liquid and its vapor are in dynamic equilibrium. If T and P define a point not on the curve, the system is not at equilibrium. See Appendix G for the vapor pressures of water at various temperatures. When a liquid–vapor equilibrium has been established, the equilibrium vapor pressure (often just called the vapor pressure) can be measured. The equilibrium vapor pressure of any substance is a measure of the tendency of its molecules to escape from the liquid phase and enter the vapor phase at a given temperature. This tendency is referred to qualitatively as the volatility of the compound. The higher the equilibrium vapor pressure at a given temperature, the more volatile the compound. As described previously (see Figure 13.14), the distribution of molecular energies in the liquid phase is a function of temperature. At a higher temperature, more 1000 800 760 mm Hg Normal BP 34.6 °C Normal BP 78.3 °C Normal BP 100 °C Pressure (mm Hg) 600 Diethyl ether 400 Ethanol H2O 200 0 20° 0° 20° 40° 60° Temperature (°C) 80° 100° 120° Active Figure 13.18 Vapor pressure curves for diethyl ether [(C2H5)2O], ethanol (C2H5OH), and water. Each curve represents conditions of T and P at which the two phases, liquid and vapor, are in equilibrium. These compounds exist as liquids for temperatures and pressures to the left of the curve and as gases under conditions to the right of the curve. See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. 13.5 Properties of Liquids 611 molecules have sufficient energy to escape the surface of the liquid. The equilibrium vapor pressure must, therefore, increase with temperature (Figure 13.18). All points along the vapor-pressure-versus-temperature curves in Figure 13.18 represent conditions of pressure and temperature at which liquid and vapor are in equilibrium. For example, at 60 °C the vapor pressure of water is 149 mm Hg (Appendix G). If water is placed in an evacuated flask that is maintained at 60 °C, liquid will evaporate until the pressure exerted by the vapor is 149 mm Hg (assuming enough water is in the flask so that some liquid remains when equilibrium is reached). See the General ChemistryNow CD-ROM or website: • Screen 13.9 Properties of Liquids (2): Vapor Pressure, to view an animation of equilibrium vapor pressure and for a simulation of vapor pressure curves Example 13.6—Vapor Pressure Problem You place 2.00 L of water in your dormitory room, which has a volume of 4.25 104 L. You seal the room and wait for the water to evaporate. Will all of the water evaporate at 25 °C? (At 25 °C the density of water is 0.997 g/mL, and its vapor pressure is 23.8 mm Hg.) Strategy One approach to solving this problem is to calculate the quantity of water that must evaporate to exert a pressure of 23.8 mm Hg in a volume of 4.25 104 L at 25 °C. Because water vapor is like any other gas, we use the ideal gas law for the calculation. Solution Calculate the amount and then mass and volume of water that fulfills the following conditions: P 23.8 mm Hg, V 4.25 104 L, T 25°C (or 298 K). P PV RT 23.8 mm Hg 1 atm 760 mm Hg 104 L2 0.0313 atm n a 0.082057 980. g H2O 1 0.0313 atm21 4.25 L atm b1 298 K2 K mol 54.4 mol 54.4 mol H2O 18.02 g 1 mol H2O 1 mL 0.997 g 980. g H2O 983 mL Only about half of the available water needs to evaporate to achieve an equilibrium water vapor pressure of 23.8 mm Hg at 25 °C in the dorm room. Exercise 13.6—Vapor Pressure Curves Examine the vapor pressure curve for ethanol in Figure 13.18. (a) What is the approximate vapor pressure of ethanol at 40 °C? (b) Are liquid and vapor in equilibrium when the temperature is 60 °C and the pressure is 600 mm Hg? If not, does liquid evaporate to form more vapor, or does vapor condense to form more liquid? 612 7 Chapter 13 Intermolecular Forces, Liquids, and Solids Exercise 13.7—Vapor Pressure If 0.50 g of pure water is sealed in an evacuated 5.0-L flask and the whole assembly is heated to 60 °C, will the pressure be equal to or less than the equilibrium vapor pressure of water at this temperature? What if you use 2.0 g of water? Under either set of conditions, is any liquid water left in the flask, or does all of the water evaporate? 6 5 ln P (P in mm Hg) 4 3 Vapor Pressure, Enthalpy of Vaporization, and the Clausius-Clapeyron Equation Plotting the vapor pressure for a liquid at a series of temperatures results in a curved line (Figure 13.18). However, the German physicist R. Clausius (1822–1888) and the Frenchman B. P. E. Clapeyron (1799–1864) showed that, for a pure liquid, a linear relationship exists between the reciprocal of the Kelvin temperature (1/T ) and the natural logarithm of vapor pressure ( ln P ). ln P (¢H°ap/RT ) v C (13.1) 2 1 0 0.0025 0.0030 0.0035 1/T (K 1) 0.0040 Figure 13.19 Clausius-Clapeyron equation. When the natural logarithm of the vapor pressure (ln P) of water at various temperatures (T ) is plotted against 1/T , a straight line is obtained. The slope of the line equals ¢H°ap/R. Values of T v and P are from Appendix G. Here ¢H °ap is the enthalpy of vaporization of the liquid, R is the ideal gas constant v (8.314472 J/K mol ), and C is a constant characteristic of the compound in question. This equation, now called the Clausius-Clapeyron equation, provides a method of obtaining values for ¢H °ap. The equilibrium vapor pressure of a liquid can be v measured at several different temperatures, and the logarithm of these pressures is plotted versus 1/T. The result is a straight line with a slope of ¢H °ap/R. For v example, plotting data for water (Figure 13.19), we find the slope of the line is 4.90 103, which gives ¢H °ap 40.7 kJ/mol. v As an alternative to plotting ln P versus 1/T, we can write the following equation that allows us to calculate ¢H °ap knowing the vapor pressure of a liquid at two difv ferent temperatures. ln P2 or ln P2 P1 ¢ H° vap R c 1 T1 1 d T2 (13.2) ln P1 c ¢ H° vap RT2 Cd c ¢ H° vap RT1 Cd For example, ethylene glycol has a vapor pressure of 14.9 mm Hg (P1) at 373 K (T1), and a vapor pressure of 49.1 mm Hg (P2) at 398 K (T2). ln a 49.1 mm Hg b 14.9 mm Hg ˇ 1 c 0.0083145 kJ/K mol 373 K ¢ H° vap 0.0083145 kJ/K mol 59.0 kJ/mol ¢ H° vap 1 d 398 K 1 K 1.192 ¢ H° vap 1 0.000168 2 See the General ChemistryNow CD-ROM or website: • Screen 13.9 Properties of Liquids (2): Vapor Pressure, for three tutorials on using the Clausius-Clapeyron equation 13.5 Properties of Liquids 613 Exercise 13.8—Clausius-Clapeyron Equation Calculate the enthalpy of vaporization of diethyl ether, (C2H5)2O (see Figure 13.8). This compound has vapor pressures of 57.0 mm Hg and 534 mm Hg at 22.8 °C and 25.0 °C, respectively. If you have a beaker of water open to the atmosphere, the mass of the atmosphere presses down on the surface. As heat is added, more and more water evaporates, pushing the molecules of the atmosphere aside. If enough heat is added, a temperature is eventually reached at which the vapor pressure of the liquid equals the atmospheric pressure. Larger and larger bubbles of vapor form in the liquid and rise to the surface; the liquid boils (Figure 13.20). The boiling point of a liquid is the temperature at which its vapor pressure is equal to the external pressure. If the external pressure is 760 mm Hg, this temperature is designated as the normal boiling point. This point is highlighted on the vapor pressure curves for several substances in Figure 13.18. Normal boiling points of other liquids are included in Table 13.4, where you can also see the relationship between normal boiling point and enthalpy of vaporization. The normal boiling point of water is 100 °C, and in a great many places in the United States, water boils at or near this temperature. If you live at higher altitudes, however, such as in Salt Lake City, Utah, where the barometric pressure is about 650 mm Hg, water will boil at a noticeably lower temperature. The curve for the equilibrium vapor pressure of water in Figure 13.18 shows that a pressure of 650 mm Hg corresponds to a boiling temperature of about 95 °C. Cooks know that food has to be cooked a little longer in Salt Lake City or Denver to achieve the same result as in New York City at sea level. Charles D. Winters Boiling Point Figure 13.20 Vapor pressure and boiling. When the vapor pressure of the liquid equals the atmospheric pressure, bubbles of vapor begin to form within the body of liquid, and the liquid boils. See the General ChemistryNow CD-ROM or website: • Screen 13.10 Properties of Liquids (3): Boiling Point, to watch a video of the boiling of a liquid in a partial vacuum and to explore the relationship of molecular composition and structure and boiling point in two simulations ■ Cooking under Pressure To shorten cooking time, a pressure cooker can be used. This sealed pot allows water vapor to build up to pressures somewhat greater than the external or atmospheric pressure. At pressures greater than 760 mm Hg, the boiling point of the water is higher than 100 °C, and foods cook faster. Critical Temperature and Pressure The vapor pressure of a liquid will continue to increase if the temperature is raised above the normal boiling point. On first thought it might seem that vapor pressure–temperature curves (such as shown in Figure 13.18) should continue upward without limit, but this is not so. Instead, when a specific temperature and pressure are reached, the interface between the liquid and the vapor disappears. This point is called the critical point. The temperature at which this phenomenon occurs is the critical temperature, Tc, and the corresponding pressure is the critical pressure, Pc (Figure 13.21). The substance that exists under these conditions is called a supercritical fluid. It is like a gas under such a high pressure that its density resembles that of a liquid, while its viscosity (ability to flow) remains close to that of a gas. For most substances the critical point is at a very high temperature and pressure (Table 13.5). Water, for instance, has a critical temperature of 374 °C and a critical pressure of 217.7 atm. Consider what the substance might look like at the molecular 614 Figure 13.21 Critical temperature and pressure for water. The curve representing equilibrium conditions for liquid and gaseous water ends at the critical point; above that temperature and pressure, water becomes a supercritical fluid. Chapter 13 Intermolecular Forces, Liquids, and Solids Pc Pressure 217.7 atm Critical point Super critical fluid Liquid Vapor 374.0 °C Tc Temperature Table 13.5 Critical Temperatures and Pressures for Common Compounds* Compound CH4 (methane) C2H6 (ethane) C3H8 (propane) C4H10 (butane) CCl2F2 (CFC-12) NH3 H2O CO2 SO2 Tc (°C) 82.6 32.3 96.7 152.0 111.8 132.4 374.0 30.99 157.7 Pc (atm) 45.4 49.1 41.9 37.3 40.9 112.0 217.7 72.8 77.8 *Data taken from D. R. Lide: Basic Laboratory and Industrial Chemicals, Boca Raton FL, CRC Press, 1993. level under these conditions. At this high pressure, water molecules have been forced almost as close together as they are in the liquid state. The high temperature, however, means that each molecule has enough kinetic energy to exceed the forces holding molecules together. As a result, the supercritical fluid has a tightly packed molecular arrangement like a liquid, but the intermolecular forces of attraction that characterize the liquid state are less than the kinetic energy of the particles. Supercritical fluids can have unexpected properties, such as the ability to dissolve normally insoluble materials. Supercritical CO2 is especially useful. Carbon dioxide is widely available, essentially nontoxic, nonflammable, and inexpensive. It is relatively easy to reach its critical temperature of 30.99 °C and critical pressure of 72.8 atm. The material is also easy to handle. CO2 is highly useful because it does not dissolve water or polar compounds such as sugar, but it does dissolve nonpolar oils, which constitute many of the flavoring or odor-causing compounds in foods. As a result, food companies now use supercritical CO2 to extract caffeine from coffee, for example. To decaffeinate coffee, the beans are treated with steam to bring the caffeine to the surface. The beans are then immersed in supercritical CO2, which selectively dissolves the caffeine but leaves intact the compounds that give flavor to coffee. (Decaffeinated coffee contains less than 3% of the original caffeine.) The solution of caffeine in supercritical CO2 is poured off, and the CO2 is evaporated, trapped, and reused. ■ Supercritical CO2 and the Environment Supercritical CO2 has properties that make it attractive as a solvent, so not surprisingly other uses are being sought for this substance. For example, more than 10 billion kilograms of organic and halogenated solvents is used worldwide every year in cleaning applications. These cleaning agents can have deleterious effects on the environment, so it is hoped that many can be replaced by supercritical CO2. (For more about supercritical CO2 see page 632.) Surface Tension, Capillary Action, and Viscosity Molecules in the interior of a liquid interact with molecules all around them (Figure 13.22). In contrast, surface molecules are affected only by those molecules located below the surface layer. This phenomenon leads to a net inward force of attraction on the surface molecules, contracting the surface area and making the liquid behave as though it had a skin. The toughness of the skin of a liquid is measured by its surface tension—the energy required to break through the surface or to disrupt a liquid drop and spread the material out as a film. Surface tension causes water drops to be spheres and not little cubes, for example (Figure 13.23a), because the sphere has a smaller surface area than any other shape of the same volume. Capillary action is closely related to surface tension. When a small-diameter glass tube is placed in water, the water rises in the tube, just as water rises in a piece of paper in water (Figure 13.23b). Because polar Si ¬ O bonds are present on the surface of glass, polar water molecules are attracted by adhesive forces between the two different substances. These forces are strong enough that they can compete 13.5 Properties of Liquids 615 Figure 13.22 Intermolecular forces in a liquid. Forces acting on a molecule at the surface of a liquid are different than those acting on a molecule in the interior of a liquid. Water molecules on the surface are not completely surrounded by other water molecules. Water molecules under the surface are completely surrounded by other water molecules. with the cohesive forces between the water molecules themselves. Thus, some water molecules can adhere to the walls; other water molecules are attracted to them and build a “bridge” back into the liquid. The surface tension of the water (from cohesive forces) is great enough to pull the liquid up the tube, so the water level rises in the tube. The rise will continue until the attractive forces—adhesion between water and glass, cohesion between water molecules—are balanced by the force of gravity pulling down on the water column. These forces lead to the characteristic concave, or downward-curving, meniscus seen with water in a drinking glass or in a laboratory test tube (Figure 13.23c). In some liquids, cohesive forces (high surface tension) are much greater than adhesive forces with glass. Mercury is one example. Mercury does not climb the walls of a glass capillary. In fact, when it is in a glass tube, mercury will form a convex, or upward-curving, meniscus (Figure 13.23c). a, S. R. Nagel, James Frank Institute, University of Chicago; b and c, Charles D. Winters (a) A series of photographs showing the different stages when a water drop falls. The drop was illuminated by a strobe light of 5-ms duration. (The total time for this sequence was 0.05 s.) Water droplets take a spherical shape because of surface tension. (b) Capillary action. Polar water molecules are attracted to the OH bonds in paper fibers, and water rises in the paper. If a line of ink is placed in the path of the rising water, the different components of the ink are attracted differently to the water and paper and are separated in a process called chromatography. (c) Water (top layer) forms a concave meniscus, while mercury (bottom layer) forms a convex meniscus. The different shapes are determined by the adhesive forces of the molecules of the liquid with the walls of the tube and the cohesive forces between molecules of the liquid. Figure 13.23 Adhesive and cohesive forces in liquids. 616 Chapter 13 Intermolecular Forces, Liquids, and Solids ■ Viscosity Another factor in determining viscosity may be the presence of long chains of atoms in substances such as oils. These long chains are floppy and could become entangled with one another in the liquid; the longer the chain, the greater the tangling and the greater the viscosity. One other important property of liquids in which intermolecular forces play a role is their viscosity, the resistance of liquids to flow. When you turn over a glassful of water, it empties quickly. In contrast, it takes much more time to empty a glassful of olive oil or honey. Olive oil consists of molecules with long chains of carbon atoms (see Chapter 11), and it is about 70 times more viscous than ethanol, a small molecule with only two carbons and one oxygen. Longer chains have greater intermolecular forces because there are more atoms to attract one another, with each atom contributing to the total force. In contrast, honey is a concentrated solution of smaller sugar molecules. However, these molecules have numerous ¬ OH groups and are thus capable of hydrogen bonding. See the General ChemistryNow CD-ROM or website: • Screen 13.11 Properties of Liquids (4): Surface Tension, Capillary Action, and Viscosity, to watch videos on these three topics Exercise 13.9—Viscosity Glycerol (HOCH2CHOHCH2OH) is used in cosmetics. Do you expect its viscosity to be larger or smaller than the viscosity of ethanol, CH3CH2OH? Explain briefly. Glycerol 13.6—The Solid State: Metals Many kinds of solids exist in the world around us (Figure 13.24). Solid-state chemistry is one of the booming areas of science, especially because it relates to the development of interesting new materials. As we describe various kinds of solids, we hope to provide a glimpse of the reasons this area is exciting. Solid-state chemistry can be organized by classifying the common types of solids (Table 13.6). This section describes the solid-state structures of common metals, and the next section takes up ionic solids. Next, we examine molecular and network solids in Section 13.8. Finally, Section 13.9 outlines important properties of solids. Table 13.6 Structures and Properties of Various Types of Solid Substances ■ Chemistry of Materials For a glimpse into scientists’ latest efforts to create new materials and to find new uses for old materials, see “The Chemistry of Modern Materials,” page 642. Type Ionic Examples NaCl, K2SO4, CaCl2, (NH4)3PO4 Structural Units Positive and negative ions; no discrete molecules Metal atoms (positive metal ions with delocalized electrons) Molecules Atoms held in an infinite two-, or three-dimensional network Covalently bonded networks with no long-range regularity Metallic Iron, silver, copper, other metals and alloys Molecular Network H2, O2, I2, H2O, CO2, CH4, CH3OH, CH3CO2H Graphite, diamond, quartz, feldspars, mica Amorphous Glass, polyethylene, nylon 13.6 The Solid State: Metals Silicon, a network solid 617 Aluminum, a metallic solid Photo: Charles D. Winters NaCl, a crystalline ionic solid Polyethylene, an amorphous solid Figure 13.24 Some common solids. Crystal Lattices and Unit Cells In both gases and liquids, molecules move continually and randomly, and they rotate and vibrate as well. Because of this movement, an orderly arrangement of molecules in the gaseous or liquid state is not possible. In solids, however, the molecules, atoms, or ions cannot change their relative positions (although they vibrate and occasionally rotate). Thus, a regular, repeating pattern of atoms or molecules within the structure—a long-range order—is characteristic of the solid state. The beautiful, external (macroscopic) regularity of a crystal of salt (see Figure 13.24) suggests that it has an internal symmetry, a symmetry involving the ions that Forces Holding Units Together Ionic; attractions among charges on positive and negative ions Metallic; electrostatic attraction among metal ions and electrons Dispersion forces, dipole–dipole forces, hydrogen bonds Covalent; directional electron-pair bonds Typical Properties Hard; brittle; high melting point; poor electric conductivity as solid, good as liquid; often water-soluble Malleable; ductile; good electric conductivity in solid and liquid; good heat conductivity; wide range of hardness and melting points Low to moderate melting points and boiling points; soft; poor electric conductivity in solid and liquid Wide range of hardness and melting points (threedimensional bonding two-dimensional bonding); poor electric conductivity, with some exceptions Noncrystalline; wide temperature range for melting; poor electric conductivity, with some exceptions Covalent; directional electron-pair bonds 618 Chapter 13 Intermolecular Forces, Liquids, and Solids Figure 13.25 Unit cells for a flat, two-dimensional solid made from circular “atoms.” Many unit cells are possible, with two of the most obvious being squares. The lattice can be represented as being made up of repeating unit cells. (That is, this two-dimensional lattice can be built by translating the unit cells throughout the plane of the figure. Each cell must move by the length of one side of the unit cell.) In this figure all unit cells contain a net of one large sphere and one small sphere. make up the solid. Structures of solids can be described as three-dimensional lattices of atoms, ions, or molecules. For a crystalline solid, we can identify the unit cell, the smallest repeating unit that has all of the symmetry characteristic of the way the atoms, ions, or molecules are arranged in the solid. To understand unit cells, consider first a two-dimensional lattice model, the repeating pattern of spheres shown in Figure 13.25. The yellow square at the left is a unit cell. The overall pattern can be created from a group of these cells by joining them edge to edge. It is also a requirement that unit cells reflect the stoichiometry of the solid. Here the square unit cell at the left contains one-fourth of each of the four larger spheres and one smaller sphere, giving a total of one large and one small sphere per two-dimensional unit cell. You may recognize that it is possible to draw other unit cells for this two-dimensional lattice. One option is the square in the middle of Figure 13.25 that fully encloses a single large sphere and parts of small spheres that add up to one net small sphere. Yet another possible unit cell is the parallelogram at the right. Other unit cells are possible, but it is conventional to try to draw unit cells in which atoms or ions are placed at the lattice points, that is, at the corners of the cube or other geometric object that constitutes the unit cell. The three-dimensional lattices of solids can be built by assembling threedimensional unit cells much like building blocks (Figure 13.26). The assemblage of these three-dimensional unit cells defines the crystal lattice. To construct crystal lattices, nature uses seven three-dimensional unit cells. They differ from one another in that their sides have different relative lengths and their edges meet at different angles. The simplest of the seven crystal lattices is the cubic unit cell, a cell with edges of equal length that meet at 90° angles. We shall Figure 13.26 Cubic unit cells. (a) The cube, one of the seven basic unit cells that describe crystal systems. (b) Stacking cubes to build a crystal lattice. Each crystal face is part of two cubes, each edge is part of four cubes, and each corner is part of eight cubes. b a c a g b Cubic a b ab (a) c g 90° Each face is part of two cubes (b) Each edge is part of four cubes Each corner is part of eight cubes 13.6 Simple cubic Body-centered cubic The Solid State: Metals 619 Figure 13.27 The three cubic unit cells. The top row shows the lattice points of the three cells, and the bottom row shows the same cells using space-filling spheres. The spheres in each figure represent identical atoms (or ions) centered on the lattice points. Because eight unit cells share a corner atom, only 1/8 of each corner atom lies within a given unit cell; the remaining 7/8 lies in seven other unit cells. Because each face of a fcc unit cell is shared with another unit cell, one half of each atom in the face of a face-centered cube lies in a given unit cell, and the other half lies in the adjoining cell. Face-centered cubic look in detail at just this structure, not only because cubic unit cells are easily visualized but also because they are commonly encountered. Within the cubic class, three cell symmetries occur: primitive or simple cubic (sc), body-centered cubic (bcc), and face-centered cubic (fcc) (Figure 13.27). All three have identical atoms, molecules, or ions at the corners of the cubic unit cell. The bcc and fcc arrangements, however, differ from the primitive cube in that they have additional particles at other locations. The bcc structure is called “body-centered” because it has an additional particle, of the same type as those at the corners, at the center of the cube. The fcc arrangement is called “face-centered” because it has a particle, of the same type as the corner atoms, in the center of each of the six faces of the cube. Metals may assume any of these structures. The alkali metals, for example, are body-centered cubic, whereas nickel, copper, and aluminum are facecentered cubic (Figure 13.28). H Li Be Na Mg K Ca Sc Ti Y V B Al C Si N P O S F He Ne Figure 13.28 Metals use four different unit cells. Three are based on the cube, and the fourth is the hexagonal unit cell. See “A Closer Look: Packing Oranges,” page 621. Cl Ar Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr I Xe Rb Sr Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te Re Os Ir Cs Ba La Hf Ta W Fr Ra Ac Pt Au Hg Tl Pb Bi Po At Rn Simple cubic Cubic close packing (Face-centered cubic) Body-centered cubic Hexagonal close packing 620 Chapter 13 Intermolecular Forces, Liquids, and Solids Figure 13.29 Atom sharing at cube corners and faces. (a) In any cubic lattice, each corner particle is shared equally among eight cubes, so one eighth of the particle is within a particular cubic unit cell. (b) In a face-centered lattice, each particle on a cube face is shared equally between two unit cells. One half of each particle of this type is within the given unit cell. (a) (b) ■ Atoms per Unit Cell Cell Type Simple cubic (sc) Body-centered cubic (bcc) Face-centered cubic (fcc) Atoms Per Unit Cell 1 2 4 When the cubes pack together to make a three-dimensional crystal of a metal, the atom at each corner is shared among eight cubes (Figures 13.26, 13.27, and 13.29a). Because of this, only one eighth of each corner atom is actually within a given unit cell. Furthermore, because a cube has eight corners, and because one eighth of the atom at each corner “belongs to” a particular unit cell, the corner atoms contribute a net of one atom to a given unit cell. Thus, the primitive or simple cubic arrangement has one net atom within the unit cell. (8 corners of a cube)(1/8 of each corner atom within a unit cell) 1 net atom per unit cell In contrast to the simple cube, a body-centered cube has an additional atom wholly within the unit cell at the cube’s center. The center particle is present in addition to those at the cube corners, so the body-centered cubic arrangement has a net of two atoms within the unit cell. In a face-centered cubic arrangement, there is an atom in each of the six faces of the cube in addition to those at the cube corners. One half of each atom on a face belongs to the given unit cell (Figure 13.29b). Three net particles are therefore contributed by the particles on the faces of the cube: Photo: Charles D. Winters (6 faces of a cube)(1/2 of an atom within a unit cell) 3 net face-centered atoms within a unit cell Thus, the face-centered cubic arrangement has a net of four atoms within the unit cell, one contributed by the corner atoms and another three contributed by the atoms centered in the six faces. An experimental technique, x-ray crystallography, can be used to determine the structure of a crystalline substance (Figure 13.30). Once the structure is known, this information can be combined with other experimental information to calculate such useful parameters as the radius of an atom (Study Questions 13.54–13.55). Aluminum metal. The metal has a face-centered cubic unit cell with a net of four Al atoms in each unit cell. Figure 13.30 X-ray crystallography. In the x-ray diffraction experiment, a beam of x-rays is directed at a crystalline solid. The photons of the x-ray beam are scattered by the atoms of the solid. The scattered x-rays are detected by a photographic film or an electronic detector, and the pattern of scattered x-rays is related to the locations of the atoms or ions in the crystal. Photographic film Sample X-ray source X-ray beam 13.6 The Solid State: Metals 621 A Closer Look Packing Oranges It is a “rule” that nature does things as efficiently as possible. You know this if you have ever tried to stack some oranges into a pile that doesn’t fall over and that takes up as little space as possible. How did you do it? Clearly, the pyramid arrangement below on the right works, whereas the cubic one on the left does not. crystal, the layers repeat their pattern in the manner ABABAB. . . . Atoms in each A layer are directly above the ones in another A layer; the same holds true for the B layers. In the ccp arrangement, the atoms of the “top” layer (A) rest in depressions in the middle layer (B), and those of the “bottom” layer (C) are oriented opposite to those in the top layer. In a crystal, the pattern is repeated ABCABCABC. . . . By turning the whole crystal, you can see that the ccp arrangement is the facecentered cubic structure (Figure 2). Photos:Charles D. Winters (a) Hexagonal close-packing (hcp) (b) Cubic close-packing face centered cubic (fcc) Top layer Open space between balls A C Middle layer B B If you could look inside the pile, you would find that less open space is left in the pyramid stacking than in the cube stacking. Only 52% of the space is filled in the cubic packing arrangement. (If you could stack oranges as a body-centered cube, that would be slightly better; 68% of the space is used.) However, the best method is the pyramid stack, which is really a face-centered cubic arrangement. Oranges, atoms, or ions packed this way occupy 74% of the available space. To fill three-dimensional space, the most efficient way to pack oranges or atoms is to begin with a hexagonal arrangement of spheres, as in this arrangement of marbles. Bottom A layer A Figure 1 Efficient packing. The most efficient ways to pack atoms or ions in crystalline materials are hexagonal close packing (hcp) and cubic close packing (ccp). Charles D. Winters Charles D. Winters Succeeding layers of atoms or ions are then stacked one on top of the other in two different ways. Depending on the stacking pattern (Figure 1), you will get either a cubic close-packed (ccp) or hexagonal close-packed (hcp) arrangement. In the hcp arrangement, additional layers of particles are placed above and below a given layer, fitting into the same depressions on either side of the middle layer. In a three-dimensional (a) (b) Figure 2 Models of close packing. (a) A model of hexagonal closepacking, where the layers repeat in the order ABABAB. . . . (b) A facecentered unit cell (cubic close-packing), where the layers repeat in the order ABCABC. . . . (A kit from which these models can be built is available from the Institute for Chemical Education at the University of Wisconsin at Madison.) 622 Chapter 13 Intermolecular Forces, Liquids, and Solids See the General ChemistryNow CD-ROM or website: • Screen 13.12 Solid Structures (1): Crystalline and Amorphous Solids, to view an animation of unit cells 13.7—The Solid State: Structures and Formulas of Ionic Solids The lattices of many ionic compounds are built by taking a simple cubic or facecentered cubic lattice of spherical ions of one type and placing ions of opposite charge in the holes within the lattice. This produces a three-dimensional lattice of regularly placed ions. The smallest repeating unit in these structures is, by definition, the unit cell for the ionic compound. The choice of the lattice and the number and location of the holes that are filled are the keys to understanding the relationship between the lattice structure and the formula of a salt. This is illustrated with the ionic compound cesium chloride, CsCl (Figure 13.31). The structure of CsCl has a primitive cubic unit cell of chloride ions. The cesium ion fits into a hole in the center of the cube. (An equivalent unit cell has a primitive cubic unit cell of Cs ions with a Cl ion in the center of the cube.) Next consider the structure for NaCl. An extended view of the lattice and one unit cell are illustrated in Figures 13.32a and 13.32b, respectively. The Cl ions are arranged in a face-centered cubic unit cell, and the Na ions are arranged in a regular manner between these ions. Notice that each Na ion is surrounded by six Cl ions. An octahedral geometry is assumed by the ions surrounding an Na ion, so the Na ions are said to be in octahedral holes (Figure 13.32c). The formula for NaCl can be related to this structure by counting the number of cations and anions contained in one unit cell. A face-centered cubic lattice of Cl ions has a net of four Cl ions within the unit cell. There is one Na ion in the center of the unit cell, contained totally within the unit cell. In addition, there are 12 Na ions along the edges of the unit cell. Each of these Na ions is shared among four unit cells, so each contributes one fourth of an Na ion to the unit cell, giving three additional Na ions within the unit cell. (1 Na ion in the center of the unit cell) (1/4 of Na ion in each edge 12 edges) net of 4 Na ions in NaCl unit cell Cs , radius 165 pm ■ Lattice Ions and Holes When ions are packed into a lattice, the holes in the lattice are usually smaller than the ions used to create the lattice. Therefore, a lattice is usually built out of the larger anions, and the smaller cations are placed into the holes in the lattice. For NaCl, for example, an fcc lattice is built out of the Cl ions (radius 181 pm), and the smaller Na cations (radius 98 pm) are placed in appropriate holes in the lattice. Figure 13.31 Cesium chloride (CsCl) unit cell. The unit cell of CsCl may be viewed in two ways. The only requirement is that the unit cell must have a net of one Cs ion and one Cl ion. Either way, it is a simple cubic unit cell of ions of one type (Cl on the left or Cs on the right). Generally, ionic lattices are assembled by placing the larger ions (here Cl ) at the lattice points and placing the smaller ions (here Cs ) in the lattice holes. Cl , radius 181 pm Cl ions at each cube corner 1 net Cl ion in the unit cell. One Cs ion at each cube corner 1 net Cs ion in the unit cell. Cl lattice and Cs in lattice hole Cs lattice and Cl in lattice hole 13.7 (a) (b) The Solid State: Structures and Formulas of Ionic Solids (c) Na in octahedral hole Na in octahedral hole 623 Cl Na Photo: Charles D. Winters NaCl unit cell (expanded) 1 hole of this kind in the center of the unit cell 12 holes of this kind in the 12 edges of the unit cell (a net of 3 holes) Octahedral holes in a face-centered lattice Figure 13.32 Sodium chloride. (a) Cubic NaCl is based on a face-centered cubic unit cell of Na and Cl ions. (b) An expanded view of a sodium chloride lattice. (The lines represent the connections between lattice points.) The smaller Na ions (silver) are packed into a face-centered cubic lattice of larger Cl– ions (yellow). (c) A close-up view of the octahedral holes in the lattice. This accounts for all of the ions contained in the unit cell: four Cl and four Na ions. Thus, a unit cell of NaCl has a 1:1 ratio of Na and Cl ions, as the formula requires. Another common unit cell has ions of one type in a face-centered cubic unit cell. Ions of the other type are located in tetrahedral holes, wherein each ion is surrounded by four oppositely charged ions. As illustrated in Figure 13.33, there are eight tetrahedral holes in a face-centered unit cell. In ZnS (zinc blende), the sulfide ions (S2 ) form a face-centered cubic unit cell. The zinc ions (Zn2 ) then occupy one half of the tetrahedral holes, and each Zn2 ion is surrounded by four S2 ions. The unit cell consists of a net of four S2 ions and four Zn2 ions, which are contained wholly within the unit cell. In summary, compounds with the formula MX commonly form one of three possible crystal structures: 1. Mn ions occupying all the cubic holes of a simple cubic Xn lattice. Example: CsCl. 2. Mn ions in all the octahedral holes in a face-centered cubic Xn lattice. Example: NaCl. 3. Mn ions occupying half of the tetrahedral holes in a face-centered cube lattice of Xn ions. Example: ZnS. Chemists and geologists, in particular, have observed that the sodium chloride or “rock salt” structure is adopted by many ionic compounds, most especially by the alkali metal halides (except CsCl, CsBr, and CsI), the oxides and sulfides of the alkaline earth metals, and the oxides of formula MO of the transition metals of the fourth period. Finally, the formulas of compounds must always be reflected in the structures of their unit cells; therefore, the formula can always be derived from the unit cell structure. 624 Chapter 13 Intermolecular Forces, Liquids, and Solids Single tetrahedron with a tetrahedral hole shown as a white sphere. Tetrahedral hole (a) (b) Fcc lattice of S2 ions Zn2 ions in half of the tetrahedral holes Figure 13.33 Tetrahedral holes and two views of the ZnS (zinc blende) unit cell. (a) The tetrahedral holes in a face-centered cubic lattice. (b) This unit cell is an example of a face-centered cubic lattice of ions of one type with ions of the opposite type in one half of the tetrahedral holes. See the General ChemistryNow CD-ROM or website: • Screen 13.13 Solid Structures (2): Ionic Solids, to view an animation of ionic unit cells Example 13.7—Ionic Structure and Formula Problem One unit cell of the mineral perovskite is illustrated here. This compound is composed of calcium and titanium cations and oxide anions. Based on the unit cell, what is the formula of perovskite? Ca2 O2 Ti4 Strategy Identify the ions present in the unit cell and their locations within the unit cell. Decide on the net number of ions of each kind in the cell. Solution The unit cell has Ca2 ions at the corners of the cubic unit cell, a titanium ion in the center of the cell, and oxide ions on the cube edges. Number of Ca2 ions: (8 Ca2 ions at cube corners) Number of Ti 4 (1/8 of each ion inside unit cell) 1 net Ca2 ion ions: 1 net Ti4 ion One ion is in the cube center 13.8 Other Kinds of Solid Materials 625 Number of O2 ions: (12 O2 ions in cube edges) (1/4 of each ion inside cell) 3 net O2 ions Thus, the formula of perovskite is CaTiO3. Comment This is a reasonable formula. A Ca2 ion and three O2 ions would require a titanium ion with a 4 charge. Titanium is in Group 4B of the periodic table, so Ti4 is a predictable ion. Exercise 13.10—Structure and Formula If an ionic solid has a fcc lattice of anions (X) and all of the tetrahedral holes are occupied by metal cations (M), is the formula of the compound MX, MX2, or M2X? 13.8—Other Kinds of Solid Materials So far we have described the structures of metals and simple ionic solids. Now we will look briefly at the other categories of solids: molecular solids, network solids, and amorphous solids (Table 13.6). Molecular Solids Molecules such as H2O and CO2 are found in the solid state under appropriate conditions. In these cases, it is molecules, rather than atoms or ions, that pack in a regular fashion in a three-dimensional lattice. We have commented already on one such structure, the structure of ice (Section 13.3, Figure 13.10). The way molecules are arranged in a crystalline lattice depends on the shape of the molecules and the types of intermolecular forces. Molecules tend to pack in the most efficient manner and to align in ways that maximize intermolecular forces of attraction. Thus, the water structure was established to gain the maximum intermolecular attraction through hydrogen bonding. As illustrated in Figure 13.9, organic acid molecules often assemble in the solid state as dimers, with two molecules being linked by hydrogen bonding. The greatest interest in the structures of molecular solids focuses on the structures of the molecules themselves and not on the way they pack into the solid. It is from structural studies on molecular solids that most of the information on molecular geometries, bond lengths, and bond angles discussed in Chapters 9 through 11 was assembled. See the General ChemistryNow CD-ROM or website: • Screen 13.14 Solid Structures (3): Molecular Solids, to view an animation of the packing of molecules in a molecular solid Network Solids Network solids are composed entirely of a three-dimensional array of covalently bonded atoms. Common examples include two allotropes of carbon: graphite and diamond. Elemental silicon is a network solid with a diamond-like structure. 626 Sinclair Stammers/Science Photo Library/Photo Researchers, Inc. Chapter 13 Intermolecular Forces, Liquids, and Solids Figure 13.34 A mixture of natural and synthetic industrial diamonds. The colors of diamonds may range from colorless to yellow, brown, or black. Poorer-quality diamonds are used extensively in industry, mainly for cutting or grinding tools. Industrial-quality diamonds are produced synthetically by heating graphite, along with a metal catalyst, to 1200–1500 °C and a pressure of 65–90 kilobars. Diamonds have a low density (d 3.51 g/cm3), but they are also the hardest material and the best conductor of heat known. They are transparent to visible light, as well as to infrared and ultraviolet radiation. Diamonds are electrically insulating but behave as semiconductors with some advantages over silicon. What more could a scientist want in a material—except a cheap, practical way to make it! In the 1950s, scientists at General Electric in Schenectady, New York, achieved something alchemists had sought for centuries: the synthesis of diamonds from carbon-containing materials, including wood or peanut butter. Their technique was to heat graphite to a temperature of 1500 °C in the presence of a metal, such as nickel or iron, and under a pressure of 65–90 kilobars. Under these conditions, the carbon dissolves in the metal and slowly forms diamonds (Figure 13.34). More than $500 million worth of diamonds are made this way annually. Most are used for abrasives and diamond-coated cutting tools. Silicates, compounds composed of silicon and oxygen, represent an enormous class of chemical compounds. You know them in the form of sand, quartz, talc, and mica, or as a major constituent of rocks such as granite. The structure of quartz is illustrated in Figure 13.35, and other silicates are described in Chapter 21. Most network solids are hard and rigid and are characterized by high melting and boiling points. These characteristics reflect the fact that a great deal of energy must be provided to break the covalent bonds in the lattice. For example, silicon dioxide melts at temperatures higher than 1600 °C. The solid consists of tetrahedral silicon atoms covalently bonded to oxygen atoms in a giant three-dimensional lattice (Figure 13.35). A high temperature is required to break the covalent bonds between silicon and oxygen and thereby disrupt this stable structure. See the General ChemistryNow CD-ROM or website: • Screen 13.15 Solid Structures (4): Network Solids, to view an animation of the structures of network solids. Amorphous Solids A characteristic property of pure crystalline solids—whether metals, ionic solids, or molecular solids—is that they melt at a specific temperature. For example, water melts at 0 °C, aspirin at 135 °C, lead at 327.5 °C, and NaCl at 801 °C. Because they are specific and reproducible values, melting points are often used as a means of identifying chemical compounds. Another property of crystalline solids is that they form well-defined crystals, with smooth, flat faces. When a sharp force is applied to a crystal, it will most often cleave to give smooth, flat faces. The resulting solid particles are smaller versions of the original crystal (Figure 13.36). Many common solids, including ones that we encounter everyday, do not have these properties, however. Glass is a good example. When glass is heated it softens over a wide temperature range, a property useful for artisans and craftsmen who can create beautiful and functional products for our enjoyment and use. Glass also possesses a property that we would rather it not have: When glass breaks, it leaves randomly shaped pieces. Other materials that behave similarly include common polymers such as polyethylene, nylon, and other plastics. Figure 13.35 Silicon dioxide. Common quartz, SiO2, is a network solid consisting of silicon and oxygen atoms. Photo: Charles D. Winters 13.9 The Physical Properties of Solids 627 Charles D. Winters (a) A salt crystal can be cleaved cleanly into smaller and smaller crystals that are duplicates of the larger crystal. (b) Glass is an amorphous solid composed of silicon and oxygen atoms. It has, however, no long-range order as in crystalline quartz. (c) Glass can be molded and shaped into beautiful forms and, by adding metal oxides, can take on wonderful colors. Figure 13.36 Crystalline and amorphous solids. The characteristics of these amorphous solids relate to their molecular structure. At the particulate level, amorphous solids do not have a regular structure. In fact, in many ways these substances look a lot like liquids. Unlike liquids, however, the forces of attraction are strong enough that movement of the molecules or ions is restricted. 13.9—The Physical Properties of Solids The shape of a crystalline solid is a reflection of its internal structure. But what about physical properties such as hardness and the temperatures at which solids melt? These and many other physical properties of solids are of interest to chemists, geologists, engineers, and others. Melting: Conversion of Solid to Liquid The melting point of a solid is the temperature at which the lattice collapses and the solid is converted to a liquid. Like the liquid-to-vapor transformation, melting requires energy, called the enthalpy of fusion (given in kilojoules per mole) (Chapter 6). Heat energy absorbed on melting enthalpy of fusion Heat energy evolved on freezing enthalpy of crystallization ¢H° fusion (kJ/mol) ¢H° fusion (kJ/mol) Enthalpies of fusion can range from just a few thousand joules per mole to many thousands of joules per mole (Table 13.7). A low melting temperature will certainly mean a low value for the enthalpy of fusion, whereas high melting points are associated with high enthalpies of fusion. Figure 13.37 shows the enthalpies of fusion for the metals of the fourth through the sixth periods. Based on this figure and Table 13.7 we can make two statements: (1) Metals that have notably low melting points, such as the alkali metals and mercury (mp 39 °C), also have low enthalpies of fusion; and (2) transition metals have high heats of fusion, with those of the third transition series being extraordinarily high. This trend parallels the trend seen with the melting points for these elements. Tungsten, which has the highest 628 Chapter 13 Intermolecular Forces, Liquids, and Solids Table 13.7 Melting Points and Enthalpies of Fusion of Some Elements and Compounds Compound Metals Hg Na Al Ti W Melting Point (°C) 39 98 660 1668 3422 Enthalpy of Fusion (kJ/mol) 2.29 2.60 10.7 20.9 35.2 0.440 0.510 6.41 10.8 1.99 2.41 2.87 6.02 33.4 28.2 26.1 23.6 Type of Interparticle Forces Metal bonding; see page 643 Molecular Solids: Nonpolar Molecules 219 O2 220 F2 102 Cl2 7.2 Br2 Molecular Solids: Polar Molecules HCl 114 HBr 87 HI 51 H2O Ionic Solids NaF NaCl NaBr NaI 0 996 801 747 660 Dispersion forces only All three HX molecules have dipole–dipole forces. Dispersion forces increase with size and molar mass. Hydrogen bonding All ionic solids have extended ion–ion interactions. Note the general trend is the same as for lattice energies (see Section 9.3 and Figure 9.3). melting point of all the known elements except for carbon, also has the highest enthalpy of fusion among the transition metals. These properties affect the uses of this metal. For example, tungsten is used for the filaments in lightbulbs; no other material has been found to work better since the invention of the lightbulb in 1908. The melting temperature of a solid can convey a great deal of information. Table 13.7 provides some data for several basic types of substances: metals, polar and nonpolar molecules, and ionic solids. In general, nonpolar substances that form molecular solids have low melting points. Melting points increase within a series of related molecules, however, as the size and molar mass increase. This happens because dispersion forces are generally larger when the molar mass is larger. Thus, increasing amounts of energy are required to break down the intermolecular forces in the solid, a principle that is reflected in an increasing enthalpy of fusion. The ionic compounds in Table 13.7 have higher melting points and higher enthalpies of fusion than do molecular solids. This trend is due to the strong ion–ion forces present in ionic solids, forces that are reflected in high lattice energies (see Section 9.3). Because ion–ion forces depend on ion size (as well as ion charge), there is a good correlation between lattice energy and the position of the metal or halogen in the periodic table. For example, the data in Table 13.7 show a decrease in melting point and enthalpy of fusion for sodium salts as the halide ion increases in size. This parallels the decrease in lattice energy seen with increasing ion size illustrated in Figure 9.3. Sublimation: Conversion of Solid to Vapor Molecules can escape directly from the solid to the gas phase by sublimation (Figure 13.38). Solid ¡ gas heat energy required ¢ H° sublimation 13.9 The Physical Properties of Solids 629 40 W Re Ta 30 Nb Heat of fusion, kJ/mol Hf 25 Ti 20 Y Sc 15 Sr 10 Rb K Cs 0 1A 2A 3B 4B 5B 6B 7B 8B 8B Periodic group 8B 1B 2B 3A 4A 5A Hg Ca Ba 5 La V Cr Mn Fe Co Zr Tc Ru Rh Pd Pt Ni Ag Cu Au Cd Zn In Ga Tl Sn Pb Bi Mo Ir Os Fourth period Fifth period Sixth period 35 Figure 13.37 Heat of fusion of fourth-, fifth-, and sixth-period metals. Heats of fusion range from 2–5 kJ/mol for Group 1A elements to 35.2 kJ/mol for tungsten. Notice that heats of fusion generally increase for B-group metals on descending the periodic table. Sublimation, like fusion and evaporation, is an endothermic process. The heat energy required is called the enthalpy of sublimation. Water, which has a molar enthalpy of sublimation of 51 kJ/mol, can be converted from solid ice to water vapor quite readily. A good example of this phenomenon is the sublimation of frost from grass and trees as night turns to day on a cold morning in the winter. Figure 13.38 Sublimation. Sublimation entails the conversion of a solid directly to its vapor. Here, iodine (I2) sublimes when heated in warm water. If an ice-filled test tube is inserted into the flask, the vapor condenses on the cold surface. Iodine sublimes when heated Charles D. Winters 630 Chapter 13 Intermolecular Forces, Liquids, and Solids 13.10—Phase Diagrams Depending on the conditions of temperature and pressure, a substance can exist as a gas, a liquid, or a solid. In addition, under certain specific conditions, two (or even three) states can coexist in equilibrium. It is possible to summarize this information in the form of a graph called a phase diagram. Phase diagrams are used to illustrate the relationship between phases of matter and the pressure and temperature. Water A phase diagram for water appears in Figure 13.39. The lines in a phase diagram identify the conditions under which two phases exist at equilibrium. Conversely, all points that do not fall on the lines in the figure represent conditions under which only one state exists. Line A-B represents conditions for solid–vapor equilibrium, and line A-C for liquid–solid equilibrium. The line from point A to point D, representing the temperature and pressure combination at which the liquid and vapor phases are in equilibrium, is the same curve plotted for water vapor pressure in Figure 13.18. Recall that the normal boiling point, 100 °C in the case of water, is the temperature at which the equilibrium vapor pressure is 760 mm Hg. Point A, appropriately called the triple point, indicates the conditions under which all three phases coexist in equilibrium. For water, the triple point is at P 4.6 mm Hg and T 0.01 °C. The line A-C shows the conditions of pressure and temperature at which solid–liquid equilibrium exists. (Because no vapor pressure is involved here, the pressure referred to is the external pressure on the liquid.) For water, this line has a negative slope; the change for water is approximately 0.01 °C for each one-atmosphere increase in pressure. That is, the higher the external pressure, the lower the melting point. The negative slope of the water solid–liquid equilibrium line can be explained from our knowledge of the structure of water and ice. When the pressure on an object increases, common sense tells us that the volume of the object will become smaller, giving the substance a higher density. Because ice is less dense than liquid water (due to the open lattice structure of ice), ice and water in equilibrium respond to increased pressure (at constant T ) by melting ice to form more water because the same mass of water requires less volume. Ice Skating and the Ice-Liquid Equilibrium Ice is slippery stuff. It was long assumed that you can ski or skate on ice because the surface melts slightly from the pressure of a skate blade or ski, or that surface melting occurs because of frictional heating. This has always seemed an unsatisfying explanation, however, because it does not seem possible that just standing or sliding on a piece of ice could produce a pressure or temperature high enough to cause sufficient melting. Recently, surface chemists have studied ice surfaces and have come up with a better explanation. They have concluded that water molecules on the surface of ice are vibrating rapidly. In fact, the outermost layer or two of water molecules is almost liquid-like. This arrangement makes the surface slippery, explaining why we can ski on snow and skate on ice. Phase Diagrams and Thermodynamics Let us explore the water phase diagram further by correlating phase changes with thermodynamic data. Suppose we begin with ice at 10 °C and under a pressure of 500 mm Hg (point a on Figure 13.39). As ice is heated, it absorbs about 2.1 J/g • K in warming from point a to point b at a temperature between 0 °C and 0.01 °C. At 13.10 Phase Diagrams 631 C 760 mm Pressure (mm Hg) Normal freezing point Normal D boiling point a Solid b Liquid c Vapor A 4.58 mm Triple point B 0° 0.01° Temperature (°C) 100° Active Figure 13.39 Phase diagram for water. The scale is intentionally exaggerated to be able to show the triple point and the negative slope of the line representing the liquid–solid equilibrium. See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. this point the solid is in equilibrium with liquid water. Solid–liquid equilibrium is maintained until 333 J/g has been transferred to the sample and it has become liquid water at a temperature slightly greater than 0 °C. If the liquid, still under a pressure of 500 mm Hg, now absorbs 4.184 J/g, it warms to point c. The temperature at point c is about 89 °C, and equilibrium is established between liquid water and water vapor. The equilibrium vapor pressure of the liquid water is 500 mm Hg. If 2260 J/g is transferred to the liquid–vapor sample, the equilibrium vapor pressure remains 500 mm Hg until the liquid is completely converted to vapor at 89 °C. Carbon Dioxide The basic features of the phase diagram for CO2 (Figure 13.40) are the same as those for water. There are some important differences, however. In contrast to water, the CO2 solid–liquid equilibrium line has a positive slope. That is, increasing pressure on solid CO2 in equilibrium with liquid CO2 will shift the equilibrium to solid CO2. Thus, adding pressure to solid CO2 will cause it to move to the more dense phase, the solid phase. Because solid CO2 is denser than the liquid, the newly formed solid CO2 sinks to the bottom in a container of liquid CO2. Another feature of the CO2 phase diagram is the triple point that occurs at a pressure of 5.19 atm (3940 mm Hg) and 216.6 K ( 56.6 °C). CO2 cannot be a liquid at pressures lower than this. Thus, at a pressure of 1 atm, solid CO2 is in equilibrium with the gas at a temperature of 197.5 K ( 78.7 °C). As a result, as solid CO2 632 Chapter 13 Intermolecular Forces, Liquids, and Solids Figure 13.40 The phase diagram of CO2. Notice in particular the positive slope of the solid–liquid equilibrium line. Pc 73 atm Critical point “Supercritical fluid” Pressure (atm) 16 14 12 10 8 6 4 2 0 100 Solid Liquid Gas 5.19 atm Triple point 56.6°C 60 20 0 20 Temperature (°C) Tc 31°C 60 warms to room temperature, it sublimes rather than melts. CO2 is called dry ice for this reason; it looks like water ice, but it does not melt. From the CO2 phase diagram we can also learn that CO2 gas can be converted to a liquid at room temperature (20–25 °C) by exerting a moderate pressure on the gas. In fact, CO2 is regularly shipped in tanks as a liquid to laboratories and industrial companies. Finally, the critical pressure and temperature for CO2 are 73 atm and 31 °C, respectively. Because the critical temperature and pressure are easily attained in the laboratory, it is possible to observe the transformation to supercritical CO2 (Figure 13.41). See the General ChemistryNow CD-ROM or website: • Screen 13.17 Phase Changes, to view animations of phase changes and to do an exercise on phase diagrams The separate phases of CO2 are seen through the window in a highpressure vessel. As the sample warms and the pressure increases, the meniscus becomes less distinct. As the temperature continues to increase, it is more difficult to distinguish the liquid and vapor phases. Once the critical T and P are reached, distinct liquid and vapor phases are no longer in evidence. This homogenous phase is “supercritical CO2.” Figure 13.41 Phase changes for CO2. Dr. Christopher M. Rayner/University of Leeds. Chapter Goals Revisited 633 Chapter Goals Revisited When you have finished studying this chapter, you should ask if you have met the chapter goals. In particular you should be able to Describe intermolecular forces and their effects a. Describe the various intermolecular forces found in liquids and solids (Sections 13.2 and 13.4). General ChemistryNow homework: Study Question(s) 4, 6 b. Tell when two molecules can interact through a dipole–dipole attraction and when hydrogen bonding may occur. The latter occurs most strongly when H is attached to O, N, or F (Sections 13.2 and 13.3). General ChemistryNow homework: SQ(s) 8, 87 • • See the General ChemistryNow CD-ROM or website to: Assess your understanding with homework questions keyed to each goal Check your readiness for an exam by taking the exam-prep quiz and exploring the resources in the personalized Learning Plan it provides c. Identify instances in which molecules interact by induced dipoles (dispersion forces) (Section 13.2). Understand the importance of hydrogen bonding a. Explain how hydrogen bonding affects the properties of water (Section 13.3). Understand the properties of liquids a. Explain the processes of evaporation and condensation, and use the enthalpy of vaporization in calculations (Section 13.5). General ChemistryNow homework: SQ(s) 12, 53 b. Define the equilibrium vapor pressure of a liquid, and explain the relationship between the vapor pressure and boiling point of a liquid (Section 13.5). General ChemistryNow homework: SQ(s) 14, 17, 19, 20, 43, 47, 79 c. Describe the phenomena of the critical temperature, Tc, and critical pressure, Pc, of a substance (Section 13.5). d. Describe how intermolecular interactions affect the cohesive forces between identical liquid molecules, the energy necessary to break through the surface of a liquid (surface tension), and the resistance to flow, or viscosity, of liquids (Section 13.5). e. Use the Clausius-Clapeyron equation, which connects temperature, vapor pressure, and enthalpy of vaporization for liquids (Section 13.5). General ChemistryNow homework: SQ(s) 22 Understand cubic unit cells a. Describe the three types of cubic unit cells: primitive or simple cubic (sc), body-centered cubic (bcc), and face-centered cubic (fcc) (Section 13.5). b. Relate atom size and unit cell dimensions. General ChemistryNow homework: SQ(s) 54 Relate unit cells for ionic compounds to formulas (Section 13.7) General ChemistryNow homework: SQ(s) 26, 28, 83, 85 Describe the properties of solids a. Characterize different types of solids: metallic (e.g., copper), ionic (e.g., NaCl and CaF2), molecular (e.g., water and I2), network (e.g., diamond), and amorphous (e.g., glass and many synthetic polymers) (Sections 13.6–13.8 and Table 13.6). b. Define the enthalpy of fusion and use it in a calculation (Section 13.9). 634 Chapter 13 Intermolecular Forces, Liquids, and Solids Understand the nature of phase diagrams a. Identify the different points (triple point, normal boiling point, freezing point ) and regions (solid, liquid, vapor) of a phase diagram, and use the diagram to evaluate the vapor pressure of a liquid and the relative densities of a liquid and a solid (Section 13.10). General ChemistryNow homework: SQ(s) 34 Key Equation Equation 13.2 (page 612) The Clausius-Clapeyron equation relates the equilibrium vapor pressure, P, of a volatile liquid to the molar enthalpy of vaporization (¢H °ap) at a given temperav ture, T. (R is the universal constant, 8.314472 J/K • mol.) Equation 13.2 allows you to calculate ¢H °ap if you know the vapor pressures at two different temperatures. v Alternatively, you may plot ln P versus 1/T; the slope of the line is ¢H °ap/R. v ln P2 P1 ¢ H° vap R c 1 T1 1 d T2 Study Questions ▲ denotes more challenging questions. ■ denotes questions available in the Homework and Goals section of the General ChemistryNow CD-ROM or website. Blue numbered questions have answers in Appendix O and fully worked solutions in the Student Solutions Manual. Structures of many of the compounds used in these questions are found on the General ChemistryNow CD-ROM or website in the Models folder. Assess your understanding of this chapter’s topics with additional quizzing and conceptual questions at 3. What type of intermolecular force must be overcome in converting each of the following from a liquid to a gas? (a) liquid O2 (c) CH3I (methyl iodide) (b) mercury (d) CH3CH2OH (ethanol ) 4. ■ What type of intermolecular forces must be overcome in converting each of the following from a liquid to a gas? (a) CO2 (c) CHCl3 (b) NH3 (d) CCl4 5. Rank the following atoms or molecules in order of increasing strength of intermolecular forces in the pure substance. Which exist as gases at 25 °C and 1 atm? (a) Ne (c) CO (b) CH4 (d) CCl4 6. ■ Rank the following in order of increasing strength of intermolecular forces in the pure substances. Which exist as gases at 25 °C and 1 atm? (a) CH3CH2CH2CH3 (butane) (b) CH3OH (methanol ) (c) He 7. Which of the following compounds would be expected to form intermolecular hydrogen bonds in the liquid state? (a) CH3OCH3 (dimethyl ether) (b) CH4 (c) HF (d) CH3CO2H (acetic acid) (e) Br2 (f ) CH3OH (methanol ) Practicing Skills Intermolecular Forces (See Examples 13.2–13.4 and General ChemistryNow Screens 13.3–13.7.) 1. What intermolecular force(s) must be overcome to (a) Melt ice? (b) Sublime solid I2? (c) Convert liquid NH3 to NH3 vapor? 2. What type of forces must be overcome within the solid I2 when I2 dissolves in methanol, CH3OH? What type of forces must be disrupted between CH3OH molecules when I2 dissolves? What type of forces exist between I2 and CH3OH molecules in solution? ▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O Study Questions 635 8. ■ Which of the following compounds would be expected to form intermolecular hydrogen bonds in the liquid state? (a) H2Se (b) HCO2H (formic acid) (c) HI (d) acetone O H3C C CH3 16. Refer to Figure 13.18 as an aid in answering these questions: (a) You put some water at 60 °C in a plastic milk carton and seal the top very tightly so that gas cannot enter or leave the carton. What happens when the water cools? (b) If you put a few drops of liquid diethyl ether on your hand, does it evaporate completely or remain a liquid? 17. ■ Which member of each of the following pairs of compounds has the higher boiling point? (a) O2 or N2 (c) HF or HI (b) SO2 or CO2 (d) SiH4 or GeH4 18. Place the following four compounds in order of increasing boiling point. (a) SCl2 (b) NH3 (c) CH4 (d) CO 19. ■ Vapor pressure curves for CS2 (carbon disulfide) and CH3NO2 (nitromethane) are drawn here. (a) What are the approximate vapor pressures of CS2 and CH3NO2 at 40 °C? (b) What types of intermolecular forces exist in the liquid phase of each compound? (c) What is the normal boiling point of CS2? Of CH3NO2? (d) At what temperature does CS2 have a vapor pressure of 600 mm Hg? (e) At what temperature does CH3NO2 have a vapor pressure of 60 mm Hg? 800 Vapor pressure (mm Hg) 700 600 500 400 300 200 100 0 30 10 10 50 70 30 Temperature (°C) 90 110 CS2 CH3NO2 9. In each pair of ionic compounds, which is more likely to have the greater heat of hydration? Briefly explain your reasoning in each case. (a) LiCl or CsCl (c) RbCl or NiCl2 (b) NaNO3 or Mg(NO3)2 10. When salts of Mg2 , Na , and Cs are placed in water, the positive ion is hydrated (as is the negative ion). Which of these three cations is most strongly hydrated? Which one is least strongly hydrated? Liquids (See Examples 13.5 and 13.6 and General ChemistryNow Screens 13.8–13.11.) 11. Ethanol, CH3CH2OH, has a vapor pressure of 59 mm Hg at 25 °C. What quantity of heat energy is required to evaporate 125 mL of the alcohol at 25 °C? The enthalpy of vaporization of the alcohol at 25 °C is 42.32 kJ/mol. The density of the liquid is 0.7849 g/mL. 12. ■ The enthalpy of vaporization of liquid mercury is 59.11 kJ/mol. What quantity of heat is required to vaporize 0.500 mL of mercury at 357 °C, its normal boiling point? The density of mercury is 13.6 g/mL. 13. Answer the following questions using Figure 13.18. (a) What is the approximate equilibrium vapor pressure of water at 60 °C? Compare your answer with the data in Appendix G. (b) At what temperature does water have an equilibrium vapor pressure of 600 mm Hg? (c) Compare the equilibrium vapor pressures of water and ethanol at 70 °C. Which is higher? 14. ■ Answer the following questions using Figure 13.18. (a) What is the equilibrium vapor pressure of diethyl ether at room temperature (approximately 20 °C)? (b) Place the three compounds in Figure 13.18 in order of increasing intermolecular forces. (c) If the pressure in a flask is 400 mm Hg and the temperature is 40 °C, which of the three compounds (diethyl ether, ethanol, and water) are liquids and which are gases? 15. Assume you seal 1.0 g of diethyl ether (see Figure 13.18) in an evacuated 100.-mL flask. If the flask is held at 30 °C, what is the approximate gas pressure in the flask? If the flask is placed in an ice bath, does additional liquid ether evaporate or does some ether condense to a liquid? 20. ■ Answer each of the following questions with increases, decreases, or does not change. (a) If the intermolecular forces in a liquid increase, the normal boiling point of the liquid ______. (b) If the intermolecular forces in a liquid decrease, the vapor pressure of the liquid ______. (c) If the surface area of a liquid decreases, the vapor pressure ______. (d) If the temperature of a liquid increases, the equilibrium vapor pressure ______. 21. The following table gives the equilibrium vapor pressure of benzene, C6H6, at various temperatures. Temperature (°C) 7.6 26.1 60.6 80.1 Vapor Pressure (mm Hg) 40. 100. 400. 760. Blue-numbered questions answered in Appendix O ▲ More challenging ■ In General ChemistryNow 636 Chapter 13 Intermolecular Forces, Liquids, and Solids (a) What is the normal boiling point of benzene? (b) Plot these data so that you have plot resembling the one in Figure 13.19. At what temperature does the liquid have an equilibrium vapor pressure of 250 mm Hg? At what temperature is this pressure 650 mm Hg? (c) Calculate the molar enthalpy of vaporization for benzene using the the Clausius–Clapeyron equation (Equation 13.1, page 612). 22. ■ Vapor pressure data are given here for octane, C8H18. Temperature (°C) 25 50. 75 100. Vapor Pressure (mm Hg) 13.6 45.3 127.2 310.8 Ca2+ Ti 4+ O2– 26. ■ Rutile, TiO2, crystallizes in a structure characteristic of many other ionic compounds. How many formula units of TiO2 are in the unit cell illustrated here? (The oxide ions marked by an x are wholly within the cell; the others are in the cell faces.) Ti4 Use the Clausius-Clapeyron equation (Equation 13.1, page 612) to calculate the molar enthalpy of vaporization of octane and its normal boiling point. x Metallic and Ionic Solids (See Example 13.7, Exercise 13.10, and the General ChemistryNow Screens 13.12–13.13.) 23. Outline a two-dimensional unit cell for the pattern shown here. If the black squares are labeled A and the white squares are B, what is the simplest formula for a “compound” based on this pattern? x O2 27. Cuprite is a semiconductor. Oxide ions are at the cube corners and in the cube center. Copper ions are wholly within the unit cell. (a) What is the formula of cuprite? (b) What is the oxidation number of copper? Copper O2 24. Outline a two-dimensional unit cell for the pattern shown here. If the black squares are labeled A and the white squares are B, what is the simplest formula for a “compound” based on this pattern? 28. ■ The mineral fluorite, which is composed of calcium ions and fluoride ions, has the unit cell shown here. (a) What type of unit cell is described by the Ca2 ions? (b) Where are the F ions located, in octahedral holes or tetrahedral holes? (c) Based on this unit cell, what is the formula of fluorite? Ca2 25. One way of viewing the unit cell of perovskite was illustrated in Example 13.7. Another way is shown here. Prove that this view also leads to a formula of CaTiO3. F ▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O Study Questions 637 Normal MP Normal BP Other Types of Solids (See the General ChemistryNow Screens 13.14–13.16.) 29. A diamond unit cell is shown here. (a) How many carbon atoms are in one unit cell? (b) The unit cell can be considered as a cubic unit cell of C atoms with other C atoms in holes in the lattice. What type of unit cell is this (sc, bcc, fcc)? In what holes are other C atoms located, octahedral or tetrahedral holes? Pressure (atm) 1.0 Solid 0.5 0.37 atm Triple point 0 125 121° 120 115 Temperature (°C) 112° 110 108° 105 Liquid Gas 30. The structure of graphite is given in Figure 2.15. (a) What type of intermolecular bonding forces exist between the layers of six-member carbon rings? (b) Account for the lubricating ability of graphite. That is, why does graphite feel slippery? Why does pencil lead (which is really graphite in clay) leave black marks on paper? Physical Properties of Solids 31. Benzene, C6H6, is an organic liquid that freezes at 5.5 °C (see Figure 13.1) to form beautiful, feather-like crystals. How much heat is evolved when 15.5 g of benzene freezes at 5.5 °C? (The heat of fusion of benzene is 9.95 kJ/mol.) If the 15.5-g sample is remelted, again at 5.5 °C, what quantity of heat is required to convert it to a liquid? 32. The specific heat capacity of silver is 0.235 J/g K. Its melting point is 962 °C, and its heat of fusion is 11.3 kJ/mol. What quantity of heat, in joules, is required to change 5.00 g of silver from a solid at 25 °C to a liquid at 962 °C? Phase Diagrams and Phase Changes (See the General ChemistryNow Screen 13.17.) 33. Consider the phase diagram of CO2 in Figure 13.40. (a) Is the density of liquid CO2 greater or less than that of solid CO2? (b) In what phase do you find CO2 at 5 atm and 0 °C? (c) Can CO2 be liquefied at 45 °C? 34. ■ Use the phase diagram given here to answer the following questions: (a) In what phase is the substance found at room temperature and 1.0 atm pressure? (b) If the pressure exerted on a sample is 0.75 atm and the temperature is 114 °C, in what phase does the substance exist? (c) If you measure the vapor pressure of a liquid sample and find it to be 380 mm Hg, what is the temperature of the liquid phase? (d) What is the vapor pressure of the solid at 122 °C? (e) Which is the denser phase—solid or liquid? Explain briefly. 35. Liquid ammonia, NH3(/), was once used in home refrigerators as the heat transfer fluid. The specific heat of the liquid is 4.7 J/g K and that of the vapor is 2.2 J/g K. The enthalpy of vaporization is 23.33 kJ/mol at the boiling point. If you heat 12 kg of liquid ammonia from 50.0 °C to its boiling point of 33.3 °C, allow it to evaporate, and then continue warming to 0.0 °C, how much heat energy must you supply? 36. If your air conditioner is more than several years old, it may use the chlorofluorocarbon CCl2F2 as the heat transfer fluid. The normal boiling point of CCl2F2 is 29.8 °C, and the enthalpy of vaporization is 20.11 kJ/mol. The gas and the liquid have specific heats of 117.2 J/mol K and 72.3 J/mol K, respectively. How much heat is evolved when 20.0 g of CCl2F2 is cooled from 40 °C to 40 °C? 37. The critical temperature and pressure of chloromethane are 416 K and 66.1 atm, respectively. (Chloromethane’s triple point is at 175.4 K and 0.0086 atm.) Can CH3Cl be liquefied at or above room temperature? Explain briefly. 38. Methane (CH4) cannot be liquefied at room temperature, no matter how high the pressure. Propane (C3H8), another alkane, has a critical pressure of 41.8 atm and a critical temperature of 369.9 K. (The triple point for propane is at 85 K and 1.7 10 9 atm.) Can propane be liquefied at room temperature? General Questions These questions are not designated as to type or location in the chapter. They may combine several concepts. 39. Rank the following substances in order of increasing strength of intermolecular forces: Ar, CH3OH, CO2. ▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O 638 Chapter 13 Intermolecular Forces, Liquids, and Solids 40. What types of intermolecular forces are important in the liquid phase of (a) C2H6 and (b) (CH3)2CHOH? 41. Construct a phase diagram for O2 from the following information: normal boiling point, 90.18 K; normal melting point, 54.8 K; and triple point, 54.34 K at a pressure of 2 mm Hg. Very roughly estimate the vapor pressure of liquid O2 at 196 °C, the lowest temperature easily reached in the laboratory. Is the density of liquid O2 greater or less than that of solid O2? 42. ▲ A unit cell of cesium chloride is shown on page 622. The density of the solid is 3.99 g/cm3, and the radius of the Cl ion is 181 pm. What is the radius of the Cs ion in the center of the cell? (Assume that the Cs ion touches all of the corner Cl ions.) 43. ■ If you place 1.0 L of ethanol (C2H5OH) in a room that is 3.0 m long, 2.5 m wide, and 2.5 m high, will all of the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethanol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785 g/cm3. 44. Select the substance in each of the following pairs that should have the higher boiling point. (a) Br2 or ICl (b) neon or krypton (c) CH3CH2OH (ethanol ) or C2H4O (ethylene oxide, structure below) H2C O 45. Which salt, Li2SO4 or Cs2SO4, is expected to have the more exothermic enthalpy of hydration? 46. In which salts does the cation bind most strongly to water molecules? In which is the binding less strong in comparison? Explain your reasoning. (a) Fe(NO3)3 (c) NaCl (b) CoCl2 (d) Al(NO3)3 47. ■ Use the vapor pressure curves illustrated here to answer the questions that follow. 900 800 Vapor pressure (mm Hg) (b) Considering only carbon disulfide (CS2) and ethanol, which has the stronger intermolecular forces in the liquid state? (c) At what temperature does heptane (C7H16) have a vapor pressure of 500 mm Hg? (d) What are the approximate normal boiling points of each of the three substances? (e) At a pressure of 400 mm Hg and a temperature of 70 °C, is each substance a liquid, a gas, or a mixture of liquid and gas? 48. What quantity of energy, in joules, is evolved when 1.00 mol of liquid ammonia cools from 33.3 °C (its boiling point ) to 43.3 °C? (The specific heat capacity of liquid NH3 is 4.70 J/g K.) Compare this value with the quantity of heat evolved by 1.00 mol of liquid water cooling by exactly 10 °C. Which evolves more heat on cooling 10 °C, liquid water or liquid ammonia? Using intermolecular forces, explain briefly why one liquid should evolve more energy than the other. 49. ▲ Silver crystallizes in a face-centered cubic unit cell. Each side of the unit cell has a length of 409 pm. What is the radius of a silver atom? (Hint: Assume the atoms just touch each other on the diagonal across the face of the unit cell. That is, each face atom is touching the four corner atoms.) 50. ▲ Tungsten crystallizes in the unit cell shown here. CH2 316.5 pm (a) What type of unit cell is this? (b) How many tungsten atoms occur per unit cell? (c) If the edge of the unit cell is 316.5 pm, what is the radius of a tungsten atom? (Hint: The W atoms touch each other along the diagonal line from one corner of the unit cell to the opposite corner of the unit cell.) 51. ▲ The unit cell shown here is for calcium carbide. How many calcium atoms and how many carbon atoms are in each unit cell? What is the formula of calcium carbide? (Calcium ions are silver in color and carbon atoms are gray.) 700 600 500 400 300 200 100 0 10 Carbon disulfide Ethanol Heptane 20 30 40 50 60 70 Temperature (°C) 80 90 100 110 (a) What is the vapor pressure of ethanol (C2H5OH) at 60°C? ▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O Study Questions 639 52. Rank the following molecules in order of increasing intermolecular forces: CH3Cl, HCO2H (formic acid), and CO2. 53. ■ Rank the following compounds in order of increasing molar enthalpy of vaporization: CH3OH, C2H6, HCl. 54. ■ Calcium metal crystallizes in a face-centered cubic unit cell. The density of the solid is 1.54 g/cm3. What is the radius of a calcium atom? Pictured here is a face-centered cubic lattice of atoms. Measuring the distance shown allows a scientist to estimate the radius of an atom A B 62. ▲ Assuming that in a simple cubic unit cell the spherical atoms or ions just touch along the cube’s edges, calculate the percentage of empty space within the unit cell. (Recall that the volume of a sphere is (4/3)pr 3, where r is the radius of the sphere.) 63. Equilibrium vapor pressures of dichlorodimethylsilane, SiCl2(CH3)2, are given below. (The compound is a starting material to making silicone polymers.) A distance equivalent to 4 times the radius of an atom Temperature (°C) 0.4 17.5 51.9 70.3 Vapor Pressure (mm Hg) 40. 100. 400. 760. 55. ▲ The very dense metal iridium has a face-centered cubic unit cell and a density of 22.56 g/cm3. Use this information to calculate the radius of an atom of the element. 56. ▲ The density of copper metal is 8.95 g/cm3. If the radius of a copper atom is 127.8 pm, is the copper unit cell simple cubic, body-centered cubic, or face-centered cubic? 57. ▲ Vanadium metal has a density of 6.11 g/cm . Assuming the vanadium atomic radius is 132 pm, is the vanadium unit cell simple cubic, body-centered cubic, or facecentered cubic? 58. ▲ Iron has a body-centered cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g/cm3. Use this information to calculate Avogadro’s number. 59. ▲ Calcium fluoride is the well-known mineral fluorite. It is known that each unit cell contains four Ca2 ions and eight F ions and that the Ca2 ions are arranged in a fcc lattice. The F ions fill all the tetrahedral holes in a facecentered cubic lattice of Ca2 ions. The edge of the CaF2 unit cell is 5.46295 10 8 cm in length. The density of the solid is 3.1805 g/cm3. Use this information to calculate Avogadro’s number. 60. Mercury and many of its compounds are dangerous poisons if breathed, swallowed, or even absorbed through the skin. The liquid metal has a vapor pressure of 0.00169 mm Hg at 24 °C. If the air in a small room is saturated with mercury vapor, how many atoms of mercury vapor occur per cubic meter? 61. ▲ You can get some idea of how efficiently spherical atoms or ions are packed in a three-dimensional solid by seeing how well circular atoms pack in two dimensions. Using the drawings shown here, prove that B is a more efficient way to pack circular atoms than A. A unit cell of A contains portions of four circles and one hole. In B, packing coverage can be calculated by looking at a triangle that contains portions of three circles and one hole. Show that A fills about 80% of the available space, whereas B fills closer to 90% of the available space. ▲ More challenging 3 (a) What is the normal boiling point of dichlorodimethylsilane? (b) Plot these data as ln P versus 1/T so that you have a plot resembling the one in Figure 13.19. At what temperature does the liquid have an equilibrium vapor pressure of 250 mm Hg? At what temperature is this pressure 650 mm Hg? (c) Calculate the molar enthalpy of vaporization for dichlorodimethylsilane using the the ClausiusClapeyron equation (Equation 13.1). 64. ▲ The following data are the equilibrium vapor pressures of limonene, C10H16, at various temperatures. (Limonene is used as a scent in commercial products.) Temperature (°C) 14.0 53.8 84.3 108.3 151.4 Vapor Pressure (mm Hg) 1.0 10. 40. 100. 400. (a) Plot these data as ln P versus 1/T so that you have a plot resembling the one in Figure 13.19. (b) At what temperature does the liquid have an equilibrium vapor pressure of 250 mm Hg? At what temperature is this pressure 650 mm Hg? (c) What is the normal boiling point of limonene? (d) Calculate the molar enthalpy of vaporization for limonene using the the Clausius-Clapeyron equation (Equation 13.1). ■ In General ChemistryNow Blue-numbered questions answered in Appendix O 640 Chapter 13 Intermolecular Forces, Liquids, and Solids Summary and Conceptual Questions The following questions may use concepts from the preceding chapters. 65. Acetone, CH3COCH3, is a common laboratory solvent. It is usually contaminated with water, however. Why does acetone absorb water so readily? Draw molecular structures showing how water and acetone can interact. What intermolecular force(s) is (are) involved in the interaction? 66. Cooking oil is not miscible with water. From this observation, what conclusions can you draw regarding the polarity or hydrogen-bonding ability of molecules found in cooking oil? 67. Liquid ethylene glycol, HOCH2CH2OH, is one of the main ingredients in commercial antifreeze. Do you predict its viscosity to be greater or less than that of ethanol, CH3CH2OH? 68. Liquid methanol, CH3OH, is placed in a glass tube. Predict whether the meniscus of the liquid is concave or convex. 69. Account for these facts: (a) Although ethanol (C2H5OH) (bp, 80 °C) has a higher molar mass than water (bp, 100 °C), the alcohol has a lower boiling point. (b) Mixing 50 mL of ethanol with 50 mL of water produces a solution with a volume slightly less than 100 mL. 70. ▲ Why is it not possible for a salt with the formula M3X (Na3PO4, for example) to have a face-centered cubic lattice of X anions with M cations in octahedral holes? 71. Can CaCl2 have a unit cell like that of sodium chloride? Explain. 72. Rationalize the observation that CH3CH2CH2OH, 1-propanol, has a boiling point of 97.2 °C, whereas a compound with the same empirical formula, methyl ethyl ether (CH3CH2OCH3) boils at 7.4 °C. 73. Cite two pieces of evidence to support the statement that water molecules in the liquid state exert considerable attractive force on one another. 74. During thunderstorms in the Midwest, very large hailstones can fall from the sky. (Some are the size of golf balls!) To preserve some of these stones, we put them in the freezer compartment of a frost-free refrigerator. Our friend, who is a chemistry student, tells us to use an older model that is not frost-free. Why? 75. Refer to Figure 13.13 to answer the following questions. (a) Of the three hydrogen halides (HX), which has the largest total intermolecular force? (b) Why are the dispersion forces greater for HI than for HCl? (c) Why are the dipole–dipole forces greater for HCl than for HI? (d) Of the seven molecules in Figure 13.13, which involves the largest dispersion forces? Explain why this is reasonable. 76. A “hand boiler” can be purchased in toy stores or at science supply companies. If you cup your hand around the ▲ More challenging ■ In General ChemistryNow bottom bulb, the volatile liquid in the boiler boils and the liquid moves to the upper chamber. Using your knowledge of kinetic-molecular theory and intermolecular forces, explain how the hand boiler works. 77. ▲ The photos illustrate an experiment you can do yourself. Place 10 mL of water in an empty soda can and heat the water to boiling. Using tongs or pliers, quickly turn the can over in a pan of cold water, making sure the opening in the can is below the water level in the pan. (a) Describe what happens and explain it in terms of the subject of this chapter. Charles D. Winters (a) (b) (b) Prepare a molecular-level sketch of the situation inside the can before heating and after heating (but prior to inverting the can). 78. A fluorocarbon, CF4, has a critical temperature of 45.7 °C, a critical pressure of 37 atm, and a normal boiling point of 128 °C. Are there any conditions under which this compound can be a liquid at room temperature? Explain briefly. 79. ■ ▲ The following figure is a plot of vapor pressure versus temperature for dichlorodifluoromethane, CCl2F2. The heat of vaporization of the liquid is 165 kJ/g, and the specific heat capacity of the liquid is about 1.0 J/g K . Blue-numbered questions answered in Appendix O Charles D. Winters Study Questions 641 8 7 Vapor pressure (atm) 6 5 4 3 2 1 40 30 20 10 0 10 20 30 Using Molecular Models to Explore Intermolecular Forces and the Solid State On any screen of the General ChemistryNow CD-ROM or website, click the Molecular Models menu item. For each of the following questions, locate the required model in the indicated folder. 82. ▲ Examine the model of the structure of CaO (Inorganic folder, Ionic Solids). (a) Describe the structure. What type of lattice is described by the Ca2 ions? What type of hole do the O2 ions occupy in the lattice of Ca2 ions? (b) How is the formula of CaO related to its unit cell structure? (How many Ca2 ions and how many O2 ions are in one unit cell?) (c) How is the CaO structure related to the NaCl structure? Temperature (°C) (a) What is the normal boiling point of CCl2F2? (b) A steel cylinder containing 25 kg of CCl2F2 in the form of liquid and vapor is set outdoors on a warm day (25 °C). What is the approximate pressure of the vapor in the cylinder? (c) The cylinder valve is opened, and CCl2F2 vapor gushes out of the cylinder in a rapid flow. Soon, however, the flow becomes much slower, and the outside of the cylinder is coated with ice frost. When the valve is closed and the cylinder is reweighed, it is found that 20 kg of CCl2F2 is still in the cylinder. Why is the flow fast at first? Why does it slow down long before the cylinder is empty? Why does the outside become icy? (d) Which of the following procedures would be effective in emptying the cylinder rapidly (and safely)? (1) Turn the cylinder upside down and open the valve. (2) Cool the cylinder to 78 °C in dry ice and open the valve. (3) Knock off the top of the cylinder, valve and all, with a sledge hammer. 80. ▲ Two identical swimming pools are filled with uniform spheres of ice packed as closely as possible. The spheres in the first pool are the size of grains of sand; those in the second pool are the size of oranges. The ice in both pools melts. In which pool, if either, will the water level be higher? (Ignore any differences in filling space at the planes next to the walls and bottom.) 81. Figure 13.41 is a series of photos of CO2 as it changes from a mixture of liquid and vapor at equilibrium to the supercritical fluid. Draw a representation of the situation at the molecular level of the liquid–vapor equilibrium in the photo at the left and of the supercritical fluid at the right. 83. ■ Examine the structure of ZnS using the model in Inorganic: Ionic Solids. (a) Describe the structure. What type of lattice is described by the Zn2 ions (silver color)? What type of holes do the S2 ions (yellow color) occupy in the lattice of Zn2 ions? (b) How is the formula of ZnS related to its unit cell structure? 84. Lead sulfide, PbS (commonly called galena), has the same formula as ZnS. Does it have the same solid structure? If different, how it is different? How is its unit cell related to its formula? (The model is found in Inorganic: Ionic Solids.) 85. ■ Examine the structure of CaF2 using the model in Inorganic: Ionic Solids. (a) Describe the structure. What type of lattice is described by the Ca2 ions? What type of holes do the F ions occupy in the lattice of Ca2 ions? (b) How is the formula of CaF2 related to its unit cell structure? (How many Ca2 ions and how many F ions are in one unit cell?) (c) How is the CaF2 structure related to the ZnS structure? 86. Acetaminophen is used in analgesics. (A model is in Organic: Alcohols.) (a) Draw the structure of acetaminophen. (b) Is the molecule capable of hydrogen bonding? If so, what are the sites of hydrogen bonding? 87. ■ Aspartame is an artificial sweetener. (A model is in Organic: Amines.) (a) Draw the structure of aspartame. (b) Is the molecule capable of hydrogen bonding? If so, what are the sites of hydrogen bonding? Do you need a live tutor for homework problems? Access vMentor at General ChemistryNow at for one-on-one tutoring from a chemistry expert ▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O ...
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