Chapter 15: Liquids and Solids PDF

15. Liquids and Solids 15-1 Molecules in Liquids and Solids 15-8 Phase Diagrams 15-2 Heating Curves 15-9 Crystal Structures 15-3 Fusion and Vaporization 15-10 Crystal Forces 15-4 Intermolecular Forces...

15. Liquids and Solids 15-1 Molecules in Liquids and Solids 15-8 Phase Diagrams 15-2 Heating Curves 15-9 Crystal Structures 15-3 Fusion and Vaporization 15-10 Crystal Forces 15-4 Intermolecular Forces 15-11 Electrons in Metals 15-5 Properties of Liquids 15-12 Liquid Crystals 15-6 Vapor Pressure 15-13 Colloids 15-7 Relative Humidity Y our study of chemistry has taken you from a fundamental understanding of how atoms combine to form compounds to calculations involving chemi- cal equations. Now we begin to build on our picture of molecular structure in order to predict chemical changes and the properties of matter. In this chapter, we discuss how the behavior of liquids and solids depends on the attraction between molecules. If molecules did not attract one another, then gases would not condense into liquids. Without an orderly arrangement of atoms in a solid, a diamond would not retain its hardness and the “lead” in pencils its softness. In this chap- ter we shall discuss these and other properties of liquids and solids in terms of intermolecular forces. We shall examine the processes of melting, freezing, and vaporization and then introduce phase diagrams that give the temperatures and pressures at which a substance occurs as a solid, a liquid, or a gas. We then discuss how various types of crystal structures are described and how these con- trast with other solids. Finally, we briefly discuss liquid crystals and colloids. 15-1. The Molecules in Solids and Liquids Are Close to One Another We learned in Chapter 13 that the molecules in a gas are in constant motion, traveling distances more than a hundred molecular diameters between colli- sions at one bar. In addition to traveling in straight lines (translational motion), the molecules rotate and vibrate freely as they travel between collisions (Figure 15.1). Because the average distance between the molecules is so large, gases have low densities and can be readily compressed. In addition, the volume of a gas is always the same as the volume of its container. By contrast, the molecules of a solid are close together and restricted to fixed positions in space. The molecules vibrate about these fixed positions, but they usually do not rotate or move easily from site to site. The molecules in a liquid are also close together, but they are not restricted to fixed positions; they rotate and move throughout the fluid. The molecules in a liquid are es- sentially in constant interaction with one another, and the distance that a mol- ecule travels between collisions is usually less than a molecular diameter. The 519 520 15. LiquiDS AnD SOLiDS TAbLe 15.1 Characteristics of solids, liquids, and gases Phase Translation Rotation Vibration Average distance between particles solid none none about fixed less than one or hindered positions in space molecular diameter liquid hindered hindered free less than one molecular diameter gas free free free about 10 molecular diameters at one atm translational and rotational motions of molecules in a liquid are hindered by frequent collisions between the molecules. The molecular distinctions be- tween a solid, a liquid, and a gas are summarized in Table 15.1. Because the particles in solids are fixed in space and are not able to move from site to site, a solid body has a fixed shape, and the volume and shape of a solid are independent of its container. Although the molecules in a liquid are essentially in continual contact with one another, they are able to move past one another and move throughout the liquid. Thus, unlike a solid body, a liquid can be poured from one container to another. Furthermore, a liquid has a fixed volume because the molecules are constantly attracting one another. But, un- like a solid, a liquid assumes the shape of its container because its constituent molecules are not held in fixed positions in space. Liquids and solids have densities that are about a thousand times greater than the densities of gases at one atm. The densities of gases are often ex- pressed in units of grams per liter, whereas the densities of liquids and solids are expressed in units of grams per milliliter or grams per cubic centimeter. The relative densities of gases, liquids, and solids imply that the distances between molecules in liquids and solids are much smaller than they are in gases. The molar volume of a substance is a direct indication of the average sepa- ration between the molecules. The molar volume of the liquid phase of a com- pound is approximately equal to the molar volume of the solid phase of the same compound. Yet the molar volume of a gas is much larger than that of a liquid or a solid at the same pressure and temperature. The molar volumes for solid, liquid, and gaseous water at 0°C and 100.0°C at one atm are shown in Table 15.2. Note that the molar volume of water increases by a factor of 1630 Figure 15.1 Molecular motion may be translational, rotational, or vibrational. (a) Translation is movement of the entire molecule through space. (b) Rotation is the spinning of a molecule in space. (c) Vibration is the back-and-forth movement of nuclei about their equilibrium positions. (a) (b) (c) 15-2. THe PrOCeSSeS OF MeLTing AnD bOiLing On A HeATing Curve 521 TAbLe 15.2 Molar volume of solid, liquid, and gaseous water at 0°C and 100°C and one atm (1.0135 bar) Phase Molar volume at Molar volume at of H 2 O 0°C/mL?mol –1 100°C/mL?mol –1 solid 19.8 — liquid 18.0 18.8 gas — 30 600 (30 600 mL/18.8 mL) upon vaporization at 100.0°C. The average separation between the molecules on going from the liquid to the gas phase at one atm increases over 10-fold. Title General Chemistry - 4th ed Author McQuarrie/Gallogy 15-2. The Processes of Melting and boiling Appear as Horizontal Artist George Kelvin Lines on a Heating Curve Figure # 015-003 Let’s consider an experiment in which a pure substance(476) is converted from a 08/18/09 solid to a liquid to a gas by the application of heat at a constant rate at one Date atm. Figure 15.2 shows howCheck the temperatureApproved if revision of the substance—water, in this case—varies with time when it is heated at a constant rate. Such a plot is called a heating curve. Initially, the water is at a temperature of –10°C, so it is in the form of ice. As energy in the form of heat is added, the temperature of the ice 125 Heat required to vaporize 1 mole of liquid water H2O(g) ΔHvap 100 H2O(l) + H2O(g) 75 t/°C 50 H2O(l) Figure 15.2 The heating curve for one mole of water at one atm starting with ice at –10°C. The energy is added as heat at a constant rate of 100 J?min–1. 25 Heat required to The most noteworthy features of ΔHfus melt 1 mole of ice the heating curve are the horizontal 10 portions, which represent the molar H2O(s) + H2O(l) enthalpy of fusion, Hfus, and the molar 0 H2O(s) enthalpy of vaporization, Hvap. You –10 can see that the molar enthalpy of 0 100 200 300 400 500 600 vaporization is much larger than the Time/min molar enthalpy of fusion. 522 15. LiquiDS AnD SOLiDS increases until 0°C, the melting point of ice, is reached. At 0°C, the temperature remains constant for 60 min even though the ice is being heated at a constant rate of 100 J?min–1. The energy as heat being added at 0°C melts the ice to liquid water. The temperature of the ice-water mixture remains at 0°C until all the ice is melted. The energy absorbed as heat that is required to melt one mole of any substance is called the molar enthalpy of fusion and is denoted by DHfus. The experimental data plotted in Figure 15.2 show that 60.1 minutes are required to melt one mole of ice when heat is added at the rate of 100.0 J?min–1; so the molar enthalpy of fusion for ice is given by DHfus = (60.1 min?mol–1)(100.0 J?min–1) = 6010 J?mol–1 = 6.01 kJ?mol–1 After all the ice is melted, and not until then, the temperature increases from 0°C. The temperature continues to increase until 100.0°C, the boiling point of water, is reached. At 100.0°C, the temperature remains constant for 406.5 minutes even though heat is being added at a rate of 100.0 J?min–1. The energy being absorbed at 100.0°C vaporizes the liquid water to water vapor. The temperature of the liquid-vapor mixture remains at 100.0°C until all the water is vaporized. The energy absorbed as heat that is required to vaporize one mole of any substance is called the molar enthalpy of vaporization and is denoted by DHvap. According to Figure 15.2, the value of DHvap for water is DHvap = (406.5 min?mol–1)(100.0 J?min–1) = 40 650 J?mol–1 = 40.65 kJ?mol–1 Once all the water is vaporized, the temperature increases from 100.0°C, as shown in Figure 15.2. The horizontal portions of the heating curve of a pure substance repre- sent the enthalpy of fusion and the enthalpy of vaporization. The other regions represent pure phases being heated at a constant rate. We learned in Chapter 14 that it requires energy to raise the temperature of a substance. The heat absorbed in raising the temperature of a substance from T1 to T2 at constant pressure without a change in phase is given by q P = nC P(T2 – T1) = nC PDT (15.1) (see Equation 14.30), where C P is the molar heat capacity at constant pressure and n is the number of moles. Recall that the heat capacity is the measure of the ability of a substance to take up energy as heat. Different rates of tempera- ture increase for H2O(s), H2O(l ), and H2O(g) appear in Figure 15.2 (steep seg- ments of the heating curve) because C P[H2O(l )] > C P[H2O(s)] > C P[H2O(g)] 15-3. energy is required to Melt a Solid and to vaporize a Liquid The melting, or fusion, of one mole of ice can be represented by H2O(s) → H2O(l ) DHfus = 6.01 kJ?mol–1 15-3. energy iS requireD TO MeLT A SOLiD AnD TO vAPOrize A LiquiD 523 or, in general, we write X(s) → X(l ) DHfus where X stands for any element or compound. The energy as heat that is re- quired to melt n moles of a substance is given by q fus = n DHfus (15.2) The enthalpy of fusion is necessarily positive because it requires energy to break up the crystal lattice. Recall that a positive value of DH means that energy as heat is absorbed in a process. The vaporization of one mole of water can be represented by H2O(l ) → H2O(g) DHvap = 40.65 kJ?mol–1–1 or, in general, X(l ) → X(g) DHvap The energy that is absorbed as heat and is required to vaporize n moles of a substance is given by q vap = n DHvap (15.3) The value of DHvap is always positive because energy is required to separate the molecules in a liquid from one another. Gas phase molecules are so far apart from one another that they interact only very weakly relative to liquid phase molecules. Essentially, all the energy put in as enthalpy of vaporization is re- quired to separate the molecules of the liquid from one another. Because the temperature does not change during vaporization, the average kinetic energy of the molecules does not change. Table 15.3 gives the molar enthalpies of fu- sion and vaporization of several substances. eXAMPLe 15-1: How long would it take to convert 50.0 grams of ice at 0°C to water at 100.0°C if the heating rate is 250.0 J?min–1? The molar heat capacity of H2O(l ) is 75.3 J?mol–1?K–1. Solution: A mass of 50.0 grams of water corresponds to 2.77 moles. First we calculate how long it takes to melt the ice at 0°C. The value of DHfus for ice given in Table 15.3 is 6.01 kJ?mol–1. Then we calculate how long it takes to heat the water from 0°C to 100.0°C. The energy required to melt 2.77 moles of ice is given by Equation 15.2: q fus = n DHfus = (2.77 mol) (6.01 kJ?mol–1) = 16.6 kJ The time it would take to add this much energy is given by q fus divided by the rate of heating, or 524 15. LiquiDS AnD SOLiDS TAbLe 15.3 normal melting points, boiling points, and molar enthalpies of vaporization and fusion at these temperatures and one atm (1.01325 bar) pressure Compound Chemical Melting Boiling Hfus Hvap formula point/K point/K /kJ?mol –1 /kJ?mol –1 ammonia NH3 195.42 239.82 5.66 23.33 argon Ar 83.79 87.30 1.18 6.43 bromine Br 2 266.0 332.0 10.57 29.96 carbon dioxide CO2 216.59 (195) 9.02 (25.2) sublimes sublimes chlorine Cl2 171.7 239.11 6.40 20.41 chloromethane CH3 Cl 175.5 249.06 6.43 21.40 dichloromethane CH2 Cl2 176.0 313 4.60 28.06 helium He — 4.22 0.014 0.08 hydrogen H2 13.81 20.28 0.12 0.90 hydrogen bromide HBr 186.4 206.77 2.41 17.6 hydrogen chloride HCl 158.98 188 2.00 16.15 hydrogen sulfide H2S 187.7 213.60 2.38 18.67 iodine I2 386.9 457.6 15.52 41.57 krypton Kr 115.77 119.93 1.64 9.08 lithium bromide LiBr 825 1583 17.6 148.1 mercury Hg 234.32 629.88 2.29 59.11 methanal H 2 CO 181 252 — 24.5 (formaldehyde) methane CH4 90.68 111.67 0.94 8.19 neon Ne 24.54 27.07 0.328 1.71 nitrogen N2 63 77.36 0.71 5.57 oxygen O2 54.36 90.20 0.44 6.82 sulfur dioxide SO2 198 263.10 7.4 24.94 water H2O 273.15 373.15 6.01 40.65 xenon Xe 161.36 165.04 2.27 12.57 15-3. energy iS requireD TO MeLT A SOLiD AnD TO vAPOrize A LiquiD 525 16.6 kJ 16.6 × 103 J time = — = — = 66.4 min 250.0 J·min–1 250.0 J·min–1 The energy required to heat the water from 0°C to 100.0°C is (Equation 15.1) q P = nC PDT = (2.77 mol)(75.3 J?mol–1?K–1)(100.0 K) = 2.086 × 104 J The time it takes to add this much energy as heat is given by 2.086 × 104 J time = — = 83.4 min 250.0 J·min–1 The total time it would take to convert 50.0 grams of ice at 0°C to water at 100.0°C at this rate of heating is time = 66.4 min + 83.4 min = 149.8 min Pr ACTiCe PrObLeM 15-1: How long would it take to convert 100.0 grams of solid sodium at 20.0°C to sodium vapor at 1000.0°C if the heating rate at a pressure of one atm is 8.0 kJ?min–1? The melting point of sodium is 97.8°C; its boiling point is 883°C; its molar enthalpy of fusion is 2.60 kJ?mol–1; its molar enthalpy of vaporization is 97.4 kJ?mol–1; and the heat capacities of solid, liquid, and gaseous sodium are 28.2 J?mol–1?K–1, 30.8 J?mol–1?K–1, and 20.8 J?mol–1?K–1, respectively. Answer: 70 minutes eXAMPLe 15-2: Compute the energy released as heat at a pressure of one atm when 28 grams of liquid water at 18°C is converted to ice at 0°C. (An ice cube contains about 28 grams, or one ounce, of water.) The molar heat capacity of H2O(l ) is C P = 75.3 J?mol–1?K–1, and DHfus = 6.01 kJ?mol–1 for ice. Solution: The overall process must be broken down into two steps. We must first bring the H2O(l ) from 18°C to 0°C (the freezing point of water) and then consider the process H2O(l ) → H2O(s) at 0°C: step 1 step 2 28 g H2O(l) → 28 g H2O(l) → 28 g H2O(s) at 18°C at 0°C at 0°C We will use three significant figures Twenty-eight grams of water corresponds to 1.55 moles. From Equation for the number of moles of water 15.1, we have to avoid later round-off errors. 526 15. LiquiDS AnD SOLiDS q P = nC P(T2 – T1) = (1.55 mol)(75.3 J?mol–1?K–1)(273K – 291K) = –2100 J = –2.1 kJ The negative sign for q P reflects the fact that energy must be removed to lower the temperature of the water. For step 2, qP = n(–Hfus) = (1.55 mol)(–6.01 kJ·mol–1) = –9.3 kJ where the minus sign in front of DHfus arises because freezing is the reverse of melting, that is, DHfreezing = –DHfus. The total amount of energy that must be removed as heat from the 28 grams of water is (–2.1 kJ) + (–9.3 kJ) = –11.4 kJ. Pr ACTiCe PrObLeM 15-2: Using the data in Table 15.3 and Table 14.6, compute the energy released as heat at one atm when 25.0 grams of liquid mercury at 300.0°C is converted to solid mercury at –60.0°C. Take the heat capacity of Hg(s) to be 28.3 J?mol–1?K–1. H H Answer: –1.54 kJ H C C H C C C It is possible for a solid to be converted directly to a gas without passing C C C H C C H through the liquid phase. This process is called sublimation. The molar en- thalpy of sublimation, DHsub, is the energy absorbed as heat when one mole of H H a solid is sublimed at constant pressure. Essentially all the energy put in as heat naphthalene in the sublimation process is used to separate the molecules in the solid from one another. The larger the value of DHsub, the stronger the intermolecular at- tractions in the solid are. The best-known example of sublimation is the conversion of dry ice, CO2(s), to carbon dioxide gas according to CO2(s) → CO2(g) DHsub = 25.2 kJ?mol–1 The name “dry ice” is used because the CO2(s) does not become a liquid at one atmosphere. Dry ice at one atmosphere has a temperature of –78°C and is widely used as a one-time, low-temperature refrigerant. The sublimation of 44.0 grams (one mole) of dry ice requires 25.2 kJ of heat. vcchvv Ice, H2O(s), sublimes at temperatures below its melting point (0°C). The equation for the sublimation of ice is H2O(s) → H2O(g) DHsub = 46.7 kJ?mol–1 Figure 15.3 Naphthalene, C10H8 (s ), Snow often sublimes, and so does the ice in the freezer compartment of your is one of a number of substances that refrigerator. Perhaps you’ve noticed that ice cubes left in the freezer get smaller readily sublime at room temperature. Here we see crystals of naphthalene as time passes. Another substance that sublimes at temperatures below its melt- condensing directly from vapor to ing point is naphthalene, C10H8(s) (margin), a principal component of one type crystals on the surface of a cold tube. of mothball (Figure 15.3). 15- 4. vAn Der WA ALS FOrCeS Are ATTr ACTive FOrCeS beT Ween MOLeCuLeS 527 15-4. van Der Waals Forces Are Attractive Forces between Molecules In the process of vaporization or sublimation, the molecules of the liquid or solid, which are in contact with one another, become separated and widely dis- persed. The value of DHvap or DHsub reflects how strongly the molecules attract one another in the liquid or solid phase. The more strongly the molecules at- tract one another, the greater the value of DHvap or DHsub. The simplest force at the atomic or molecular level is that between ions. We studied this force in Chapter 6 when we discussed ionic compounds. Ions with opposite charges attract each other; ions with like charges repel each other. The electrical force between ions is relatively strong; it requires a rela- tively large amount of energy to separate ions of opposite charge. Therefore, the molar enthalpies of vaporization of ionic compounds are much larger than those of most nonionic compounds. Molar enthalpies of vaporization of ionic compounds are typically at least 100 kJ?mol –1. The boiling points of ionic compounds also are higher than those of nonionic compounds. This is the reason that all ionic compounds are solids at room temperature. Most of the compounds listed in Table 15.3 are covalent compounds. In Chapter 7 we learned that some molecules have dipole moments; in other words, they are polar. An example of a polar molecule is methanal, CH2O(g), better known by its common name, formaldehyde, whose Lewis formula is H C O H The bond dipole moments in a formaldehyde molecule can be illustrated by recall that we define the direction H of a dipole moment to be from C O the negative charge to the positive H charge, in agreement with the iuPAC convention. because an oxygen atom is more electronegative than a carbon atom, which in turn is more electronegative than a hydrogen atom. The net dipole moment is illustrated by H C O H Thus, even though a formaldehyde molecule is electrically neutral overall, it has a positively charged end and a negatively charged end and therefore is a polar molecule. eXAMPLe 15-3: Recall Section 8-9 where we discussed how to use the symmetry of a molecule to predict whether or not it has a net dipole mo- ment. Use VSEPR theory to predict which of the following molecules has a net dipole moment: (a) SF6 (b) SF4 (c) PF5 528 15. LiquiDS AnD SOLiDS F Solution: See margin drawings. (a) Sulfur hexafluoride, SF6, is an AX6 molecule; so it is octahedral and has no net dipole moment. (b) Sulfur tetrafluoride, SF4, is an AX4E molecule; so it is seesaw-shaped and has a net dipole moment. (c) Phosphorus pentafluoride, PF5, is an AX5 molecule; so S it is trigonal bipyramidal and has no net dipole moment. Pr ACTiCe PrObLeM 15-3: Which of the following molecules are polar? (a) C2Cl4 (b) PCl3 (c) SO2 (d) CH3Cl (e) CCl4 (a) Answer: (b), (c), and (d) are polar. F Polar molecules attract one another (Figure 15.4), and this attraction is called dipole-dipole attraction. The charge separation in polar molecules is S considerably smaller than the full electronic charges on ions, so dipole-dipole forces are weaker than ion-ion forces. Molar enthalpies of vaporization for typi- cal polar compounds are around 10–20 kJ?mol–1. We must emphasize here that the interaction energies between neutral molecules are much weaker than the (b) covalent bonds within molecules, which have energies of hundreds of kilojoules per mole. The forces between neutral molecules are collectively referred to as intermolecular forces; the prefix inter means between. The forces within a mol- F ecule (i.e., chemical bonds) are called intramolecular forces; the prefix intra means within. A particularly important dipole-dipole attraction occurs when one or more Title General Chemistry - 4th ed hydrogen atoms in a molecule are directly bonded to a highly electronegative P atom, as in the case of water and ammonia. Let’s consider the molar enthalpies Author McQuarrie/Gallogy of vaporization of water, ammonia, and methane: Artist George Kelvin Figure # 015-004 (482) Molecule: H2O NH 3 CH4 (c) Date 08/18/09 Hvap /kJ?mol–1 40.7 23.4 8.2 Check if revision Approved These three compounds have approximately the same molecular mass (18, 17, and 16, respectively), but the amount of energy required to separate the mol- ecules of these three liquids are very different. Water molecules in liquid water attract one another through the electro- static interaction between a hydrogen atom and the oxygen atom on different + – + – molecules: δ– δ– Hδ+ Figure 15.4 Although polar molecules O H O δ+ are electrically neutral overall, they δ+ H Hδ+ hydrogen attract one another by a dipole-dipole bond force. The molecules orient themselves as shown because the positive end of one attracts the negative end of The electrostatic attraction that occurs between molecules in which a hydrogen another. The dipoles are said to be atom is bonded to a highly electronegative atom, such as an oxygen, nitrogen, oriented head-to-tail. or fluorine atom, is called a hydrogen bond. Because a hydrogen atom is so 15- 4. vAn Der WA ALS FOrCeS Are ATTr ACTive FOrCeS beT Ween MOLeCuLeS 529 small, the charge on it is highly concentrated, so it strongly attracts electronega- tive atoms in neighboring molecules. Hydrogen bonds are a particularly strong form of dipole-dipole attraction yielding molar enthalpies of vaporization around 20–40 kJ?mol–1. The pattern shown in Figure 15.5 extends throughout liquid water and gives water its large value of DHvap (40.7 kJ?mol–1) and its high boiling point. We shall see that hydro- gen bonds give water many special properties. Hydrogen bonding Title greatly affects the structure - 4th edof ice (Figure 15.6), which General Chemistry is described as an open structure because of the significant amount of empty Author McQuarrie/Gallogy space between the molecules. The open structure of ice is a direct consequence of the fact that eachArtist George hydrogen atomKelvin is hydrogen-bonded to the oxygen atom of an adjacent molecule. Note Figure the tetrahedral arrangement # 015-007 (483) of the oxygen atoms in ice. Every oxygen atom sits Date 08/18 /09 in the center of a tetrahedron formed by four other oxygen atoms. Approved Check if revision The structure of liquid water is less open than the structure of ice because, when ice melts, the total number of hydrogen bonds decreases. Unlike most other substances, the density of water increases upon going from solid to liq- uid because of a partial breakdown of the hydrogen-bonded structure (that’s why ice floats on water). The extent of hydrogen bonding in liquid water is only about 80%, whereas in ice nearly 100% of the oxygen atoms are hydrogen- bonded. The extent of hydrogen bonding in water decreases as the temperature increases. Figure 15.5 There are many hydrogen bonds in liquid water; each oxygen atom can form two hydrogen bonds because each oxygen atom has two lone pairs of electrons. Thus, each water molecule has the ability to form four hydrogen bonds. At 25°C about 80% of the hydrogen atoms in water are hydrogen bonded. Figure 15.6 The crystalline structure of ice. Each water molecule can form four hydrogen bonds. Each oxygen atom is located in the center of a tetrahedron formed by four other oxygen atoms. The entire structure is held together by hydrogen bonds. 530 15. LiquiDS AnD SOLiDS HyDrOgen bOnDing HAS iMPOrTAnT COnSequenCeS FOr LiFe Water contains two hydrogen atoms covalently bound to a central oxygen atom. We know that oxygen falls into Group 16 in the periodic table. If we compare the boiling points of other substances from this group with the formula H2X, we find an interesting trend: Molecule H2Te H2Se H2S Boiling point/°C –2.0 –41.25 –59.55 Title General Chemistry - 4th ed By extrapolation water, H2O(l ), should have a boiling point around Author McQuarrie/Gallogy –90°C (these data are plotted in Figure 15.10). The remarkable difference Artist George Kelvin between this and the actual boiling point of water of 100.0°C is due to hydrogen bonding. Were it not for hydrogen bonding, most of the water Figure # 015-008 (484) on earth would be gaseous and life as we know it could not exist. Moreover, Date 08/18/09 unlike most other substances, the density of water increases upon going Check if revision Approved from a solid to a liquid because of a partial breakdown of its hydrogen- bonded structure. Because of this ice floats on water, whereas for most other substances the solid form is the more dense and sinks when placed into its liquid phase. The fact that ice floats on water causes the surfaces of lakes and ponds to freeze before the bottom in winter. This surface ice forms an insulating layer that allows fish and plants to survive below. Were this not the case, most rivers and ponds would freeze solid from the bottom up, altering aquatic life and the food chain. These amazing properties of water that arise from hydrogen bonding are what makes finding water one of the criteria used when searching for life on other planets. Hydrogen bonding also occurs in liquid ammonia, NH3(l ), but the indi- vidual hydrogen bonds are weaker than those in water. This is because nitrogen is less electronegative than oxygen and the fractional charges on the nitrogen and hydrogen atoms in an ammonia molecule are less than those on the oxygen and hydrogen atoms in a water molecule. Furthermore, there are fewer hydro- gen bonds in NH3(l ) because each nitrogen atom can form only one hydrogen bond (Figure 15.7). Consequently, the value of DHvap for ammonia is only about half that for water. Methane is a nonpolar molecule because it is tetrahedral and the four bond moments cancel. Thus, of H2O(l ), NH3(l ), and CH4(l ); CH4(l ) has the lowest value of DHvap. We must consider one more attractive force. Even though methane is nonpo- lar, it liquefies at 91 K and has a molar enthalpy of vaporization of 8.2 kJ?mol–1; therefore, methane molecules must attract one another. Even the noble gases, which consist of single, spherical atoms, can be liquefied. How neutral, nonpo- lar molecules attract one another was not understood until quantum theory Figure 15.7 There are fewer hydrogen was developed. Let’s first consider a single argon atom. As the electrons move bonds in NH3(l ) than in H2O(l ) about in the atom, there will be instants of time when the distribution of the because an NH3 molecule has only one total charge will not be perfectly symmetric. For example, perhaps there will lone electron pair, so each nitrogen be more electrons on one side of the atom than on the other side, as shown in atom can form only one hydrogen bond. Figure 15.8. Over a period of time, these little fluctuations in the electronic  Check if revision 15- 4. vAn Der WA ALS FOrCeS Are ATTr ACTive FOrCeS beT Ween MOLeCuLeS 531 charge distribution will average out, thereby yielding a perfectly symmetric dis- tribution. But at any instant of time, there will be an instantaneous asymmetry in the electronic charge distribution, resulting in an instantaneous dipole mo- ment (Figure 15.8). Now let’s consider two argon atoms separated by a small δ– δ+ distance. The motion of the electrons in one atom influences the motion of the electrons in the other atom in such a way that the instantaneous dipole moments line up head-to-tail (Figure 15.9). This synchronized motion of the electrons leads to an effective dipole-dipole attraction that accounts for the at- tractive force between atoms, nonpolar molecules, and a sizeable fraction of the Figure Title 15.8 If it were General possible -to4th Chemistry take ed attractive forces between large molecules as well. This force was first explained an instantaneous view of an atom, by the German physicist Fritz London and is now called a London force. itAuthor McQuarrie/Gallogy might look like this drawing. The instantaneous position of the electrons Because London forces are due to the motion of electrons, their strength Artist Kelvin leads to an instantaneous dipole depends on the number of electrons. The more electrons there are in the two Figure # 015-010 moment. The negative charge is due interacting molecules, the stronger their attraction for each other. Therefore, to a greater-than-average electronic Date 12/03/09 for purely nonpolar substances, which attract each other only by way of London charge density, and the positive charge Check if revision is due to Approved a less-than-average electronic forces, we expect the value of DHvap to increase with the number of electrons, or even with the size of the molecules. The data in Table 15.4 support this pre- charge density. As the electrons move around, the dipole moment points in diction. Notice that, within each group, the value of DHvap increases with the all directions at different times and number of electrons. averages to zero. Figure 15.10 shows plots of the boiling points of the noble gases and the hy- drides of the nonmetallic elements. The hydrogen-bonded compounds H2O(l ), NH3(l ), and HF(l ) have unusually high boiling points. Except for the hydrogen- bonded compounds, there is a general increase of boiling point with increasing molecular mass. This increase in boiling point is due to the increase in London δ– δ+ δ– δ+ forces, which increases with the number of electrons in a molecule. In Figure 15.10, we see that the boiling points of the noble gases (plotted in purple) and the hydrides of Group 14 (plotted in green) are lower than those of the other molecules. This is because the noble gases and Group 14 hydrides are nonpo- Figure 15.9 When two atoms are near each other, the motions of the electrons in the two atoms affect each other so that the instantaneous dipole TAbLe 15.4 relation between number of electrons in some moments are aligned head-to-tail. We atoms and nonpolar molecules and their enthalpies of vaporization say that the motion of the electrons is correlated. This effect leads to an instantaneous dipole-dipole attraction Substance Number of electrons Hvap/kJ?mol –1 between the two atoms. He 2 0.08 Ne 10 1.71 Ar 18 6.43 Kr 36 9.08 Xe 54 12.57 F2 18 6.62 Cl2 34 20.04 Br 2 70 29.96 I2 106 41.57  Figure # 015-011 (486) Date 07/03/09 Check if revision Approved 532 15. LiquiDS AnD SOLiDS 100 H 2O HF 0 H2Te NH3 H2Se HI H2S SbH3 HBr SnH4 HCl Temperature/°C PH3 GeH4 –100 AsH3 Xe SiH4 CH4 Kr –200 Ar Ne Figure 15.10 Boiling points of the noble gases and hydrides of the nonmetallic He elements. Note the abnormally high values for hydrogen fluoride, water, –300 and ammonia, which are the result of 1 2 3 4 5 hydrogen bonding. Row in the periodic table lar and have no dipole-dipole forces contributing to their total intermolecular forces. The attractive forces between molecules, be they dipole-dipole forces or London forces, are collectively called van der Waals forces. Table 15.5 com- pares the various intermolecular forces that we have discussed in this section. TAbLe 15.5 various types of attractive forces between ions and molecules Type Examples Typical value of Hvap/kJ?mol –1 ion-ion NaCl, KBr ~100 hydrogen bonding H2O, NH3 20–40 London Ar, CH4 5–40 dipole-dipole H2 CO, HCl 10–20 15-5. viSCOSiT y, SurFACe TenSiOn, AnD CAPiLL Ary ACTiOn Are PrOPerTieS OF LiquiDS 533 eXAMPLe 15- 4: Without referring to any of the tables in the book, rank the following liquids in order of increasing molar enthalpies of vaporization and boiling points: NaCl(l ) H2(l ) C2H4(l ) CH3OH(l ) Solution: The only ionic compound listed is NaCl(l ). Thus, we predict that H NaCl(l ) has the largest value of DHvap and the highest boiling point. The H H H C O H Lewis formula for a methanol molecule, CH3OH (margin), shows that there C C is a hydrogen atom bonded to an oxygen atom. Thus, we predict that in H H H CH3OH(l ) there will be hydrogen bonding between the molecules. The methanol ethene (ethylene) Lewis formula for an ethylene molecule, C2H4 (margin), indicates thatHC2H4 is a nonpolar molecule. H2 is also a nonpolar molecule. However, there are H C O H H H many more electrons in a C2H4 molecule, so we predict that C C H H H DHvap[NaCl(l )] > DHvap[CH3OH(l )] > DHvap[C2H4(l )] > DHvap[H2(lmethanol )] ethene (ethylene) The boiling points are in the same order. The actual values of DHvap are 170 kJ?mol–1, 35.2 kJ?mol–1, 13.5 kJ?mol–1, and 0.90 kJ?mol–1, respectively; and the boiling points are 1738 K, 338 K, 169 K, and 20.3 K, respectively. Pr ACTiCe PrObLeM 15- 4: Which of the following molecules (common names given in parentheses) would you predict to exhibit hydrogen bond- ing? Explain why. O O C H2 H H HF H3C C O H hydrogen methanal hydrogen ethanoic acid (formaldehyde) fluoride (acetic acid) Answer: A hydrogen molecule contains no nitrogen, oxygen, or fluorine atoms, and so does not exhibit hydrogen bonding. Although an acetone molecule contains hydrogen and oxygen atoms, the hydrogen atoms are not directly bonded to the oxygen atom, and so hydrogen bonding is not observed in acetone. Hydrogen fluoride and acetic acid both exhibit hydro- gen bonding because in each molecule there is a hydrogen atom directly bonded to a highly electronegative atom. 15-5. viscosity, Surface Tension, and Capillary Action Are Properties of Liquids Several properties of liquids that depend on intermolecular forces are of practi- cal importance to chemists. The viscosity of a liquid is a measure of its resis- tance to flow. Substances that flow very slowly, such as molasses, are said to be viscous. Viscosity can be measured experimentally by timing how long it takes a certain volume of a liquid to flow through a thin tube. Many engineering processes involve the flow of liquids through pipes, so engineering design is 534 15. LiquiDS AnD SOLiDS TAbLe 15.6 Physical properties of a variety of liquids at 25°C Compound Formula Viscosity Surface Polarity (relative to water) tension/mJ?m –2 benzene C 6H6 0.68 28 nonpolar bromine Br 2 1.1 41 nonpolar ethanol C 2H5OH 1.2 22 slightly polar glycerol C 3H5 (OH) 3 1050 63 slightly polar hexane C 6H14 0.34 18 nonpolar methanol CH3OH 0.89 22 slightly polar octane C 8H18 0.57 19 nonpolar 2-propanone (CH3) 2 CO 0.34 23 slightly polar (acetone) sulfuric acid H2 SO4 27.6 — polar water H2O 1.00 72 polar Figure 15.11 A metal paper clip floats on water because of the distribution of its mass and the high surface tension of the water. The surface of water is like an elastic skin that resists penetration. often concerned with the viscosities of liquids and liquid mixtures. Viscosity decreases with increasing temperature because, as the temperature increases, molecules have greater kinetic energy and therefore can more readily overcome the intermolecular interactions. The viscosities of a variety of compounds are given in Table 15.6. If a paper clip is carefully placed on a water surface, it floats even though the density of the paper clip is greater than that of water (Figure 15.11). The clip is held up by surface tension. Water striders and some other insects can walk on water, being supported by the surface tension of the water, which resists penetration of the surface (Figure 15.12). What is the cause of the surface tension of a liquid? A molecule in the body of a liquid is subject to attractive forces in all directions, but a molecule at the surface experiences a net attractive force toward the interior of the liquid (Fig- ure 15.13). Thus, molecules at the surface of a liquid experience a net inward force. This force tends to minimize the number of molecules at the surface and so minimize the surface area of the liquid. This force is the surface tension. Any liquid whose molecules attract one another strongly has a high surface tension; water is a good example. The surface tension of a liquid tends to hold a drop of liquid in a spherical shape because a sphere is the shape that has the smallest surface area for a given volume (Figure 15.14). The higher the surface tension is, the more nearly spherical the drop is (Figure 15.15). Table 15.6 lists the surface tension for a variety of liquids. Note that surface tension has units Figure 15.12 (top) Some insects like of energy per unit area. Surface tension can be thought of as the energy that it this water strider use surface tension takes to create an area of surface. to walk on water. (bottom) Researchers at Carnegie Mellon University are Certain compounds, such as sodium dodecylsulfate, NaC12H25SO3(s), lower working to develop a microrobot that the surface tension of a liquid by concentrating at the liquid surface. Such mol- uses surface tension to walk on water. ecules are called surfactants (surface active agents). A 0.1% sodium dodecyl-  Date 08/19/09 Check if revision Approved 15-5. viSCOSiT y, SurFACe TenSiOn, AnD CAPiLL Ary ACTiOn Are PrOPerTieS OF LiquiDS 535 sulfate solution has a surface tension of 20 mJ?m–2, whereas pure water has a surface tension of 72 mJ?m–2. Reduction of the surface tension of water by surfactants is the basis of detergent action. In a detergent solution the surface tension of the water is comparable to that of the grime (oils) on clothing, for example, thereby allowing the solution to wet the grime. Closely related to surface tension is capillary action, the rise of a liquid in a thin tube. Capillary action occurs when the adhesive forces between the molecules of the surface of a capillary wall and the molecules of the liquid are sufficiently great that the liquid adheres to, or wets, the solid surface. The ad- hesive force pulls the liquid up into the capillary. The liquid column rises until the upward adhesive force is balanced by the downward gravitational force. Capillary action plays a major role in the movement of water in plants, animals, and soil. The water is pulled by the capillary action up into and through living structures (Figure 15.16). Figure 15.13 Molecules in the interior of One important consequence of surface tension and capillary action is the a liquid are attracted in all directions, formation of a meniscus, which is the shape formed by the surface of a liquid but the molecules at the surface in a capillary (Figure 15.17). For a liquid that adheres to glass (for example, wa- experience a net inward attraction that ter), the liquid rises in a glass capillary and the meniscus is concave. For a liquid minimizes the surface area of the liquid and results in surface tension. that does not adhere to glass (for example, mercury), the liquid is lower where it contacts the glass and the meniscus is convex. Figure 15.14 Surface tension causes drops of liquid to assume a spherical shape because a sphere is the shape with the smallest surface area for a given volume. Figure 15.15 Shapes of equal volumes of Hg(l ), H2O(l ), (CH3)2SO3(l ) (dimethylsulfoxide), and (CH3)2CO(l ) (acetone), from left to right. Surface tension holds the drops in a spherical shape, and gravity flattens them. The effect of gravity is the same for all the drops; thus, the higher the surface tension, the more nearly spherical is the drop. Surface tension has units of energy per unit area, usually expressed as millijoules per square meter (mJ?m–2). 536 15. LiquiDS AnD SOLiDS Figure 15.16 Capillary action is shown here as colored water rises in a celery stalk and in these glass tubes. The cells in a celery stalk form a capillary structure. H2 One final property of liquids that we shall discuss is called polarity. Gener- C ally speaking, liquids comprised of molecules that have large dipole moments H2C CH2 form polar liquids. For example, water is a very polar solvent because individual O H2C CH2 water molecules have a large dipole moment. On the other hand, cyclohexane, C H H H2 C6H12(l ), is a nonpolar liquid because individual cyclohexane molecules have water is polar cyclohexane no dipole moment (margin). is nonpolar Water and other polar liquids tend to be good solvents for polar species such as ionic compounds. In contrast, cyclohexane and other nonpolar liquids are poor solvents for ionic compounds, but tend to be good solvents for nonpolar species, such as oils, which do not dissolve readily in water or other polar sol- vents. A general rule of thumb to remember is, “like dissolves like.” Table 15.6 lists the relative viscosities, the surface tensions, and polarity of a variety of compounds. The most important thing to notice is the relative values of these quantities. 15-6. A Liquid Has a unique equilibrium vapor Pressure at each Temperature Let’s look more closely at the process of vaporization. Suppose that we cover a beaker containing a liquid with a bell jar, as shown in Figure 15.18, and main- tain a constant temperature. Now suppose that we evacuate all the gas from the bell jar, so that the space above the beaker and its contents is a vacuum. The molecules in the liquid are in constant motion; some at the surface will break away from the liquid and form a vapor phase above it. The pressure of the vapor is observed to increase rapidly at first and then progressively more slowly until a constant pressure is reached. Let’s see why. Figure 15.17 A meniscus formed by water (left) in a glass tube is concave, In order to break free of the liquid, a molecule at the surface must have whereas the meniscus formed by enough kinetic energy to overcome the attractive force of its neighbors and be mercury (right) is convex. moving in the right direction. The number of molecules that leave the surface  Date 08/19/09 Check if revision Approved 15- 6. A LiquiD HAS A unique equiLibriuM vAPOr PreSSure AT eACH TeMPer ATure 537 is proportional to the surface area of the liquid. Because the surface area is con- stant, the rate of evaporation is constant (Figure 15.19). There are no molecules in the vapor phase initially, so there is no condensation from the vapor phase to Title General Chemistry - 4th ed the liquid phase. As the concentration of molecules in the vapor phase increases, theAuthor pressure McQuarrie/Gallogy of the vapor increases and the number of vapor phase molecules thatArtist collideGeorge with the liquid surface increases. As a result, the rate of conden- Kelvin sation of the vapor increases. Eventually, a state is reached where the rate of Figure # 015-021 evaporation from the liquid surface (491)is equal to the rate of condensation from the vapor 08/19/09 Datephase. The pressure of the vapor no longer increases but takes on a con- stant value. Check The evaporation-condensation if revision Approved process appears to have stopped, and we say that the system is at equilibrium, meaning that no change appears to be taking place (Figure 15.20). The pressure of the vapor is now a constant value. The equilibrium between the liquid and the vapor is a dynamic equilibrium; that is, the liquid continues to evaporate and the vapor continues to condense, Figure 15.18 When a liquid is placed in a closed container that has been evacuated, the pressure of the vapor above the liquid eventually reaches a constant value that depends upon the particular liquid and the temperature. Title General Chemistry - 4th ed Author McQuarrie/Gallogy Artist George Kelvin Figure # 015-020 (491) Date 08/19/09 Check if revision Approved (a) (b) Figure 15.19 When a liquid is placed in a closed container, (a) the rate at which molecules escape from the surface is constant, but the rate at which molecules enter the liquid from the vapor is proportional to the number of molecules in the vapor. (b) When the number of molecules in the vapor is such that the rate of escape from the surface is equal to the rate of condensation from the vapor, the liquid and vapor are in equilibrium with each other. Rate of evaporation of liquid A Equilibrium attained Rate Rate of condensation of vapor Time Figure 15.20 Equilibrium is attained (at A) when the rate of evaporation of the liquid equals the rate of condensation of the vapor. At equilibrium the pressure of the vapor is constant and is called the equilibrium vapor pressure. 538 15. LiquiDS AnD SOLiDS Rate of evaporation at T2 T2 Equilibrium attained Rate of evaporation at T1 Figure 15.21 The change in the rate T1 Rate of condensation with time for the T2 > T1 vapor over a liquid as it approaches equilibrium. Because the rate of Rate of condensation at T2 evaporation increases with increasing temperature, the equilibrium vapor Rate of condensation at T1 pressure increases with increasing temperature (T2 > T1). Time but the rate of evaporation is exactly equal to the rate of condensation and thus there is no net change. In such a case, we have the equilibrium condition rate of evaporation = rate of condensation The pressure of the vapor at equilibrium is called the equilibrium vapor pres- sure. We shall show that the value of the equilibrium vapor pressure depends on the particular liquid and the temperature. Let’s consider the approach to a dynamic liquid-vapor equilibrium at two dif- ferent temperatures. The higher the General Chemistry Titletemperature, the more-rapidly 4th ed the molecules in the liquid phase move and theAuthor higherMcQuarrie/Gallogy the rate of evaporation is. Figure 15.21 shows that, because the rate of evaporation at T2 is greater than the rate of evapo- Artist George Kelvin ration at T1 (given that T2 > T1), the equilibrium vapor pressure at T2 is greater than Figure that at T1. Thus, we see that the value 015-024 of#the equilibrium vapor pressure of a liquid increases with increasing temperatureDate 08/19/0915.22). (Figure At each temperature, a liquidCheck has aif definite revision equilibrium Approvedvapor pressure. The equilibrium vapor pressures of water at various temperatures are given in Table Figure 15.22 Solid iodine in equilibrium 15.7. The equilibrium vapor pressure of water plotted as a function of the tem- with gaseous iodine at 40°C (lighter- perature is called a vapor pressure curve. Figure 15.23 shows the equilibrium colored vapor) and at 90°C (darker- colored vapor). The equilibrium vapor vapor pressure curve for water and for ethanol, CH3CH2OH(l ). In Chapter 23 we pressure increases with increasing discuss a mathematical expression that describes the equilibrium vapor pressure temperature, as shown by the more as a function of temperature. intense color. 1520 CH3CH2OH H2O Figure 15.23 The equilibrium vapor pressure curves for water, H2O(l ), and ethanol, CH3CH2OH(l ), over 78°C (351 K) 100°C (373 K) P/Torr the temperature range 270 to 420 K 760 (–3° to 147°C). Observe the rapid increase in vapor pressure with increasing temperature. The equilibrium vapor pressure curve for ethanol lies above that for water because ethanol has a higher 0 equilibrium vapor pressure than water 270 290 310 330 350 370 390 410 at the same temperature. T/K 15- 6. A LiquiD HAS A unique equiLibriuM vAPOr PreSSure AT eACH TeMPer ATure 539 TAbLe 15.7 equilibrium vapor pressure of water as a function of temperature t/°C P/bar P/atm P/Torr t/°C P/bar P/atm P/Torr 0 0.00611 0.00603 4.59 60 0.199 0.197 149.5 5 0.00873 0.00861 6.55 65 0.250 0.247 187.7 10 0.0123 0.0121 9.21 70 0.312 0.308 233.8 15 0.0171 0.0168 12.8 75 0.386 0.381 289.2 20 0.0234 0.0231 17.5 80 0.474 0.467 355.3 25 0.0317 0.0313 23.8 85 0.578 0.571 433.6 30 0.0425 0.0419 31.8 90 0.701 0.692 525.9 35 0.0563 0.0555 42.2 95 0.845 0.834 634.0 40 0.0738 0.0728 55.4 100 1.013 1.000 760.0 45 0.0959 0.0946 71.9 105 1.208 1.192 906.0 50 0.123 0.122 92.6 110 1.432 1.414 1074.4 55 0.158 0.155 118.0 120 1.985 1.959 1488.7 eXAMPLe 15-5: A 0.0896-gram sample of water is placed in a 250.0-mL sealed container. Is there any liquid present when the temperature is held at 70.0°C? Solution: First, we calculate the pressure assuming that all the water is gaseous. If this pressure is greater than the equilibrium vapor pressure of H2O(l ) at 70.0°C, then H2O(g) will condense until the pressure is equal to the equilibrium vapor pressure. If the pressure is less than the equilibrium vapor pressure, then no liquid will be present. The pressure is given by   1 mol (0.0896 g) — (0.08206 L·atm·mol–1·K–1)(343.2 K) nRT 18.02 g P= — = — V 0.250 L = 0.560 atm According to Table 15.7, the equilibrium vapor pressure of water is 0.308 atm at 70.0°C, so some of the water vapor will condense until its vapor pressure is 0.308 atm. Pr ACTiCe PrObLeM 15-5: What mass of water will condense in Example 15-5? Answer: 0.0403 g 540 15. LiquiDS AnD SOLiDS The vapor pressure of a liquid depends on the attractive forces between its constituent molecules. Substances with relatively strong intermolecular attrac- tions will have relatively low vapor pressures. At a given temperature, relatively few of the molecules of such a substance will have sufficient kinetic energy to overcome the attractive forces of the other molecules and enter the vapor phase. The normal boiling point of a liquid The boiling point of a liquid is the temperature at which its vapor pressure is the temperature at which its vapor equals the external pressure. The normal boiling point, the boiling point at pressure is equal to exactly one exactly one atm, of water is 100.0°C. If the external pressure is less than the one atmosphere. The normal boiling point atm, then the temperature at which the vapor pressure of liquid water equals of water, for example, is 100.0°C. We this pressure is less than 100.0°C. For example, the elevation at Vail, Colorado, saw in Chapter 13, however, that the is about 2500 meters (8200 feet). The atmospheric pressure at this elevation is pressure unit of the bar is replacing about 0.75 atm, so water boils at 92°C. In a pressure cooker, on the other hand, the atmosphere, so some people when the pressure is 2.0 atm, water boils at 120°C. Because atmospheric pres- prefer to define the normal boiling sure decreases with increasing elevation and because the rate at which food point of a liquid as the temperature cooks depends on the temperature, it requires a significantly longer time to at which its vapor pressure is equal cook food by boiling in an open container at high elevations than at sea level. to 1.01325 bar (which is equal to one An egg must be boiled for almost 5 minutes at Vail, Colorado, in order to be atmosphere). cooked to the same extent as one boiled for 3 minutes at sea level. 15-7. relative Humidity is based on the vapor Pressure of Water The vapor pressure of water in the atmosphere is expressed in terms of relative humidity. Relative humidity is the ratio of the partial pressure of the water va- por in the atmosphere to the equilibrium vapor pressure of water at the same temperature expressed as a percentage, as given by   PH2O relative humidity = — × 100 (15.4) P°H2O where P H2O is the partial pressure of the water vapor in the air and PH̊2O is the equilibrium vapor pressure of water at the same temperature. At 20.0°C, the equilibrium vapor pressure of water is 17.5 Torr. If the partial pressure of the water vapor in the air is 11.2 Torr, then the relative humidity is   11.2 Torr relative humidity = — × 100 = 64.0% 17.5 Torr If the temperature of the air is lowered to 13.0°C, where the equilibrium vapor pressure of water is 11.2 Torr, then the relative humidity is   11.2 Torr relative humidity = — × 100 = 100% 11.2 Torr At 13.0°C, air that contains water vapor at a partial pressure of 11.2 Torr is saturated with water vapor. At this temperature, the water vapor begins to con- dense as dew or fog, which consists of small droplets of water. The air tempera- ture at which the relative humidity reaches 100% is called the dew point. Most people begin to feel uncomfortable when the dew point rises above 20°C, and air with a dew point above 24°C is generally regarded as extremely humid or muggy. 15- 8. A PHASe DiAgr AM DiSPL AyS regiOnS OF ALL PHASeS OF A Pure SubSTAnCe 541 eXAMPLe 15-6: Calculate the relative humidity and the dew point for a day when the partial pressure of water vapor in the air is 22.2 Torr and the temperature of the air is 30.0°C. The equilibrium vapor pressure of water at 30.0°C is 31.8 Torr (Table 15.7). Solution: The relative humidity, given by Equation 15.4, is   PH2O   22.2 Torr relative humidity = — × 100 = — × 100 = 69.8% P°H2O 31.8 Torr The dew point is the temperature at which the equilibrium vapor pressure of water is equal to 22.2 Torr. According to Table 15.7, this is about 24°C. Such a day would be considered very uncomfortable. Pr ACTiCe PrObLeM 15-6: Calculate the dew point for a day when the relative humidity is 78% at 20.0°C. Answer: approximately 17°C 15-8. A Phase Diagram Displays the regions of All the Phases of a Pure Substance Simultaneously The vapor pressure curve of a pure substance can be combined with two other useful quantities, the sublimation pressure curve and the melting point curve, into a single diagram called a phase diagram. The phase diagram of water is shown in Figure 15.24. Along the vapor pressure curve (the red curve in Figure 15.24), liquid and vapor exist together at equilibrium. To the left of this curve, at lower temperatures, the water exists as a liquid. To the right of this curve, at higher temperatures, the water exists as a vapor. The equilibrium vapor pressure of a liquid increases with temperature up to the critical point (Figure 15.24), where the vapor pressure curve terminates abruptly. Above the critical tem- perature, the gas and the liquid phases become indistinguishable. A gas above its critical temperature cannot be liquefied no matter how high a pres- sure is applied. The critical point for water occurs at 218 atm (221 bar) and 647 K. Water vapor above 647 K cannot be liquefied by the application of pressure. Along the sublimation pressure curve (gray curve in Figure 15.24), solid and vapor exist together at equilibrium. To the left of this curve, at lower tem- peratures, water exists as a solid (ice). To the right of this curve, at higher tem- peratures, water exists as a vapor. Along the melting point curve (blue curve in Figure 15.24), solid and liquid exist together in equilibrium. To the left of this curve, water exists as a solid, and to the right of this curve, water exists as a liquid. The melting point at a pressure of exactly one atm is called the normal melting point or the normal freezing point. However, melting points are only weakly dependent on pressure, so a melting point curve is an almost vertical line (Figure 15.24). For almost all  Check if revision Approved 542 15. LiquiDS AnD SOLiDS Critical point 218 221 Solid Liquid water water (ice) P/atm P/bar Water vapor Normal freezing ≈ point ≈ 1.00 1.01 ≈ Triple ≈ Normal boiling point 0.0060 point 0.0061 ≈ 273.16 373.15 647 T/K Figure 15.24 The phase diagram of water (not to scale; see the breaks in both the horizontal and vertical axes), displays simultaneously the sublimation pressure curve (gray), the vapor pressure curve (red), and the melting point curve (blue). The triple point, the critical point, the normal boiling point, and the normal freezing point are indicated in the figure. The phase equilibrium lines are the boundaries between the regions of stability of the solid, liquid, and vapor phases, which are labeled solid water (ice), liquid water, and water vapor. substances, the melting point increases with increasing pressure at a rate of 0.01 to 0.03 K?atm–1. Water is anomalous because its melting point decreases with increasing pressure. The melting point of ice decreases by about 0.01 K?atm–1 of applied pressure. Consequently, unlike most other solids, ice can be melted by the application of pressure. We discuss why this is so in Section 19-7. Notice that the three curves in Figure 15.24 separate regions in which water exists as a solid, a liquid, or a vapor. Let’s use Figure 15.24 to follow the behavior of water as it is heated from –50.0°C to 200.0°C at a constant pressure of one atm. At –50.0°C (223.2 K) and one atm, water exists as ice. As we heat the ice at one atm, we move horizontally from left to right along the dashed line in Figure 15.24. At 0°C (273.2 K), we cross the melting point curve and pass from the solid region into the liquid region. At 100.0°C (373.2 K), we cross the equi- librium vapor pressure curve and pass from the liquid region into the vapor region. The three curves in Figure 15.24 intersect at a point called the triple point. At the triple point, and only at the triple point, all three phases—solid, liquid, and gas—coexist in equilibrium. The triple point for water occurs at 4.58 Torr (0.0060 atm) and 273.16 K. Notice that if we heat ice at a constant pressure less than 4.58 Torr, then the ice sublimes rather than melts. The sublimation of water is used in the process of “freeze-drying.” 15- 8. A PHASe DiAgr AM DiSPL AyS regiOnS OF ALL PHASeS OF A Pure SubSTAnCe 543 eXAMPLe 15-7: Use the phase diagram of water given in Figure 15.24 to predict the result of increasing the pressure of water vapor initially at one atm and 500 K, keeping the temperature constant. Solution: At one atm and 500 K, water exists as a vapor. As the pressure is increased, we cross the liquid-vapor curve below the critical point at a pres- sure of about 150 atm, and the vapor condenses to a liquid. Pr ACTiCe PrObLeM 15-7: Given the following data for iodine and the fact that solid iodine is more dense than liquid iodine, sketch the phase diagram for iodine. triple point 113°C 0.12 atm critical point 512°C 118 atm normal melting point 114°C 1 atm normal boiling point 184°C 1 atm Solution: Your sketch should be similar in shape to that in Figure 15.25 (page 544) for carbon dioxide, but with the values for the various points taken from the table above. eXAMPLe 15-8: The vapor pressures (in Torr) of solid and liquid argon are given by 919.0 K ln Ps = 17.283 – — (solid) T 700.5 K ln Pl = 15.236 – — (liquid) T where T is the Kelvin temperature. Calculate the temperature and pressure of the triple point of argon. Solution: The vapor pressures of the solid and the liqu